LIBRARY OF
WELLESLEY COLLEGE
PRESENTED BY Th* McMillan Company
ELEMENTARY ALGEBRA BY
H.
S.
HALL, M.A., and
S.
R.
KNIGHT,
REVISED AND ENLARGED
FOR THE USE OF AMERICAN SCHOOLS BY F.
L.
SEVENOAK,
A.M.
Assistant Principal of the Academic Department, Stevens Institute of Technology
THE MACMILLAN COMPANY LONDON: MACMILLAN & 1896 All rights reserved
CO., Ltd.
B.A.
Copyright, 1896,
By
the MACMILLAN COMPANY.
First edition, 1895.
Second
edition, revised
l^)A
and enlarged, 1896.
H
NorfajooU ^vfss J. S.
— Berwick & Smith. Norwood Mass. U.S.A.
Cushing & Co.
;
'li.j
PKEFACE. Within ment
a comparatively short time, the algebra require-
for admission to
of Science has
many
of our Colleges
and Schools
been much increased in both thoroughness of
preparation and amount of subject-matter.
This increase
has made necessary the rearrangement and extension of
elementary algebra, and revision of Hall
it is
for this reason that the present
and Knight's Elementary Algebra has been
undertaken.
The marked
success of the work, and the hearty endorse-
ment by many
of our ablest educators of the treatment of
the subject as therein presented, warrant the belief that the
present edition, with
its
additional subject-matter, will be
found a desirable arrangement and satisfactory treatment "of
every part of the subject required for admission to any
of our Colleges or Schools of Technology.
Many
changes in the original chapters have been made,
among which we would proof,
call attention to the following:
A
by mathematical induction, of the binomial theorem
for positive integral index has been
added
to
Chapter xxxix.
a method of finding a factor that will rationalize any bino-
mial surd follows the treatment of binomial quadratic surds
Chapter xlii. has been re-written in part, and appears as a chapter on equations in quadratic form
;
and the chapter
PREFACE.
VI
on logarithms
lias
been enlarged by the addition of a four-
place table of logarithms with explanation of
its use.
Chapters xxi., xxv., xxx., xxxiii., xxxviii.,
and
XLiv., XLV., XLVi., XLVii., XLix.,
the
A
subject that
L.
xlii., xliii.,
treat of portions of
have not appeared in former
chapter on General Theory of Equations
is
editions.
not usually
found in an Elementary Algebra, but properly finds here a place in accordance with the purpose of the present
and
revision;
its
introduction makes the
for use in college classes.
work
available
Carefully selected exercises are
given with each chapter, and at the end of the work a large miscellaneous collection will be found.
The Higher Algebra
of Messrs. Hall
drawn upon, and the works
DeMorgan consulted I gratefully
Burkitt
Webb
and Knight has been
of Todliunter, Chrystal,
in preparing the
acknowledge
my
new
and
chapters.
indebtedness to Prof. J.
of the Stevens Institute of Technology both
for contributions of subject-matter,
and valuable suggestions
My
thanks are also due to
as to
Prof.
methods of treatment.
W. H.
as to the
Bristol, of the
same
institution, for suggestions
arrangement of the chapter on General Theory of
Equations.
FRANK June,
1896.
L.
SEVENOAK.
PEEFACE TO SECOND EDITION. The
printing of the present edition from entirely
plates lias enabled us to correct a
found in the friends, a
first
edition,
somewhat
material to Chapters
XLviii.,
and of several
will render the
book
give, at the suggestion of
fuller explanation of the
cult parts of the subject.
new
and
We iii.,
sets
still
new
few typographical errors
more
diffi-
hope that the addition of iv.,
of
v.,
x.,
xx., xxi., xlii.,
Miscellaneous Examples,
more acceptable
to those
whose
commendation of the former edition has given us much pleasure. June, 1896.
CONTENTS. PAGE
CHAPTER I.
II.
III.
Substitutions
Definitions.
Negative Quantities.
Simple Brackets.
1
Addition of Like Terms Subtraction
Addition.
Commutative Law
23
Multiplication
24
for Addition
for Addition
.
Commutative Law
Law
Associative
Index
Law
.
.
for Multiplication
.
Law
Multiplication of
25 25
for Multiplication
26
Compound Expressions
27
Products written by Inspection
35
Index
Law
Division of
by Detached
Coefficients
......
29 31
37 38
for Division
39
Compound Expressions
40
Important Cases in Division Horner's Method of Synthetic Division
44
Removal and Insertion of Brackets Miscellaneous Examples II.
48
Simple Equations
55
.
46
53
.
IX.
17
28
Division
VIII.
15
Rule of Signs Note on Arithmetical and Symbolical Algebr Distributing the Product Multiplication
VII.
15
25
for Multiplication
for Multiplication
Distributive
VI.
9 13
and Subtraction and Subtraction Dimension, Degree, Ascending and Descending Power Miscellaneous Examples I.
Law
Associative
IV.
.
....
Symbolical Expression
Problems Leading
to
.
.
.
Simple Equations
,
60 65
,
CONTENTS. CHAPTEK
X.
Resolution into Factors Trinomial Expressions Difference of
Two
Important Cases
Sum
.
Squares
.
.
or Difference of
Two Cubes
Miscellaneous Cases
Converse Use of Factors The Factor Theorem
XL XII. XIII.
Highest Common Factor Factor Theorem Employed
Lowest
Common
.
Multiple
....
Fractions
Reduction to Lowest Terms Multiplication
and Division
.
.
Addition and Subtraction
....
Rules for Change of Sign Cyclic Order
XIV.
Complex
Fractions.
Mixed Expressions
Miscellaneous Exercise on Fractions
XV. XVI.
Fractional and Literal Equations
.
Problems Miscellaneous Examples III.
XVII.
Simultaneous Equations Elimination by Addition or Subtraction Elimination by Substitution
Elimination by Comparison
.
Unknown
.... ....
Equations involving Three Reciprocal
.
Quantities
Literal Simultaneous Equations
XVIII.
XIX.
Problems
Indeterminate and Impossible Problems. suits
Meaning
XX.
of
^
Involution
To square a Multinomial To cube a Multinomial .
Application of Binomial
Theorem
Negative
Re
CONTENTS. CHAPTER
XXI.
Evolution
XI
....
Square Root of a Multinomial Cube Root of a Multinomial Some Higher Roots The nth Root of a Multinomial Square and Cube Root of Numbers .
XXII.
The Theory of Indices Meaning of Fractional Exponent Meaning of Zero Exponent Meaning of Negative Exponent .
XXIII.
Surds (Radicals)
Reduction of Surds Addition and Subtraction of Surds Multiplication of Surds
Division of Surds
Compound Surds Factor which
will rationalize
Any
Binomial Surd
Properties of Quadratic Surds
Square Root of a Binomial Surd Square Root of a Binomial Surd by Inspection Equations involving Surds
XXIV.
....
Imaginary Quantities Imaginary Unit
.
Fundamental Algebraic Operations
XXV.
Problems Miscellaneous Examples IV.
XXVI.
Quadratic Equations Pure Quadratic Equations Affected Quadratic Equations Solution by Formula .
.
.
.
Solution by Factoring Formation of Equations with Given Roots Values found for the Unknown Quantity which do not satisfy the Original Equation .
XXVII. XXVIII.
Equations in Quadratic
Form
Simultaneous Equations involving Quadratics of the Same Degree Symmetrical Equations Miscellaneous Cases
Homogeneous Equations
.... .....
Xll CHAPTER
CONTENTS.
xm
CONTENTS.
Advantages of Common Logarithms Logarithms transformed from Base a to Base Logarithms in Arithmetical Calculation Four-Place Table of Logarithms Use of the Table Cologarithms Exponential Equations
.... .
.
XXXIX. XL. XLI.
Interest
and Annuities
.
Limiting Values and Vanishing Fractions
Convergency and Divergency of Series Tests for Convergency
Auxiliary Series
XLII.
Undetermined
Coefficients
Functions of Finite Dimensions Functions of Infinite Dimensions
Expansion Expansion
of Fractions into Series of
Surds into Series
Reversion of Series Partial Fractions
The General Term XLIII.
.
.
Continued Fractions Formation of Successive Convergents Limits to Error
Recurring Continued Fractions
XLIV.
Summation
of Series
Scale of Relation
The Sum of n Terms The General Term The Method of Differences .
Any
Required Term of n Terms Piles of Shot and Shells
The Sum
Interpolation
XLV.
XL VI.
.
.
Binomial Theorem
:
Proof for
Any Index
Exponential and Logarithmic Series
h
CONTENTS.
XIV
XLVII.
......
Determinants Minors Vanishing of a Determinant Multiplication of a Determinant A Determinant expressed as the Sum of Two )tlicr Determinants Simplification of Determinants Solution of Simultaneous Equations of the First Degree .
....
Signs of the
....
Terms of Lower Order
Determinant
XLVIII.
,
Theory of Equations Horner's Method of Synthetic Division
Number of the Roots Depression of Equations Formation of Equations
.
.
Relations between the Roots and the Coefficients Fractional Roots
....
.
Imaginary Roots Transformation of Equations Standard Form of Reciprocal Equations Descartes' Rule of Signs Derived Functions Equal Roots Location of the Roots Sturm's Theorem and Method Graphical Representation of Functions Solution of Higher Numerical Equations Newton's Method The Cube Roots of Unity Cardan's Method Biquadratic Equations Incommensurable Roots Horner's Method of Approximation Any Root of Any Number .
.... .
.
.
Miscellaneous Examples VII.
.
ALGEBRA.
CHAPTER Definitions.
I.
Substitutions.
Algebra treats of quantities as in Arithmetic, but with for while the quantities used in arithmetical processes are denoted by figures, which have a single definite value, algebraic quantities are denoted by 1.
greater generality
symbols,
;
which may have any value we choose
to assign
to them.
The symbols of quantity employed are usually the letters own alphabet and, though there is no restriction as
of our
;
may represent, it is understood that in the same piece of work it keeps the same value throughout. Thus, when we say " let a equal 1," we do not mean that a must have the value 1 always, but only in the particular example we are considering. Moreover, we may operate with symbols without assigning to them any parto the
numerical values a symbol
ticular numerical value
that Algebra
is
;
indeed
it
is
with such operations
chiefly concerned.
We
begin with the definitions of Algebra, premising that x, h-, (), = will have the same mean-f, — ings as in Arithmetic. Also, for the present, it will be assumed that all the algebraic symbols employed represent integral numbers. the symbols
2.
may
An
,
algebraic expression is a collection of
consist of one or
B
symbols;
more terms, which are the parts 1
it
sepa-
ALGEBRA.
2
Thus, from each other by the signs + and — 7a-j-ob — Sc — x-{-2y is an expression consisting of dve rated
.
terms. Note. 3.
When
no sign precedes a term the sign
+
is
understood.
Expressions are either simple or compound.
A
simple
A
compound expresexpression consists of one term, as 5 a. Compound expressions sion consists of two or more terms.
may
be further distinguished.
terms, as 3 a
— 2 b,
is
2a
+
three terms, as
Thus an expression
+
and compound
sions are frequently spoken of as monomials, expressions as multinomials or polynomials.
4.
When two
or
of two
often called a binomial, and one of Simple expresc, a trinomial. 36
more quantities are multiplied together
the result is called the product. One important difference between the notation of Arithmetic and Algebra should be In Arithmetic the product of 2 and 3 is here remarked.
2x3, whereas in Algebra the product of a and b be written in any of the forms a x b, a b, or ab. The form ab is the most usual. Thus, if a = 2, 6 = 3, the product ab = a xb = 2 x3 = 6; but in Arithmetic 23 means 3. "twenty-three," or 2 x 10 written
may
-
+
5. Each of the quantities multiplied together to form a product is called a factor of the product. Thus 5, a, b are the factors of the product 5 ab.
Note. The beginner should carefully notice the difference between term and factor.
6.
When
one of the factors of an expression
cal quantity, it is
is
a numeri-
called the coefficient of the remaining
Thus, in the expression oab, o is the coefficient. coefficient is also used in a wider sense, and it is sometimes convenient to consider any factor, or factors, of a product as the coefficient of the remaining factors. Thus, in the product G abc, G a may be appropriately called
factors.
But the word
DEFINITIONS. the coefficient of
numerical
is
A
be.
the coefficient
instead of
a,
which
coefficient
sometimes called a
When
Note.
write simply
SUBSTITUTIONS.
not merely
literal coefficient.
unity
is
is
3
it is
we
usually omitted, and
1 a.
7. A power of a quantity is the product obtained by repeating that quantity any number of times as a factor, and is expressed by writing the number of factors to the right
and above
of the quantity
Thus,
it.
a X a is called the seco7id power of a, and is written a^ a X a X a is called the third power of a, and is written
and
;
a?
;
so on.
The
index or exponent
indices of
Thus
When
the index
8.
Thus
o}.
is
"« squared" ;
unity
a, 1
rt,
a^,
and it
and
Ex.
1.
By By
3a a^
Thus,
What
is
and
The beginner must be
coefficient
2, 5,
7 are respectively the
a^, a^, dJ.
Note, a^ is usually read a^ is read " a to the fourth " instead of
number which expresses the
the
is
power of any quantity.
;
a^ is read
"« cubed"
we write simply have the same meaning.
omitted, and 1 o} all
the difference in meaning between 3 a and a^
?
we mean the product of the quantities 3 and a. we mean the product of the quantities a, «, a. if
a
= 4, = 3xa = 3x4 = 12; a'^=:ax«xa = 4x4x4 —
3ff
Ex.
2.
If 6
Here
4&'^
whereas
2 6*
Ex.
3.
Here Note. Ex.
4.
If
ic
5x'*
64.
= 5, distinguish between 4 6^ and 2 h^. = 4x6x& = 4x5x5 = 100 =:2x&x&x?>x/) = 2x5x5x5x5 =
1250.
= 1, find the value of 5x^. = 5xxxajxxxx = 5xlxlxlxl =
5.
;
The beginner should observe that every power If
a
= 4, =
5x"
X
=
1,
5xx''
find Uie value of 5x«.
=
a,
careful to distinguish between
index. is
;
so on.
5xl'*
= 5xl =
5.
of 1 is
1.
ALGEBRA.
4 9.
10. *'
The Sign
of Continuation,
The Sign
of
Deduction,
is
•••,
is
.-.,
read **and so on." read
**
therefore" or
hence." 11.
In arithmetical multiplication the order in which the
factors of a product are written
is
immaterial.
Thus
3x4 = 4x3. In like manner in Algebra ab and ha each denote the product of the two quantities represented by the letters a and h, and have therefore the same value. Although it is immaterial in what order the factors of a product are written, it is usual to arrange them alphabetically. Fractional coefficients which are greater than unity are usually kept in the form of improper fractions. Ex.
If
a
=
6,
Here if ax^
a:
=
=
7,
s
=
if x 6 x
5, find
7
x 5
=
the value of i| axz. 273.
EXAMPLES If
a
=
7,
I.
a.
SUBSTITUTIONS.
DEFINITIONS. Ex.
If
2.
a
=
=
4, ^
9,
=
8
#
27 X
the value (M e of
6, find
8x9x6^
hoc^
27 a^
— 2' 27 a'
X 9 X 36
11.
27 x 64
If one factor of a product is equal to 0, the product
13.
must be equal
A
a:
to 0, ivhafever values the other factors
may have.
usually called a zero factor. For instance, if x 0, then alfxy^ contains a zero factor. when x 0, whatever be the values of Therefore cOfxy^ factor
is
=
=
=
«, b, y.
Again, values
rt
if c
and
= 0,
then
may
b
Every power
Note.
=
c^
;
therefore alrc^
of
=
=
b
=
whatever
is 0.
EXAMPLES a
= 0,
have.
X
=
y
=
o^
b.
I.
find the value of
2,
c
1.
4ax2.
3.
Sh-^y.
5.
f&'^x.
7.
^xy^.
2.
a^b.
4.
3xy'\
6.
^b^y^.
8.
«3^'.
0,
g
6,
find the value of
If
If
11
=
a
7,
2,
=
6
§^.
3,
14.
Sb
^i^.
12.
15.
c
=
0,
1,
5,
p=
i^'. 9a^
17.
=
4,
r
=
^.
20.
3''2^
9.
10.
23.
18.
5«*c"-.
Sx^y.
^^^. 64r«
9a'-
3«2i!)^.
a^cy.
21.
2'-rt5.
24.
?^. 32
9^/ 13.
^^.
16.
19.
iba>:
^^.
22.
c^b^.
25.
^.
The square root of any proposed ex14. Definition. pression is that quantity whose square, or second power, is equal to the given expression. Thus the square root of 81 is 9,
because
The square
9^
= 81.
root of a
is
denoted by -^a, or more simply ya.
Similarly the cube, fourth, fifth, etc., root of any expression is that quantity Avhose third, fourth, fifth, etc., power is
equal to the given expression. roots are denoted by the symbols -^, ^,
The
Examples
:
^27 = ^32 =z
3
;
because 3^
2
;
because 2^
= =
27.
32.
^,
etc.
ALGEBRA.
6
The symbol Ex.
1.
^
is
sometimes called the
Find the value
= = Ex.
2.
Fmd
when «
of 5y'(6 a^b^c),
i)y/{6a^b*c)=o X
the value of
6
V(6 x X V(6 X
5
X 36
=
^ I'^X
M/a6*\_^/9x 34\_3i/9x 9x ^
33
x
If 1.
a
=
8, c
V(2«)-
=
0,
k
=
9,
X
=
4,
?/
=
1,
1*
27 X 8)
=
5
3,
=
1,
c -
x 8)
=
5
X ^1296
180.
^hen
\_
81 \
I.
a
=
^ ^
/9 / O x X
AV
.8x125/
EXAMPLES
radical sign.
9,
b
=
x
S,
=
9\ 9x 9\_
1000
c.
find the value of
/
5.
9 10*
:
SUBSTITUTIONS.
DEFINITIONS. given,
7
and by combining the terms the numerical value of
the whole expression
is
obtained.
16. We have already, in Art. S, called attention to the importance of carefully distinguishing between coefficient and index; confusion between these is such a fruitful source of error with beginners that it may not be unnecessary once more to dwell on the distinction. Ex.
1.
Here
find the value of c* - 4 c + 2 c^ - 3 c^. = 5^ = 5 x 5 x 5 x 5 = 625 4 c = 4 X 5 = 20 2 c3 = 2 X 53 = 2 X 5 X 5 X 5 = 250 3 c2 = 3 X 52 = 3 X 5 X 5 = 75. the value of the expression = 625 — 20 + 250 — 75 = When p = 9, r = 6, k = 4, find the value of
When
c
=
5,
c*
;
;
;
Hence Ex.
2.
= lX 1+9-2=:: 17.
By
itself zero,
Ex.
1.
If
which contains a and may be called a zero term.
=
2,
6
= 0,
The expression
aj
=
3,
^ab^ +
y 2
=
1,
zero factor
is
find the value of
+ 3 abx. = (4 x 2^) - + (2 x 3 x ^ 32 - + 6 + = 38.
4 a3
Note.
7i.
Art. 13 any term
a
780.
xy'^
The two zero terms do not
1)
+
affect the result.
+ 7 abx - | y^, when = 7, y = \. 3= 4 X 72 - 52 X 1 + - f X 13 = 29f-25-2t = l^. a^y
-
Note. 18.
The
5,
&
f
2/3
= 0,
X
zero term does not affect the result,
In working examples the student should pay atten-
tion to the following hints 1.
Too much importance cannot be attached to neatness and arrangement. The beginner should remember
of style
that neatness
is
in itself conducive to accuracy.
ALGEBRA.
8
The
=
should never be used except to connect Beginners should be particularly careful not to employ the sign of equality in any vague and inexact sense. 3. Unless the expressions are very short the signs of equality in the steps of the work should be placed one under the other. 4. It should be clearly brought out how each step follows from the one before it for this purpose it mil sometimes be advisable to add short verbal explanations the importance of this will be seen later. 2.
sign
quantities ^vhich are equal.
;
;
EXAMPLES a
li
.
=
b = S, c = \ d = + 56 -8c+ 9f?. - 46 + 6c + 5
2,
,
Ort
2.
3a
3.
6ab-Scd-]-2da-5cb-\-2db.
c7.
+
+
+
abc
5.
Sabc-2bcd-{-2cda-4dab.
If
a
=
bed
cda
4.
6 = 2, c = 3, + b^+ + #.
1,
11.
a^
13.
Sabc-b'^c-Qa^
=
+
6.
2bc
7.
Sbcd-\- 5 cda
8.
dab.
d
d.
the numerical value of
0, find
1.
I.
9.
10.
Scd--ida-\-
-
c'^
b'^
the numerical value of
0, find
12.
c'3
4 6c3
_
a^
-
-
6^
19.
If a
=
16. 17. 18.
6-^
2 a2
c-2
|
abh.
fr-2
6-^
a2
C-*
?>*
8,
6
=
6, c
=
l,
a;
+ abc.
+ + + cR 2a^ + Sb^-i c*. a^ + b* - cK a^
+ 2 + 2 + 2 - 2 6c - 2 cd -2 da -2 ab. + 2 + 2 c2 + d^ + 2 a6 + 2 6c + fed 2c2 + 2a2 + 2 62-4c6 + 6rt6aZ. 13 a2 + -V + 20 rt6 - 16 ac - 16 6c. - 3 + I cK 6 rt6 - I ac2 - 2 « + i 125 6*c-9d5 + 3a6cU
i4. 15.
l^ab.
dab
7
(Z
4, find
24.
the value of
^{bxij)-lb''
+ ^. b^y
CHAPTER Negative Quantities.
II.
Addition of Like Terms.
work the student has been accuswith numerical quantities connected by the signs + and — and in finding the value of an expression such as If + 7| — 3|- H- 6 — 4i he understands that the quan19.
tomed
In
liis
arithmetical
to deal
;
which the sign -f- is prefixed are additive, and those which the sign — is prefixed are subtractive, Avhile the first quantity, 1|, to which no sign is prefixed, is counted among the additive terms. The same notions prevail in Algebra thus in using the expression 7 a + 36 — 4c — 2 c? we understand the symbols 7 a and 3 6 to be additive, while 4 c and 2 d are subtractive.
tities to
to
;
20. But in Arithmetic the suui of the additive terms is always greater than the sum of the subtractive terms and if the reverse were the case, the result would have no arithmetical meaning. In Algebra, however, not only may the sum of the subtractive terms exceed that of the additive, but a subtractive term may stand alone, and yet have a ;
meaning quite intelligible. Hence all algebraic quantities may be divided into positive quantities and negative quantities, according as they are expressed with the sign -f or the sign — and this is quite irrespective of any actual process of addition and sub;
traction.
This idea
may be made
clearer
by one or two simple
illus-
trations.
Suppose a man were to gain $ 100 and then lose $ 70, would be $30. But if he first gains $ 70 and then loses f 100 the result of his trading is a loss of $ 30. (i)
his total gain
9
.
ALGEBRA.
10
The corresponding
algebraic statements would be
^100-^70 = + $30, $70 -$100 = -$30, and the negative quantity
in the second case is interpreted
money opposite in character to the positive quantity, or gain, in the first case in fact it may be said to possess a subtractive quality which would produce its effect on other transactions, or perhaps wholly as a debt, that
is,
a
sum
of
;
counterbalance a
sum
gained.
Suppose a man starting from a given point were to walk along a straight road 100 yards forwards and then 70 yards backwards, his distance from the starting-point would be 30 yards. But if he first walks 70 yards forwards and then 100 yards backwards his distance from the startingpoint would be 30 yards, but on the opposite side of it. As before we have (ii)
100 yards 70 yards
— 70 yards = + 30 — 100 yards = — 30
yards, yards.
In each of these cases the man's absolute distance from the starting-point is the same but by taking the positive and negative signs into account, we see that — 30 is a distance from the starting-point equal in magnitude but opposite in ;
+
direction to the distance represented by 30. Thus the negative sign may here be taken as indicating a reversal of direction.
Many
other illustrations might be chosen
but it will be here to remind the student that a subtractive quantity is always opposite in character to an additive quantity of equal absolute value. ;
sufficient
Note. signs
-I-
Absolute value
and
is
the value taken independently of the
—
When any number of quantities are 21. Defixitiox. connected by the signs -h and — the resulting expression Thus 11a — 27 a -\- ISb — 5b is called their algebraic sum. This expression, however, is not, as is an algebraic sum. will be shown, in its simplest form. ,
ADDITION OF LIKE TERMS. 22. Addition
algebraic
sum
is
of
11
the process of finding in simplest form the
any number of
quantities.
23. Like terms, or similar terms, do not differ, or differ only in their numerical coefficients. Other terms are called 5 a'^h, 2 a^h 3 a^h^, Thus 3 a, 7 a unlike, or dissimilar. — 4 cC'h- are pairs of like terms and 4 a, 3 6 7 a^, 9 d-h are pairs of unlike terms. ;
;
;
;
ADDITION OF LIKE TERMS. Rule
I.
The sum of a number of
Rule
II.
If all the terms are positive, add the
like
terms
is
a like term.
coefficients.
Find the value of 8 a + 5 a. Here we have to increase 8 like things by 5 like things same kind, and the aggregate is 13 of such things Ex.
of
the
;
8 lbs.
for instance,
Hence
+
=
5 lbs.
8a+5a =
also,
8a
Similarly,
+5a +
r<
+
13 lbs.
13«.
+
2 «
6a
=
22
a.
Rule III. If all the terms are negative, add the numericcdly and pjrefix the minus sign to the sum.
To
Ex.
find the
sum
of
—3
a:,
—
6x,
—
7 x,
coefficients
— x.
Here the word sum indicates the aggregate of 4 subtractive quantiIn other words, we have to take away succesties of like character. sively 3, 5, 7, 1 like things, and tlie result is the same as taking away 3
+
+
5
7
+
1
such things in the aggregate.
Thus the sum
Rule IV.
of
—3
a:,
—5
a*,
—
—
7 x,
If the terms are not
cdl
a-,
is
—
16 x.
of the same
sign,
together separately the coefficients of cdl the positive terms the coefficients
two
of
all the
results, p)receded
coefficient
Ex.
and 8
is
negative terms; the difference of these loill give the
by the sign of the greater,
of the sum required.
The sum of 17 aj and — 8 a: 9, and the greater is positive.
1.
add and
is
9x, for the difference of 17
Ex. 2. To find the sum of 8 a, — 9«, — «, 3 a, 4 a, —11a, a. The sum of the coefficients of the positive terms is 16. The sum of the coefficients of the negative terms is 2L The difference of these is 5, and the sign of the gieater is negative lience the required sum is — 5 a.
;
ALGEBRA.
12
We
need
terms
may
not, however, adhere strictly to this rule, for the be added or subtracted in the order we find most
convenient.
This process Ex.
called collecting terms.
is
Find the sum
3.
=
The sum The sum
Note.
opposite signs
of | «, 3 «,
3f «
—
=
1^ «
EXAMPLES Find the sum
5 a, Trt, 11a,
2.
Ax,
3.
7 6, 10 6,
4.
6
c,
8
23
rt,
Sx,lx,9
c,
116, 9
9.
a.
X. 6,
2
ft.
2c, 15c, 19c, 100c,
c.
6.
-7 _ 3 X, - 5 - 11 -56,-66,-116,-186.
7.
—'3y,
8.
—
5.
c,
y,
a:.
:r,
.'•,
—'
=
2
a.
f «.
II.
of
1.
X,
—
i a,
two quantities numerically equal but with The sum of 5 a and — ba is 0.
of
zero.
is
-
2ia
—y, —2y, —Ay.
-2 c, —50 c, —13
c.
- Uh, - ob, - Sb, - b. - X, -Sx, 2x, - x.
10.
ox,
11.
26y, -lly, -15y,
12.
5/,
13.
2
-
9/,
-
-Sij,'2y.
y,
-
3/; 21/,
30/.
14.
1 y,
- 3 s, s, - s, - 5 s, 5 s. - Uy, 16y, - Sy, - 2y.
15.
5x,
—Ix, —Ix^lx,
16.
7a6,
19.
3a3-7a3-8a3+2a3-lla3.
s,
—
3a6,
2.r,
—oab,
—5a-.
2ab, ah.
Find the value of
+
11x2
+ 3x2-
17.
_9a;2
18.
3a2x- 18a2x + 21. 22. 23. 24. 26. 27.
28.
29. 30.
4a;2.
- 5x^ - Sx^ - a262. -9x^-4x4- 12x^+13x4 - 7x*. 7 abed — 11 abed — 41 abed + 2 abed. |x-Jx + x+fx. 25. P.a + ^a - ^a. -56 + i6-|6 + 26-^6 + t&. - f X2 - 2 X2 - f X2 + X2 +i X2 + i^X^. - a6 - h ab - lab - \ab - I ab + a6 + y^ a6. fx-|x + |x-2x + Ya;-i:e + -.5 a;2 - |x2 - Ax2 - ix2 - x2.
4 a262
_
20.
a2x.
a^l/2
_
7
cr-b'^
+
4x^
5 a262
a;.
1
x^
:
CHAPTER Sesiple
Brackets.
III.
Subteactiox.
Addition.
24. When a number of arithmetical quantities are conthe value of the result is the and nected by the signs same in whatever order the terms are taken. This also'
—
+
,
holds in the case of algebraic quantities. Thus a — 6 c is equivalent to a -\- c — b, for in the first of the two expressions b is taken from a, and c added to the result in the second c is added to a, and b taken from the result. Similar reasoning applies to all algebraic expres-
+
;
Hence we may write the terms
sions.
any order we please. Thus it appears that the expression a in the equivalent form — b -\- a.
of
an expression in
— b may
be written
To illustrate this we may suppose, as in Art. 20, that a represents a gain of a dollars, and & a loss of b dollars it is clearly immaterial whether the gain precedes the loss, or the loss precedes the gain.
—
25. Brackets ( ) are used, as in Arithmetic, to indicate that the terms enclosed within them are to be considered
The
as one quantity.
sidered in Chap. vi. simpler cases.
8
+ (13 4- 5)
their
full
here
;
use of brackets will be conshall deal only with the
we
means that 13 and 5 are
sum added
to
8.
It is clear that
to be added and 13 and 5 may be
added separately or together without altering the
Thus
8
+ (13 + 5) = 8
Similarly a -\- {b be added to a.
Thus
a
-\-
-\-
c)
(b
4-
13
means that the sum -\-
c)
= 13
a
-\-
result.
+ o = 26.
b
+
c.
of b
and
c is to
:
ALGEBRA.
14
— 5)
means that to 8 we are to add the excess of if we add 13 to 8 Ave have added 5 too much, and must therefore take 5 from the result. 8 4- (13 13 over 5
now
;
Thus
8
+
(13
- 5) = 8 + 13 - 5 =
10.
Similarly a-\-(b—c) means that to a we are to add ished by c.
Thus
a
+
(b~c)
= a-\-b — c
.
.
.
b,
dimin-
.
(1).
In like manner,
a
+ b-c +
(cl-e
-/) =
a
+b-c
-\-
d
- e -f.
(2).
Conversely,
+ b-c-\-d-e -f= a + b-c^{d-e -/). — b-\-c = a + c — b, [Art. = the sum of a and c — b, = the sum of a and —b-\-c, [Art. therefore a — b-{-c=a-\-{—b-{-c) a
(3).
Again, a
21.]
.
By
considering the results the following rule Rule. the sign
When an -{-,
(1), (2), (3), (1),
.
we
.
.
21.] (4).
are led to
expression
ivitliin brackets is preceded by can be removed without making any
the brackets
change in the expressioyi.
Conversely
:
tvithin brackets
Any and
p)art
of an expression
the sign
-\-
may
be enclosed
prefixed, the sign of every term
within the brackets remaining unaltered.
Thus the expression a — b-\-c—d-\-e may be written in any of the following ways,
+ — 6 + c-fZ + e) a — b (c — d 4- e), a — + c+ (— + e). c(
(
-\-
?>
(?
The expression a — {b -{- c) means that from a we are sum of b and c. The result will be the same whether b and c are subtracted separately or in one sum. Thus a — ip -{• c) = a — b — c. 26.
to take the
:
SIMPLE BRACKETS.
15
Again, a — (b — c) means that from a we are to subtract If from a we take b we get a — b\ excess of b over c. but by so doing we shall have taken away c too much, and tlie
do a — Thus a — (b — c) = a — b In like manner, a — b — (c —d — e)
must therefore add
b.
Accordingly the following rule
When an
Rule. the sign
term
—,
-\- c.
a
=^
may
—b—c
may
luithin the brackets be
be
d
-\- e.
be enunciated
expression within brackets
the brackets
-{-
removed
is
^yreceded by
if the sign of every
changed.
Any
part of an expression may be enclosed within brackets and the sign — prefixed, provided the sign of every term ivithin the brackets be changed.
Conversely
:
Thus the expression a — b-\-c-\-d any of the following Avays, a a a
We I.
— e may
be written in
— (-\-b — c — d-\-e), — b — {— c — d e), — b -Y c — {— d e). -\-\-
have now established the following results
Additions and subtractions
Thus a-\-b
may
be
made
in
:
any
order.
— c-\-d — e —f= a — c -\-b d — /— e = a — c — /+ d-{- b — -\-
e.
This
is
known
as the Commutative
Law
for Addition
and
Subtraction. II.
The terms of an expression may be grouped
i)i
any
manner.
+ b-c + d-e -f = (a + l))-c-^(d - e)-f = a +(6 - c) + - e)~f= a + b -{c - d)-{e +/).
Thus a
(c?
This
is
known
as the Associative
Law
for Addition
and
Subtraction.
ADDITION OF UNLIKE TERMS. 27.
When two or more
we have seen
that they
Wke terms are to be added together may be collected and the result
;
;
:
ALGEBRA.
16
expressed as a single like term. If, however, the terms are unlike, they cannot be collected. Thus in finding the sum of two unlike quantities a and b, all that can be done is to connect them by the sign of addition and leave the result in the form a + h. Also, by the rules for removing brackets, a-\-{ — h) = a — h\ that is, the algebraic sum of a and — 6 is written in the
—
form a
h.
It will be observed that in Algebra the word sum used in a wider sense than in Arithmetic. Thus, in the language of Arithmetic, a — h signifies that h is to be subtracted from a, and bears that meaning only but in Algebra
28.
is
;
means the sum of the two quantities a and without any regard to the relative magnitudes of a and
it
also
- 4 a 45 & + 2c 2« + 3& Ex. 1. Find the sum of 3 a 5 + 2 c) + (2 The sum = (3 a + 3 6 - d) + ( - 4 a + 2 6) =:3a-56 + 2c + 2a + 35-cZ-4rt + 26 = 3a + 2a-4a-56-l-36 + 26 + 2c-(Z = a + 2 c — d, -
c?
;
;
—h h.
2
&.
rt
?)
like terms.
by collecting
The addition
is
more conveniently
effected
the expressions
lines
by the
fol-
lowing rule
Arrange
Rule.
may
terms
he
in the
column, beginning „
-
-
-,a
+
4«
-f
7
,
The
—
c
2 &
,
,
8ab „
-\-
,
,
.
,
— — ~-2ab + lbc ,
-^
be
that the like
left. fii*st
change.
3
aft -f
Here we
7
,
c
J
,
-\-
=-;
A
Aad
—ad
z -\-
t
ji
;
8 ab
-\-
Sac
-2 ad; -2ab
4 ad. ^^
-2o ad
+iac + bad
,
-2ab o -Sab ;
o Sac
so
algebraic sura of the terms in the
Add together — 5ab + Gbc -7 ac
2.
+ 4tac + bad; be — -dab + Qbc-7 ac „
on the
in
columns: then add each
column is a, that of the terras in the second column is zero. The sint^le tei-ms in the third and fourth columns are brought down without
+2r-d
Ex.
vertical
ivith that
o
•
a
same
,
,.,
^
,
„ ^ first
rearrange ^^ the expressions °,
^^
so that like terms are in the .
tical ,
^
same
^^
ver-
columns, and then add up each ^
'
column separately. ^ '^
ADDITION.
EXAMPLES
17
III.
a.
Find the sum of
5.
a + 2/j-3^; -3a + 5+2c;2rt-36 + c. 3a + 26-c; -a + 3& + 2c;2a-6 + 3c. -Sx-\-2y + z; x-Sy-h2z; 2x + y -3z. -x + 2y + Sz; Sx- y + 2z; 2x-hSy -z. 4« + 36 + 5c; - 2a + Sb -Sc a - b + c.
6.
-
1.
2. 3.
4.
;
-
-
+ 15 6 + 8 c a-h 51 ±9c. 25 a - 15 & + c 13 « ~ 10 & + 4 c a + 20 6 - c. -16a-105 + 5r; lOa + 56 + c; 6a + 55-c. 5ax — 7by+cz; ax + 2by — cz; —oax+2by-\-o cz. 20 p + q- r p -20q-{- r; 2^ + q -20r.
7.
15 a
19 6
18
c
;
14 a
;
;
8.
9.
10.
;
;
Add
together the following expressions
11.
-
12.
loa6-275c-6ca
13.
14. 15. 16. 17.
18. 19.
20.
6ab -{-6bc-7 ca; 8 r/6 - 4 6c + 3 ca 14rt6-186c+10cff
—
;
-
;
2 «6
-
2 6c
+
4 ca.
-49 a6 + 45 6c-3ca.
;
— a6 + 6c + 2 cd. «6 — 6c + ca qr rp + pq + pq + i)g - qr + rp. z; 2x + Sy -2z; Sx- 4:y + z. x-\- y 2a-36 + c; 15 a- 21 6 -8c; 3a + 24 6 + 7c. — 25 xy -\-2\yz — zx 23 xy — lo yz + zx. ixy — 9yz 2 zx - 5 a6 + 9 6c - 7 ca 2 a6 - 7 6c - ca. 17 a6 - 13 6c + 8 ca 47a;-63y + ^; -25x+15y-32; -22x + 48y + Ibz. 23a-176-2c; -9a + 156 + 7c; -13a+36-4c. 5 a6
+
6c
qr
-
3 ca
rp
;
;
;
;
-\-
-\-
;
;
;
;
DIMENSION, DEGREE, ASCENDING
AND DESCENDING
POWERS. 29. Each of the letters composing a term is called a dimension of the term, and the number of letters involved is called the degree of the term. Thus the product abc is said to be of three dimensions, or of the third degree; and a.v* is said to be of five dimensions, or of the fifth degree. A numerical coefficient is not counted. Thus ^o/W and
each of seven dimensions, or of the seventh degree. sometimes useful to speak of the dimensions of an expression vrith regard to any one of the letters it in-
a-h^ are
But
it is
;
ALGEBRA
18
For instance, the expression
volves.
in a,
4-
which
of
is
be said to be of three dimensions of fonr dimensions in b, and of one dimension in c.
A
30.
when
Sa^b^c,
may
eight dimensions,
compound expression
is said to be homogeneous terms are of the same degree. Thus 8 a^ — a^b'^ a homogeneous expression of six dimensions, or of
all its
9ab^
is
the sixth degree.
31. Different powers of the same letter are unlike terms thus the result of adding together 2 x"^ and 3 x^ cannot be expressed by a single term, but must be left in the form ;
Similarly, the algebraic
5
a^b^
— 3 ab^ —
sum
of 5 a^b^
This expression
b^.
is
— 3 ab^
— b^
and
is
form
in its simplest
and cannot be abridged. 32. In adding together several algebraic expressions containing terms with different powers of the same letter, it will be
found convenient to arrange
all
the expressions in
descending or ascending powers of that letter.
made
clear
Ex. 1. Add 4x - 2x3 + 3x2 3 x3
-
5 x2 2 x2
—
2x3 4-3x2
3 x3
— — —
Ex.
2.
3 x3
—
x^
x2 x2
2 x2
together
3x3
;
-
Add
3 x^
+
-
x2
9X
.
5
5
a-
6x
_
- 5 ic2 - x3 +
x2
2 x2
;
-
8
-
9
a:-
4.
the
column we have like powers of the same letter. The result is in descending powers of x.
a'2b
(^ij)
5 6^
-ab'^-3a^; Sa^ + 6b^; 9a'^b-2 a^
+ —
a&2
_|_
ab"-
+
Q^2b
14
a-'h
— +
+
ab^.
a3
Sa^
-1-8 a^
S
+
together
_ 2 &3 4. ^3 -2b^ + 3ab^ — «62 ^_
+
7 ;
In writing the first expression we put in first term the highest power of x, in the second term the next highest power, and so on till the last term, in which x does not appear. The other expressions are arranged in the same way, so that in each
+ 6X + 7 — 9X — 8 + 4x — 9X + X +4 — 7X+3
3 a&2
" Sb^
This will be
by the following examples.
2a^ 4 a^
Here eacli expression is arranged according to descending powers of 6, and ascending powers of a,
ADDITION.
EXAMPLES Find the sum 1.
2
rt6
2.
2
cc^
3.
4. 5.
6. 7. 8.
9.
10.
11.
12. 13.
14. 15. 16.
17. 18. 19.
20.
Add 21. 22. 23. 24..
25. 26."
27. 28. 29. 30.
19
III. b.
of the following expressions
+ + — 2 xi/ + 3 - 7 rt6 - 4
6 ahc
3 ac
?/2
—
;
4
;
2/2
5 a6
+
5
+
2 6c
-
.r?/
—
2 x-
:
ahc
-5
a:-
;
-
— 2 6c — -6
3 «&
;
2
.i-y
3 «c.
y'^.
- 6 «2 + 9 ^^^ _ 3 4 ^2 _^ ^ 5 ^2. _ ^2 _^ ^^. ^ ^2_ ^^ + ^2 — X^ — 3 + 3 ^^ 3 ^^ + 4 ^py _ 5 4- x^ + - 3 x^ + 5 x + 1. 2 x2 - 2 X + 2 x3 - x2 + X - 1 — 6 x^ — 6 + x. 2 x^ — x2 — X 4 x^ + 8 x2 + 7 X 9x2-7x + 6; -14x2+15x-6; 20x2-40x-17. - 6 2 - 2 x - 3. 10 x3 + 5 X + 8 3 x^ - 4 he ah a^ — ah h^ — ac; ac — he c^. - 2 a^ + 3 d^ 4 c3 _ 2 «3 - 3 #. 5 a3 - 3 c3 + # 6x3 - 2x 4- 1 2x3 + X + 6 x2 _ 7x3 + 2 x - 4. rt3_^2 + 3^. 3^3 _^ 4^2+ 8«; ba^ - Qa"^ -l\ a. - 2 x?/ 2 ^2 _ 3 ^2 _ 4 y-. 2 x2 _ 2 - 3 x^. x2 + — 2 x3 + —x+ _^ 2 x3 — 2 +X -'Sx-y -6 x?/2 - x3 3 xhj + 4 x?/2. x3 + 3 x2?/ + 3 xif 53 _ 10 ah- - a^ 4. 6 aV^ - 2 63 + 2 a%. a3 4- 5 3 x4y 4- 2 x3?/3 - 6 x?/* 3 x3y3 4- 6 x?/4 x5 - 4 x4?/ - 5 x3^3 63 + 3 ^^2^ 4. ^^c. 4. e a6c orh - 10 a6c + c3 a3 _ 4 2 x2y - 3 x?/2 + 2 y3 yS 3 ^2^ 4. 4 ^^2. x3 - 4 yi^y + 6 xy2 3 a-
a;2
+
a:y
-
if-
?>2
-
;
&-2
;
^^
a;?/
.|_
2/2
;
;
(-(/,
;>;2
;
2/2.
;
;
x'^
;
x'^
;
-\-
.
.
x'^
;
-{-
;
-i-
6'^
;
;
?/2
;
;
^-2
.
;
?/3
?/3
?/
;
?/2
a;2
?/3.
;
;
rt?^2
//i
.
;
.
;
a'^jy
?/5.
;
;
.
_|_
.
together the following expressions
ia-i6;-« + f6;fa-6. -i«-i6; -f a + f6; -2a-
:
6.
- ia-26; |6-3c. - -V a - -V c 2 a - 3 6 -U- & - c i -^ - X?/ - I - X2 - I X?/ + 2 I x2 f X2 + X?/ 62 2 _ - ah + 62. _ _ 52 (^2 3 a2 + 2 a6 1 ^ f a2 1:^2 i|x?/-2/2; -x?/ + i fx2-ix2/+f'V?/2; -|x2 + - f x3 4- 5 «x2 - f a^x x3 - ^ (2x2 4- ^2-;^ _ ^ ^3 + | a^x. - f x2 4- 4 f X2/ + -V-2/2 f x2 - f x?/ - 7 i a3 __ 2 a26 - | # + ^62 4- \ 63 ^26 63 | a62 4- 2 63 f f - 2a +
f c;
;
;
2/2
?/2
;
?/2.
;
3
.
f^^,
;
2/2.
1
.
;
2/2
2/2.
;
;
;
;
a a
:
ALGEBRA.
20
SUBTRACTION. The simplest 33. Subtraction is the inverse of Addition. cases have been considered under the head of addition of like
terms, of which some are negative.
5a-3a=
Thus
2
[Art. 23.]
a,
3 a — 7 a = — a, — 3a — 6a = — da. 4:
Also, by the rule for removing brackets [Art. 2G],
— (— S a)= 3 a S = 11 a, — 3 a — (— S a)= - 3 a + S 3a
and
-\-
SUBTRACTION OF UNLIKE TERMS. 34.
The method
Ex.
Subtract
The
It
is,
is
shown
Sa — 2b —
result of subtraction
c
in the following
from 4 a
—
3&
+
5
example
c.
= ia — Sb-{-5c—(oa — 2b — c) = 4a-36 + 5c-3a + 2& + c = 4a-3rt-3 6 + 2& + 5c + c = a-b -\-Qc.
however, more convenient to arrange the work as follows, the
signs of all the terms in the lower line being changed.
4«-36 + by addition
Rule.
3a
a
+
2
—
5c
7>
+
c
?>
+
6c
Chavge the sign of every term in the expression and add it to the other expression.
to
he subtracted,
Note.
not necessary that in the expression to be subtracted the operation of clianging signs ought to be performed mentally. It is
the signs should be actually changed
;
-x3
ALGEBRA.
22
EXAMPLES
III. d.
From 1.
3x^
-
5 y^
2.-8 x-^/2 + 3.
_
4.
«25c
+
8 xz take
15
+
x3?/
-
+
4 x^
2
13 xi/ take 4
-
?/^
-
10 xz.
+
x2?/2
7 x^?/
-
8 xy^.
-
+ 6 a6 + a^62 take 4 3 «6 5 a'^h'^. + h-ca + c2a& take 3 a%c - 5 &%« - 4 d^ah. _ 7 ^26 + 8 a62 + cd take 5 a^^ _ 7 «5'2 ^ e ^d. 5. 6.-8 x^y + 5 X2/2 - x2?/2 take 8 x^^/ - 5 + x:hf. 7. 10 a262 + 15 a&2 + g a^h take - 10 ^2^2 + 15 (^52 _ 8. 4x2 - 3 X + 2 take - 5 x2 + 6 x - 7. 9. x3 + 11 x2 + 4 take 8 x2 - 5 x - 3. 10. - 8 ^2x2 4. 5 x2 + 15 take 9 a2x2 _ 8 x2 - 5. 8
a:?/^
g a%.
Subtract
-
x2
12.
3 x^2
_
13.
&3
11.
14.
15. 16. 17.
18. 19.
20.
x3
+
X
3
r^2y
-
c3
+ 1 from + x3 -
+
2 ahc
?/3
x^
+
x2
from
from a^
+
ft^
-
x
+
1.
x^
+
3
x2?/
_
3
+
3 x?/2
+
y^.
«?>c.
- 3 x2y + 5 x3 from 8 x^ + 7 x2?/ - 3 x^2 _ ^3^ _ 3x3 from 5 x* - 8x3 - 2 x2 + 7. ^4 _^ 5 53 _^ c3 - 3 a6c from 7 a?>c - 3 «3 ^ 5 53 _ ^3. a3 1 — X + x^ — x"^ — x3 from x* — 1 + x — x2. 1 a"^ -^c(? + ^a^ + a from «2 _ 5 ^^3 _ 7 + 7 ^^5. 10 ci^h + 8 a&2 _ 8 a3&3 _ &4 from 5 a'h -6ab^-l a^b^. a3 -63 + 8 ah'^ - 7 a2?, from - 8 a&2 + 15 a'^b + &3. 7 x?/2
-
?/3
_f. ;^
_j.
From
- i x?/ - I take - | x"^ xy - y^. _ I ^ _ 1 take - |a2 + « - i. ix2 - ix + i take x - 1 + i x2. ^"^ take «x x2 f ci^i i f f _ xy2. A x?/2 take x22/ _ | x3 | f i a2x take i a^x + ^ a^ - | ax2. a'^ - 2 ax2 i
21.
i
22.
I a2
23. 24.
25.
26.
a;2
«/2
-\-
1
?/2
?/2
1
35. AVe shall close this chapter with an exercise containing miscellaneous examples of Addition and Subtraction.
SUBTRACTION.
23
MISCELLANEOUS EXAMPLES To
1.
a
-4c+
4
—
2.
-
6.
3x
—
From
5 x^
+
+
take the
1
sum
of 2
7 c
—
x
5
add the sum
of
and 3 x"^
+
+
7 x^
2
+
X.
—
+
7 a^
from the sum
5 «2
of
8 a-
—
a^
and
Subtract
5
+
x"^
3x
—
and add the
from 2 x^,
1
result
to
+ 3x-l. 2y'^ 6.
sum
of
7.
result
-}-
Find the sum from c — 4 6.
Add
8.
of
5a — 76 +
together 3x2
by 3
What
9.
—
1
5
remainder when
to the
?/3
Take x"^ — y^ from 3 x^/ — 4 ?/2, and add the remainder ixy — x^ — 3 y'^ and 2 x^ + 6 ?/2.
the result
-
the sum of 2 y — Sy^ and yis subtracted from 5 y^.
Add
5.
—
x^
+
—
x
7
+
5
c
and 3
—
6
and 2x3
to the
9 «, and subtract the
+ 5x — 3,
and diminish
2.
expression must be added to 5 x^
—
x
+
2 to produce
3 x^
+
2 to produce
7
1 ?
10.
4x3
6
3 a2
4.
3x2
7 x2
a— 36 — 2c and 2 6 — a +
-
c
Subtract 3 a + a _ 2.
3.
1
of 2
and
2 x^
2 «3
sum
the 7 6
I.
What
expression must be added to 4 x^
—
+ 7x-6?
What expression must be subtracted from 3 « — 5 6 + c so as 2a — 46 + c? 12. What expression must be subtracted from 9x-+ llx — 5 so to leave 6x-^— 17x + 3? 13. From what expression must 11 a^ — 6 ab — 7 6c be subtracted 11.
to leave
as
so as to give for remainder 5 a^
+
7
a6
From what
+
expression must 3 a6 as to leave a remainder 6 ca — 5 6c ? 14.
7 6c ?
15. To what expression must 7 x^ — make 9x3-6x-7x2? 16. To what expression must 5 a6 — 11
produce zero 17.
If
b he
-\-
—
6 ca be subtracted so
x^
—
5 x be
6c
—
7 ca
added so as to
be added so as to
?
3x2
— 7x +
2
be subtracted from zero, what will be the
result ? 18.
Subtract 3x3
difference 19.
result.
from
zero,
Subtract 3 x-
—
7
x
+
1'
and add
—
5 x 4- 1
from 2
x^
—
5x
—
3,
this last result to 2 x^
then subtract the
—
from unity, and add
2 x^ 5 x^
—
—
4.
6 x to the
,
I
CHAPTER
IV.
Multiplication. 36.
Multiplication in
its
primary sense
signifies repeated
addition.
3x4 = 3 taken 4 times = 3 + 3 + 3 + 3.
Thus
Here the multiplier contains 4 units, and the number we take 3 is the same as the number of units in 4.
of
times
Again
a x
= a-\-a -^a +
b
•••,
= a taken
b times
the number of terms being
b.
= 4x3; and so long as a and b denote posAlso whole numbers, it is easy to show that a x b = b x a.
3x4
itive
37. When the two quantities to be multiplied together are not positive whole numbers, we may define multiplication as an operation performed on one quantity which ivhen ^jerformed on unity produces the other. For example, to multiply 4 by f we perform on f that operation which when performed on unity gives -f-; that is, we must divide |- into 7 equal parts
and take 3 of them.
Now
each part will be equal to <->
X
,
7
4x3 and the result of taking 3 of such parts is expressed by 5x7 Hence
r x 7
5
=
—
-^
-•
5x7
24
;
MULTIPLICATION.
25
Also, by the last article,
4x3 ^ 3x4 ^3,,4 5x7 7x5 7 5 • 5'^7~7'^5 The reasoning is clearly general, and we may now say X h= b X a, where a and b are any positive quanti-
that a ties,
integral or fractional.
The same
is
true for any
factors of a product
Commutative Law
38.
may
number
be taken in
of quantities, hence the
any
This
order.
is
the
for Multiplication.
Again, the factors of a product
may
be
grouped in any
way we please.
Thus
This
= axbxcxd = (ab) X (cd) = a
the Associative
is
Since,
39.
that
abed
X
by
a^
is,
the index
a^
definition, a^
= aaa
.'.
Law
X
(be)
x d
=a
x
(bed).
for Multiplication.
= aaa, and a^ = aaaaa, = aaaaaaaa = a^ = a^+^
x aaaaa
of a
letter
sum of its the Index Law
in the product is the
indices in the factors of the product.
This
is
for Multiplication.
Again, .'.
When
= 5 aa, and 1 a^ = 1 aaa. 5 a^ X 7 a^ = 5 X 7 X aaaaa = 35 a^.
5 a^
the expressions to be multiplied together contain letters, a similar method is used.
powers of different Ex.
5 a362
Note.
X 8 a%x^
=
5
aaabb x 8 aabxxx
The beginner must be
=
40 a^b^x^.
careful to observe that in this proc-
ess of multiplication the indices of one letter cannot combine in any way until those of another. Thus the expression 40a^b^x^ admits of
no further simplitication.
ALGEBRA.
26
To multiply two simple expressions together, and prefix their product to the product of the different letters, giving to each letter an index 40.
Rule.
multiply the coefficients together
equal
to the
sum of
the indices that letter has in the separate
factors.
The rule may be extended to cases where more than two expressions are to be multiplied together. Ex.
1.
Find the product
The product
= x^
The product
x
x^
y.
of x^, x^,
x^
=
and
x^+s x x^
of three or
x^.
= xr+^+^ =
x^^.
more expressions
called the
is
continued product. Ex.
2.
Fmd
The product
the continued product of 6x^y^,
=
5
x:^ x
8
ij'h^
xSxz^
=
Sij-z'^,
and Sxz^.
120 xh/z^.
MULTIPLICATION OF A COMPOUND EXPRESSION BY A SIMPLE EXPRESSION. 41.
By (a
definition,
+
b)
m = m m m-\= (»i + m -\-m-\-
together with
Also
-\-
(m
-\-7n
-\-
-\-m
-\-
-f-
diminished by
(m -\-m-\-m
taken a times),
-\-
the 2)roduct
times
taken
b times)
• • •
taken
a — b
•••
taken a times),
(1).
-\- •••
taken
= am — bm {a — b-\-c) m = am — bm + cm.
Similarly,
Hence
-\-
taken a
'•'
-\
= am + bm — m -f m-\b) m = m (a = {in m m
+b
••-
times
b times) (2).
of a compound expression by a single
factor is the algebraic sum of the partial products of each term of the compound expression by that factor. This is known as
the Distributive
Law
for Multiplication.
MULTIPLICATION. Ex.
3(2 « (4
+ 3 - 4 c) = G - 7 - 8 ^3) X ?>
.x2
?/
rt
+
96
3 xif
=
27
-
12
12
Thf - 21 xy^
c,
-
24 xy^^s.
Note. It should be observed that for the present «, 5, c, «i denote positive whole numbers, and that a is supposed to be greater than
6.
EXAMPLES
IV.
a.
Find value of 1.
5x2x7x5.
6.
2«&cx3«c3.
11.
xhfxQa-x^.
2.
4«3x5a8.
7.
2a353x2a3&3.
12.
ahc x xyz.
3.
7
a6 x 8
a352.
3,
5
^i^i
^ 2
13.
3 a^b'x'^
4.
G
X//2
x^
9.
4
«2;>3
X
7
14.
4 a^bx
5.
8
a% X
10.
5
a'^b^
x
x'?/2.
15.
5 «2x
X
5
&5.
«.
a^
x
x 8
x 5 a^bx. 7 62x4. ex.
Multiply 21.
5x + 3?/by2x2.
by x^.
22.
«2
by
23.
by
16.
5x3^/3
17.
2x2^/
18.
3 «3ic4^7
19.
+ 6c by a^. 5 «6 - 7 6x by 4 a2;,a:3.
20.
6«3a;3.
^2x5^9.
ab
_ ^2 by a3&. he + ca - «& by a?>c. _ 2 c2 by 4a26c3. 5 «2 ^ 3 _ 7 .^2^2 by 3 x'^. 5 x2?/ + ^
^2
^-2
24. 25.
ar?/2
MULTIPLICATION OF COMPOUND EXPRESSIONS. 42.
If in Art. 41
(a
we
+ 6)(c +
+ for m in (1), = «(c + d) + 5(c + d) = (c-\- d)a 4- (c + fZ)^ = ac + ad -\-bc bd.
write f?)
c
cZ
-\-
Again, from (2)
- b)(c + d)= a(c + d)- b{c + d) = {G + d)a-{c + d)b. — ac + ad — (be + bd) = ac ad ~bc — bd. by writing c — d for m in (1) + &)(c - d) = a{c - d) + b(c = (c-fZ)a+(c-f/)6 = ac — ad bc~ bd.
(a
-\-
Similarly,
(rt
rZ)
-\-
we have [Art. 37.]
ALGEBRA.
28 Also, from (2) (a
-b)(c-
d)
=a(c- d) -b{c-d)
= (c-d)a-(c-d)b = ac — ad — (be — bd) = ac — ad — bc-{- bd. term on the right-hand side of and the way in which it arises, we lind that
If Tve consider eacli last result,
this
(-\-a)x(+c)=-{-ac,
(-b)x(-d)=-^bd, (-b)x(-^c) = -bc, (-\-a)x{-d) = -ad. These results enable us to state what Rule of Signs in multiplication. Rule
is
known
as the
The product of two terms icith like signs is product of two terms icith unlike signs is neg-
of Signs.
positive;
the
ative.
43.
The
rule of signs,
and especially the use of the negasome difficulty to the
tive multiplier, will probably present
Perhaps the following numerical instances may be useful in illustrating the interpretation that may be given to multiplication by a negative quantity. To multiply 3 by — 4 we must do to 3 what is done to unity to obtain — 4. Now — 4 means that unity is taken 4 times and the result made negative; therefore 3 x(— 4) implies that 3 is to be taken 4 times and the product made beginner.
negative.
But 3 taken 4 times gives .-.
3
-f 12.
x(-4)=-12.
— 3 x — 4 indicates that — 3 is to be taken 4 and the sign changed the first operation gives — 12,
Similarly,
times,
and the second
Thus
;
+ 12. (_3)x(-4)=-f-12.
MULTIPLICATION.
29
Hence, multiplication by a negative quantity indicates that we to proceed just as if the multiplier were positive, and then
are
change the sign of the product. 44.
]S"oTE
ox Arithmetical axd Symbolical Algebea.
Arithmetical Algebra
is
that part of the science wliicli deals
symbols and operations arithmetically intelliStarting from purely arithmetical definitions, we are gible. enabled to prove certain fundamental laws. Symbolical Algebra assumes these laws to be true in every case, and thence finds what meaning must be attached to symbols and operations which under unrestricted conditions no longer bear an arithmetical meaning. Thus the results of Arts. 41 and 42 were proved from arithmetical definitions which require the symbols to be positive whole numbers, such that a is greater than h and c is greater than d. By the principles of Symbolical Algebra we assume these results to be universally true when all restrictions are removed, and accept the interpretation to which we are led thereby. Henceforth we are able to apply the Law of Distribution and the Eule of Signs without any restriction as to the symbols used. solely with
45. To familiarize the beginner with the principles we have just explained we add a few examples in substitutions where some of the symbols denote negative quantities. Ex. 1. If « = — 4, find tlie value of a^. Here a^ = {- 4)^ =(- -4) x (- 4) x (- 4) = Ex. 2. Here
If
a.
If a
=-2,
- 64. = 1, & = 3, c = 2, find the value of _ 3a45c3 =_ 3 x (- l)^x 3 x (- 2)^ z::-3 X 1 X 3 x(-8)= 72.
3 a^hc^.
ALGEBRA.
30
= - 4,
a
If
=-
6
=-
c
3,
22.
Sa^+bx-4: cy. 2 «?>2 _ 3 &c2 + 2/x.
23.
fa^
21.
-2b^28.
46.
Ex.
By
cx^.
c
V(3
6c)
-
o
3
is
•.
(
is
-
by
a
X
(
-
ab'^x)
5 a-b^xr^
is
=
Multiply
-^j.^
—
down
3 a26 X 2 is
1
.-,
-,
and by the
a^
_
5 ^25
_
the algebraic
is
4
ab'^
sum
'i.a^^^
6
a2j)
_
4 ab^) x
—
at once
a^ft^
-
by
(-
3
ab'^)
x
=-
EXAMPLES
may be
for
= 6 a^ft^
aft* ^
^
;
•
^1
rule of signs the reis
positive.
3 «62.
formed
;
18
IV.
a^b'^
+
15 a^b^
+
12 a^b*.
c.
Multiply together 1.
ax and
-
Sax.
2.-2 abx and 3. a2^ and - a62. 4.
6 x2y and
-
7.
- 3 a-b^c^d^. and 5 x~y^z. Sxy + 4:rjz and - 12 xyz.
8.
a6
5.
7 a6x.
10 xy.
rule
ab*.
of the partial products
according to the rule enunciated in Art. 40
_
12 ab.
and by the
This result, however,
quired product
The product
=
5 a%H'^.
of 3a26,
^
^
x 36
positive.
- 5 abH)
q^ox
also 4 a
;
«6%.
g
thus (6 «3
2c\
c?/i.
b^ - 4 &-/ - 6 c%.
negative
written
4.
7
2 cy V(3 ^c^).
of the product
Thus the complete product
Ex.
+
3 63
6.
^
^
&% -
further illustrate the rule of
Find the continued product
/
5
4a x(-36) = -12«6.
— 5 abH
Multiply
3.
^
3
-
find the value of
1,
indices.
Multiply 4 « by
.
91,
26.
- 5 V(4 c'V') -
the rule of signs the product
of signs the product
o
2 a^
7V(a-^)-3V(^'c2)+5V(/-':K).
Here the absolute value
Ex.
25.
3
The following examples
2.
= -
?/
a^^y^
30.
.-.
Ex.
4,
3
29.
and the law of 1.
=
:/;
24.
2V(«c)-3V(^y) + V('->'c'). 3 V(acx) - 2 V(?>'^y) - 6 V(t%).
27.
signs
/= 0,
1,
6.
-
abed and
:<;*/;«;
-
6c and a26c3.
MULTIPLICATION. 9.
-x-y - z and -
a-
10.
16. 17.
-
3
ah
+
he
—
x'^?/
ca and
Find the product
2« -36 + 4c and -
22.
3X
-2y -
-
47.
I xy
and
The
-
4 and
-
25.
-
and
—
ah.
of
21.
26.
12.-2 a% - 4 a6'^ and - la%^-. - 6 x?/^ + 8 x2^2 and 3 xy. 13. 5 — 7 xhj— 5 x?/3 and — ^xhj^. 14. — 8 x^?/^: and xyz. 18. 8 x?/;^ - 10 xhjz^ and - x?/^. 19. a6c - a''-hc - ah'^c and - ahc.
x.
and ahc.
c-
— ahc. — 5 x?/--s + 3 xyz^ 15. 4 x2y2s2 - 8 xyz and - \2x^yz^. - l^xy^ - 15x2?/ and - Ix^y^. 20. — a^^c + b'^ca — ckib —
11.
+
l)^
31
f
3 x2
+
24.
| x.
a2ic2
f
|« - ^6 -
23.
f a.
and -
f
;;t-y.
results of Art. 41
f
^
^;;g
27.
-
^2
may
_
«2;^2
_
3
|
c
and f «x.
^^3 and
-
| a^x.
^2^
| x3?/2
and
-
i-
x2
+2
?/2.
be extended to the case
where one or both of the expressions to be multiplied together contain more than two terms. For instance (a — 6 + c) m = am — bm cm m hj x — y, we have (a - & + c)(x -y)= a(x - y)- b(x + c(x - y) = {ax - ay) - (bx - bij) + (ex - cy) = ax — ay — bx + by + ex — cy. -\-
;
replacing
7/)
48. These results enable us to state the general rule for multiplying together any two compound expressions. Rule. Multiply each term of the first expression by each term of the second. When the terms multiplied together have like signs, prefix to the product the sign -{-, ivhen unlike prefix
—;
the algebraic
sum of
the
particd
products so formed
gives the complete product.
This process Ex.
1.
is
Multiply X
called Distributing the Product.
+
8
by X X X
+ + +
7.
8 7
x2+ 8x + 7x + by addition
x-
+
15x
+
56
56
ALGEBRA.
32
Note. We begin on the left and work to the right, placing the second result one place to the right, so that like terms may stand in the same vertical column. Ex.
Multiply 2
2.
./:
-
3 ^
by 4 x
-
7 y.
2x -Sy ix - 7 y 8 x2
by addition
8 x-
-
12
:nj
14 xy
+21 y2
—2(jxy
EXAMPLES
+
21 y^
IV.
d.
Find the product of 21.
2x-3andx + 8,
b.
22.
2x
+
28.
- 8. x - 5 and 2x - 1. 2 x - 5 and x - 1. 3 x - 5 and 2 x + 7. 3x + 5 and 2x - 7. 5 x - 6 and 2 x + 3. 5 x + 6 and 2 x - 3.
29.
3x
-by
and 3 x
30.
3x
-
and ^x -by.
1.
x+oandx + lO.
2.
X
4.
+ X X -
5.
X
6.
X
3.
7. 8.
9.
10.
11. 12. 13.
14. 15. 16. 17.
18. 19.
20.
5
and x -
7
and x
-
10.
23.
7
and x
+ 10. - 10.
24.
+ 7 and x + 7 and x 4- 10. X + 6 and x - 6. X + 8 and x - 4. X - 12 and x - 1. X + 12 and x - 1. X - 15 and x + 15. X - 15 and - x + 3. ^ X - 2 and - x - 3. _ X + 7 and x — 7. _ 4- 5 and - x - 5. X - 13 and x + 14. X - 17 and x + 18. X + 19 and x - 20. _ X - 16 and - x + 16. - X + 21 and x - 21. y^
25. 26.
27.
3 and x
5 y
+
5
35.
-2b and « + 3 6. - 7 b and a + Sb. 3a - 6 6 and a - Sb. a — 9b and ^ + 5&. x+ a and x - b.
36.
x-
37.
x
31.
Z2. 33.
34.
38. 39. 40.
y.
a
a
a and x + 6. - 2 a and x + 3 6. ax - by and ax + by. xy - ab and xy + ab. 2pg - 3 r and 2j?g + 3 r.
MULTIPLICATIOX.
We
49.
now
shall
33
a few examples
give
greater
of
difficulty.
Ex.
Zx2
Find the product
1.
— 2x — _ r
—
T-^
-7—^
— 6 xJ^ Ex.
o.r
,,
ox~
— 2x —
o
and 2
a;
—
5.
Each term of the first expression is mulby 2 x, the first term of the second
5
tiplied
expression then each term of the first expression is multiplied by — 5 like terms ^^g placed in the same columns and the results added.
7777-
+
\[
X
19 x2 2.
of
Multiply
;
+
^.
+
25
;
-o
a-b + Schja-\-2b a a a-
- &+ + 2b — o.b +
3c
3 «r
2ab a?
+
a?)
+
3
-
2
?>2
+
6 be
—
2 5-
+
6 be
r/c
When the coefficients are fractional, we use the ordinary process of Multiplication, combining the fractional coefficients by the rules of Arithmetic. Ex.
3.
_ «7) + 1 &2 by \a^\b. ia2- \ab +f?)2
Multiply 1 «2
1
ha + \b
+ i«25_ irt52+ A d^b + i a?>2 +
253 I 63
50. If the expressions are not arranged according to powers ascending or descending of some common letter, a rearrangement will be found convenient.
+ X 2 a;3 + 2
- z- + 2 X- - o yz + 2 .r^ - 3 2/0 — 0-
Multiply 2xz
Ex. 3-2
— IX
xy y a;2?/
+2z + 2 x^^ - 3 xyz -
2 x~y
x-?/
—2 'ix^z + 2 xyz + 6 x-2 — 3 X2/5 +
xy by
x-y -{-2z.
xz-
—
a-?/^'
-f-
-\-
xy-
+
3
?/-5r
4 xz^ 3 xz-
—
xy-
+
3 y-z
+ —
yz"^
6
^/^'^
5 yz-
-
2 z^
—2z^
34
ALGEBRA.
EXAMPLES
IV.
e.
Multiply together
+
+
1.
a
2.
a-2h +
3.
4. 5. 6. 7.
8. 9.
2 x3
13.
-
19.
20.
21. 22. 23.
24.
25.
.26. 27. 28.
29.
30.
31. 32.
c.
?/.
x'2/
12.
18.
—
2/2,
a;2
x2
17.
b
and a -\-2b
c
2/2
a^x
16.
+
- c. - ah + 62 and d^ + ah + b^. and x + 4 x'^ + 3 x3-2a;2 + 8 and x + 2. x^ - x^y^ + y^ and a;^ + + + y^ and x — y. «2 _. 2 ax + 4 a:2 and «2 + 2 ax + 4 x^. 16 a2 + 12 a6 + 9 62 and 4 a - 3 6.
11.
15.
and a
c
cfi
10.
14.
6
—
+
X
-
ax2
-
and
3 x2
«5 4- ^45
+
—
x^
4-
2
and x
a^
x2
+
x
-
-\-
a.
6.
+ 2 X and 2 x2 -f 3 x + _ ^^352 and - a-b.
2.
5 and x2 - 2 x + 3. + 2 r<62 and a2 _ 2 a6 + 2 62. 4 x2 + G x?/ + 9 and 2 x - 3 and — x2 + xy + x2 — 3 X2/ — 63 - a2&2 4. «3 and a^ + a^/^^ + ^3, and x2 + 2 xy + x2 - 2 x?/ + «6 + cd + «c + 6fZ and ab + cd — ac — bd. - 3 a262 + 4 ^^3 + 15 a% and 5 ^2^,2 + ^^53 _ 3 ^4. 27 x3 - 36 ax2 + 48 a2x - 64 a^ and 3 x + 4 «. _ ^2 and a- + 5 rt6 + 62. «2 _ 5 x2 — x?/ + X + 2/^ + + 1 and x y — \. a2 + 62 + c2 - 6c - c« - ab and a + 6 + c. - x^y + + ^""^y^ + X* - X2/3 and x-\-y. xi2 _ xV^ + a^^^* - a;3?/6 + y^ and x^ + 3 a2 + 2 a + 2 a'5 + 1 + a* and a2 - 2 « + 1. — ax2 + 3 axif — 9 ay^ and — ax — 3 a2/2. - 2 x^y + + 3 x22/2 + X* - 2 xi/^ and x2 + 2 xy + ^a2 + ia + i and ]a- J. ^ x2 - 2 X + } and ^ x + ] x3
a^
+
7
X
2
cfib
?/2
?/.
?/2.
2/2
?/2
2/^-
(-(5
-\-
2/
?/*
y'^.
2/*
•
y^.
.
:
MULTIPLICATION. 33.
f x^
34.
I x2
36.
1 x2
36.
f
+ i y^ «ind \x — \ y. - f rt2 and f a;^ - ^ «x + i a^. - I and J x2 + I X - f f + f x2 + 1^2 and f a2 + ^2 _ 3
+ -
rtx
35
xy
«x
a;
3
qj;;^.
Note. Examples involving literal, fractional, and negative exponents will be found in the chapter on the Theory of Indices.
Although the result two binomial factors, such as x-\-d> and X — 7, can always be obtained by the methods already explained, it is of the utmost importance that the student 51. Products Written by Inspection.
of multiplying together
shoidd learn to write the product rapidly by inspection. This is done by observing in what way the coefficients of the terms in the product arise, and noticing that they result from the combination of the numerical coefficients in the two binomials which are multiplied together thus ;
+ 7)== + 8a^ + 7.x + m = x^ + 15x4- 56. (x- 8) (x - 7)= - 8x - 7x + m = .x2-15x + 56. (x + 8) (x - 7)= + 8x - 7x - 56 = x^ + X — 56. _ 8x + 7x - m (x - 8) (x + 7)= = x' — X — ^Q. {x
+ 8)
(.^•
a;2
.X-
.x2
.^•2
In each of these results we notice that 1. 2.
The product consists of three terms. The first term is the product of the first terms
of the
two binomial expressions. 3. The third term is the product of the second terms of the two binomial expressions. 4. The middle term has for its coefficient the sum of the numerical quantities (taken with their proper signs) in the second terms of the two binomial expressions.
:
ALGEBRA.
36
The intermediate step in the work may be omitted, and the products written at once, as in the following examples
+ 3)=x'-{-5x-^ 6. (x 3) (x + 4)= af-\-x- 12. (x + 6) (x -c))=x^- 3x - 54. (x -4.y)(x- 10y)=x^ - Uxy + 40/. (x - 6?/) (x + 4?/)= x^ -2xy- 24/. (x-\-2)(x
By an
easy extension of these principles
we may
write
the product of any two binomials.
Thus
+ 3y) (x - y)= 2x^ + 3xy-2xy- 3/ = 2x^-\-xy-3y\ (3x - 4.y) (2x + y)= 6x' -Sxy + 3xy - ^f = 6x^ — 5xy — 4:y^. (2x
MULTIPLICATION. (3x+8)(3x-8). (2x-5)(2x-5).
34.
31.
(Sx-2i/)(3x +
y).
35.
32.
(3x
2?/).
36.
29. 30.
+
2?/)(3x
+
33.
37
+ 7 i/)(2x - 5y). + 3rt)(5x - 3a). (2x - 5«)(a^ + 5a). (2x + a)(2x + «). (2x
(5x
MULTIPLICATION BY DETACHED COEFFICIENTS. 52.
cation (i.)
In the following cases we lessen the labor of multipliby using the Method of Detached Coefficients When two compound expressions contain but one :
letter. (ii.) When two compound expressions are homogeneous and contain but two letters. Ex. 1. Multiply 2 a;3 - 4^2 + 5»: - 5 by 3x2 + 4a: - 2.
Writing coef&cients only,
2- 4+ 53+ 4- 2
5
CHAPTER
V.
Divisiox. 53.
When
quotient
produces
a quantity a
The operation
a.
in each of these
and
is
divided by the quantity h, the when multiplied by b
defined to be that which
is
modes
is
denoted by a
of expression
a
is
-r- b,
-, or
a/6;
called the dividend,
b the divisor.
Division
is
thus the inverse of multiplication, and
= a.
(a-r-b)x^ 54.
The Rule
rpi
of Signs holds 7
Thus
ab
-^
a
ctb =a ——
J
-T-
a
X
b
=
.
b.
a
a
— ab
for division.
= — ab = a x(—b) = — ^
^^
^ (-«)x(-6) ^ —a —a — ab -^(— a)= — ab = (— a) X b = b. —a —a ab^(-a) a)= 1
Hence
/
,
b.
a
a ''^'
\
^
,
-^^
^
in division as well as multiplication like signs
produce +,
unlike signs produce
—.
55. Since Division is the inverse of Multiplication, it follows that the Laws of Commutation, Association, and Distribution, which have been established for Multiplication, hold for Division.
:
:
39
DIVISION.
DIVISION OF SIMPLE EXPRESSIONS.
The metliod
56.
shown
is
in the following
Ex. 1. Smce the product of 4 and x is divided by x the quotient is 4, 4 x h- x — 4. or otherwise,
examples
follows that
la-, it
when 4x
is
Ex.2. Divide 27 ^-
a5
by 9
1=
3 aa
—
3.
„
=
iThe quotient 4.
We
see, in
the
common
factors
to
both, as in Arithmetic.
.,
3 o}.
by 7 ^b nan hh
—
.
a.hh.
^
9
r/^
=3
ffi,
alf-c^.
«-36-'c3
1
quotient
t remove ^from ., the divisor and .
e
^^^-^^^^^
27 «5
Divide 35
rr>i
U
<-^aaaxi
Therefore
Ex.
^^
a-3.
J.
^
= ^f^ ^ ^^
The quotient
.
— = 5 a«
err
r
.'
.
c
=
-
o
o «%.
cc
the index of any letter in the of the indices of that letter in the This is called the Index Law for
each case, that
the difference
is
dividend and divisor. Division.
AYe can now state the complete rule
The index of each
Rule.
letter in the
by subtracting the index of that in the dividend.
To tient
quotient
letter in the
is
divisor
obtained
from
that
proper sign the quoof the dividend by that of the divisor.
the result so obtained prefix icith its
of the
Ex.
4.
coefficient
Divide ^bn%h:^ by -Qcfihyi^.
The quotient Ex.
5.
Note.
= ( - 5) x «6-352-i^4-2 = — 5 a%xr-
- 21 cfih'^ -^(-7 cfih'^) =SbIf
same power
we apply
the rule to divide any power of a letter by the we are led to a curious conclusion.
of the letter,
Thus, by the rule but also
^ (^s ^ ^^s-s ^^ ^^0 053^^3-^ _ 1_ «3
.
«3 .-.
rt'^
=
l.
This result will appear somewhat strange to the beginner, but its be explained in the chapter on the Theory of
full significance will
Indices.
:
ALGEBRA.
40
A COMPOUND EXPRESSION BY A SIMPLE
DIVISION OF
EXPRESSION. To divide a compound expression hy a single and take the
Rule.
57.
factor, divide each term separately by that factor, cdgehraic sum of the particd quotients so obtained.
This follows at once from Art. 40.
(9x-12y + 3z)-^-S=-Sx^4:ij -z. (36 a^b'^ - 24 a^b^ - 20 a^b'^) - 4 a^Z) = 9 a6 - 6 6* - 5 a-b. (2 x^ - bxy + ^x^-if^) -^ - ix = - ^x + lOy-Sxtf.
Ex.1. Ex.
2.
Ex.
3.
EXAMPLES
V.
a.
Divide 15.
-
-
9 x^
16.
x^-Sx^ +
35x6 by
7 x^.
17.
x^
18.
lOx^
19.
15x5-25x* by -5x3.
by
3 x3
1.
27 X*
2.
-
3.
x2.
hj
4.
xV
by
5.
a^x^
by -a^x^.
x^y.
by -Sa^b'^c.
6.
12a6?,6c6
7.
-
8.
15x5?/V by 5x^22.
-
by
«5c9
-
10.
-
11.
63a"68c3 by 9rt565c3.
9.
16 xhj^
by
48 a9 by
-
7 a2&c
13.
28a453 by -4^36.
14.
16 &2^x2
by
- 2 xy.
DIVISION OF 58.
We
8x6
4.^4
+
x.
by
4xV^-8x3?/2+6x?/3 by -2x7/.
26.
|
x^y'^
_
f x2+
f xy
-
2 a5x3
+
28.
2.
x.
- |x3?/2. + J^x by - f x.
| a^x* by | a3x.
COMPOUND EXPRESSIONS.
employ the following rule
Divide the term on the left
3
3 ^3^4 by
^^. -
Rule. 1. Arrange divisor and dividend according ing or descending poicers of some common letter.
on the
x"^.
3x* by xK
25.
23.
by -la^bc.
12.
-
+
24.
22.
8 a^.
1 x^
5 x^y.
34
21.
4 xy^.
-
-
x by
-24x6-32x^by -8x3. x^ _ 51x2^3 by n xy. a^ - ab - ac hy -a. ^3 _ ^^^ _ ^^-2^2 by a^. 3 x3 - 9 x^y - 12 x^-^ by -
20.
ac3.
50 ifx^ by
of the divisor, and
left
to
ascend-
of the dividend by the term
p>ut the residt in the quotient.
DIVISION. 3.
Multiply the
whole
41
divisor by this quotient,
and put
the
product under the dividend. 4.
Subtract and bring
terms as
may
Repeat
down from
the dividend as
many
be necessary.
these operations
terms from the dividend
all the
till
are brought down. Ex.
1.
Divide x^
+
11
a;
Arrange the work thus
+
30 by x
+
6.
:
+
x
6)x-2
+
lla:
+
30(
by a-, the first term of the Multiply the whole divisor by a;, and put under the dividend. We then have
divide x^, the first term of the dividend, divisor
;
the quotient
the product
ocr
+ Qx
is x.
x
+
+
llx
ic2+
Qx
6)x2
by subtraction
5x
+
30(x
+
30
On
repeating the process above explained we find that the next term in the quotient is + 5. The entire operation is more compactly written as follows :
a;
+
4-
llx
a;2+
Qx
6)^2
+
30(a;
+
5
5x + 30 5X + 30
The reason for the rule is this: the dividend may be divided into as many parts as may be convenient, and the complete quotient is found by taking the sum of all the partial quotients.
Thus
a? -f
11 x -f 30
process into two parts, namely,
each of these is divided by complete quotient x + 5. Ex.
2.
Divide 24 x^ 8X
-
-
05 xy
+
3 ?/)24 x2
21
-
a;
?/2
65
a.*^
is
-f-
+6
;
6
divided by the above x, and 5 x + 30, and thus we obtain the
by 8 x - 3 y.
X2/
+
21 if{^ x
+ +
21?/2
24x2- 9xy -56xi/ -50x^/
21i/2
-ly.
5
ALGEBRA.
42
EXAMPLES
V.
b.
Divide
18.
9. 4x- + 23 x + 15 by 4x + 3. xM-3x + 2 by x+ 1. X- - 7 X + 12 by X - 3. 10. 6 x2 - 7 x - 3 by 2 x - 3. 11. 3 x^ + x - f4 by x - 2. a2 _ 11 a + 30 by a - 5. 12. 3 x2 - x - 14 by x + 2. a2 - 49 a + 600 by a - 25. 13. 6x2 - 31x + 35 by 2x - 7. 3x2 + 10x + 3 by x + 3. 14. I2a2-7ax-12x2 by 3a-4x. 2x2 + llx + 5 by 2x + 1. 15. 15a2+i7 «:c_4a;2 by 3a + 4x. 5x2 + llx + 2 by x + 2. 16. 12a2_ii ac-36c2 by 4 rt-9c. 2x2 + 17X + 21 by 2x + 3. 17. -4x^-15^/2 + 96x2 by 12x- 5?/. 7x3 + 96x2-28x by 7x-2. 20. 27xH9x2 -3x-10 by 3x-2.
19.
100x3-3x-13x2by3+25x.
1.
2.
3. 4.
5. 6. 7. 8.
2L 16a3_46«2+39a_9by8a-3.
59. The process of Art. 58 is applicable to cases in which the divisor consists of more than two terms. Ex.
2x2-
1.
Divide 6 x^
-
x*
+
4 x^
-
5 x2
-
x
-
15
15(3x3
by 2 x2 - x + + x2-2x-
3.
43
DIVISION. 60.
Sometimes
will be found convenient to arrange the
it
expressions in ascending powers of some
+
Ex. Divide 2 a^ 2
-
4a
-
-
10
5 «2)10
10
-
16 a
1(3
-
39 a^
+
common
letter.
- 4 « - 5 cA + 15 (^4,^5 + 2 a - 3 a^
15 a^
by
2
- 39 a2 4. 2 a^ - 25 a^ - 14 a^ -f 2 #
a
20 g 4a
4a- Sa^-\Oa^ 61.
When
process Ex.
6 a2
+
12 a3
+
15 ^i
6a'2+ 12a3+ 15
g^
the coefficients are fractional^ the ordinary be employed.
may
still
yV xy^ + ^^ y^hj ^x + I y. ix + iy)i^^ + i2^y^ + ikv^i^^^ - \^y +
Divide \x^
-\-
-
\ 7-r
x'^y
+
^^y
~
\y''-
^\xy^ -9-
xy-^
+ ^^t
\xy'^
In the examples given hitherto the divisor has been exactly contained in the dividend. When the division is not exact, the work should be carried on until the remainder is of lower dimensions [Art. 29] than the divisor.
EXAMPLES
V.
C.
Divide
-
1.
x3
2.
2
?/3
3.
6
m^
4.
6 a5
5.
X*
6.
a4
a;2
-
3
-
- 12 by x2 + 3 X + 3. - 6 - 1 by 2 _ 5 _
9X
?/2
+ „
+ ^3 _
x3
7 x2
-
6X
8 a2
+
12 a
1.
+
8
-
by
a;2
9 by
+
a'-2
2X
+
2
+ rt
a.
8.
- 3.
+ 6 a3 + 13 a2 + 12 a + 4 by a^ + 3 « + - a;3 + 4 x2 + 7 X + 1 by x2 - X + 3.
7.
a-i
2 X*
9.
x5-5x4 + x^
?/
— m^ — 14 m + 3 by 3 7n^ + 4 to — 1. - 13 a^-\-4.a^ + 3a^\)ySa^-2a^-
8.
10.
?/2
2/
-
4 x*
+
9x3-6x2-x+2 3 x3
+
3 x2
by
-3X+2
2.
x2-3x + 2. x2 - X - 2.
by
ALGEBRA.
44
30 y
12 13
?/2
?/2
^.
X;3
14
^•
?/i
15
?)i'^.
-
16
a:
17
18
+ Ilic3-82x2-5u: + 3by 2a:-4 + 3x2, + 9 - 71 ^3 + 28 y* - 35 by 4 _ 13 ^ + 6. 6 A;5 - 15^-* + 4 + 7 ^-2 _ 7 + 2 by 3 A;3 - + 1. 15 + 2 m^ — 31 + 9 m'^ + 4 m^ + w^ by 3 — 2 «i — 2ic3-8ic + x4 + 12-7x2by x2 + 2 -3x. x5 _ 2x4 _ 4x3 + l9a;2 by x3 - 7 + 5. 192-0:4+ 128x + 4x2-8x3by 16 - x^. 14x4 + 45x3^ + 78XV 4- 45x?/3 + Uy^ by 2x2 + 5x?/ + 7 X^ — X4?/ + X3y2 _ x^ if by x3 — X x^ + x4?/ — x3?/2 + x3 — 2 x?/2 + 1/ by x2 + X?/ — 69 by a3 _ ^a. 22. x^ - if by x2 + XIJ + y^. x7 - 2 2/14 - 7 x5^4 _ 7 ^^12 4_ 14 ^3^8 by X - 2 a3 + 3 a25 + 53 _ 1 + 3 «52 by a + 6 - 1. 26. ai2 _ ^12 by cfi by x3 + ^^y + x?/2 + x8 _ ai2 + 2 a6^,6 4. 512 by a4 4. 2 ^252 + 54. 1 - a3 _ 8x3 - 6 ax by 1 - « - 2 x. 30^4
11
.
19.
20. 21. 23. 24.
25. 27.
28.
?/2
?/2.
2/2.
?/3.
2/8
Find the quotient 29.
30. 31. 32. 33. 34.
62.
of
^ _2^^x2 _ 27 x3 by 1 a - 3 x. 27 «^ - T2 cfi + tV « - 6? ^y\o,-\' A f a2c3 + ^§3 a^ by a2 ^ q^c. — Iff ax4 by f a — I x. 27 a^ |«2 + ia + i^by fa^ 36x2 + i?/2+ -4x2/-6x + i2/by6x-i2/-|• i a3
_
9
^2^
1
1
Important Cases in Division. The following examples may be easily verified tliey are of great importance and should be carefully noticed. 62.
in division
iC"
X
;
—y_ -
45
DIVISION.
and so on the divisor being x — y, the terms in the quotient all positive, and the index in the dividend either odd or even. ;
+ if = x^- xy + 7f, +y xP + — x^y + x'^y- — xy^ + x+y ^^_±yL = x^- x^y + xHf - ^y^ + xhf - xf + y\ x+y
a^
x
?/^
y"^,
x'^
II.
and so on the divisor being x + y, the terms in the quotient alternately positive and negative, and the index in the dividend alivays odd. ;
x-y. x-\-y
M-=Qi^
III.
+y x^ — y^ x+y
—
o?y
x
V
and
so
on
x^
— x^y
+ xy'^ — y^, -\- Qt?y'
— x^y^ -f ^11^ —
?/^?
the divisor being x-\-y, the terms in the quotient and negative, and the index in the divi-
;
alternately positive
dend always
even.
The expressions
IV.
the index
is even,
v^^-y"", a)*^-?y^ x^-\-y^
and the terms hotU
••.
(Avhere
positive), are never
—
by x-{-y or x y. All these different cases
divisible
may
be more concisely stated
as follows: (1)
x'^
-/"
is
divisible
hy x
-y
(2)
;f»
+ /"
is
divisible
by
;f
(3)
jf"
-y^
is
divisible
by
jf
+/ +/
(4)
jr''
+ /"
is
never divisible by x
ii
n be any whole number.
if
/?
be any odd whole number.
if
/;
be any
+/
or
jf
ej^e/?
whole number.
— / when
n
is
an even
whole number.
Note.
General proofs of these statements will be found in Art. 106.
:
ALGEBKA.
46 DIVISION BY 63. In Art. 52
DETACHED COEFFICIENTS.
we considered
certain cases of compound expressions in which the Avork of multiplication could be shortened by using the Method of Detached Coefficients. In
the same cases the labor of division can be considerably abridged by using detached coefficients, and employing an
arrangement of terms known as Horner's Method of Synthetic The following examples illustrate the method Division. Ex.
1.
Divide
3ic5-8xt-5x3+26x2-28x + 24 by ^3-2 3-
a;2-4
a;
+ 8.
DIVISION.
47
+
3 a'^b^ + ab^ is the complete the remainder.
Inserting literal factors, a^ -i-2 a^b
and — Explanation.
quotient,
5 a^b*
+ 3 a%^ +
The term ab^
for the coefficient of this of the vertical line.
We
«&•'
is
in the divisor
is
missing, so
we
write
term in the column of figures on the left add the columns as in Ex. 1, but as the
sum by 2 before placing then use these quotients as each case — 3, 0, and 1, and form the horizontal lines as in Ex. 1. Having obtained the required number of terms in the quotient.^ the remainder is found by adding first
term of the divisor
is 2,
im
divide each
We
the result in the line of quotients. multipliers, the multiplicand being in
the rest of the columns and setting
down
the results vjithotit dividing
By
continuing the first horizontal line (dividend), as shown in this example, we at once see what literal factors the remainder by
2.
must contain.
EXAMPLES Divide 1.
2. 3.
4. 5.
V.
d.
:
- 4 a"^ + 2 a^ -h ^ a + 1 hj cfi - 2 a - 1. - 4 a^b + 6 ^252 + 54 _ 4 ^53 by «2 + _ 2 ab. «5 - 10 aH) + 16 «3?)2 _ 12 ^253 + + 2 65 by (a- by. x^ - 2 6^4 4. 58 by x^ + bx"^ + b^x + b^. x^ - 3 x'^y^ + Sxy^ — 6y^ by x"^ -4^xy y^ to four terms a^ a*
quotient.
?,2
rt/j4
-\-
in tlie
CHAPTER
VI.
Removal axd Insehtion of Brackets.
We
frequently find it necessary to enclose within of an expression already enclosed within For this purpose it is usual to employ brackets brackets. The brackets in common use are ( ), of different forms. Sometimes a line called a vinculum is drawn II, [ ]. 64. brackets
i)art
over the symbols to be connected thus a — b -\- c with the same meaning as a —(b -\- c), and hence ;
a
— b-\-c = a — b —
is
used
c.
65. To remove brackets it is usually best to begin with the inside pair, and in dealing with each pair in succession we apply the rules already given in Arts. 25, 26. Ex.
1.
Simplify,
by removing brackets, the expression
a-26-[4a-6&-{3«-c+(5a-26- 3a-c + 26)}]. Removing the brackets one by
one,
we have
- 6b - {Sa - c-\-(5a -2b - Sa c - 2 /j)}] c -2b]^ -[4,a - Qb - {^a - c + 5a -2b - Za = a-2b-lia-Gb - Sa + c- 5a + 2b + oa - c + 2b] = a-2b - 4« + 66+ 3«-c+ 6a -2b - Sa-\- c -2b = 2 by collecting like terms.
-2b = a -2b a
-I4:a
-\-
-\-
(7,
Ex.
2.
Simplify the expression
-l-2x-{3y-(i2x-Sy) + (3x-2y)}-]-2xl =_[_2a; -{Sy-2x + Sy + Bx - 2y] 2:>:] = -[-2x-Sy + 2x-Sy - Sx-\-2y + 2 »:] = 2x + Sy -2x + Sy -\-Sx-2y -2x = x + 4y.
The expression
-i-
48
:
.
.
REMOVAL AND INSERTION OF BRACKETS. EXAMPLES
VI.
49
a.
Simplify by removing brackets 1.
2. 4. 5.
6.
8. 9.
-(b - c)+ a+(b - c)+ b -(c + a). 3. «-[2«-{3&-(4c-2a)}]. «-[& + {<-«-(& + «)}]. {a - (b - c)] + {& -(c - a)} -{c-(a - b)}. 2«-(5&+[3c-a])-(5a- [6 + r]). a
-{-[-(^a-b^cm
7.
-(-(-(- x)))-(-(-y)).
-[«-{& -(c - «)}]-['> - {c -(« - 6)}]. -[-{-(& + c-«)}] + [-{-(c + a-6)}].
10.
-5.r-[3?/-{2.r. -(2?/-a:)}].
11.
13.
-(-(-«))-(-(-(-:>-0)). 3 a - [a + 6 - {« + + c -(« + & + c + -2«-[3a: + {3c-(4y + 3.x + 2«)}].
14.
Sx-l5y-{6z-(ix-7y)}^.
15.
_[5.'*:-(ll2/-3x)]-[5^-(3.r,-6y)].
12.
?;
c?)}].
16.
-[15x-{14?/-(150+12?/)-(lOx-15^)}].
17.
8.r.
18.
19. 20.
-{16?/-[3x-(12?/-x)-8?/]+4.
-lx-{z +(x- z)-(z -x)- z}-x^. — [« + {a —(a — x) — (« + x) — «} — «] - [« - {« + (x - a) -(x -«)-«}- 2 «]
66. A coefficient placed before any bracket indicates that every term of the expression within the bracket is to be multiplied by that coefficient.
The
Note. fraction
is
between the numerator and denominator
line
a kind of vinculum.
Thus
^"
~
is
equivalent to i(x
of a
—
5).
Again, an expression of the form y/{x + y) is often written vx + ?/, the line above being regarded as a vinculum indicating the square root of the
compound expression x + y taken as a
Thus whereas
ivhole.
= ^169 = 13, ^26 + ^144 = 5 + 12 = 17.
67. Sometimes the "work.
\/25
it is
+
144
advisable to simi^lify in the course of
.
ALGEBRA.
50 Find the value
Ex.
84
of
-7[- llx-4{-
The expression
17.r
+ 3(8-"9
-5.r.)}].
= 84 - 7 [- 11 x - 4{ - 17 x + 3 (8 - + 5x)}] ^ 84 - 7 [ - 11 - 4 { - 17 X + 3 (5x - 1)}] :^84-7[-lla;-4{-17x+ 15x-3}] = 84-7[- ll:c-4{ -2x-3}] = 84-7[- llx-f 8x+12] = 84-7[-3.r + 12] = 84 + 21x-84 = 21 X. .X
AYhen the beginner has had a steps
may
little practice,
EXAMPLES Simplify by removing brackets 1.
2. 3.
4. 5.
6. 7. 8.
9.
10. 11. 12. 13. 14. 15.
16. 17.
18.
the
number
of
be considerably diminished.
VI. b.
:
- [2 5 + {3 c - 3 « - (a + 6)} + 2 a - (?) + 3 c)]. « 6 - (c + « - [?) + c - (« + 6 - {c + a - (6 + c - «)})]). a -(?>-(')- [a - & - c - 2 {6 + c - 3 (c - a) - d}]. 2x-(Sy-4:z)-{2x- (3 y + 4 s)} - {3 - (4^ + 2 x)}. b + c - (a -h b -[c a - (b + c - {a + b -(c + a - 6)})]). a-(b-c)-[a-b-c~2{b + c}-]. 3 a^ - [G a2 _ {8 &2 _ (9 c2 _ 2 «2)}] 6 - (c - a) - [6 - « - c - 2{c + a - 3 (a - 6) - fZ}J. _ 20 (« - d) + 3 (& - c) - 2 [6 + c + - 3{c + - 4 (rZ - a)}]. - 4 (a + cZ) + 24 (?) - c) - 2 [c + + a - 3 (fZ + a - 4 (?) + c)}]. - 10(a + ?>)-[c+ + &-3{a + 2&- (c + «- 6)}]+ 4c. a - 2 {b - c) -[- {- (ia - b - c - 2{a+6 + c})}]. 2(3&-5a)-7[rt-6{2-5(a-&)}]. 6 {a - 2 [6 - 3 (c + (Z)]} - 4{a - 3 [5 - 4 (c - (Z)]}. 5 {a - 2 [a - 2 (a + x)]} - 4{« - 2 [a - 2 (a + x)]}. - 10 {rt - 6 [« - (6 - c)]} + 60 - (c + a)}. - 3 { - 2 [- 4 (- «)]} 4- 5{ - 2 [- 2 (- «)]}. _2{-[-(x-2/)]} + {-2[-(x-2/)]}. «
-I-
?/
-\-
fZ
fZ
rt
{Z)
fZ
.
.
REMOVAL AND INSERTION OF BRACKETS.
51
i9.H«-5('.-<.«-|{i(.-i)-i[«-f(.-^)]}. '^*-~^^^
35
20.
-
tV {3 X
[
(«_ 5)_
-
f (7 X
-4
+
8
(2/
-
2
a;)
^)}J
(^,--c)}-{^-'-^}-i{c-« -!(«-&)>
21.
3.[|
22.
lx-^Qy-lz)-[x-{lx-ily~lz)}-Qy-^z)-].
8
INSERTION OF BRACKETS. 68.
The converse operation of inserting brackets is imThe rules for doing this have been enunciated in for convenience we repeat them. 25, 26
portant. Arts.
;
Rule
Ally part of an expression
I.
brackets
and
the brackets
Ex.
Rule
II.
may
be enclosed ivithin
prefixed, the sign of every term ivithin
remaining unaltered.
—
a
brackets
+
the sign
b-{-c
—
Any part
and
the sign
—
d
e
=
a
— b+(c —
d
—
e).
of an expression may be enclosed icithin — prefixed, provided the sign of every
term ivithin the brackets be changed.
—
Ex.
a
69.
The terms
b
-\-
c
—
d
—
e
=
a
—(b —
c)
— (d +
e).
of an expression can be bracketed in various
ways. Ex.
may
4- ex — aij + hy — cy — hx) + (ex — ay) + (by — cy), (ax — bx + ex) — (ay — by + cy), (ax — ay) — (bx — by) + (ex — cy)
The expression ax — hx
be written
or
or
(ax
A factor, common to every term within a bracket, be removed and ]3laced outside as a multiplier of the expression within the bracket. 70.
may
Ex.
1.
In the expression ax^
-
cx
+
7
-
dx^
bracket together the powers of bracket.
-\-bx-c cc
-
dx^
-\-bx'^
-2x
so as to have the sign
+
before each
Z
: :
ALGEBRA.
52
= {aj^- (W) + (hx^- dx"^) + (bx -ex -2 x) + (7 - c) = x3(« - d) + x^{b - d) + x{h - c - 2) + (7 - c) = (a - cZ)x3 + (6 - d)x2 + (5 _ c - 2)x + 7 - c. result the compound expressions a — d, h —d^ b — c — 2
The expression
In this
last
are regarded as the coefficients of
x^,
x'^,
and x
In the expression - a'^x - 7 a -]- ahj together the powers of a so as to have the sign The expression = - (a% - a'^y) - (7 « -f «6)
Ex.
2.
respectively.
S - 2x - ab bracket — before each bracket. - (2 x - 3) -\-
= _a\x-y)-aO + ^))-(^x-S) = -{x-y)a^-0 + b)a-{2x-Z).
EXAMPLES
VI.
c.
In the following expressions bracket the powers of x so that the signs before all the brackets shall be positive
+ 5 + 2 6x - 5 x2 + 2 X* - 3 x. - 2 X + «6 + 5 ax^ + ex - 4 x2 - 7 x3 + 5 rtx2 - 2 ex + 9 ax3 + 7 X - 3
+
1.
ax*
2.
3 6x2
3.
2
6x2
-
7
bx^.
x'^.
In the following expressions bracket the powers of x so that the signs before all the brackets shall be negative 4.
«x2
+
5.
7 x^
-
6.
3
62x'^
- «2x4 _ 2 6x3 - 3x2 - ^x*. - abx^ + 5 ax + 7 x^ - abcxK - 6x - ax* - ex* - 5 e^x - 7 x*. 5 x3
3 e2x
Simplify the following expressions, and in each result regi'oup the terms according to powers of x :
7. 8.
9.
10.
- 2 ex - [6x2 _ ^^x - dx - (6x3 3 ^^^2^^ _ (ex2 - 6x)]. 5 ax3 _ 7(6x - cx2) - {6 6x2 _ (3 ^^^2 ^_ 2 ax) -4 ex^}. ax^ - 3 {- ax3 + 3 6x - 4 [i cx^ _ f(ax - 6x2)]}. x5 - 4 6x* - i [12 ax - 4 1 3 6x* - of— - 6xA - | ax* 1 1. ax3
_)_
71. In certain cases of addition, multiplication, etc., of expressions which involve literal coefficients, the results may be more conveniently written by grouping the terms
according to powers of some Ex.
1.
letter.
together ax^ — 2 6x2+3,
6x— cx3-x2, and x^ — ax24cx. + 6x — cx^ — x"^ \- x^ — ax^ + ex bx ex ex? + x^ - a.x2 - 2 6.x2 - x^ z=.{a-e^ l)x3 _ (a + 2 6 + l)x2 + (6 + c)x + 3.
Add
The sum
common
= ax^ — = ax^ -
2 6x2
+
3
-\-
-\-
-\-
EXAMPLES
JNIISCELLANEOUS Ex.
— 2 &aj + 8 c by px - q. — 2 5x + 3 c) (px - q) apx'^ — 2 bpx^ + o cpx — aqx^ + 2 &gx — 3 eg apx^ — (2 6^ + ag)x2 + (3 cp + 2 bq)x — 3 cq.
Multiply ay^
2.
= = =
The product
(rr-x'^
EXAMPLES Add
VI.
d.
together the following expressions, and in each case arrange
the result according to powers of x
:
— 2 ex, bx^ — cx^, and cx^ — x. X2 - X - 1, ax2 - 5x^ 6x + x\ «%^ — 5 X, 2 «x2 — 5 «x3, 2 X'5 — ?>x2 — ax. ax'^ + ^)x — c, (/x — r — px^, x'^ + 2 x + 3. px-^ + gx^px^ — qx, gx^ — px, g — ax^
1.
2. 3.
4. 5.
x-^,
and
Multiply together the following expressions, arrange the result according to powers of x: «x2
7.
cx"-^
12. 13.
?>
14.
x'-^
(?
MISCELLANEOUS EXAMPLES 1.
Find value of (a
2,
c
2.
Find the sum
=
each case
in
9.2 x^ - 3 x - 1 and ?>x + c. + bx + 1 and ex + 2. — 2 X + 3 and ax — b. 10. c/x^ — 2 ?>x + 3 c and x — 1. — — 11. px^ — 2 x — g and ax — 3. axbx e and jjx + q. — — — and x^ + «x2 bx x^ «x2 — bx + c. c and x- + 3 X «x^ ax^ + + 3 x + &. and x* + ax^ — bx^ — ex + d. X* — ax^ - ?>x2 + ex +
6.
8.
&
63
II.
=-
-
b)- -\-(b
-
e)-
+ (a -
b)
II.
-\-
2 e^
when a =
1,
3.
of
2 x,
3
x^,
5,-3 x^, - 4,
-
x,
6
8 x^,
x^,
arranging result in descending powers.
Diminish the sum of
3.
+
1162
4
Show
5.
Simplify {a
6.
Subtract
+
7.
4 a2
— 8.
+
7 ^^
_
5
and 4
2 ?«2
_
What 5 «,
b^
+ ?/)2 = 2 - (a - b) (a - c). 3 wi^ — 4 + 1 and
that (l+x)2(l + ?/2)-(l + x2)(l
4.
iH^
6^
52
_
3 5
by
4. 7
2.
+
6) (a
+
c)
(x-y)(l-x?/).
—
3
m
expression must be taken from the sum of 2 a, 2 a — 3 &2 in order to produce a^ — b'^?
«
7
the
sum
of
?>?,
on-
from
y,^
+
3
&,
+
Find value
of
a^+^c+d)
IJI~\+d\
when
«: :2, c
= 0,
f?
= 8.
.
ALGEBRA.
54 Multiply a-
9.
10.
Divide 343 x^
11.
If
=
a
=
c
+
I X?/2
-
+
x2?/
+ x
=
4,
+
c
+
7
8
1.
?/.
find the value of
—
be added to 2 x^
_4,|
3 x^^
.f
6 to produce
5 ?
-
Simplify (X
13.
d
3,
What number must
12.
&
512 y^ by
+ [(6-c)(2,;-^)]-( V(2«^-&')
«
7 X2
+
=2,
h
1,
- cy by
(b
-\-
-
2/)
{3
x
- (x + y)} + {(2 x - 3 y) - (x - 2 y)}. - 2) (a _ 3) = («2 _ 3 + iy2 _ 1.
- 1) (a ^ x2 - 2 X - 3. - c) + 2 - (5 a + 3 6) + 4 c - 2 16. Simplify 9 a - (2 find its value when « = 7, & = — 3, c = — 4. 17. Multiply 3 a-b - 4 ab^c + 2 a-^'-^c^ by - 6 aVy^c^ and U. Show that a (a x*
15.
-
+
10 x2
^^
9
fZ
/>
the result by 3
2,
c
=
0,
f?
=
1,
-
cd
+
c^
-
a-\-2c
Find the sum of 3 a
+
2
-
5 c
a
=-
1,
-
2 a
-
20.
of 2 x^
3 &
+
5
/>
c
-
5 X
?;,
find the value of
-
—
+
i
-
by
Divide 6 a^-a'^b + 2 a%'^ + lo a63+4
1
2/2
Simplify 5 x* - 8 x^ - (2 x^ subtract the result from 4 x* — x +
-
j\
-
7)
Simplify by removing brackets 5[x
25.
If
1,
6 (c
e
= 2, +
fi
c
=
3,
^^
and d ^
6*
by
2
(x*
+
5)
=
4,
^bd(b
—
4{x
+ +
2 a^. 5/, b
-
a
-^
2 d,
sum
f
2/^.
a2_3 ah + ^
+
(3 x^
-
b^.
x),
and
— 3(2x — 3x
f 2)}]
find value of
+ c + d-a)
Express by means of symbols
(1) 6's excess over c is greater than a by 7. (2) Three times the sum of a and 2 6 is less of b and c.
by 5 than the product
Simplify
3«2_(4 a^b'^)-{2 28.
3
2.
24.
27.
a^b
x^-\ xy +
22.
x?/
23.
26.
rZ,
4 from zero, and add the difference to the
Multiply i x2
=
+
2
and unity.
21.
a
divide
2/.
Subtract ax^
—
=
ac-
ad-}19.
and
ab'^c^.
a^
If
18.
rZ,
0-2- (3
b-rr-)-2 b-B rt}-{5 b-7 a-ic^-b^)}.
Find the continued product x2
+
x?/
+
?/2,
x2
-
x^/
of
+ f,
y^
-
ofiy'^
+
y^-
CHAPTER
VII.
Simple Equations. 72. An equation asserts that two expressions are equal, but we do not usually employ the word equation in so wide a sense. Thus the statement x-\-3-\-x = 2x-\-3, which is always true whatever value x may have, is called an identical equation, or is
an
identity.
The
sign of identity frequently used
=.
The
parts of an equation to the right and left of the sign members or sides of the equation, and
of equality are called
are distinguished as the right side 73.
and
left side.
Certain equations are only true for particular values Thus 6 is only true when
3x=
of the symbols employed.
x = 2, and is called an equation of condition, or more usually an equation. Consequently an identity is an equation which is always true whatever be the values of the symbols involved; whereas an equation, in the ordinary use of the word, is only true for particular values of the symbols. In the above example 3x = 6, the value 2 is said to satisfy the equation. The object of the present chapter is to explain how to treat an equation of the simplest kind in order to discover the value which satisfies it. 74. The letter whose value it is required to find is called the unknown quantity. The process of finding its value is called solving the equation. The value so foinid is called the root or the solution of the equation. 75.
The
sidiary to
solution of equations, and the operations subit,
form an extremely important part of Mathe55
:
ALGEBRA.
5(J
All sorts of mathematical problems consist in the some quantity by means of its
matics.
indirect determination of
which are known, and these means of equations. The operation in general of solving a problem in Mathematics, other than a transformation, is first, to express the conditions of the problem by means of one or more equations, and secondly, to solve these equations. For example, the problem which is expressed by the equation above given relations to other quantities relations are all expressed
l3y
the very simple question, '^ What is the number such In the present that if multiplied by 3, the product is 6 ? " chapter, it is the second of these two operations, the soluis
tion of an equation, that
is
considered.
equation which involves the unknown quantity is called a simple equation. The process of solving a simple equation depends upon the following axioms
An
76.
in the first degree
we add equals, the sums are equal. from equals we take equals, the remainders are
1.
If to equals
2.
If
equal. 3.
If equals are multiplied
by equals, the products are
equal. 4.
If
equals
by equals, the quotients are
are divided
equal.
Consider the equation 1 x = 14. what numerical value x must have consistent with this statement. Dividing both sides by 7, we get (Axiom 4). ic = 2 77.
It is required to find
Similarly,
-
if
multiplying both sides by
= — 6,
2,
we
get
= -12 (Axiom equation 7 — 2 — = 23 + 15 — 10, we have 4 = 28. X = 7. i«
Again, in the collecting terms,
a;
.-.
.
.
.^•
a;
.t
.
.
3).
by
:
SIMPLE EQUATIONS.
67
TRANSPOSITION OF TERMS. 78.
To
- 8 == + 12.
solve 3aj
a;
Here the unknown quantity occurs on both sides of the equation. We can, however, transpose any term from one This we proside to the other by simply changing its sign. ceed to show. Subtract x from both sides of the equation, and
3a;-a;-8 Adding 8
to both sides,
= 12
.
.
.
we
get
(Axiom
2).
(Axiom
1).
we have
3a;-x =
12
+8
.
.
Thus we see that + x has been removed from one side, and appears as — a? on the other and — 8 has been removed from one side and appears as + 8 on the other. It is evident that similar steps may be employed in all cases. Hence we may enunciate the following rule ;
Rule.
equation 79.
Any to
may
term
the other by
be transposed from one side of the changing its sign.
We may
for this
is
change the sign of every term in an equation; equivalent to multiplying both sides by 1,
—
which does not destroy the equality (Axiom
3).
— 3 aj — 12 = x — 24. Ex, Take the equation Multiplying both sides by — 1, 3 x + 12 = — x -f- 24, which is the original equation with the sign of every term changed. 80. We can now give a general rule for solving a simple equation with one unknown quantity.
Rule.
the terms containing the unknoimi of the equation, and the known quanCollect the terms on each side ; divide
Transpose
all
quantity to one side the
other.
tities
to
both
sides by the
coefficient
the value required
Ex.
1.
Solve 5(x
Removing
-
brackets,
transposing, collecting terms,
is
of the unknown
cpiantity,
obtained.
- 7(6 - x) + 3 = 24 - 3(8 - x). 5x- 15-42 + 7x + 3=:24-24 + 3x; 5x + 7x-3x = 24-24 + 15 + 42-3; 9 x = 54.
3)
.-.
x
=
6.
and
;
ALGEBRA.
58 Ex.
Solve
2.
Simplifying,
5x - (ix we have
- C12 x'^ -
5X
-
7)(Q X
41 X
+
i:>)
o5)
=
- S(4x -
a
=6-3(4 x^ -
9)(x
13 x
+
-
1).
9),
and by removing brackets, 5X
Erase the term
—
12 x2
+
41
a:
-
12x2 on each
collecting terms,
7 .-.
Note. it,
G
-
12
x'^
+
39 x
-
27.
and transpose;
= 6-27 +35; = 14. x = 2.
5x + 41x-39x
thus
Since the
—
we have formed
Ex.
Solve 7 X
3.
-
x
sign before a bracket affects every term within
work
in the first line of
until
=
35 side
of Ex. 2,
we do not remove
the brackets
the products.
5[x
- {7 -
(3(x
-
3)}]
=
3x
+
1.
Removing brackets, we have 7
X
- {7 - 6 X + 18}] = 3 X + 1, - 5[x - 25 + 6 .x] = 3 X + 1, 7x- 5x+ 125-30x = 3x-f 1 7x — 5x — 30x — 3x= 1 — 125 — 31 x = — 124 x = 4.
-
5[x 7
transposing, collecting terms,
X
;
;
.-.
81. It is extremely useful for the begimier to acquire the habit of occasionally verifying, that is, proving the truth Proofs of this kind are interesting and conof his results. and tlie habit of applying such tests tends to vincing ;
make
the
student self-reliant and confident in his
own
accuracy.
In the case of simple equations we have only to show when we substitute the value of x in the two sides of the equation Ave obtain the same result. that
Ex. To show that x = 2 satisfies the equation 5x-(4x-7)(3x-5)=G-3(4x-9)(x-l).
When x=2, The
right sideC,
= 6-3(Thus, since these equation.
Ex.
2,
Art. 80.
5x-(4x-7)(3x-5)=10-(8-7)(C-5) = 10-1=9. - 3(4x - 9)(x _ l) = - 3(8 - 9)(2 - 1)
the left side
two
results are the
1)
same, x
=9.
=
2
satisfies
the
:
SIMPLE EQUATIONS.
EXAMPLES
59
VII.
Solve the following equations
+
1.
3X
3.
3x +
5. 6.
7. 8.
9.
10. 11.
12. 13. 14. 15. 16. 17.
18.
19.
20. 21.
22. 23. 24.
25. 26.
27. 28.
29. 30.
31. 32. 33.
34. 35. 36. 37. 38. 39.
40.
= + 25. = 5(x-2).
15 4
a:
-
3
2.
2
4.
2x-\-S
a-,
= 3 x - 7. = lQ-(2x-S).
8(x-l)+17(x-3)=4(4.x-9)+4. 15(x - 1) + 4(x + 3) = 2(7 + 5x-6(x-5) = 2(a; + 5)+5(x-4). 8(x-3)-(6-2x)=2(x + 2)-5(5-x). 7(25-x)-2x = 2(3x-25). 3(169 - x) - (78 + X) = 29 X. 5 X - 17 + 3 X - 5 = 6 X - 7 - 8 X + 115. 7 X - 39 - 10 X + 15 = 100 - 33x + 2t>. 118-65x-123 = 15x + 35-120x. 157 -21 (x + 3)= 163 -15(2 X- 5). 179-18(x-10)=158 -3(x- 17). 97 -5(x + 20)=lll-8(x + 3). X - [3 + {x - (3 + X)}] = 5. 5 X - (3 X - 7) - {4 - 2 X - (6 X - 3)} = 10. 0,-).
14x-(5x-9)-{4 -3x-(2x-3)} = 30. 25X-19- [3-{4x-5}J = 3x-(6x-5). (X + 1) (2 X + 1) = (x + 3) (2 X + 3) - 14. (X + 1)2 - (x2 _ 1) =3 x(2 X + 1) - 2(x + 2) (X +
1)
+
20.
2(x+l)(x + 3)+8=(2x+l)(x + 5). 6(x2-3x + 2)-2(x2- l)z=4(x+ l)(x + 2)-24.
- 4) - (x2 + X - 20) = 4 x2 - (5 X + 3) (X - 4) - 64. + 15)(x-3)-(x2-6x + 9)=30-15(x-l). 2x-5{3x-7(4x- 9)} = 66. 20(2 - x) + 3(x - 7) - 2[x + 9 - 3 {9 - 4(2 - x)}] = 22. X + 2 - [x - 8 - 2 {8 - 3(5 - x) - x}] = 0. 3(5-6x)-5[x-5{l -3(x-5)}]=23. (x+l)(2x + 3)=2(x+l)2 + 8. 3(x - 1)2 - 3(x2 - 1) = X - 15. (3x + l)(2x-7) = 6(x-3)2 + 7. x2 - 8 X + 25 = x(x - 4) - 25(x - 5) - 16. x(x+ l) + (x + l)(x + 2) = (x + 2)(x + 3)+x(x-i- 4) - 9. 2(x + 2)(x-4)=x(2x + 1)-21. (x + l)2 + 2(x + 3)2zr:3x(x + 2)+35. 4(x + 5)2 - (2 X + 1)2 = 3(x - 5) + 180. 84 + (X + 4)(x - 3)(x + 5) = (x + l)(x + 2)(x + 3). (X + 1) (x + 2) (x + 6) = x3 + 9x2 + 4(7 X - 1). 2(x (x
CHAPTER
VIII.
Symbolical Expression. 82. In solving algebraic problems the chief difficulty of the beginner is to express the conditions of the question by means of symbols. A question proposed in algebraic
symbols will frequently be found puzzling, when a similar Thus, the answer to the question "find a number greater than x by a " may not be self-evident to the beginner, who would of course readily answer an analogous arithmetical question, "find a number greater than 50 by G." The i)rocess of addition which gives the answer in the second case supplies arithmetical question would present no difficulty.
the necessary hint; and, just as the number which is greater than 50 by 6 is 50 6, so the number which is greater than
+
X by a
is
ic
+ a.
83. The following examples will perhaps be the best introduction to the subject of this chapter. After the first we leave to the student the choice of arithmetical instances,
should he find them necessary. Ex.
1.
By how much does x exceed
17 ?
Take a numerical instance " by how much does 27 exceed The answer obviously is 10, which is equal to 27 — 17. Hence the excess of x over 17 is cc — 17. ;
Similarly the defect of x from 17
Ex.
2.
If
X
is
is
17
—
17 ?
"
x.
one part of 45 the other part
is
45
—
x.
Ex. 3. How far can a man walk in a hours at the rate of 4 miles an hour ? In 1 hour he walks 4 miles. In a hours he walks a times as far, that is, 4 a miles.
.
SYMBOLICAL EXPRESSION. Ex.
A
4.
and B are playing
with q dimes after What B has won :
B has won A has lost,
for
$ x,
61
money A begins with $p and B how many dimes has each ? ;
.-.A has 10 {p — x) dimes, B has q + lOx dimes.
EXAMPLES
VIII.
a.
2.
What must be added to x to make y ? By what must 3 be multiplied to make
3.
AVhat dividend gives b as the quotient when 5
1
6.
What is the defect of 2 c from By how much does 3^ exceed
6.
If
4.
3
d.
k
?
a
?
the divisor
is
100 be divided into two parts, and one part be
m,
?
what
is
the other ?
What number is less than What is the price in cents
7.
-^^
8.
If the difference of
9.
what
c ?
of a oranges at ten cents a dozen ?
two numbers be
11,
and
if
the smaller be
x,
the greater ?
is
10.
20 by
If
the
sum
of
two numbers be
c,
and one
of
them
is
20,
what
is
the other ?
12.
What is the excess of 90 over x ? By how much does x exceed 30 ?
13.
If
14.
What
15.
In X years a
11.
16.
17.
100 contains x 5 times, what is
is
the value of x
.
the cost in dollars of 40 books at x dimes each
?
'
man will be 36 years old, what is his present age ? How old will a man be in a years if his present age is x years ? If X men take 5 days to reap a field, how long will one man
take? 18.
19.
dozen 20.
hour
What What is
value of x will is
x cents
make
5 x equal to 20 ?
the price in dimes of 120 apples,
when
the cost of two
?
How many
hours will
it
take to walk x miles at 4 miles an
?
21.
How far can I walk How many miles is
in x hours at the rate of y miles
an hour
?
between two places, if a train travelling p miles an hour takes 5 hours to perform the journey ? 22.
it
ALGEBRA.
62 23.
A man
24.
If I
has a dollars and h dimes,
how many
spend x half-dollars out of a sum of
cents has he
.s20,
how many
?
half-
dollars have I left ? of a purse containing Z(t and b dimes a express in cents the sum left.
Out
25.
cents
;
26.
27.
By how much does 2 x — 5 exceed x+\2 What number must be taken from « — 2 6
a four dimes, 28.
If
29.
If I give
31.
A
—
c
3 & ?
shared equally amongst x persons, and each pays many cents does the bill amount to ?
how
away
c
dimes out of a purse containing a dollars and
how many dimes have
I left ?
spend x quarters a week, of a yearly income of $1/ ? If I
to leave «
spends
bill is
h half-dollars, 30.
man
how many
dollars do I save out
bookshelf contains x Latin, ?/ Greek, and z English books how many are there in other languages ?
;
there are 100 books,
if
32.
I
have x dollars in
cents in another
;
if
I give
my
purse,
away a
tj
dimes in one pocket, and z how many cents have I
half-dollar,
left? 33. In a class of x boys, ij work at Classics, z at Mathematics, and the rest are idle what is the excess of workers over idlers ? ;
We
84.
add a few harder examples worked out
in full.
What is the present age of a man who x years hence will 1. times as old as his son now aged ij years ? In X years the son's age will be ?/ + ic years hence the father's age therefore now the father's age is m{]i + x) — will be m{y + x) years Ex.
be
m
;
;
X years. Ex.
2.
Find the simple interest on $ k in n years Interest on $ 100 for
1
year
is
at
/ per
cent.
-•?/,
f 100
^k Interest on
.*?
k
§ for n years
is
100'
8 loo"
x yards long, ij feet broad, and a feet high find Ex. 3. A how many square yards of carpet will be required for the floor, and how many square yards of paper for the walls.
room
is
;
;;
;
;
;
SYMBOLICAL EXPRESSION. The area
(1) .
of the
the area of the walls
•.
1
-.
.
number
Sxy square
feet
of square yards of carpet required
The perimeter
(2) .
number
the
•.
of the floor is
63
is
is
—-^ = ^^
room is 2(3 x + y) feet 2a(Sx -{- y) square feet
It
.
1
•
of square yards of paper required
•
is
2a(3x + ^^— ^r
y) -^'
Ex. 4. The digits of a number beginning from the left are a, 6, c what is the number ? Here c is the digit in the units' place b standing in the tens' place ;
represents b tens
The number
similarly a represents a hundreds.
;
is
= 100a+
10 6
+
is
the number are inverted, a symbolically expressed by
100c
Ex.
5.
numbers
What
is
(1) the least
which the
of
•.
is
the
the product
=
+
n
new number
is
formed
a.
?
sum = n
n(n
+
106
-\-
+ 1, n 2. + 1) + 0^ + ^)
to n are n
= And
c units
sum, (2) the product of three consecutive
The numbers consecutive .
+
c.
If the digits of
which
+
b tens
therefore equal to a hundreds
-\-
3 u
1) (u
(n
-{-
+ 3. + 2).
We may remark here that any even number may be denoted by 2n, where n is any positive whole number; for this expression is exactly divisible by 2. Similarly,
any odd number may be denoted by 2 7i + 1 when divided by 2 leaves remainder 1.
for this expression
EXAMPLES
VIII.
b,
1.
Write four consecutive numbers of which x
2.
Write three consecutive numbers of which y
3.
Write
4. 5.
What What
five
consecutive numbers of which x
is
the next even
is
the odd
number
number next
is is
is
the least.
the greatest.
the middle one.
after 2 n ?
before 2
r/:
+
1 ?
Find the sum of three consecutive odd numbers of which the middle one is 2 « + 1. 6.
ALGEBRA.
64
A man makes a journey of x miles. He travels a miles by coach,
7.
h
by
and
train,
carry him
A
8.
week
How
by boat.
finishes the journey
far does the boat
?
horse eats a bushels and a donkey h bushels of corn in a bushels will they together consume in n weeks ?
how many
;
9. If a years hence
A
10.
man was
boy
old will he be y
x years old, and
is
How
that of his father.
What
11.
how
x years old 5 years ago,
?
old
man who
the age of a
is
five years hence his age will be half the father now ?
is
y years ago was
m
times as old
as a child then aged x years ?
A's age
12.
old
:
double B's, B's
is
is
three times C's, and
C
is
x years
find A's age.
13.
AVhat
14.
What What What
15. 16.
annum
is
the interest on $1000 in h years at
is
the interest on
is
the interest on $50 a in a years at a per cent. ?
is
^x
the interest on
in
c
per cent.
a years at 5 per cent.
$2-4.t?/ in
x months at y per cent, per
?
17. A room is x yards in length, and y feet in breadth square feet are there in the area of the floor ?
18. A square room measures x feet each yards of carpet will be required to cover it ? 19.
A
of carpet 20.
?
?
way
;
how many
how many square
;
room is p feet long and x yards in width how many yards two feet wide will be required for the floor ? ;
What
is
cost in dollars of carpeting a
broad, with carpet costing
c
room a yards
dimes a square yard
long, h feet
?
in the middle there 21. A room is a yards long and h yards broad a carpet c feet square how many square yards of oil-cloth will be required to cover the rest of the floor ? ;
is
;
22.
How
miles in 23.
many 24.
c
A
long will hours ? train
miles will
is it
it
?
if
he walks 20
running with a velocity of x feet per second travel in y hours ?
How many men
do in xz hours
take a person to walk h miles
will
be required to do
in
;
x hours what y
how men
:
CHAPTER
IX.
Problems Leading to Simple Equations. 85.
The
i)rinciples of the last chapter
may now
be em-
ployed to solve various problems.
The method
of procedure is as follows
Represent the unknown quantity by a symbol, as x, and express in symbolical language the conditions of the question we thus obtain a simple equation which can be solved by the methods already given in Chapter vii. ;
Unknown
Note.
are usually represented by the last
quantities
letters of the alphabet.
Ex. 1, Find two numbers whose sum is 28, and whose difference is 4. Let X represent the smaller number, then x 4- 4 represents the greater.
Their
sum
is
a:
+ (x +
Hence
4),
x
which
is
x
-f-
4
2
a;
-|-
.
and so that the
numbers are 12 and
The beginner that
•.
X
it satisfies
is
+
to be equal to 28.
= 28 = 24
x=
12,
=
16,
4
;
;
16.
advised to test his solution by proving
the conditions of the question.
Ex. 2. Divide 60 into two parts, so that three times the greater exceed 100 by as much as 8 times the less falls short of 200. Let X represent the greater part, then 60 — x represents the less
may
Three times the greater part
is
3 x,
and
its
excess over 100
is
3x-100. Eight times the less
is
8(60 200
F
— -
x),
and
8(60 65
-
its
X).
defect from 200
is
ALGEBRA.
(36
Whence
the symbolical statement of the question
Sx-
100
3a:- 100 480
-
-
100
200
5x= a:
and
60
—
x
180;
= 36, = 24,
the greater part,
the
less.
A may
Ex. 3. Divide 847 between A, B, C, so that than B, and B $ 8 more than C.
Suppose that C has x dollars X
+
8
+
;
then
B
has x
+
have $10 more
8 dollars,
and
A has
10 dollars.
Hence
x
+ X
C has
so that
is
= 200-8(60-0;); = 200 - 480 + 8 X, = 8 X - 3 X,
$7,
B
(./:
+
+
a;
$15,
8)
+
+ (x + 8 + 10) = 47 + X + 8 + 10 = 47, 3.r = 21; .-. x=7, ;
8
A $25.
A
person spent $112.80 in buying geese and ducks if each goose cost 14 dimes, and each duck 6 dimes, and if the total number of birds bought was 108, how many of each did he buy ?
Ex.
4.
;
In questions of this kind it is of essential importance to have all quantities expressed in the same denomination in ;
the present instance
money
it
will be convenient to express the
in dimes.
Let X represent the number of geese, then 108
number
—
x represents the
of ducks.
Since each goose cost 14 dimes, x geese cost 14 x dimes. since each duck cost 6 dimes, 108 — x ducks cost 6(108 dimes. Therefore the amount spent is
And
14a;-f 0(108
-x)
but the question states that the amount
Hence
14 2,
a:
7
+ 6(108 - x)= + 324 - 3 x = 4a; = = 108 — x =
a:
.-.
and
a;
x,
dimes; is.
also $112.80, that
is,
dimes.
dividing by
—
1128
;
564,
240; 00, the 48, the
number number
of geese
;
of ducks.
1128
PROBLEMS LEADING TO SIMPLE EQUATIONS. Ex. old
A
5.
is
what are
;
twice as old as
B
;
ten years ago he
67
was four times
Let X represent B's age in years, then 2 x represents A's age. Ten years ago their ages were respectively, x — 10 and 2 5c 2x — 10 = 4(x — 10) years thus we have ;
—
10
;
2
a;
-
10 2
.-.
so that
as
their present ages ?
B
is
15 years old,
A
a:
x
= 4 X - 40, = 30 = 15, ;
30 years.
Note. In the above examples the unknown quantity x represents a number of dollars, ducks, years, etc, and the student must be careful to avoid beginning a solution with a supposition of the kind, " let X = A's share," or "let x = the ducks," or any statement so vague and inexact. ;
EXAMPLES 1.
One number exceeds another by
IX. 5,
and
their
sum
is
29
find
;
them, 2. The difference between two numbers is 8 greater the result will be three times the smaller
Find a number such that 11 than its defect from 89, 3.
4.
What number
exceeds 20 5.
is
its
if
;
;
2 be added to the
find the
excess over 50
may
numbers,
be greater by
by as much as
that which exceeds 8
its
double
?
Find the number which multiplied by 4 exceeds 40 as much as
40 exceeds the original number. 6. A man walks 10 miles, then travels a certain distance by train, and then twice as far by coach. If the whole journey is 70 miles, how far does he travel by train ? 7.
What two numbers are those whose sum is 58, and
difference 28
?
288 be added to a certain number, the result will be equal to find the number, three times the excess of the number over 12 8.
If
;
^ 16
Twenty-three times a certain number above seven times the number find it.
9. is
is
as
much above
14 as
;
a^
10. Divide 105 into two parts, one of which diminished by 20 shall be equal to the other diminished by 15. 11.
Divide 128 into two parts, one of which
is
three times as large
as the other. 12.
Find three consecutive numbers whose sum
shall equal 84.
ALGEBRA.
68 The
13.
difference of the squares of
two consecutive numbers
is
find them.
35
;
14.
The sum
to
it is
five
15.
Find two numbers
of
two numbers
times the other
;
and one of them with 22 added numbers.
is 8,
find the
differing
by 10 whose sum
is
equal to twice
tlieir difference.
A
16.
money
and B begin to play each with $60. double B's, what does A win ?
is
If
they play
till
A's
17. Find a number such that if 5, 15, and 35 are added to it, the product of the first and third results may be equal to the square of the second.
The
18.
bers
121
is
The
19.
squares
is
between the squares of two consecutive num-
difference
find the numbers.
;
two numbers
difference of
27
;
is 3,
and the difference
of their
find the numbers.
20. Divide $380 between A, B, and C, so that B may have $30 more than A, and C may have $20 more than B. 21.
A sum
or dimes 22.
;
of
$7
made up
is
how many
If silk costs five
buying 22 yards of
of 46 coins
are there of each
silk
which are either quarters
?
times as much as linen, and I spend $48 in and 50 yards of linen, find the cost of each
per yard. 23. A father is four times as old as his son only be twice as old find their ages.
;
in
24 years he will
;
24.
as B's 25.
old as 26.
A
is
25 years older than B, and A's age
below 85
is
A's age
B
is
;
much above
as
20
three times B's, and in 18 years
A
will
be twice as
find their ages.
;
A
is
four times as old as B, and in 20 years will be twice as find their ages. is 5 years older than B
who
old as C,
;
A's age
is
six times B's,
times as old as
B
;
27.
is
find their ages.
and
fifteen years
hence
A will
be three
find their ages.
28. A sum of $16 was paid in dollars, half-dollars, and dimes. The number of half-dollars used was four times the number of dollars and twice the number of dimes how many were there of each ? 29. The sum of the ages of A and B is 30 years, and five years ;
hence
A
will be three times as old as
B
;
find their present ages.
spend $69.30 in buying 20 yards of calico and 30 yards of silk the silk costs as many quarters per yard as the calico costs cents per yard find the price of each. 30.
I
;
;
PROBLEMS LEADING TO SIMPLE EQUATIONS. 31.
I
purchase 127 bushels of grain.
If the
number
wheat be double that of the corn, and seven more than
number of bushels of corn equals the number of bushels of each.
the
number
69
of bushels of five
times the
of bushels of oats, find
if the 32. The length of a room exceeds its breadth by 3 feet length had been increased by 3 feet, and the breadth diminished by 2 find the dimensions. feet, the area would not have been altered ;
;
33.
The length
of a
had been increased by 60 feet
room exceeds
2 feet, the area
breadth by 8 feet if each would have been increased by
its
;
find the original dimensions of the room.
;
A and B start from the same place walking at different rates when A has walked 15 miles B doubles his pace, and 6 hours later if A walks at the rate of 5 miles an hour, what is B's rate passes A 34.
;
;
at first ? 35.
B
A sum
of
money
have together |20,
is
among A, B, and C, C $30, and B and C $40
divided
A and
;
so that
A
and
find the share
of each. 36.
A man sold two pieces of cloth, losing $ 6
on the other. loss, find
the
If his entire loss
amount
lost
was $4
on each
less
more on the one than than four times the smaller
piece.
Two men
received the same sum for their labor but if one had received f 10 more, and the other $ 8 less, then one would have had three times as much as the other. What did each receive ? 37.
;
38. In a certain examination the number of successful candidates was four times the number of those who failed. If there had been 14 more candidates and 6 less had failed, the number of those who passed would have been five times the number of those who failed. Find the number of candidates. 39.
A purse
rest dimes. of each 40.
If
kind
An
contains 14 coins,
some
of
which are quarters and the
the coins are worth .|2 altogether,
how many
are there
?
estate
the share of the
was divided among was three times
first
three persons in such a way that that of the second, and the share
of the second twice that of the third. The than the third. How much did each receive
first ?
received 1 900
more
CHAPTER
X.
Resolution^ into Factors.
When an algebraic expression is the 86. Definition. product of two or more expressions, each of these latter quantities is called a factor of it, and the determination of these quantities is called the resolution of the expression into its factors.
87. Rational expressions do not contain square or other any term.
roots (Art. 14) in
88. Integral expressions do not contain a letter in the denominator of any term. Thus, x^ -\- ?> xy -{- 2 y'^, and \ x^ -\- ^ xy — ^ y"^ are integral expressions.
89. In this chapter we shall exj)lain the principal rules by which the resolution of rational and integral expressions into their component factors, which are rational and integral expressions,
may
WHEN EACH
be effected.
OF THE TERMS IS DIVISIBLE BY A COMMON FACTOR.
90. The expression may be simplified by dividing each term separately by this factor, and enclosing the quotient within brackets; the common factor being placed outside as a coefficient. Ex.
1.
The terms
of the expression Z a^
—
Q ah have a
common
factor Za. .-.
Ex.
2.
5 a-bx^
-
Z(fi-Qab = 3aia-2b).
15 abx^-
-
20 b^x^ 70
^
5 bx'^Ca'^x
-
3 a
-
4 &2).
:
RESOLUTION INTO FACTORS.
EXAMPLES
X.
71
a.
Resolve into factors
-ax.
1.
a^
2.
x^
6.
3.
2
7.
4.
7^)2
5.
- x:\ a - 2 «2. + p.
8.
13.
^a^-2>a%-\-Q
14.
2
x2«/3
-
15.
6 x3 -
9
6
x2?/
+
c'%
cfib'^.
+
x2?/2
- 2 x\ - 5 a^^ 15 + 25 x^. 16 X + 64 xhj.
12
2
9.
15
cfi
-
225 aK
10.
54
-
81
x.
11.
10 x^
12.
3 x^
- 25 - x2 +
_
15 a^x^
8X
5
xy/3.
x?/2.
-
10 as^s
16.
5 x^
17.
7
18.
38 a^x^
-
«
7 a^
+
WHEN THE TERMS CAN BE GROUPED 91. Ex.
1.
terms a factor
b,
«4,
57 ff%2.
SO AS TO
two terms contain a factor x, and the last two first two terms in one bracket, and
first
we
enclose the
Thus,
the last two in another. x2
—
ax
+
&^
—
(lb
= (x2 — ax) + (bx — ab) = x(x — a)-h b(x— a) = (x- a)(x + b),
since each bracket of (1) contains the 2.
14
x.
CONTAIN A COMMON FACTOR. Resolve into factors x2 — «x + bx — ah.
Noticing that the
Ex.
+
a;*?/.
Resolve into factors 6 x2
—
same 9 ax
factor x
+
4 bx
—
—
.
.
.
(1)
a.
6 ab.
- 6 «6 = (6 x2 - 9 ax) + (4 6x - 6 ab) = 3 x(2 X - 3 a) + 2 &(2 X - 3 a) = (2x-3«)(3x + 2?>). Ex. 3. Resolve into factors 12 a^ — iab — S axP- + bx:^. 12 d^-4:ab-^ ax2 + &x2 = (12 a2 - 4 ab) - (3 «x2 - 6x2) = 4«(3« -6)-x2(3« - b) = (3a-Z>)(4a-x2). 6 x2
-
9 «x
+
4
?>x
Note. In the first line of work it is usually sufficient to see that each pair contains some common factor. Any suitably chosen pairs will bring out the same result. Thus, in the last example, by a different arrangement, we have 12 a2
_
The same
4
a?>
-
3 ax2
+
6x2
^ (12 ^2 _ 3 ^^^^2) _ (4 a6 - 6x2) = 3a(4a - x2)-6(4«-x2) = (4rt-x2)(3a-6).
result as before, for the order in
product are written
is
of course immaterial.
which the factors of a
:
:
ALGEBRA.
72
EXAMPLES
X.
\
/
b.
Resolve into factors 1.
ifi
^
a2
3.
4. 5.
6.
ah
+
ac-\- he.
ac
-\-
ab
aV + acd +
-
he.
+ hd. + Sc. 2 X + ex + 2 c + c2. x"^ - ax + 5 X - 5 a. 5a + a6 + 56 + ah -hy-ay + if. mx — my — nx + ny. ?nx — ma + wx — na. 2 ax + ay + 2 &x + hy. a^ -\-Sa
+
ahc
ac
6-^.
7. 8.
9. id^!
11.
+ -
— hx — S ay + hy. + Sxy — 2ax- ay. 14. mx — 2 my — + 2 ?iy. 15. aa;2 -Sbxy - axy + 3 6y2_ 16. x^ + mxy — 4:xy — 4 my'^. 17. 2 x* - x^ + 4 x - 2. 18. 3x3 + 5x2 + 3x + 5. 19. x* + x^ + 2 x + 2. 20. y^ — y^ + y — \. ^L axy + 6cxy — az — hcz. 22. /^x'-^ + g^x'^ - af - a/^. 12.
3 ax
13.
ox^
7iaj
TRINOMIAL EXPRESSIONS. *
92. When the Coefficient of the Highest Power is Unity. Before proceeding to the next case of resolution into factors the student is advised to refer to Chap. iv. Art. 51. Attention has there been drawn to the way in which, in forming the ]3roduct of two binomials, the coefficients of the different terms combine so as to give a trinomial result. Thus, by Art. 51, (x
+ 5)(x + 3)=x' + Sx-\-15
(x-5)(x-3)=x^-Sx-\-15 (x-\-5)(x-S)=x^-{-2x-W (oj - 5)(aj + 3)= x2- 2a; -15
.... .... .... ....
We
(1),
(2),
(3), (4).
now propose to consider the converse problem namely, the resolution of a trinomial expression, similar to those which occur on the right-hand side of the above identities, into its :
component binomial factors. By examining the above results, we notice that
The first term of both the factors is x. The product of the second terms of the two
1.
2.
factors is
equal to the third term of the trinomial thus in (2) above we see that 15 is the product of — 5 and — 3 while in 15 is the product of -f 5 and — 3. (3) ;
;
—
RESOLUTION INTO FACTORS.
73
3. The algebraic sum of the second terms of the two factors equal to the coefficient of x in the trinomial thus in (4) the -^ gives — 2, the coefficient of x in the trisum of — 5 and nomial. In showing the application of these laws we will first consider a case where the third term of the trinomial is positive.
is
;
+
Ex.
Resolve into factors x^
1,
The second terms
+
and
24,
sum +
their
x2
.-.
+
11
a:
+
24.
must be such that their product clear that they must be + 8 and +
of the factors It is
11.
+ llx +
24=(ft;
+
8)(x
+
3).
Resolve into factors x'^ —10x-\- 24. The second terms of the factors must be such that their product + 24, and their sum — 10. Hence they must both be negative, and is easy to see that they must be — 6 and — 4. Ex.
2.
Ex.
3.
Ex.
4.
Ex.
5.
+
a2,
Resolve into
and
their .'.
Note. his results,
Sl
+
10 X-
The second terms 10
25
must be such that their product is Hence they must be — 10 a and a.
of the factors
sum —
x2
—
11 a.
11 ax
+
10 a2
—
=(x —
10 a) (x
—
a).
In examples of this kind the student should always verify by forming the product {mentally, as explained in Chapter
IV.) of the factors
he has chosen.
EXAMPLES Resolve into factors 1.
it
=(x-9)(x-9) = (x-9)2. = (x2 + 5) (x'^ + 5) = (x^ + 5)2. factors x'^ — 11 ax + 10 «2.
x2- ISx + X*
is
x2- 10x + 24=(x-6)(a;-4).
...
+
is 3.
:
X.
c.
.
:
ALGEBRA.
74
+ G00y'^. + 34 + 289. a^b^ + 37 + 300. a2 - 29 a?> + 54 62. x^ + 162 x2 + 6561. 12 - 7 X + X-.
x^-\-49xy
33.
20
x2?/2
a:?/
34.
132
a-^fta
35.
29. 30.
31. 32.
Ex.
is
35,
37. 38.
Resolve into factors x2
1.
iK2.
the third
term of the
tri-
negative.
The second terms signs,
36.
Next consider a case where
93. 7iomial
—
+ 9x + - 23 x + aj2. 88 + 19 x + x^. 130 + 31 x^ + x'^y^. 204 - 29 x2 + x*. 216 + 35 x + x2.
27. 28.
+
—
2 x
35.
must be such that their product is and their algebraic sum + 2. Hence they must have opposite and the greater of them must be positive in order to give its of the factors
sign to their sum.
The required terms . •.
Ex.
+
2 X
-
Resolve into factors
2.
and — 5. + 7) (x - 5) x2 — 3 x — 54.
+
are therefore
x2
35
=
7
(x
must be such that their product is Hence they must have opposite them must be negative in order to give its
Tlie second terms of the factors
—
54,
signs,
and and
their algebraic tlie
greater of
sum —
3.
sign to their sum.
— 9 and + 6. a:2_3x- 54= (x-9)(x + 6).
Tlie required terms are therefore ...
Remembering opposite signs,
Ex.
if
that in these cases the numerical quantities must have
Resolve into factors x2^2
3.
method may be adopted.
preferred, the following _|_
23
x?/
-
420.
Find two numbers whose product is 420, and whose difference is 23. These are 35 and 12 hence inserting the signs so that the positive may predominate, we have ;
x-2/2
+
23 xy
-
420
=
(xy
EXAMPLES
+ X.
35) (xy
-
12).
d.
Resolve into factors 1.
x2
2.
x2
3.
x2
4. x2
- 2. - 2. X - 0. 2 X - 3.
+ 2 X - 3. + X - 56. x2 - 4 X - 12. «2 _ a _ 20.
-
X
5.
x2
9.
«2
+ -
X
6.
x2
10.
a-
11.
x2
12.
x2
7. 8.
+ + +
«
_
4 a 9x
x
-
20.
-
117. 36.
156.
:
RESOLUTION INTO FACTORS. 13. 14. 15.
x^^x- 110. x^-9x- 90. x^-x- 240.
16. a2 17.
_ -
28.
35.
-
85.
xV -
5xy
x2
+
-
ax
20. x'-2-32
+ x2 +
21. x2
11 «
«-
When
94.
-
42
24.
23.
-
25.
115.
27. 7/^+6
31. x^
-
a^^^i
x^
-
a%-^
33. x^
+
x^.
(3x
iv.
x^
38.
-
_ -
xV-27 x^.
132
(^4.
462
870.
-
120
7
ax
-
a^x^.
x^.
the Coefficient of the Highest
Again, referring to Chap. following results:
ft^
26. x^
260.
32.
- U a - 26. + U ay-2i0. - a'^&s _ 5^ 54. - 14 x2 - 51.
a^
24. a'^y-
a"-.
x^-105 y\
18 x
- 152. 22. 16 X + 12 abx - 28 b'-x^. 29. a2 _ 18 axy - 243 x^K 30. x^ + 13 a-x:^ - 300 a\ 2 + X - x\ 36. 110 - X 6 + X - x2. 37. 380 - X (fi
34.
12 «
18. 19.
75
Art. 51,
Power
is
not Unity.
we may
write the
+ 2)(x-\-4:)=3x^ + Ux-{-S
(3x-2)(x-4:)=3a:^-Ux +
S
(3x-\-2)(:x-4:)=Sx^-10x-S l3x-2)(x + ^)=3x''-{-10x-S The converse problem presents more
.
.
.
(1),
.
.
.
(2),
.
.
.
(3),
.
.
.
(4).
difficulty
than the
we have yet considered. Consider the result 3 x~-l'ix + S = (3x- 2) (x - 4). The first term 3 x^ is the product of 3 x and x. The third term + 8 is the product of — 2 and — 4. The middle term — 14 a; is the result of adding together the two products 3x x — 4 and x x —2. Again, consider the result 3 a^^— 10 0^—8 = (3 x4-2)(;r— 4). The first term 3 x^ is the product of 3 cc and x. The third term — 8 is the product of + 2 and — 4. The middle term — 10 x is the result of adding together the two products 3 a: x — 4 and x x2; and its sign is negative because the greater of these two products is negative. cases
Considering in a similar manner results
(1)
and
(4),
we
see that 1.
If the third term of the trinomial is positive, then the its factors have both the saine sign, and this the same as that of the middle term of the trinomial.
second terms of sign
is
ALGEBRA.
76 2.
If the third term of the trinomial is negative, then the
second terms of
95.
its
factors have opposite signs.
The beginner
will frequently find that
to select the proper factors at the first trial.
it is
not easy
Practice alone
will enable him to detect at a glance whether any pair he has chosen will combine so as to give the correct coefiicients of the expression to be resolved.
Ex. Resolve into factors 7 x~ — 19 x — 6. Write (7 x 3) (x 2) for a first trial, noticing that 3 and 2 must have opposite signs. These factors give 7 x~ and — 6 for the first and third terms. But since 7 x 2— 3 x 1 = 11, the combination fails to give the correct coefficient of the middle term. Next try (7 x 2) {x 3). Since 7x3 — 2x1 = 19, these factors will be correct if we insert the signs so that the negative shall predominate.
7x2- 19a;-6=(7x + 2)(x-3).
Thus
[Verify by mental multiplication.]
96. In actual all
work
it
these steps at length.
will not be necessary to put
The student
down
will soon find that
the different cases may be rapidly reviewed, and the unsuitable combinations rejected at once. Ex.1.
X- 15
(1),
Ux2-29x-15
(2).
Resolve into factors 14x2
+
29
we may
write (7 x 3) (2 x 5) as a first trial, noticiug that 3 and 5 must have opposite signs. And since 7 x 5 — 3 x 2 = 29, we have only now to insert the proper
In each case
signs in each factor.
In (1) the positive sign must predominate. In (2) the negative sign must predominate.
Therefore
Ex.2.
In (1) In (2)
+ 29 x - 15 = (7 x - 3) (2 x + 5). 14x2-29x-15=(7x + 3)(2x-5).
14
x'^
+ 17 X + 6 5x2-17x + 6
Resolve into factors 5 x2
we we
(1), (2).
notice that the factors which give G are both positive. notice that the factors which give 6 are both negative.
:
RESOLUTION INTO FACTORS. And
we may we may
therefore for (1) (2)
Q
5x3+1
And, since
/
Q x^
5 x2
write
(5x+ —
write (5x
)(ic )(a:
+ —
77 ). ).
= 17, we see tliat + n X + Q =(5 X + 2)(x + S). - 17 + 6 = (5 X - 2) (X - 3). x
2
a:
In each expression the third term 6 also admits of factors 6 and 1, but this is one of the cases referred to above which the student would reject at once as unsuitable.
Ex.
9a;-
3.
Ex.4.
- 48 xy + 6
64 y^
= {3x - Sy)(Sx = (i3x-Syy.
+ 7a:-5a:2=(3 +
Sy)
5a-)(2 -a;).
In Chapter xxvi. a method of obtaining the factors of aiiy + 6x + c is given.
Note.
trinomial in the form ax^
EXAMPLES
X.
e.
Resolve into factors
12.
+ 3a:+ 1. + 5 x + 2. 2x2 +5 + 2. 3a;2+ lOx + 3. 2x2 + 9x + 4. 3 + 8 X + 4. 2x2 + llx + 5. 3x2 + llx + 6. 5:c2 + llx + 2. 3x2 + -2. 4x2 + llx-3. Sx- + 14x - 5.
13.
2 X-
1.
2a;2
2.
3
3.
14.
a;^
a:
4.
5. 6.
a:2
7. 8.
9.
10. 11.
a:
97.
+
15 X
We
-
8.
add an exercise containing miscellaneous exam-
ples on tlie preceding cases.
—
:
ALGEBRA.
5
6.
x^-\-x-6.
6.
2 ax^ + 3 axy
7.
x'^-\-7xy
8.
a2
9.
x2
10. 11.
12. 13. 14. 15. 16. 17.
18. 19.
_ -
rt|/
— 3 by^.
2 6a;y
-60y^.
-
210
+ 110. 24 + 37 X - 72 x2. 98-7x-x2. 3 .x2 + 23 X 4- 14. 2 x2 -f 3 X - 2. x2 -20xy - 96 a2 - 20 abx + 75 a2 _ 24 « + 95. 7 + 10 X + 3 x2. «2 _ 4 « _ 21. x2 + 43 xy + 390 y\ 21 X
amx^ -|- bmxy — anxy — bny\
22.
6x2 -7a: -3.
26.
27.
3 x2
28.
x2
30. 31. 32.
33.
34.
IS
11 x
-
19 x - 14.
- 19x + 90. x2 + 3 x - 40. x2y2 _iQxy + 39. 204 - 5 x - x2. 15 x2 + 224 x - 15. 3x2 + 41 X + 26. 65 + 8 x?/ - x^y^.
29.
WHEN AN EXPRESSION
+
12x2
25.
b'^x'^.
102.
- 4 x^. _ 23x?/ + 10?/2. 3 x2 + 7 x + 4. a2_ 32^ + 256. 3
24.
?/2.
23rK+
a;'-^
23.
?/2.
+
20. 21.
THE DIFFERENCE OF TWO
SQUARES.
By
98.
multiplying a (a
a result which
may
The product of
-\-
hj a — h we obtain — b) = a- — b%
-\-b
b){a
the identity
be verbally expressed as follows
the
sum and
the difference
of any two quan-
equal to the difference of their squares. Conversely, the difference of the squares of any two quan-
tities is
tities is
equal to the jjroduct of the
sum and
the difference of
the two quantities.
Ex.
1.
— 16 = (5 x)2 - (4 yy.
Resolve into factors 25x2 25 x2
-
16
?/2
t/-.
first factor is the sum of the difference of 5x and 4 y.
Therefore the factor
is
.:
25x2-
The intermediate Ex.2.
1
16
?/2
steps
=(5x4-
may
-49c0=(l +
4
5x and
2/)
(5
4?/,
and the second
X- 4?/).
usually be omitted. 7c3)(l
-
7c3).
:
RESOLUTION INTO FACTORS. The
79
difference of the squares of two numerical quantities sometimes conveniently found by the aid of the formula
is
Ex.
(329)2 -(171)2
=(329 +171)(329- 171)
=
500 X 158
EXAMPLES Eesolve into factors 1.
X.
=
79000.
g-.
ALGEBRA.
80 Ex.
Resolve into factors x^ — (2b — S c)2. of x and 2b —Sc is x + 2 6 — 3 c, and their difference
2.
The sum
is
x-{2b-Sc)=x-2b + Sc. x2
.-.
- (2 & - 3 c)2 = (x +
2 &
-
3 c) (x
-
2 5
+
3 c).
If the factors contain like terms, they should be collected so as to give the result in its simplest form.
(3x + 7?/)2-(2x-3!/)2 = {iSx + 7 y) + {2x - Sy)}{(Sx + 7 y)-{2x = (3x + 7y + 2x-3?/)(3x+7?/-2x + 3?/) = (5x + 4?/)(x + 10?/).
Ex.3.
EXAMPLES Resolve into factors
+
-
+ 2 yy - a\ + 3&)2-16x2.
1.
(a
c2.
4.
(x
(a-6)2-c2.
5.
(a
6.
(x+
3.
(x
+
X. h.
:
2.
6)2
?/)2-4s2.
-(2«- 36)2.
5a)2-9?/2.
+
(a-2x)2-62. (2x-3a)2- 9c2.
9.
1
19.
(4 a
12.
c2
20.
(a
21.
l_(7«-36)2.
22.
(a
23.
(a-3x)2-16?/2.
14. 15.
(7x
16.
(a
17.
(a
+
+ -
2/)2-l.
6)2
n)2
- (to - ?i)^- (6 + my.
\.
(?>_c)2-(a-x)2.
11.
- (« - 6)2. - (5 a - 3 6)2. (a + 6)2 -(c + cO^. (a - 6)2 - (x + yy.
-
(x
8.
9x2
18.
5 c)2
7.
10.
13.
Sij)}
+ x)2 - (6 + z/)2, + 2 6)2 - (3 x + 4 yy. -
24.
(2 a
25.
(a
+
6)2
6
- (x -
5 x)2
-
yy.
- 1. - (x -
c)2
?/
+
zy.
Resolve into factors and simplify: 26.
29. 30. 31. 32.
33.
- x2. 27. x2 -{y - xy. 28. (x + 3 ^)2 - 4 34. 16 rt2 - (3 a + 1)2. + yy- (23 x - yy. 35. (2a + 6 - c)2 -(a - 6 + c)2. (5x + 2?/)2-(3x-^)2. 9x2-(3x-5?/)2. 36. {x -7 y + zy -{7 y - zy. 37. (x 4- y - 8)2 - (x - 8)2. (7 X + 3)2 - (5x - 4)2. 38. (2x + « - 3)2 -(3 - 2x)2. (3a + l)2-(2a-l)2. (X
+
?/)2
?/2.
(24 x
100. Compound Expressions Arranged as the Difference of two Squares. By suitably grouping the terms, compound expressions can often be expressed as the difference of two squares, and so be resolved into factors.
:
RESOLUTION INTO FACTORS. Ex.
— 2ax + x^ — = (a^ - 2 ax + x^) - 4:b^ =r (« - X)2 - (2 &)2 = la - X + 2b)(a - X - 2b). 2. Resolve into factors 9a^ — c^ + icx — i x^. 9 a2 _ c2 + 4 ex - 4 x2 = 9 a2 - (c2 - 4 ex + 4 x^) = (3«)2-(c-2x)2 = (3 « + c - 2 x) (3 a - c + 2 x). 3. Resolve into factors 12 xy + 25 — 4x2 — 9 = 25 - (4 x2 - 12 xy + 9 if-) 12 x?/ + 25 - 4 x2 - 9 = (5)2-(2x-3?/)2 = (5 + 2 X - 3 (5 - 2 X + 3 ?/). 4. Resolve into factors 2 &d - a2 _ c2 + &2 ^^ 2 ac. 1. '
Ex.
Ex.
81
Resolve into factors a^
a^
- 2 ax +
x^
-
4c h'^.
4tb^
?/2,
]f-
?/)
Ex.
_|_
_l_
Here the terms 2bd and 2ac suggest the proper preliminary rangement of the expression. Thus
2bd-
a2
_
c2
+
62
^
t^2
_^
2 ac
= + 2 bd + f?2 _ ^2 + 2 «c - c^ = &2 + 2 &cZ + (^2 _ (a2 _ 2 «c + c2) = {b-\- ay -(a- c)2 = (b + d a - c)(b d - a + c). ?)2
-\-
EXAMPLES
X.
-\-
k.
Resolve into factors
VI. 2.
x2+2xy+
?/2-a2.
a--2ab + b^-x^.
6.
x2-Gax+9a2_i6 62. 4a2 + 4rt6 + &2_9c2. x2 + a2 + 2ax-?/2. 2ay+a2 + 2/2-x2.
7.
x2-a2_2a&-&2. 2ab
3.
4. 5.
ar-
8. 9.
?/2-c2 + 2cx-x2.
l-x'^-2xi/-if.
10.
c2-x2-y2 + 2x?/.
H,
x2
12.
rt2_4(^?,
13.
14.
+ ?/2+2xy-4x2?/2. + 4 52_9(^2c2, x2 + 2x?/+2/2-a2-2«6-62. (^2_2rt6 + &2_c2_2cd-c?2.
.
:
ALGEBRA.
82
101. Important Cases. By a slight modification some expressions admit of being written in the form of the difference of two squares, and may then be resolved into factors by the
method of Ex.
1.
Art. 98.
Resolve into factors x^
+
cc^
+ y^= (X*
x2?/2
+
-\-
x^y'^
2
x2«/2
+ +
?/*.
y^)
-
x^f^
= {%'' + y-^y-ixyY = (x2 + + xy) (x2 + y^- xy) = (x2 + xy + ?/2) (x2 - xy + 2/2) Resolve into factors x* - 15 2. +9 = (x^ - 6 x22/2 + 2/4) _ 9 x4 - 15 x2?/2 + 9 = (x2-3 2/2)2-(3x2/)2 = (x2 - 3 + 3x2/)Cx2 - 3 - 3x2/). ?/2
Ex.
x'^y'^
2/*.
.-k-22/2
2/*
2/2
2/2
EXAMPLES
X.
1.
Resolve into factors
+
1.
x4
2.
81 a*
3.
x*
4.
5,^4
5.
x4
+
16 x2
+
9
+
is
+
2/^-7
]ji,
4 x4
7.
4 m*
9.
10.
Z/'^-
+ 9 - 93 x^y'^. + 9 ?i4 _ 24 ?>i2u2. 9 x* + 4 2/4 + 11 x22/2. x4 - 19 x:hj^ + 25 16 a'^ + b^- 28 a252,
6.
8-
x^if-
+ ,^4 _ 18 m'^n-. - 6 xY' +
WHEN AN
a^
256.
«2//2
2/4
2/*.
EXPRESSION IS THE SUM OR DIFFERENCE OF TWO CUBES.
102. If we divide a^ + b^ by a -\- b the quotient is — ab -f />- and if Ave divide a" — b^ hj a — b the quotient «^ + a 6 4- ^'"• We have therefore the following identities ;
:
+ 6^ = (a + b){cr - ab + b") «' = (a - b){cr + ab +
a^
^"^
;
Ir").
These results are very important, and enable us to resolve any expression which can be written as the sum or the difference of two cubes. into factors
Ex.
1
Note.
8 x3
-
27
2/^
The middle term
= (2 x) ^ - (3 yf = (2x-3y) (4 x2 +
6 xy
is
6 xy
+
9
2/2).
the product of 2 x and 3
y.
:
RESOLUTION INTO FACTORS. 64^3+
Ex.2.
1
=(4
«)-^
+(1)3
= (4a+ l)(i6a'-^-4a+
We may
83
1).
usually omit the intermediate step and write the
factors at once. Ex.
3.
343 «6
Ex.4. 8x9
+
-
27
729
ic3
= (7
«2
_ 3 x) (49 a^ + 21 ah: + + 81).
=(2x3 + 9)(4r6_ 18x3
EXAMPLES Resolve into factors 1.
X. m.
9 x^).
.
ALGEBRA.
84 Ex.
Resolve into factors 28
3.
28 x^y
Ex.
+
64 xhj
-
60 yhj
Resolve into factors
4.
= {x-2 4 x2
-
25
4-
2/2
2 X
+
—
x^iil^
y)(:c^
+
2
(2 X
:=
= (2x
1.
—
x^
3.
6 x2
4.
729 2/6-64x6.
5. x5
-
^2_/y2_c2 + ^-2_2(af?_&c).
11.
14. 15.
16. 17.
18. 19.
20. 21.
22.
4
or?q'^
+
32 y^g^^
8 y^)
_(_
?/
?/)
X.
o.
:
4096.
13.
10.
—
77.
12.
8.
9.
—
y'^z^.
-X-
x?/ + + n^ — x^ - 16 X - 65. aH6*-c4-# + 2a2^y2_5 — ^^^ym^x + JH^?/ — (a -f 6 + c)"2 -(a-b - c)"^. 4 + 4 X + 2 a2/ + x2 - a2 _ x2 - 10 X - 119.
7.
yhj.
ai-y'2-2yz-z\
2.
6.
60
x?/ + 4 y2)(p + 2 ri)(p - 2 gr). — 25 y^ 2 x + 5 y. + 5 ?/) (2 X - 5 + 2 X + 5 + 5?/)(2x-52/+ 1).
EXAMPLES Resolve into two or more factors
A
8 y^p^
4 q^{x^
5 y
-
ofi
—
8
Resolve into factors 4 x^
5.
64 xMj
= 4 v^ii{l + 16 x - 15) = 4.x2?/(7x-5)(.x + 3).
?/3)
The expression = p^{y? —
Ex.
x% +
2 )H w + 2
33
?Ji-
x-^
'i^-^
?/2.
- a^ + - 2 xy. «2 + x^ - (2/2 4- ^2) _ 2 (2/2 - «x) 21 x2 + 82 X - 39. 1 - «2x2 ^ 2 a6x2/. «2. chl^ - r2 - a-ic%Z3 a^x^ - ahf - b'^x^ + ?)2y6. x2 - 6 X - 247. «^x2 - c3x2 _ aY + c'^y-acx2 - bcx + adx - bd. x2
2/'-^
//2_;y2
_|.
«%3 _ 35 ^:5,.2 _^ 14 ^ix. _ 1, - (m'^ + «-) 4 2 7hm.
32.
14
33.
^,8
34.
1
35.
75x4-48rt^.
37.
8
x^?/ 4-
38.
3
x22/2 4-
39.
720 a~b
40. 41. 42. 43. 44.
45.
+
26 axy
-
5«4/>4-5«&.
36.
52 xy
60
+
2/.
35
«2.
nb'.
- 64 rt2y6. _ ^12. - 36 y*. 24 x'^y"^ - 30 (a + &)* - 1. «* _ (/> + <-)4. (c + d)^ - 1. a^x^>
^12
.r2/3
RESOLUTION INTO FACTORS.
l-(x-
52.
(a
47. 250 (a
63.
a^
48.
64. «2
46.
yy.
- by + 2. (c + fZ)H (c - ciy. 8(x + yy-(2x- yy. — 4 +X—2 a2 _ + _
49. 50.
?/-
ic""
51.
56.
?/.
57.
f^
58.
Resolve
x^^
—
?/i6
53
+
«
_
«
+
+
&•
5.
4. 3 ^, 9 52 + - y)''^- (x - y). x^y — x^!/^ — x?iy + x?/*. 4 «2 _ 9 52 4 2 « - 3 6.
65. 4
jr>.
^,2
+ by + .|,
85
(^
(a;
into five factors.
CONVERSE USE OF FACTORS. 104. The actual processes of multiplication and division can often be partially or wholly avoided by a skilful use of factors.
It should be observed that the formulae which the student has seen exemplified in the preceding pages are just as useful in their converse as in their direct application. Thus the formula for resolving into factors the difference of two squares is equally useful as enabling us to write at once the product of the sum and the difference of two quantities.
Ex.
1.
+ 3 6 - c by 2 « - 3 may be arranged thus
Multiply 2
rt
These expressions 2 a
Hence
Ex.
2.
+
5X
3 x2
?>
+
c.
:
+ (3 - c) and 2 a - (3 b-c). = {2 a + (3 & - c)}{2 « - (3 b = (2 a)-2 - (3 & - c)2 = 4 «2 __ (9 52 _ 6 &c + c2) = 4 «2 _ 9 52 4 G &c - c2. ?>
the product
Divide the product of 2 x^
-
+
x
-
6,
c)}
and 6 x^
[Art. 98.]
-
5x
+
1
by
2.
Denoting the division by writing the divisor under the dividend between them, the required quotient
(Art. 53), with a horizontal line
+ x-6)(6x2-5x+l) 3 x2 + 5 X - 2 - 3)(x + 2)(3x - l)(2x (3x-l)(x + 2)
_ ~
(2x2
_
(2x
= (2x-3)(2x-l).
1)
:
ALGEBRA.
86 Ex.
Prove the identity
3.
+
17(5 a;
3 ay^
-
2(40 x
Since each term of the the
first
+
first
27 a) (5 x
+
3 a)
=
25
x"^
-
9
cfi.
expression contains the factor 5 x
+
3 «,
side
= (5 X + 3 «){17(5 + 3 «) - 2 (40 x + 27 a)} = (5x4-3 a) (85 X + 51 a - 80 X - 54 a) = (5 X + 3 «) (5 X - 3 a) = 25x2-9a2. :k
/
Ex. 4. Show that (2 x + 3 ?/ - ^)3 + (3 x 5(x + 2?/). The given expression is of the form A^ divisor of the form A-\- B.
+ (2 x +
Therefore is
divisible
that
is,
(2
by
x
3
«/
3
?/
by
or by
1.
2x-
2.
3 x2
3. 4. 5.
6. 7.
8.
9.
10. 11.
13. 14.
+
B^, and therefore has a
y
+
^)3 is divisible
//
X.
p.
of
+ 3^ and 2x + 7?/- 3^;. + 7 and 3 x^ + 4 xy + 7 y^. and 6x^ — 5xy — 9 5 x2 + 5 xy — 9 and 7 x2 + 8 xy - 3 y2. 7 x2 - 8 xy + 3 x3 + 2 x2?/ + 2 x^2 + ^3 and x^ - 2 xMj + 2 x?/2 + 4. (X + ?/)2 + 2(x + ?/) + 4 and (x + ?/)2 - 2(x + Multiply the square of a + 3 6 by a^ _ 6 «& + 9 b^. Divide(4x + 3y-2s;)2_(3x-22/ + 3^!)2byx + 5?/-5^. Divide x« + 16 «%* + 256 «» by x2 + 2 ax + 4 Divide (x2 + 7 x + 10) (x + 3) by x2 + 5 x + 6. Divide (3x + 4?/ - 2;2)2 -(2x + 3?/ - 45r)2 by x-{-y + 2z. 7?/
—
4
x?/
1/2
y'^.
?/2
?/2
?/3.
?/)
a"-.
Prove the following 12.
7
- s)^ + (3 x + 7 + zy - ^) + (3 x + 7 ^ + .s), 5 x + 10 y, 5(x + 2y).
EXAMPLES Find the product
+
identities
+ by - (« - &)2(« -\-b) = ab{a + ft). # - (c - d)3(c + d) = 2 C(^(c2 - tZ2). c4 (X + yy-Z xy(x + ?/)2 = (x + y) (x^ + y^). (a
4:
by
:
;
RESOLUTION INTO FACTORS. Show
15.
Show
16.
+
+ 1 exactly divides + x2 + 4)3 -(x^ -2X + 3)3. - 7 x + 2) ^ _ 8 + 8) Ms di visible
that the square of x (a;3
and by X
87
that (o
:k-
.x
(.r;-^
by
2 x
- 3,
2.
If any
105. The Factor Theorem.
i-atioyial
expression containing x hecoynes equal to
K^lien
and integral a is ivritten
X, it is exactly divisible by x — a. Let stand for the expression. Divide P by x — a until Let li denote this the remainder no longer contains x.
for
P
remainder, and
Q
Then
the quotient obtained.
P= q(x-a) + R. Since this equation is true for all values of x,
we
will
assume that x equals a. By hypothesis, the substitution of thus, a for X makes P equal to ;
.-.
As by X
the remainder
i?
is 0,
= 0. the expression
is
exactly divisible
— a.
The following examples
illustrate the application of this
principle Ex.
By X
—
3
1.
Resolve into factors x^ + 3aj"^ — 13 x — 15. we find that this expression becomes when x = 3 a factor. Dividing by x — 3, we obtain the quotient
trial is
x2
The
+
6X
+
;
hence,
5.
factors of this expression are easily seen to be x
+
1
and x
+
5
hence, ^3
Ex.
2.
+ 3x2- 13x-15^(x-3)(x+
Resolve into factors x3
+
6x2
+
It is evident that substituting a positive
11 x
l)(x
+
+
5).
6.
number
for
x
will not
make
By substituting — 1, however, for x, the hence, expression becomes — 1 + 6 - 11 + 6, or x3+ 6x2+ llx + 6 is divisible by x+1. Dividing by x+1, we obtain the quotient x2 + 5 X + 6, and factoring this expression we have
the expression equal to
0.
;
x3_|-(5x2
+ llx + 6=(x+
l)(x
+
2)(x
+
3).
::
ALGEBRA.
88
Note. The student should notice that the only numerical values that need be substituted for x are the factors of the last term of the expression, and that we change the sign of the factor substituted before connecting divisor
Ex.
by X
—
a;
it
and
in
with x. Thus, in Ex. 1, the factor 3 gives a Ex. 2, the factor — 1 gives a divisor x -\- \.
Without actual
3.
-
3,
division,
show that 5x^—6 a^^+i
is
divisible
1.
expression is divisible by x — \, it will become 0, or "vanish," Making this substitution, we obtain substituted for x. hence the division is possible. If the
when
1 is
;
EXAMPLES Without actual
X.
show that x
division,
—
r.
2
is
a factor of each of the
following expressions 1.
x^-bx +
2.
x^-\-x'^-'^x-A.
Determine by
3.
2.
4.
inspection whether x
+
x3- 7^2 + 16a;x3- 8x2+ 17a;3
is
12. 10.
any
a factor of
of the
following expressions
- 7 X + 6. + 6x2 + llx +
5.
x3
6.
a;3
9.
Show
that 32xio
Resolve into factors
2x3
10.
6.
-33x^ +
1
7.
x3
8.
x3
is
+ +
6X
+ 6. + x + 3. by x - 1.
3x2
divisible
.
+ 4x2- 2x-
4.
11.
+
3^^3_6^2 _3a;
We
shall employ the Factor Theorem in 106. general proofs of the statements made in Art. 62.
We
suppose n
(I.)
By
a;"
—
a positive integer. always divisible by x —
Art. 105, x"—y'^ is exactly divisible
stitution of y for X in the expression
Making
a result.
giving
to he
is
?/"
6.
this substitution rf.n
x"^
y.
x—y
by
—
y''
if
the sub-
gives zero as
we have
_yn_ yn _ yn ^ Q
always divisible by x — y. by x + y ichen n is even. If this be true, the substitution of — ?/ for x in the ex-
Therefore x" (II.)
X"
—
—
?/"
y""
—
is
is divisible
pression
x""
stitution
we have
?/"
gives zero as a result.
a;"
-2/"
= (-?/)"-?/".
Making
this sub-
:
.
RESOLUTION INTO FACTORS.
When n (III.)
— +
x""
.«'•
is divisible
?/"
?/"
— 2/", or zero. becomes hj x -\- y when n is even.
this expression
is even,
Therefore
is
?/"*
never divisible by x
Here the substitution of y
89
for
— y.
in the expression
a?
a;"-}-?/"
gives ^'"
As
+
?/"
this expression is
= r 4-r = 2?/'. not zero, +
y'^
x""
is neve?' divisible
by x-y. (IV.) + is divisible by a? Here the substitution of — y a^''
^,n _|_ yri
when n
?/
for
a;
= (-2/)" +
2/"
is odd.
the expression
in
gives ^"
When
?i
+
2/"
+
is
?/**
divisible
—
2/"
+
2/"?
hjx-\-y when n
is
odd.
odd, this expression
is
Therefore x^
The
+
T/'*
becomes
may
results of the present article
oi'
zero.
be conveniently
stated as follows (i.) X"-
(ii.)
For
-
all positive integral
2/"
When
^.n _j_
yn
values of «,
= {x — y) (x^~^ + a;"-'2/ + x''-'^y- H n
is
^ ^^
h 2/""^
odd,
_|_ ^^)
(^^n-1
_
r,fi-2
y
_|_
a.n-3^2
(_
yu-ly
(iii.)
When
107.
We shall now discuss some cases of greater difficulty,
and
n
is
even,
show how certain expressions of frequent occurwhich are not integral, may be separated into factors.
also
rence,
The student may omit this portion of the chapter until reading the subject a second time.
—
a^
= (a6_64)(a3_i) = («3 + 8)(a3-8)(«3_ 1) _ 2 a + 4) (a = (a + 2)
2)
Ex.
1.
Resolve
cfi
—
Q^a^
-\-
64 into six factors.
The expression
= «3 (^6 _ 64) - (^6 _
(a--2
64)
{cfi
+
2 a
+
4) (a
-
1) (a2
+
«
+
1)
ALGEBRA.
90 Ex.
b'^
- h (h + 1) {a- 1) y^-{n-h-\) = {ax-(b-^l)y}{(n-l)x + hi,]. ocij
From Ex.
108. a^ 4-
a
2.
2,
+ c^ — 3 abc
Thus
Art. 59,
we
?/2
see that the quotient of
+ 6 + c is a^ + U' + c^'-Sabc = (a + ^ + c)(o2 ^2 ^2 _ _ by «
-\-
b'-
-\- c'
— be — ca — a6.
aJ'
_|.
_|_
^^^
^^^
_
^^^^^
_
_
(ly
This result is important and should be carefully remembered. A¥e may note that the expression on the left consists of the sum of the cubes of three quantities a, b, c, diminished by three times the product abc. Whenever an expression admits of a similar arrangement, the above formula will enable us to resolve it into factors. Ex.
1.
^3
= (a —
b taking the place of b in
Ex.
— 6^ + c^ + 3 abc. (— &)3 + c3 - 3 a (- &) c — b + c) (a- + + ^ i>c -
Resolve into factors a^ 53 + c3 + 3 abc = «3 +
_
b'^
formula
c'-^
ca
-\-
«6),
(1).
2.
x^-Sy^-21
_18x?/
= x3 + (-2?/)3+ (_ 3)^ - 3:k (= (x - 2y -S)(x^ + 4:y^ 9 - 6 y -{-
2 ?/)(- 3) Sx 2:ry).
-\-
109. Expressions which can be put into the form
+
x"
±— ?f
may sum
be separated into factors by the rules for resolving the [Art. 102.] or the difference of two cubes.
Ex.1.
l_27&« = f^-Y-
(362)^
—
S_a cc^
H
into four factors.
If a-2^3
= X^((fi-l)-(rt2 _^_x^-{-^ yS yZ yS
1)
= {a'^-l)lxi= (, + ,)(„_,)(.._2)^..+ ^^^4)
•
•
RESOLUTION INTO FACTORS.
EXAMPLES Resolve into two or more factors
X.
91
s.
:
— 3 x^ — i/. 5. n^ -\-{a + h)ax + hx^. — 20 7i^ + 45 nm- — 9m^. 6. jm{'ni- + 1) — w(p- + m-). G hx(a^ + 1) - a(4 x^ + 9 &2). 7. «6(x2 + 1) + a;(«2 + ?/2). i/^z\xi - 1) + ^'^(!/* - ^^^) 8. (2 a2 + 3 2/2)x + (2 x^ + 3 a2)?/. 9. (2 x2 - 3 a^)>j + (2 «-' - 3 y^)x. 10. a(a - l)x- + (2 a- - l)x + a(a 4- 1). 11. 3x2-(4rt + 2?>)x + «-+ 2a&. - 4 c) (/^ - c) + ahxy. 12. 2 a'-'x2 _ 2(3 13. («2 _ 3 « + 2)x2 + (2 «2 _ 4 a + l)x + a(a - 1). x-ij
1.
4
2. 3.
4.
+
3 x}/
mn-
^/^
/>
+ l)x2 + (a + + 3 be. a^ + 8 c3 + I - 6 ac. _ 6 abc. 4. 8 a'^ + «(«
14.
63
15. 16. 17.
+
-
c3
^.3
?>3
21.-^-1.
—+
23.
125
a3&3
216
22.
b)xi/
1
a3_^.
-b(b_
18.
a^
19.
«3
20.
8 a^
1)^'-
27
+
27
+ c^ Z^^
c^ + 9 a6c. _ 3 «?>c. + c^ - 18 a5c.
~+
25.
^3. -"
—-—
26.
729
8
1000.
125
Z^_i.
24.
b'^
_
?^3
512
x3
+ 81 x^ + 6561 into three factors. (rt^ _ 2 a'^b'^ - b^y - 4 a'^b^ into four factors. — (V'Y into four factors. 4(ff?> + cd)'^ — (a^ + b- —
27.
Resolve x^
28.
Resolve
29.
Resolve
30.
Resolve x^
(fi
into four factors.
256
Resolve
31.
x^^
—
y'^^
into six factors.
Resolve into four factors 32.
^'-8x-«3 +
:
36.
8x3.
34.
+ xhf - 8 x^i/ ^ -Sy^. ^ ^9 + + 64 x3 + 64.
35.
4«-964-i|-^^
33.
x^
-^
32
37.
a;6-25x2
39.
16 x^
9 x2
+
:kG
Resolve into 38.
^|/i_^_J_ + ^. 72
x2
x^
+
x4
-
five factors
16 x3
-
16.
4 1
^4
6i:-ix4.
:
-
81 x^
-
16 x*
+
81.
;
CHAPTER XL Highest Common Factor. 110. Definition. The Highest Common Factor of two more algebraic expressions is the expression of highest dimensions (Art. 29) which divides each of them without remainder. or
The abbreviation H. C. F. is sometimes used instead of the words highest commoyi factor.
SIMPLE EXPRESSIONS. The H.
111. Ex. Ex.
C. F.
can be written by inspection.
1.
The
highest
common
factor of «*, «3, a^^ a^
2.
The
highest
common
factor of a%^^
a-Wc
ab^c'^,
the highest power of a that will divide a-^, «, a^ power of h that will divide &*, h^^ h"' and c is not a
a
is
;
is a^.
;
is
«&*
;
for
h^ is the highest
common
factor.
112. If the expressions have numerical coefficients, find by Arithmetic their greatest common measure, and prefix it
as a coefficient to the algebraic highest
Ex. The highest 7 a^xy
;
for
it
common
factor of
common
21 rt%3^,
factor.
35«%'V, 28«3r?/4
is
consists of the product of
(1) the numerical greatest common measure of the coefficients (2) the highest power of each letter which divides every one of the
given expressions.
EXAMPLES Find the highest common factor
XI.
a.
of
1.
4a62, 2cfih.
3.
6xy%,
2.
3x2^2^ xhf-.
4.
ahc, ^aV^c.
^x'^y^zK
92
5.
ba%^, Voahd^.
6.
^x-y'h'^,
12 xy^z.
HIGHEST COMMON FACTOR.
93
7.
4a253c2^ 6a%''G^.
12.
26xfz, lOOx^yz, 126
8.
7 a-b'^c^,
liab'^c^.
13.
a^bpxy, b-qxy^ a%xr^.
9.
QOaWc^
49 «x'^ eSmf, SGaz"^.
14.
16a^b^c\
10.
liable, 34a%c, 51 «&c2.
15.
SSaVft, 42
11.
a-%-V,
16.
24a352c3^
&%^^
c%2?/,
a^c^a,
xy.
26
a^b^c'^.
30 «c%3.
Wa%^d^, iOa'b^c^
COMPOUND EXPRESSIONS. 113. H. tored
by
Compound ExpresHHns which can be facThe method employed is similar to
C. F. of
Inspection.
that of the i^receding article. Ex.
1.
Find
highest
tlie
common
factor of
4 cx^ and 2 cx^
4 c%2.
common
be easy to pick out the
It will
are arranged as follows
+
factors
if
the expressions
:
= 4 cx^, + 4 c2x2 = 2 cx^{x + 4 cx3
2 cx^ therefore the H, C. F.
Ex.
is
Find the highest common factor
2.
3 a2
+
2c);
2 cx^.
9 ab, a^
-
9
ab'',
Resolving each expression into
(fi
of
+
d^b
+
9
ah'^.
we have
its factors,
+ 9a6 = 3rt(« + 36), _ 9 ab^ := a(a + 3 6) (a «3 + 6 a2?> + 9 ab'- = «(« + 3 6) (« + H. C. F. is «(« + 3 &). 3a2 «3
therefore the
114.
3
?>),
36);
When
different
there are two or more expressions containing powers of the same coinpouud factor, the student
should be careful to notice that the highest common factor must contain the highest power of the compound factor which is common to all the given expressions. Ex.
1.
The highest common x(a
Ex.
2.
—
x)'^,
a{a
—
x)^,
Find the highest ax2
+
2 a2x
+
a^,
factor of
and 2 ax{a -
common
2
ax'^
-
xy
is
{a
—
xy.
factor of
4 a2x
-
6
a^,
3 (ax
+
a'^y.
ALGEBRA.
94
Resolving the expressions into factors,
2 ax-
we have
= a{x- -\-1ax-\- or) = '^•'• + '0— 4 a-x — G a^ = 2 a {x- — 2 ax — S a-) = 2a(x + a)(x-Sa) S (ax + a^y = S a^(x ay
ax~
+
2 a-x
+
a^
(1),
.
.
.
(2), (3).
-j-
Therefore from (1), (2), (3), by inspection, the highest is a(x + «)•
common
factor
EXAMPLES Find the highest common factor
10.
+ ab, a- - b\ (x + y)2, x2 - y\ 2 x2 — 2 a:y, a;^ — x~y. 6 x^ - 9 a-y, 4 x2 - 9 a;3 + x^ + a^ft - ab% a^b'^ - a-b^. a^ — a-a-, «3 — ax-, a^ — ax^. 2 aa;. a- - 4 x2, rt2 a- — a2 _ q^, a-x — ax^. 4 + 2 xy, 12 a.% - 3 y\
11.
20
1.
2. 3.
4. 5.
6. 7. 8.
9.
23. 24.
25. 26. 27.
28. 29. 30.
31.
of
a^
12.
?/2.
Gbx +
13.
x^
14.
a'?/
15.
x^
i by, 9 ex
+
6 ry.
+ x, (x + l)-2, r^ + 1. — y, x>y — xy. - 2 a:^ + ^^^ (a; _ y^,
16.
i/^-
a;2y/,
XI. b.
_|.
a;2,
a;2
a;
-4, 50x2-2.
x^ — xy'\ x^ + x'^y + xy + a^-g - a%x - 6 a&2^^ ^•2^)^2 _ 4 «;j2a32 + 3 ?^3^2. 2 x2 + 9 X + 4, 2 x2 + 11 X + 5, 2 x2 - 3 x - 2. 3 X* + 8 x3 + 4 x2, 3 x5 + 11 + 6 x^, 3 x* - 16 x^ - 12 x2. - 7 x^ - 15 x2. 2 x^ + 5 x3 + 3 x2, 6 x4 + 13 x3 + 6 x2, 2 12 X + 6 x2 + 6, 6 X + 3 x2 + 3, 18 x + 3 x2 + 15. x4 + 4 x2 + 3, X* + 5 x2 + 6, 3 x* + 11 x2 + 6. 2 a2 + 7 «fZ + 6 cP, 2 «« + 9 «^/ + 9 a\ a2 + 11 ad + 3 (P. 2 2 x2 4- 8 xy + 4 14 C x2 + 10 xy + 12 x^/ + + y\ y'-.
a;4
x-^
./,-2
115. H. C. F. factored
by
of
?/2.
Compound Expressions which cannot be
Inspection.
in such cases,
2/2,
we adopt
To find the highest method analogous
a
Arithmetic for finding the greatest or more numbers.
coiuinoii factor
to that used in
common measure
of
two
HIGHEST COMMON FACTOR. The term
Note.
instead of highest greatest ties
;
common measure
greatest
common
factor
95 sometimes used the term
is
but, strictly speaking,
;
common measure ought to be confined to arithmetical quanticommon factor is not necessarily the greatest
for the highest
common measure
in all cases, as will appear later.
(Art. 121.)
116. AVe begin by working out examples illustrative of the algebraic process of finding tlie highest common factor, postponing for the present the complete proof of the rules
But we may conveniently enunciate two principles, use. which the student should bear in mind in reading the examples which follow.
we
I. If an expression ccmtains a certain factor, any multiple of the expression is divisible by that factor. II. If two expressions have a common factor, it will divide
their
sum and
difference of
Ex.
their difference
;
and
also the
sum and
the
any multiples of them.
Find the highest common factor of 3 x2 - 24 X - 3 x2 - 24 X 4x3-ox2-21x
4 x3
-
9 and 8
4 x3
2x
2x2 2x2
yJ^
8.'C3
3x 6x
2 x2
is
x
G x2
48 X
4x2 4x2
-
53 X 53 X
3x-9 3x-9 Therefore the H. C. F.
-
2x2
-
39.
—
21
6xX-
18
5X
3
3.
Explanation. First arrange the given expressions according to descending or ascending powers of x. The expressions so arranged having their first terms of the same order, we take for divisor that whose highest power has the smaller coefficient. Arrange the work in parallel columns as above. When the first remainder 4x2— 5 x— 21 is made the divisor we put the quotient x to the left of the dividend. Again, when the second remainder 2 x2 — 3 x — 9 is in" turn made the divisor, the quotient 2 is placed to the right ; and so on. As in Arithmetic, the last divisor x — 3 is the highest common factor required.
117. This method is only useful to determine the comfactor of the highest common factor. Simple factors
pound
of the given expressions
must be
first
removed from them,
ALGEBRA.
96
and the highest common factor of tliese, if an}^, mnst be observed and multiplied into the compound factor given by the rule. Ex. Find the highest 24 x4
We have and
-
2 x^
24 x^
18x4
common
factor of
- 60 x2 - 32 X and 18 x^ - 6 x^ - 39 x2 - 18 x. - 2 x'^ - GO x2 - 32 x = 2 x(12 x^ - x2 - 30 x - 16), -Qx^ - 39x2 - 18x = 3x(6x^ - 2x2 - 13x - 6).
Also 2 X and 3 x have the common factor x. Removing the simple factors 2 x and 3 x, and reserving their common factor x, we continue as in Art. 116.
2x 6x3-2x2-13x-6 6a:3-8.r2_ 8x
HIGHEST COMMON FACTOR.
97
have no simple factors, therefore their H. C. F. can have therefore reject the factor 9 and go on with divisor Resuming the work, we have 10 X + 7.
pressioiis
We may
none. 3x'^
—
X 3x3 3x3
13 x2
10 x2
+ +
23 X 7
21
X
3x2+16x-21 3 + 10 X - 7
3 x2 - 10 X + 3x2- 7x 3x +
x'2
X3x-
2 )6
3x+
7
X
7
7
14 7
Therefore the highest common factor is 3 x — 7. The factor 2 has been removed on the same grounds as the factor 9 above.
Ex.
2.
Find the highest
common
factor of
2x3+ X2-X-2 3x3-2x2 + .'C-2
and
(1), (2).
As the expressions stand we cannot begin to divide one by the other without using a fractional quotient. The difficulty may be obviated by introducing a suitable factor, just as in the last case we found it useful to remove a factor when we could no longer proceed with the The given expressions have no common division in the ordinary way. simple factor, hence their H. C. F. cannot be affected if we multiply either of them by any simple factor. Multiply (2) by 2, and use (1) as a divisor: 2 x3
+
7
2x 14x3+ 7x214 x3 17 X
14
-
10 x2
-
ALGEBRA.
98
Note. If, in Ex. 2, the expressions had been arranged in ascending powers of x, it would have been found unnecessary to introduce a numerical factor in the course of the work.
119. The use of the Factor Theorem (Art. 105) often marked degree, the work of finding the highest common factor. Thus in Ex. 2 of the preceding article it is easily seen that both expressions become equal when 1 is substituted for x, hence x — 1 is a factor. to Dividing the first of the given expressions by x — 1, we It is evident that this obtain a quotient 2 x^ -j-Sx -{- 2. will not divide the second expression, hence a; — 1 is the lessens, in a very
H.
C. F.
120.
When
efficients
the following Ex.
is
We of the
+
by Detached CoH. C. F.,
Division
of
63) is employed in finding the a convenient arrangement.
Find the H. a:4
Method
the
(Art.
C. F. of
3 x3
+
12 X
-
16 and x^ -lSx-\- 12.
write the literal factors of the dividend until
same degree as the
first
x^ x^
+
term of the
+ -
13
13
+
12
-
36
+ J9 -
52
39
12
X -\-S;
The addition
a term
0+12-16
Sx^-{-
+
we reach
divisor.
13
of the terms in the third
column gives 13
which is we can proceed no further with the division and have for a remainder Removing from this remainder the factor 13, as 13 x'^ + 39 X — 52. it is not a factor of the given expressions, we have for a second diviof lower degree than the first
term
of
x'^,
the divisor, hence
The first divisor, as written before the signs loere x'-^ + 3 x — 4. changed, forms the second dividend
sor
:
X-'
x3
+
0x'-^- 13
X
-
3
-3 +
+
12
+9-12
4 ;
no remainder, the last divisor, as written before the changed^ is the H. C. F. Thus x^ + 3 x — 4 is the H. C. F..
since there ivere
-3+4
is
signs,
;
:
HIGHEST COMMON FACTOR. 121. Let the two expressions in Ex. written in the form
Then
their
If,
118, be
a;
a;-
2 x^ -[-^x-{-2 and 3 divisor.
Art.
+ - - 2 =(.^' - l)(2a^2 + 3.^ + 2), - 2.r -\-x-2 ={x - 1)(3.T- + x + 2). highest common factor is x — 1, and therefore
2x' ^x"
2,
99
x^
however,
-\-
x
have no algebraic
-{-2
common
we pnt x=&, then
+ a;--.T-2 = 460, - 2 ^ - 2 = 580
2.«3
and
3
iv'^
i«2
a.-
and the greatest common measure of 460 and 580 is 20; whereas 5 is the numerical vahie x — 1, the algebraic highest
common factor. Thus the numerical values of the algebraic common factor and of the arithmetical greatest common measnre do not in this case agree. The reason may be explained as follows when x = 6, the
highest
:
and 3 x^ -{- x + 2 become equal to 92 and 116 respectively, and have a common arithmetical factor 4; whereas the expressions have no algebraic comexpressions 2
mon
ic^
+3 +2 ic
factor.
happen that the highest common factor two expressions and their numerical greatest common measure, when the letters have particular values, are not the same for this reason the term greatest common measIt will thus often
of
;
nre
is
inappropriate
when applied
EXAMPLES Find the highest common factor 1.
2. 3.
4. 5.
6. 7.
8.
to algebraic quantities.
XI.
c.
of the following expressions
+ 2x2-13ic+10, x3 + ic2- lOx + 8. x^ - 5a:2 _ 99x + 40, x^ - Gx^ - 86 x + 35. x3 + 2 x2 - 8 X - 16, x'^ + 8 x-^ - 8 X - 24. x3 - x^ - 5x - 3, x3 - 4 x2 - 11 X - 6. x3 + 3x'^ - 8x - 24, x3 + 3 x2 - 3x - 9. «3 _ 3 «;;c2 + 2 x3. a^ _ 5 cfir^ + 7 <^a;2 _ 3 2x3-5x2+ llx + 7, 4x3- 11x2 + 25x+ 7. 2 x3 + 4 x2 - 7 X - 14, 6 x3 - 10 x2 - 21 X + 35. x3
r^Z^
ALGEBRA.
100
12.
- 3 x3 - 2 x2 - X - 1, 9 x4 - 3 x^ - x - 1. 3 ax2 + 2 «% - 2 a^, 3 x^ + 12 «x'2 + 2 3 x^ + 8 a^. 9 ax2 + 9 a% - 7 «^ 4 x^ - 20 ax^ + 20 a^x - 16 a^. 2 x3 10 x^ + 25 ax2 - 5 a3, 4 x^ + 9 ax^ _ 2 a% - a^.
13.
24x4y4-72x3?/2-6x2?/3-90x!/4, 6xV+13a;5|/3-4x2?/4-15x?/5.
14.
4x5a2+i0xia3_60x%4+54x2a5, 24x5aH30 x%5_i26x2a6.
15.
4x5
16.
72x3-12ax2+72«2x-420a3, 18xH42ax2 - 282
17.
x5
x'^
18.
1
1
3 x*
9.
10.
«2.r,
11.
+
14x*
+ 20x3 +
- x3 - X + 1, + X + x3 — x^,
70x2, Sx^
+
28x6
-
Sx^
-
12x4
+
56x3.
a2;;c+270a3.
+ x^ + x* - 1. - x* - x^ + x^.
122. The statements of Art. 116 lows
may
be proved as
fol-
:
I.
If
F
Thus II.
i^ is a factor of
If
F
F
mA.
A and B, then it will divide mA ± A = aF, B = bF, mA ± nB = 7naF ± nhF = F(ma ± nb).
divides
For suppose then
Thus
A it will also divide mA. A = ciF, then mA = maF.
divides
For suppose
divides
7nA
±
nB.
nB.
We
may now enunciate and prove the rule for find123. ing the highest common factor of any two compound algebraic expressions.
We
suppose that any simple factors are
first
removed.
(See Example, Art. 117.) be the two expressions after the and Let
A
B
simple
Let them be arranged in descending or ascending powers of some common letter also let the highest power of that letter in B be not less than the highest power in A. Divide B hj A-, let j) be the quotient, and C the remainSuppose C to have a simple factor in. Eemove this der. Further, suppose factor, and so obtain a new divisor D. that in order to make A divisible by D it is necessary to factors have been removed.
;
:
:
HIGHEST COMMON FACTOR.
101
A
multiply by a simple factor n. Let q be the next quotient the remainder. Finally, divide and hj E; let r be the quotient, and suppose that there is no remainder. Then will be the H. C. F. required. The work will stand thus
E
D
E
pA m)C D)nA(q
qP E)D{r
vE
E is a common factor
show that
First, to
By examining
the steps of the work,
divides D, therefore also 7iA
;
therefore A, since
E
Again
E
divides
E
Hence
7i
qD is
;
of
qD
therefore
A and
is
it
-f-
and
C,
it
a simple factor. mD, that is, C.
p^
also divides
E
E, therefore
divides D, therefore
A
B.
clear that
And
since
-f C, that is,
B.
A and B. that E is the highest common factor.
divides both
Secondly, to show not, let there be a factor
If
X
of higher dimensions
than E.
X divides A
Then
therefore that
is,
Thus esis,
X
B — pA,
and B, therefore
D (since m is
a simjyle factor)
;
that
is,
C
;
— qD,
therefore 7iA
E.
X divides E is
-,
which
is
impossible, since by hypoth-
of higher dimensions than E.
Therefore
E is the
highest
common
factor.
The highest common
factor of three expressions be obtained as follows First determine the highest common factor of and B next find G the highest common factor of and C then G will be the required highest common factor of A, B, 0. For contains every factor which is common to and B, and G is the highest common factor of and C. Therefore G is the highest common factor of A, B, C.
124.
A, B,
C may
F
A
F
F
A
F
;
;
;
CHAPTER
XII.
Lowest Common Multiple. 125. Definitiox. The Lowest Common Multiple of two more algebraic expressions is the expression of lowest dimensions which is divisible by each of them without remainder. The abbreviation L. C. M. is sometimes used instead of the or
words lowest common
multiple.
SIMPLE EXPRESSIONS. 126. The L. C. M. can be written by inspection. Ex.
The lowest common multiple
1.
of
a^
«^,
(fi,
««
is a^.
2. The lowest common multiple of a"5'^, ah'', a^b' is a^b'^ for «3 is the lowest power of a that is divisible by each of the quantities and b"^ is the lowest power of b that is divisible by each of a^, rr, «2
Ex.
;
.
the quantities
&*, b^,
h"'.
127. If the expressions have numerical coefficients, find
by Arithmetic their
least
common
a coefficient to the algebraic low^est
multiple,
common
Ex. The lowest common multiple of 21 420 a^x'^y^ for it consists of the product of
and prefix
35
a^xhj.,
it
as
multiple. cfix^^y,
28 aHij^
is
;
(1) the numerical least (2) the lowest
common
power of each
multiple of the coefficients
letter
which
is
divisible
by every power
of that letter occurring in the given expressions.
EXAMPLES Pind the lowest common multiple
XII.
a.
of
1.
x^y-, xijz.
4.
V2ab,Sxy.
7.
2x,Sy,4z.
2.
Sxhjz,4x^y\
5.
ar, be, ab.
8.
3x^
3.
5 aV>c% 4 ab-c.
6.
a'^c,
9.
7 a-, 2 ab, 3
bc% 102
cb^.
4 y2,
3z^
R
LOWEST COMMON MULTIPLE.
103
10.
a%c, b^ca, c^ab.
13.
S5 a'^c^ 42 aHb^, SO ac^b^
11.
5 «%, 6 C&2, 3 &c2.
14.
66 a*&%^ 44 a354c2^ 24
12.
2
15.
7
3 xy,
x2|/3,
4xY'
a%
4
ac'^
6 ac^, 21
a'^b^c^
be.
COMPOUND EXPRESSIONS. 128. L. C. M. of Compound Expressions which can be factored
by
The method employed
Inspection.
is
similar to
that of the preceding article. Ex. 1. The lowest common multiple of Qx^(a — x)'^, 8<:i^(a — x)'^ and 12 ax{a — xy is 24 a^x:^(^a — x)^. For it consists of the product of (1) the numerical L. C. M. of the coefficients (2) the lowest power of each factor which is divisible by every power of that factor occurring in the giveii expressions. ;
Ex.
2.
Find the lowest common multiple 3 a^
+
9 ab, 2 a^
_
is
+
aS
«6-2,
of
e a^b
+
9
ab'^.
= 3a(« + 3 5), 2 a^ - 18 a&'2 = 2 rt(a + 3 (« - 3 6), + Qa-b^^ab'^ = a{a + Zb){a^Sb) = a(a + 3 6)2. Sa^
+
^:)ab
/>)
a^
Therefore the L. C. M.
is
6a(«
+
3
EXAMPLES Find the lowest common multiple 1.
5.
6. 7. 8.
9.
XII.
3 b).
to.
of
3. a- + ab, ab + — x)'^, cfi — 11. 6x2 -2x, 9x2 -3a;. (1 + x)3, 1 + r^. 12. x^ + 4 x + 4, x2 + 5 x + 6. x2 + 2 X, x2 + 3 X + 2. x2 - 3 X + 2, x2 - 1. 13. x2 - 5 X + 4, x2 - 6 x + 8. 14. 1 - x^ (1 - x)3. (« + x)3, cfi + x3. «2 + ^2^ ((^ + ^)2^ 15. (^ _ ^;)3^ ^3 _ 16. x2 + X - 20, x2 - 10 X + 24, x2 - x - 30. - 42, x2 - 11 X + 30, x2 + 2 X - 35. 17. x2 + 18. 2 x2 + 3 X + 1, 2 x2 + 5 X + 2, x2 + 3 X + 2. 19. «2 _ (« _ ^)2^ ^3 _ ^3.
x^, x^
4.
-
by\a -
3
4 x^y
X.
—
2.
y, 2 x2
+
x.
21
x^
7 x:\x
+
1).
10.
(a
fe^.
xP-.
a;3.
a;
rj;l^
ALGEBRA.
104
Ux + e,
3 + 8 X + 4, x- + 5 + 6. + b x + 6. + 2, 5 X- + 10 X + 3, 1 + x^. 1 + x^ (1 + 2 a:2 + 3 X - 2, 2 x- + 15 x - 8, x^ + 10 x + 16. - 4. 3 x2 - X - 14, 3 x2 - 13 X + 14, 12 x2 + 3 X - 42, 12 x^ + 30 x^ + 12 x, 32 x^ - 40 x - 28. 3 X* 4- 26 x3 + 35 x^, 6 + 38 x - 28, 27 x^ + 27 x^ - 30 x. 5 x2, 60 60 X* + 5 x3 + 32 xy + 4 40 x^^ - 2 x^y - 2 x?/. 4 x^ - x?/ - 5 2 x^ - 5 x?/ - 7 /-, 8 X? - 38 x?/ + 35 4 x2 - 9 x?/ + 5 y^, 3 x2 - 5 x?/ + 2 12 x2 - 23 xy + 10 3 rt-2j;2+ 14 o^x - 5 «^ 6 x2+39 «x+45 «2, 6 ax3+7 «2j;2_3 — 4 ax'V" + 11 axy- 3 ay-, 3 x^^=^ + 7 x-^=^ — 6 x^^, 24 ax- — 22 ax + 4 a.
Sx^
20.
5 x2
21.
22.
+ +
11
.r
:c2
:*;-
ic
a;)--2,
23.
24.
x"^
25. 26.
.x^
27.
x'^ij
28.
?/,
?/"',
29.
?/2,
?/2.
i/^,
30.
^^3^;,
31.
Compound Expressions which cannot be AYhen the given expressions are such that their factors cannot be determined by inspection, they must be resolved by finding the highest common factor. 129. L.
M.
C.
by
factored
of
Inspection.
Ex. Find the lowest 2 x*
+
x3
-
20 x2
-
The highest common
By
division,
2 x4 2
a:*
+ +
we
common 7
+
factor
multiple of
24 and 2 x^ is
x2
+
2 x
+ 3 x^ — 3.
-^
13 x2
-
7
x
+
15.
obtain
- 20 x2 - 7 3 x3 - 13 x? -
a;3
Therefore the L. C. M.
We may
130. the lowest
X
3)(^ x2 - 3 X - 8). 7 x + 15 = (x^ + 2 x 3) (2 x2 - x - 5). is (x2 + 2 x-3)(2 x- - 3 x-8)(2 x2-x-5).
X
+
24 =(x2
now give the common multiple
+
2 X
-
proof of the rule for finding of
two compound algebraic
expressions.
A
and
F
be the two expressions, and their highest Also suppose that a and h are the respective quotients when A and B are divided by F\ then = aF, B = hF. Therefore, since a and h have no comand B is mon factor, the lowest common multiple of ahF, by inspection.
Let
common
JB
factor.
A
A
;
LOWEST COMMON MULTIPLE.
105
an important relation between the highest and the lowest common multiple of two expressions which it is desirable to notice. the lowest Let F be the highest common factor, and common multiple of A and B. Then, as in the preceding 131. There
common
is
factor
X
article,
A = ciF, B = bF, X = abF. product AB = aF x hF = Fx abF = FX
and Therefore the
(1).
Hence the product of tico expressions is equal to the product of their highest common factor and lowest common multipAe. Again, from
X=
(1)
^=^
x
iJ
=|
x
J
hence the lowest common multiple of two expressions may be found by dividing their product by their highest common factor ; or by dividing either of them by their highest common
and multiplying
factor,
the quotient by the other.
The lowest common multiple
132.
C may
A, B,
of three expressions
be obtained as follows and B. Next find T, the First find X, the L. C. M. of will be the required L. C. M. L. C. M. of and C; then of A, B, For is the expression of lowest dimensions which is divisible by and C, and is the expression of lowest is the exdimensions divisible by and B. Therefore :
A
X
Y
a
Y
X
X
Y
A
pression of lowest dimensions divisible by
EXAMPLES Find the
1.
ple of 2.
-
a:-
5
.-«
6,
x^
Find the lowest ab(x~
3.
liighest
+ +
1)
+
commou
-
4,
x^
common x(«-
-
XII.
factor
3
a;
-
— Sby + 2
b'^,
xy
c.
and the lowest common multi2.
multiple of
-K b')
-
1)
+
of xy
—
hx,
and ab^x-
Find the lowest common multiple y'^
all three.
— 2bx —
ay
+
2 ab, xy
—
- b-). — ay, — ay + ab.
x{a-
xy bx
ALGEBRA.
106 4.
ple of 5.
Find the highest common factor and the lowest common multi+ 2 x-^ - 3 a:, 2x^ + bx^ -^x.
a:^
Find the lowest common multiple of
\-x, 6.
-
(1
(1
a:-2)2,
+
xy.
Find the lowest common multiple of x^
-
+
lOx
24,
- 8x +
a;'2
x^-Qx-\-
12,
8.
Find the highest common factor and the lowest common multiple of 6x3 + x2 - 5x - 2, 6x3 + 5x2 - 3x - 2, 7.
8.
(&c2 9.
- aUy,
multiple of
h-\ac^-
-
a^c^
a-3),
+
2ac^-\- c\
Find the lowest common multiple of X-'
-
x3|/
2/3,
-
y\x -
y-i,
common
Also find the highest 10.
common
Find the lowest
+
y)-, x^
factor of the
first
x?/
+
?/2.
three expressions.
Find the highest common factor of
2x2+ 5x- 12, 6x2 -x- 12. common multiple is the product of the three quantities divided by the square of the highest common factor. 11. Find the lowest common multiple of x^ + ax^ + a^x + «4^ ^.4 ^ (-^2^2 ^
+
6x-^- 13x
6,
Also show that the lowest
(-^4_
12.
Find the highest common factor and the lowest
ple of 3 x^
—
7 x-y
+
5
X2/2
—
3 x^ 13.
+ 3 xy- — 3 x^ — 5 x'^y + xy- — y^. x^y
y^^ -f
Find the highest common factor
4x3-
10x2
+ 4x +
14.
Find the lowest common multiple of
15.
Find, the highest
a2
ple of (2 16.
ple of 17.
x2-
3 a2)y
_
rt3_ ^3^ ^3
/>2,
common
+ (2 a2 _
3
(^j2/,
factor
^2)^.^
2.
_al2_2 &3.
and the lowest common multi-
(2 a^
+
3 y^)x
+ (2 x2 +
3 a^)y.
Find the highest common factor and the lowest common multix3 - 9x2 + 26x - 24, x^ - 12 x2 + 47 x - 60. Find the highest common factor of x3
18.
_
multi-
of
3x4-2x3-3x +
2,
common
y^,
—
15 «x2
-f
48 «2x
+
64
a^,
x2
—
10 ax
+
16 a-.
+
xyy\
Find the lowest common multiple of 21 x(x^
-
^2)2^
35 (xV
_ xY)y
15
«/(x2
;
CHAPTER
XIII.
Fractions. 133. Definition. If a quantity x be divided into h equal parts, and a of these parts be taken, the result is called the fraction - of
x.
If X be the unit, the fraction
fraction - "
;
- of x
b so that the fraction
-
is
called simply " the
represents a equal jxirts,
h
h
make up the unit. The quantity above the
b ofivhich
horizontal line is spoken of as the numerator, and that below the line as the denominator of the fraction.
134.
A
Simple Fraction
one of which the numerator
is
and denominator are whole numbers.
REDUCTION OF FRACTIONS. 135. To reduce a fraction changing its value.
136. To prove that integers.
By - we mean a
is
= ^, mo
to
change
where
d^ b^
^ we '''^
mean ma equal
form without
m
equal parts, b of which
unit,
by
its
are positive
make up
the (1)
parts,
the unit,
mb
of
which make up (2).
107
;
ALGEBRA.
108 But
that
= mb parts in (2) = m parts in (2) (1) = ma parts in (2) a _ ma
b parts in (1) .-.
1 part
.'.
a parts in
in (1)
is,
;
j
mb
b
ma _ a
Conversely,
mb
b
Hence, The value of a fraction is not altered if ive multiply or divide the numerator and denominator by the same quantity. 137.
shown
As
a Fraction to its Lowest Terms.
Reduction of
in the preceding article an algebraic fraction
may
be changed into an equivalent fraction by dividing numerator
and denominator by any common factor
;
Ex.
1.
Reduce
factor
if this
be the highest common factor, the resulting fraction to be reduced to its lowest terms.
is
said
24a%2x3
to lowest terms
36 «5x2
24 aH'^x^
^ 23 X 3 aH^x^ _ 2c^^ ~ 22 X 32 a^x? ~ 3 a2 *
36 a^x^
Ex.
2.
Reduce
24 a3c%2
to lowest terms
18 dhi^
24 a%2x2 18^3x2
Ex.
3.
Reduce
_ to
12
24
aH^
-
12
6 a'^xX^
4 ac^
a-2x)
^^^~^^-^ lowest terms 9xy -12 y^
6 x2
—
8 xy
a%3
(fii^x:^
3a
-2x
>
_ 2x(3a; — 4?/) _ 2x - 4y) ~ Sy
9xy - \2y^~ Sy(Bx Note.
The beginner should be
careful not to begin cancelling
numerator and denominator in the most convenient form, by resolution into factors where necessary.
until he has expressed both
EXAMPLES .
Reduce to lowest terms 12^mn^^
2
ax
g
15 mhip^
4Gx^y^z \
69 x'^y^z*
'
4
XIII.
a.
:
— ax 3 a2 _ 6 ab 2 a-^b - 4 ab'^
g '
a^x^
g
abx arx
_
+ bx"^ + cx^
l^a^-b'^c
100 (a^
-
a:^b)
3
2
FRACTIONS. 4a;2-9y2
^ '
4
4-
a;^
6 x^
+
^g *
j4
^Y
a(4 ?/3)
^Q
+5 3 x^ + 6 X 4* x2 + 4 X + 5a36 + 10a^6^ 3 a-b-^ + 6 a&3
5
1/'^
x?/
x4-14x'-^- 51 '
X* j^g
-
x^
- 3 aa;) — 9 aj^) x3 - 2 X7f — 4 x^y'^ + 4
x(2 g^ '
6 xy
20 (x'^
g
g
- 2 x2 - 15 + xy - 2 y2 x3 - ^3
109
X*
(a;?/
^^
_
'
xhj'^
a'-^x
j^g '
-3?/)
— 27
2/^
a;2-5x
^2
—4x— ^ 4 ^^y
x^
2/*
a;3y
+
2
a;^^;
x3
-
5
8
+ 9 a^b + 6 ^252 Ig a^^a^b-2 dW 2x2 + 17X + 21 g^ 3 x^ + 23 x + 14 3 x2 + 26 X + 35* 3 x- + 41 x + 20* «%2 _ 16 27 a + g^ ^^ «' 18 « - 6 g2 + 2 d^' ax?' + 9 «x + 20 3 g^
"
jg
'
'
'
g^
(^2
•
138. When the factors of the numerator and denominator cannot be determined by insj^ection, we find the highest common factor, by the rules given in Chapter xi. Ex.
Reduce
to lowest terms
The H.
First Method.
^x^ - 13x2 + 23x 15x3 - 38x2 -2x
C. F.
of
+
21 21
numerator and denominator
is
3x-7. Dividing numerator and denominator by 3x — 2 x + 3 and 5 x2 — x — 3.
—
7,
we obtain
as re-
spective quotients x2
Thus
3x3-13x2 + 23x-21 15x3
- 38x2 -2x + 21
x2-2x + 3
(3x-7)(x2-2x + 3) (3x-7)(5x2-x-3)
5x2-
x-3
This is the simplest solution for the beginner but in this and similar cases we may often effect the reduction without actually going through the process of finding the highest ;
common
factor.
By Art. 116, the H. C. F. of numerator and Second 3fethod. denominator must be a factor of their sum 18x3 — 51x2 + 21 x, that is, of 3x(3x — 7)(2x — 1). If there be a common divisor it must clearly be 3 x — 7 hence arranging numerator and denominator so as to show 3 X — 7 as a factor, ;
the fraction
^ ^^(3x - 7)- 2x(3x - 7)+
3(3x
-
7)
5x'^(3x-7)-x(3x-7)-3(3x-7) _ (3 X- 7)(x2-2x + 3) ^ x2-2x4-
-(3x-7)(5x2-x-3)
5x2
-x-3*
ALGEBEA.
110
139. If either numerator or denomiiiator can readily be resolved into factors we may use the following method.
x^-hox^-4x
Ex. Reduce to lowest terms
- 18 x2 + 6 + 5 = x{x^ + 3x — 4) = x(x + 4) (x — 1). 7
The numerator
a:3
ic
Of these factors the only one which can be a
the fraction
x(x
-
(x
-
+
4) (x
1) (7 X-
-
-
to lowest
terms
2
a^b - ah-^ -2b^ a^ a^ + 'Sa^b + S ab'^ + 2 6^
x3-5x2
4-
-
g
7x-3
g
7
x3-3x + 2 ft3 + 2 gg - 13 « + 10 a^ + a'^-10a-\-S 2 x3 + 5 x^j/ - 30 X//2 + 27 ?/ 4x3 + 5x^/2-21 _ 15 4 fl3 +12 a25 _ 6a3 + 13a26_4a62- 15 63 1 + 2 x2 + x3 + 2 x^ 1 + 3x2 + 2x3 + 3x4'
jQ
'
jj '
^g
jr^s
aj^-2
x2-2x+l + 7 X - 10 3 a3 _ 3>a25 + ^^52 _ 4 a2 - 5 «6 + 62
4)
11
X
3 ax^
+
^^ ^
-
6'
ax^
'
jg
««x
a^
+
a*
+ 4x + 2 3x4-2x3-3x + 2 16x^-72x%^ + 81a^ 4x2 + 12 ax + 9 a^ 6x3 + x2-5x-2 6x3 + 5x2-3x-2' Sx^ + 2x2 - 15x - 6 7x3 - 4x2 - 21 x + 12 4x4 + 11x2 + 25 4x4- 9x2 + 30x-25" 3 x3 - 27 rtx2 + 78 a23; _ 72 ^a 2 x3 + 10 «x2 - 4 rt2x _ 48 a^ ax3 - 5 a^x^ - 00 + 40 gi a:* - 6 ax3 - 80 a-x'^ + 35 a^x 10x2
'
ff3a;
^^ • '
+
-
j^
3 x3
53
+
+
4x3-
2/3
g
+
-
7 x2
4 x^ x^
'
^
5)
1)
x(x
XIII. b.
"
3
is
:
-
J
_ -
1)
11 X
EXAMPLES Reduce
divisor
+ 4)(x-l)
7x2(x-l)-llx(x- l)-5(xx{x
common
MULTIPLICATION AND DIVISION OF FRACTIONS. 140. Rule I. To multiply a fraction by an integer. 3Iultiply the numerator by that integer ; or, if the denominator be divisible by the integer, divide the
denominator by
it.
;
;
:
FRACTIONS.
The proof
unit unit is c
;
as follows
is
- represents a equal
(1)
Ill
— represents
which make up the
parts, b of
ac equal parts, b of which
make up
the
and the number of parts taken in the second fraction times the number taken in the first ;
that
- X
is,
c
= —' b
b
= ~ = -. ±xd bd bd b
(2)
Hence - xb
=—=
a; that
is,
[Art. 136.1
the fraction -
is
the quantity
which must be multiplied by b in order to obtain a. Now the quantity which must be multiplied by b in order to obtain a is the quotient resulting from the division of a by 6 therefore
[Art. 53]
;
fraction -
is the
141. Rule
quotient of
To
II.
the numerator, if
we may
it
define a fraction thus
a divided by
divide a fraction
:
the
b.
by an
integer.
be divisible, by the integer ; or, if the
Divide numer-
ator be not divisible, midtiply the denominator by that ii\^ger.
The proof (1)
unit
is
as follows
:
^ represents ac equal parts, b of
;
- represents a equal parts,
b of
which make up the
which make up the
unit.
The number of number taken
the
tion
is
that
is,
parts taken in the
first fraction is c times Therefore the second fracfraction divided by c
in the second.
the quotient of the
first
^^c=^-. h
b
ALGEBRA.
112 But
(2)
if
the numerator be not divisible by
a
.'.
-
-J-
c
=
— -^ = — c
we have
,
by the preceding
,
c,
case.
be
be
h
_ ac
142. Rule III. To multiply together two or more fractions. the mimerators for a new numerator, and the denominators for a new denominator.
MidUphj
To X
find the value of
- x
-•
b
d a
^
Let
a;
=
c -•
X
-
d
b
Multiplying each side by
X X
X d
b
x
6
=-
d,
we have
X- X
b
d
b
X d
= -x/^X-Xc? xbd
.-.
[Art. 37.]
d
b
= axc. = ac.
[Art. 140.]
Dividing each side by bd, we have
x=^; bd'
Similarly, b
_aG
a
c
b
d~bd
x- X
-
d
f
=
—
and so
;
for
any number of
bdf
fractions.
To divide one fraction by another. 143. Rule IV. and proceed as in multiplication.
Invert
the divisor,
Since division
is
define the quotient
the inverse of multiplication,
x,
when y
is
divided by
a
c
a
d
b
we may
^, to be such that
FRACTIONS. Multiplying by
we have
-',
xx-X- = -X-', d
c
d
b 1.
Simplify
2.
2
Simplify
+
3 «
4a3
^^^
b
x 3 x 5 x «6c3 X 4 X 6 X a^b-^c
^^'^
+
'^^
4 «3 2ffi2
be
Ex.
c
[Art. 142.] -^
G
_ a +
4 «2
12
^^
18
12 a
^
Sc^ 12 a^b
x ii^!^:!^. ^^(2
g
+
18
2a(2a-3) _ 2(?-.3 6(2« + 3) ~ 12 a common to both numerator and
4- .S)
4«3
by cancelling those factors which are denominator.
b
^x^x^^.
The expression
Ex.
c
^-^^=.^ = ^x^.
Hence Ex.
113
^
'
ALGEBRA.
114 g
jQ
16x^-9cfi ^ x-2 4 X - 3 a' x^-4:
25«2-62 ^ x(3a+2) -4x2 5a + 6
, -
6
x2
a;
3 a2
16
-j-
2
irt
ly.
ax- 3 x2 _ ^2 x2-12x-45 x2-6x-27*
-
^'^
-
x^
+
6 x2
x2
-
+
X^'
^ 4x2 + x - 14 X 16 x2 - 49 +2 - 3 62 _^ ^2,4^5 +3 62
+
(a
g.4
-4 +
a3_,.53
+ 216 X x2 - X - 42' x2 - 4 (x - 2)2 Spg + Z^ (X + 2)2'
36 X
49
3
x>
.
_
^
(^2_2a-
^
'
^2-
a-
12
a2 4. 9
3)2
-X-2 + 2X- 8 x2 - 6 x a;2 - 18 X + 80 ^ x2 - 15x + x2 - 5 X - 50 l-^' s.l-3^' + 2x3 1 - x3 (1 - x)3 ^2
26
_
1
6)3
_
64 p2^2 x2
X
5
2 a&
(a
•
4x2
.
•
20
25
.
-X- 1
+
2 x2
21.
15
+ 5x + 3x2 (1 + X)3
2x2
?>*
a2
x2_4x-45
18.
jc
3 ax ^2a^-\-
- x)2 x2- 14x(a
17.
"
'
"
2 x2
. _
x2
-2X-3 j^ x2-9 + 13 X + 15 ^ 2x2 + 11 4x2-1 4x2-9
+5X+ x2-l
x2
j3
a;
'
*
9 «2x2
+ 3 + 2 ^ x^ + lx-^ 12 + 9 + 20 x2 + 5 x + 6 x2 + 4x 2x2 + 5x + 2 ^ 4* 2x2 + 9x + x2 - 4 - 27 6 ^ 4 62 _ 25 2&2 + 56 2 62_ii?, + i5* +5 x2
j2
_
X
-
x2
-
25
20
,
x2
x2
+1 + 5X
7
^
X
.
x2
52_5x_14^x2-llx+18 (x
—
x2
2)2
—
x+ x
56
-
5 l'
x2-16x+63 — 4 a2x + 4 ax
.
4
ax^
ADDITION AND SUBTRACTION OF FRACTIONS. fli
i/i/i^. 144. To prove
b
We
have
+
cf
a,^ml b
bd
—bd
ad -j- be
^^^^^
73
c^hc d
bd
^^^^ ^3^
-j
4
S
ALGEBRA.
116
is 6 ax{x — d)(x + a). must, therefore, multiply the numerators by 3 x(x + a) and 2 a
The lowest common denominator
We
respectively.
Hence the equivalent
fractions are
16xHx + 6 ax{x
—
a)
a) (x
We may now
146.
Sa^ a){x
^^^
+
—
6 ax(x
«)
-\-
a)
enunciate the rule for the addition or
subtraction of fractions.
Rule
II.
To add
Ex.
1.
Reduce
or subtract fractions.
common denominator ; add or and retain the common denominator. Find the value of
^^+^ + bx-4:a 3«
=
3(2a;
_ ~
n
.r
^
9a
The lowest common denominator Therefore the expression
+
9
is
a.
+ a)+ .3 ft
+
-
5x.
5y
—
4 a
_ \\x — ~ 9a
4g
9a Ex.
2.
Find the value of
xy
a(x
-
2
?/)
+ x(S y -a)- y(Sx-2a)
1
2
2
a;
1
-3
x-6 .r
+
x-l— x-9 ,
3.
a:-
5.r
2
25
+
12
x
+
-\
15
— 3xy +
ay
2
= 0,
other.
XIII. d.
of
6
9
S xy
numerator destroy each
364*
2.X-
— ax axy
— 2ay +
EXAMPLES Find the value
ax
ay axy.
is
axy ax
since the terms in the
a
^—^ + ^^^^^ - §-^-=1^.
The lowest common denominator Thus the expression
tliem to the
subtract the numerators,
lowest
45
S
2
^
+
.r
x
5
8
^ *
a
-
?)
'ah
r.
o
27
-\-
2x
X
-2h
&
-
c
Sx^' c
-
a
ca
be
a-bh
b.
.a + 7b-•
h
2a
4 a
8a
FRACTIONS. Y
a-x
a
X g
2 a^
+
-
Ij^
h'^
~
^
3.
'
c^
^''
—
_
^
10 x^
-
gg
—
2
^^
—
g^
j2
3
j/
6c
_
2x-a X,
The lowest common denominator
—
is
-
+
y^
xrif-
Sx-2z
5
xz
x
—
ac
6"^
_ ab —
ac
^c
^x-?.a _ x — 2a
xy
x?-
xy
x^
Therefore the expression
-
xjf
a;
15a;3
Simplify
f-
xij
c"'^
8
_3
2_
^^
^
-c-2
I
Ex.
x?
?>'-
^
5x
-
a^
2 ax
a-
9
_
X
a
117
(?-
a6
^
a
(.->:
—
2 «)(a^
—
«).
- a)-(2 x - a)(a: - 2 g) - 2 «) (.r - a) 2 x^ - 5 g.y + 3 cfi - (2 x? - 5 «x + 2 ^2) _ ~ (x — 2 a) (x — «) ^ 2 x2 5 «a- + 3 g^ _ 2 a;2 + 5 gx - 2 g^ (x — 2 g) (x — g) (^ ^
=
^ «)(^:
(.'>;
-
g-
(x
Note.
—
—
2 g) (x
g)
In finding the value of such an expression as
-(2x-g)(x-2g), the beginner should
first
remove the brackets,
as
express the product in brackets, and then done. After a little practice he will
we have
be able to take both steps together.
The work will sometimes be shortened by the fractions to their lowest terms. ^^
Ex.4. Simplify
The expression
=
^-^
+
2xy
5x//-4^/^ x2-16?/2
,
2x2
+
+ 5x^-4.^2 _ y x2 - 16 X + 4 — + 5 xy — 4 — ?/2
_ ~ x2
?/2
?/(.!•
X2^- 16
_ x2 + 5 xy -
4
x2
-
x2
+ 4 X?/ _
x2-16?/2
?/2
8 xy
?/
4:
y)
?/2
-
xy
16 ^2
X
x-iy
+
4 y-
first
reducing
2
'
5
ALGEBRA.
118
EXAMPLES
XIII. e.
Find the value of 1
1
1
4
. I
+ _2
2
x
+
x+3
x
+
x
'
2
]__^
-A- +
x-^-4
g
X
—
a
4-
a—
2
—
x+
a
r^
g
4
6x +
X— 10.
x-j-
1__^
3
Q
h__^
a '
3
_
^
—
?)
g
—
X
-
12.
—
2X
1
^
14 '
16.
4
^^
13.
15.
3y
X
a(x"-^
x-^
—
9
- x3
21.
-
(1
+ _ - 2a ^<
x)3
a:^
+
2 g^
x2
-
4
a:(x-^
—
X
y'^
__x l,j
1
+
1
+
Some
-\-
— 2X r<3 - 63 + a/j +
+
6-3
+
^2
_a + a)^ X + 4q X + 2« (x
X"
?/2
(a
a ab
y
?/(x-2 4- ?/2)
y
1
25.
-
+
-
xy
+
x
—
y
I
x^
^3
a)
.^^
— 4 g'^ — 2 «x + xy + X
a^
147.
a^)
+
^
2/2)
+
y
2g
5x
L^+ -
(x2
x^
24.
y)
^
^/^
^^
x^
oc_
— x^
+
x(x
+
—
_|.
x'^
23.
^£i__^^.
^
y{x
«'^)
3-.
^^
^
a'^'
i«l±_^_2«^^. 2 « + 6 4 «- - 52 X-
—
25 x^
?/2
?^
a
2
^y
20.
.
+
«-^
—
2
19.
^+^
«
X
x(x-?/)
9 ^
a:
g
18.
(x-2)2
£±l_£Lzil. x + 2 —4 x—7 X — x —
'x-2
g
X
x« + x
.^_ + ^iL_.
11.
h
.
a X
^
b
a^
«-^
//^
+ 2x)2 — 8 x^
6^
modification of the foregoing general methods
may sometimes
be used with advantage. The most useful explained in the examples which follow, but no general rules can be given which will apply to all cases.
artifices are
Ex.
1.
Simplify
a±S _a±4 a
Taking the
first
—
4
rt
—
3
«2
8_. _ 1(3
two fractions together, we have the expression
8 8 7 ^ ^ cfi_Q^(a2_iQ) _ (a-4)(«-3) (a + 4)(a-4) a^ - IG (a-4)(a-3) 62 - a 7(a + 4;)-8(a-3) "" " (a + 4) (a - 4) (a - 3) (a + 4) (a - 4) (« _ 3;
:
FRACTIONS. Ex.
^
Simplify
2.
2 x^ -f
The expression
119 ^
•
a:
-
3
1
+
a:'-^
(2x-l)(;«+l)
4
+
a;
1
(3x+l)(x +
l)
3x+l+2a;-l +
(2a;-l)(ic
(2
Ex.
.
Simplify
3.
\J
Here L. C. M.
—
a^
X*,
Hence
-J a —
l)(3a:+l)
a:- l)(ic+
l)(3x+
a
-\-
X
cfi -\-
it
=a
X
-\-
—{a —
-
— _
«4
x)
2x x^
(fi -\-
^4
a^
1
1
l
4.
,
+ 2x
2« 2a + 3
l-4.x2'
35
6.
I
+
X
J^--
9 - a2 7.
(a
^-?! 6(a:2-l)
3
+
x)2
1_. - x2
a
3
-
L_+ 2(x-l)
.
3x +
-^ +
6x2-4
r_.i ^!L±^.
?/3
x^
-
1 a;^
3(j;+I)
5x
2 ^
j3^
1
x)^
(3a + 2) 3(4«2_9)' 4
+9
6rt ^
+
(1
x"-^
5
3
x^
a
3
,
-
1
x-2 12
2
2«
^^^
«2_52
2(« + &)
,
2«-3
1_.
2
+
jQ
4a-^-db-
a^
1
l+x
?/2
8&-2-
1
1
«
„
l-2x 2a-3 6
5
f.
2(a-6)
3x ^
x8
1
o
4-13x 4 x^
y
5. '
2x x^-y^ 3x
1^1 2x-y +
__5
XIII.
x"^
-
«8
of
x-y
y
8
^ x*
EXAMPLES Find the value
x^
4 x3
4 x3
3
x^
first
x2
2x a^
2x
+
a^
be convenient to proceed as follows
will
a^
+
x^
two denominators give a'^ — ic2, which readily combines with a^ + x"^ to give L. C. M. which again combines with a* + x* to give L. C. M. a^ — x^. should be evident that the
it
The expression
x
4^
^x
^
X
1)
- 9 x + 20
^e
yS
^
x^
-
7
a;
yd
x---Ux +
1
14.
_
1
1
+
12
a-2
-
6x
+
80
ALGEBRA.
120 15.
—J-x-l
1
16.
1
19.
20
1
x+1
+ 35 +
« '
22.
— X-
^
-
3 X
+
+
x"^
+
5 X
+
26
27
30.
31.
33.
+ 2)(x+3) 9(x + 3) lG(a: + l)(x-2)' « + 2 + + 6)(« + 3/>) 4(a + 2 6)(a + 3 l)(x
f<
+
x2
?>
6)'
-2
x2
-
1
12
+
9X
14
+
4 X
x2
+
10 X
+
21
2
x2-l 2x+l 2x2+3x+l 5(2x-3) 7x 12(3x+l) ll(6x2 + x-l) 6x2+ 7x- 3 11(4x2 + 8x + 3)* 1 8« ^ ~ ^ _^ ~ ^ oq1+2^« l-2a x + 2 x + 3 x-1 l-2« l + 2rt (l-2rt) ^" 5 -3 x+4 ^ ^ 29 x + 3 X - 4 x2 - 16 1 + X 1 + x3 (1 - x)3 24 X 3 + 2 X 3 - 2 X 9- 12 x + 4 x2 3-2 X 3 + 2x ^ 4« I ^^. -J: 32. + 1 3 - X 3 + X 9 + x2 2 a + 3 2 a - 3 4 (/2 + 9 I
a;
^
_^+ _^+ -x) + (a
^_+ - x)
+
(rt
2 «
6 «2
+
+ 54
^1^ 8-8x
3 a2
+
4-
—-
(rt2
+
a:"-)
3-6x +
48
^ 9
3
,-1-+ + 8x
r/2
_ 27
+
+
+ 6x
2
+
8 x2
8x
2
4 x2
2
+
2 x*
+
^^'
"^^
2^
39.
.
^ 4
3
1
2 a
ic2
1
3 «
8
x2)
x
^ 8 X
^-
38.
4
x)
2
x2
-
+
^
^-
37.
2(l
1
8
^^+1 2-x + 4
1
4(1
x)
34.
40.
X
4
4(l
35.
+
15
6
+ 5x + 2x2
2
1
2
,
3
25
(a
-
x2
X
23.
2&)
2
10 a^
1
(x
2)
+ a:-18x-
5
-
2
/>
4(« + 6)(a
a
15(x-l) lG(x-3)(x-2)
2(x+l)(x-3)
21
-
18.
,
(x+l)(x + 5x
3
''
-x-1 Qx^-x-2
2x2
-1 a-2cfi
A
§
I
*
17.
+ x-^
2x^
2oi^
x
(2
+
x)2
4
-
x2
:
.
.
FRACTIONS.
121
r
148. We have thus far assumed both numerator and denominator to be positive integers, and have shown in Art. 140 that a fraction itself is the quotient resulting from But in algethe division of numerator by denominator. bra division
and we
a process not restricted to positive integers, The this definition as follows
is
shall
now extend
algebraic fraction -
of a by
the quotient resulting
where a and b
b,
By
149.
is
:
may
from
the division
have any values ivhatever.
the preceding article -^^—
—a
—
is
the quotient result-
by ing from the division of & and this is obtained by dividing a by b, and, by the rule of signs, prefixing ;
+
:=4*
Therefore Again, ^^^
—b
is
=+^=^ b
(1).
b
the quotient resulting from the division of
b
— a by
b
;
and
this is obtained
the rule of signs, prefixing
Therefore
11^ =
b,
and, by
-^
b
Similarly,
by dividing a by
— (2).
b
-^ = -^
(3).
These results may be enunciated as follows (1) If the signs of both numerator and, denominator of a fraction be changed, the sign of the whole fraction changed.
ivill
be un-
If the sign of either numerator or denominator alone ivill be changed.
(2)
be changed, the sign of the ivhole fraction
The
principles here involved are so useful in certain cases
we quote them in another form, which will sometimes be found more easy of application. of reduction of fractions that
1.
We may
change the sign of every term in the numerator
and denominator of a fraction
We may
tvithout altering
its
value.
change the sign of a fraction by simjily changing the sign of every term in either the numerator or denominator. 2.
— ALGEBRA.
122
Note. The student should keep clearly in his mind the distinction between tenn and factor. The rule governing change of sign for factors will be given in Art. 150.
Ex
''
1 11
X
Ex.
2.
Ex.
3.
Ex.
Simplify
4.
2y
3x
3X
—
a;-
may
4
h If
x-
-{-
2y
step
-
a
X
-{-
X-
_
a
^^
-X
—
4
The intermediate
= -h + -y — >•- ~ __ —X ~
__ ~
2y
+
x~ -^
:c-
4:
usually be omitted.
_1^ + a(Sx-a)^
X
X
a
'
3X
-^L_ +
+
—X
./•-
—
a
—
a-
x-
Here it is evident that the lowest common denominator of the first two fractions is x- — a-, therefore it will be convenient to alter the sign of the denominator in the third fraction.
—^ + -^-^ - ^K^a;-^) — —
=
Thus the expression
+
X
_
a(x
—
X
a
g) +
a
X?
_ ax ~
Ex.5.
a'^
—^
Simplify
3
+ 2 ax-
_^
Sx-l _^ \
-x^
x^-l
\0(x+
1)
—
a)
S ax
+
_
a'^
2
a;^
1
2X
'iAzil
S{x-l)
cfi
a(Sx
a2
2x^
=
The expression
-
^
3
a:
X-
+ a)—
2 x(x
+
+
2 "^
2(:>:+l)
-0(3x- l)+3(x- 1) +6+ - 1)
1)
Q(x^ 10
.r
+
10
-
18
6(a:2
a;
EXAMPLES
3x
-
3
13
-
G(x^
5
-
a:
1)
XIII. g.
Simplify 1
1
1.
4j;
—
5a
4
+
rt
1
—
1
5
1
—
5«
2
3 1
-f
«
rt'-^
—
g
.
1
x-\-2a
2(a'^-4ffx)
x-\-a
a;"^
x-a x+a
or tt'
—
+ Sax — x'^
a'^
x:^
_ Sa x — U'
x+a X—a
9
FRACTIONS. 5.
_l_ + -J_ + -i^ 2
—^-+ 6
10.
1-4 x^
x-1
2
.!•+ 1
_
^
g '
9
2-5.r.
3 + a;
:c+3
3— X
3
-2a;
2
1-
+
15.
16
-
X ^^
./:
11
Q-6b-^
^^^ .
255"-
— 4 «-
ic-
62
1
2« +
—
{x
-
(b
«)
_
(«-2
1
X
+
a;
+
a:'-
a
a;
—
3
4 a'^(a
+
a
-
y-
X'
+
-
x
2
+
.V
+
x
-{-
4
y a;2
y
—x
-
2
a;
+
2-x
2
6)
_
,,2>)
{x
a)
—
a) (x
a
—
(^.2
«2)
+
a^
1
y^
+
x
—
+ -
j/^
x^
+
a;-)
««
-
xy
_^
{x
+
y) {x-
+ ab(b* ff-^
b-^)
S a
2 «-(«-
«)
b) '
+
^j.2
+
x-
x
-
a-^
b{a-
x+ y -
?/
—
2a
a b-^)
+
y
^
^
y'^
1,1
_|_
y-
b (((a-^
+
6
.c
—
rr^^.x
|/^
a-
X-
4
a-2
x
if
c
+_J^+ 3 «
:>•)
+ —
'
—
a
«'-
1
24.
X-
1*
-
1
1
+
x'^
c
b) (6
^5-2
^-2)
4a
,
rt
—
+
+
;/2) (^.2
+
ff
(x
b)
'
21.
—
(6-a)(6-x)' ,
—
y^
xy
a-6
0:
(a;
—
b
y
a){a
+
+
b ,
—r (a-b){x-a)
20
-
'
2x +
CT
19.
xy
r/x^- 6
fY.i-^)
a)
j^
6
^
-
y-
-2a
b
+ - 4a:-
2(6x
1
+
-
1
- ^) _ ^2
6
6) {X
g-^
+2 a — ob
^^'(^
^^
^_
x^
9
3
b
-
(rt
—
46-4
«
jg
4
3
_
^g^
^
-
I
^3
icy
12
3
2 X
3 a-
xh/-
-
^g^
2
+ -
^a--
2
+
—
«
:>;=5
ic-—
3
56
^-
-
2y,(2.-c-ll)
2a;
-^— + + a
1
3
_5
'26 +
+
^
1
1
-
<)
jr
11.
_
^
:>;2
+
«
o 1
123
+
6-^
a^)
y-)
a^ b^
-
a^
«2)
(;/2
+
/,2)
4
:
ALGEBRA.
124
From
150.
Art. 149
it
follows that
(1) Changing the signs of an odd number of factors of numerator or denominator changes the sign before the
fraction. (2) Changing the signs of an even number of factors of numerator or denominator does not change the sign before
the fraction.
Consider the expression
+
I
(a
- b)(a -
By changing nator,
we (a
Now nators
{b
-
I r){h
- a)
+
1 (e
- a)(c - h)
the sign of the second factor of each denomi-
obtain
- 6)(c - a)
it is is
c)
(rt
(h
- c)(a -
h)
(c
readily seen that the L. C.
- a){b - c)
^
^
M. of the denomi-
— b)(b — c)(g — a), and the expression _ —(b — c) — {g — a) — {a — b) (a — ?>)(5 — c)(c — a) — b-\-c — c-{-a — a + b 0. (a-b)(b-c)(c-a)
151. There is a peculiarity in the arrangement of this example which it is desirable to notice. In the expression (1) the letters occur in what is known as that is, b follows a, a follows Cyclic Order ;
Thus, if a, b, c are arranged round the circumference of a circle, as in the annexed diagram, if we start from any letter and move round in the direction of c,
c
follows
b.
the arrows, the other letters follow in cyclic order,
namely
abc, bca, cab.
The observance
of this principle
in a large class of examples in
Thus we
three letters are involved.
order
when we
write b
—
is
r,
c
especially important
which the differences
— a,
a
of
are observing cyclic
—
b;
whereas we are
«a
a
125
FRACTIONS.
violating cyclic order by the use of arrangements sucli as It will always c. a, b c, b c, a~h, or a c, a h be found that the work is rendered shorter and easier by
—
—
—
—
—
following cyclic order from the beginning, and adhering to it throughout the question.
EXAMPLES
XIII. h.
Find the value of a
-
h) {a
(a
-
b) (a
{X
-
II)
(X
-
(a
-
(b-
-c)
^
a)
(c
-
a) (c
-
b)
(c
-
a) (c
-
b)
y)
r + - c)A. {b -a)
z)
ill
-
z) (y
-X)
(z
-
y x) {z
-
-
z)
(y-
z){y
-
x)
{z
-
x)iz
- y)
-
c)
(&
-
c) (5
-
r «)
(c
-
{x-y)(x-
z)
(y
-
z) {y
-
x)
{z
-
+-,
-
c)
{X
-
y){x
z
•
X
,
—c ——
b
0.
7
,
-
c) (&
(b
3.
g
c
b ,
{a
b) (a
-+--:
-
—
b
-
a) (c
b)
z-xy
1 + {b-c){b-a)
-
+
q r)
(q
-
r
+
{,r-p){:r-q)
-p
+p - q - p) (r - q)
r ^
r) (q
y)
r-a
.
(q-r){q-p)
q-r
q) (P
-
^-
I
q-
,
x) {z
1 + (c-«)(c-6)
^>
I
pp-^
,
,
{p-q)(p-r) (P
a
+
y'^zx
1 + (a-6)(a-c)
g
a
7-7-.
x^yz
Q
—
c
,
,
- p)
(r
'''
10, (rt^
«' (a2
-
&-2)
x (P
-
(X
-
+
-
y
q) (P
-
(&2
(-2)
(q
-
q^r y) (X
r) (q
r
-z)
c-2) (6-2
x+y
^ r)
_
iy-
- p)
+P
z) (y
-
_
(c2
{r
X + y - p) (r -
_^
P
_^
x)
-
(fi)
{z
-
+
a') (r2
q)
q
x) (z
-
y)
-
//2)
CHAPTER
XIY.
Mixed Expressions.
Complex Fractions.
We
now propose to consider some miscellaneous 152. questions involving fractions of a more complicated kind than those already discussed. In the previous chapter, the numerator and denominator have been regarded as- integers but cases frequently occur in which the numerator or denominator of a fraction is itself ;
fractional.
153. A Complex Fraction is one that has a fraction in the numerator, or in the denominator, or in both.
a a
Thus
-, -,
X c
In the
-
Complex
are
Fractions.
c
d
last of these types, the outside quantities,
a and d, two
are sometimes referred to as the extremes, while the
middle quantities,
154.
By
h
and
are called the means.
c,
definition (Art. 148)
-
is
the quotient resulting
d from the division of - by h
^
and
;
this
d'
a h
~
c
d 126
ad be
by ^ Art. 143
is
-^• he
COMPLEX FRACTIONS. 155.
127
From the preceding article we deduce an easy method down the simplified form of a complex fraction.
of writing
Midtiply the extremes for a neiv numerator, and the means for a new denominator. a
+
X
_ab(a
b a^
—
X?
~
b{a^
-^
—
x)
_
a a
.r'-)
—
x'
ah
by cancelling common factors
numerator and denominator.
in
156. The student should especially notice the following and should be able to write off the results readily.
cases,
1
_
a
.
a
'
_
^^
a
b
a ~~^ = a T= 1 b
X
b
_
^^
a'
= ab.
b 1
a_l^l_l 1
a
'
b
a
b_b I
a
157. We noAv proceed to show how complex fractions can be reduced by the rules already given. a ,c
Ex.1.
Ex.
2.
b
d
a
c
b
d
ALGEBRA.
128
«2
Ex.
3.
_yi
Simplify a.
(I
The numerator
(^2
+ —
^2 _^
/^2
h
_a —
h
h
a
b
-\-
_ ((^2 _ + + h^) («^ ^2)2
(«--^
4
fraction (a2
+
(«2
4
a'-^?>2
&--2)
(«2
-
/>2)
4:aVy^
(^2+
62)
4 ^252
_ +
62) (^2
_
52)
4r<7>
^
Similarly, the denominator
Hence the
^^2) 2
?>-^)
(«2_
62)
«?)
+ j,^ ^a - b) {a + b){a-b) (^,
4 ^6
ab «24. 62
Note. To ensure accuracy and neatness, when the numerator and denominator are somewhat complicated, the beginner is advised to simplify each separately as in the above example. Ill
the case of complex fractions like the following, called we begin from the lowest fraction, and
Continued Fractions,
simplify step by step.
„
9ic2-64
.
r
1
5.
ALGEBRA.
130 Sometimes
158.
it is
convenient to express a single frac-
tion as a group of fractions.
10
~
xY
xy^
10
~
^J^_l 2y
10
^
x'Y
10
xY
3^
X
2x;^'
MIXED EXPRESSIONS.
We
may often express a fraction in an equivalent 159. It is then called form, partly integral and partly fractional. a Mixed Expression. Ex.
x+7 ^ (a;+2)4-5 ^ x+2 x+2
1.
-
3 X 2. Ex "':*•.+
2
^ Pj(x +
5
3.
Show
x^ that 2
- 15 +5
-
—
7
x
may
-
1
x-3
Performing the indicated a remainder
x+2
5) ;/•
In some cases actual division
Ex.
-^
2
17
^
17
3
x
a;
+
be advisable.
^
division,
^ 3(a- + 5) +5
2
.x
we
-
—
1
obtain a quotient 2 x
—
1,
and
4.
2x'--lx-l ^
Therefore
-
2 x
^^
1
160. If the numerator be of lower dimensions (Art. 29) than the denominator, we may still perform the division,
and express the result and partly fractional. Ex.
in a
^^
Prove that
l+3x-^
By
division
1
+
=
2x
form which
-
6 x^ (2 x
+
- 6 xH
C x3
-
18
x^>
18x5 18 x5
whence the
result follows.
+
+
18 x^
6 x3
-6x3
-
partly integral
^^
18 x^ 1
3 x2)2 x
2X
+
is
54 x7
^'
3x2
:
MIXED EXPRESSIONS.
131
Here the division may be carried on to any nuinber of terms in and we can stop at any term we please by taking for our remainder the fraction whose numerator is the remainder last found, and whose denominator is the divisor. Thus, if we carried on the quotient to four terms, we should have the quotient,
+
l
is
1 iV
l+Sx^
3x-^
The terms
in the quotient
divided by
or
— a^,
+ ^' + ^ + ^, tV
tAj
the
may
first
be fractional thus if x^ four terms of the quotient are ;
and the remainder
is ^'.
tV
*v
161. Miscellaneous examples in multiplication and diviswhich can be dealt with by the preceding rules for the reduction of fractions. ion occur
Ex.
——
Multiply x-\-2a
The product
=lx+2a __
2 x^
+
+
2x +
_2
x'-^
+
7
_
ax
+
2X
6 «"
-
+
+
- -^^\
ax X
- o?- - 2 g^ +a
- 3 a^ +a (2x + 3a)(xx + «
5 a2 ^ 2 x^
+
ax
X
3a
+
g)
—
a).
,,
2x + 3a 5 a) (x
2 x?
a^
3rt
(2 X 4- 5 g) (x
= (2 X +
^
a
^^^"j x (2 x - a
ax
7
^^
-
by 2 x
EXAMPLES
XIV.
a)
b.
Express each of the following fractions as a group of simple fractions in lowest terms ^
3 x^y
+ xxp' -
2
3 g^x
—
4
a'^x^
g
^
if-
'
Ox?/
+
G ax^
+
12 ax 3
g3
-
3
b-hS ab- + + 3 g62a6
g2?)
?>
+
c
abc he
g
+
ca
+
ab
abc 6=^ b^
g
gS^c a^bc
-
3 abh-
'
6 abc
+
2 g6c
'
:
132
ALGEBRA.
Perform the following
remainder after four
divisions, giving the
terms in the quotient 7.
x---(l+x).
9.
8.
a^(a-b).
10.
13.
14.
15.
Show Show Show
that
+ x)^(l-x).
J^—-^ ^^^2b (a
-
that x2
-
that
(l
l~{l-x + x'^).
Show
that
^^^^
a-b
- -^^ = t^lzJ/l.
xy +ij^
x+
x-\-y
y
60^''-17x^-4x+l ^
+ 9x-2 ^'+^'-^' = 1 +
12
17.
Divide x
18.
Multiply
cfi
-
19.
Divide 52+ 3 6
20.
Divide a2
+
-
+
(a
6
1
+2 + c)(«+6-c).
Multiply 4 x2+ 14 x
^^
'
-
+
+
- ^^^ « + 2X 36
6-3
a2 21.
-
X 4 x2
4._656i_
9;/2
49
+ a:
- -i^ by
2
25
2a6
by x
+
2 ax
_
^
2a&
+ ^^/^ ~ ^^ x2 - 16
x2--(x+3).
l-(l-x)2.
'
6)2
5x2
16.
11.
12.
-
3
+
^
C
2x-7
+
2 «x
^'^'^ .
+
4 x2
^^
1362 «
by
+
6-3
"^
^^^"^"^
^>-^(^^ «'"
^_^3^^
|3y
9 62
4
by
-
36
'^
^
'^
+
^^
12x2+18x +
6
27
162. We add an exercise in which most of the processes connected with fractions will be illustrated.
EXAMPLES Simplify the following fractions
3 6(c2
_ ^2 - x2)
x(x+
ff)(x-|- 2ct)
y2 "
"^
Vbc
+
6x ^
x(x
3« 3.
V^ b\a-b
X
r I
c.
:
-
2 ex
+
X2J
«)(2a;+ g)
6a
L_^ a-\-2bJ
.2^2 - + X+1
5.
c2
+
XIV.
2 a2
+ a6-262
4x
X2-X+1
4
f
•!!+]/ V^_/:lzi?/Y.
\x-y)
\x
+
yJ
FRACTIONS.
^11 (X
g l
a
—
^:'^
f I
—
x2
+
a2
+ ax) x2 + qx ) V X — « /
"^
g
.
+
a2
a)
V
— +
^^
3 x5
Z-^'
f^\
«
y]
Q^^-^'"^
«^
x^
V
a)
\
«3 _ x^ + x2 1 x*-2x2+ ^ - 10 x3 + 15 X - 8 rtx
,.
^^
13.
2x3-9x2 + 27 3x3-81x+162
l>\h
«^
2 «x
a2-x2 12
X2
'
P
6
11
2
+ 1)2 X + 1 1 + X + l + x3 + 2x + 2x2 + x3 (^2 - &2)a; ff(a2 - 62) ^^2
X
133
gg
14.
+
ff(l
a^
+ -
ff)y
+
i/
y^
4 ,, 15.
-
a
«
3
2
—^^ + «+2 «+l
1
+
111 «
+
2C
16.
„
— 3
1
1
2 2x-3 +,^9x+r^9 + oX
.,
2x2
9
'
,.
4x 17.
+
X3
18
+X+ 1-^^
-
3X
+
(x
-
l/«2 / «^
23.
24.
X
+
x
L2 Vx
?/
,2
a
/ /
n a
\a
+
'>•
\2 \
x/
x2-?/2 X2?/
y
^ 4-
x-Sa
+ 6 «2
a-x
+
jyt
1
—
^—
2\x-rj
^'^-6x + 4 x3
2
^f-i
x/
20
^'
\a'^-x^)
+
1
.
-L 1 l^ «4-x
x2\\ x2
V
1 I
1
X
+
1
1
2
0:2-5 ax
2 «)-
-
X
2V1-X
x)2
2 a
21.
+
X2
W
2x3-x2-2x+ x3
-
X3
1
+ ax)2-(a +
(1
jg
X2
I2/
X
+
1
+ ^'-^ 2 X
Xy2
Mx
-
2/
x^y
/
?/
-{-
1
x?/2j
25.
(3x-5-^)(3x + 5-|).(x-^).
26.
/?-^_ + -J_Wf«+^-«JZ^Y Ix
rt
+
x
a
— xi
\a
—
X
a-{-xJ
-
21 x2
+
15x
+
20
ALGEBRA.
134 1
2ic—
4x^—1
1
+ 1
(« 1
30
+ «/> — b)b + «6
"
1
- «6 — 6) — ab
a(a 1
_J^__y x^+ij'i a
+
-
b
1 4^
_
a
- a) X + «5
a{b 1
x-\1
'if^
—
ij
+
do.
y)
1—^2/
35.
1
a)
2x+l + b a-h —b a b a — b _ a + a-j- b a — b a
.. •
x(l^x){l-xy
1+i
_
a;?/
?/(a;
-^
b
"*•
1_1
Qo 04i.
I
^•24.1_1
a
-{-
?)
X
ab^ a'^
— a% + ft'-^
•
FRACTIONS.
Xj\lJ^- X^J
\y (X
+ zy -
y^
(X
45.
46.
+
<--
(«
-
51.
52.
53.
^2
'
if
(^y^ ^yi
_
^2
4^-2-(x-y
x)'-2
+
-
?/2
)-^
-
c)
"^
ff
6) («
(.X
«
^ (&
-
+
?>
c) (&
a)
+
&
,
-
(c
-
c
«) (c
-
?>)
?/
9y2_(4g_2x)2 16^2_(2x-3y)2 (2x + 3|/)2- 16 2;2+ (3^4-40)2-4x2 "^ 1
50.
+[yy -
-
x2-(2y-3^02 4?/2_(3g-x)2 9^'2-(x-2y)2 (3 + x)2-4?/2'^ (x + 2?/)2-902+ (2 + 3 2:)2 - x2' 2;
•
XIJ
XIJ
+ y)2 _ 4 ^2 + (y + 2 ^)2 - x2' (a^ -y)(y-^) + (y - ^) (^ - a;) + (g - a;) (x - ?/) x(^ - x) + ?/(x - ?/) + ^(?/ - ^) « — & — c h — c — a c — a — h {a-b)(a-c) (b-c)(b-a) {c - a){c - h) (2 S
47.
'
+
y2_(2^_x)2
a;2-(y-2^)2 •
X^
135
4 x2
(4^
- (3 y - 4 ^)2 + 2x)2 - 9 ^2'
CHAPTER XV. Fractional and Literal Equations. 163. In this chapter
we propose to give a miscellaneous Some of these will serve as a
equations.
of
collection
useful exercise for revision of the methods already explained in previous
more
but we also add others presenting the solution of which will often be facili-
chapters;
difficulty,
tated by some special artifice.
The following examples worked
Ex.
1.
Solve 4
- ^^^ = ^ - 1. 22
8
Multiply by 88, which nators,
in full will sufficiently
most useful methods.
illustrate the
is
2
the least
common
multiple of the denomi-
and we get
352- ll(x-9)=4x-44; 352 — llx + 99 = 4x — 44;
removing brackets, transposing, collecting terms
— llx — 4aj = — 44 — 352 and changing signs, 15 a; = 495 x
.-.
Note.
—
99
;
;
a*
In this equation
—
9 is
= 33.
regarded as a single term with
8
minus sign before it. In fact it is equivalent to — i (a: — 9) the line between the numerator and denominator having the same effect the
,
as a bracket.
Ex.2.
Solve
8x+23_5x+_2^2x+j_^^ 20
Multiply by 20, and 8X
+
3
a:
23
transposition,
4
5
- ^^C^^-^^) = 3a;
By
+
we have
-f-
31
8
a-
+
12
4
= 20(5
a;
4-
Sx + 136
i
2)
^
-
20.
,
FRACTIONAL AND LITERAL EQUATIONS. Multiplying across, 93
.r
+
124
84 .-.
Ex.3.
Solve
a:
= = =
20(5 r 7
X
+
137
2),
;
12.
A:zA + ^zii = ^ZL^ + ^-7 10
x-Q
x-Q
x-1
This equation might be solved by clearing of fractions, but the work v^ould be very laborious. The solution will be much simplified by proceeding as follows :
Transposing, "
^11^ _ ^.H^ = ^jzl _ ^lli. a;- 10
x-9 x-6
x-1
Simplifying each side separately^
- 8) (x -
{x
.
- (x -
we have
(X - 7) (x - 6) - (x - 4) (x - 9) (x-10)(x-7) (a:-9)(x-6) x2-15 x + 56- (x2-15 x+50) ^ x^-lS x+42- (x^-lS x+36) (x-10)(x-7) (x-9)(x-6) 7)
5) (x
-
10)
^
.
'
.
'
(x
-
10) (X
-
(X
7)
-
9) (X
-
6)*
Hence, since the numerators are equal, the denominators must be equal
;
that
is,
(x
- 10) (x - 7) = (x - 9) (x - G) - 17 X + 70 = x2 - 15 X + 54, 16 = 2x;
x2
.-.
Ex.
4.
We
have
x
=
8.
11 ^ 4X-55 _ x-6 X— 13 X —6 X — 14 X — 7 /2 + -i-W4+— l\ 5 + -i a:-13V x-14V + x-6J
Solve
5x-64 _ 2x-
1^
X—
13
X
—
6
Simplifying each side separately^
X
—
14
X
we have
(x-14)(x-7) - 13) (X - 6) = (X - 14) (X - 7), x2 - 19 X + 78 = x2 - 21 X + 98, 2 X = 20 x = 10.
(x-13)(x-6) (x
;
.-.
ALGEBRA.
138
164. To solve equations Avliose coefficients are decimals, we may express the decimals as common fractions, and proceed as before but it is often found more simple to work ;
entirely in decimals.
Ex.
1.
- 1.875 = .l'2x + 1.185. .375 x - .12x = 1.185 + 1.875 (.375 — .l'2)x = 3.06,
Solve
.375
Transposing, collecting terms,
that
X-
.255
is,
=
x
;
3.06;
.•.:.=?:0«=12. .255
Ex.2.
Solve
+
.ere
-
.25
transposing,
24 x
a:
-
+ 4x
1.8
- .75x-
i-
9-*'
!•
^0*
-"-g
3>
i"*^
—
27 x
9
12
1.
we have
fractions,
— 1^ V — — 4 x = 68 — + 27 x = 68 47 X = 47
X A- ^ 3**'T^4
2.
24
clearing of fractions,
=
ia-
common
Expressing the decimals as
-
12
;
9,
;
.-.
x=l.
EXAMPLES XV. 1.
4(x
+
2)
5x
^^
2
^
^
'L+A + 14
=
'LzlA
2
3
5
4^5
"^
+ 6
3)
2V
3V
2]
4
f3x-2iLz_^\:=l(2a;-57)-^. x-f3x^
^
~^^'
7
Ol
1/
5
2x-5
2(x-_3)_
5)
3
2
4
10.
+
^Q^
21
8
4
5 ^
3
5(x
^
'
7
8.
^x-S
7
6
9
11
,--S
13
5
a.
.
x-3 ^ 4.r-3 2X-15
10
10 ^
^
'"
^3
6^
yl
4(x + 3)
^3 '
9
^ 8x + 37 7x-29 18
5X-12*
97
FRACTIONAL AND LITERAL EQUATIONS. 13.
(2a;-l)(3x+8)
5a: 15.
+
-i :c+3 7
16.
17.
31
3
6x +
_-,
2
_^
^=— + 2x+6
^i-.
1
60
_
2a;
+2
^_2
G0 + 8a; ^-^^
a:-l
x-Q
X
X-
x+4
»^
21
lOi a;
^- + -^^ = -^+
^
„ x+5
• "
^
ic
30 + 6x
^g
—7
a:-9
139 48 ^
x-l
—
X— a;-ll
x-15
a;-17
ALGEBRA.
140 Ex.
2.
Solve
-^ X
Simplifying the
—
X
a
^ = ^^^. —
—
a(x —
X
b
c
we have
left side,
—
— a) _a — b — b) X — c (a — b)x _a — b {x — d){x—b) X— c ^ 1 {x — a) (ic — 6) X — c across, x^ — ex = x^ — ax — bx -^ ab, ax bx — ex = ab, {a + b — c)x = ab {x
b)
—
b (x
a)(x
,
*
a;
•
Multiplying
_
-\-
;
«^
:.x^. a
+
b
—
e
EXAMPLES XV. 1.
2.
4. 5.
6.
7. 8.
g
10. '
ax-2b = bbx-^a. a^{x - a) + &2(x -b) = abx. (x-a){x + b) = {x-a + bY. a(x-2)+2x = 6 + a. m^(m — x) — imix = n^(n + x). (a + x){b + x) = x(x — e). {a-b){x-a) = (a—c)(x—b). 2x + Sa _ 2(Sx + 2a) x + a Sx-\-a 2(x- b) ^ 2x+ b
Sx-c
b)x^
-
+
l_i = l_l. x~ x b
'a
'Aa « x
..
9a
Sx
b
b
=
]
+ ^.
_4b
2x
a
a
- a ^ x-b b - x~ a-x x-a ^ (x- by x
^g '
^g '
2
2x-a'
^f(«-f).
= bx{x -a) + ax{x -
(rt
19.
b{a
20.
b{a-x)-^{b^xY-\-ablr^ + \y=().
21.
a;2+a(2rt-X)-
22.
(2a;-a)fx +
a2)
Aa
c(«-Z>)
x
18.
a(Jjx
J
^^
3(x-c)
n.lx(.-
^^
b.
h).
+ x?)-{a-Vx)(h-x) = x'^^~-
— rr^X-^'V
^^^\
+«2.
= 4x(|-x^-l(a-4a;)(2a +
3a;).
;
CHAPTER
XVI.
Problems leading to Fractional and Literal Equations. 166. We here give some problems which lead to equations with fractional and literal coefficients. by 4, and such that one-half by 8. Let a; represent the smaller number, then x + 4 represents the greater. One-half of the greater is represented by h{x + 4), and one-sixth of the less by ^ x. i(x + 4)-ix = 8; Hence multiplying by 6, 3 x -f 12 — x = 48 Ex.
Find two numbers which
1.
differ
of the greater exceeds one-sixth of the less
;
2 X . '
and Ex.
X
;
number,
18, the less
22, the greater. ;
certain sum,
win
4-
X 4
36
has •$ 180, and B has $ 84 after B has won from A a A has then five-sixths of what B has how much did B
A
2.
•.
= = =
;
?
Suppose that B wins x -f X dollars.
dollars,
A
has then 180
—
x dollars, and
B
has 84
180 - x 1080 -6x
= f(84 -f x) = 420 + 5x, 11 X = 660 x = 60.
Hence
;
.-.
B
Therefore
wins 1 60.
EXAMPLES
Find a number such that the sum of
1.
may be equal to 15. 2. What is the number whose together
make up
There
3.
by 3
XVI.
:
find
is
a
13
eighth,
its
sixth
sixth,
and ninth parts
and fourth parts
?
number whose
fifth
it.
141
part
is
less
than
its
fourth part
::
ALGEBRA.
142 4.
Find a number such that six-sevenths of by 2.
it
shall exceed four-
fifths of it
5. The fifth, fifteenth, and twenty-fifth make up 23 find the number.
parts of a
number together
:
6. Two consecutive numbers are such that one-fourth of the less exceeds one-fifth of the greater by 1 find the numbers. :
7.
Two numbers
differ
by
and one
28,
eight-ninths of the other
is
find them.
There are two consecutive numbers such that one-fifth of the 8. greater exceeds one-seventh of the less by 3 find them. :
9.
by
10,
Find three consecutive numbers such that if they be divided 17, and 26, respectively, the sum of the quotients will be 10.
A
begin to play with equal sums, and when B has lost has gained $ 6 more what he had to begin with, than half of what B has left what had they at first ? 10.
B
and
A
five-elevenths of
:
From a certain number 3 is taken, and the remainder is divided the quotient is then increased by 4 and divided by 5, and the find the number. result is 2 11.
by 4
;
:
In a cellar one-fifth of the wine is port and one-third claret besides this it contains 15 dozen of sherry and 30 bottles^ of hock how much port and claret does it contain ? 12.
13.
B's
is
:
Two-fifths of A's money is equal to B's, and seven-ninths of equal to C's, in all they have $ 770 what have they each ? :
A, B, and C have $1285 among them: A's share is greater than five-sixths of B's by $ 25, and C's is four-fifteenths of B's find 14.
:
the share of each. 15.
A man
16.
The width
had been
of a
3 feet more,
been square
:
17.
What
18.
I
$35 and half as much as he gave what did he pay for the horse ?
sold a horse for
and gained thereby $ 10
:
room
is
two-thirds of
and the length 3
its
If the
length.
feet less, the
for
it,
width
room would have
find its dimensions.
the property of a person whose income is 8 430, when he has two-thirds of it invested at 4 per cent, one-fourth at 3 per cent, and the remainder at 2 per cent ? is
bought a certain number of apples at three for a cent, and number at four for a cent by selling them at six-
five-sixths of that
teen for six cents I gain 3^ cents 19.
:
:
how many
apples did
I
buy
?
Find two numbers such that the one may be n times as great and their sum equal to b.
as the other,
PROBLEMS.
143
A man
agreed to work a days on these conditions for each to receive c cents, and for each day he was idle he was to forfeit d cents. At the end of a days he received m cents. How many days was he idle ? 20.
:
day he worked he was
21.
A
sum
money
of
divided
among
three persons
:
the
first
;
receives h dollars
receives
is
more than a third of the whole sum the second more than a half of what remains and the third the amoimt which is left. Find the original sum.
receives a dollars
;
c dollars,
22. Out of a certain sum a man paid .$96; he loaned half of the remainder, and then spent one-fifth of what he had left. After these deductions he still had one-tenth of the original sum. How much had he at first ?
A man moves 12 miles in an hour and a half, rowing with the and requires 4 hours to return, rowing against a tide one-quarter
23. tide,
as strong 24.
requires
the
:
find the velocity of the stronger tide.
A man
first
c :
moves a miles in b hours, rowing with the tide, but hours to return, rowing against a tide d times as strong as
find the velocity of the stronger tide.
25. A has a certain sum of money from which he gives to B 8 4 and one-sixth of what remains; he then gives to C $5 and one-fifth of what remains, and finds that he has given away half of his money. How many dollars had A, and how many dollars did B receive ? 26.
The fore-wheel
feet in circumference.
fore-wheel has
made
c
is a feet, and the hind-wheel is h the distance passed over when the revolutions more than the hind-wheel ?
of a carriage
What
is
27. In a certain weight of gunpowder the nitre composed 10 pounds more than two-thirds of the weight, the sulphur 4| pounds less than one-sixth, and the charcoal 5i pounds less than one-fifth of the nitre. What was the weight of the gunpowder ? 28. Two-thirds of A's money is equal to B's, and three-fourths of B's is equal to C's together they have §650. What amount has each ? ;
29. A dealer spends $1450 in buying horses at $100 each and cows at § 30 each through disease he loses 10 per cent of the horses and 20 per cent of the cows. By selling the remainder at the price he gave for them he receives $ 1260 find how many of each kind ;
:
he bought.
\
A, B, C start from the same place at the rates of c, c + d, an hour respectively B starts k hours after A how long after B must C start in order that they may overtake A at the same instant, and how far will they then have walked ? 30.
c -\-2d miles
:
;
ALGEBRA.
144
1.
MISCELLANEOUS EXAMPLES III. — 4p^ + 8 from unity, and Sp"^ — p —
Subtracters
and add the
from
1
4.
- yY + {x - zy + 2{{x - y){z - x)+ yz}. 6{x-2[x-S{x-l)]} = 70. Divide xf^ ^- x^ -2ix^ - Sox + 57 by x'^+2x- 3.
5.
Find the factors of
2. 3.
Simplify (x
Solve
(a
(i.) 6.
zero,
results.
+
-
&)2
121
Find the H. C. F. of
7.
A man
8.
Solve
a^
-
_
«*
(ii.)
;
2
a'^
54
+
(iu.)
.
^^2
and 2a^
1
-
5
a:
14.
+ ia -
a^
-\-
-
7.
being asked his age said: "Ten years ago I was five tiriies as old as my son, but 20 years hence, I shall be only twice as old as he." How old was he ?
6- x-l
(i.)
x-2
3-.r
3
-
4X
(ii.)
x
9
-
27 9.
4 5X
3
4
By how much does
—
y^
Show that (ax + byy + {ay - hxY +
~
'
3
Sy"^
Sy
-\-
6
x-\-
6
2 9 exceed y — 4 y^-\-Q
-\-
— y^ ?
10.
11.
Solve
(i.) ^
^
3
(ii.)
12.
If
x
=
1,
y
= 2,
Simplify
(i.)
52
+
+
79 a
c2).
'
a2
4
2/2
^
(ii.)— rtft
+
+
ft2
15.
Find the L. C. M. of x^
16.
The
difference
of ?/
(y
a:^+\l ab-\-mi)^;
—^^+ —
(i.)
^•)
(ii.)
-
zy.
10a2
-
8.
^^ Qjc
+
2c?/
6
a
,
_
c^^
(^-2
_
52
6^2
+
11 x
+
6 and x^
^2
+
+
-
7
x
+
6.
between the numerator and the denominator of 8, and if each be increased by 17, the fraction
a proper fraction is becomes equal to | find :
12
5;
14.
Divide
+
i/)(a'2.
6
= 3, find the value + z)-V{y- z)']- x2 +
Find the factors of
18.
= (^2 +
4
13.
Solve
c2?/2
(3x- l)2+(4x-2)2zr(5x-3)2.
{X -y)\_(x
17.
+
02^2
2x-l_3x-_2^5£-4_7_x + 6
it.
=
(i.)
20(7x4- 4)- 18(3x+4)-
(ii.)
J(x+l)+K^ + 3)=Ka^ + 4)+16.
(i.)
2x(x2
(ii.)
5x(x
5
25(x
- l)(x+ 2) by x2 + - 2 - 11) (x2 - X - 156) by x^ + a:
+
6)
-
132
;
;
x2
x.
MISCELLANEOUS EXAMPLES
A
19.
boy
-
;
Find the continued product
20.
x^
one-sixth the age of his father, and five years older the united ages of all three being 51, how old is
is
than his sister each ?
cfix"
+
of
x^
+
+
ax
a^,
x'^
—
ax
-{ o?,
a*.
Show without
21.
pression r^
145
III.
—
—
2 x2
22. Simplify
?
(i.)
+
+?+
^—
a
3
a factor of the ex-
is
6.
^ ^ 3 (x - 6)
^
X
3 (ii.)
—
actual division that x
5X
-
6
'
^+1.
2a — X
X
x
Find the H. C. ¥. oi 2a^ + a;^ - a -2 and a^ - a^ -'ia^ + 1a by the usual method. Is the work shortened by proceeding as in 23.
Art. 119 ?
Show
that the square of the H. C. F.
is
contained in the
second expression. Divide x^
25.
Simplify
+
24.
(i.)
-
19 x^
216 by (x^
3x
+
2±1 _ .^ _ l^jzM: 2
Find the factors
26.
-
27
+
of (i.) 14
2
9) (x
^
a^-ll a-15
;
2).
_^
1 + ^x2-4 (x-2)2 (ii.) + b a'^b'^+dbK
(ii.)
2i)2_8
-
a'^
Of a party 5 more than one-third are Americans, 7 less than one-half are Englishmen, and the remainder, 8 in number, are Ger27.
mans 28.
:
find the
Solve
number
(i.)
2(5
in the party.
x- 2)- 3(5 x - 8)= 5(x + l)-(2x-
2x^-9
^.._^ ^ ^
11);
^_x-J^ 18
27
'
4 ^
1±^±.^^
29.
Simplify
30.
Find three numbers whose sum is 21, and of which the greatest 4, and the middle one is half the sum of the
(i.)
;
(ii)
exceeds the least by other two. 31.
Employ
the Factor «3
32.
Show
that
_
2 a2
^(^^ x2
_,.
Theorem 1
and 2
in finding the H. C. F. of a3
+
a2
^
4 ^
_
7.
+ ^-^)-^(^^-^-^)--l^^JL^. -X-2 x2 + X - 2 x2 - 4 x2 - 1 from P to Q by different routes, one of which
33. Two trains go 15 miles longer than the other. train on the shorter route takes 6 hours, and a train on the longer, travelling 10 miles less per hour, takes 8^ hours. Find length of each route. is
A
ALGEBRA.
146 34.
Find the factors of 4 x2
(i.)
35.
-
4 xy
-
15 \f
;
9 x*
(ii.)
-
82
x.V
17-3x_2+^^14 + 7x
Solve
3
5
_^
9
^4.
g^
3
36. The number of months in the age of a man on his birthday in 1875 was exactly half of the number denoting the year in which he was born. In what year was he born ?
37.
38.
^
^ ^
(i.) ^-
•>'
Divide
^
(X
Simplify
^
'
find the value of the quotient
39.
_l_
- ^ _ 2(r. +.2) - 3)2 x2 - 9 x^-]-x^y'^ + x^y^-\-Qc?y^-\-y^ by
Simplify
2 x2
^
_
i-
x^—x^y-{-x'^y'^
when x =
and y
=
I
+ 1)2 — xy^-\-y^ and (2
X
\.
(Dj-,,^^-^^; X^ 1 + X + X2 1+3x4-3 x2 +
X
.
1
+
2 x3
X
but if it were regiment has sufficient food for m days reinforced by p men, would have food enough for n days only. Find 40.
the
A
number
;
of
men
in the regiment.
X
X 41.
Solve
(i.) ^ ^
(ii \
42.
—~
X
—-^ + -—^ + d
-^^
Simplify (1
h
^
baa
-
(^
+
a)-
^-
I
+
= 0;
c
^->
_«+
6
r"^i-a+ I
^ ^
1
+
a
[
+
a2 J
—
CHAPTER
;
XVII.
Simultaneous Equations. 167. Consider the equation 2 x tiao
unknown
From
this
quantities.
we
get
y
=
-{-
5y
= 23, which
———
-
contains
(1).
Now for every vahie we give to x there will be one corresponding value of y. Thus we shall be able to find as many pairs of values as we please which satisfy the given equation. Such an equation is called indeterminate. For instance, if x = 1, then from (1) y = -2J-. Again, if x == — 2, then ?/= 2_l and so on. But if also Ave have a second equation of the same kind expressing a different relation between x and y, such as •
we have from If
now we
this
?/
= t^-^I
-
(2).
seek values of x and y which satisfy both ?/ in (1) and (2) must be identical.
equations, the values of
23-2.^24-3..
Therefore
4
5
Multiplying across, 92
- 8 . = 120 - 15 . 7. = 28; = 4. .-.
a7
Substituting this value in equation
y Thus,
(1),
we have
= 23-2. = 23-8 p— = 3.o
—
5
5
both equations are to be satisfied by the same values of . and y, there is only one solution possible, if
147
ALGEBRA.
148
Defixition. When two or more equations are by the same vahies of the unknown quantities, they
168. satisfied
are called simultaneous equations.
In the example already worked, we have used the
169.
method
of solution
which best
illustrates the
meaning
of
the term simultaneous equations; but in practice it will be found that this is rarely the readiest mode of solution. It must be borne in mind that since the two equations are
simultaneously true, any equation formed by combining will be satisfied by the values of x and y which satisfy the original equations. Our object will always be to obtain an equation which involves one only of the unknown
them
quantities.
170. The process by which we cause either of the unkno^vn quantities to disappear is called elimination. It may be effected in different ways, but three methods are in general use (1) by Addition or Subtraction (2) by Sub:
stitution
;
and
;
(3)
by Comparison.
ELIMINATION BY ADDITION OR SUBTRACTION. Ex.1.
171.
Solve
7x +
2y=i7
5x-4?/ = Here
it
will be
more convenient
Multiplying (1) by
and from
2,
Ux -\-4ty = 9i, 1;
= 95 x = 5.
19 x
adding,
.*.
To
(2).
to eliminate y.
6x — iy =
(2)
(1),
l
;
find y, substitute this value of x in either of the given equations.
Thus from
(1)
35
+
2
.-.
and
?/
?/
X
= 47 = 6,| = 5. ;
i
In this solution we eliminated y by addition.
Ex.2.
Solve
= + 2y =
3x-\-7y
21
(1),
5a;
10
(2).
;
SIMULTANEOUS EQUATIONS. To
eliminate x
we multiply
by 5 and
(1)
the coeliicients of x in both equations equal.
(2)
by
149 so as to
3,
make
This gives
= 135, = 48; 29 y = 87 y = 3.
lbx-\-Zby 15
+
a:
subtracting,
6
1/
.-.
To
find X, substitute this value of y in either of the given equations.
Thus from
+
3x
(1)
21
.•..
and
2/
In this solution
;
)
we eliminated x by
subtraction.
Multiply, ivhen necessary, in such a
Rule.
make
= 27 = 2,, = 3.
the coefficients of the
Add
equal in both equations. coefficients
miknown quantity
manner as
to he
to
eliminated
the resulting equations if these
are unlike in sign; subtract if like in sign.
ELIMINATION BY SL^STITUTION. 172.
Ex.
2x-5y=l
Solve
7x + Transposing
—by
in (1),
3?/
=
(1),
24
(2).
and dividing by x
2,
we obtain
= ^JL±l. 2
Substituting this value of x in (2) gives
Whence
35
?/
+
7
and
= 48, = 41 .-.y = l.
+
6y
41
?/
;
This value substituted in either (1) or (2) gives
x
=
3.
Rule. From one of the equations, find the value of the vnknoicn quantity to be eliminated in terms of the other and known quantities; then substitute this value for that quantity in the other equation,
and
reduce.
150
ALGEBRA. ELIMINATION BY COMPARISON.
173.
x+15y = 53 2/+3x = 27 = 53 -
Ex. Solve
From
(1)
and from
a;
x
(2)^
= ^^ ~
^
(1), (2). 15?/,
^
.
3
Placing these values of x equal to each other,
we have
53-15^^ = 2^-1/. 3
Whence
159
and
-
45
?/
44
?/
y
.-.
= = =
-
27
132
?/,
;
3.
Substituting this value in either (1) or (2) gives
=
X
8.
Froyn each equation find the value of the unknown Rule. quantity to he eliminated in terms of the other andknoivn quantities;
then form
an equation
ivith these values,
EXAMPLES Solve the equations 1.
3
a:
+
4?/
4x + 2.
?/
=10,
=
+
2?/
8.
4.
9.
10.
11.
?/
5.
6.
7.
15
a:
+7?/ =20,
9x+15?/ =
?/
3.
reduce.
a.
:
9.
= 13, 3x + = 14. 4x + 7?/ = 29, x + 32/ = ll. 2x-?/ = 9, = 19. 3X - 7 5x + 6?/=17, 6x + 5?/ = 16. 2x + ?/=10, 7x + 8?/ = 53. 8x-?/ = 34, x + 8?/ = 53. x
XVII.
and
= 39, = 35. 6ic 28 - 23 = 33, 63 X- 25?/ = 101. 35x+17?/ = 86, 56 X - 13 = 17. 15x + 77?/ = 92, 55x-33?/ = 22. 5x-7?/ = 0, 7x + 5?/ = 74. 21x-50?/ = 60, 28x-27?/ = 199. -3?/ + 17y
14a;
a:
?/
15.
39. 16.
17.
18.
13.
14.
19.
21.
80.
6?/-5x=18, 12x-9?/ = 0. 8x = 5y,
=
8
?/
+
1.
3x=7?/, 12?/
20.
99,
bx=ly-2l, 21x-9?/ = 75.
13 x
?/
12.
39x-8?/ = 52x-15?/ =
= 5x-l.
19x+17?/ = 2x-?/ = 93x 15x
+ +
15?/
93?/
0,
53.
= 123, = 201.
.
;
SIMULTANEOUS EQUATIONS.
We
174.
151
add a few cases in which, before proceeding
to
solve, it will be necessary to simplify tlie equations.
+ 2?j)-{Sx + nij)=U 7x-dy-S(x-4:i/) = SS 6x-hl07j-Sx-Utj = U; .-. 2x-7j = 14
Ex.1. Solve
From
(1)
From
(2)
From
(3)
5(x
-9y - 3x +
x
7
= 4:X + 3y = 6x-3^ =
.'.
and hence we may find x Ex.
2.
=
S,
S8
(2).
(3).
;
SS
(4).
42;
and y =2.
- ^^Lizl = ^ « -
Solve 3 X
12y
(1),
^ ^^^^
3l_ti_ 1(2.^-5)=?/ Clear of fractions.
from
(2).
Thus 10 = 28a:-2i; \4x-2y =-^\ - lOx + 25 = 15?/ 10x + 6y = 37
42ic-2y +
(1)
.-.
From
9?/
(2)
+
12
(3).
.-.
(4).
Eliminating y from (3) and (4), we find that _— _ 14 ^ — X Y3
Eliminating x from (3) and (4)
,
we
find that
Note. Sometimes, as in the present instance, the value of the second unknown is more easily found by elimination than by substituting the value of the unknown already found.
EXAMPLES 1.
^+2/-16, 'x
2
3.
+ =
^
2
1^
4.
5
'
'
x-y = 4.
^-y=3,
^-f =
+ l = u.
^
XVII.
'
-
= ^-y = ?/
5,
45
2.
a:
b.
5.^ + ^=10,
8.
l 6.
a-.
+
=
y 3
= m. ?/,
^+y 3-^ = U.
ALGEBRA.
152
lx-^y = 3,
7-
lx--ly = 4, }x + j'^y = S. = 0, 9. 2a; + i?/-3x = 8.
^^y = U,
^Q
4x-y = 20. 8.
^^
3
+
x
^
= 4f.
+
4^
+ ?^==3x-7y-37 =
f?/
'
4
2
= 7.
15.
0.
y^7
3a;-l
^^ fa;
54 y = n.
3x +
3ic-7?/=zO,
11.
^-^ =
12.
3
•
?/
14.
5^
7
x
+
2
3?/=9,
^±i ^ ^y - 5 ^g^-^, 10
2
8
SIMULTANEOUS EQUATIONS INVOLVING THREE
UNKNOWN
QUANTITIES.
175. In order to solve simultaneous equations wliicli contwo unknown quantities we have seen that we must have two equations. Similarly, we find that in order to solve simultaneous equations which contain three unknown quantities we must have three equations. tain
Eliminate one of the unhnoKjns from any pair of the and then eliminate the same unknoivn from another pair. Two equations i7ivolving two unknowns are thus obtained, ivhich may he solved by the rides already given. The reynaining unknoivn is then found by substituti7ig in any one of the Rule.
equations,
given equations.
2y-6z = lS -2z = lS 7x + 5y-Sz=26
6x +
Ex.1. Solve
Sx-\-Sy
(1), (2), (3).
Choose y as the unknown to be eliminated. Multiply (1) by 3 and (2) by 18
cc
+
6
2, ?/
-
6x + 6y subtracting,
12
a;
—
15 ^
11^
Again, multiply (1) by 5 and (3) by
subtracting,
=
39,
4^= 26; =
13
'(4).
2,
30x+ I0y-25z = Ux+lOy - Qz = 16x-19;s =
65, b2',
13
(5).
SIMULTANEOUS EQUATIONS. Multiply (4) by 4 and (5) by
3,
-
48 X 48
cc
44 ^ 57 ^ 13 s
subtracting,
and from
153
x
(4)
from (1)
2/
= 52, = 39 = 13
;
;
= 2,V = 3.]
Note.
After a little practice the student will find that the solution often be considerably shortened by a suitable combination of the proposed equations. Thus, in the present instance, by adding (1) and
may (2)
and subtracting (3) we obtain 2x — 4:Z = 0, or x = 2z. and (2), we have two easy equations in y and z.
Substi-
tuting in (1)
Ex.2.
^_i=L/+i=£ + 2,
Solve
2
From
6
the equation
7
^ 3
+ = 13.
-
— 1=^ +
^ 2
Sx—y =
we have
And, from the equation
| o
Eliminating z from (2)
and
2
(3),
(1)
whence x
=
10, y
12 x
=
18.
-
4
78
(3).
we have
21x4-4?/
and from
(2).
+ ^ = 13,
2y4-3z =
we have
?/
= 282; = 48 ;
Also by substitution in (2)
we obtains =
14.
Consider the equations
6x-Sy-
z
=
13x-72/ + 3s = 7x-4:y =
6
28
X7
X
16?/
—
4
?/
(1),
14
(2),
8
(3).
Multiplying (1) by 3 and adding to (2),
or
(1).
7x-2z = 42
we have
3.
12
^ — 1=^ + 2,
Also, from the equation
Ex.
1,
= 32, = 8.
we have
ALGEBRA.
154
Thus the combination of equations (1) and (2) leads us to an equation which is identical with (3) and so to find x and y we have but a single equation 7 x — 4 ?/ = 8, the solution of which is indeter[Art. 167.] minate. In this and similar cases the anomaly arises from the fact that the equations are not independent; in other words, one equation is deducible from the others, and therefore contains no relation between the unknown quantities which is not already implied in the other ,
equations.
EXAMPLES x+
3.
7.
2?/
+ 20 =
11,
XVII.
c.
;
SIMULTANEOUS EQUATIONS. because a
= -,
and
^
=-
The
is
related to b
a
h
exactly as h
and consequently a
;
155
is
related to a.
and y are - and - respectively, and ^ y i 1 in solving the following equations we consider _ and - as reciprocals of x
X the
unknown
y
quantities.
^-^ = 1
Ex.1. Solve
X 1^ X
(1),
y
+^=7
(2).
y
Multiply (1) by 2 and (2) by 3
X
;
thus
y
§2 X
18 y
+
= 21,
adding,
— = 23
multiplying across,
46
X
= 23 x=2; y = 3.
.-.
and by substituting
Ex.2. Solve
in (1),
~ ~ + Ay
a;,
-=-
2x
Sz
n), ^' ^
4:
^'''
l=ry
i-F, + ! = ^A clearing of fractional coefficients,
X
y
^
z
§-1 =
(2)
from (3)
obtain
6_^5_4^3
from(l) from
we
(•^)^
X
l^_§ + ^=,32. X
y
(5),
y
z
.........
(Q).
156
ALGEBRA.
Multiply (4) by 15 and add the result to (6)
105
,
42
X dividiiig
by
6
7,
X
= 11
(7)
y
§5=11;
adding,
X .: (5)
2=3,^ |/=1
from (4)
EXAMPLES
2-5 = 3,
we have
y 15
from (5)
from
;
—
XVII.
d.
;
:
157
SLMULTANEOUS EQUATIONS. LITERAL SDrTLTAXEOUS EQUATIONS. 177. Ex.
aj:-ryy
Solve
1.
a'x^h'y
=
(1),
c
=d
(2).
one that the student will freqtiently meet with in the course of his reading. In the first equation we choose certain letters as the coeflBcients of x and y. and we choose corresponding letters irith accents to denote corresponding quantities in the second equation. There is no necessary connection between the values of a and a', read " a and a prime,'" and they are as different as Tlie notation here first
a and b
;
but
nsed
is
often convenient to use the
it is
assist the
coefficients of
z
:
b, b'
as being coefficients of
«-2? (f3
To
thtis slightly
',
?>i> ^^2, t'z,
read
etc.,
letters are
••
return to the equation ax -\-by
a'x-hb'y Multiply (1) by
b'
and
(2)
»v.
used with a suj^Xj such as a sub one, a sub two." etc.
Sometimes instead of accents oi,
same letter
mark some common meaning of such letters, and thereby memory. Thus a and a' have a common property as being
varied to
by
=c = c'
1
.
,2;.
Thus
b.
= b'c^ = be' Qab' — a'b)x = b'c — be' b'c - be' ab'z -\-bb'y
a'bz
by subtraction,
-f-
bb'y
;
;
a'h
C3).
previously explained in Art. 171, we might obtain y this value of j: in either of the equations (1) or i^D but y is more conveniently found by eliminating x, as follows
As
by substituting ;
Multiplying
(1)
by
a'
and
(2)
by
a,
we have
+ al>y = a'c, = ac'; (a'h — ab") y = a'c — ac' _ a'c — ac' aa'x
aa'x -\-ab'y
by subtraction,
'''
'^'a'b-ab^
ALGEBRA.
158 or,
changing signs in
.the
y ^
= ac' — a'c
terms of the numerator and denom-
same denominator
inator so as to have the
and x
^^zJl c — a
= y^ —
+
y
(1)
x{c
by clearing
—•
i
(1),
b
c
—a_a
a—h
c
x{c
be'
ab'-a'b
ab'-a'b'
Ex.2. Solve
From
= b'c —
-,
,
as in (3),
(2).
c
we have
of fractions,
-b)- a(c-5)+ vie- a)- b(c- a) = (c - a){c - 6), — &) + y(c — a) — ac — ah -^bc — ah + c^ — ac — bc-{- ab, x(c-b)-hy(c-a) = c^-ab
x{a -h)-\- a(a
-b)+ cy x{a — b)
x{c{c
—
b)
c
and
(4)
by
c
= =
ca cy
-\-
Multiply (3) by
(3).
we have
Again, from (2),
—
a(a
-
b),
ac
(4).
a and subtract,
— (c — a)(a — b)]= c^ — abc — ac{c — a), - ac + a^ _ ab) = c(c^ - ab -ac-\- a"^);
ic(c2
x
.'.
and therefore from
y
(4)
=c = b.
EXAMPLES = = m. Ix + my = n, px -{ qy = r. ax = by, 3. bx ay = c. 4. ax + by = a\ hx + ay = b^. x + ay = a', 5. ax + a'y = l. Q.px-qy = r, rx — py = q-
I.
ax
-\-
bx
-\-
by ay
I,
'
;
XVII.
^ — y_J_ ±^1
e.
10.
I
a
^_]L— a'
f--(^+!)'
1
a'b'
b'
11.
-{-
a bx q ^'
b
+
ay
= ^ab.
3x,2y_o " "^
T T
^_^ = a
qx-rb=p{a-y),
ab
b
b
12.
'^^
X
y
'ni'
m
px lx
^^'
s
^+y-=i, m m' ^'
qy = 0, + my = n. -\-
(«-^0^=C« + ^')^> x
-^
y
=
c.
CHAPTER
XVIII.
Pkoblems leading to Simultaneous Equations. 178. In the Examples discussed in the last chapter we have seen that it is essential to have as many equations as there are unknown quantities to determine. Consequently in the solution of problems which give rise to simultaneous equations, it will always be necessary that the statement of the question should contain as
many
independent con-
ditions as there are quantities to be determined.
Ex. 1. Find two numbers whose difference is whose sum is 9. Let X represent the greater number, y the less. Tlien
x-y=i\\
Also,
^+-^ =
11,
and
one-fifth of
(1).
9,
5
x
or
By Ex.
and 25
^
=
45
56 and by subtraction, 2 numbers are therefore 28 and 17.
addition, 2 x
The
=
+
2.
;
and 10
If 15 lbs. of tea
lbs. of
tea
and 13
lbs. of coffee
lbs. of coffee
(2). ?/
=
34.
together cost $ 15.50, )$ 24.55, find the
together cost
price of each per pound.
Suppose a pound of tea to cost x cents and a pound of coffee to cost y cents.
Then from
the question,
we have
15 x
25
.^
+ +
10?/
13
?/
Multiplying (1) by 5 and (2) by 75 X 75 X Subtracting,
= =
1550
(1),
2455
(2).
3,
we have
+ 50 = 7750, + 39 = 7365. = 385, 11 ?/
?/
?/
y
=
160
35.
;
;
PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS. 161 And from
+
15 x
(1)
Whence
350
=
1550.
15 x
= =
80.
.-.
x
1200
Therefore the cost of a pound of tea of coffee is 35 cents.
;
80 cents, and the cost of a
is
pound
A
person spent $0.80 in buying oranges at the rate of 3 for Ex. 3. 10 cents, and apples at 15 cents a dozen if he had bought five times as many oranges and a quarter of the number of apples, he would have spent $ 25.45. How many of each did he buy ? ;
Let X represent the number of oranges and y the number of apples. cost X oranges ""
15 V ^ cents
4
or
1^
^
—3' ,
^^
Tr
;
12
Again, 5 x oranges cost 5 x x
L^x^,
cents,
3
—
y apples cost
3
10 X
or
^
—
- cents, and ^ apples cost ^
3
cents;
48
12'
.-.^ + 1^^=2545
(2). ^ ^
48
3
Multiply (1) by 5 and subtract (2) from the result
285 y ^ 48
or
.-.
or.
=
855
= x —
;
144;
y
and from (1) 150. Thus there were 150 oranges and 144 apples. Ex. 4. If the numerator of a fraction is increased by 2 and the denominator by 1, it equals f and if the numerator and denominator are each diminished by 1, it equals h find the fraction. Let X represent the numerator of the fraction, y the denominator X then the fraction is -• ;
:
y
From
the
first
supposition,
^-tl = 2/ y
from the second, These equations give
Thus the
fraction
M
cc
is y%.
=
8,
+
1
^ 8
(1)
=-
(2)
ALGEBRA.
162
The middle difjjit of a number between 100 and 1000 is 5. and the sum of the other digits is 11. If the digits be reversed, the number so formed exceeds the original number by 495. Find it. Let X rej)resent the digit in the units' place Ex.
zero,
;
y represent the digit in the hundreds' place then, since the digit in the tens' place is 0, the number will be repre[Art. 84, Ex. 4.] sented by 100 1/ + X. ;
And if the digits are reversed, the sented by 100 X + y. .-.
100 X
so
formed
be repre-
will
- (100 y + v) = 495, + y- 100?/-x = 495; 99x- 99^ = 495, x — y = h
+
100x
or
number
2/
.-.
that
is,
Again, since the we have
From
(1)
and
sum
of the digits
11,
is
x
(2)
Hence the number
we findx is
=
8,
?/
=
-\-
y
(1).
and the middle one
is 0,
=
(2).
\\
3.
308.
EXAMPLES
XVIII.
1.
Find two numbers whose sum
2.
The sum
and whose difference
is 34,
is
10.
of
two numbers
is 73,
and
their difference
is
37
find
:
the numbers. 3.
One-third of the
their difference 4.
4
:
sum
find the
two numbers numbers.
of
is
14,
One-nineteenth of the sum of two numbers find the numbers. is 30
difference 5.
is
is
is
4,
and
of
their
:
Half the sum of two numbers
ference
and one-half
18
:
20,
is
and three times their
dif-
find the numbers.
Six pounds of tea and eleven pounds of sugar cost $ 5.65, and 6. Find the eleven pounds of tea and six pounds of sugar cost $ 9.05. cost of tea and sugar per pound.
Six horses and seven cows can be bought for $250, and thir7. teen cows and eleven horses can be bought for $461. What is the value of each animal
?
A, B, C, D have $ 290 between them A has twice as much as C, and B has three times as much as D also C and D together have $ 50 less than A. Find how much each has. 9. A, B, C, D have $270 between them; A has three times as much as C, and B five times as much as D also A and B together have $ 50 less than eight times what C has. Find how much each 8.
;
;
;
has.
:
PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS. 163 Four times B's age exceeds A's age by twenty years, and onefind their ages. is less than B's age by two years 11. One-eleventh of A's age is greater by two years than oneseventh of B's, and twice B's age is equal to what A's age was thir10.
third of A's age
teen years ago
;
and
A walks
How many
;
and
How many
hours.
C walks
hours
in five
A
B
12.i miles less than D does in twelve walks 3;^ miles less than C does in seven miles does each walk per hour ?
In eleven hours
13.
twelve miles more than B does in seven walks seven miles mqve than does miles does each walk per hour ?
in thirteen hours
in nine hours.
hours
find their ages.
In eight hours
12.
hours
:
:
D
Find a fraction such that if 1 be added to its denominator it reduces to |, and reduces to | on adding 2 to its numerator. 15. Find a fraction which becomes | on subtracting 1 from the numerator and adding 2 to the denominator, and reduces to i on subtracting 7 from the numerator and 2 from the denominator. 14.
If 1 be added to the numerator of a fraction it reduces to i be taken from the denominator it reduces to |. Required the
16. if
1
;
fraction. 17. If f be added to the numerator of a certain fraction the fraction will be increased by Jy, and if J- be taken from its denominator
the fraction becomes | 18.
The sum
of a
by reversing the
find
it.
number
of
:
digits is 1 10,
two digits and of the number formed and the difference of the digits is 6 :
find the numbers. 19. The sum of the digits of a number is 13, and the difference between the number and that formed by reversing the digits is 27 find the numbers. 20. A certain number of two digits is three times the sum of its find the digits, and if 45 be added to it the digits will be reversed number. 21. A certain number between 10 and 100 is eight times the sum of its digits, and if 45 be subtracted from it the digits will be reversed find the number. 22. A man has a number of silver dollars and dimes, and he observes that if the dollars were turned into dimes and the dimes into but if the dollars were turned into halfdollars he would gain f 2.70 How many dollars and the dimes into quarters he would lose $ 1.30. of each had he ? 23. In a bag containing black and white balls, half the number of white is equal to a third of the number of black and twice the whole number of balls exceeds three times the number of black balls by four. How many balls did the bag contain? :
:
;
;
;
ALGEBRA.
164
A
number consists of three digits, the right hand one being hand and middle digits be interchanged, the number is diminished by ISO if the left hand digit be halved and the middle and right hand digits be interchanged, the number is diminished by 24.
If the left
zero.
;
4o4
find the
:
number.
of 10 men and S boys amount to S 22.30 if 4 men together receive $ 3.40 more than 6 boys, what are the wages of each 25.
The wages
;
man and boy ? 26.
A grocer wishes to mix sugar at 8
sort at 5 cents a
What
pound.
pound
to
cents a
make 60 pounds
quantity of each must he take
27.
A
28.
A man walks
pound with another
to be sold at 6 cents a ?
cenain distance had he gone half a mile an hour faster, he would have walked it in. four-fifths of the time had he gone half a mile an hour slower, he would have been 2^ hours longer on the road find the distance. traveller walks a
;
:
3o miles panly at the rate of 4 miles an hour, he had walked at 5 miles an hour when he walked versa, he would have covered 2 miles more in the same find the time he was walking.
and partly at at 4, and rice time
:
5;
if
Two persons. 27 miles apart, setting out at the same time are together in 9 hours if they walk in the same direction, but in 3 hours find their rates of walking. if they walk in opposite directions 29.
:
When
a certain number of by 10. the result is the same as if doubled, and then diminished by times the sum of its digits by 18 30.
:
two
digits is doubled,
and increased
number had been reversed, and also the number itself exceeds 3
the 8;
find the
number.
lend a sum of money at 6 per cent, the interest for a certain time exceeds the loan by $ 1«» but if I lend it at 3 per cent, for a fourth of the time, the loan exceeds its interest by •$425. How much do I lend ? 31.
If I
;
A
but if he takes 3 hours longer than B to walk 30 nules 32. doubles his pace he takes 2 hours less time than B find their rates of walking. ;
:
CHAPTER XIX IXDETEKMTS'ATE AXD IMPOSSIBLE PbOBLEMS. MEA^-I^"G OF
TIVE ReSITLTS.
-.
— X
.
-. (»
XeGA-
— X
IXDETER^nXATE A>T) BIPOSSIBLE PEOBLEMS.
By
179.
reference to Art. 167,
single equation involving
it
will be seen that a
two unknown quantities
is satis-
by an indefinitely great number of sets of values of the unknowns involved, and that it is essential to have as many fied
equations expressing different, or independent conditions, If the as there are unknown quantities to be determined. conditions of a problem furnish a less number of independent equations than quantities to be determined, the problem is If, however, the conditions give said to be indeterminate.
us a greater numl;>er of independent equations than there are unknown quantities involved, the problem is impossible. Suppose the problem furnishes
From (2)
and
not
all
and
(1)
(3)
we
(2)
2x-\-y=
5,
x+i/=
3.
= o and y = — o. From and y = 1. These values canthe same time, hence the problem is we
obtain x
be true at
obtain x
=2
impossible.
XEGATTTE EESrXTS. 180.
When
A iciU
>>
A
old. and B*s age f^ three-fifths of A"s. be five times as old as B ?
40 years
165
ALGEBRA.
166
Let X represent the number of years that
Then
+ = 5 (24 + a;); 40 + = 120 + 5ic, x = - 20. this analysis, A will be five 40
.-.
or
B
According to in — 20 years.
loill
elapse.
a;
cc
The meaning
times as old as
of this result ought to be
at once evident to a thoughtful student.
Were
the result
any positive number of years, we would simply count that number forward from the present time (represented by the word '' is " in the problem) manifestly then the — 20 years Hence the problem should read, " A is refers to i)ast time. 40 years old, and B's age is three-fifths of A's. When ivas ;
" as old as B ? the problem read Suppose is 40 years old, and B's age
A five times A
is three-fifths
of A's
:
find
the time at which A's age is five times that of B. Let us assume that x years will elapse.
Then
40
+ = 5(24 -fa;); x = -20. a;
.-.
Interpreting this result,
we
see
that
we should have
assumed that x years had elapsed.
The student will notice that the word " will " in the first statement suggested that we should assume x as the number of years that ivould elapse, and that the negative result showed a fault in the enunciation of the problem but that the problem, as given in the next discussion, permitted us to make one of two possible suppositions as to the nature of the unknown quantity, so that the negative result indicates simply a wrong choice. Hence in the solution of problems involving equations of the first degree, negative residts indicate fault in the enunciation of the problem, or (1) ;
(2)
A A ivrong choice
heticeen tico possible suppositions, as to
the nature of the unlcnoimi quantity, allowed by the problem.
Generally it will be easy for the student to make such changes as will give an analogous possible problem.
INDETERMINATE AND IMPOSSIBLE PROBLEMS. 167
EXAMPLES
XIX.
Make such
necessary changes in the statements of the following problems as will render them possible arithmetically. 1.
A
27 years old and
is
as old as
B
B
15
;
how many
in
years will
A
be twice
?
2.
What
3.
If to the
are the two
sum
numbers whose
of twice a certain
difference
number and
is 50, ^
and sum 40 ? same num-
of the
is equal to twice the number. $ 400, and then finds that 6 times what he had at first is equal to 5 times what he has left. 5. What fraction is that which becomes | when 1 is subtracted from its numerator, and h when 1 is subtracted from its denominator? find the date 6. A is to-day 25 years old, and B's age is f of A's when A's age is twice that of B.
ber 10 be added, the result 4.
A man
loses
:
MEANING OF
^, O'
181. Meaning of -•
^, CO'
^,
^^ °o
Consider the fraction - in whicli
X
the numerator a has a certain fixed value, and the denominator X is a quantity subject to change; then it is clear that the smaller x becomes, the larger does the value of the fraction - become.
iL^iOa, 1
To
By making
For instance,
-^ = 1000
'
a, ^
1
— —= -
100000 a.
i_
10
T¥"00'
the denominator x sufficiently small, the value
of the fraction - can be
made
as large as
we
i^lease
;
that
x is, as the denominator x approaches to the value 0, the fraction becomes infinitely great. The symbol go is used to express a quantity infinitely great, or more shortly infinity. The given above, is sometimes written
full verbal statement,
a -= 182. Meaning of tor
—
•
If,
00.
in the fraction -, the denomina-
X gradually increases and finally becomes infinitely large,
—
^
ALGEBRA.
168
the fraction - becomes infinitely small
;
that
is,
as the de-
nominator of a fraction approaches to the value infinity, the fraction itself approaches to the value 0. This full verbal statement is sometimes written
« 00
183. Meaning of
form
in
The symbol - may be indeterminate
-•
Thus the value
or in fact.
when x
of
but by putting the fraction in the form
is -,
we
= 0.
see that the expression
is
equivalent to x rf^
= 2.
becomes 4 when x
V^'
Again,
X
(-^3
—a
=-
by putting the fraction in the form ^^^^
+ ''^^^~
we
see that the a~,
expression
when x =
is
''^
+ 2, which = a, but
when x
——
^^
X or 3
=2
equivalent to x^
tt
+ xa + a-,
These fractions assumed the form -
a.
under particular conditions, but it is evident that they do not necessarily have the same value.
On
the other hand, the symbol -
may show
that a value
Thus, solving in the regular is really indeterminate. the equations x-\-y + 2 = 0,
we
cret x ^
= ^^— = -,
2-2
and we can
0'
have any value whatever that the value of
a;
is
184. Meaning of -• 00
-
is
equally so of
—
if
we
easily ^ see that
way
x can
give y a value to suit, so
indeterminate.
Inasmuch
as
— = 0, OC
what
is
true of
;
;
;
CHAPTER XX. Involution. 185. Definition. Involution is the general repeating an expression as a factor, so as to find third, fourth, or any other power.
name its
for
second,
Invohition may always be effected by actual multipliHere, however, we shall give some rules for writing at once (1) any power of a monomial (2) the square and cube of any binomial (3) the square and cube of any multinomial (4) any power of a binomial expressed by a positive cation.
integer.
186. It (1)
(2)
is
evident from the Rule of Signs that
no even power of any quantity can be negative; any odd power of a quantity will have the same sign
as the quantity itself.
Note. It is especially worthy of notice that the square of every expression, whether positive or negative, is positive.
INVOLUTION OF MONOMIALS. 187.
From
definition
we
have, by the rules of multi-
plication,
(a^y
= (]C^.cC-.a^= a2+2+2 = a\
(- xy = (- y?) (- y?) = a;3+3 = x\ (- ay = (- a') (- a^)(- a') = - a'+'+'
(-3a'y=(- 3y(aY = 81 169
a''.
= - a'\
:
:
ALGEBRA.
170
Hence we obtain the following rule for raising a simple expression to any proposed power Rule.
(1)
Arithmetic,
liaise the coefficient to the required poioer by
and j)rejix
(2) Multiply the
the exponent
the 2yroper sign found by Art. 42. index of every factor of the expression by
of the power required.
Examples.
(-
(1)
2 x^)^
(2)
(-3a63)6
^^)
(S)^
= _ 32 a;^ = 729aG?;i8. 16
a4&i-2
81 xhj^
It will be seen that in the last case the numerator and the denominator are operated upon separately.
EXAMPLES XX.
a.
Write the square of each of the following expressions 1.
3a63.
5.
^xyz^.
2.
5xV-
6.
-f«263. '
3.
-2a6c2.
^
2^^.
_
g *
4.
11
6V.
3
^. 3a:V
*
9.-1^-
-2x1/2.
^o
_ 3«! 5 x3'
Write the cube of each of the following expressions 13.
2al)\
14-
3^^-
15.
-2a?c'^.
16.
-3a3&.
^^^ '^-
J_.
17 '
4c5x**
11.
3
?/3
:
§-^.
lo
-5^
:
19.
7:^3^4.
20.
-fa^.
3?/'2
Write the value of each of the following expressions 21.
{ZaVy.
24.
(-ly.
23.
Ui). -
(II)-
j
-^-^\\
26.
C^iL'V.
28.
-
(-f)-
- [-wr
I
TO SQUARE A BINOMIAL. 188.
r>y multiplication {ci
we have
+ hf = {ii-^h){a + h) = a- ^-2 ah + b= (a-6)(a-6) = a'^-2«6 +
Rule to the
I.
The square of
sum of their squares
.
.
.
.
(1),
6^'
(rt-6)2
the
sum of two
quantities
is
(2).
equal
increased by twice their product.
:
INVOLUTION. Rule equal
The sqnare of the difference of tim qucmtities is sum of their squares diminished by twice their
II.
to
171
the
product. Ex.1.
Ex.
2.
(x
(2 a3
+ 2yy = x^ + 2^x-2y-\-{2yy = x2 + 4ri/ + 4?/2.
_
3
^^2^2
189. These rules
^=(2 «3)2
_2
= 4 a6 _
12 a362
•
2 a^ 3 62 •
+ (3 52)2
+ 9 54.
may sometimes
be conveniently applied
to find the squares of numerical quantities.
Ex.
1.
The square
of 1012
= (1000 + 12)2 = (1000)2 ^ 2 1000 12 + (12)2 = 1000000 + 24000 + 144 = 1024144. .
Ex.
2.
The square
of 98
= (100 - 2)2 = (100)2 -2-100.2 = 10000 -400 + 4 = 9604.
.
+(2)2
TO SQUAEE A MULTINOMIAL. 190.
We may now
extend the rules of Art. 188 thus
= (a + by + 2(a + b)c + c' [Art. = a' +b'- + r-\-2ab-{-2ac-\-2 be. In the same way we
may
188,
Eule
1.]
prove
- b + cy = a^ + b' c^ - 2 ab 2ac - 2bc (a + 6 + c + dy =a' + b' + c'-^d' + 2 ab + 2 ac (a
-{-
-\-
-\-2ad-\-2bc-\-2bd-\-2cd.
In each instance we observe that the square consists of (1) the sum of the squares of the several terms of the given expression; (2) twice the sum of the products two and two of the several terras, taken with their proper signs that is, in each ;
:
ALGEBRA.
172
product the sign is + or — according as the quantities composing it have like or unlike signs. Note.
The square terms
are always positive.
The same laws hold whatever be the number
of terms in
the expression to be squared. Rule. To find the square of any multinomial : to the sum of the squares of the several terms add twice the product (with the proper sign) of each term into each of the terms that follow it. Ex.
1.
{x-1y-^zy = x'^+^y'^-\-^z'^-2'X'2y-2-x-Zz+2.1y'^z = x^ 4:if + 9 z^ - 4:xy - 6 xz + I2yz. -\-
Ex.2.
(l
+ 2ic-3a;2)2z=l
+ 4a:2+9xH2.1.2x-2.1. 3x2-2. 2ic. 30^2
= 1 +4x2 + 9x4 + 4x- 6x2 -12x3 = 1 + 4 X - 2 x2 - 12 x^ + 9 x*.
EXAMPLES XX.
b.
Write the square of each of the following expressions a + 36. a-Sb. a - h -
1.
2. 9.
3.
4. c.
1^.
12.
«+6-c. « + 26 + c. 2a-36 + 4c.
13.
x2-2/2-2;2.
10.
11.
x-5^. 2x + 3y. ,_
xy
+
—y
5.
Sx-y.
T.
9x-2y.
6.
3x +
S.
6ab-c.
yz
+
16.
2x +
17.
m-n-p-q.
3^/
«
'2 _
+
2 6
+
«
-
1Q 26. ^»20.
-.
4
— b.
X
-{-
^^
,
15.
,
a
zx.
5?/.
_§
«_q7,
2
3
|x2-x+|.
TO CUBE A BINOMIAL. 191.
By
actual multiplication, (a
we have
+ by = (a+h)(a-\-b)(a-{-b) = a^-{-3a'b-}-Sab'-^b^
(a-by=a'-3a'b-\-3ab''-b^ From these results we obtain the following
.
.
.
(1),
.
.
.
(2).
rule
:
—
of any binomial: take the cube of the first term, three times the square of the first by the second, three times the first by the square of the second, and the cube of Rule.
the last.
To find
the cube
:
:
173
INVOLUTION.
sum of the quantities, all signs will the difference of two quantities, the signs will be and , commencing with the first. alternately If the binomial be the
+
be
;
if
—
+
EXAMPLES XX.
c.
Write the cube of each of the following expressions
+ a. x-a. x-2y. 2 «& - 3 c. x
1.
2.
Z.
4.
5. 6. 7. 8.
x2
+
4y2.
_5 2 a3 _ 3 5 x^ - 4 w*. 4
^
^_2b^
i^-i,
^^
3
^/2.
ic2
^^
«
^
j^_^^ 3
^^
^
6
TO CUBE ANY MULTINOMIAL. 192. Consider a trinomial
(^ci-^b-^cy=la + (b + c)y = a'-i-3d\b + c)
+ 3a(b + cf + + cf = a^j^h^ + c^-{-3 a\h + c) + 3 lr{a + c) -\-3c\a-\-h) + Q>abc.
To cube a multinomial
(?>
: take the cube of each term, of each term by every other term, and The signs six times the product of every three different tenns. are determined by the laiv of signs for multiplication.
Rule.
three times the square
4.
a
+
bx
+
x^.
:
ALGEBRA.
174 In these (1)
Tlie
results,
spoken of as expansions, we notice that
number of terms equals
the index
of the binomial
plus one. (2) The exponent of a in the first term is the same as the index of the binomial, and decreases by one in each succeeding
term. (3) The quantity b appears for the first time in the second term of the expansion ivith an exponent 1, and its exponent increases by one in each succeeding term. (4) The coefficient of the first term is 1. (5) TJie coefficient of the second term is the same as the
index of the binomial. (6) The coefficient of ayiy term may be found by midtiplying the coefficient of the pireceding term by the exponent of a in that term, and dividing the result by the exponent of b plus 1. Ex.
1.
Expand
(a
+
hy.
[
:
INVOLUTION.
175
Performing the operations indicated, we have + 6a'b-3a'c-^12 ah'' - 12 ahc + 3 ac^ + 8 6^ - 12 bh
a^
A
Note.
full
Integral Index
is
discussion of the Binomial
given in Chapter
EXAMPLES XX. Expand
(x-Vy)^.
2-
{a-h)^.
3.
(2 a
4.
(3x+2?/)5.
5.
(2x-3&)5.
6.
(C2
+
for Positive
e.
the following expressions
1.
+
Theorem
XXXVII.
+
9.
(^x2
^^ 10.
ia%
_ 11
ahc [2 / 2a&c
^3)5.
14,
(^_^4.c)3.
16.
(a
+ &-2c)4.
17.
{a
+
18.
(a + 2
19.
(«
d^\^
6)6.
'
V ..
^3)6,
7.
(2a6-c2)5.
8.
(3a2&2_2c#)5.
2 c¥ 2c¥y '
3
/9,T
5 7.2
j
\ 5
-
z) (V"-!y Vx
13.
(a
+ 26 +
3c)3.
+
20.
Find the middle term
21.
Find the two middle terms of {x
22.
Find the term independent
of (a
h
+ c-
64
_2 6 + c-(^)4.
l)io.
—
of a in
(
ay.
c-2f?)3.
?/)ii.
—-—
•
J
;
CHAPTER
XXI.
Evolution'. 195. The root of any proposed expression is that qnantitr which being repeated as a factor the requisite number of
times produces the given expression. The operation of finding the root is :!ir iziTerse
196 .1
^
"
(Art. 14.) is
called Evolution:
it
of Involution.
^ Rule
of Signs vre see that
._y iven root of a ^positice
quantity
may
be either
positive or negative; (2)
(3)
no negative quantity can have an even root every odd root of a quantity has the same sign
quantity
as the
itself.
XoTE. It is especially worthy of notice that every positive quantity has two square roots equal in magnitude, but opposite in sign. ^(9<^Qfi) = -Zaj-\
Ex.
In the present chapter, however, we shall confine our attention to the positive root
EVOLmOX
OF MONOMIALS.
197. From a consideration of the following examples we be able to deduce a general rule for extracting any proposed root of a monomial will
Examples.
(1)
(2)
(3) (4)
V(«*^)= «*^
because
{(i^y=
cfib*.
^(- 2?) = - x» because (-ac»)» = -z®. ^(<«>) = c* because (c*)» = c». ^(81 x«) = 3 z5 because (3 x»)* = 81 x^. 176
:
:
:
EvoLUTioy. Rule.
and
(1)
itrf^jix
ITT
Find
the
the root of the coeMcient hy Arithmetic, proper sign found hy Art. 42.
(2) Divide the exponent of every factor of the expression by the index of the proposed root.
Example?.
(1)
(2)
^(_61z«)= - 4^2. ^(I6a^ = 2a^.
It will be seen that in the last case we numerator and the deniminator.
oi)erate separately upon, liie
EXAlklPLES XXI Write the square root of each of :lr 1.
Aa-h\
5.
81(i«&8.
2.
9x^y^.
6.
100 a*.
3.
2.5 jr*^.
7.
c^b^c^.
4.
IQa^l^.
8.
a?^Ari2,
:
:
11
a. :
^ng
expressions
324 r'^
g
2-56
^^
jQ
:?V
2S9p^
I'^Qjr
^
8lQ^g
"36?^
400
&l;riy^^
Write the cube root of each of the following expressions 13.
'2:€t^h^c\
14.
-8ai-^6^
15.
Mi^y^z^.
16.
-Z^Za^j^.
jg
87^
'II
2^
,.
17.
- :ri5v« ^^
125
x^
^-^
''^^"^ ':filP
„, 21.
19-
^i^^;:^
:>4:3 ;ri5
-T^^-^-
Write the value of each of the following espressions 22. 23. 24.
<(729a^-=?^).
^
-zi^^i^).
J28^
.
'^ "^
^^(2.56aSx«). ^
:
^^
\^rr'
^i^
^___ ^-
A^Sr
zvoLm<:)x of multes-o^iials. Since the 198. The Square Root of Any Multinomial. 6^. we have to discover a prosquare of a 2 a5 & is o^ cess by which a and h. the terms of the root, can be found when a- -{- 2 ab -\- Ir is g:iven. The first term, a, is the square root of o^. Arrange the terms according to powers of one letter a. and its square root is a. Set this
+
+
+
:
:
ALGEBRA.
178
down as the first term of the required root. Subtract a^ from the given expression and the remainder is 2 ab + b^ or (2a + b)xb. Thus, b, the second term of the root, will be the quotient when the remainder is divided by 2 a + b. This divisor consists of two terms 1. The double of a, the term of the root already found. 2. b, the new term itself. The work may be arranged
as follows
o^
2a-\-b
2 ab 2 ab
Ex.
1.
Find the square root
of
+ b^ +
?>2
9x^
9x^-i2xy +
49
—
42xy
-\-
i9y\
y^{Sx-7y
9^ 6x-7 y Explanation. term of the root.
-42xy + -42xy +
The square
49 y^
49 y^
root of 9 x^
jg
3
x,
and
By
this is the first
doubling this we obtain 6x, which is the first term of the Divide — 42 xy, the first term of the remainder, by 6 cc and we get —7y, the new term in the root, which has to be annexed both to the root and divisor. Next multiply the complete divisor by — 7 ?/ and subtract the result from the first remainder. There is now no remainder and the root has been found. The process can be extended so as to find the square root of any multinomial. The first two terms of the root will be obtained as before. When we have brought down the second remainder, the first part of the new divisor is obtained by doubling the terms of the root already found. We then divide the first term of the remainder by the first term of the new divisor, and set down the result as the next term in the root and in the divisor. We next multiply the complete divisor by the last term of the root and subtract the product from the last remainder. If there is now no remainder the root has been found if there is a remainder we continue the process. divisor.
;
Ex.
2.
Find the square root of 25 x2a2
_
12 xa^
+
IG x*
+
4 a^
-
24:X^a.
:
EVOLUTION. Rearrange in descending powers of 16x4
179
x.
-24x% + 25x2a2_
i2xa-3
+
4a4(4x2
12x«3
+
4a4
12
+
4 «4
- 3x«+
2a2
16x4 :2
-
3 xa
24
x% +
24x%f +
xa
25x2^2 9x2^2
16x-V-
2 a2
16
x2rt2
_
a-«3
Explanation. When we have obtained two terms in the — 3 xa, we have a remainder
root,
4x2
16x2a2_ i2xa3
+
4«4.
Double the terms of the root already found and place the
result,
8x2—
6x«, as the first part of the divisor. Divide 16x2^2^ the first term of the remainder, by 8x2, ^he first term of the divisor we get + 2 «2 which we annex both to the root and divisor. Now multiply There is no remainder and the complete divisor by 2 or- and subtract. ;
the root
is
found.
EXAMPLES Find the square root 1.
2. 3.
4. 5.
x2-10x?/
XXI.
b.
of each of the following expressions
+ 25?/2.
6.
4x4-12x3+29x2-30x + 25.
9x4 -12x3- 2x2 4-4X + 1. 8. x4 -4x3 + 6x2-4x+ 1. + ^2. 9. 4«4 + 4^3 _ 7^,2 _ 4^ 4. 25x2-30xy + 9?/2. 10. 1 - 10 x + 27 x2 - 10 x^ + x*. a4 - 2 a3 + 3 (^2 _ 2 a + 1. 11. 4x2 + 9^/2 + 25^2 4- I2x?/ -30?/^-20x^. 12. 16 x6 + 16 x^ - 4 x8 - 4x9 + x^. 13. x6 - 22 x4 + 34 x3 + 121 x2 - 374 x + 289. 14. 25 x4 - 30 ax3 + 49 dP-x'^ - 24 ciH + 16 a*. - 6 ?/202 + 9 z\ 15. 4 X* + 4 x2?/2 - 12 X202 + 4 a^hc + arl)^ + 4 «2c2 + 9 /^2c2 _ 12 a&c2. 16. 6 obH _ 12 c2a2 + 4 a4 + 4 a?-b'^. 6 62c2 + 9 c4 + 17. 18. 4 X4 + 9 + 13 x22/2 _ 6 X^3 _ 4 a:3y. 19. 1 -4x+ 10x2 -20x3 + 25x4-24x5 + 16x6. 20. 6 acx^ + 4 62^:4 + «2a;io + 9 c2 _ 12 6cx2 - 4 a&x^.
4x2-12x?/+ 81 x2
+
18
9?/2.
7.
xi/
_f.
?/4
?)4
?/4
199.
When
the expression whose root is required contains we may proceed as before, the fractional
fractional termS;
ALGEBRA.
180 part of the
Chapter
work being performed by the
rules explained in
xiii.
200. There is one important point to be observed when an expression contains powers of a certain letter and also powers of its reciprocal. Thus in the expression 2.r
+ i + 4+.r^+^+7ar^-f|, X
or
the order of descendinr/ powers
af
is
X
x^
X-
and the numerical quantity 4 stands between x and
-•
X
The reason
arrangement will appear in Chapter
for this
XXII. Ex. Find the square root of 24
-r
^^ -^^^- §?!. X'
Arrange the expression
->••-
X
in descending
y
'
X-
§J/_4
^+24 X 32?/
X
8^-8 + ?
8-
y
y-\ X
X
y-
powers of
y.
y
EVOLUTION.
EXAMPLES Find the square root 1.
of
XXI.
181
c.
each of the following expressions
:
ALGEBRA.
182
1. Three times the square of a, the term of the root already foimd. 2. Three times the product of this first term a and the
new term b. 3. The square
of
b.
The work may be arranged
as follows
a'
+ Sa'b -^ Sab' + b\a + b
r,3
3 (ay
3 X a (by
=3 a' xb =
=
1.
+ 3ab-^b'
Find the cube root
of
Sx^
SGx^y + 54 x?/'^ - 27 y\ - 36 x^y + 54 X7j^ - 27 y^(2 x-'Sy ?/3
-ISxy
±M
(-^yy= 12x2Explanation. term of the root.
+ 3ab'-\-b^
- 36x2?/ + 54x2/2- 27
=12x2
3x2xx(-3.v)=
3a'b
-
8 x3
S(2xy
+ 3ab'' + b^
±b^ 3a'
Ex.
3a'b -\-3ab
18x^/
The cube
+ 92/2 -36x2^ + 54x2/2-272/3 root of Sx-^
is
2x, and this
is
the
By
first
taking three times the square of this first term we obtain 12x2, is the first term of the divisor, and is called the "trial divisor." Divide — SQx^y, the first term of the remainder, by 12x2 and we get — 3 2/, the new term in the root. To complete the divisor, we first annex to the trial divisor three times the product of 2 x, the part of the root already found, and — 3 y, the new term of the root this is — 18x2/. We then annex tlie square of — oy, the new term, and the divisor is complete. We next multiply this divisor by the new term, and subtract the result from the first remainder. There is now no remainder and the root has been found. The process can be extended so as to find the cube root of any The first two terms of the root will be obtained as multinomial. When we have brought down the second remainder, we form before. the trial divisor by taking three times the square of the two terms of the root already found, and proceed as is shown in the following example.
which
:
EVOLUTIOJS".
183
.
ALGEBRA.
184
EXAMPLES Find the cube root 1.
3. 4. 5.
6. 7.
8. 9.
10. 11. 12.
:
XXI.
d.
of each of the following expressions
2. a^x^ - Z a^x'^y- + S axy* - y^. + I. + 108 ab^ - 27 b^. 1 + 3 + 6 x2 + 7 x3 + 6 x* + 3 ics + a;6. 1 - 6 X + 21 x2 - 44 x3 + 63 X* - 54 x5 + 27 x^. a^-\-6 a%-Z a^c-{-\2 ab'^-l^ a6c + 3 «c2+8 63_ 12 62^ + 5c2-c3. 8 a6 - 36 a5 + 66 a* - 63 ^3 + 33 ^^2 _ 9 + ]. 8 x6 + 12 x^ - 30 x* - 35 x^ + 45 a;2 + 27 x - 27. 27 xc - 54 x^a + 117 x^rt^ _ \\Q xhi^ + 117 xSa* _ 54 xcv' + 27 a^. 27 x6 _ 27 x5 - 18 X* + 17 x3 + 6 x2 - 3 X - 1. - 56 x^?/^. 24 x*!/2 + 96 x2?/i - 6 x^^ + x6 - 96 x?/^ + 64 216 + 342 x2 + 171 x* + 27 x^ - 27 x^ - 109 x^ - 108 x.
a^ -h S
64 «3
a"^ -\-
_
S a
144 «25
a;
^^
2/6
202. We add some examples of cube root where fractional terms occur in the given expressions. Ex.
Find the cube root
of 54
-
27 x^
+--
—
X^
X3
Arrange the expression in ascending powers of 8
36
+
x,
&4
-
27 x3 Vx2
12
3 X
x4
Vx2J
3xX2
(-3x)2
x(-3x)=
-
,
EVOLUTION.
'
?/3
27 „
g
10
12x2
a^
a^
64 gs
54 X
,
27
,io
a:2
x^
240 a
a
&
^^ -
108 a
48 a^
8 ^3
x
x^
x^
^q^
60 x
12 x^
a
«2
^
^
^Q^^
_
a3
3^2
6^
&2
90x2
yS
,
y2
3 52
53
(^2
(1(3
8x^
,
x^ ^
X
x^
6a
,
a
192 a2
6 6
y\
27
X
3
x3
X^
X^
18
x-^
9
X
y
2/2
r2
r^
185
108 X
yQ
_
27
j_
(jj3
48x_^ yb
y
The fourth root of an expres203. Some Higher Roots. is obtained by extracting the square root of the square root of the expression. Similarly by successive applications of the rule for finding the square root, we may find the eighth, sixteenth ••• root. The sixth root of an expression is found by taking the sion
cube root of the square root, or the square root of the cube root.
Similarly by combining the two processes for extraction and square roots, other higher roots may be ob-
of cube tained.
Ex.
1.
Find the fourth root of 81 x*
-
216 xhj
+
216 x'V
_
96 xy^
-f
16 yK
Extracting the square root by the rule we obtain 9 x^ — 12 x?/-f4y2 and hy inspection, the square root of this is 3 x — 2 2/, which is the required fourth root. ;
Ex.
By
2.
Find the sixth root of
inspection, the square root of this is
(--^3)-3(x-i), *
which may be written
x^
— 3xH
3
;
X
and the cube root of
this is
x
—
X
which
is
the required sixth root.
1 x=^
:
ALGEBRA.
186
We conclude the subject of higher roots by giving a rule, which depends upon the Binomial Theorem, for finding the nth root of any multinomial. (1) Arrange the terms according to the descending powers of some letter. (2) Take the nth root of the first term, and this ivill be the term of the
first
root.
When any number of terms
of the root have been found, subtract from the given midtinomial the nth x)ower of the part of the root cdready found, and divide the first term of l)th power of the first the remainder by n times the (ii (3)
—
TERM of the
root,
and
this will be the next
term of the
root.
204. AYhen an expression is not an exact square or cube, process of evolution, and obtain as
we may perform the
many terms To
Ex.
we
of the root as
please.
find four terms of the square root of 1 1
+
2
a:
-
2 ^2(1
2
ic
-
2 x2
X
-
3 x3
+
+
f x2
+
+
2x
—
2
x^.
f x3
1
2
+
a;
2x+ 2
+ 2x-
2
+
ic2
3x2 3 x2
2ic- 3^2
+
-
1x3
3x3-f.x4 3x3
Thus the required
result
is
1
+
x
—
EXAMPLES
|
+ x2
3x4
+
XXI.
| x^.
f.
Find the fourth roots of the following expressions
2.
- 28 x3 + 294x2- 1372 X + 16-32+24 8 ^J_
3.
a*
4.
1
5.
1
1.
x4
m
m'^
m^
240L
m*
+ 8 a^x + 16 x* + 32 ax^ + 24 a^x^. + 4 X + 2 x2 - 8 x3 - 5 x* + Sx-"^ + 2 x« - 4 x" + x8. + 8x + 20x2 + 8x3 -2Gx*-8x5 + 2Ux'i - Sx^ + x«
:
+
:
EVOLUTION.
187
Find the sixth roots of the following expressions
+6x+
6.
1
7.
x6
8.
a6
_
+
15x-^
+ 18 a^x +
12 ax5
+
20x3
135 «%2 _
240 ah:^
6x5 + «6. + 60 a^x^ - 160 a^^H 64 ««. 540 a'^x'^ + 1215 a%4 _ 1458 ax^ + 729 x^.
15x*
+
192 a^x
Find the eighth roots of the following expressions x8
9.
10.
- 8 x^?/ + 28 x6?/2 - 66 x^y^ + 70 x*?/* - 56 x^y^ +28 x^^e _ 8 x?/" + 2(p - l)x3 + (p2 _ 2p - l)x2 - 2(p - l)x + 1}K
y^.
{x4
Find to four terms the square root of a2
11.
_
Find
to three
l-6x + 21x2.
18.
Find the
fifth
aW _
ai'^
20.
«i'^
+
^2.
«4
13.
_
3^2.
14.
9
rt2
+
12 ax.
terms the cube root of
15.
19.
x^
12.
^,
16.27x6-27x5-18x4.
17.
64-48x +
9x2.
root of
_ -
-
+ 720 a* + 32. - 5 a9 + 20 aS - 50 «7 + 105 a^ - 161 «5 + 21O «* - 200 a^ + 160 a2 - 80 a + 32. 15 ^i - 10 a^ + 5 a9 + 5 «§ - 10 a^ _ 15 a6 4. 11 — 5rt2 -^ 5 a _ 1. 10 «9
^
50 «8
160 a7
+
S60 a^
640 a^
+
400 «2
592 «& 160 «
0^5 _f.
Before leaving 205. Square and Cube Root of Numbers. the subject of Evolution it may be useful to remark that the ordinary rules for extracting square and cube roots in
Arithmetic are based upon the algebraic methods we have explained in the present chapter. 1. Find the square root of 5329. Since 5329 lies between 4900 and 6400, that is between (70)2 and (80)2, its square root consists of two figures and lies between 70 and Hence, corresponding to «, tlie first term of the root in the alge80.
Ex.
we here have 70. The analogy between the algebraic and arithmetical methods be seen by comparing the cases we give below.
braic process of Art. 198,
a2
+
2 a6
+ b%a +
5329(70 4900
b
a2
2« +
6
2ah +
&2
+
&2
2 ah
140
+3=
143
429 429
+
3
=
will
73.
ALGEBRA.
18(8
Ex.
Find the square root
2.
Here 53824 and (300)-2.
lies
of 53824.
between 40000 and 90000, that a
2rt
+
6
.
.
.
.
.
.
+
30
=
53824(200 40000 430113824
460+
2
=
462
400
b
+
c
30
+
2
=
232
12900
1
+ 6)+c
2(a
between (200)^
is
1924 1924
Ex.
Find the cube root of 614125.
3.
lies between 512000 and 729000, that and (90)3. therefore its cube root consists of two between 80 and 90.
Smce 614125
3 a- = 3 X (80)2 = 3xrtx6 = 3x80x5= 5x5= b- =
is
between (80)3 and lies
figures
a
-r
b
614125(80 512000
+
5
=
85.
19200 102125 1200
25 20425 102125
206. We shall now show that in extracting either the square or the cube root of any number, when a certain number of figures have been obtained by the common rule, that number may be nearly doubled by ordinary division. 207.
If
figures,
the square root of a number consists of 2/7-1-1
when
the first
/?
+1
of these
ordinary method, the remaining n
Let
have been obtained by the
may be
X denote the given number
;
obtained by division.
a the part of the square
root already found, that is the first n -\-l figures found
the
common
rule,
with n ciphers annexed
;
by
x the remaining
part of the root.
Then
^y=a-^x', .-.
2a
X=a- + 2ax-hx^; 2a
^
EVOLUTION.
189
y—
a- is the remainder after n -\-l figures of the represented by a, have been found; and 2« is the We see from (1) divisor at the same stage of the work. — a- divided by 2 a gives x, the rest of the quotient that
Xo^v
root,
X
required, increased
—
We
bv -
shall
—
show that
is
a
2a p rope r fraction, so that by neglecting the remainder arising "
2a
from the division, we obtain x, the rest of the root. For X contains n figures, and therefore oi? contains 2 n figures at most also a is a number of 2 + 1 figures (the last n of which are ciphers) and thus 2 a contains 2 -f- 1 /<
;
/<
figures at least
From
;
and therefore
^
is
a proper fraction.
the above investigation, by putting a
=
1,
we
see
that two at^least of the figures of a square root must have been obtained in order that the method of division, used to obtain the next figure of the square root,
may
give that
figure correctly.
Ex.
Find the square root of 290 to
five places of decimals.
290(17.02 1
271190 1
3402
189 i
I
10000
6804 3196
Here we have obtained four figures in the square root by the ordinary method. Three more may be obtained by division only." using 2 X 1702, that is 3404, for divisor, and 3196 as remainder. Thus 3404)31960(938 30636 13240
10212
30280 27232
3048
ALGEBRA.
190
And
therefore to five places of decimals \/290
=
17.02938.
be noticed that in obtaining the second figure of the root, the division of 190 by 20 gives 9 for the next figure this is too great, and the figure 7 has to be obtained tentatively. It will
;
208.
when
If the
the
first
+ 2 figures, have been obtained by the ordi-
cube root of a number consists of 2/7 /?
+2
of these
may
nary method, the remaining n
be obtained by division.
Let" ^denote the given
number; a the part of the cube root already found, that is, the first n + 2 figures found by the common rule, with n ciphers annexed x the remaining ;
part of the root.
Then
-^N=a-\-X',
N= d" -\-3a-x + 3aaj2 -f
x^-,
^
— a^ is the remainder after n -\- 2 figures of the represented by a, have been found; and 3a- is the divisor at the same stage of the work. We see from (1) — a} divided by 3 a^ gives x, the rest of the quotient that Now
root,
N
2
required, increased by
3
— + -^ a
expression
is
We
.
shall
show that
this
3cr
a x>roper fraction, so that by neglecting the
remainder arising from the division, we obtain
x,
the rest of
the root.
By
supposition, x
102*^+^;
—1
therefore
and
;
^
—
than
is less is
less
than
1
and a
-——
:
is
greater than less
than
11 —
than
that
is,
10^"
QT^
is
less
than
o cl"
i-U
10",
77—77; l^ence
o X
— + —x^x^
is
3 X 10"+!' a 3ais therefore a proper fraction.
less
.,
-Lv'
;
that
is,
less
than ---\ 10 3 x 10"+^
,
and
;
CHAPTER
XXII.
The Theory of
Indices.
209. Hitherto all tlie definitions and rules with regard have been based upon the supposition that they were positive integers for instance,
to indices
;
(1)
(2) (3) (4)
The
= a- a- a-" to a'' X a" - a''+^ = a''. a'' ^ a' = a''-^ = a'\ (ay = a''''' = a''. a"
object of this
chapter
is
fourteen factors.
twofold;
first,
to
give
general proofs which shall establish the laws of combination in the case of positive integral indices
secondly, to
;
explain how, in strict accordance with these laws, intelligible meanings may be given to symbols whose indices are fractional, zero, or negative.
We
shall begin
by proving, directly from the definition
of a positive integral index, three important propositions.
m
is a 2Jositive integer, 210. Definition. When factors each equal to a. stands for the product of
a"*
m
211. Prop.
I.
To prove that
a*^
x
a"
= a"'+'\
when
m
and
n are positive integers.
= a a a to m factors = a- a- a-" to n factors rr X «"= (a- a- a-" to m factors) x (a a a = a a a to m n factors ^ «"*+", by definition.
By
definition,
a""
•
•
• • •
;
a""
.-.
•
'
•
• • •
-|-
191
•
• • •
to
n factors)
ALGEBRA.
192 Cor.
p
If
also a positive integer, then
is
cr X a" X
and
so for
any number of
c(F
= cr+'^+p
;
factors.
212. Prop. II. To prove that a"* -^ a" = a"""", when and n are positive integers, and m is greater than n. ^ _^
„
_ a'" _«•«•«•••
= a- a To prove
213. Prop. III.
7?i
factors
a' a- a-'- to n factors
cC"
/?
to
m
•
m—n
a •" to
that
(a'")"
factors
= a""S
when
m
and
are positive integers.
(ccy
= a'" = (a
•
•
a"'
•
a a •
a"* ••• • • •
to
to n factors
m factors) (a
the bracket being repeated
it
•
a
•
(t
• • •
to
m
factors)
• • •
times,
= a- a- a--= cr'*.
to m?i factors
214. These are the fundamental laws of combination of and they are proved directly from a definition which intelligible only on the supposition that the indices are
indices, is
positive
But
and integral. found convenient
it is
indices, such as
a%
a~'
-,
or,
to use fractional
more generally,
and negative a'',
these have at present no intelligible meaning.
and For the
a~";
definition of a"^ [Art. 210], upon which we based the three propositions just proved, is no longer applicable when is
m
fractional or negative.
Now
it is
whether positive or by the therefore determine meanings for symbols
important that
all indices,
negative, integral or fractional, should be governed
same laws.
We
p
such as
a',
a~", in the following
way
conform to the fundamental law,
a'"
:
x
we assume a"
= a"'"^%
that they
and accept
;
;
THE THEORY OF INDICES.
193
the meaning to which this assumption leads us. It will be found that the symbols so interpreted will also obey the other laws enunciated in Props, ii. and iii. p
215. To
meaning
find a
a% p and q being positive
for
inte-
gers.
Since n,
x
cC^
= a"*+"
a"
to be true for all values of
is
m
by replacing each of the indices
X
a' P
Similarly,
a'^
a'^
= a^
a^
P
x
P
X
a'^
X
that
a'
^,P
p
4, 5,
X
we have
^
2p
x
-,
and
= a'. a'^
= a^
^
'JP
=
>" q factors,
P
p
a^
*
= a''
Proceeding in this way for p
and n by
p.p
p
p
m
a''.
we have
gp
rt^ •••
to q factors
=
(a')'
is,
Therefore, by taking the
= a^
c(F.
qt\\ root,
p
p
or, in
words,
a'^
is
equal to 5
Examples.
=
(1)
x^
(2)
J=
l/a.
(4)
a
2
(5)
A:2
X k^
(6) 3«3?>^
n,
216. To
find a
Since
x
a"
a,
4«W =
X
meaning
by replacing the index a^
X
•^"+'*
2
=^2 + 3=^
= ft"'+" is
.-.
ofct/'.''
-l/x^.
4^=V4^ = V64 = 8. 5 2 2,5 3 a^ X a^ = «3 + 6 = «2.
(3)
a'"
qth root
'^tlie
for
30
.
12«^^^62+f
^ 12««6i
a'\
to be true for all values of
m a"
by
0,
we have
= a°+'' = a"
ft"z= — =
1.
a"
Hence any quantity with Ex.
x*-<^
o
X
a;*'^?'
zero index is equivalent to
= x*~'^+*^~* =
x*^
=L
1.
m
and
ALGEBRA.
194
n,
217. To
find a
Since
x
a'"
a"
meaning
= «"*+" is
by replacing the index a~
for a~'\
to be true for all values of
m
by
—
ii,
we have
??i
and
THE THEORY OF
195
INDICES.
220. The following examples will illustrate the different principles w^e have established.
Examples.
.-,.
—- = —Sx—•
3a-2
(1)
-
.^.
2a^ xa^
x6 oT^ ^ 4 ^
'
^
^
9a f,
(4) 2
-%3
V« + -^ +
i
+|-i+f - f
«^
=
+ 3 a^ +
2
a^
=
a^(5
a~2
=:5a2
+
EXAMPLES Express with positive indices 1.
2x i
:
(fi
^ 4 ^_i ^ ^ 3
3
% X a2
XXII.
a.
+
a^).
a2
3a
^
196
THE THEORY OF
n
;
197
INDICES.
222. To prove that (a6)" = a"6", whatever be the value a and b being any quantities whatever. Case
Let n be a positive
I.
Now
integer.
(ahy =:ab ab ab '•' to n factors = (a a a to n factors) (b -b -b •-•ton factors) = '
•
Case
•
•
• • •
Let n be a positive
II.
where p and q are positive
Now
we have {aby
integers, p
=
(ab)
*.
p
the gth power of {abf =l(aby\^={ab)p, [Art. 221.]
= a^6^ = (a'6*). Taking the gth
(ab)
root,
^
—=
= (aby = -J— = (aby
Hence the proposition
is
I.]
= «" 6^
Let n have any negative where r is positive,
(abY
[Case
p p
Case III.
— r,
a''b'\
Replacing n by -,
fraction.
p
by
of
value.
a'^-b-"
Replacing n
= a''b\
a^'b''
proved nniversally.
may
be expressed in a verbal form by saying that the index of a product may be distributed over its factors.
This resnlt
Note. sion.
An
index
Thus (a2
equal to ^cfi
Examples.
+
(1)
(2)
+
6^^
is
not distributive over
&2)2
is
not equal to a
and cannot be further
(yz')'*-''(zxy(xy)-^
{{a
- hYY^
X {{a
= +
tlie
+
h.
terms of an expres-
Again
y'^-'^ z'^-'^ z'^x'^x-'^y-'^
=
{a
-
6)(a
+
&)}-*'
?>)-^}'
= {(« -
{a?-
+
&"2)^
is
simplified.
—
?/«-2c^«.
+ 5)_ 52y-w
hy^^ x {a
=
(a2
*'
223. Since in the proof of Art. 222 the qnantities a b are ivholly imrestricted, they may themselves involve
and
indices.
Examples.
114 (x^if^)^ -^(x^y'^)
(1)
=
_i ^
2.2 _2.i4. =xHj~^ -^x ^y^=x^ij-'^.
(a3& 2-f-Va2&-3)6= (a35 2^^^^ 2)6
^
(a.3)6^(,^.
198
ALGEBRA.
EXAMPLES
XXII.
b.
Simplify and express with positive indices 1.
(Va^y.
2.
(Vx^^)-K
3.
(a:«i/-»)3x(rV)-«.
:
^/^/^K
8.
8a-3J 10.
W(^"¥: 13.
Vo^
14.
V^aft-ic--^
15.
y/c^^ X («4-i)-^
16.
^x-^^y^
17.
(«"^^r.)-3 X Vx-^y/a-^-
(cc-h«/a;)~.
11.
(X X
12.
(^x*--
:!yx n)l-n.
^xy^'
X Vrt6^. 27.
_1
X (a-i6-2c-4)
28. -f-
vV^J^.
f
5/^^l:^x^^/^«^V'
v/(66
-
^
^6)
a353^-i n
1
V^a^^+^P^^
19.
^(a +
20.
{(a;-?/)-3}«-{(x
2L
6)5
(a«'-i)'^i
+
30.
(.xw+i)'*^-!
_]_
(an 5"n)ft.
18.
--
29.
1
X (a
+
?>)"i
+
«^-
?/)"}3.
X
31.
«2(p-?) }"•
{ i/«^
f^-%fii^y.
2/"
fxr^\ 23.
(a"M^aa;"^^a;^)i
24.
v'Ca
+
6)6
X («2
-
lf~y^.
(i/x-^y^
'^
34.
X' Z 25.
(S)'-(%^)""
35.
-I *x 26.
2n X (2"-i)" X 2«+i X 2"-i 4-" 2»»+i
\^;-
(2h)"-i
4n+l »t— lAn-f (2«-1)
1
224. Since the index-laws are universally true, all the ordinary operations of multiplication, division, involution, and evolution are applicable to expressions which contain fractional
and negative
indices.
33
THE THEORY OF
INDICES.
199
225. In Art. 200, we pointed out that the descenclin< powers of x are i
.3
A
reason
seen
is
we
if
Ex.
Multiply 3 x~3
1.
Arrange
+
Divide 16 «-3
2.
2a-i
+
/\%L
X
+
/)»0
2
1 XT
/)»— 1
a:*
powers
in descending
/->»—-
by x^
+2x'^ -\-Sx~^
x^
-2
-
+2x +3 -2x - 4 4a:3 +3 6 «-2
+
/)>
-
—
2.
of x.
X
x^
Ex.
1 X-
write these terms in the form
/\»-
r\iO
1
X
a^s
_
.J-
6a;
5a-i
+
5 a-i
1)16«-
6
6
by
3
1+2 a-\
+
a-i
+6
16a-3+ 8a-2 14
«-^- 7a-i 12a-i 12 a-i
Ex.
Find the square root
3.
4x2 y
Use
+6 +6
of
^^ _2x + ^+a;3-4 4
y-',
fractional indices,
and arrange
V(^^^"^)-
in descending
x^-ixHj~^-\-ixhj--^+x^y^-2x
2x2
-2xy
2
—
4 x?/
2
+ ^{(x^-2xf^+^
1
?/
x^^/2-2x+| -f 3
1
x^l/2-2x Note.
x.
4 X - ?/ 2-1-4 x2?/-i 5 _i 4x-y 2^ 4^2^-1 3
2 x'-^
powers of
In this example
it
?/
+4
should be observed that the introduction
of negative indices enables us to avoid the use of algebraic fractions.
—
•
ALGEBRA.
200
EXAMPLES -5+
1.
Multiply 3x3
2.
Multiply
3.
Find the product of
4.
Find the product of
5.
Divide 21 x
6.
Divide
3
7.
Divide
8.
Divide
9.
Divide
10.
Divide
1
XXII.
c.
+ 3x"i _1 _3 + a ^ -Ga ^. c* + 2 c^* — 7 and 5 — 3 c^* + 5 + 2 x^<^ + 3 a;~2« and 4 x« — 3
8x~3 by 4x^
3a^-4a^-a
_1 ^
1
by 3a^
2 a*
c=^.
-«•
+ x^" + x^ + 1 by 3x3 + 1. 15 a - 3 a^ - 2 a~3 + 8 a-i by 5 a^ + 4. 16 cr^ + 6 a-2 + 5 «-i - G by 2 a-i - 1. 5 63 _ 6 6^ - 4 6"3~ - 4 &"3 - 5 by 6^-2 6"i 21 a^x + 20 - 27 a* - 26 a'^^ by 3 a^ - 5. 8 c-" - 8 c» + 5 c^'* - 3 c-^" by 5 c» - 3 c-».
Find the square root of
9x-
14.
+ 10-4x"^ + x-i. + 16-30a-24a^ + 49ai 4x« + 9x-« + 28-24x~^- 16x^. 12 a=* + 4 - 6 «3x + f^4x + 5
15.
Multiply a^
16.
Multiply
17.
Multiply 2 \/a5
11.
12. 13.
12x^
25a^
^20:,
1
-
-
+ 4 a"^ - 2 a^ by ^x - 2 x^ by 1 - {/x.
4
-
- a "3.
8 a"^
2
--
rt^
by
-
2 a
3^/-
Ma
a 18.
Divide \/x^
19.
Divide
20.
Divide 4
1
+
-
2 x^
- V« -
W-
16 x~^
1
8 x3
-
by
a:^
X
A+
9,
-—
2 «2 by
1
+
4 x"^
a^
+
+
— V^
5
2
+ ir. + 3x~3
by 2 x^^
x/x
Find the square root of 21.
9x-4-18x-3Vy+^-6^(^2) +
22.
4Vx3 - 12^(.x3|/)+ 25Vi/ - 24i^(^^) + 16x"'>
23.
8lf^^+l'\ + V
24.
y-
^"%l + 16
J
^
^y^
36^(xV^-l)y/y
+ L=jiA^-0^y. 2x
4 rt"i
- rA
1ft
5
«"^' 4-
2''-
158^^y
Q
- '^'x - -^. ^x
"
THE THEORY OF INDICES.
201
The following examples will illustrate the formulce when applied to expressions involving
226.
of earlier chapters fractional
and negative h
indices.
h
p
h
p
h
h
p
^
h p
=
-
1
h
P
a^b~^
-
1
__h p
_p ^
-a
n>^.
- a;? + 3 by 2 x^-p + xp - S. = {2 x^p - {xp - 3)} {2 x^p + (x^' - 3)} = (2 x^p)'^ - (xi' - 3)2 =z 4 x^p -x~p + GxP-
Multiply 2 x^p
2.
The product
Ex.
P
«-Z>^~«
p
h
+
a'h'^
= a% Ex.
_P
= a''~''-a''b^-\-ah
Ex.1. {a^-¥){ci"^-^b~~^)
3.
The square
^
9a;
+ 4 + x-i -
=
9X
+
= 9X -
4
+
-
of 3 x^
2
-
12 x^
-
2
+
6
+
4 x"^"
+
x-i.
+a
The quotient
=:{a^
^ ) -^
a
n
= (a^)2 _
+
(a^
a
.
-4-
«
sc"^
^)
_n
(aVa
')
+ (a'^)'-^ =
EXAMPLES
.
-.
w
^^ rt~^
2
.
_
n
= {(«')' + («" 2)3}
+2
n
+
"
3n
+
•
4 x"^
by a^
^
Sn
3x* x"^
n
3n
a^
.
-
3n
Ex.4. Divide
-2
.
x-i
12 x2
- x~^
2
3x^ 2
.
9.
XXII.
an
-
1
+
«-«.
d.
Write the value of
2.
(x^-7)(x^ + 3). (4x-5x-i)(4x +
5.
{a--2a-'y.
8.
(5 x«?/*
9.
(ia3-a~^)2.
1.
3x-i).
3.
(7x-9y-'^)(lx
4.
(x»»
-
i/'OC^^'*"
+
-
3 x-«y-*) (4 x«^^
+
10.
(«-
+
9y-^).
2/"")-
a
1
6.
+
ax)2.
7.
(a:2_ix-«)2.
5 x-«?/-*). (3x«?/-^
+
5x-''«/^)(3x«i/*- 5x-«2/-*).
—
.
202
ALGEBRA. «"")^1
12.
13.
{(a
+
Z>)^
+ (a-6)^2.
14.
{(a
+
6)^
- (a - h)~^
1
(xa -x~~a
+
xy.
Write the quotient of 15. 16.
20.
1-8 a-3
xt-27byx^-3.
21.
a^^-x6bya2x +
^2.
a:"*
23.
x^
24.
x^"
a:
-
by
9a
17.
«2^-
18.
x^^
19.
c2x
ax
by
16
(ic
+
ic^
-
4)
1
26.
(2a;3+4 + 3a;
27.
(2-a:3
28.
(ax
+
7
a^ 2
a3
-4.
+ 8 by a:« + 2. _ c-x by c- - c-1.
Find the vahie 25.
+ 3 a^
cc^
+
(a;
+
+
x^
_1
4).
g^ 1
1
3)(2x3+4-3x
3 a-x) (ax
3,
by
:«-i
-Wij +
x^^
+
-
32 by x^
2 a-i. ^3. 1.
1.
+
2.
_
7
_
2
+
4&3
x-Ta;^
^
32.
v""
x^
-
4
^^-^
^x^
+
4
+ 4 x~^
g^ 3 cr^)
U
X — b-^/X —
^).
-SaH
2\/a6
1
1
of
+x)(2+xHx). +
-
by
^,,7 + ah - h^
ft-
ah
y/a
V«
y/x)
CHAPTER
XXIII.
Surds (Radicals). 227.
A
surd
is
an indicated root which cannot be exactly
obtained.
Thus V^j
By
a/5? i/(^^
Va-
+
0'
are surds.
reference to the preceding chapter
it Avill
be seen that
these are only cases of fractional indices; for the above quantities might be written
may always
be expressed as quantities with they are subject to the same laws of combination as other algebraic symbols. Since surds
fractional indices
228. A surd is sometimes called an irrational quantity and quantities which are not surds are, for the sake of distinction, termed rational quantities. ;
This Surds are sometimes spoken of as radicals. also applied to quantities such as Va^, V9, V27, which are, however, rational quantities in surd form.
229.
term etc.,
is
230. The order of a surd or surd index.
Thus
-s/x,
is
^a
indicated by the root symbol, are respectively surds of the
third and nth. orders.
The surds
of the
the second order
Thus
-y/3,
;
most frequent occurrence are those of
they are sometimes called quadratic surds.
^a, a/x
+ y are
231. A mixed surd can be extracted.
is
quadratic surds.
one containing a factor whose root 203
ALGEBKA.
204
This factor can evidently be removed and its root placed before the radical as a coefficient. It is called the rational factor, and the factor whose root cannot be extracted is called the irrational factor.
232. AVhen the coefficient of the surd
is
unity,
it is
said
to be entire.
233.
When the irrational factor is integral, and all rational
factors have been removed, the surd
234.
When
is
in its simplest form.
surds of the same order contain the
same
irrational factor, they are said to be similar or like.
^^3 are like 2^3 are imlike
Thus
5V3, 2^3,
surds.
But
3^2
surds.
and
235. In the case of numerical surds such as ^2, ^5, •••, although the exact value can never be found, it can be determined to any degree of accuracy by carrying the process of evolution far enough.
V^ = 2.236068..-;
Thus that
is
V^
lies
between 2.23606 and 2.23607; and therefore
the error in using either of these quantities instead of y'5 is By taking the root to a greater number less than .00001. of decimal places we can approximate still nearer to the true value. It thus appears that it will never be absolutely necessary to introduce surds into numerical work, which can always be carried on to a certain degree of accuracy but we shall in the present chapter prove laws for combination of surd quantities which will enable us to work with symbols such as ^2, ^5, -^a, ... with absolute accuracy so long as the symbols are kept in their surd form. Moreover it will be found that even where approximate numerical results are required, the work is considerably simplified and shortened ;
by operating with surd symbols, and afterwards substituting numerical values,
if
necessary.
205
SURDS.
REDUCTION OF SURDS. 236. Transformation of Surds Different Order having the
may tiie
Same
of
Any
Order into Surds of a surd of any order
A
Value.
be transformed into a surd of a different order having Such, surds are said to be equivalent. value.
same
Examples.
(1)
^2 =
=
2^2
^z \/2^.
(2)
p/a^aP-
api
= ^^a?.
2^
may
Surds of different orders
237.
therefore be trans-
This order may be multiple of each of the given orders, but it is usually most convenient to choose the least common mul-
formed into surds of the same order.
ayiy
common
tiple.
Ex. Express ^a^,
The
least
^a^
as surds of the
multiple of
given surds as surds of
to
same lowest order. and expressing the the twelfth order they become ^^a^, ^^h^,
-^b'^,
common
4,
3,
6
12
is
;
238. Surds of different orders may be arranged according magnitude by transforming them into surds of the same
order.
Ex.
Arrange ^3, ^6, ^10 according to magnitude.
and, expressing the least common multiple of 2, 3, 4 is 12 given surds as surds of the twelfth order, we have
The
;
V3 =736 =1^729, ^10 = ^^103 = Hence arranged
3^1000.
in ascending order of
V3,
EXAMPLES
magnitude the surds are
^6.
XXIII.
a.
Express as surds of the twelfth order with positive indices 1.
x^.
2.
«-i
4.
-L.
5.
-J-.
-^
a~2,
3.
\/«a-3
6.
Ill-,-
:
^ Vcr^x'^.
ALGEBRA.
206 Express as surds of the
V^K
7.
8.
J
10.
11.
^^„^-
-13.
12.
^.
14.
1-
aK
9.
ie«.
order with positive indices
wtli
-,r
Express as surds of the same lowest order
:
x"2-
-^-^.
:
15.
V«>
v^«^-
18-
y/^S
21.
V5, ^11, ^13.
16.
^d\ ^x\
y/a.
19.
\/«3p, \/a6.
22.
^8, V^,
20.
V^^
23.
^2, ^8, ^4.
17.
^x6,
-^0:5.
'V^^""-
'^^9^.
239. Reduction of a Surd to of
any expression
is
tile
= ^a'^b. Vahc = ^a -y/b
Similarly, so for
•
any number of
Examples.
root
expression.
^b = (aby =ah\
For
The
Simplest Form.
equal to the product of the roots of
the separate factors of
and
its
v/6.
•
-^c
[Art. 222.]
;
factors.
(1)
^15
=^3
(2)
^a^b
= -^a^ .^b =
(3)
y/60
=y/25.y/2=by/2.
•
*/5.
a'^^b.
appears that a surd may sometimes be expressed as the product of a rational quantity and a surd when the surd factor is
Hence
it
;
integral
and as small as
possible, the surd is in its simplest
form
[Art,
233].
Thus the shnplest form
of y/\2^ is 8y'2. Conversely, the coefficient of a surd may be brought under the radical sign by first raising it to the power whose root the surd expresses, and then placing the product of this power and the surd factor under the radical sign.
Examples.
(1) (2)
In this form a surd
By
is
•
said to be an entire surd [Art. 232].
same method any rational quantity may be expressed a surd. Thus 2 may be written as Vi, and 3 as v27.
the
form of
V^ = V^^ V^ = \/245. a^h = -^a^ -^b^Vcfih. 7
in the
SURDS.
207
When
the surd has the form of a fraction, we mulnumerator and denominator by such a quantity as will make the denominator a perfect power of the same degree as the surd, and then take out the rational factor as
240.
tiply both
a coefficient. Examples.
VJ = Vf = V2
(1)
Imx
^oN
_
jabnix
_
1.
V288.
2.
^m.
3. >.
5.
i\/2.
1
EXAMPLES Express in the simplest form
=
x i
,
XXIII.
b.
:
6.
2V720.
10.
v^-
Y
5V245. ^
11.
\/36^.
2187.
i^ 15.
^256.
3V150.
17.
v/- 108 x*r^.
20.
Va3 +
2
ff'-^6
8-
v'1029.
12.
V27«355.
9-
13.
V|.
+
21.
\/8
x-^?/
23.
14 V5.
25.
5^6^
26.
tVV-V-
27.
V2. 29.
«
30. ""•
33.
34.
\ ^2
24
x'^i/S
-
8
xij^.
:
«-c/^^ ^
a
^27x4.
3^
Vx»+^?/2i'.
19.
- 24 x^ +
J3^.
x2 \
?>
.^^^
16.
\/x3»?/-"+^-
18. a6--^.
>'
Express (1) as entire surds, (2) in simplest form 11
Vl^
,
T..OO
22.
^.
14.
^«^-'
l_^E:r X»*
'
?/3
2c >'9a25
28.
ADDITION AND SUBTRACTION OF SURDS. 241. To add and subtract like surds simplest form,
sum
and
prefix to their
o/l the coefficients.
:
Reduce them
common
to their
irrational part the
ALGEBRA.
208
Ex.
1.
The sum
of 3
= Ex.
2.
The
vo +
6
—
V20, 4 ^5, 4
5
Vo + J V5 = ^ 5
V5.
+ y y/- y^a - z yf^ = x-2x-^a-^y{-ij)^a- z- z -^'a = (2x'^-y2-z')^a.
sura of x y/8x^a
242. Unlike surds cannot be
collect ed.
Thus the sum of 5-^2, -2^3, and v6 and cannot be further simplified.
5^2-2^3
is
-{-\/6,
EXAMPLES Find the value
V45 - V20 + 7 y/o. V63 + 5 V7 - 8V28.
1.
3
2.
4
3.
^U - 5 V 176 + 2 V^O.
5.
2 V363 - 5 V243 + V192. 2^189 + 3 ^875 - 7 ^56.
6.
5
4.
XXIII.
^81 -
7
c.
of
+
3 */l62
8.
5
^=^54 -
9.
4
Vl28 + 4 V75 -
7 */S2
^1250.
2 ^^^"^^16
+4
5 vl62.
V24 -
2 V54 - V^- V294 - 48 Vi3 Vl47 - | Vj - ViV 5
10. 11.
^192 4 4 ^648.
-
7.
V2'^2
12.
MULTIPLICATION OF SURDS. 243. To multiply two surds separately the rational factors
of the
and
1
11
by""
= abx'Y
1
For
a ^x x
b
^y = ax''
same order: Multiply
the irrational factors.
x
1
= ab (xyy = ab Vxy. Examples.
V^
x 3
^7 =
15 V21-
(1)
5
(2)
2^x-x2>y/x = Qx.
(3)
VcT^ =
^(a + 6)(a-&)= ^«2 _
If the surds are not in their simplest form,
labor to reduce
=
them
to this
.
will save
form before multiplication.
Ex. The product of 5^32, V^S, 5 4 V2 X 4 V3 X 2 3 ve = 480 •
it
/>2.
2V54 •
V2
•
V3
•
V<5
=
480 X 6
=
2880.
SURDS. 244. To multiply surds
different orders
Reduce them
:
surds of the same order, and proceed as before.
to equivalent
Multiply 5
Ex.
of
209
The product
=
^2 by ^22 x
5
2 y/5. 2
^o^
=
10
EXAMPLES
^2^ x
5^
=
10 ^500.
XXIII. d.
Find the value of 1.
2V14XV21.
2.
3V8XV6.
6.
^e/^T2 X ^:^^2.
7.
Vmx^Ul.
4.
2V15X3V5.
8.
5VI28X2V^.
5.
8Vl2x3v'24.
9.
aVb^ x
b'^V^.
2^xv/3
j^j '
'
'
,-
x^x
3
—
\2
a*
1 aM x -^''^•
12.
DIVISION OF SURDS. 245. Suppose it the quotient when .
At
first
root of 5,
which
is
is
required to find the numerical value of is divided by ^7.
^5
would seem that we must find the square 2.236- ••, and then the square root of 7, 2.645..., and finally divide 2.236-.. by 2.645-..; sight
which
it
is
three troublesome operations.
But we may avoid much of this labor by multiplying both numerator and denominator by ^7, so as to make the denominator a rational quantity. Thus
V7"
7
Now
= .845... 246. The great utility of this artifice in calculating the numerical value of surd fractions suggests its convenience in the case of all surd fractions, even where numerical p
:
ALGEBRA.
210
Thus
values are not required.
it is
as follows
usual to simplify ^!l^^ v^
a^b _ a-y/b x ^c _ aVbc The process by which surds are removed from the denominator of any fraction is known as rationalizing the denominaIt is effected by multiplying both numerator and denominator by any factor which renders the denominator
tor.
rational.
We
shall return to this point in Art. 250.
247. To divide surds
:
Express
the result as
a fraction and
rationalize the denominator.
Ex.
1.
Divide4V75by 25V56.
^JVIJ,
The quotient
4 x
SyS
_2y3
25x2Vl4~5Vl4 _ 2 yS X V14 _ 2 V42 _ V42 ~5V14 X V14~5 X 14~ 35 ^h _ ^hx ^c _ y/bG__ ^/bc 25V56
*
Ex.
2.
EXAMPLES XXIIL
e.
Find the value of
V10-V2.
4.
21V384--8V98.
2.
3V7-^2\/8.
5.
5V27--3V24.
3.
2V120 - v3.
6.
-13V125--5V65.
1.
^
Given
^7 = 11.
3V48 5V112
^2 =
2.64575
ii.
:
_
6V84^
1.41421,
^.
3^11 *
5
2V98^7V22*
_3_^f2^_,r^8^ a-b'^a-b ^a — b >
v^ =
Gy/U^2y/2l.
1.73205,
V^ =
>'((a
2.23607,
-
6)^
^6 =
2.44949,
find to four decimal places the numerical value of 14.
V2 12.
10.
V392
7.
ii.
18.
V6 15.
V^'
-i-.
21.
2v3
-^^. \/5
19. '"'
^_.
^
22 *""
v'oOO"
^
16.
144
V7*
17.
V2-V3.
^V«.
2, "
V243'
25
V252
23.
/25C '256 \1575'
_A,. 2^96
SURDS.
211
COMPOUND SURDS. 248. Hitherto
we have
confined our attention to simple
surds, sucli as -^5, -^a, V.^•
two or more simple surds
— 3-y/b;
2-y/a
-^'a
+ ^b
+ y- An
is
are
called a
expression involving
compound surd
compound
A
surds.
which has a surd in one or both of the terms,
;
thus
binomial, is
called a
binomial surd.
249. Multiplication of Compound Surds.
compound
in the multiplication of
Ex.
1.
Multiply
The product Ex.
2.
=
2^x-6
3v'x(2 V»^
by S^x.
-5) = 6x- 15^x.
+ 3 ^x by V^ - V-^= (2 ^5 + 3 ^x) ( V^ - V-'^") = 2^5 V5 + 3v'5 ^x - 2 V5 '
= 3.
•
•
y/x
-
Find the square
=
of
2^x + \/7 —
7
1.
3.
4. 5. 6. 7.
•
^x
4x. •
\/7-4x
+ 4\/7x-4x2.
EXAMPLES Find the value
3y/x
10 -3x-\- V5x.
(2^x + V7 -4x)2 = (2^xy + (V7-4x)2 + 4 ^x = 4x + 7 -4x + 4V7x -4x=2
2.
proceed as
Multiply 2 ^5
The product
Ex.
We
algebraic expressions.
XXIII.
f.
of
8. (S^a-2y/x)(2^a + S-^x). - v'«) X 2 V^9- (V^ + Vx- 1) X Vx- 1. 10(v/a + V^) X ^«^( Vx + « - Vx - «) X Vx + «. (Vx + - l)x Vx + 11. (Vrt + X - 2Va)2. 12. (2V«-Vl + 4«)2. (2V3 + 3V2)2. 13. (V^Tf^ - Vl^^)2. + (V7 5V3)(2V7-4V3). 14. ( V^Tf^ - 2)( V^+^ - 1). + (3v'5-4v2)(2Vo 3V2). 15. (V2 + V3-V^)(V2+V3+V5)16. (V5 + 3V2+V7)(V^'> + 3V2-V7).
(iS^x-5)x2^x.
(
V»'
2/
?/.
ALGEBRA.
212 Write the
s(iuare of
- \/2 - «. y/v? + 2 + Vwi - n.
17.
V2 +
18.
Vx2 -2y^-{-
19.
V?/i
«
a;
+
?i
250. If
+ _ 2y/a^ - b^. 3xV2 - 3V7^V4 x2 + 1 - V4 x2 - 1. SVa'^
20.
a;
y'^.
21.
2a:'-2.
22.
we multiply
])2
together the
sum and
the difference
of any two quadratic surds, we obtain a rational ^product. This result should be carefully noted.
V« + \/?>) ( V« - V&) = ( V«)'^ - (V^O^ ^a-h. - 48 - - 3. (2) (3V5 + 4V3)(3V5 - 4V3) = (3V5)2 -(4V3)2 = 45 Similarly, (4- Va + 6)(4+ Va + 6) = (4)2-( V« + ?>)2 = 16-«-6. Examples.
(1)
(
251. Definition. When two binomial quadratic surds only in the sign which connects their terms, they are said to be conjugate. differ
Thus
o-yjl
+ 5 Vll is conjugate to 2>^1 — 5 ^11. — Va^ — x^ is conjugate to a + Va^ — x^.
Similarly, a
The product of two conjugate surds
is
Ex,
[Art. 250.]
rational.
(S^a-\-Vx-9 a) (3 ^a - Vx-9a) z=(3^ay -(y/x-9ay = 9a -(x - 9a)=^
ISa
-
x.
252. Division of Compound Surds. If the divisor is a binomial quadratic surd, express the division by means of a fraction, and rationalize the denominator by multiplying numerator and denominator by the surd which is conjugate to the divisor.
Ex.
1.
Divide 4
The quotient ^
Ex.
2.
+ 3 V2
by
5
- 3^2.
= 1+^ = '-±^
x |+
|^
5 + 3V2 + 20+18 12^2 15^2 + ^ 38 + 27 V2 ^ 25-18 1
5-3V2
5-3V2
nationalize the denominator of
V«2 ^b--i- a Tlie expression
=
—^t= Va^
4- 62
a
—:== _ y/(fi
+
^-2
rt
+ 62_,,|^ V^zM^2 _ + b'^) - a2
?/2|V^2 (a2
x _i_
a.
' ,
I
SURDS. Ex.
8.
213
^ by --—^.
Divide ^^
^^^xf^:^
Thequotient
(V3)-2-(V2)^
.
14_12 + 8V3- 7V3 = 2 — y'S, on rationalizing. 2 + ^3 Ex.
4.
Given
^6 =
87
2.236068, find the value of
7-2V5
Kationalizino; the denominator, Dininator,
_87(7 + 2V5)
87
7-2^5
49-20
= 3(7 + 2V5) = 34.416408. It will
be seen that by rationalizing the denominator we have
avoided the use of a divisor consisting of
7 figures.
253. In a similar manner, where the denominator involves we may by two operations render that
three quadratic surds,
denominator
rational.
V2 V2 + V3 - V5 V^C V2 + V^ + V^) The expression( V'2 + V3 - V5)(V2 + V3 + V^) ^ 2 + v6 + yio ^ (2 + V6 + yio) V6 2V6 (2V6;V6 ^ 3+V6 + Vl5
Ex.
6
EXAMPLES Find the value 1.
2. 3.
XXIII.
g.
of
(9V2-7)(9V2+7). (3 + 5V7)(3-5V7).
4. 5.
(2Vll + 5V2)(2Vll-'V2). -}- 2^b)(^a - 2^b). (3c - 2 Va:)(3c + 2 Vx).
(^a
6. (5V8-2v7)(5v'8 + 2v7). 7. (Va + x—^a)(Va-{-x-]-y/a). 8. (V2i9 + 3g - 2^q)(V2p +3g +
9.
XO.
2 V^).
+ V« — ic) ( Va + x — V« — a-). (5 Vx2 _ 3 + 7 «) (5 Vx2 - 3 -7a).
(
V« +
X
2/2
1/'-^
ALGEBRA.
214 11.
12.
13. 14. 15.
20-(lH-3v7). 17- (3V7 + 2V3).
16.
(3
v«
17.
(3V2-1)^(3V2+1). (2V3 + 7V2)^(5V3-4V2). {2x-y/xy)^{2\'xy - y).
+ V5)(V5-2)^(5-V5).
18.
Kationalize the denominator of 19
20.
25V3-4V2 7V3-5V2' 10v6-2V7 3V0 + 2V7'
V^ + V2
21 '
22
9
+
2V14"
2V3 + 3V2 5
+
2v<>
'
^"
Va:2+a2+rt
\
;
SURDS.
Case Let
Suppose the given surd is -^a -\- -^/b. n have the same meanings as before then
II.
X, y,
;
— y" is divisible n is even, — y''=(x-\-y) (a;"-i — x'^'^y
(1) If
X"
-\
rationalizing factor ^n^2y _^ r^n-l
n
If
odd,
is
+ y"
.t"
X" 4- y" =z (x~{-y)
Thus the
is
_ ^n-1
^yn-2
^
—
is af^
_
Ex.
1.
Find the factor which
=
Let X
:c*5
1 S"^,
-
y^ =(^x
5
_ 5
.
and
Ex.
?/'*.
will rationalize -^S
+ ^5.
x*?/
+
— x^y^ +
x^y^
x?/*
3253 _ 322353 + 3?14 _ 13 14 32 53 _ 53 53 + 32 53 _ 15 .
.
.
.
Express
_
+
(5^
=
53
9^)
58^
5
4.
.
is 3"2
y^)
5
53
2
9
—
is
--
_
_
52
-
9*)
33
(5^
.
J
2.
an equivalent fraction with a rational denominator.
To 5^
2.
n-l.
the required factor
y,
32
^.
and the rational product
as
—
y) (x^
-\-
4153
32
_
32
or
y
5^; then x^ and y^ are both rational, and
thus, substituting for x
32
_|_
1
=
?/
and
+ ?/"-i).
xy"""'
rationalizinsr factor is ^t) ^n-l ^n-2y _j ^yn-2
+
+ y,
by x
divisible
is x"
.
?/".
— x''~-y H
(x''-'^
and the rational product
xy""-^
\-
is
_
and the rational product
+ y, and — y'''^).
by x
ic"
Thus the
(2)
215
=
which
rationalize the denominator, X,
S^
=
y
;
equal to 5^
is
then since x^
-y^=(x3
the required factor
is
5^
2
+
5^'
•
y) (x^ i
3^
+ 4
and the rational denominator ninator the expression
(51
5^
j^
+
is
5-
3?) (5!
2.
+
x'^y
1
2
5^
f
•
3
—
4 3^^
+
5I
xy"-
+
?/),
3
= .
+ +
3^
5^
—
3?
-{-
;
3
5!
= .
22
3I
-|,
3
J)
—
3^,
put
;
ALGEBRA.
216
PROPERTIES OF QUADRATIC SURDS. 255. The square root of a rational quantity cannot be partlyrational and partly a quadratic surd.
-^n
If possible let tljen
n
by squaring, ,
that
a surd
is,
is
= a-\- ^m = a- + m + 2 a ^/m n — a? — m ;
equal to a rational quantity, which
is
impossible.
256.
For
that
If if
X
jr
is
+ V/ = « + V^'
= a and/ = 6. = a m then a + m \-^y = a+^b; y'6 = m + ^y
not equal to
is,
which
^^^^ will
«, let
x
jr
-\-
;
;
is
[Art. 255.]
impossible.
= a, = x -f ^y = a + ^b, x — ^y = a — -y/b. x
Therefore
and consequently,
we must
also
^•
2/
If therefore
have
257. It appears from the preceding article that in any equation of the form
«'+V?/
we may equate
= «H-V^
(1)?
on each side, a)id also the irrational imrts; so that the equation (1) is really equivalent to tiDo independent equations, x = a and y = b.
258.
If
the rational parts
Va+V^==V^+V/'
then will
^a-yb
For by squaring, we obtain a
+ V^ = ^ + a=X + .-.
Hence and
2Va5y/ ?/,
+
^b =
a — -y/b = x — 2\/xy + Va — ^b = ^/x — ^y.
2/;
2Vxy. ?/,
[Art. 257.]
;
SURDS.
217
259. To find the square root of a binomial surd.
Va + ^b = ^x -f
Suppose
V^
5
then as in Art. 258,
+ y^a
x
(1);
2V^-V^ Since
{x
(2).
— yf = (x -^ yy — = a^-b — = X Va" — b. y
Combining
this
with
(1),
we
from
.
.
.-,
xy
4:
(1)
and
(2)
find
^^a+Va^-b^ ^,,,^y^a-Va^~b^ V(ct'-b)
v^rrv-^=V^±^^^+V-
260. The vahies just found for x and y are compound surds unless or — b is a perfect square. Hence the method of Art. 259 for finding the square root of a -^b is of no practical utility except Ex.
when
Find the square root
VlG +
Assume Then
16
+2
+
of 16
2 ^^55
y/bb
=
y/x
=:
a:
+
—
a^
&
is
a perfect square.
2 ^55.
+ ^y.
+
2 \/xij
+
y
;
= \Q 2 V^ = 2^55 {x - y)'^ = (a: + y)'^ - ixy = 162 - 4 X 55 = 4x9. x
.'.
Since
+
y
.
.
.
.
From
(1)
and
(3)
X
That
is,
.:
x-y=±6
we
obtain
=
11, or 5,
In the same
way we may show Vl6 -
2
(2).
.
.
by
(1)
and
(2),
(3).
and
the required square root
(1),
.
?/
is
=
5,
or 11.
^11 +
y'5.
that
v55 = Vll - V^-
ALGEBRA.
218
Note. Since every quantity lias two square roots equal in magnitude but opposite in sign, strictly speaking we should have
+ 2 ^55 = ± ( Vl^ + V^)' — 2 ^^55 = ± ( V^ — V^)-
the square root of 16 the square root of 16
However,
it is
usually sufficient to take the positive value of the
square root, so that in assuming Va — ^h = y/x — -yjij it is understood With this proviso it will be unnecessary in that X is greater than y. any numerical example to use the double sign at the stage of work corresponding to equation (3) of the last example.
261. When the binomial whose square root we are seeking consists of two quadratic surds, we proceed as explained in the following example. Ex. Find the square root of y/ll^ Since
— V147.
^Wo - V147 = V^C V^S - V^l) = V^C^ - V^l)> ^^Yib-^wi = */7 v^r:^!. •
V5 - V21 =VI VV175-V147 =
But .-.
To
262.
find the square root of a binomial surd
by
inspec-
tion.
Find the square root of 11 + 2 ^30. 1. have only to find two quantities whose sum product is 30, thus Ex.
We
11
Ex.
2.
+
2
V30 = 6 + 5 + 2 V6 x 5 = ( V^ + .-. Vll + 2V30 = v^ + V^-
Find the square root
of 53
—
is 11,
and whose
V^)^-
12 y'lO.
First write the binomial so that the surd part has a coefficient 2
53
thus
-
VIO = 53-2 V360. two quantities whose sum
We have now to find product is 360 these are 45 and 8 ;
53
hence
.-.
\/53
-
-
12 VIO
12
;
12
=
;
45
+
8
-
2\/45 x 8
=(V45-V8)2; VlO = V4f> - \/8 = 3^5-2^/2.
is
53 and whose
SURDS.
219
Ex. 3. Find the square root of a -\- b +V2 ah + h'^. Rewrite the binomial so that the surd part has a coefficient 2
+
a
&
+ V2 a& +
have
now
whose product
is
to find
^
—
two quantities whose sum
"^
;
^
these are
thus
4
^
We
;
"^
and -
,
;
is
(a
+
h)
and
hence
2
.-.
A/;7TiTv2"«M^^=\rt + i^+J-
Note. Tlie student should observe that when the coefficient of the surd part of the binomial is unity, he can make this coefficient 2 if he will also multiply the quantity under the radical by \.
263. Assuming v « 258 gives us Vet — -^b
+ V^ = x = x — -y/y.
EXAMPLES
-\-
^y,
XXIII.
tlie
method
h.
Find the square roots of the following binomial surds 1.
2. 3.
4.
7-2^10. 13 + 2^30. 8-2V7. 5 + 2V0.
7. 8.
9.
10. 11.
75+12V21.
5.
6.
19.
18-8^5. 12. 2 a - V3 a^ -2ab- bK Find the fourth roots
21.
17
22.
56
+ 12V2. + 24V5.
41-24V2. 83+12V35. 47-4V33. 21+V5. 41-fV^16 + 5V7. 20. 1
V27 + 2^6. ^Z2 - ^-l^t.
15.
3-^/5
16.
2« +
17-
(ix-2ay/ax-a;K a + x+^2ax+:>:^.
18.
+
23.
IV5 +
24.
14
+
8 V3.
:
14.
13.
«2
+ (1 +
of the following binomial surds 31.
of Art.
+ V40. 2\/ a-^- &2
«-
+
«^)^
:
- 20 V6. 248 + 32V60.
25. 49
26.
Find, by inspection, the value of
28.
\/3 - 2^2. V4 + 2V3.
29.
V6-2V5.
27.
30.
Vl9 + 8^3.
33.
31.
V8 + 2V15. VQ - 2 Vl4.
34.
32.
35.
Vu + 4^6. ViS - 4 ^14. V29 + 6^22.
.
.
:
ALGEBRA.
220
Express with rational denominator -
36.
— i— ^3-1
37.
42.
^^' ^^
39.
V^±v^.
40.
.
—
^^3_. V'^+
41.
V8-^4
— 2
38.
43.
Find value of
+ ^7
J
6
+ 2V3
,
>'33-19v3
EQUATIONS INVOLVING SURDS. 264. Sometimes equations are proposed in which the the radical sign. Such equations are varied in character and often require special artifices for their solution. We shall consider a few of the simpler cases, which can generally be solved by the following method
unknown quantity appears under
Bring to one side of the equation a single radical term by on squaring both sides this radical will disappear. By repeating this process any remaining radicals can in turn be itself:
removed. Ex.
1.
2y/x- V4a:-ll =
Solve
Transposing,
Square both sides
— — ^y/x + 2 ^/x
;
then 4a;
1 1
1.
= \/4a3 — 11. = 4 x — 11,
4vx=:12, y/x .'.
Ex.
2.
Solve
x
= Z; = 9.
.
SURDS. 9.
v'2a;
+
ll
=^6.
\/4x2-7x+
10.
1
=2X-
13.
+ 25 = 1 + V»;V8x + 33-3 = 2v/2x. Vx + 3 + Va; = 5-
14.
10
11.
If.
V'ic
12.
-\/25
265.
+ 9x = 3V^.
When
equation,
221
-4+
= Vx +
15.
Vic
16.
V9x-8 = SVxTl - 2.
17.
\/4x
3
18.
+ 5 - V^ = \/x + 3. \/25a:-29-V4x-ll = S^ic.
19-
Vx +
20.
V^ + V4 «
=
4 a&
+ ^x. + x = 2\/?> + 2 a
radicals appear in a fractional
we must
11.
x.
form in an
clear of fractions in the ordinary way,
combining the irrational factors by the rules already explained in this chapter. Ex.
VO +
Solve
2 X
- V2x = V9 + 2x
Clearing of fractions,
- V2x(9 + 2x) = 5, + 2x= V2x(9 + 2x). + 16 x + 4 x"^ = 18 x + 4 x^, 9
+
2X 4
Squaring,
16
16=:2x, X
= 8.
EXAMPLES
XXIII.
1.
^
S^x-U 4VX-13 2.
9V^zi23^6Vx-n 2^3^-6
3v»5-8 3
Vx + 3 ^
3
Vx-
g_
6
7.
"^
Vx" 11-
yx + 3 ^ yx + 9 yx + 2 ^x v^ + l' 2yx-i ^ yx-2 2yx + | ya;-| 6yx-7 5^ 7vx-26
o
yx-1
Vx_VS^--l—
5
Vx-2 S^x-Vi 3VX-13' V^-2 ^
^^
^x
7yx-2r |
y;^
2yx-y4x-3=-—^.
13.
3yx =
+y9x-32.
^
^^^"^^ 14.
^x-7=
15.
(yx+ii)(yx-ii) + iio=o.
16.
2yx = l^^^-^^-
5
VrT^+y«;=—1^. yiTx
10
12.
^
I2yx- 11 _6yx + 4yx-4f 2yx +
^^ + ^+V^"^
1
yx +
2yx-3
7
CHAPTER XXIV. Imaginary Quantities. 266. An imaginary quantity is an indicated even root of a negative quantity. In distinction from imaginary quantities all other quantities are spoken of as real quantities. Although from the rule of signs it is evident that a negative quantity cannot have a real square root, yet quantities repre-
V—
V—
a, sented by symbols of the form 1, are of frequent occurrence in mathematical investigations, and their use therefore proceed to explain leads to valuable results. in what sense such roots are to be regarded. When the quantity under the radical sign is negative, we can no longer consider the symbol as indicating a j)ossible arithmetical operation; but just as ^a may be defined as a symbol which obeys the relation ^a x ^a = a, so we shall
We
^
V— a
V—
V—
a x a = — a, and we meaning to which this assumption leads us. It will be found that this definition will enable us to bring imaginary quantities under the dominion of ordinary algebraic rules, and that through their use results may be obtained which can be relied on with as much certainty as others which depend solely on the use of real quantities. define
to be such that
shall accept the
Any imaginary
expression not involving the operby an exponent that is an irrational or imaginary expression, can be reduced to the
267.
ation of raising to a j)ower indicated
V—
form a -{- 6 1, which may be taken as the general type of Here a and b are real quantities, all imaginary expressions. but not necessarily rational. An imaginary expression in
1
IMAGINARY QUANTITIES.
223
form is called a complex number. If a = 0, the form becomes 6V— 1, which is called a pure imaginary expression. this
268.
that
By
V^H! x V"^^
definition,
- 1.
V—iy = — a.
(-y/a-
is,
==
V— 1
may
be regarded as equiv-
alent to the imaginary quantity
V—
a.
^a
Thus the product
.
269. It will generally be found convenient to indicate the imaginary character of an expression by the presence of the — 1 which is called the imaginary unit thus symbol
V
;
V^^ = V4x(-1)= 2 V"=^. V- 7 a' = -V7a'x(-l) = a^7^/^T. 270. We shall always consider that, in the absence of any statement to the contrary, of the signs which may be prefixed before a radical the positive sign
is
to be taken.
But in the use of imaginary quantities the following point deserves notice.
Since
by taking the square
(— a)x( — b)= we have
ab,
root,
V— a X V— b = ± -Vab. Thus in forming the product of V— a and V— & appear that either of the signs before Vab.
This
is
+
or
—
it would might be placed
not the case, for
V— a X V— b=^a. V— 1
X V^'
V—
271. In dealing with imaginary quantities we apply the laws of combination which have been proved in the case of other surd quantities.
Ex.1. Ex.
3.
a
+ bV^^±(c + dV^^) =
The product
of a
^
+
^
ac
V— —
bd
1
a
and
±c + (b ±d)V^. c
+ (6c +
+ dV— ad)
V—
1
1.
;
:
ALGEBRA.
224
V—
i;
i is often rex^resentecl by the letter 272. The symbol but until the student has had a little practice in the use
of imaginary quantities he will find it easier to retain the
V—
symbol
The
1.
successive powers of
V—
1,
or
i,
are as
follows
(V3T)^ = -l, (
= -l; e = -i; ;^
V^^)' = - V^^,
and since each power is obtained by multiplying the one it by 1, or i, Ave see that the results must now
V—
before recur.
273. If a For,
+ 6 V^^ = a
if
-f-
then
.-.
Now
a^
and
?>^
ft2_^62
6
= 0.
= 0.
are both positive, hence their
be zero unless each 6
= 0, and & V^^ = 0, &V— 1 = — a
0, then a
is
separately zero
;
that
is,
sum cannot a
= 0,
and
= 0.
+ 6V^^1 = c +
If
a
f?)
;
0-^
two imaginary expressions may be necessary and sufficient that the real 2mrts sJiould be equal, and the imciifmary 2mrts sliould be equal. The student should carefully note this article and make use of it as opportunity may offer in the solution of equations involving imaginary expressions.
Thus
equal
in order that
it is
IMAGINARY QUANTITIES. 275.
When
two imaginary expressions
225
differ
only in the
sign of the imaginary part, they are said to be conjugate.
—
Thus a
Similarly
?>
V— 1
is
conjugate to a
^2 + SV^l
is
-|-
&V— 1.
conjugate to
^2 — 3V— 1.
276. The sum and the product of two conjugate imaginary expressions are both real.
For
+ bV^^ -{-a — 5 V^^ = 2 a. - 6 V^^) = a^-(- b') = + a
Again («
bV~^l)(a
cr
+
b'.
277. If the denominator of a fraction is of the form a -\-bV~l, it may be rationalized by multiplying the numerator and the denominator by the conjugate expression 1. For instance, a— 6
V—
c
a
+ cW^l ^ (c + (W^^) (a - V"^T) + b V -~1 (a + b V - 1) (a - b V^^) ^ ac + bd + (ad — be) V"^^ a' + ad — bc^ __ ac + bd j />
—
b''
/
b'
Thus, by reference to Art. 271, Ave see that the sum, difand quotient of tioo imaginary exjjressions is in each case an imaginary expression of the same form. ference, product,
278. Fundamental Algebraic Operations
upon
Imaginary
Quantities.
Ex.
Find value of
1.
5V ~cr^ - 2V-4a-t Ex.
Multiply
2.
(2
V- «*
4.
:=5V9(-<4(
5V-9a4 _ _
1)
=- 2\/4'r7H-
1)
2\/-4a4.
^ 15ctV- 1 = - 4ff2V^ri
2V^^ by SV^^. 2V^^ = 2V3\/^=i;;
3\/^^ = 3V2\/^l VS \/^) (3 \/2 V^) = 6 V6 (
V^ ^ ;
2
6 V6.
5
:
ALGEBRA.
226 Ex. 2
4-
.
V^
by 2 + V^^T. +3 + 3V^n;)(2 - V"^) _ 7 + 4\/^n; _ (-1) (2+V-l)(2 V31)
Divide 2
3.
3V^n. _
2+V-
(2
The method
279.
Find the square root of
-
We
now
have
product
is
—
144
7
...
24
—
7
—
24
V—
1.
V^^=- 7-2 V-144.
two quantities whose sum
these are 9 and
-7-24
hence
-
to find ;
be used in finding
+ 6 V— 1.
the square root of a Ex.
may
of Art. 262
+ 4\/^
7
—
16
is
v/^n =
9 + (- 16)-2 V9 x(= (V9-V- 16)2;
V- 7-24 v^^n
—
and whose
7
;
16)
±(3-4V^no.
EXAMPLES XXIV. Simplify
:
1.
VZTs + V^TS.
2.
4V^^27 + 3V-12.
5.
2
V - 4 a%2 +
5\/^l6-2-
4.
2V^20 +
3-
6.
V - 36 a'^x^. V^-V^+V^+V3T. 9. (2v::2+V^)(V^-V^).
7.
(\/^^)(\/-12).
8.
(2
12 13
,
15
V-
3.
a^x^
+
7
12
10. (2 + \/^r7^)(3-v^^:^). n. (4: + V^2)(2~SV^^). + V^^)(l - V^3). (2\/^^ + 3V^^)(4\/^r3 - 5V^2). V27-4-V^^. 14. -V^^-CV^^ + V"^). ( _ V3^ + V^2) - (2 v^:=l - V^^)
Express with rational denominator 18.
16.
3+ 2V-I
^
3-2v/^ri
- 5\/^T 2 + 5\/^n; + xV — 1 _ g — xV— 1 a — .xV— 1 a + V— 1
2 17.
^^ + 2-
f(
19.
3V-2 -2V-
a;
Find the square root of 20.
-5+12\^^^.
21.
Express in the form a 0.«
'l±li. 2 — 3i
-\-
9A
-
11_60V-1.
22.
-47+8V^
ib:
^ 2v/3-
iV2
V^
25.
1+r 1-i'
CHAPTER XXV. Problems. 280. In previous chapters we have given collections of problems which lead to simple equations. We add here a few examples of somewhat greater difficulty. Ex.
A
1.
f 2.G0; by
grocer buys 15
lbs. of figs
and 28
lbs.
of currants for
and the currants
selling the figs at a loss of 10 per cent,
how much at a gain of .30 per cent, he clears 30 cents on his outlay per pound did he pay for each ? Let x^ y denote the number of cents in the price of a pound of figs and currants respectively then the outlay is :
;
15 X .-.
The
loss
currants
is
upon the
— x 28?/
15x
figs is
+ 28 y cents. + 28?/ = 260
—x
15 x cents,
(1).
and the gain upon the
cents; therefore the total gain
is
^ -Scents; 5 ...
2
42l/_3x^3Q
From (1) and (2) cost 8 cents a pound,
we
.2). ^
^
2
5
find that x
= 8,
and the currants
and
?/
=
5
;
that
is,
the figs
cost 5 cents a pound.
2. At what time between 4 and 5 o'clock will the minute-hand watch be 13 minutes in advance of the hour-hand ? Let X denote the required number of minutes after 4 o'clock then, as the minute-hand travels twelve times as fast as the hour-hand, the
Ex.
of a
;
hour-hand
will
move over
—
minute divisions
in
x minutes.
At
4
12 o'clock the minute-hand finally is 13 divisions in
over 20
-f 13,
is
20 divisions behind the hour-hand, and
advance
or 33 divisions
;
therefore the minute-hand
more than the hour-hand. 227
moves
ALGEBRA.
228 Hence
a:
=—+
33,
12
Hx = 33; X
.-.
Thus the time
is
= 36.
36 minutes past
4.
"At what ^?"mes between 4 and between the two hands ? " we must also take into consideration the case when the minute-hand is 13 In this case the minute-hand gains divisions heMnd the hour-hand. If
the question be asked as follows
:
5 o'clock will there be 13 minutes
20
—
13, or 7 divisions.
Hence which gives
—
x
—
-
x
=
7/^.
—
|-
7'
Therefore the times are
7
past
4,
7,
and
36' past 4.
persons A and B start simultaneously from two places, and walk in the same direction. A travels at the rate of p miles an hour, and B at the rate of q miles how far will A have walked before he overtakes B ? Suppose A has walked x miles, then B has walked x — c miles.
Ex.
3.
Two
c miles apart,
;
A, walking at the rate of hours
;
and
B
will travel
x
—
c
miles an hour, will travel x miles in miles in
X
_x —
p
q
qx
= px — pc
x
Therefore
A
A
^ hours
:
these two times
^
whence
4.
r-c
we have
being equal,
Ex.
p
has travelled
c
;
p-q
~
—
pc
p-q
miles.
train travelled a certain distance at a uniform rate.
Had
the speed been 6 miles an hour more, the journey would have occupied
and had the speed been 6 miles an hour less, the journey would have occupied 6 hours more. Find the distance. Let the speed of the train be x miles per hour, and let the time 4 hours less
occupied be
by xy
miles.
;
?/
hours
;
then the distance traversed will be represented
;
;
PROBLEMS. On time
the
supposition the speed per hour
first
tal?:en is
?/
represented by
On
229
— 4 hours. (x + 6) (?/ —
In this case
is
x
+
6 miles, and the
distance traversed will be
tlie
4) miles.
the second supposition the distance traversed will be repre-
— 6) (?/ + 6) miles. All these expressions for the distance
sented by (x
From
must be equal
xy=(x-^Q)(iy -^) = {x-Q){y
.-.
=
xy
xy
Qy
-\-
6?/-4ic
or
and
+ Q).
we have
these equations
=
xy
— Qy
xy
— 4iX — 2ii, = 24 -\-
Qx —
(1)
Qx-Qy = m
or
From
(1)
and
(2)
we
Hence the distance
is
obtain
cc
=
30,
?/
;
ZQ, (2).
= 24.
720 miles.
A person
invests $ 3770, partly in 3 per cent Bonds at f 102, Railway Stock at $ 84 which pays a dividend of 4J pe? cent if his income from these investments is $ 136.25 per annum, what sum does he invest in each ? Let x denote the number of dollars invested in Bonds, y the number of dollars invested in Railway Stock then
Ex.
5.
and partly
in
;
;
+
x
The income from Bonds Stock
is
or $^, 84
x
(2)
and by subtracting
and from
(1)
$
is
3770
—^, or $—
(1).
;
and that from Railway
66
5. 34
whence
=
I ?-^.
Therefore
From
?/
(1)
+
+
^= 56
136i
= 4632 J, ^ ff = 862^ y = 28 x 37i = x = 2720.
fi
(2). ^
*
^
?/
1050
;
Therefore he invests $ 2720 in Bonds and $ 1050 in Railway Stock.
EXAMPLES XXV.
A sum of f 100 is divided among a number of persons if the 1. number had been increased by one-fourth each would have received a half-dollar less find the number of persons. ;
:
:
ALGEBRA.
230
I I bought a certain number of marbles at four for a cent 2. kept one-fifth of them, and sold the rest at three for a cent, and gained a cent how many did I buy ? ;
:
if I bought a certain number of articles at five for six cents 3. they had been eleven for twelve cents, I should have spent six cents how many did I buy ? less ;
:
A man
wins twice as much as he had to begin with, $ 16 he then loses four- fifths of what remained, and afterwards wins as much as he had at first how much had he originally, if he leaves off with 1 80 ? 4.
and then
at whist
loses
;
:
A number
5.
by
and
9,
of
two
digits exceeds five times the
ten-digit exceeds its unit-digit
its
by
1
:
sum
find
of its digits
it.
The sum of the digits of a number less than 100 is G if the be reversed the resulting number will be less by 18 than the
6.
;
digits
original
number
:
find
it.
A man
being asked his age replied, " If you take 2 years from ray present age the result will be double my wife's age, and 3 years ago her age was one-third of what mine will be in 12 years." What were their ages ? 7.
At what time between one and two
8.
watch
first at
o'clock are the hands of a
right angles ?
9. At what time between 3 and 4 o'clock minute ahead of the hour-hand ?
When
10.
6
and
the minute-hand one
are the hands of a clock together between the hours of
7 ?
will be as is
o'clock, and in 10 minutes the minutebefore the hour-hand as it is now behind it
between 2 and 3
It is
11.
hand what
is
the time
much ?
At an election a majority of 162 was three-elevenths of the whole number of voters find the number of votes on each side. if there had been 10 13. A certain number of persons paid a bill 12.
:
;
more each would have paid $ 2 less if there had been 5 less each would have paid $2.50 more find the number of persons, and what each had to pay. 14. A man spends $ 100 in buying two kinds of silk at $4.50 and $4 a yard by selling it at $4.25 per yard he gains 2 per cent how much of each did he buy ? ;
:
:
;
15.
Ten years ago the sum
their father's age: one
present
how
sum
is
of the ages of two sons was one-third of two years older than the other, and the
of their ages is fourteen years less than their father's age
old are they ?
:
PROBLEMS.
231
16. A basket of oranges is emptied by one person taking half of them and one more, a second person taking half of the remainder and one more, and a third person taking half of the remainder and six
How many
more.
A person
did the basket contain at
swimming
first ?
stream which runs 1^- miles per hour, finds that it takes him four times as long to swim a mile up the stream as it does to swim the same distance down at what rate does he 17.
in a
:
swim
?
At what
18.
watch be
same
tiines
between
7
and
8 o'clock will the
When
at right angles to each other ?
hands
will they
of a
be in the
straight line ?
The denominator
19.
of a fraction exceeds the numerator by 4 taken from each, the sum of the reciprocal of the new frac-
and
if
tion
and four times the
5
is
;
original fraction
is
5
:
find the original fraction.
Two
persons start at noon from towns 60 miles apart. One walks at the rate of four miles an hour, but stops 2J hours on the way the other walks at the rate of 3 miles an hour without stopping when and where will they meet ? 20.
:
;
21. A, B, and C travel from the same place at the rates of 4, 5, and 6 miles an hour respectively and B starts 2 hours after A. How long after B must C start in order that they may overtake at the same instant ? ;
A
A
dealer bought a horse, expecting to sell it again at a price 22. that would have given him 10 per cent profit on his purchase but he ;
had to sell it for $ 50 less than he expected, and he then found that he had lost 15 per cent on what it cost him what did he pay for the :
horse
?
23. A man walking from a town, A, to another, B, at the rate of 4 miles an hour, starts one hour before a coach travelling 12 miles an hour, and is picked up by the coach. On arriving at B, he finds that
his coach journey has lasted 2 hours
:
find the distance
between
A
and B. 24. What is the property of a person whose income is $ 1140, when one-twelfth of it is invested at 2 per cent, one-half at 3 per cent, one-third at 4^ per cent, and the remainder pays him no dividend ? 25. A person spends one-third of his income, saves one-fourth, and pays away 5 per cent on the whole as interest at 7|^ per cent on debts previously incurred, and then has f 110 remaining: what was the
amount
of his debts ?
Two
vessels contain mixtures of wine and water in one there three times as much wine as water, in the other five times as much water as wine. Find how much must be drawn ofl: from each to fill a third vessel which holds seven gallons, in order that its contents may 26.
is
be half wine and half water.
;
ALGEBRA.
232
There are two mixtures of wine and water, one of which conmuch water as wine, and the other three times as much wine as water. How much must there be taken from each to fill a pint cup, in which the water and wine shall be equally mixed ? 27.
tains twice as
same time to walk, one from A to B, A, a distance of a miles. The former walks at the rate of p miles, and the latter at the rate of q miles an hour: at what distance from A will they meet ?
Two men
28.
set out at the
and the other from B
A train runs from A to B
29.
A to
to
in 3 hours a second train runs from beyond B, in 3^ hours, travelling at a speed mile per hour. Find distance from A to B. ;
C, a point 15 miles
which
is less
by
1
bought at 36 cents and chicory at 9 cents per lb. in what proportion must they be mixed that 10 per cent may be gained by selling the mixture at 33 cents per lb. ? 30.
Coffee
31.
A man has
is
:
one kind of coffee at a cents per pound, and another pound. How much of each must he take to form a a — b lbs., which he can sell at c cents a pound without
at b cents per
mixture
of.
loss? 32. A man spends c half-dollars in buying two kinds of silk at a dimes and b dimes a yard respectively he could have bought 3 times as much of the first and half as much of the second for the same money. How many yards of each did he buy ? ;
A man rides
one-third of the distance from A to B at the rate an hour, and the remainder at the rate of 2 6 miles an hour. If he had travelled at a uniform rate of 3 c miles an hour, he could have ridden from A to B and back again in the same time. Prove 33.
of a miles
= +r cab
-
that
-
A, B, C are three towns forming a triangle. A man has to to the next, ride thence to the next, and drive thence to his starting-point. He can walk, ride, and drive a mile in «, b, c minutes respectively. If he starts from B he takes a + c — b hours, if he starts from C he takes b + a — c hours, and if he starts from A he takes c + b — a hours. Find the length of the circuit. 34.
walk from one
MISCELLANEOUS EXAMPLES 1.
Distinguish between like and unlike terms.
IV.
Pick out the like
— S ab b^ — 2 a^ S -^ o ab -^ 7 a^. 2. Subtract - 2a^ + S a'^b + bb^ - Aab^ from -l-2ab'^ + and multiply the result by — l-f2a — &. 3. Divide Sx^-Sx^y -{ 4 xy^ - y^hy 2x- y.
terms in the expression a^
-{-
-\-
b'^
3b^
:
'
MISCELLANEOUS EXAMPLES 4.
will
—
5.
If
the
number
a denote
Factor the following expressions
_
a3
—x-1 ^
^
^
7.
Solve
2x-f 3x-l lI^^^-2Jli^ = li±I^+5-6x.
8.
There
is
of
3
5
increased by 36
but
;
Find the H.
C. F. of 4
Simplify
11.
Find the value
12.
Simplify
13.
Find the
x
V~c^)
(
of
which when divided by the unit
the digits be inverted the
if
_ -
a;3
and 3 x3 10.
digits
16 x^
13 x2
^
= 3,
c
value of
fa^
-
^V\^ Vj-^.
^.
^_ X -±- ^ %
Solve
X
-
(3 &
|
+
c)
1
+
62
_ i^ | _
(f\
-
xf.
= 4.
_i^ + _A_ = _£_.
«x 6x — a
15.
X — + Find the L. C. M. of
16.
Solve
1
-
-
x, 1
x2, 1
3^ +
^=
2x
2?/_,2
5
T"
-
x^,
and
(1
32,
3
5
"
^•
^-2/-{2x-|-7-(^-4)+2-^}.
17.
Simplify
18.
Find the square root
19.
Solve the equations
of
x*5
+
8 x^
-
2 x^
+
16 x^
-
3£-5 = 21x-f2^ + li5V ^^27 (i.)
V 3
(ii.)
Expand
42
y
+ 5i^ = ?^-7. 2f^-lUll 15 5 5 3 /
V
20.
is
+ 13 x - 3 + 13 x - 3.
^a^ 20 V| + 14 Vf + 2 V2I -1^1
(
number
number.
find the
:
h
3
number of two
a
digit gives a quotient 6
2, &
what
27.
Find the value
14.
a,
:
6.
=
+
a2-64,
(ii.)
«
by
of dollars I possess is represented
?
(i.)
9.
233
IV.
the following binomials (i.) (x + 3«)4, (ii.)
(2.-1)1
8x
+L
when
ALGEBRA.
234 The sum
21.
two
of the
digits of a
number
ence if the digits be inverted, the number the number. ;
22. (iii.)
Find the factors of (I) - 7 a%-^x^ -\- xK
x^-9x -
is
8 times their differ-
diminished by 18
is
36, (ii.)
2x2
- 3x -
find
:
and
14,
a^b^
23.
Rationalize
denominator
the
—^
of
V 13 + 4V3. 24.
^^
Simplify
+
For what values 25.
- 11 x^ - 3x + + 3x2-6x-8
3^3 a:3
simplify
10.
numerator and denominator vanish
of x will both
2x + Sy + 4z = X
+
4
3x+ (ii.)
?
^
y
+
Sl,
= + 2z =
+
?/
z
lS, 16.
^^ = |,. + ,_6 = |(,-.).
V- 8 + V^r - V- 18 +V - 2 + 2 V- 3.
26.
Simplify
27.
Simplify (^2x
28.
Find the middle term of the expansion
30.
and
Solve the equations (i.)
29.
:::,
'-'^^
- ^1^')
(s^/
of
/a Vx
_^
63
a3
X\ -
)
1*^ •
aj
a
b
(«_^V«+??_i\
i + i + JL
\b
«-
a/ \b
+
+ i^^
5
11
«!
Simplify
+ ^i±X') -^ (^ +
a
J
b'^
ab
Solve the equations (i.) (ii.)
— a)— + 6)x —
a(x
b(x
(«
«//
— =
&) a'2,
= (« + &)(x — « - ?>). («- + 62)x — ahy = a^.
the 31. A sum of f 10.10 is divided among 7 women and 10 men same sum could have been divided among 23 women and 4 men. Find how much each woman and man receives. ;
32.
Find the cube root of 8xC+ 12x5+ 18 xH 13x3+ 9x2+ 3x
and the square root of
y-
-{-
iy
-\-
10
-\
\-
y
34.
Simplify
VsiT- 24 ^6 +
[(
+ 1,
9 9^
12 12
y^'
V^^ + V^3)(\/^^ + 2\/^^)].
:
CHAPTER XXVI. Quadratic Equations. 281. Suppose the following problem were proposed for solution
A dealer bought a number of horses for f 280. bought four less, each would have cost $ 8 more did he buy ? We should proceed thus
If he ;
had
how many
:
Let X
= the
number
of horses
— = the number of X
then -
;
dollars each cost. If
he had bought 4
—
and each would have cost
.-.
he would have had x
less,
8
X
whence
x(x .-.
or
_
horses,
280 a;
-4'
— 4) + 35(x — 4) = 35 x + 35 -140 = 35 x:'-Ax = UO.
4a.'
4,
dollars.
x-4. 280
—
;
.t;
a;
.-.
This equation involves the square of the unknown quanand in order to complete the solution of the problem we must discover a method of solving such equations. tity
;
282. Definition. of the
unknown
An equation
which contains the square
quantity, but no higher power,
is called a quadratic equation, or an equation of the second degree. If the equation contains both the square and the first
power of the unknown,
it is
called an affected quadratic 235
;
if it
:
:
ALGEBRA.
236
unknown
contains only the square of the
it is
said to be a
pure quadratic.
Thus and
2x^
— 5x = 3 is an affected quadratic, 5 or = 20 is a pure quadratic.
PURE QUADRATIC EQUATIONS. 283. A 2^ure quadratic may be considered as a simple equation in which the square of the unknown quantity is to be found. Ex.
—-27 —=
^^
^
Solve
x-2
Multiplying across, 9 x^ transposing,
—
99
16
x"^
.-.
and taking the square root
x2
-
.
11
= 25 a;^ — = 576 x^ = SQ;
675
;
;
of these equals,
we have
x=±6. Note.
hand
We
prefix the double sign to the
number on
the right-
side for the reason given in Art. 196.
284. In extracting the square root of the two sides of the equation x^ 36, it might seem that we ought to prefix the double sign to the quantities on both sides, and write ±6. But an examination of the various cases shows this to be unnecessary. For ± x = ± G gives the four cases
=
±x=
-\-x
= -\-6,
-\-x
= — 6, —x = -\-6, — = — 6, a;
included in the two already given, namely, .T = 4- G, X = — 6. Hence when we extract the square root of the two sides of an equation, it is sufficient to put the
and these are
all
double sign before the square root of one
EXAMPLES XXVI.
side.
a.
Solve the following equations 1.
4a;-^+ 5=:x'-^
+
17.
o^-i
3.
ix+ l)ix-l) = 2x^-i.
^
2x^-6 '2
x^-4:
6x'^-10
4
7
a;
+
2
^q
;
;
QUADRATIC EQUATIONS. g '
X
—
a
X
-\-
a
^ X
+ —
«
3(a;2-i)
8.
-1 (2 X - c) (x +
^
_5
a
7
x'^
237
3(9x^-1) ^^ (x2-l)(x2 + 3) +3 (0 + (2 X + c) (x - cZ) = 2 c(Z(2 C(Z -
4(a;2-4) X-
1).
AFFECTED QUADRATIC EQUATIONS. The equation
285.
x^ ==
36
may
be solved in a similar
of both sideSj
we have two
an instance of the simplest (x — 3)^ = 25
is
The equation
form of quadratic equations.
way
for taking the square root
;
simple equations,
x-3 = ±5. — 3 = + 5, whence = 8 — taking the lower sign, x 3 = — 5, whence x = — 2. x = 8, or — 2. the solution is Now the given equation (x — Sy = 25 may be written x^ — 6x-{- (3)^ = 25, — X = 16. or Taking the upper
sign,
.^
a.*
.-.
ic^
Hence, by retracing our
steps,
x^
()
we
learn that the equation
~ 6x = 16
can be solved by first adding (3)^ to each side, and then extracting the square root and we add 9 to each side because this quantity added to the left side makes it a ;
square.
l')erfect
Now
whatever the quantity a 05^
and
x^
may
be,
+ 2 ax a^ = {x + a)^, — 2 ax + a^ = (x — ay -\-
so that, if a trinomial is a perfect square,
power,
ocF,
has unity for a
coefficient,
be equal to the square of half the Ex.
1.
Solve
and
highest
coefficient
of x.
7 X = x2 - 8.
Transpose so as to have the terms involving x on one square term positive.
Thus
its
the term without x must
x2
-
7
X
= 8.
side,
and the
;
ALGEBRA.
238 Completing the square, y? that
—
.-.
We
Note. Ex.
x
+ (D" =
2.
-
t)2
^ + = V-
x= |± § = 8,
V"
or
32
Solve 3
Divide throughout by
Thus
a:^
-3^2 = 10 + 10 x = 32.
3, so as to
is
make
Simplifying, 7
side.
;
-^g^-
•
;
Ex.3. Solve
x=:-f ±-V =2, or -5i. 7(a;f 2rt)-2 + 3«2 = 5a(7x+ 23o). + 28 ax + 28 a^ + 3 ^2 = 35 4- 115 ^a^ 7 x^ — 7 ax = 84 or.
a:2
Whence completing the square, tliat is,
1.
the coefficient of x^ unity.
.-.
is,
-
a;.
+ V- X - ^fx? + -f x + (f )2 = ^~ + {x + f )2 = J-fi x^
completing the square,
that
I
do not work out (|)- on the left-hand
Transposing,
that
7
(x
is,
oja;
X2-
X
.
QUADRATIC EQUATIONS. Thus
tion.
=1
if jy
— 2x -\-l = 0,
then {x
—
239 1)^
=
0,
whence
the only sohition. Nevertheless, in this and similar cases we find it convenient to say that the quadratic has two equal roots.
X
is
EXAMPLES XXVI. 1.
2.
3. 4. 5.
6.
19.
20. 21. 22.
2^
b.
= 17x + 4ic2. 13. 21 a;2+22x+5=0. 3 14. 50a:2 - ISr. = 27. + 121 = 44 X. 8. 21+x = 2x2. 25x = 6x2 + 21. 9. 9x2-143-6x = 0. 15. 18»:2_27ic-26=:0. 10. 12x2 = 29x-14. 8x2 + x = 30. 16. 5x2 = 8x + 21. 3x2 + 35=22,x. 11. 20x2 = 12 -X. 17. 15x2-2«x = x+22-6x2=0. 12. 19x = 15-8x2. 18. 21 x2 = 2ax + 3a2, x2 = 11 Jcx + 7 23. (x + 1)(2 x + 3) = 4 x2 - 22. = 0. 12 x2 4- 23 A:x + 10 24. (3x-5)(2x-5)=x2 + 2x-3. 12 x2 - ex - 20 c2 0. 25. a^x^ -2ax + a^ = h. 2(x - 3) = 3(x + 2) (x - 3) 26. cc?x2 = c2x + (Px - cd. 1 _ 3 5X gg 5x-7 _ x-5 g^ x + 4 x-2 _^,^' 2' x+1 7x-5 2X-13' x-4 x-3 3x-8_5x-2 ort ^+3 2x-l_Q 32 _^ ^ 1 5x2+14x =
55.
7.
15
ic2
rt2.
A;2.
^'2
zr:
3
288. Solution by Formula.
After suitable reduction and
transposition every quadratic equation can be written in the form ax^ -\-hx -\- c (),
=
where
c, may have any numerical values whatever. we can solve this quadratic, we can solve any.
a, b,
therefore
If
,
ALGEBRA.
240
= — c^ c x^ -\--x =
Transposing, dividing by
aa? -\-hx
»
,
.
.
.
.
(1)
b
a,
a
a
Completing the square by adding to each side
f
—
]
\2aJ a that
a
4a2
\2aJ Ix-i
IS,
2aJ
4.a'
'
extracting the square root,
±V(^'-4ac)
h
a.-
+ 2a
2a
'
2a Note. The student will observe that b, the first term of the numerator of the fraction, is the coefficient of x in equation (1) loith its sign changed, and that 4 ac, under the radical, is pZws or minus according as the signs of a and c in equation (1) are like or unlike.
289. Instead of going through the process of completing the square in each particular example, we may now make use of this general formula, adapting it to the case in question by substituting the values of Ex. Solve Here a = 5,
5x2 &
=
11, c
=-
..X.
+
11 x
12
=
a, b,
c.
0.
12.
-ll±V(ll)2-4.5(-12) ^^
_ -ll
J=V361 10
290. In the result x
-
~
-11±10 _4
~5'
10
= -b±Vf-^''')
,
^^._3
2a it must be remembered that the expression y/(b^ — 4:ac) is the square root of the compound quantity b^ 4 ac, taken as cannot simplify the solution unless we know a ivhole. the numerical values of a, b, c. It may sometimes happen 4 ac a perfect square. In that these values do not make b~ such a case the exact numerical solution of the equation
—
We
—
cannot be determined.
;
241
QUADRATIC EQUATIONS. 5x2- 15x +
Ex.1. Solve
We
^"
have
15
± V^C-
=0.
11
15)^
-
4
5
.
.
11
2.5
^ 15±V5
.
10
V^ =
Now
2.236 approximately. 15
±
2.236
^
-^.,^3^
^^,
^2764.
10
These solutions are correct only to four places of decimals, and them will be found to exactly satisfy the equation. Unless the numerical values of the unknow^n quantity are required is usual to leave the roots in the form
neither of it
15
+ V5 10
Ex.
We
Solve
2.
x2
^
have
15
- V5 10
'
-
3X
^ 3 ± V(-
"
+ 5 = 0. -4 1
3)^
^ 3 ^ V9 _ 20 ^ 3
:jz
.
5
V^^TT 2
2
But — 11 has no square root exact or approximate [Art. 196]; so In such that no real value of x can be found to satisfy the equation. a case the roots are said to be imaginary or impossible [Art. 266].
291. Solution by Factoring. The following method will sometimes be found shorter than either of those already given.
Consider the equation x^ -\-^x = 2. Clearing of fractions, 3x^-\-7x — 6 = by resolving the left-hand side into factors,
(3i^-2)(a;
Now
if
product
the factors
either of is zero.
by either
From
a;
.
.
(1)
be zero, their is satisfied
of the suppositions a;
- 2 = 0,
roots are
this
simplified
+ 3) = 0. 3 — 2, + 3, ic
.
Hence the quadratic equation 3
Thus the
.
we have
we
|,
or
to the
can be readily obtained
if
+ 3 = 0.
— 3.
see that ivhen
and brought
iK
a quadratic equation has been
form of equation
(1), its
solution
the expression on the left-hand
.
ALGEBRA.
242
Each of these factors side can be resolved into factors. equated to zero gives a simple equation, and a corresponding root of the quadratic. Ex.
Solve
1.
2x'^
Transposing, so as
we have
Now
to
- ax-^2bx =
have
ah.
terms on one side of the equation^
all the
2 x'2 — ax -\-2hx — ah = 0. + 2hx — ah = x{2 x - «) + 6(2 x — = (2x-a)(x + h). (2 x - a) (x + 6) = 2x-a = 0, orx + 6 = 0.
—
2x'^
Therefore
ax
a)
;
whence
.-.
=-,
X
-h.
or
2
Ex.
2(x2
Solve
2.
We have that
2 x2
- 6) = 3(x - 4). - 12 = 3x - 12 ;
2x2= 3x
is,
2 x^
Transposing,
-
3x
•
X(2X-3)z:iO. .-. x=:0, or
Thus the
roots are
.
(1)
= 0.
0,
2x- 3 = 0.
|
Note. In equation (1) above we might have divided both sides by X and obtained the simple equation 2 x = 3, whence x = |, which But the student must is one of the solutions of the given equation. be particularly careful to notice that whenever an x is removed by division from every term of an equation it must not he neglected^ since the eqiiation
is satisfied hij
x
=
0, v^hich is therefore
one of the
roots.
Formation of Equations with Given Roots. to form an efiuation whose roots are known.
292.
now easy
It is
Ex.
Form
the equation
whose roots are 3 and x
Here
= 3,
or x
.-. a: -3 = 0, or x both of these statements are included in
^
=
^
},.
;
=6;
(x-3)(x-.\)=0, 2 x2
or
From
this
it
-
7
X
+
3
=
0.
also appears that the factors of a trinomial, in the
form
ax2 + bx + c, can be obtained by placing the expression equal to zero, solving the resulting quadratic equation (Art. 288), and subtracting .shall return to the subject of this each root separately from x.
We
article in
Chapter xxx.
QUADRATIC EQUATIONS.
243
Unknown Quantity which
293. Values found for the
do not
satisfy the Original Equation.
From
following example
tlie
it
will be seen that in solv-
may
ing certain equations values
be obtained which will
not satisfy the original equation. Ex.
Solve \/x
+
5
+ V3x +
4
= Vl2x+
1.
Squaring both sides, a:
+
5
+ 3x +
+ 2V(x +
4
Transposing and dividing by \/{x
+
5) (3
a;
{x
+
5) (3
x
Squaring,
+ +
13x2 -51a:
or
5)(3x
+ 4)=
12r.
+
1,
2,
4)
=4 -4
4)
:=
a;
16
xP-
-
(1).
32 x
+
16,
-4 = 0,
(x-4)(13x +
l)
.-.
X
= 0; = 4,
or
—
Jg.
If we proceed to verify the solution by substituting these values in the original equation, it will be found that it is satisfied by x = 4, but not by X = — -^^. But this latter value will be found on trial to satisfy the given equation if we alter the sign of the second radical thus, ;
VX+ 5 - V3x+4 = Vl2x+ On
1.
squaring this and reducing, we obtain
-V(x +
5)(3x
+
4)
= 4x-4
(2);
and a comparison of (1) and (2) shows that in the next stage of the work the same quadratic equation is obtained in each case, the roots of which are 4 and — Jg, as already found.
From
this
it
appears that
when
the solution of an equation requires
we cannot be certain without trial found for the unknown quantity will satisfy the
that both sides should be squared,
which
of the values
original equation.
In order that all the values found by the solution of the equation be applicable, it will be necessary to take into account both signs of the radicals in the given equation.
may
EXAMPLES XXVI. Solve by the aid of the formula in Art. 288 1.
c. :
ALGEBRA.
244
Solve by resolution into factors
= 7 + x.
:
= T-4^x + 3. x2-2 = f|x. 7x2 = 28-96x. 96x2 = 4x + 15.
4x2
13.
6x2
14.
18.
15.
2H-8x2=:26x. 26x-21 + llx2=0.
16.
5x2+26x4-24=0.
20.
25.
35
h'^
28.
x2
26.
36 x2
29.
3 x2
27.
x2
30.
= 5x + 6. 35-4x = 4x2.
25x2
21. 22. 23.
12x2-nax=36a2.
24.
12x2 4-36a2=43ax.
- 2 ax + 8 x = 16 a. - 2 ax - 6x = 0. ax- + 2 x = &x.
:
J^ + ^ = l(x+5).' X+ 4
\/3 X 33.
19.
= 9x^ + 6 hx. - 35 62 = 12 hx. - 2 ax + 4 a& = 2 hx.
Solve 31.
17.
3
X
10
-4_
6x2
35.
x2
+ Vx + X
-
1 I
3x+4
x+1
34.
3
11
+
= VIO X + _
,
'
37 ^'-
2 cd
— +
-
fR
_ x2
?i?.
x2
1
x+a4-6 =^h
42.
^
44 "
45. 46. 47.
48.
50.
^^
1
a;"^6
+
x
2 cx2 + 2 fr2(x+c)
4- j»2
x2
—
m'~
—
x2
+
m'^
?}i2
^
= V6 X +
=
1
=2V3x-
-1
^•
1.
-^+^ = ^-+ — — — 6
X
Vx -p + Vx -
7>i
A^
a
g
c
=
0.
1
h
—
P
Vx —
c
+
Q
,
(7
V(x-2)(x-3)+5^^|^^=
Vx Vx2
+ 6x +
'
dx(x+5c).
^'
^^
^^^
2
4.
?>
a6
|.
+ +X +X (a - 6)x2 + (6 - c)x + c - a = 0. a(6 -c)x2+ 6(c-a)x + c(a- 6) = Va -X + V6 - X = Va + 6 - 2 X. X
^«+
1
,
+
«
3x- 11 ^ 4x+ 13 x+1 x-4 ~ _ +w—2n
2 «i
a 49.
~ +X
= Vx -
4
x
a
VxTf + V3 X + Vx + 3 + V2x+
^-2 43 'x-3
6
+ Ul.
,
41.
38.
???
- Vx +
11^;
>/2x+
'
cx2.
X X
16.
^^-
2
r 72 - -li^ = cZx h + c + 2 ex - 2 dx = „g X + m X — m 40.
2
1
8.
CHAPTER XXVII. Equations in Quadratic Form.
=
294. An equation in the form a.r^" bx'' c, n being a positive or negative integer or fraction, is in quadratic form.
+
Thus
^^
+ 4 x^ =
117,
x'^
+ 7 x'^ = 44,
and
x~'^
-\-
x'i
=a
are
equations in quadratic form. We give a few examples showing that the ordinary rules for quadratic equations are applicable to those in quadratic form. Ex.
1.
Solve
By formula
a:*
-
ISx^
=-
[Art. 288] x^ .
36.
18±\/(13)2
ALGEBRA.
'246
EXAMPLES
5.
+36 = 0. = 8. x^- 19x3 = 216. 8x6 + 65x3+ 8=0. 3V»^-3x"^ = 8.
6.
27 x^
1.
2.
3.
4.
7. 8.
.r4- 13x2 x6
+
7 x3
-
1
=
26
xi
x4-74x2 = - 1225. x-2-2x-i = 8.
XXVII. 10.
a.
EQUATIONS IN QUADRATIC FORM.
247
Squaring and solving the resulting quadratics, we obtain from the first
=
X
G or
—
1
=
and from the second x
;
~-
—
'—
The
first
pair of values satisfies the given equation, but the second pair satisfies
the equation a;2
Ex.
Solve 3x2
2.
Transposing, 3 ^2
Add
-5x-2Va:2-6x + 3 =
- 7 + SVSx'^ - 16x + 21 = Wx. - 16 x - 7 + 3 V3 x^ - 16 x + 21 =
28 to each side
3x2
-
+
16 x
y^
-{-
Sy =
+
Squaring and solving,
.
The
values 5 and
21
+ 3V3x2- 16X +
21
=
28.
we have
1,
\/3x2- 16X
0.
then
;
Proceeding as in Ex.
Thus
12.
2S
we
]
whence
=4
21
or
?/
=
4 or
\/3x2- 16x
— 7. + 21 = -
7.
obtain
= 6,l,or«±|yil.
i satisfy the original equation.
The other values
satisfy the equation
3x2-7 -3\/3x2- 16x4-21 =
16x.
296. Occasionally equations of the fourth degree may be arranged in expressions that will be in quadratic form. Ex.
Solve x*
This
may
-
or
(x2 r
,
8 x3
T
by formula, •^
+ 24 X + 5 = 0. + 16 x- — 6 x2 + 24 x = — — 4x)2 — 6(x2 — 4x) = — 5 +
10 x2
be written x*
'
x2
—
A 4 x
—
8x^
6 ± V36 = —^
20
The student
2'
± =6 ^
2
whence
5,
;
x
=
5,
—
1,
or 2
4
oy^
.
,
5 or 1
;
± V5.
will notice that in such
divide the term containing
=
examples he should
by twice the square root
of
term and then square the result for the third term. In this case a third term of 16 x'^ is required, therefore we write the term 10 a;- of the original equation in the form
the
16
first
0.*^
— 6x-.
297. Equations like the following are of frequent occurrence.
ALGEBRA.
248 Ex.
^^
^!ll^+
Solve
x'^-6
—
—
Write y for
-
;
thus
y
+^=
Q,
or 2/2_6?/
=
^i
or
1.
or
a;2
or
a:
whence
y
is,
a:2
-
- 6 =: 0, x=6, - 1 5
Thus
X
;
EXAMPLES Solve the following equations 1.
a;2
+
a;
+
1
•
a:-
+
X
_
=
.^
3,
XXVII.
_ G = 0. - 2. b.
5.
^2
+
2
Va;--^
a7 3.
4
^)-(-^r-(-^) a^'^-S
3
a;
=
:
42
=
5
X
a;
that
+
+
G
:c
= 24 - 6 X
a;.
CHAPTER
XXVIII.
Simultaneous Equations, Involving Quadratics. 298. We shall now consider some of the most useful methods of solving simultaneous equations, one or more of which may be of a degree higher than the first but no fixed rules can be laid down which are applicable to all cases. ;
299. Equations solved by finding the Values of (x
Ex.1. Solve
= 16 = S6 = 225 2 xy 4iXy = 144 —2 xy. y^ = 81; x — y = ±9. +
x
+/)
(1),
ij
xy
From
(1)
by squaring,
from (2), by subtraction, by taking the square Combining
oc^ -\-
x"^
root,
this with (1)
x x
from which we find
= — y= x = y =
+
y
-\- ij^
(2). ;
;
-\-
we have
to consider the
x X
l5,l 9.
/
12,
\
o.
+ y= —y=—
x= y
i
x-y =
Ex.2. Solve
two
9.
i
S,\
= l2J
12
(1),
= 8b x^-2xy + = 144 ^xy = SiO; = 484 x^ 2xy -{ root, x y = ± 22. ;
from (2), by addition,
y'^
-{•
;
by taking the square Combining
this
-]-
with (1) we have the two cases. x + y = 22,}^ x + y = -22,^ = 12J 12. /
x-y=
x-y
Wlience
(2).
y'^
(1),
a;
=
y=
x
17, ) 5.
=—
5,
\
y=-nJ
i
249
cases,
16,\
xy
From
and
ALGEBRA.
250
300. These are the simplest cases that arisfe, but they are sj^ecially important since the solution in a large number of other cases is dependent upon them.
As
a rule our object is to solve the proposed equations by finding the values of x -\- y and x y. the foregoing examples it will be seen that we can
—
symmetricaUfj,
From
always do this as soon as we have obtained the product of the unknowns, and either their sum or their difference. Ex.1. Solve
a:2
+
//2^74
(1),
=
(2).
xy Multiply (2) by
2,
then by addition and subtraction we have
_ 144, a;y + -2xy + i/= 4. x + y = ± 12, X — y=± 2.
^2 -f 2 a-2
Whence
Srj
?/2
We have now four cases to consider namely, x + y= x+y = -V2,\ y = 12,-) 12, | x-y= 2.) x-y = --2.l x-y= 2.) ;
x
+
From which
the values of x are
and the corresponding values Ex.2. Solve
a:2
x
By
of y are
7,
5,
5,
7,
+ ^- = 185. + y=n
— —
.
2 xy
xy
= =
104
— 7 — 5.
+
?/
x
-
y
==-
12,
|
2. )
;
(1),
.
(2).
subtracting (1) from the square of (2)
.'.
5, 7,
a-
we have
;
62
(8).
Equations (2) and (3) can now be solved by the method of Art. 299, Ex. 1 and the solution is ;
X ?/
= =
13, or 4,
EXAMPLES Solve the following equations 1.
x
+
y
=
2S,
Z.
x
xy=lSl. 2.
X
+
?/
xy
= =
51,
518.
4,
or 13.
\ J
XXVIII.
a.
:
+
y
=
74,
5.
x-y = S,
.
X
-
y
xy
= =
5,
126.
6.
x
:/•?/
=
xy
=
1075,
=
IS.
-
y
513.
SIMULTANEOUS EQUATIONS. 7.
.x 8.
ft;
xy = 92S, + y = Si. - = - 8,
15.
=
x?/
=
17.
3848.
11.
12.
+
13.
2/2
14.
18.
?/2
29.
nA
o
w'^
X 25.
3,
=-
y^
19. ^/.
•?
,
+V= ^^
14.
yV(^ -!/)=!, a:2-4xj/+y2 = 52. 1
26.
=6.
+ = 185, X — ^ = 3. 21. X + = 13, x2 + ^2 = 97. 22. X + = 9, x2 + xy + y''— 61. x2 +^x?/ + =i> + 2, gx2 + x?/ +
=
-{-
^
?/
x-y 20.
?/
- S xy
2/-2
19.
-
a:
x^
?/
?/
?/
23.
65,
?/2
= -8. x-v/ = -18, x?/ = 1363. a;y = - 1914, X + = — 65. = 89, x2 + x?/ = 40. = 170, x2 + xy = 13. x
x'^
1353.
xy=-219S,
10.
=
= 178, = 16. x+?/ = 15, = 125. x2 + x - = 4, = 106. x2 + x2 + y2^180, + X +
16.
:c-?/=-22,
9.
^/^
xy=2S.
?/
xy
+
a;2
251
1
+ = x^ -
x
+
1
+
if-
27.
X
-
2,
y
= 2.
1-1
?/
2/
xy 28.
12
y
wx
=
12.
+by = 2, = 1.
a^x^
2/2
301. Equations which can he reduced to One of the Cases Any pair of equations of the form
already considered.
X-
± pxy -\-y^ = a^
(1),
x±y = b
(2),
where p is any numerical quantity, can be reduced to one of the cases already considered for, by squaring (2) and combining with (1), an equation to find xy is obtained; the solution can then be completed by the aid of equation (2). ;
Ex.1. Solve
x3-?/3
=
999
(1),
x-y = 3 By from
division,
x^ x2
(2),
by subtraction,
From
(2)
and
(4),
-
-\-
xy
2
X2/
(2).
= 333 ?/ = 9 3xy = 324, xy = 10S
+ +
y-
(3);
;
^ y
= =
(4).
1^'
°^'
"
^'
9,
or
—
12.
ALGEBRA.
252
+ x^-y^ + ?/ = 2G13 + + = 67 = 39 Dividing (1) by (2), a;-2-a;^ + = From (2) and (3), by addition, x- + xy = 14:; by subtraction, Ex.
2.
Solve
(1),
x"^
x2
a:?/
(2).
2/2
(3).
2/2
?/2
y
3.
;
x=±7, ±2,\
wlience
Ex.
53
Solve
= ±2, ±7.i
r-j^^^^ •-
^^^ Ex.
1.]
SIMULTANEOUS EQUATIONS. 11
1
,
1
x2'^2/2
_
481 576'
253
ALGEBRA.
254
X,
The student will notice that, having found the values of we obtained those of y from the equation y = mx, using,
m
in each case, the value of
employed in finding those par-
ticular values of x.
303. Equations of which One is of the First Degree and the may from the simple equaOther of a Higher Degree. tion find the value of one of the unknowns in terms of the
We
other,
and substitute in the second equation.
Ex.
=5 = 2\
3:i:-4y
Solve
Zx'^-xy -oif
From
(1)
we have
x
=^
(1),
(2). •'
"^
;
o
and substituting in .
•.
75
^il+M! _ y^^M _ 3 ,^ = 21
(2),
+
120
?/
+
48
2/2
9?/2+ 105?/-
-
15
?/
-
?/2
+
35
?/2
27
=
;
189
;
114=0;
- 38 = (2/-l)(32/ + 38)=0; .-. = i, = 3, and by substituting in (1), 3
12
-
?/
;
.'.
?/
or -^v-;
cc
or
—
i|^.
304. Symmetrical Equations. The following method of may always be used when the given equations are symmetrical, that is, when the unknown quantities in each equation may be interchanged Avithout destroying the The same method may generally be employed equality. with advantage where the given equations are symmetrical except with respect to the signs of the terms. solution
Ex.
Solve
x^
+
y^
=
S2
(1),
x-y = 2 Put then from (2)
x
we
=
u
+
V,
and y
obtain
Substituting in (1),
v
{u .-.
= =
u
(2).
—
v;
1.
+ {u - 1)* = 82 2(1*4 + 0^2 + 1) =82; ?(2 _ 40 = n^ +
+
1)*
;
;
,
SIMULTANEOUS EQUATIONS.
255
= 4, or — 10 = ± 2, or ± V— 10. x=u-{-v = S, -1, 1±V3k); = u~v = l, -3, - liV- 10.
whence
u^
and
;
w
Thus,
7j
We may assume x + y = 2u and x — y = unknown quantities, whence we obtain
Note. ing
y
=z
««?/
—
u
V,
2v, u and v be-
x=
u
v^
and
The examples we have given
305. Miscellaneous Cases.
will be sufficient as a general explanation of the
be employed;
to
-\-
the values used in the above.
but in some cases special
methods
artifices
are
necessary.
xhj"-Qx=S4:-Sy Sxy + y = 2(9+x)
Ex.1. Solve
From
(2).
xV _6x + 3?/ = 34;
(1),
3 y = 54 — 9xy=z- 20,
(2)
Qxy -6x-h
by subtraction,
xhj^
from
(1),
^2
;
9±vsrEM=l±l = ,ov^.
(i.)
Substituting xy
From
(ii.)
these equations
Substituting xy
From
=
we
=
these equations
2
— 2x = 3. x = 1, or — f 1 y = 5, or — 2. /
5 in (2) gives y
4:
obtain
,
we
obtain x
and
y
Ex.2. Solve
?/
—
in (2) gives y
=— =
S
2 x
=
6.
—^^—
±^l v/17.
J
^ 49 + x2 = 19 + x?/ + = 39
+
2/^ 4-
^2
(1)^
^2 -f ^.x
x2
(2),
?/2
(3).
Subtracting (2) from (1), y2
that
is,
(y
_r^2^ ^^y _x)=30; -x)(x + y z) = SO
(4).
-\-
Similarly from (1) and (3),
{z~x')(x-\-y
+ z)=lO
.......
(5).
ALGEBRA.
256 ^
Hence from
(4)
and
by
(5),
division,
—
z
whence
y
Substituting inequation (3),
we
x2- 3x2 + From
X
a:^
(2),
j^
xz
3^2
^
=
?jz
z"-
= =
or
X
= ±2,
=±
,
^
z
= ±3; =.±
1x.
13. 10.
Solving these homogeneous equations,
x
—
obtain
we
obtain
and therefore y ;
=±
and therefore y
5
=^
;
SIMULTANEOUS EQUATIONS. 31.
257
CHAPTER XXIX. Problems leading to Quadratic Equations. 306.
We
shall
now
discuss
some problems which give
rise to quadratic equations.
Ex. 1. A train travels 300 miles at a uniform rate if the rate had been 5 miles an hour more, the journey would have taken two hours ;
less
:
find the rate of the train.
Suppose the train travels at the rate of x miles per hour, then the 300 ^ ^ hours. time occupied is ^' 300 On the other supposition the time is ^ ^ hours .
.
—
;
...
^00_^300_2 a-
whence (x
or
x2
+
+
30) (x
5x
-f
- 750 = - 25) = .-.
x
(1).
X
5
=
0, 0,
25, or
-30.
Hence the train travels 25 miles per hour, the negative value being inadmissible. It will frequently
happen that the algebraic statement of the queswhich does not apply to the actual problem But such results can sometimes be explained by
tion leads to a result
we
are discussing.
In the a suitable modification of the conditions of the question. present case we may explain the negative solution as follows Since the values x = 25 and —30 satisfy the equation (1), if we write — X for x, the resulting equation, :
_J00_^300_2 — x+ 5 — X will
be satisfied by the values x
signs throughout, equation (2)
=-
^2),
25 and 30.
becomes -
—
-
=
Now, by changing
'^ 1-
2
;
and this is the algebraic statement of the following question if the rate had been train travels 300 miles at a uniform rate 5 miles an hour less, the journey would have taken two hours more ; The rate is 30 miles an hour. find the rate of the train. :
A
;
258
'
;
PKOBLEMS.
259
2. A person, selling a horse for $72, finds that his loss per one-eighth of the number of dollars that he paid for the horse what was the cost price ?
Ex.
cent
is
:
Suppose that the cost price of the horse
on $100
is
Hence the
loss
on $x
-^,
x x
is
or
800 .
•
the selling price ""
.
is
x dollars
;
then the loss
$-.
is
-^ dollars 800
'
— dollars.
x
800
Hence
x
—=
-
72,
800
or
x2
that
- 800x + 57600 = - 80) (x - 720) = X = 80, ;
(x
is,
;
.
and each of these values problem. Thus the cost Ex.
3.
A
•
.
or 720
;
found to satisfy the conditions of the either $ 80, or $ 720.
will be is
cistern can be filled
by two pipes
in 33i
minutes
larger pipe takes 15 minutes less than the smaller to find in
what time
it
will be filled
by each pipe
fill
if the the cistern, ;
singly.
Suppose that the two pipes running singly would fill the cistern When running together they will fill in X and x — 15 minutes. - H
X
)
x-15/
of the cistern in
of the cistern in
one minute.
But they -^
one minute
X
;
fill
—
33i
,
or
-— 100
hence
x-15
100
100(2x-15) = 3x(x-15), 1500 = 0, (x-75)(3x-20)=0; .-. x = 75,
3x2-245x-f
or 6f.
Thus the smaller pipe takes 75 minutes, the The other solution, 6f is inadmissible.
larger 60 minutes.
,
Ex. 4. The small wheel of a bicycle makes 135 revolutions more than the large wheel in a distance of 260 yards if the circumference of each were one foot more, the small wheel would make 27 revolutions more than the large wheel in a distance of 70 yards: find the circumference of each wheel. ;
ALGEBRA.
260
Suppose the small wheel to be x feet, and the large wheel y feet in circumference. 780 780 In a distance of 260 yards, the two wheels make and revo-
— — X
y
lutions respectively.
Hence
Z80_I82=135, a;
y
119 X
y
Similarly, from the second condition,
210
210
^
52
we
obtain
^*
;
:
:
PROBLEMS.
261
Hence we have the following equations
^+^ = +
5
(1),
+ -t^ =
7
(2),
X
z
y
12
-
y
oc
z
— +-=5f From
(1)
and
(3),
^=—
by subtraction, y
y Similarly,
From
18 z 9z
(5) (6)
and
by
(7),
+
—^ y-z
(3),
(4)
and from
From
from (2) and
(3).
2
division,
whence
y
= =
y(y y(y
z 1 y
+ —
.
(4).
....
(5).
.
.
.
18
=9
z)
(6);
z)
(7).
= ^^-i^
y-z
= o z; = 4|, x =
from (4) z = 11', and hence ?/ 4. Thus the rates of walking and rowing are 4 miles and 4^ miles per hour respectively; and the stream flows at the rate of 1^ miles per .'.
hour.
EXAMPLES XXIX. 1. Find a number whose square diminished by 119 times the excess of the number over 8.
A man is five times
2.
of their ages
The sum
3.
as old as his son,
equal to 2106
is
of
tlie
:
is
and the sum
equal to ten
of tlie squares
find their ages.
reciprocals of
two consecutive numbers
is
^f
find them. 4. Find a number which when increased by 17 the reciprocal of the number.
Find two numbers whose sum
5.
the difference of whose squares 6.
higher 7.
hour
is
:
If
find
equal to 60 times
9 times their difference, and
is 81.
number and
of a
its
square
is
nine times the next
it.
a train travelled 5 miles an hour faster, it would take one 210 miles what time does it take ?
less to travel
:
Find two numbers the sum
8.
sum
The sum number
is
is
of
whose squares
is
74,
and whose
is 12.
9. The perimeter of a rectangular field is 500 yards, and 14400 square yards find the length of the sides. :
its
area
:::
ALGEBRA.
262
The perimeter of one square exceeds that of another by 100 and the area of the larger square exceeds three times the area find the length of their sides. of the smaller by 325 square feet 11. A cistern can be filled by two pipes running together in 22i minutes the larger pipe would fill the cistern in 24 minutes less than the smaller one find the time taken by each. 12. A man travels 108 miles, and finds that he could have made the journey in 4| hours less had he travelled 2 miles an hour faster at what rate did he travel ? had they cost a dollar 13. I buy a number of foot-balls for $ 100 apiece less, I should have had five more for the money find the cost 10.
feet
;
:
;
:
;
:
of each.
He broke 4 on his 14. A boy was sent for 40 cents' worth of eggs. way home, and the cost therefore was at the rate of 3 cents more than the market price for 6. How many did he buy ? 15. What are the two parts of 20 whose product times their difference ?
is
equal to 24
16. A lawn 50 feet long and 34 feet broad has a path of uniform width round it if the area of the path is 540 square feet, find its ;
width.
A
can be paved with 200 square tiles of a certain size if were one inch longer each way it would take 128 tiles find the length of each tile. outside this 18. In the centre of a square garden is a square lawn If is a gravel walk 4 feet wide, and then a flower border 6 feet wide. the flower border and lawn together contain 721 square feet, find the 17.
each
hall
;
tile
:
;
area of the lawn. 19. By lowering the price of apples and selling them one cent a dozen cheaper, an applewoman finds that she can sell 60 more than At what price per dozen did she sell she used to do for 60 cents.
them
at first ?
Two
20.
rectangles contain the
difference of their lengths
is
same
area, 480 square yards.
The
10 yards, and of their breadths 4 yards
find their sides.
when multiplied by 21. There is a number between 10 and 100 the digit on the left the product is 280 if the sum of the digits be multiplied by the same digit, the product is 55 find it. ;
;
:
A
22.
farmer having
dollars apiece,
outlay 23.
more
how
him x has realized x per cent profit on his
sold, at
finds that he
$75
each, horses which cost
find x.
:
If
a carriage wheel 14f
ft.
fast
is
one second be 2| miles less
in circumference takes
to revolve, the rate of the carriage per
the carriage travelling ?
hour
will
PROBLEMS.
263
he 24. A merchant bought a number of yards of cloth for $ 100 kept 5 yards and sold the rest at $ 2 per yard more than he gave, and received $ 20 more than he originally spent how many yards did he ;
:
buy? 25. A broker bought as many shares of stock as cost him $ 1875 he reserved 15, and sold the remainder for $ 1740, gaining $4 a share on their cost price. How many shares did he buy ? ;
Two trains start 26. A and B are two stations 300 miles apart. simultaneously from A and B, each to the opposite station. The train from A reaches B nine hours, the train from B reaches A four hours after they meet find the rate at which each train travels. :
A
A
from P to Q, two stations 240 miles and travels uniformly. An hour later another train B starts from P, and after travelling for 2 hours, comes to a point that A had passed 45 minutes previously. The pace of B is now increased by 5 Find the rates miles an hour, and it overtakes A just on entering Q. at which they started. 27.
train
starts to go
apart,
A
28. cask P is filled with 50 gallons of water, and a cask Q with 40 gallons of brandy x gallons are drawn from each cask, mixed and replaced and the same operation is repeated. Find x when there are 8 1 gallons of brandy in P after the second replacement. ;
;
A
and B have 30 cows between them they sell same sum. If A had sold and if B had sold his his at B's price, he would have received $320 How many had each ? at A's price, he would have received $ 245. 29.
Two
farmers
;
at different prices, but each receives the
;
30. A man arrives at the railroad station nearest to his house \\ hours before the time at which he had ordered his carriage to meet him. He sets out at once to walk at the rate of 4 miles an hour, and, meeting his carriage when it had travelled 8 miles, reaches home How far exactly 1 hour earlier than he had originally expected. is his house from the station, and at what rate was his carriage driven ?
;
CHAPTER XXX. Theory of Quadratic Equations. MISCELLANEOUS THEOREMS. 307. In Chapter xxvi. it was shown that after suitable reduction every quadratic equation may be written in the
form ax^-^bx-\-c
=
and that the solution of the equation
(1), is
^^ -b±Vb''~^ac
^2).
We shall now prove some important propositions connected with the roots and coefficients of all equations of which (1) is the type. NUMBER OF THE 308.
A
For,
if
ROOTS.
quadratic equation cannot have more than two roots. possible, let the equation ax~
three different roots
values must satisfy
-\-
bx
-{-
7'i,
r^.
+ 6ri + c = + hr,-\-c = ari+br^ + c = ari2
(1)
and
divide out by
(2),
ri
—
=
have
of these
(1),
ari
From
c
Then since each the equation, we have r^,
(2),
(3).
by subtraction,
r^,
which, by hypothesis,
then
aOW
i\)+b 264
= 0,
is
not zero
THEORY OF QUADRATIC EQUATIONS. Similarly from (2) and
.-.
by
which Ti is
is
(3),
a(7\
siibtra.ction,
265
— ^3)=
;
impossible, since, by hypothesis, a
is
not zero, and
Hence there cannot be three
not equal to ?v
different
roots.
309. The terms 'unreal,' 'imaginary,' and 'impossible' same sense; namely, to denote expressions which involve the square root of a negative quantity,
are all used in the
such as
V—
1,
V—
3,
V—
a.
important that the student should clearly distinguish between the terms real and rational, imaginary and irraThus -^25 or 5, 3i, — |- are rational and real; -yjl tional. 7 is irrational and also is irrational but real; while It is
V—
imaginary.
CHARACTER OF THE ROOTS. 310. In Art. 307 denote the two roots in (2) by
then
-b-\--Vb^-4.ac 2 a
2 a
we have
(1) If
r^
-6-V6'-4ac
the following results ac, the quantity under the radical,
and
r^y
:
6-— 4
is positive,
the roots are real and unequal. (2) If 6^ — 4 ac is zero, the roots are real and equal, each
reducing in this case to (3) If b^
—
2 a 4:
ac
is
negative, the roots are imaginary and
unequal. (4)
If
6^—4
ac
is
a
x>erfect square,
the roots are rational
and unequal.
By
ap]3lying these tests the nature of the roots of
quadratic
may
any
be determined without solving the equation.
:
ALGEBRA.
266 Ex.
1.
Show
x-—G a:+7=
that the equation 2
by any real values of x. Here a = 2, 6 = - 6, 6'-2
-
c
4 «c
=
7
;
cannot be
satisfied
so that
=(- 0)-^ - 4
.
2
.
7
=-
20.
Therefore the roots are imaginary.
Ex. 2. For what value of k have equal roots ?
The condition
will the equation 3 x-
(_6)•2-4.3.^' =
0,
=
3.
k
3.
Show
is
cfi
-
62
+
2 &c
roots will be rational provided
a perfect square.
or 4(6
—
c)2.
:c
+
=
Jc
that the roots of the equation
x^-2ax-h The
G
for equal roots gives
wiience
Ex.
—
But
Hence the
-
c2
=
are rational.
(— 2 a)2— 4 («2_ 52^2
this expression reduces to 4(6^
_
&c — c2)
2 6c
+
c'-)
roots are rational.
RELATIONS OF ROOTS AND COEFFICIENTS. 311. Since
,^^-h+VW^^c^
,,
= r±^VF^I^-
2a
we have by
2
rt
addition
_ — h+ Vb^ — 4 ac — & — Vb' — 4 ac _
b
2a
a
...
and by multiplication we have *'''
4(r
be—
By
writing the equation in the form x^-^-x-\ =0, these a a results may also be expressed as follows
In a quadratic equation is
2vhere the coeffieient of the first term
unity,
the
(i.)
with
its
(ii.)
sum
of the roots
sign changed
is
the product of the roots
Note.
unknown
equal to the coefficient of x
;
is
equal to the third term.
In any equation the term which does not contain the quantity is frequently called the absolute term.
.
;
THEORY OF QUADRATIC EQUATIONS.
267
FORMATION OF EQUATIONS WITH GIVEN ROOTS. 312.
— - = ^i +
Since
r^,
and - =
the equation x^
-\-
may
- x -\--
a'2-(-ri
—
(sum of
Again, from
(1)
be written
+ r2).r +
Hence any quadratic may x^
?Y/-2
Ex.
1.
Form
The equation
roots) x
+ product
(x
Form
The equation that
.
(2).
(3).
(x
—
—
2.
—
f
+ 2) = 0, — X — Q = 0.
3) (x
the equation whose roots are f and is
=
=
whose roots are 3 and
-
x'^
2.
of roots
form an equation with given roots.
or
Ex.
(1).
form
we have
the equation is
=
also be expressed in the
(x-n)(:x-r.;)
We may now
7\r2,
ft
ft
= f) (x + |)
;
(7x-3)(5a; + 4)=0,
is,
35x2+13x-12=:0.
or
When
the roots are irrational
it
is
easier to use the following
method.
Ex.
3.
Form
the equation whose roots are 2
We have
sum
of roots
product of roots .-.
the equation
is
by using formula
oc^
— 4:X +
1
= =
+ V^
and
2
- y/S.
4,
1
=0,
(2) of the present article.
313. The results of Art. 311 are most important, and they are generally sufficient to solve problems connected with In such questions the roots should the roots of quadratics. never he considered singly, but use should be made of the relations obtained by writing down the sum of the roots, and their product; in terms of the coefficients of the equation.
;
;
ALGEBRA.
268 Ex.
1,
of (1) a2
We
If
^
a and b are the roots of
1,2^
+
(2) a^
— px
x'^
-{-
q
=
find the value
0,
63.
have
a
=p,
b
-{-
— q. by- 2 ab = p'^ - 2 q. a^ + b^ ={a a^ -\-b^ = (a + b) (a^ + 6^ _ ab) = P{(« + by - 3«&} =p{p'^ ab
.'.
Again,
Ex.
-\-
+ mx +
>i
= 0,
find
a
b
sum
have
Ix;^
—
the equation whose roots are -,
We
equation
If a, b are the roots of the
2.
3^).
= ^-\-^ = ^!±-^,
of roots
=
x b a
.-.by Art. 312 the required equation
ab
a
b
=?
product of roots
1
is
^._,«^+^-'\x-hl=0 ab or
As .'.
- (a^ +
abx^ in the last
the equation
example a^
^
is
or
nlx^
Ex.
3.
-
x^
+
b^)x
=
ab
-\-
0.
= V^^zl^^ and
6^
=-,
ab
P
~
^^^^
— (m^ —
^^^
^
^
2 nl)x
-
= 0,
nl
= 0.
+
+
I
Find the condition that the roots of the equation ax^
-{-
bx
+
c
=
should be (1) equal in magnitude and opposite in sign, (2) reciprocals. The roots will be equal in magnitude and opposite in sign if their
sum
is
zero
;
=
therefore
0,
or 6
=
0.
a Again, the roots will be reciprocals when their product therefore a
Ex.
4.
=
1,
or c
is
Find the relation which must subsist between the
cients of the equation px^
+
qx
-\-
b
+
r
=
when one
the other.
We have
a
= — ~,
ab
=
3
6,
we obtain by 46
=
~;
p
p but since a
unity
= a,
substitution
= - ^, p
3
6-^
= -. p
root
is
coeffi-
three times
THEORY OF QUADRATIC EQUATIONS. ,
From
b-^
= X.
the
first of
these equations
3p
_
g2
.
16i)2
=
Sq^
which
is
and from the second
r
Sp
*
or
= t^-^,
b'^
269
16 pr,
the required condition.
The following example
314.
illustrates a useful applica-
tion of the results proved in Art. 310.
—— /V.2
Ex.
If
aj
a real quantity, prove that the expression
is
.
can have all numerical values except such as lie between 2 and Let the given expression be represented by y, so that
x2
+
2x(l
6.
^'
2(x-3) then multiplying across and transposing,
11
2x
I
we have
-y)+6ij- 11=0.
This is a quadratic equation, and if x is to have real values, 4(1 — ?/)2— 4(6?/ — 11) must be positive; or simplifying and dividing by 4, ^2 — 8 2/ + 12 must be positive that is, (y — Q)(y — 2) must be positive. Hence the factors of this product must be both positive or both negative. In the former case y is greater than 6 in the latter y is less than 2. Therefore y cannot lie between 2 and 6, but may have any other value. ;
;
In this example y^
— Sy -\-12
is
it will
be noticed that the expression
positive so long as y does not lie between
the roots of the corresponding quadratic equation
f-Sy + 12 = 0. This is a particular case of the general proposition investigated in the next article. 315.
For
all real
has the same sign as ax-
-f-
+c =
6jr
Case
I.
values of a,
except
x the when
are real and unequal, and
;
ax"^
-\-
bx
-j-
e
x
lies
between them.
Suppose that the roots of the equation ax^ -^bx-\- c
are real
expression
the roots of the equation
denote them by
i\
and
7*2,
= and
let Vi
be the greater.
;
:
ALGEBRA.
270
Then
^ bx
ax^
c
-i-
= afx^ + ^.^ + -^ a
\
a)
= a \x'-{i\ + r.:)x + i\r.^ = a (X — (x — To).
[Art. 311.]
7',)
Now X
X
if
— Ti,
—
X
greater than
is
or less than
7*i
are either both positive
7'2
rg,
the factors
or both negative
expression (x — Vi) {x — r^ is positive, and has the same sign as r^. But if x lies between the expression (x — r^ (x — r^ is negative, and
therefore the ax-^-{-'bx-\-c ri
and
r^,
the sign of
Case
cidir
If
II.
+ 6a; + c is r^
and aoi?
and
{x
—
?'i)^
is
opposite to that of
-[-hx
-{-
c
= a{x — ri)^,
positive for all real values of x
has the same sign as
-\-hx-\- c
a.
are equal, then
r^
;
hence ax^
a.
Case III. Suppose that the equation ax^+ imaginary roots then
hx-\- c
=^
has
;
ax^ -^ hx
-\-
c
=a\
Q?
+ -X a
(
but since
6^
— 4 ac
is
- y a)
negative [Art. 310], the expression 2
(-^) is
-\-
4ac-62
positive for all real values of x
has the same sign as
;
therefore ax^ -\-hx
-\-
c
a.
EXAMPLES XXX.
a.
Find (without actual solution) the nature of the roots of the following equations
+ X - 870 = + 6x = 5
1.
a;2
2.
8
7. 8.
:*;2.
Form the 6, -3.
-
9,
-
0.
= 14 _ Sx\ + 7 = 4x.
3.
i 3;2
5.
4.
a:-
6.
2x = (.r
x'^-i- 5.
+ 2)--2=4:t;+ 15.
equations whose roots are 9.
11.
10.
a-j-h, i,
f.
a-h.
11.
|«,
12.
0,
f
-|ff.
THEORY OF QUADRATIC EQUATIONS. the equation
If
13. is
the vakie of
A:
+
2(1
k)x
+
o has equal roots, what
=;
A;2
?
Prove that
14.
sc^ _f.
271
equation
tlie
3
— (2 m +
7?ix-
+
.3ii)x
2w
=
has
rational roots.
Without solving the equation Sx^ — 4.r — 1 = 0, and the sum of the squares of the roots.
15.
find the sum,
the difference,
Show
16.
Form
c)x are always
real.
the equations whose roots are
+ V5, 3-V5.
17.
3
20.
1(4
-2^v'3, -2-^3-
18.
"-J^-
^ii-^,
21.
±V7).
h are the roots of
If «,
~\) = (J) —
that the roots of a(a:-
+
-~,l' 5 o
^, A.
22.
the equation px^
19-
(/x
+
=
r
find the
0,
values of
^
52,
23.
rt2
24.
(«-6)2.
29.
If
rt25 _^ ^52.
27.
a^^^
26.
aM-ft*.
28.
^ + ^.
X-
?)
— Px + ^ = 0, If
30.
rt,
—
in
=
and p and g.
terms of
—
of x^
ax
+
0,
= 0,
6
find
the roots of
the equation
—
,
double the other.
Form an
of the equation
equation whose roots shall be the cubes of the roots
2x(x
— a)=
a^.
Prove that the roots of the equation («
+
6)x2
- (a +
&
+
c)x
+^= ^
are always real. 34.
a^, b^
Find the condition that one root of the equation ax2+6x+c =
31.
may he
33.
P and ^
find
5 are the roots
whose roots are
32.
— px +
are the roots of x^
a^fes.
a
6 rif,
+
25.
Show
that
+
-
+
(a
+
b
x
is
real the expression
c)x?
2 (a
&)
x
+
(rt
+
&
-
c)
=
lie
between
has
rational roots. 35.
o
and1
Show
that
if
r 5.
—
X is real, prove that such as lie between 2 and — f ^ 36.
If
.
3
'>'2
^
"~
— I
2
^~
— 15 ^ 2x-8
r2
cannot
can have
all
values except
;
:
ALGEBRA.
272
MISCELLANEOUS THEOREMS. 316. The Remainder Theorem.
+ pi«"^i + divided by x — a, the
sion be
X"
If any algebraical expres-
+ Ih^""'^ H
2^2!^"'''
remainder
h i\- 1^
+ P«
be
ivill
a" +PiCt''~^ +P2ft""- 4-P3«"^^ H
hpn-ltt
Divide the given expression by x — a obtained which does not involve x. Let and R the remainder then
till
Q
+iV
a remainder
is
be the quotient,
;
aj«
+
piic"-^ 4- p2a5"-2 _^
. .
Since jR does not contain ever value we give to x.
Put x
= a,
.
_|_
+ i?„ =
^^_^a;
x, it
will
Q(.^'
-a)-\-R.
remain unaltered what-
then
+ 7>n-ic^ +p„ = Q X + R = a" + i^ift"-! + ^2^" H h Pn-i^^ + J\
a- +pia'»-^ +i92«''^-' +••• .:
i?^
'
which proves the proposition.
From this it appears that when an algebraic expression with integral exponents is divided hjx — a, the remainder can be obtained at once by writing a in the place of x in the given expression. Ex. The remainder when x* — 2x^-{-x~7
is
divided by
a;
+2
is
(_2)4_2(-2)3 + (-2)-7; that
16
is,
+ 16-2-7,
or 23.
317. In the preceding article the remainder is zero when the given expression is exactly divisible by x—a-, hence we
deduce If any rational and integral expresiclien a is written for x, exactly divisible by x — a. (See Art. 105.)
The Factor Theorem.
sion containing x becomes equal to it is
318.
To find
the
condition that
x^
-\-
j^x
-\-
q
may
be
a
perfect square. It must be evident that any such general expression cannot be a perfect square unless some particular relation
MISCELLANEOUS THEOREMS.
273
To find the between the coefficients p and q. necessary connection between p and q is the object of the present question. Using the ordinary rule for square root, we have subsists
x^
px
2x-^§
f
P
^
^
+q
px-\-^
f If therefore y?-\-px+q be a perfect square, the remainder,
q — ^,
must be
by placing
this
Hence the condition
zero.
is
determined
remainder equal to zero and solving the
resulting equation.
An
expression is said to be symmetrical it contains when its value is unaltered by the interchange of any pair of them; thus x-\-y -\-z, hc-\- ca-\- ah, a? ?/ + s^ — xyz are symmetrical functions of the first, second, and third degrees respectively. It is worthy of notice that the only symmetrical expres-
319. Symmetry. with resx3ect to the
letters
+
sion of the first degree in
where
M
is
x, y, z is
independent of
of the
form
M(x-\-y-\-z),
x, y, z.
320. It easily follows from the definition that the sum, and quotient of any two symmetrical The expressions must also be symmetrical expressions. recognition of this principle is of great use in checking the accuracy of algebraic work, and in some cases enables us to dispense with much of the labor of calculation. In difference, product,
the following examples
which
we
shall
assume as true a principle
will be demonstrated in Chap. xlii.
Ex. 1. Find the expansion of (x + y -\- zy. We know that the expansion must be a homogeneous expression of three dimensions, T
ALGEBRA.
274
of the form x^ }- y^ z^ + Aijic^y + xy^ + y'^z + yz"^ + z'^x + zx^) + Bxyz^ where A and 5 are quantities independent of Put = 0, then A, the coefficient of x^y, is equal to 3, the coefficient of x^y in the expansion of {x + yY. Put x = y = z=\, and we get 27 = 3 + (3 x 6) + ^ whence ^ = 6.
and therefore
-\-
a-,
?/,
2;.
;?
;
Thus {x-Vy
=
x3
Ex.
+
+
?/3
-\-
+
^3
zY 3
a:2?/
+
3
Pind the factors
2.
(53 4. c3) (6
-
c)
x?/2
+
3
y'^z
+
Z
yz"^
^ Z zH
^-
^ zx^
+
6 xyz.
of
+ (c3 +
«3) (c
-
a)
+
(«3
+
63) (a
-
6).
Denote the expression by E then E is an expression involving a, which vanishes when a = b, and therefore contains a — b as a factor thus Similarly it contains the factors 6 — c and c — « [Art. 317], ;
;
a) (a — b) as a factor. is of the fourth degree, the remaining factor must be Also since of the first degree ; and since it is a symmetrical expression involving [Art. 319.] a, b, c, it must be of the form ?n(a + 6 + c).
^ contains
(&
—
f)(c
—
E
.-.
E=m(b -c)(c-d){a-
b){a
+
b
+
c).
find most To obtain m 6, c convenient thus by putting a = 0, 6 = 1, c = 2, we find m = 1, and we have the required result. Note. Por further information on the subject of Symmetry, the reader may consult Hall and Knight's Higher Algebra, Chap, xxxiv.
we may
any values that we
give to a,
;
EXAMPLES XXX. Without actual division 1.
2. 3.
find the
b.
remainder when
— 5x- + 5 is divided by x — 5. 3 x5 + 11 x4 + 90 x2 - 19 X + 53 is divided by x + x3 — 7 x?a + 8 xcfi + 15 «3 is divided by x + 2 a. x^
5.
Without actual division show that 4.
32 x^^
5.
3 x4
6.
x>
+
+
-
+ 1 is divisible by x - 1. - 13 x2 - 20 X + 4 is divisible by x" - 4. - 5 x2 - 36 X - 36 is divisible by x'^ - x - 6.
33 x5
5 x3
4 x3
Resolve into factors 7.
x3
8.
x3
9.
x3
10.
x3
+ -
:
+ 11 X - 6. 5 x2 - 2 X + 24. 9x2 + 26x + 24. x2 - 41 X + 105. 6x2
-
11.
x3
12.
x3
13.
6x3
14.
6x3
+ 70. - 31 x - 22. + 7 x2 - x - 2. + x^ - 19x + 6. 39 X 8
x'^
MISCELLANEOUS THEOREMS. Find the values
16. + Qx^ + l^x^+l^x-l. - 2 ax3 + (a2 + 2 h)x? -Zahx +
x^
17.
x4
18.
4p%4 _ 4pqx^ + (g2 + Q'"^^
_
x4
+
2i)2)a;2
2316
«^^*
9 20.
make each
will
2 gca;^
I
2 «x3
+
Find the values
3
+
rt2x2
of x
sions a perfect cube
9
ex
which
+
6a:3
??i%'5
24.
If
?>c.x
g
26.
+ 3x +
31.
^^
d.
make each
of the following expres-
^^i2jix^
_|.
39 ,^^j2^2
22.
_
51
^-
^
+ 4a%2 _
28a6.
,^3.
24.
25.
llx2
^pqx +^-.
be any positive integer, prove that 52«-l
Find the factors
+
2
will
23.
9
a:4
1 b'\
:
8x3-36x2+56x-39.
_
+
_
_5
b'^x'^
21.
11
of the following expres-
:
15.
19
by
which
of x
sions a perfect square
275
of
«(&-c)3 + 6(c-a)3 + c(a-6)3. a(6-c)2 + 6(c-a)2 + c(a-6)2 + 8a6c.
is
always divisible
—
CHAPTER XXXI. Indeterminate Equations of the First Degree.
is
we saw
if the number of unknown number of independent equabe an unlimited number of solutions, and
321. In Art. 167 quantities
that
greater than the
tions, there will
By
the equations will be indeterminate. ditions,
When
however,
we can
introducing con-
number
limit the
positive integral values of the
of
solutions.
unknown
quantities
are required, the equations are called simple indeterminate equations.
The introduction of this restriction enables us to express the solutions in a very simple form. Ex. 1. Solve 1 X -{- 12 ?/ = 220 in positive integers. Transpose and divide by the smaller coefficient thus, ;
Since x and y are to be integers, ^
Now
we must have
= niteger.
multiplying the numerator hy such a number that the division may give a remainder of unity in this case 3, we
of the coefficient ofy
^
have
Dthat
is,
and therefore
1
—
2
?/
15?/ r^
2
—
2
—
H
•
?/ •-
= niteger
;
= integer
;
.
^=
,
?/
276
integer.
INDETERMINATE EQUATIONS. 2
Let
II
then
= m,
=
y
2
an integer
a:
+
24
-
m = 220
84
=
x
.-.
;
-7 m
(1).
Substituting this vahie in the original equation, 7
277
we
obtain
;
+ 12m
28
(2).
or have any negative integral Equation (1) sliows that m may be value, but cannot have a positive integral value. Equation (2) shows in addition that m may be 0, but cannot have a negative integral value greater than 2. Thus the o\\\y positive integral values of x and y are obtained by placing m = 0, — 1, — 2. The complete solution may be exhibited as follows :
m= X
= 28,
?/=
Ex.2.
Solve 5
ic
—
14//
=
-2,
1,
16,
4,
9,
16.
2,
=
Proceeding as in Example
x
-
0,
11 in positive integers 1,
we
2
+ 2?/+^i^;
....
(1).
obtain
5 ...
x-2y -2= ii^-lLl =
integer.
5
Now
multiplying the numerator by
llUii^ that
3
is,
?/
+
^'
4,
we
obtain
integer;
= integer.
"*"
5 1/
Let
^
_L ^
This to
m
is
any
(1),
=
m, an niteger
=
5
x=
14
y
.'.
and from
4
7)1
_
;
4, )
m — 9.
i
called the general solution of the equation,
positive integral value,
values for x and y
:
thus
we
we have
w=
l,
= l, X = 5, ?/
2,
3,
4...
6,
11,
16...
19,
and by giving
obtain an unlimited
33, 47 »-
number
of
:
:
ALGEBRA.
278
From Examples
1 and 2 the student will see that there a further limitation to the number of solutions according as the terms of the original equations are connected by or'—. If we have two equations involving three unknown quantities, we can easily combine them so as to eliminate is
+
one of the
unknown
and can then proceed as
quantities,
above. Ex. 3. In how many ways can $ 5 be paid in quarters and dimes Let X — the number of quarters, y the number of dimes then
?
;
4
.-.
10 2
?/
=
100
2x4-- +
?/
=
50;
/
= 1>,
5X
or
+
.
•
.
.-.
and
X ?/
;
an integer
= 2i7, = 50 -
;
hp.
and 1, 2, 3, •••, 9 however, the sum be paid and 10. If either in quarters or dimes, p may also have the values p = 0, then X = 0, and the sum is paid entirely in dimes if j? = 10,
by giving therefore the number of ways is Solutions are obtained
to
p
9.
the values
;
If,
;
then
?/
=
values of
Thus
and the sum is paid entirely in quarters. x and y are admissible, the number of ways is 0,
if
zero
11.
EXAMPLES XXXI. Solve in positive integers 1.
2.
of
3x + 6x +
8?/
2?/
= =
103.
3.
7x+12?/ =
53.
4.
13x+
11?/
Find the general solution in positive X and y which satisfy the equations 7. 8.
13.
5x-7?/ = 3. 6x-13?/ = l.
9.
10.
A farmer spends
152.
=
414.
5.
23 x
6.
41x
integers,
8x-21?/ = 33. 17?/-13x = 0.
+ 25 = + 47y = ?/
and the
11. 12.
1/
2191.
least values
19y-23x = - 30 x =
77
915.
7.
295.
if each horse •$ 752 in buying horses and cows and each cow $23, how many of each does he buy ? In how many ways can $ 100 be paid in dollars and half-dollars, ;
costs $37, 14.
including zero solutions
16,
?
Find a number which, being divided by 39, gives a remainder and, by 56, a remainder 27, How many such numbers are there ?
15.
CHAPTER XXXII. Inequalities. 322. Any quantity a is said to be greater than another thus 2 is greater than quantity b when a — h is positive —3, because 2 — (—3), or 5, is positive. Also h is said to be less than a when 6 — a is negative thus —5 is less than —2, because —5 — (—2), or —3, is negative. In accordance with this definition, zero must be regarded as greater than any negative quantity. ;
;
323. The statement in algebraic language that one expression is greater or less than another is called an inequality.
324. The sign of inequality is >, the opening being placed Thus, a>b is read " a is towards the greater quantity. greater than 6." 325. The first and second members are the expressions on the left and right, respectively, of the sign of inequality.
326. Inequalities subsist in the same sense when corresponding members in each are the greater or the less. Thus, the inequalities a > & and 7 > 5 are said to subsist in. the
same
sense.
In the present chapter, trary
is
we
shall suppose (unless the con-
directly stated) that the letters always denote real
and positive quantities. 327. Inequality Unchanged.
An
inequality will
still
hold
after each side has been increased, diminished, midtiplied, or
divided by the same positive quantity. 279
;
ALGEBRA.
280 For,
if
> h,
a
then
evident that
it is
+ >6+c — c>b — c; ac > he
rt
a
;
c
328. Term Transposed. trans^wsed from one side
7«
any term may be changed.
-c>b,
a c to
inequality
a?*
to the other if its siyn be
If
by adding
c
each side, a >b-\-
329. Members Transposed.
c.
If
the sides
of an inequality be
transposed, the sign of inequality must be reversed.
For
if
> b,
a
then evidently
330. Signs Changed.
b
If the signs of all the teiims of an of inequality must be reversed.
inequ'dlity be changed, the sign
AYhen a tive
that
;
then a — b is positive, and b — a is — « — (— b) is negative, and therefore — a < —b.
> b,
is,
331. Negative Multiplier. rrodtiplied by the
nega-
If the sides of an inequality be quaidity, the sign of inequality
same negative
must be reversed. For,
if
a
> 5,
then
— a < —b, and — rtc < — be.
therefore
332. Inequalities Combined. If inequalities, subsisting in same sense, be either added, or multiplied together, the results ivill be unequal in the same sense.
the
For
if
«!
and
tti
>
bi,
+ «2 -h «3
a.2
-f-
>
> 63 a„ > b^, it + a^ > bi + bo -f 6. -h
h,,
•••
a^a^ci^
eta
• • •
«m
• • •
> ^^i^A '"
K-
is •••
clear that -f 6„;
INEQUALITIES.
281
333. It folloAvs from the preceding article that
then
a"
and
a~"
where n
is
if
a
> h,
> h'\
< h~'\
any positive quantity.
334. The subtraction of two inequalities subsisting in the same sense does not necessarily give an inequality subsisting in the
same
sense.
335. The division of an inequality by another subsisting same sense does not necessarily give an inequality subsisting in the same sense. The truth of these last statements is readily seen by considering the inequalities in the
5>4, 3>2. Subtracting member for member would give 2 > Dividing member by member would give | > 2. Ex.
1.
2.
Find hmit of x in the Inequality 3
Clearing of fractions,
15
5
we have
15:c-25>3a:+
11.
Transposing and combining, 12 .-.
a:
> 36
;
x>3.
Note. The word "limit" is here used as meaning the range of values that x can have under the given conditions. Ex.
2.
If «, h, c
denote positive quantities, prove that a2
_i_
+ c2 >hc-\+ c2 > 2 he, c2 + a2 ;> 2 ca
52
For
ca-\- ah.
&2
a'^-\-h^>2ah.
Whence by
addition a-
+
h-
+
c-
>
Z>c
+
c«
+
ah.
ALGEBRA.
282
x^
Ex.
3.
If
X
may have any
+
or ^2
+
X.
1
^
x^
\
-{x^
-
x')= x^
\-
(a;
—
1)2
^\
according as x
<—
a:> or If
X
=—
1)2(X+
+1>
x3
<
or
1,
becomes an
the inequality
EXAMPLES
that
;
is,
according as
equality.
XXXII. :
llx-4^<^+3i. (a;
+
2)(a:
3.
&x
4-
5 ax
4.
Prove that {ah
5.
Prove that
6.
Show
reciprocal
is
7.
If a2
8.
If a2
+ 3)>(a;-4)(a;-5). — 5 ah > V^ when « > &. (6
+ a:^) (aa: + hy) > 4 «?)a;?/. + c) (c + «) (a + 6)> 8 a?>c. sum
that the
never less than
+ +
52
^
1^
52
_|_
c2
and
=
a;2
^
and
1,
Which
is
quantity and
its
^2
_
+ by + ^2
show
1^
+
?/2
^2
that ax -\-by
_
g^ow
1^
that
cz<:^l.
^-^ ?
+
Show
11.
Find which
12.
Prove that
13.
Prove that 6 a6c
14.
Show
that
15.
Show
that 2(a^
that
positive
real
the greater ^^-±-^ or 2 «
10.
Note.
any
of 2.
ax-}-
reader
^
in the following three inequalities
2.
9.
+
a;2
1.
Find limit of x 1.
1)
1).
or negative
positive
is
the greater,
hence
positive,
is
is
-{x - Y)^ {x? -r){x-
x"^
= {XNow
which
real value, find
{x^-y is
?/22
+
^
«^3
<
6t'(/>
ft^
qj-
?/02 4. ^^.2^
+
a^
2
;> 9 x2^2^2.
?>3.
< «* + 54.
62c2 4. c2a2 _^
+
?>
zH) (xi/ +
the greater, 3 «62
«3?j
+
+
-\-
c)
f^2?^2
-\-
ca{c
+
-> ^^^c(a
c3)> 5f (?>
+
c)
+
a)
+ +
6
+
«?>(«
+
b).
r).
ca(c
+
«)
+
a6(«
+
&).
For further information on the subject of Inequalities the consult Hall and Knight's Higher Algebra, Chapter xix.
may
MISCELLANEOUS EXAMPLES.
MISCELLANEOUS EXAMPLES a
=-
= 2, c 0, d 1, e = a^(d — c) — VS ae + cib
6
1.
If
2.
Simplify [3(a
3.
Solve 3 X -
4.
A
1,
=
=
of
whom
is
V.
find the value of
3,
dic-a)-2ad^i-Vlab - & + c) - (a -?>)(& - c) + {(a + 4(7x-9) _4/g ^x-l\ 15
man's age
283
is
6
-
c) (3
-
&)}].
5
four times the combined ages of his two sons, one
three times as old as the other
;
in 24 years their
ages will be 12 years less than their father's age
combined
find their respective
:
ages.
Solve
6.
(i.)
SVx-l^
^
3Vx+ (ii.)
7.
Expand
8.
Simplify If
9.
i
;
if
3
3
is
is
VS X +
(2 a
-
3
— 2x+
added
17
+ 0. 7
- V2x = V2 X +
9.
62)5.
3(x
1
to the
+
^•3
6x +
i)
numerator a certain fraction
is
increased by
taken from the denominator the fraction reduces to
}
:
find
the fraction. 10.
Find the value of
(i.)
+ 2V^- + + 3V2
3\/2i3 \/3
(ii.)
4\/75
- Vf.
2+V6
11.
Solve
(i.) ^
^
(ii.)
12.
Find the
—-2x1 ^— + —^ = + 2x f(x-2x2)+
The breadth
-
4) (x
-
14.
5)
>(x -
2) (x
-
1).
is
252 square yards
:
side.
Solve
y^ = V>-, x + y = i. + x?/ = 4, 3 xy + 4 = 22. factors of (i.) x^ + 12 x'^y - 45 xy^. (ii.) 3x2-31x?/ + 56?/2.
(i.)
(ii.)
15.
-2x)=5(i-x).
of a rectangular space is 4 yards less than its
length; the area of the space
each
1(1
limit of x in the inequality
(x 13.
16.
1
Find the
x2
+
2 x2
?/2
find the length of
:
ALGEBRA.
284 16.
_^ +
Simplify
2
ic3
^
^
-
— 4 a:2
2 x^
+
1
4 x^
Solve
18.
Find two numbers whose sum
3
19.
is
5
Solve
y
and the sum
22,
is
of their
Vqi-^
^3+
+ 4x -
For what value of k
have equal roots
/
h'
3
\c
^c^
^_-^
(i.)
(ii.)
22.
X
y
2c"^.. X
Simplify
y/x-1 21.
4
250.
\h 20.
—
^±i + ^+J=6, 2j^+6_2^J^^5
17.
squares
x-
4
-6+1
^^ + ^ 2
+ Vx-^ +
4a;
will the equation
-
10
x"^
+
= 6. 2(^
+
2)x
-j-
dk =
?
Simplify the fractions
a
...
—X
1
....
1-^
a+ «^ —6
X
rt
pays $ 28 more rent for a field than A he has three-fourths of an acre more and pays $ 1.75 per acre more. C pays $ 72.50 more than A he has six and one-fourth acres more, but pays 25 cents per 23.
B
;
;
acre less 24.
:
find the size of the fields.
Solve
(i.)
Hi
i.
26.
and
^
_2x_
10_11 X
6
2x -
,
x-4
Find the value of
(
5
3
-l +
V^V^
which
will rationalize
Find the square root
of (i.) 11 (ii.)
28.
25
x-3
f
~ ^ ~ ^^^^ V2
Rationalize the denominator of - _ V'2
find a factor
27.
x + 10 X — 5
Find the factors of
(i.) (ii.)
+
+ V3 - V5 \/3 — \/2. 4\/(3.
-5+
12
V"^.
a2 _ jq - 6 ax 343xC-27«/3.
+
9
x2.
V when n =
3.
X
1
MISCELLANEOUS EXAMPLES 29.
Solve
(i.)
285
V,
x"^
A rectangular
X
^
y^
x^
2
y
If it were reduced to a by cutting an oblong piece off one end, the ratio of the piece cut off to the remainder would be less by ^^^ than the ratio of the remainder to the original field. Find the length of the field. 30.
square
31.
100 yards wide.
field is
field
Extract the square root of 4
(i.)
ic*
+
Solve
(i.)
(ii.)
33.
+
13
+6
a;V
X
a
a'2
32.
12 xhj
ic?/^
+
?/.
x^
= 13, 42: = 17, 4ic + 3?/ + 20 = 16.
2x+ + X + 2?/ + ?/
3^
-i-+—i— + \-x
Vx +
-^
^
Vx-l
1
Simplify the fractions 1
1
1_ J_
+
2
1—
2x
X—
— y^ 2x3 + 5x-^y-5x?/2 + ?/ 2 x^
(ii.)
a
i_
1
—
7 x^y
+
5
xy"'^
Two places, A and B, are 1G8 miles apart, and trains leave A B and B for A simultaneously they pass each other at the end of one hour and fifty-two minutes, and the first reaches B half an hour 34.
for
;
Find the speed
before the second reaches A. 35.
Solve
(i.)
x*
(ii.)
36.
Form
of each train.
+ 4 x + 2 Vx2 + xT3 = + = 706, X + = 8.
3 x2
the equation
?/4
30
-
x^.
?/
whose roots are the squares
of the
of the difference of the roots of
2 x2 37.
Employ {a
38.
Solve
the
+ 2{m +
method
-\-by-a^-b^ = x«/(3
x
n)x
+
iii^
+
=
n^
0.
of Arts. 319, 320 in showing that
+ y)=
5 ab(a
+
10, 27 x^
h) (a^
+
^/
+
ab
= 35.
+
6'^),
sum and
CHAPTER
XXXIII.
and Variation.
Ratio, Proportion, 336. Definition.
is the relation which one quansame kind, the comparison being
Ratio
tity bears to another of the
made by considering what
multiple, part, or parts, one quan-
tity is of the other.
The
ratio of ^4 to -B is usually written
tities
A and B are
term
is
A
:
The quanThe first
B.
called the terms of the ratio.
called the antecedent, the second term the consequent.
To find what 337. Ratios are measured by Fractions. multiple or part is of B, we divide hj B-, hence the
A
A
ratio
A B :
may
be measured by the fraction
A —
and we
,
shall usually find it convenient to adopt this notation.
In order to compare two quantities, they must be expressed in terms of the same unit. Thus, the ratio of f 2 to 15 cents is
Note.
measured by ^ the fraction
—
•
3
number of times that one quanan abstract quantity.
Since a ratio expresses the
tity contains another, every ratio is
338.
or
15
By
7='-^;
Art. 136,
mo
ma mh that of a ratio remains unaltered if the antecedent and the consequent are midtiplied or divided by the same quantity.
and thus the is,
ratio
a
:
6 is equal to the ratio
:
;
the value
339. Comparison of Ratios. Two or more ratios may be compared by reducing their equivalent fractions to a common denominator. Thus, suppose a h and x y are two :
ratios.
Now,
h
= -^,
and -
by
=^
y
by 286
;
:
hence the ratio a
\
h is
AND VARIATION.
RATIO, PROPORTION, greater than, equal as ay
is
or less than the ratio x
to,
greater than, equal
to,
287
y according
:
or less than hx.
340. The ratio of two fractions can be expressed as a ratio of
two
integers.
-
- or
—
and
d
be
-j-
b
'
\
Thus, the ratio -
:
-
is
measured by
therefore equivalent to the ratio ad
is
:
be.
341. If either, or both, of the terms of a ratio be a surd quantity, then no two integers can be found which will
measure their ratio. Thus, the ratio be exactly expressed by any two integers. exactly
^2
:
1 cannot
342. If the ratio of any two quantities can be expressed exactly by the ratio of two integers, the quantities are said to be commensurable
;
otherwise, they are said to be incom-
mensurable.
Although we cannot find two integers which will exactly measure the ratio of two incommensurable quantities, we can always find two integers whose ratio differs from that required by as small a quantity as we please.
^ = 2:236061^^
Thus,
4
553016...;
4
and, therefore,
V5
•
-^^— is
4
^ 559016 and < ^ 559017 > 1
1000000
;
1000000
it i§ evident that by carrying the decimals further, any degree of approximation may be arrived at.
and
343. Katios are compounded by multiplying together the which denote them.
fractions
Ex. Find the
ratio
compounded 2 a :3 b,
The required
ratio
When
of the three ratios :
5
5 c^
two identical
is
:
a.
^-^5c
ratios,
pounded, the resulting ratio
c
c2,
=^ y,^^ x ^a = 3 6
344.
6 «&
a^
a :
:
b
b^,
and a b, are comand is called the :
,
;
ALGEBRA.
288 duplicate ratio of
a
cate ratio of ratio of
a
:
Similarly,
b.
Also,
a'^
(c-
// is
:
called the tripli-
h^ is called the subduplicate
:
The duplicate ratio of 2 « 3 is ^a" 9b^. The subduplicate ratio of 49 25 is 7 5. The triplicate ratio of 2x 1 is 8r/^ 1.
(1)
/>
:
(2)
:
:
(3)
345.
:
h.
:
Examples
A
a h.
:
:
:
ratio is said to be a ratio of greater inequality, or
of less inequality, according as the antecedent is greater or less
than the consequent.
346. If to each term of the ratio 8 ratio 12
:
7
former because ^This
is
which we
A
we
obtained, and
is
is
3
:
see that
clearly less than
4,
a
new
than the
|.
a particular case of a more general proposition
now
shall
prove.
ratio of greater inequality is diminished,
inequality
we add it is less
is
increased,
and a ratio
by adding the same quantity
of less
to both its
terms.
Let - be the
ratio,
and
" let
b
b
by adding x
+x
be the
new
Now
— b) — — bx = x(a -^ ^
a
X a + -^— = ax
b
b-{-x
b{b-{-x)
b(b-\-x)'
and a — b is positive or negative according as a less than b. -\a >^ TT '-c - is ^ Hence, it a is > o, >a X 7
•
-r II
a
•
<^ b, 7
is
a b
— —X <^a— —X b
b T
formed
to both its terms.
^T
and
ratio
is
greater or
;
-\-
-\-
is
'
b -i-x
which proves the proposition. Similarly,
it
ity is increased,
can be proved that a ratio of greater inequaland a ratio of less inequality is diminished,
by taking the same quantity from both
its
terms.
347. When two or more ratios are equal, many useful propositions may be proved by introducing a single symbol to denote each of the equal ratios.
;
RATIO, PROPORTION,
_
;
.
AND VARIATION.
The proof of the following important theorem method of procedure.
289
will illus-
trate the
each of these ratios
where
n, are
q, r,
yo,
^(p^l±l^l±j:^
T Let 4.
any
quantities whatever.
ace =- = -=... d
b
7. =^k;
f
a = bk, c = dk, e = fk, = 7)6"^'", qc'' = qd"k'% re" = rf'k'\ pb"k'' + gd»A;^^ + ?/"^•» + jm^ + qG"" + H _ ~ pb"" + gcr* + r/" + p?>" + gc^ + ^/" H pcC' qd' 4+ ••'V _ _ « _ ~ ~b~d~ p6'' + gd" + r/'" H y
then
• • •
whence
pa**
• • •
re''
'
•'•
_
^.»
""
...
'
1
'
>'g"
-\-
By
7.
n many particular be deduced or they be proved independently by using the same method. giving different values to p,
cases of this general proposition
may
_^
For instance,
f
fractions are ecpial, each of
them
is
numerators divided by the sum of 1.
= --^—^^ b + d+f'
of these ratios
;
d which may be thus enunciated
b
Ex.
;
if
-=-:=-. each a result
q, r,
may
If
^
=-
y
4
equal
cdl
fiud the value of ^ ^
~
When a sum
:
to the
series
of the denominators,
of
all the
^
"^
.
7x + 2y
5x-3y _
—— —— 3
3
J
y_
"
y
_3^
4
Ex. 2. Two numbers are in the ratio of 5 8. If 9 be added to each they are in the ratio of 8:11. Find the numbers. Let tlie numbers be denoted by 5 a: and 8 x. :
Then
5£+9^1 8x +
9
Hence the numbers are 15 and u
11 24.
._
^
^
3^
ALGEBRA.
290 Ex. that
By
A B
If
3.
be in the duplicate ratio of
:
A
x
-{•
B
:
x,
-\-
prove
= AB.
:c2
^
the given condition,
''''
{
)
=A
B
B{A xy = A(B A'B + 2 ABx + Bx"^ = AB^ + 2 ABx + x\A - B)= AB(A - B); x^ = AB, .'.
-{- ot^"^,
-{-
Ax'^y
.'.
A—B
since
is,
by supposition, not
zero.
EXAMPLES Find the
ratio
compounded
XXXIII.
a.
of
5.
The duplicate ratio of 4:3, and the ratio 27 8. The ratio 32 27, and the triplicate ratio of 3 4. The subduplicate ratio of 25 36, and the ratio 6 25. The triplicate ratio oi x: y, and the ratio 2 ?/2 3 xJ^. The ratio 3 a 4 6, and the subduplicate ratio of b^ a'
6.
If
1.
2. 3.
4.
:
:
:
:
:
X
:
=
y
5
:
:
7,
find the value of
9.
If 6
^
If
:
a
= -,
If 7
a:
If ^
^'"^
11. 12.
If 2
5,
and ^
= ^,
+
-
a
?/
41'
:
—
y
3 6
find the value of
:
5
:
13, find the ratio
find the ratio a
P: Qhe
14.
If
b
cl
P—
the subduplicate ratio of
= - = -,
-
a,
x
:
y.
— m :Sy —
:
li
/;
b.
:
X 3 y be in the duplicate ratio of 2 x Qxy.
13.
3
~1 ax
4by
=
x.
^^^~^•'^
7
:
62
y
x-^y 2x-by
find the value of 2
y ?/
a2
=
:
— 4 3X + ~ ^ ^^ = A
10.
rri^
2
4
6
that
r=
+
find the value of x
y 8.
:
:
x
prove that each of these ratios
:
Q—
is
x,
m, prove
prove that
equal to
f ^'2a2c
^.
+
3c3e
+
4e2c
15. Two numbers are in the ratio of 3 4, and if 7 be subtracted from each the remainders are in the ratio of 2 3 find them. 16. What number must be taken from each term of the ratio 27 ;35 that it may become 2:3? :
:
:
AND VARIATION.
RATIO, PROPORTION, 17.
that
it
18.
What number must may become 8:7? If
19.
20.
If
If
be added to each term of the ratio 37 29 :
—^ = —^— = —^^,
show that p
-J— = _1_ = _^,
show
—
b
c
c
—
a
-\-
-
= - = — show
c
-\-
,
'
—
a
b
c
—
a
a
291
-]-
q-\-r
=
0.
b
that
x-y
-\-
= 0.
z
b
that the square root of
^^!^^l1|!£+1«Wjs b^ - 2 d^f + S b*cd^e'^
^.
equal to
bdf Prove that the ratio la^mc-\-ne:lb-\-md+nf ^\\\ be equal to each of the ratios a:b, c:d, e :/, if these be all equal and that it will be intermediate in value between the greatest and least of these ratios if they be not all equal. ^
21.
;
23.
bx-ay ^ ex — az by —
If
be equal to
-,
unless 6
az
^ z±ji^
ax
x
+c=
0.
cy
-{-
^.^^
^^^^^^
^^^^ ^^
^^^^^^ fractions
z
y 23.
"^^-^y = ^ZLJ/ = _^±l£.
If
oz is
equal to -
;
-\-
z
y
-X
2y
g ^^^^ g^^j^ Qf ^j^ggg j.^^.^g
-3x
=
hence show that either x
y,
=x
or z
-\-
y.
PROPORTION. Four quantities are said to be in 348. Definition. when the ratio of the first to the second is equal to the ratio of the third to the fourth. The four quantities are called proportionals, or the terms of the proportion. proportion
-
= -,
then a, b, c, d are proportionals. This is d expressed by saying that a is to & as c is to d, and the proThus,
if
b
portion
is
written
a
The terms a and d 349.
h
:
:
c
:
d,
or a
:
b
=c
:
d.
are called the extremes, b
and
If four quantities are in proportion, the
extremes
Let
\
is
a, b,
equal to the product of the means. c,
d be the proportionals.
c
the means.
product of the
;
ALGEBRA.
292
Then by
-=h d ad = bc.
definition
whence
Hence fourth
if
any three terms of a proportion are given, the
may be
Thus
found.
if a,
d are given, then
c,
h=~c
Conversely, if there are any four quantities, a, b, c, d, such that ad = be, then a, b, c, d are proportionals a and d being the extremes, b and c the means or vice versd. ;
;
Quantities are said to be in 350. Continued Proportion. when the first is to the second, as the second is to the third, as the third to the fourth and so on. Thus a, b, c, d, ••• are in continued proportion when continued proportion
;
a_b _c__
bed
If three quantities a,
b, c
a:
and
c
;
351.
and
b= b = b\
e;
:
ac
.-.
In this case
are in continued proportion, then
(Art. 349.)
be a mean proportional between a said to be a third proportional to a and b.
b is said to
c is
If three quantities are proportionals, the first is to the
third in the duplicate ratio of the first to the second.
Let the three quantities be
a, b, e;
then -
= -•
b
« = «x?!
Now
b
c
that
a:
is,
352. The products
= ^'x- = b
e
e
= a~
b :
c
^'; b'-
b^.
of the corresponding
terms
of
two
or
proportions form a proportion.
If
a:b
= c:d
and
e
:/=
For ? = ^ and b d
g
:
h,
bf = eg ^ = »;.-. ^ = ae
f
h'
:
If
then will :
dh.
%
dh'
or «e
:
bf= -^
cy •'
:
dh.
more
AND VARIATION.
RATIO, PROPORTION, Cor.
=c =d a: X = c
If
and
a: b
:
X
:
b
then
:
d, y, y.
:
may
353. Transformations that
293
made
be
in a Proportion.
d form a proportion, many other proportions may be deduced by the properties of fractions. The results of these operations are very useful, and some of them are often quoted by the annexed names borrowed from Geometry. If four quantities, a,
a ^
(1) If
:
For 3 =
:
d,
therefore
^;
a—d
then b\
d
b
b^d,
is
'
a or
b: (2) If a :b
For
ad.
=
= be
c: d,
a=d:
then a
:
'^
therefore
G
a
or
a:b
ror^ = ^; b
that
cd
—
:
c:d, then a
therefore ^
d'
c
=b
-{-
:
d.
b:b
= c-\- did
[Composition.]
+ 1=^ + 1 d
b
a-\-b
c
is,
+d '
d
b
= c-\-di d. a — b ib = c
a-\-bib
or (4) If
For that
a
:
6
a
c
b
d'
=c
:
d,
then
therefore
--1=^ b
a
—b
d
is,
b
or
[Alternation.]
= ^',
- — -• c~d''
is,
(3) If
c.
bid.
c
cd that
[Inversion.]
c.
1---a = 1-- -
d
b
that
=c
b, c,
a
d
'
— bib = c — did.
d
I
d.
[Division.]
;
ALGEBRA.
294
For by .'.
by -^
= c:
a:b
(5) If
then a
d,
1.
b
-\-
—b=c
a
:
may
2 a2
^
2 62 or
Ex.2.
(3a
-
-\-
d:
c
— d.
-^ = ^^
(4)
d
d
_
3 ^2
k; then a
=
bk, c
e2
:
— d.
e :/,
+
5
c
:
be proved in a similar way.
2 62
=
5^-2
=
dk,
ae
:
bf.
e=fk;
f
+ 3fZ2_5/2~ 2 a2 + 3 c2 -
+ 3^2-5/2 + 3 d2 _ 5/2
2 62 5 e2
6
2 62
:
3^
«e
.
/
6/'
^y.
If
+ 66 +
prove that
We
3 c2
-I-
=^=^=
b •'•
-\-
a:b = c:d =
If
Let
and by
i
Several other proportions
that
c
-;
a
Ex.
—b=
— -— = a — bc — d
division,'
or
show
a
:
[Composition and Division.]
^
^^ =
(3)
b
-\-
c
+ 2f0(3a-66 -c + 2d) = (3a-66 + c-2fZ)(3a + 66 -c-2fZ),
a, 6, r,
have
are in proportion.
tZ
3« + 66 +
^ 3a + 66-c-2.Z, 3a-66-c + 2d a -c) 2(3 a + c)' ^ — 2(3 ^ 2(6 6 + 2d) 2(6 6 -2c?) + c ^ 66 + 2(Z
+
c
2r?
3a-66 + c-2d ,.
,
.
.
Composition and division 3ff
Alternation,
^^^^
3^,^
3«-c~66-2d*
— = 4d
Again, composition and division,
;
2c
whence Ex.
3.
a
h
:
= -
whence, dividing by
a;2,
—
4
2
4-
.r2
5
dx —
2
4 5
and therefore the roots are
.r
-
0,
5
.r
.^
-
=
0,
-
6
6
6
which gives a solution x
1
x-2
d.
4a;2
a;2
X
:
x^-{-x-2
Solve the equation
Division,
c
whence, x 6
—
2.
=—
2
[Art. 291, note.]
:
AND VARIATION.
RATIO, PROPORTION,
EXAMPLES Find a fourth proportional 1.
a, ab, c.
4.
ab.
a'^b,
a^2a&,
3&2.
3.
x^,xy,6x2y.
to
2x2.
x^,
5.
b.
to
2.
Find a third proportional
XXXIII.
295
Sx,6xy.
6.
7.
1, x.
Find a mean proportional between 8.
a^b^.
2x3, 8 X.
9.
n. 27 a-b^,Sb.
12ax2, 3«3.
10.
be three proportionals, show that
If a, 6, c
+ b = a — b a — c. + 6c + c2) («c - 5c + c2) = &* + «c3 + ci prove that If a & = c 14. ab -\-cd:ab-cd = a'^ + c^:a^- c2. 15. a2 + fjc + C2 rt2 _ «C + C2 == &2 +6(^4. d^ ^2 _ 6 = V3 a2 + 5 c^ V3 62 + 5 d^. 16. a 12.
a
a
13.
(/;2
:
:
:
:
fZ,
;
:
17.
-
+ -:
,0
-
i>
+
10.
a
5^
+
c22_
:
:
,
a
=- + -:<^_^d
a
ab
6
a2 4. 52
,
c
cd
c cZ
c2
+
^2
Solve the equations 19.
3x- 1 :6x-7 = 7x-10:9x +
20.
X
2j
- 12 + 3 = 2 X - 19 x2-2x + 3 ^ x2-3x +
23.
If (a
:
?/
prove that
6
~
a, 6, c,
24.
If «, 6, c,
25.
If 6 is a
?/
5
_
10.
13
=
5
:
14.
2x- 1 _ x x2 + x + 4 x2 + 2x-l
22
- 3 d)(2 a - 2 6 - c + cZ) = (2a + 26-c-d)(a-6-3c +
3c
d are in continued proportion, prove that a d = a3 4- 63 + c3 63 + c3 + #. :
mean
proportional between a and
9 ^2
to 4 62
is
-
9
c2 in
c,
show that
the duplicate ratio of a to
d are in continued proportion, prove that 6 mean proportional between « + 6 and c + d. 26.
If a, 6, c,
27.
If
prove that a
a
=
c,
3fO
d are proportionals.
:
4 ^2
-
3x-5
2x-3 +
5
:
or a
+ 4-
6 6
6 + c = c + + c + = 0. :
cZ
(Z
:
cZ
+
a,
+
h.
c is a
:
;
•
ALGEBRA.
296
VARIATION.
One quantity A
354. Definitiox.
is
said to vary directly
two quantities so depend upon each changed, A is changed in the same ratio.
as another B, Avhen the
other that
B is
if
The word
Note,
directly
often omitted, and
is
A
said to vary
is
as B.
355. For instance
a train moving at a uniform rate it will travel 20 miles in 30
if
:
travels 40 miles in 60 minutes,
minutes, 80 miles in 120 minutes, and so on
the distance
;
in each case being increased or diminished in the
same
ratio
This is expressed by saying that when the uniform the distance is proportional to the time, or
as the time.
velocity
more
is
briefly, the distance varies as the time.
356. The Symbol of Variation. denote variation; so that cc ^
^
357.
If
A
varies as B, then
/I
The symbol is
is
read
"A
B
equal to
oc is
used to
varies as -B." multiplied
by
some constant quantity.
For suppose that A and B.
a^..., b^, bo, 63
cio,
«i,
...
are corresponding
values of
Then, by
definition,
- = -; - = -', - = bi
Cli
.*.
=^= ^=^ b2 bi
JJar^n^ -tlence
that
is,
^ ^
A = m, — B
358. Definition. versely as another
.
is
and so on
^
,
— B
^,
always the same
m is constant. A = mB.
where
One quantity
B when A
[See Art. 176.]
;
63
Cfcg
^
'
ValuC Of ^^^Y ^ -_the corresponding value of i>
.-.
of B.
bo
each being equal to
..
bs
ffco
A
is
said to vary
in-
varies directly as the reciprocal
•
RATIO, PROPORTION,
Thus
if
A
AND VARIATION.
varies inversely as B,
A=—
,
where
297
m
is
con-
stant.
The following
If 6 is an illustration of inverse variation a certain work in 8 hours, 12 men would do the same work in 4 hours, 2 men in 24 hours and so on. Thus it appears that when the number of men is increased the :
men do
;
time
is
proportionately decreased; and vice versd.
359. Definition. One quantity is said to vary jointly as number of others when it varies directly as their product. Thus'^ varies jointly as B and C when A = mBC. For instance, the interest on a sum of money varies jointly as a
the principal, the time, and the rate per cent.
360. Definition. inversely as
C when
A is said to vary A varies as —
directly as
B
and
C
361. Grouping the principles of Arts. 357-360, we have
A varies directly as B, if A varies inversely as B. if A varies jointly as B and C, if A varies directly as B and inversely
A = mB, A = —, A = mBC, A = ^^, 362.
If
when B C vary.
is
A
if
varies as
B when C
constant, then will
is
constant, and
A vary
as
A
BC when
as C.
varies as
both
B
C
and
of A depends partly on that of B and partly Suppose these latter variations to take place separately, each in its turn producing its own effect on A; also let a, b, c be certain simultaneous values of A, B, C.
The variation
on that of
C.
then A must 1. Let C he constant while B changes to b undergo a partial change and will assume some intermediate ;
value
a',
where
^=1
w.
;
ALGEBRA.
298 Let
2.
B
from
its
he constant, that
is, let it retain its value h, Avliile then must complete its change and pass intermediate value a' to its final value a, where
C changes
to c
A
;
^Jl
——
a
From that
and
(1)
(2)
c
X -=^ X ^ a' a b c
(2)
A = ~'BC,
is,
be
A varies as BC.
or
363. The following are illustrations of the theorem proved in the last article.
The amount
work done by a given number of men varies number of days they work, and the amount of work done in a given time varies directly as the number of men therefore when the number of days and the number of men are both yariable, the amount of work will vary as the product of the number of men and the number of days. Again, in Geometry the area of a triangle varies directly as its base when the height is constant, and directly as the height when the base is constant and when both the height and base are variable, the area varies as the product of the numbers representing the height and the base. Ex. 1. If ^ X B, and CocI>, then will ACcc BD. For, by supposition, A = mB, C = uD, where m and n are conof
directly as the
;
;
stants.
y
Therefore
AC = muBD
Ex.
X varies inversely as
=
10
By
2. ;
If
find x
when
supposition, x
Putting X
=
24, y
y
=
;
and
as ran
is
constant,
—
1,
and
[/'-
y-- 1 = 10, we
,
where
m
obtain 24
constant.
is
= ^, 99
m = 24
x 99 24 X 99
y^-1 hence, putting
?/
=
5,
we
equal to 24
o.
= —^—
whence
is
obtain x
=
99.
;
AC x BD. when
AND VARIATION.
RATIO, PflOPOiiTION,
299
Ex. 3. The volume of a pyramid varies jointly as its height and the area of its base and when the area of the base is 60 square feet and the height 14 feet, the volume is 280 cubic feet. What is the area of the base of a pyramid whose volume is 390 cubic feet and whose height is 26 feet ? the area of the base, and h the height; Let F denote the volume, ;
A
V= mAh,
then
m
where
is
constant.
we have
Substituting the given values of F, A, h,
280
=m
X 60 X 14 280
'
V= 390,
when
=
/i
.-.
If
2.
If
X and X X y, and a;
;
A = ¥o.
of the base is 45 square feet.
EXAMPLES 1.
26
390=ii^ x26;
.-.
Hence the area
X
?/,
= y =
y
XXXIII.
7
when x =
18, find
3
when x =
2, find
A
= 21.
when x = 18. = Q when B =
y
varies jointly as
3,
(7
= 2:
when
;
If
5.
C.
x when y
B and C; and A B = 6, 0=7. and J. = 4. A varies jointly as B and C find B when A = 54, (7 = 10. A
3.
find
;
^1 ~ 3' 14
V=\Ah.
.-.
Also
~ 60
''^
'
X
X -, and
=
?/
when x
4
=
15, find y
=
1,
9
when
^ = 5,
when x =
(7=7:
Q.
y If y
6.
X
-,
and y
= 1 when
x
find x
when
y
=
5.
X
A varies
7.
^=
15,
8.
y
=
If
10, z
9.
If
(7=6:
as
B
find
directly,
A
when
(7
(7
inversely
;
and
^=
10
when
=
14
when
= 2.
X varies as y directly, and as z inversely, and x = 14, find z when x = 49, ?/ = 45. X
X -,
and
?/
x
-,
X
prove that a"
10.
If
a
11.
li
xcc z and y
12.
If
3 «
&,
+
prove that z
x
x.
z
y
tion
and as
^ = 8,
7 &
X
between a and
cc z,
3a 6.
+
x
6".
prove that 13
&,
x'^
— y^cx. z^. = 5,
and when a
5
= 3,
find the equa-
ALGEBRA.
300 li 5
13.
— y^
x
10 X
between x and
tion
-
111/,
and when x
=
cube of x varies as the square of the equation between x and y.
If the
14.
=
15.
If the
a
=
when
X
= 8,
y varies inversely as the square of x, find x when y 2.
17.
If
a;
=z
4:
X
?/
when
y
x
If
18.
If
19.
rt
—
rt'^
+
Ij-^cc
?/
=
^
3,
=
+ a, =5
;
&
=
y
it
S
if
when
oc
where a is constant, and x find x when y = 2.
—
rt
6,
prove that
cfi
+
?>"^
x
=
1^
when
ah; and
if
«
?/
=
oc b,
1,
and
prove
ah.
be the
respectively,
X
and
and
/>,
=
35
that
x=S when
it
square root of a varies as the cube root of = 8, find the equation between a and b.
6
If
16.
the equa-
5, find
and
?/,
y
5, find
=
y
7,
//.
sum
of
and when x
18, express y in
three quantities which vary as
=
=
4,
terms of
x.
1,
^
when x =
^
2,
=
8,
a-,
x^, x^
and when
Given that the area of a circle varies as the square of its radius, and that the area of a circle is 154 square feet when the radius is 20.
7 feet 21.
find the area of a circle
:
The area
whose radius
10 feet C inches.
is
of a circle varies as the square of its diameter
:
prove
whose diameter is 2\ inches is equal to the two circles whose diameters are 1^ and 2 inches
that the area of a circle
sum
of the areas of
respectively. 22. The pressure of wind on a plane surface varies jointly as the area of the surface, and the square of the wind's velocity. The pressure on a square foot is 1 pound when the wind is moving at the rate find the velocity of the wind when the pressure of 15 miles per hour on a square yard is 16 pounds. :
23. The value of a silver coin varies directly as the square of its it also varies directly diameter, while its thickness remains the same as its thickness while its diameter remains the same. Two silver coins have their diameters in the ratio of 4:3. Find the ratio of ;
their thicknesses
if
the value of the
first
be four times that of the
second. 24. The volume of a circular cylinder varies as the square of the radius of the base when the height is the same, and as the height when the base is the same. The volume is 88 cubic feet when the height is 7 feet, and the radius of the base is 2 feet what will be the heiglit of a cylinder on a base of radius feet, when the volume is :
390 cubic feet
?
:
CHAPTER XXXIV. Arithmetical, Geometrical, and Harmonical Progressions. 364.
A
fixed law
succession of quantities formed according to some is called a series. The separate quantities are
called terms of the series.
arithmetical progression. 365. Defixitiox. cal Progression
Quantities are said to be in Arithmetiincrease or decrease by a common
when they
difference.
Thus each of the following series forms an Arithmetical Progression ...
3, 7,
11, 15,
8, 2,
-4, -10,...
a,
a
-\-
d,
a-\-2d, a
+3
....
c?,
The common
difference is found by subtracting any term from that which follows it. In the first of the above examples the common difference is 4 in the second
of the series
;
it is
—6
;
in the third
it is d.
366. The Last, or nth Term, of an A. P.
If
we examine
the series a, loe notice that in
a
+ d,
any term
a
-f
2
c/,
a
-f-
3
the coefficient
c ?,
• • •
of d
one than the number of the term in the series. 301
is
always
less
by
ALGEBRA.
302
Thus the
3d term
is
a
6th term
is
a
20th term
is
and, generally, the pth term
is
number we have
If n be the nth.
term,
+2(1; -\- od] a + ldd; a -{-(p — 1) d.
and
J
of terms,
= a + {n-
if
denote the
/
or
last,
1) d.
Let a denote the first 367. The Sum of n Terms in A. P. term, d the common difference, and // the number of terms. Also let I denote the last term, and S the required sum then ;
S=
a
+
(a
+ d)-\-(a + 2f?) + -- + - 2 d) + - d)+L (l
(?
and, by writing the series in the reverse order,
S = +{l - d)-\-(l -2 d) + ->--{-(a 1
Adding together these two
2S =
(a
-i-
l)-\-{a
+ {a 1)+ '" to S = l(a + -\-
-\- l)
•••
Since
2d) + (a
+ d)+ a.
n terms
= n (a +
I)
l
.-.
-{-
series, I),
(1).
= a-\-(n-l)d
(2);
S = '^\2a+(n-l)d\
.
.
.
(3).
368. In the last article we have three useful formulae (1), in each of these any one of the letters may denote the unknown quantity when the three others are kno^\^l. (2), (3)
Ex.
1.
;
Find the 20th and 35th terms of the 38, 30, 34,
Here the common difference .-.
— 38, or — 2. = 38 + 19 (- 2) = 30
is
the 20th term
;
= 38 + 34 ( - 2) = - 30.
and the 35th term
Ex. 2. Find the sum of the series Here the common difference is 1^
The sum
series
....
=
^^{2 x
=
V
(11
5|, 6|, 8, ;
-V-
+
•.•
to 17 terms.
hence from (3)
+
-0)
10 x li}
= ll2^ = 2031.
ARITHMETICAL PROGRESSION. Ex.
The first term of a series is 5, the last 45, and number of terms, and the common difference.
3.
find the If
)i
the
n
Avhence
d be the
common
= the
45
=
45);
16.
difference,
16th term
whence
^
=
=
5
+
15 d
;
2|.
EXAMPLES XXXIV.
a.
1.
Find the 27th and 41st terms in the
2.
Find the 13th and 109th terms
3.
Find the 17th and 54th terms in the
4.
Find the 20th and 13th terms
in the series
—3, —2, —1,
5.
Find the 90th and 16th terms
in the series
—4,
6.
Find the 37th and 89th terms
in the series -2.8, 0, 2.8,
Find the
8.
7, 3,
9.
131, 9, 41,
-1,
— d,
a
2 a
-
a &,
Find the
to 15 terms.
...
-}-
...
to 13 terms.
d,
a
-
4a
&,
6a
14, 64, 114,
16.
1, 1.2, 1.4, ...
17.
9, 5, 1, ...
...
5 5,
to 20 terms.
to 12 terms.
to 100 terms.
Find the sum 21.
5, 9, 13, ...
22.
12, 9, 6,
23.
4, 51, 6h,
24.
lOi
25.
—3,
...
2.5, 9,
•••. •••.
•••.
•••.
••..
:
10.
.6, 1.2, 1.8, ...
11.
2.7, 3.4, 4.1,
12.
x,
2x,
to 12 terms.
...
to 11 terms.
to 25 terms.
...
3:c,
...
to 40 terms. series
-i, -|,
:
18.
i,
19.
3|, 1,
-1^,
...
to 19 terms.
20.
64, 96, 128,
...
to 16 terms.
to 21 terms.
...
of the following series-: 26.
10, 9|, 9i,
to 23 terms.
27.
!>,
28.
3a,
29.
a, 0,
30.
— Sg, — g,
to 37 terms.
•••
...
••-.
series 10, 11^, 13,
to 19 terms.
...
9, Ih,
11, 17,
to SO terms.
••'
-
5,
term and sum of the following
last
15.
1, 5,
+ S d,
3
series
in the series 71, 70, 69,
in the following series
to 20 terms.
5, 7, 9,
13.
term
last ...
7.
14.
sum 400
be the number of terms, then from (1),
400=^(5 +
If
303
to 94 terms.
to 17 terms.
3;?,
a,
...
—a,
-a,
to 21 terms. to
^p, ...
.••
q,
p terms.
to a terms.
to a terms. ...
to
;.>
terms.
:
;;
;
;
ALGEBRA.
304 Find the number
of
The
first
term
is 3,
32.
The
first
term
is 79,
The sum
is 24,
34.
The sum
is
35.
The
last
36.
The
first
term
is
37.
The
first
term
is a,
38.
The sum
the last term
first
term
714, the first
term
is
the
sum
9,
term
1395.
the last term —6.
1,
the last term 58^. first
term —3.
—75, the sum —740, the
last
term
the last 13 «, and the
sum
49
term 3 x^ the
last
—320 x,
the
first
when
and the sum 1075.
7,
—16, the sum —133, the
is
difference
the last term 90, and the
31.
33.
common
terms and the
1.
a.
term —35 x.
369. If cmy two terms of an Arithmetical Progression be for the data furnish two simultaneous equations, the solution of which will give the first term and the common difference. given, the series can be completely determined
Ex.
Find the
series
whose 7th and
;
51st terms are
—3 and —355
respectively. If
a be the
first
term, and d the
—3 =
common
difference,
=«+6 —355 = = a + 50 the term 51st and = —352 subtraction, 44 whence, by = —8 and, consequently, a = 45. the 7th term
fZ
cZ
fZ
.
•
.
fZ
;
Hence the
series is 45, 37, 29
•••.
When three quantities are in 370. Arithmetic Mean. Arithmetical Progression, the middle one is said to be the arithmetic mean of the other two. Thus a
is
371. To
the arithmetic
mean between a
find the arithmetic
—d
and a
+ d.
mean between two given quan-
tities.
Let a and b be the two quantities A the arithmetic mean. Then, since a, A, b, are in A.P., we must have ;
b
each being equal to the
whence
— A = A — a, common
A=
difference „
•
;
ARITHMETICAL PROGRESSION.
305
372. Between two given quantities it is always possible any number of terms sucli that the whole series thus formed shall be in A. P. and by an extension of the definition in Art. 370, the terms thus inserted are called the arithmetic means. to insert
;
Ex. Insert 20 arithmetic means between 4 and 67. Including the extremes the number of terms will be 22 so that we have to find a series of 22 terms in A. P., of which 4 is the first and ;
67 the
last.
Let d be the
common
then
whence d
=
3,
difference
=
67
and the
;
=4+
the 22d term,
series is 4, 7, 10,
and the required means are
7, 10, 13,
•••
•••
21
tZ
;
61, 64, 67
;
58, 61, 64.
373. To insert a given number of arithmetic means between two given quantities. Let a and h be the given quantities, m the number of means. Including the extremes the number of terms will be m + 2 so that Ave have to find a series of m 4- 2 terms in A. P., of which a is the first, and h is the last. Let d be the common difference ;
= the {m + 2)th term = a + {m -hl)f?; d= m+1
then
h
whence
;
and the required means are ,
b
—a
,
m+1
— a) m+l
2(b
,
m(b
m
1. Find the 30th term of an A. P. of which the and the 100th term - 16. Let d be the common difference
Ex. 17,
— a) -{-1
;
then
.-.
The 30th term
16
= =
the 100th term 17
+
99(Z;
d=--l.
=
17
+
29( - i)
=
7i.
first
term
is
;
;
ALGEBRA.
306 Ex. is
792
2.
The
sura of three
numbers
A. P.
in
33,
is
and
their product
tind them.
;
Let a be the middle number, d the numbers are a — d, a, a + d.
common
difference
;
then the
three
Hence
whence a
a
=
11
—d+
+
a-\-a
= dS; —
d
and the three numbers are 11
;
+
11(11
.-.
-cP =
121
+
11, 11
(?.
72,
=±
fZ
and the numbers are
(7,
cr)(ll-rf)= 792,
7
;
4. 11, 18.
How many terms of the series 24, 20, 16. ••• must be taken sum may be 72 ? Let the number of terms be n then, since the common difference 20 - 24, or - 4, we have from (3), Art. 367, Ex.
3.
that the
;
is
72=5(2 x24+(n-l)(-4)}
=
-24:n
whence
-2n(n- 1) - 13 n + 36 = 0, ;
n^
(n-4)(n-9)=0; n = 4 or
or
.-.
9.
of these values satisfy the conditions of the question for if 4, write down the first 9 terms, we get 24, 20. 16, 12. 8, 4, 0,
Both
;
-
we
—8
and. as the last five terms destroy each other, the terms is the same as that of 4 terms. ;
Ex.
4.
An
A. P. consists of 21 terms
in the middle
is
Let a be the
first
237
=
a
and
129,
+
term, and d the
=
20
the
rZ
+
whence
the
;
of the last three
a
sum -f
sum
237
common
;
of the three
of 9
terms
find the series.
Then
difference.
of the last three terms
+ + 18 d = + l9d = 't9
19
a
is
sum
f?
rt
3a
+
57
r7
(1).
Again, the three middle terms are the 10th, 11th, 12th
;
129 = the sum of the three middle terms = a-\-9d-\-a + lOd + a + Ud = Sa-{- 30
hence
r?
whence
From
a (1)
Hence the
and
(2),
we
4-
10 d
obtain d
.series is 3, 7, 11,
•••
=
83.
= 4,
43 a
=
(2). S.
ARITHMETICAL PROGRESSION.
EXAMPLES XXXIV.
307
b.
Find the series in which 1.
The 27th term
2.
The
3.
The 15th term
5th term
186,
is
and the 45th term 312.
31st term — 77. - 25, and the 23rd term - 41. - 11, and the 102nd term - loOi.
and the
is 1,
is
4.
The 9th term
5.
The 15th term
is
25,
6.
The 16th term
is
214,
7.
The 3rd and
is
and the 29th term
and the
46.
term
51st
739.
and 19
7th terms of an A. P. are 7
;
find the 15th
term. 8.
The 54th and 4th terms are
—
9.
The
i
and 2nd terms are
125 and
find the
;
42nd term.
10.
and 7| find the 59th term. Insert 15 arithmetic means between 71 and 23.
11.
Insert 17 arithmetic
means between 93 and
12.
Insert 14 arithmetic
means between
—
13.
Insert 16 arithmetic
means between
7.2
31st
;
69.
— 21. — 6.4.
71 and
and
14.
How many
terms must be taken of
make 315
15.
The
series 42, 39,
16.
The
series
17.
The
series 15f, 15i, 15,
...
to
make 129
18.
The
series 20, 18f, 17 J,
...
to
19.
The
series
make 162^ ? ... to make - 42
20.
The
21.
The sum
2184
:
22.
is
66
14.
,
numbers
...
...
to
?
make -
to
100 ?
?
?
make - 52i ?
in A. P. is 39,
and
their product
is
of three
and
410
The
spectively
:
numbers
of five
least is 161
The sum is
numbers
in A. P.
is
12,
and the sum
of their
find them.
:
The sum
squares 25.
-
15,
- 10^, - 9, - 7i^, - 6|, - 6f - 6,
of three
The sum
greatest 24.
series
to
-36, ...
-
16,
find them.
squares 23.
-
:
of five
:
in A. P.
is 75,
and the product
of the
find them.
numbers
in A. P.
is
40,
and the sum
of their
find them.
12th, 85th,
find tbe
and
last
number
terms of an A. P. are
of terms,
38, 257, 395 re-
;;
:
ALGEBRA.
308
GEOMETRICAL PROGRESSION. Quantities are said to be in Geometri-
374. Dkffxitiox. cal Progression
when they
increase or decrease by a constant
factor.
Thus each of the following series forms a Geometrical Progression 3, 6, 12, 24, 1
_i
...
1
i_
a, ar, ai^, ar^,
...
factor is also called the common ratio, and it found by dividing a/
The constant
is
;
;
375. The Last, or nth Term, of a G. P.
If
we examine the
series a, ar, ai^,
ar'^,
ar^,
...
we notice that in any term the index of r than the number of the term in the series. Thus
the 3rd term
is a?*^
the 6th term
is a?*^
the 20th term and, generally,
the pth term
is
The common .
•.
ratio is ^ -4-(—
the 8th term
•^),
or
always
less
by one
is ar^^-,
arP~\
If n be the number of terms, and I = ar"~^ nth term, we have Ex. Find the 8th term of the series
is
— i, — ^
if
|,
I
—
denote the
J,
last,
or
...
;
= - i x ( - §)' _ Iv— 2187—
129
376. Geometric Mean. When three quantities are in Geometrical Progression the middle one is called the geometric mean between the other two.
; ;
;
;
GEOMETRICAL PROGRESSION. 377. To
find the geometric
309
mean between two given quan-
tities.
Let a and
Then
b
be the two quantities
G
;
the geometric mean.
since a, G, b are in G. P.,
b_^G G a common
each being equal to the
ratio
G' = ab; G = ^ab.
.-.
whence
378. To insert a given number of geometric means between two given quantities. Let a and b be the given quantities, m the number of means. There will be m + 2 terms so that we have to find a series of m + 2 terms in G. P., of which a is the first and ;
b the last.
Let r be the then
6
= the
common
ratio
(m+2)th term
— ar'"+^
1
(1).
Hence the required means the value found in
are ar,
ar"^,
• • •
ar"^,
Ex. Insert 4 geometric means between 160 and 5, We have to find 6 terms in G. P. of which 160 is the sixth.
Let r be the then
common 5
=
ratio
trials
and the means are
;
the sixth term •
whence, by
where
r has
(1).
•
'
v 80, 40, 20, 10.
=
160
— =
\;
33
>
r^
first,
and 5 the
.
ALGEBRA.
310
?•
,379.; The Sum of n Terms in G. P. Let a be the tlrg^common ratio, n the number of terms, and
first terni,
S
the
sum
Then
required.
S=a+ rS
Hence by
= ar
ar
+ ar"- H
term by
multii)lying' every
ai^
-[-
r,
h af"^^
+ aV'^
;
we have
+ af'-'^ +
h «v'"^-
-\
«?'"•
subtraction,
—a —S= (r- 1)5 = «()•»-!); ,.s = <'-"-}^ r — 1 rS
.-.
<6?'"
;
^^^
Changing the signs in numerator and denominator
^='<^-';") Note. above for than 1.
It will A',
(2).
be fomid convenient to remember both forms given all cases except when r is positive and greater
ushig (2) in
Since ar"~i written
=
I,
it
follows that «r"
=
rZ,
and formula
(1)
may be
rl — a ~ r-l'
o_ Ex.
1.
Sum
the series 81, 64, 86, ••• to 9 terms. ratio = |i = |, which is less than 1
The common hence the
sum
Ex.
Sum
2.
= ^^l - (|) !} ^
the series
The common
ratio
f,
=—
—
f
;
1
+
1,
243{1
•••
^,
;
- (f )9}
to 7 terms.
hence by formula (2 I
f
EXAMPLES XXXIV. 1.
Find the 5th and 8th terms
2.
Find the 10th and 16th terms
3.
Find the 7th and 11th terms
c.
of the series o, 6, 12,
•••.
of the series 256, 128, 64,
of the series 64,
-
32, 16,
••
•••
:
:
GEOMETRICAL PROGRESSION. 4.
Find the 8th and 12th terms
5.
Find the 14th and 7th terms of the
6.
Find the 4th and 8th terms of the series
Find
tlie hist
term
...
7.
2, 4, 8,
8.
2,
9.
2, 3, 41, ...
-6,
...
to 8 terms.
to 6 terms.
:
-
3^,
« terms.
•
12.
X, 1,
to 30 terms.
S'-^,
x
means between
15.
Insert 6 geometric
means between 56 and
—
16.
Insert 5 geometric
means between ff and
4|.
i
and
to
p
terms,
6.
128. /g.
term and the sum of the following series ...
17.
3, 6, 12,
18.
6,
19.
64, 32, 16,
to 8 terms.
18, 54,
...
...
to 6 terms.
to 10 terms.
Find the sum of the
-
to 2
X, x^, x^
means between 486 and
3,
...
11.
Insert 4 geometric
23.
....
3,
Insert 3 geometric
-
••-.
'•*•
10.
14.
last
27, 9,
.008, .04, .2,
13.
Find the
-
series ^h^ sh^ tVi
in the following series
to 9 terms.
18,
of the series 81,
311
1, 1,...
series
to 6 terms.
20. 21.
...
8.1, 2.7, .9, •
to 7 terms.
to 8 terms.
:
ALGEBRA.
312
make
—
the fraction
number
a sufficient
by as
we
little as
we
small as
-^ as
sum can be made
of terms the
please from
Sum
From
to Infinity. <;^
_
ff
is
discussed.
we have
Art. 379
— _ _a 1 — r 1 —
(1
to differ
2.
In the next article a more general case 381.
Thus by taking
please.
ar''
?•")
—r
1
r
Suppose r is a proper fraction then the greater the value of n the smaller is the value of r", and consequently of ;
——
and therefore by making n
;
make
sum
the
of
n terms of the
as small a quantity as
we
we can
sufficiently large,
from
series differ
^
_
by
please.
This result is usually stated thus the sum of an infinite number of terms of a decreasing Geometrical Progression is :
more
or
1-r'
briefly, -" the
sum
to infinity is " -
l_r
382. E-ecurring decimals furnish a good illustration of Infinite Geometrical Progressions.
Ex. Find the value of .423. .423
=
.4232323
...
=
— + -^ + -^^— + 10
1000
102
104
...
100000
4 10
,
"^
23
4
100
'99
103
10
,
"^
10
I
23
419
990
990'
-\-
10
23
ion
10
— -\-~ ^ ^
= :^ +
10^ 10^
103
1
1
_ J_ 102
which agrees with the value found by the usual arithmetical
EXAMPLES XXXIV. Sum
rule.
d.
to infinity the following series
1.
9,6,4,...
3.
i, i,
2.
12,6,3,...
4.
^,-1,1,...
i,...
5. 6.
i,
f,^,... 1,-1,1,...
Find by the method of Art. 382, the value 9.
"•
105
.3.
10.
.16.
11.
.24.
7.
.9, .03, .001,...
8.
.8,
-.4,
.2, -.
13.
.037.
of 12.
.378.
:
.
HARMONICAL PROGRESSION. Find the
series in
313
which
The 10th term is 320 and the 6th term 20. The 5th term is f and the 9th term is i. The 7th term is 625 and the 4th term - 5.
14. 15.
|^
16.
The 3d term is j\ and the 6th term - 4^ Divide 183 into three parts in G. P. such tliattlie
17. 18.
and third
is
Show
19.
sum
of the first
2^^ times the second. tliat tlie
product of any odd number of consecutive terms
of a G. P. will be equal to the nth
number of terms. 20. The first two terms
power
of the middle term,
n being
the
and every term
of
an
infinite G. P. are together equal to 1,
sum
twice the
is
of all the
terms which follow.
Find
the series.
Sum
the following series
+ 2&,
21.
2/^
22.
?jtlV?,
23.
+
?/*
1,
3-2V2 Vi W'^^ n
—
3
2
25.
The sum
1,
6,
+
2/6
+
6
6, •••
to
n terms.
...to infinity.
+ 2V2
fVf,
4n
24.
4
^-^^^
i,
of four
to infinity.
6 w
—
j\,
numbers
...
to 2
in G. P.
n terms. is
equal to the
common
ratio
and the first term is j\. Find the numbers. 26. The difference between the first and second of four numbers in G. P. is 96, and the difference between the third and fourth is 6. Find the numbers. 27. The sum of $ 225 was divided among four persons in such a manner that the shares were in G. P., and the difference between the greatest and least was to the difference between the means as 21 to 6, Find the share of each. 28. The sum of three numbers in G. P. is 13, and the sum of their reciprocals is J/. Find the numbers. plus
1,
HARMONICAL PROGRESSION. 383. Definition.
Three quantities,
be in Harmonical Progression
when - =
Any number
—
when every
Harmonical Progression.
-
are said to
•
b c of quantities are said to be in c
Progression
a, b, c,
Harmonical
three consecutive terms
are
in
ALGEBRA.
814 384. The Reciprocals
of Quantities in
Harmonical Progression
are in Arithmetical Progression.
By
definition, if a,
b, c
are in Harmonical Progression,
a—b
a_ c
— c)
a(b
.-.
,
— c^ — c{a — b), b
dividing every term by abc,
1_1^1_1 b
b
c
a
which proves the proposition. We may therefore define an Harmonical Progression as a series of quantities the reciprocals
of which are in Arithmetical Progression.
385. Solution of Questions in H, P. Harmonical x^roperbecause of their importance in Geometry and in the Theory of Sound in Algebra the proposition just proved is the only one of any importance. There is no general formula for the sum of any number of quantities in Harmonical Progression. Questions in H. P. are generally solved by inverting the terms, and making use of the properties of the corresponding A. P. ties are chiefly interesting
:
Ex. The 12th term of an H.
P
is |,
and the 19th term
is
^^
:
find
the series.
Let a be the first term, d the then
ing A. P.
5
and whence
common
difference of the correspond-
;
-2^2
f?
= _ =
the 12th term
the 19th term !»
^
= =
and the Harmonical Progression
Mean.
-{
\\d;
+
18 d
;
= t-
Hence the Arithmetical Progression
386. Harmonic
a «
is |, |, 2, |, •••
is |, |, ^, f, ••••
When
three
Harmonic Progression the middle one Harmonic Mean of the other two.
quantities is
are
in
said to be the
;
;
THE PROGRESSIONS. To
387.
find the
harmonic mean between two given quan-
tities.
Let then
«, h
-,
a
be the two quantities,
H
,
315
- are in h
H their harmonic
mean
A. P.
"
H
a
H'
h
H a H = a2 ah
b'
388. Relation between the Arithmetic, Geometric, and Harbe the arithmetic, geometric, and monic Means. If A, G, harmonic means between a and h, we have proved
H
Therefore
A = ^^
(1).
G = Vab
(2).
H=-^^
(3).
^^=^.1^ a-hb 2
= ab=G^', that
is,
G is
the geometric
mean between
A and H.
389. Miscellaneous Questions in the Progressions. Miscellaneous questions in the Progressions afford scope for much skill and ingenuity, the solution being often very The student will neatly effected by some special artifice. find the following hints useful. If the same quantity be added to, or subtracted from, the terms of an A. P., the resulting terms will form an [Art. A. P., with the same common difference as before. 1.
all
365.]
.
ALGEBRA.
816 If all the terms of
2.
an A. P. be multiplied or divided
by the same quantity, the resulting terms form an A. but with a
new common
difference.
P.,
[Art. 365.]
3. If all the terms of a G. P. be multiplied or divided by the same quantity, the resulting terms form a G. P. with the same common ratio as before. [Art. 374.]
If a,
4.
b, c,
d •" be in G.
proportion, since
by
P.,
they are also in continued
definition
a_&_c_
_1
bed
r
Conversely, a series of quantities in continued proportion may be represented by x, xr, xr, • • •
Ex.
1.
Let -,
Find three quantities
rt,
ar be the three quantities
r
then
in G. P.
we have
:
such that their product
is
^
. .
.
THE PKOGRESSIONS. Ex.
3.
The nth term
an A.
of
P.
-
is
317
2
-f
find the
:
sum
of 49
terms.
Let a be the first term, and I the = 49 respectively, we obtain
last
then by putting w
;
=
1,
and n
a
Ex. ric
4.
If «, 6, c,
(Z,
mean between a Since «,
6, c,
c?,
=
+
i
and
q
= ^^+2',
+
each ratio
d
is
^i^t_2 =: ^l±i. c e h \-(l
[Art. 347.]
+
(6
+
dy-
={a
-^
^
c){c
e).
EXAMPLES XXXIV. 1.
Find the 6th term
2.
Find the 21st term of the
3.
Find the 8th term of the
4.
Find the wth term series in
the geomet-
_ 6 _ c _ ^.
.:.
Whence
Find the
^
e.
are in continued proportion,
e
a
...
1
be in G. P., prove that h
e
A- c
2,
e.
of the series 4, 2, 1|,
•••.
series 2 J, l|f, Ij^e,
series 1^, 1|}, 2i%,
of the series 3, 1^, 1,
••• •«•
...
which
5.
The 15th term
6.
The 2d term
7.
The 39th term
is ^V?
is 2, is
and the 23d term
and the 31st term j\,
is
is ^V-
/j.
and the 54th term
is Jg-
Find the harmonic mean between 8.
11.
and
4.
- and
-.
a
b
2
9.
12.
1
and
— X
^
+
y
13.
and
— x
—
10.
I and j\.
13.
x
means between 4 and
14.
Insert two harmonic
15.
Insert three harmonic
16.
Insert
+
y 12.
means between 2| and 12. four harmonic means between 1 and 6.
y and
a:
-
?/.
:
ALGEBRA.
318 17.
G
If
show that is
equal to 18.
PI.
tlie ratio
To
the three
quantities A and B, and harmonic means of A and G the arithmetic and harmonic means of G and B.
be the geometric
mean between two
the ratio of the arithmetic of
each of three consecutive added.
is
Sliow that
tlie
terms of a G.
P., the
second of
three resulting quantities are in
P.
Sum
the following series
+ 3Jg + to 6 terms. + 2^ + to 6 terms. to p terms. 21. (2 a + a;) + 3 a + (4 « - x) + to 8 terms. 22. If - li + f 23. If + 11 + I + ... to 12 terms. 24. li X — a^ y — «, and z — a he, in G.P., prove the harmonic mean between y — x and y — z. 19.
1
20.
1
+ +
...
If
...
If
...
25.
26.
c, d be in A. P., a, e, /, d in G. P., prove that ad = ef =bh = eg.
If a, 6,
respectively
;
If a^, &2, c2
be in A. P., prove that & + c,
c
that 2(?/
«, ^,
/i,
+ a, a + &
—
a)
is
d in H. P.
are in H. P.
:
CHAPTER XXXV. Permutations axd Combixatioxs. 390. Each of the arrangements which can be made by taking some or all of a number of things is called a permutation.
Each of the cjroups or selections which can be made by taking some or all of a number of things is called a combination.
Thus the permutations which can be made by taking the namely, b, c, d two at a time are twelve in number
letters a,
;
ab,
ac,
ad,
he,
bd,
cd,
ha,
ca,
da,
cb,
dh,
dc]
each of these presenting a different arrangement of two letters.
The a, b,
c,
combinations which can be made by taking the letters d two at a time are six in number namely, ;
ab,
ac,
ad,
be,
bd,
cd-,
each of these presenting a different selection of two letters. From this it appears that in forming combinations we are only concerned with the number of things each selection contains whereas in forming permidations we have also to consider the order of the things which make up each arrangement for instance, if from four letters a, b, c, d we make a selection of three, such as abc, this single combination admits of being arranged in the following ways ;
;
abc,
and so gives
acb,
bac,
bca,
cab,
cha,
rise to six different permutations.
319
;
ALGEBRA.
820
391. Fundamental Principle. Before discussing the genthis chapter the following important principle should be carefully noticed. ways, and (when it If one operation can he performed in has been performed in any one of these icays) a second operaeral propositions of
m
tion
can then he performed in n ivays; the number of ivays of
2)erforminf/ the two operations
be
icill
m
x
n.
performed in any one way, we can associate with this any of the n ways of performing the second operation and thus we shall have n ways of performing the two operations without considering more than one way of performing the lirst and so, corresponding to each of the m ways of performing the first operation, we shall have n ways of performing the two hence the product m X )i represents the total number of ways in which the two If the first operation be
;
;
;
operations can be performed. Ex. Suppose there are 10 steamers plying between New York and Liverpool in how many ways can a man go from New York to Liverpool and return by a different steamer ? There are ten ways of making the first passage and with each of these there is a choice of jiine ways of returning (since the man is not to come back by the same steamer); hence the number of ways of making the two journeys is 10 x 9, or 90. :
;
This principle
may
easily be extended to the case in
which there are more than t^vo operations each of which can be performed in a given number of ways. Three
Ex. in
ent hotel
town where there are four hotels they take up their quarters, each at a differ-
travellers arrive at a
how many ways can
;
?
has choice of four liotels, and when he has made any one way, the second traveller has a choice of three therefore the first two can make their choice in 4 x 3 ways and with any one such choice the third traveller can select his hotel hence the required number of ways is 4 x 3 x 2, or 24. in 2 ways
The
first traveller
his selection
in
;
;
392.
To
find the
number
of
permutations of n dissimilar
things taken r at a time.
This
is
the same thing as finding the
number
of
ways
in
PERMUTATIONS AND COMBINATIONS. we can
wliicli
fill
r places
when we have n
321
different things
at our disposal.
The first place may be filled in n ways, for any one of the n things may be taken when it has been filled in any one of these ways, the second place can then be filled in and since each way of filling the first place 1 ways n can be associated with each way of filling the second, the number of ways in which the first two places can be filled ;
—
is
;
— 1). And when the first two any way, the third place can be And reasoning as before, the number
given by the product n{n
places have been filled in filled in
n
—2
ways.
which three places can be filled is n{n — l){n—2). Proceeding thus, and noticing that a new factor is introduced with each new place filled, and that at any stage the mnnher of factors is the same as the number of places filled, we shall have the number of ways in which r places can be of
ways
filled
in
equal to 71 (ii
— l)(7i — 2)
...
to r factors.
We here see that each factor is formed by taking from n a number one less than that which applies to the place filled by that factor; hence the 7'th factor is n — (r— 1), or n—r+l. Therefore the number of permutations of n things taken r at a time is 7i(7i-l)(7i-2)...(7i-r4-l). CoR.
The number
of permutations of n things taken all
at a time is
n{n
— 1)
(71
— 2)... to
71
factors,
7<7t-l)(7i-2)...3.2.1.
or It is usual to
denote this product by the symbol
is read '^factorial
393. tions of
We
?i."
Also n
!
is
r at a
that "P,
"P„
also
Y
= n{ii - l){n - 2) ...(71 - r + = |7i.
which \n.
number of permutatime by the symbol "P^, so
shall in future denote the
n things taken
\n,
sometimes used for
1)
J
;
ALGEBRA.
322
In working numerical examples it is useful to notice that the suffix in the symbol "P^ always denotes the number of factors in the formula
we
are using.
Four persons enter a carriage in which tliere are six seats in how many ways can they take their places ? The first person may seat himself in 6 ways and then the second person in 5 the third in 4 and the fourth in 3 and since each of these ways may be associated with each of the others, the required answer is 6 x 5 x 4 x 3, or 360. Ex.
1.
:
;
;
;
;
Ex. 2. How many different numbers can be formed by using six out of the nine digits 1, 2, 3, ••• 9 ?
Here we have 9 different things, and we have to find the number of permutations of them taken 6 at a time ;
the required result
.-.
= ^Pq
9x8x7x6x5x4 = 60480.
= 394. To
the number of combinations of n dissimilar
find
things taken r at a time.
Let "C^ denote the required number of combinations. of these combinations consists of a group of r dissimilar things which can be arranged among themselves
Then each
in Ij^ways.
Hence
[Art. 392, Cor.]
"(7^
X
"0,
X
[r
equal to the number of arrangements of
[r is
n things taken
time
r at a
;
that
is,
= "P, = 7i(n - 1) (n - 2)
C-
..
• • •
- r + 1)
(^n
.
(1).
.
^
This formula for "(7^ may also be written in a form for if we multiply the numerator and the denominator by \n — r we obtain CoR.
different
n (n
;
- 1) (n - 2)
—
• • •
.
since n(n
— l)(?i — 2)
- + 1) x n-r [n 1^ — r —r [r\ — r = [n. (n — r + 1) x ?•
0*,
\
>
,
\r\n •
• •
01'
•
{^y-)
)>
\7i
remember both these expressions cases where a numerical result is
It will be convenient to
for
"(7,.,
using
required,
and
(1) in all (2)
algebraic shape.
when
it
is
sufficient to leave
it
in an
;
PERMUTATIONS AND COMBINATIONS. Note.
but »0„ [0
If in
=
I,
formula (2) we put r
so that
if
must be considered Ex.
made,
the formula
is
=
we have
n,
to be true for r
as equivalent to
323
=
the symbol
n,
1.
From 12 books in how many ways can a selection of 5 be when one specified book is always included, (2) when one
(1)
specified
book
is
always excluded
?
(1) Since the specified book is to be included in every selection, we have only to choose 4 out of the remaining 11.
Hence the number
of
ways ^
=
1104
=
^^
x ^^ x ^ ^ ^
1x2x3x4
=
330.
(2) Since the specified book is always to be excluded, select the 5 books out of the remaining 11.
Hence the number
of
ways
=
nC's
=
we have
to
^ x IQ x 9 x 8 x 7 ^ ^^2. 1x2x3x4x5
395. The number of combinations of n things r at a time equal to the number of combinations of n things
n—r
is
at a time.
making all the possible combinations of n things, to group of r things we select, there is left a corresponding group of n — r things that is, the number of combinations of n things r at a time is the same as the number of combinations of n things 71 — r at a time Ill
eacli
;
This result
is
frequently useful in enabling us to abridge
arithmetical work. Ex. Out of 14
men
in
The required number
how many ways can an
=
^^Cn
=
^^€3
=
eleven be chosen ^^ x 13 x 12 ^ ^^^
?
1x2x3
use of the formula ^'^Cn, we should have had to reduce an expression whose numerator and denominator each contained 11 factors. If
we had made
396. In the examples which follow
it
is
important to
notice that the formula for permutations should not be used until the suitable selections required
been made.
by the question have
5
ALGEBRA.
324
1. From 7 Englishmen and 4 Americans a committee of 6 is formed: in how many ways can this be done, (1) when the committee contains exactly 2 Americans, (2) at least 2 Americans ?
Ex.
to be
The number
(1) is
of
and the number "^(74. Each of the
*C2
;
second
;
ways in which the Americans can be chosen is ways in which the Englishmen can be chosen
of
first
groups can be associated with each of the
hence
the required
number
of
ways =^(72 x 14 L_
_
'^C^ 17 L—
^
|_2^
17
_
[ill
_
L.
2]^Q
I^12|i
(2) We exhaust all the suitable combinations by forming all the groups containing 2 Americans and 4 Englishmen then 3 Americans and 3 Englishmen and lastly 4 Americans and 2 Englishmen. The sum of the three results gives the answer. Hence the required number of ways = ^C^ x '^d + ^Cs x '^Cs + ^d x "^Ca ;
;
^_^xi + ^xi^ + lx^ lil^l
[2^
=
210
+
+
140
21
IJIi^
=
^|_5
371.
In this example we have only to make use of the suitable formula we are not concerned with the possible arrangements of the members of the committee among themselves. for combinations, for
Ex. 2. Out of 7 consonants and 4 vowels, how many words can be made each containing 3 consonants and 2 vowels ? The number of ways of choosing the three consonants is '^Cg, and the number of ways of choosing the two vowels is ^d; and since each of the
first
groups can be associated with each of the second, the
combined groups, each containing 3 consonants and 2 vowels, is '^Cs x ^02. Further, each of these groups contains 5 letters, which may be arranged among themselves in \_5 ways. Hence
number
of
17
number
the required
of
words
= -^=^
=
5
X
|4
x ^=- x IJ
=
1
25200.
EXAMPLES XXXV.
a.
the value of sp^, "Pe, ^^5, 25C23.
1.
Find
2.
How many different arrangements can be made by taking (1) five,
(2) all of the letters of the 3.
If
"O3
:
"-iC'4
word
= 8:6,
soldier 9
find n.
PERMUTATIONS AND COMBINATIONS. 4.
How many
bag containing a
different selections of four coins can
and a penny
How many numbers 6
9, 3, 4,
In arranged
how many ways can if
?
between 3000 and 4000 can be made with
the digits 6.
be made from a
dollar, a half-dollar, a quarter, a florin, a shilling, a
franc, a dime, a sixpence, 5.
325
?
the letters of the word volume be
the vowels can only occupy the even places ?
7. If the number of permutations of n things four at a time is fourteen times the number of permutations of n — 2 things three at a
time, find n. 8.
From
5 teachers
and 10 boys how many committees can be and 6 boys ?
selected containing 3 teachers 9.
If 2ocV
= 2oc^_jo^
find
'•C12,
18(7,.
Out of the twenty-six letters of the alphabet in how many ways can a word be made consisting of five different letters two of which must be a and e ? 10.
11. How many words can be formed by taking 3 consonants and 2 vowels from an alphabet containing 21 consonants and 5 vowels ? 12.
A stage
accommodate 5 passengers on each side in how 10 persons take their seats when two of them remain one side and a third upon the other ? will
:
many ways can always upon
397. Hitherto, in the formulae we have proved, the things have been regarded as unlike. Before considering cases in which some one or more sets of things may be like, it is necessary to point out exactly in what sense the words like and unlike are used. When we speak of things being dissimilar, different, unlike, we imply that the things are visibly unlike, so as to be easily distinguishable from each other. On the other hand, we shall always use the term like things to denote such as are alike to the eye and cannot be distinguished from each other. For instance, in Ex. 2, Art. 396, the consonants and the vowels may be said each to consist of a group of things united by a common characteristic, and thus in a certain sense to be of the same kind; but they cannot be regarded as like things, because there is an individuality existing among the things of each group which makes them easily distinguishable from each other. Hence, in the final stage of the example we considered each
ALGEBRA.
326
group to consist of five dissimilar things and therefore among themselves. [Art. 392, |_5 arrangements
capable of Cor.]
398. To at a time,
find the
permutations of n things, taking them all are of one kind, q of another kind, n
when p things
of a third kind,
and the rest
all different.
suppose p of them to be a, q of them to be b, r of them to be c, and the rest to be unlike. Let X be the required number of permutations then if in any one of these permutations the p letters a were replaced there be n letters
Lei;
;
;
by p unlike
letters different
from any of the
rest,
from this any of
single permutation, without altering the position of
we
the remaining letters,
could form
\p
new
permutations.
Hence if this change were made in each of the x permutations, we should obtain x x \p permutations. Similarly, if the q letters b were replaced by q unlike the number of permutations would be x x\p X \q.
letters,
In like manner, by replacing the r letters c by r unlike we should finally obtain x x\p_x\q x\r permutations. But the things are now all different, and therefore admit Hence of [?i permutations among themselves. letters,
X X that
\p
X
\q
X
\]^=
\n^j
is,
which
is
Any
the required
number
of permutations.
case in which the things are not
all different
may
be
treated similarly. Ex.
1.
IIow
letters of the
We w.
many
different permutations
word assassination taken
have here 13
letters of
Hence the number
all
which 4 are
can be made out of the
together s,
3 are
?
a,
of permutations 113 [4[3|_2|_2
= =
13. 11 -10. 9. 8. 7. 3. 5 1001 X 10800
=
10810800.
2 are
i,
and 2 are
PERMUTATIONS AND COMBINATIONS. Ex.
2.
32T
How many numbers
The odd
odd
that the
4, 3, 2, 1, so
digits
can be formed with the digits 1, 2, always occupy the odd places ? can be arranged in their four places in
1, 3, 3, 1
-^=— [2|^
The even
digits 2, 4, 2
3,
digits
ways
(1). ^ ^
"^
can be arranged in their three places in
= ways
(2).
\1
Each
of the
ways
in (1)
can be associated with each of the ways
in (2). |4 = -^—
Hence the required number
399. To time,
when
find the
number
each thing
may
13
x
—=6
=
x 3
of permutations of
18.
n things r at a ... up to r
be repeated once, twice,
times in any arrangement.
Here we have
to consider the
places can be filled
number
when we have n
disposal, each of the
ways
of
in
which r
different things at our
n things being used
as often as
we
please in any arrangement.
The
first
place
may
be
filled in
7i
ways, and,
when
it
has
any one way, the second place may also be filled in n ways, since we are not precluded from using the same thing again. Therefore the number of ways in which the first two places can be filled is w x 7i or 7i^. The third place can also be filled in n ways, and therefore
been
filled in
first three j)laces in n^ ways. Proceeding thus, and noticing that at any stage the index of n is always the same as the number of places filled, we shall have the number of ways in which the r places can be filled equal to n''.
the
Ex. In how many ways can 5 prizes be given away to 4 boys, when each boy is eligible for all the prizes ? Any one of the prizes can be given in 4 ways and then any one of the remaining prizes can also be given in 4 ways, since it may be obtained by the boy who has already received a prize. Thus two prizes can be given away in 4- ways, three prizes in 4^ ways, and so on. Hence the 5 prizes can be given away in 4^, or 1024 ways. ;
ALGEBRA.
328
find the total number of ways in which it is possible make a selection by taking some or all of n things. Each thing may be dealt Avith in two Avays, for it may either be taken or left and since either way of dealing with any one thing may be associated with either way of dealing with each one of the others, the number of Avays of dealing
400. To
to
;
with the n things
is
2 X 2 X 2 X 2
But
to
...
71
factors.
which
this includes the case in
all
therefore, rejecting this case, the total
the things are
number
left,
of Avays is
2"-l. This
is
tions " of
Ex.
more
of
A man has G friends them
He has number
often spoken of as " the n things.
of
to
in
number
how many ways may he
some or
is 2^
—
1,
invite
one or
all of his
6 friends
;
and therefore the
or 63.
This result can be verified in the following manner. The guests may be invited singly, in twos, threes, ...
number
of combina-
dinner?
to select
ways
;
total
;
therefore the
of selections
= 401. To
G
+
find for
15
4-
20
+
15
what value
+
+
=
1
of n the
03.
number
of combinations
of n things r at a time is greatest.
n{n Since "C^
- 1) {n - 2)
...
(n
- r + 2) (n - r + 1)
1.2.3... (r-l)r
7i0i-l)0?,-2)...(?i-r and
"C,_i
1.2.3...(r-l) n
+ 2) '
— r -\-l r
The multiplying Avhich shoAvs that
factor it
^^~^"^
may
be written
decreases as r increases.
^^-t
Hence
1,
as r
receiA^es the values 1, 2, 3,... in succession, "C^ is continu-
2
;
PERMUTATIONS AND COMBINATIONS.
—^^—
71-1-1
ally increased, until
than
329
becomes equal to 1 or
1
less
1.
I^ow 'ii5t
- 1 > 1, so long as
We
> 2;
^^i±i
that
is,
r
r
have to choose the greatest value of
^^ > 2
r consistent
r.
with
this inequality. (1)
Let n be even, and equal to 2 in
n
and for r.
all
(2)
r
2m
2
r; but
all
values of r
when r=m-\-\,
=
-f-
2
up
greater than
the greatest
"(7„.
Let n be odd, and equal to 2 ?>i
and for
,
,
m inclusive this is = m = -, we find that
of combinations is
n-\-\
to 1,
4- 1
values of r up to
Hence by putting
number
2m
-h 1
then
;
m
to
?/i
-|-
1
;
then
+1
inclusive this
greater than
is
the multiplying factor becomes equal
and "^m+l
= '*^m
5
that
IS, "C,^_f^
= '^^^H-l
2
and therefore the number of combinations the things are taken
,
or
—^—
j
2
is
greatest
at a time
;
when
the result
being the same in the two cases.
EXAMPLES XXXV. 1.
b.
Find the number of permutations which can be made from
all
the letters of the words, (1) irresistible, 2.
(2)
phenomenon,
How many different numbers How many
digits 2, 3, 4, 3, 3, 1, 2 ?
(3) tittle-tattle.
can be formed by using the seven with the digits 2, 3, 4, 3, 3, 0, 2 ?
ALGEBRA.
330
3. How many words can be formed from the letters of the word Simoom, so that vowels and consonants occur alternately in each word ?
A
telegraph has 5 arms, and each arm has 4 distinct positions, 4. including the position of rest: find the total number of signals that
can be made. 5.
there 6.
In how many ways can n things be given to m persons, when no restriction as to the number of things each may receive
is
'?
How many
different
ters of the expression a^b^c^
arrangements can be made out of the
when
written at full length
let-
?
There are 4 copies each of 3 different volumes find the number of ways in which they can be arranged on one shelf. 7.
;
In how many ways can 6 persons form a ring ? Eind the num8. ber of ways in which 4 gentlemen and 4 ladies can sit at a round table so that no two gentlemen sit together. 9.
In
how many ways can a word of 4 letters be made out of the e, c, d, o, when there is no restriction as to the number of
letters a, 6,
times a letter
is
repeated in each word
?
How many
arrangements can be made out of the letters of the word I'oulouse, so that the consonants occupy the first, fourth, and 10.
seventh places
?
A boat's
crew consists of eight men of whom one can only row on bow side and one only on stroke side in how many ways can the crew be arranged ? 11.
:
13.
Show that »'+iC; = »Cr + ''Cr^i. If 2nC3:"C2=44:3, find ?i.
14.
Out
12.
be
A, B, O, p,
of the letters
made beginning with
a capital
15.
Find the number
16.
U^^Cr
=
g, r,
how many arrangements can
?
of combinations of 50 things 46 at a time.
''Wr+2, find'-Cs.
17. In how many ways is it possible to draw a sum of money from a bag containing a dollar, a half-dollar, a quarter, a dime, a five-cent
piece, a two-cent piece,
and a penny
?
.
:
CHAPTER XXXVI. Probability (Chance). 402. Definition. If an event can happen in a ways and ways, and each of these ways is equally likely, the
fail in h
probability, or the chance, of its happening is
Hence
its failing is
,
and of
to find the probability of
an
event happening, divide the number of favorable ways by the number of ways favorable and unfavorable. For instance, if in a lottery there are 7 prizes and 25 blanks, the chance that a jjerson holding 1 ticket will win a prize is -^, and his chance of not winning is |f Instead of saying that the chance of the happening of an ivhole
event
is
—a
,
it is
sometimes stated that
the
odds are a
to b
a-\-b in favor of the event, or b to a against the event. Thus in the above the odds are seven to twenty-five in
favor of the drawing of a prize, and twenty-five to seven against success.
The reason for the mathematical definition of probabe made clear by the following considerations If an event can happen in a ways and fail to happen in b ways, and all these ways are equally likely, we can assert 403.
bility
may
its happening is to the chance of its failing as a to b. Thus if the chance of its happening is re^^resented by ka, where k is an undetermined constant, then the chance of its failing will be represented by kb.
that the chance of
.-.
chance of happening
-f
chance of failing 331
= k(a + 5).
ALGEBRA.
332
Now sum
the event
is
certain to
happen or to fail therefore the and failing must rej)resent ;
of the chances of happening
certainty.
If therefore
we agree
to take certainty as our unit,
we have 1
.'.
=
A;
(a
+ b),
or k
=
+6
a
,
the chance that the event will happen
and the chance that the event
will not
is
happen
is
a-\-b
Cor. If
p
is
the probability of the happening of an event,
the probability of
its
not happening
is
1
— p.
404. The definition of probability in Art. 402 may be given in a slightly different form which is sometimes useful. If c is the total number of cases, each being equally likely to occur, and of these a are favorable to the event, then the probability that the event
Avill
happen
is
and the proba-
-, c
bility that it will not
happen
is 1
Ex. 1. (a) From a bag containing 4 white and 5 black balls a man draws a single ball at random. What is the chance that it is black ? A black ball can be drawn in 5 ways, since any one of the 5 black balls may be drawn. In the same way any one of the 4 white balls may be drawn.
Hence the chance
of
drawing a black ''
5
ball is
,
4
+
5
or
5 — 9
What are the (b) Suppose the man draws 3 balls at random. odds against these being all black ? The total number of ways in which 3 balls can be drawn is ^Cs, and the total number of ways of drawing 3 black balls is ^Cs therefore the chance of drawing 3 black balls ;
9(73
Thus the odds
9.8-7
42*
against the event are 37 to 6.
;
PROBABILITY.
333
Ex. 2. From a bag containing 5 red balls, 4 white balls, and 5 black balls, 6 balls are drawn at random. What is the chance that 3 are white, 2 black, and 1 red ? The number of combinations of 4 white balls, taken 3 at a
4.3.2
time,
In the same manner the number of combina-
or 4.
is
tions of 5 black balls, taken 2 at a time,
^^
is
or 10.
Since each of
the 4 combinations of white balls may be taken with any one of the 10 combinations of black, and with each of the combinations so
formed we may take any one of the 5 red balls, the total number of combinations will be 4 10 5 or 200. But the number of combinations •
number
of the entire
•
of balls, taken 6 at a time
is
lo
•
or 3003, hence the chance that 3 white, 2 black, and
Ex.
A C
1.2 ways
has
3
—^2 x •
i.
r
-^
ways in
9.8.7 1.2.3
,
•
10
•
9
1
red ball will
1
3 prizes in
ways
6
;
he
1
way
the
sum
A
which
of these
numbers
may draw prize
2 prizes
and 2 blanks
;
64,
which
is
'
and
1
in 3
x
is
the
number
clearly \
;
=
|4 8?
=
of in
i|. 21
therefore A's chance
:
B's
:
Or we might have reasoned thus
=
is
therefore A's chance of success
:
:
ways
1
Also he can draw 3 tickets
can win a prize.
or 84 ways -^
of success
he
;
may draw
2i
;
B's chance of success chance = if i = 16 7.
or 20
II
shares in a lottery in which there are 3 prizes and share in a lottery in which there is 1 prize and 2 that A's chance of success is to B's as 16 to 7.
B
;
Show may draw
blank in
•
A has 3
3.
6 blanks blanks.
1-^
•
1.2.3.4.5.6
the chance of which 1
—
From
^5_
_
:
A will
is ||,
get all blanks in
or ^j
;
— —'— 1.2.3'
therefore A's chance
^6^
the examples given
it will be seen that the questions in Probability requires nothing more than a knowledge of the definition of Probability, and the application of the laws of Permu-
405.
solution of the easier kinds of
tations
and Combinations.
EXAMPLES XXXVI.
A
bag contains 5 white, 7 black, and 4 red balls find the chance of drawing (b) Two white balls (a) One white ball (r) Three white balls (d) One ball of each color (e) One white, two black, and three red balls. 1.
;
:
;
;
;
;
ALGEBRA.
834 If
2.
2 lieads
chance that there should be
:
is
Thirteen persons take their places at a round table. 5 to 1 against 2 particular persons sitting together.
4.
that
tlie
tails.
One of two events must happen given that the chance of the two-thirds that of the other, find the odds in favor of the other.
3.
one
four coins are tossed, find
and 2
Show
it is
5. There are three events A, B, C, one of which must, and onlyone can, happen the odds are 8 to 8 against A, 5 to 2 against B. Find the odds against C. ;
A has 3 shares in a lottery containing 3 prizes and 9 blanks 6. has 2 shares in a lottery containing 2 prizes and 6 blanks. Compare their chances of success. B
7. There are three works, one consisting of 3 volumes, one of 4, and the other of 1 volume. They are placed on a shelf at random. Prove that the chance that volumes of the same works are all together IS
TfO-
The
8.
in a row.
letters
What
is
forming the word Clifton are placed at random the chance that the two vowels come together ?
In a hand at whist what 9. held by a specified player. 10.
is
the chance that the four kings are
There are 4 dollars and 3 half-dollars placed at random in a that the chance of the extreme coins being both half-
Show
line.
dollars
is \.
MISCELLANEOUS EXAMPLES 1.
Simplify
.,
cfi
2.
^-" -(b-
.^
c)^
+
'-'' .
b-^
- (c -
+ a)-
VI.
-^ - (« ''
c^
by^
Extract the square root of
+ 6x^ + ~\^-x^ + 15x + 25. - ^^ + 2 «4x4 + ?^ _ 2 «a-J +
4x'^
(i.)
x8
(ii. )
a^.
A
number of 3 digits exceeds 25 times the sum of the digits the middle digit increased by 3 is equal to the sum of the other digits, and the unit digit increased by 6 is equal to twice the sum of the other 2 digits find the number. 3.
by 9
;
:
4.
Find the value of 2
5.
Solve
V^ +
3
Vf -(f V\^V'i)i\ V« - V24).
(i.)
= ^^^+^+^^^^ V2 + x - V2 VSx-ll -H y/Wx = \/l2 x -
2
ic
(ii. )
23.
•
)
MISCELLANEOUS EXAMPLES Solve
6.
336
VI.
2(x + a)^3(.x + fi)^g^ X -{- a X + b
The sum of a certain number of terms of an A. P. is 45, and and last of these terms are 1 and 17 respectively. Find the number of terms and the common difference of the series. 7.
the
lirst
Solve
8.
2^-11^-1 + ^^-111 =
(i.) ^
x-2
^
Vl2 X -
(ii.)
6 5
1-x
+ V3
.r.
-
0.
= V27 x -
1
Find the value of the seventh term x)"^ when « = i, ic = |, « = 9.
(a
+
A
2.
expansion of
in the
9.
A
man starting from at 11 o'clock passed the fourth milestone at 11.30 and met another man (who started from B at 12) at 12.48; the second hian passed the fourth milestone from at 1.40: find the distance between and B, and the second man's rate. 10.
A
A
+
> 5 ax2 + 9 a^,
11.
Show
13.
Extract the cube root of
that x^
44 .r3 Solve
13.
+
63 x2
-f
x^
X
-
?/
(i.
+
x"
13 a^x
xy-\-7f
+ 27 + = 3, = 93.
6 x^
and the reciprocal
x>a.
21 x^
-}-
4 x2
v 1^
54
-\-
2 x^
(ii. )
Find a mean proportional between
14.
if
-
-
9
x.
8
+9 + 11
x?/
10 xy
V- 3
(gO
?/2 ?/2
= =
5,
35.
+ Wy^
of
16 a^b
Two
one of which sails 2 miles an hour faster than the other, start together upon voyages of 1680 and 1152 miles respectively the slower vessel reaches its destination one day before the other how many miles per hour did the faster vessel sail? 15.
vessels,
;
:
16.
Solve
x6
(i.)
Two numbers
= 8 + 7 x3. + &2 _ c2 _
x2»
(ii.)
2 bx'\
numbers obtained by adding 6 to each of the given numbers are in the duplicate ratio of 2 3. Find the numbers. 17.
are in the ratio 2:7; the
:
18.
Solve
(i.)
(n ) ^ '^ (iii.)
(iv.)
- 2 6 = 4x + 52^;. + 4 3x+10 ^ 2x + 3 2x + 3 2x X- 1 Vx + 3 + Vx + 8 = \/4 X + x2 -K xy + = 137, + xy + x = 205. 2 5x2
^
,
?/
y"^
21.
ALGEBRA.
336
[4+y^_^|l^4^] '
Simplify
19.
'
,_^6
20. Find the sides of a rectangle the area of which is unaltered if its length be increased by 2 feet while its breadth is diminished by 1 foot, and which loses f of its area if its length be diminished by 2 feet and its
breadth by 4
The
21.
sum
first
feet.
term of a G. P. exceeds the second term by
to infinity is 81
22.
Solve
(i.)
word
(11.)
which can be made from
of permutations
2
+ V4 +
i
.x
2x-3
- VOx +
2
6
+VX
—
a
—
\^x
4
\/x-2 Vx + — b= Vb — a.
24. Find the condition that one root of ax^ times the other. 25.
Find the value
Given log 648 and of 5.
26. of 3
27.
Two One
rail.
— 3x2 — 8x +
of x^
=
2.81157, log 864
15
=
3
-\-
bx
when x
them
-\-
=
c
3
= +
shall
be
ii
i
2.93651, find the logarithm
trains run, without stopping, over the of
same 36 miles
of
an hour faster than the other and 12 minutes less. Find the speeds of the
travels 15 miles
accomplishes the distance in
two
0.
1
-+
\/x
=
T
,
+
(ill.)
(iv.)
all
= 2Vx-2-l.
.2,3 Vx ,
....
and the
3Iississip2n.
Vx +
....
1,
find the series.
Find the number
the letters of the 23.
:
trains.
28.
Extract the square root of
29.
Find, by logarithms, the value of
9 x^
-
2 x^y
+ ^^ x2?/2 -
2 x^^
+
9
^4.
)^\iT / 15(.318
30.
Simplify
1+^*^"' x-i
«"-— ax-i
«~ix
x
—
((.
31. The men in a regiment can be arranged in a column twice as deep as its breadth if the number be diminished by 206, the men can be arranged in a hollow square three deep having the same number of ;
:
.
MISCELLANEOUS EXAMPLES men
in each outer side of the square as tliere
column 32.
how many men were
;
Solve
8
x'-^
(ii.)
33.
Simplify 8^
34.
A man
8 to
bought a
in the
+
+
orij
4
ic?/
i/^
y-^
x
field
4-5)
- V2 ^ 4"' - (32)"^
the length of which was to
of dollars that he paid for
1
=
U
+ xy + 5 = (ii.) 2x + 3?/ = 10, = 4f 5 x2 + X +
24
(i.)
x"^ -\-
xy
+
2 x2
S
y'^ ?/2
its
breadth as
was equal to the times the number of
acre
;
Solve
?
= 37, = 73.
number of rods in the length of the field and 13 rods round the field equalled the number of dollars that the length and breadth of the field. 35.
depth of the
the regiment
?/3
+ ^(2
The number
5.
were
first in
+ + = 152, 27 x3 + 3 x^y + xy'2 = 40.
2x^
(i.)
there at
337
VI.
-\-
2 ^2,
+
2 ^2.
it
Find
cost.
?/
Find two numbers whose sum added to their product and the sum of whose squares diminished by their sum is 42. 36.
37.
Find the sixth term
is
34,
in the expansion of each of the following
expressions (i.)
38.
and
(«
A
C=
39.
+
3&-2)T.
(ii.)
varies directly as
j\: find
Solve
B
when
(i.) (ii.)
^2a-^y.
(iii.)
B and inversely as C A = A = ^48 and C = ^75. ;
Vx + 12 + ^x + x^-^yVxy= 9, + X Vxy = 18.
12
(^^-^)'|
when
B=
^-
= 6.
2/2
40.
Form an equation whose
roots shall be
harmonic means between the roots of z
x"^
— px
-{-
the arithmetic
g
= 0.
and
CHAPTER XXXVII. Binomial Theorem. 406. It
may
be shown by actual multiplication that
+ lj)(a + c)(a + d){a + e) c d = a^ e)ct? + {he + hd + 6e + cd + ce + de)a^ H- Q)cd + hce + 5cZe + cde)a + ^ccZe (1) We may, however, write this result by inspection for the
{a
-\-
(Jj
-\-
-\-
-\-
;
complete product consists of the sum of a number of partial products each of which is formed by multii)lying together four letters, one being taken from each of the four
we examine
the
way
products are formed,
we
see that
If
factors. tial
(1)
The term
a^ is
which the various par-
in
formed by taking the
letter
a out of
each of the factors.
The terms involving a^ are formed by taking the a out of any three factors, in every way possible, and one of the letters 6, c, d, e, out of the remaining factor. (3) The terms involving a^ are formed by taking the letter a out of any tivo factors, in every way possible, and two of the letters 6, c, d, e, out of the remaining factors. (4) The terms involving a are formed by taking the letter a out of any one factor, and three of the letters b, c, d, e, out of the remaining factors. (5) The term independent of a is the x^roduct of all the (2)
letter
letters
b, c, d, e.
Ex. Find the value of (a
= =
-
2)(a
+
The product' a4-h(-2 + 3-5 + 9)a3 + (_ 6 + a4
+
6
(j3
_ 47
(j2
_
3)(«
10
-
+ (30 _ 09
(J
+
270.
338
-
18
54
5)(a
+
9).
- 15 + 27 - 45)a2 + 90 - 135)a -{ 270
.
BINOMIAL THEOREM.
339
we
407. If in equation (1) of the preceding article pose c = d = e = b, we obtain (a
We
+ by = a' + 4 a'b + 6 a'b' + 4 ab' + b'
now employ
shall
known
sup-
same method to prove a formula by which any binomial of can be raised to any assigned positive intetlie
as the Binomial Theorem,
+6
the form a gral power.
408. To
find the
expansion of (a
+ 6)'* when n
is
positive
a.
integer.
Consider the expression (ft
+ b)(a + c)(a + fO
number of factors being n. The expansion of this expression
••• (^^
+
^')?
the
is
the continued product
« + &, « + c, a-\- d, ••- a -\- k, and every term in the expansion is of ?i dimensions, being a product formed by multiplying together li letters, one taken from each of these 7i factors. The highest power of a is «**, and is formed by taking the letter a from each of the n factors. The terms involving a"~^ are formed by taking the letter a from any n — 1 of the factors, and one of the letters b, c, d, ••• k from the remaining factor; thus the coefficient of a"~^ in the final- product is the sum of the letters b, c, d, •" k; denote it by /Si. The terms involving a"~^ are formed by taking the letter a from any n — 2 of the factors, and tivo of the letters thus the coeffib, c, d, "• k from the two remaining factors cient of a'*~^ in the final product is the sum of the products of the letters 6, c, d,--- k taken two at a time denote it by JS2. And, generally, the terms involving a""'" are formed by taking the letter a from any n — r of the factors, and r of the letters 6, c, d, ••• k from the r remaining factors thus the coefficient of a""'" in the final product is the sum of the products of the letters b, c, d, -•• k taken r at a time denote it by /S,.. of the
n
factors,
;
;
;
;
ALGEBRA.
340
Tlie last term in tlie product is bed
Hence
=
+
a''
+ {a + c)(rt + 4+ /S/t"-- +
(a SycC'-^
J>)
•••
r?)
• • •
•••
k\ denote
(a
+ k)
+
S,a''-'
•••
it
by
+
4-^„-i^t
S„.
^n-
number of terms is the same as the number of combinations of n things two at a time; that is, "Co; in S^tJie number of terms is "C3; and In Si the number of terms
is
??
in
;
the
/S'o
so on.
Now
suppose
€, d,
becomes
^Sa
"(7i6;
k,
• • •
"Os^^
;
substituting for "Ci, "Co, (a
+
by
=
+
ft"
ncC'-'b
the series containing
7?
each equal to b
aS's
•••
becomes
we
;
"(73?>'^;
then Si becomes and so on; thus
obtain
^ ^<^ - ^) ^^«-2^2
+1
terms.
the Binomial Theorem, and the expression on the by. right side is said to be the expansion of (a
This
is
+
409. The coefficients in the expansion of (a + by are very conveniently expressed by the symbols "Oj, "Os, "O, ... "(7,i. shall, however, sometimes further abbreviate them by omitWith this notation we ting n, and writing Oi, Cs, Og, ... (7„.
We
have (a
+
If (a
we
=
a"
+
Cici^'-'b
write
—
b in the place of
by
+
CM^-'b''
- by = a" + 0"-X-6)+
= a- -
Cice-'b
Thus the terms
+
b,
a,a"-7j"
we
aa"-'(-
(7/6"-2^2
_
and negative, and the
last
.••
+
(7„6".
?>)'
+ (a +
...
4.(_
l)"^,^^".
6)" and (a — by — by they are alternately
in the expansion of
tive according as n is even or odd.
+
obtain
c.,ce^-^y^
are nu7nericaUy the same, but in (a positive
+
term
is
positive or nega-
BINOMIAL THEOREM. Ex.
By
Find the expansion of
1,
+
(k,
341
y/)^.
the formula, the expansion
= «6 + <;(j^^5y ^ &C2ahf- + 6 03«V + *^C'4«2?/4 + ^^Cr^aif + 6C6?/« = «6 + + 15 a^- + 20 a^V^ + 1^ «V + a^s ^o^ (v'lj
_|.
on calculating the values Ex. (a
2.
Now 7
Find the expansion
- 2 x)7 =
up C5
to
= (a
rt^
-
C'2
;
of {a
—
....
2x)'^.
+ 702a5(2 x)2 -iC^a^l + to 8 terms. that '^Gr = '"Cn-r after calculating the coefficients may be written down at once; for 7(7^ — 7(73.
70ia6(2 x)
remembering 'Cs,
"
of ^C'l, 6(7^, 6(73^
the rest
and so
on.
•••
.r)-^
Hence
-2 xy =
a?
-1 «6(2 a;) +
^
=
a^
-
21 «5(2 x)2
7 ««(2
.x)
+
cv'(2
x^ - 1-^-^
-
.35
«t(2 ^0^
+
...
a\2 rr)3 + 35 a^(2 x.y
-21a^(2xy^ -\-7a(2xf -{2xy
= a^ -
14
«% +
84
«5^;2
_ -
28O
a*x=5
672 a%5
+ +
560 a^x*
448 ax^
-
128
x^.
In the expansion of 410. The (r 4- i)th or General Term. b)", the coefficient of the second term is "(7i of the third term is "Co of the fourth term is ""C^ and so on the suffix in each term being one less than the number of the term to Avhich it applies hence "(7^ is the coefficient of the (r + l)th term. This is called the general term, because by giving to r different numerical values any of the coefficients may be found from "(7,.; and by giving to a and h their appropriate indices any assigned term may be obtained. Thus the (r + l)th term may be written (a -f
;
;
;
;
;
In applying this formula to any particular case, it should be observed that the index of b is the same as the suffix of C, and that the sum of the indices of a and b is n. * See Art. 392, Cor,
;
ALGEBRA.
342 Ex.
Find the
1.
Here
(r
1)=
-j-
term
fifth
5,
+
of (a
the required term
='^'^C4,a^^(2x^y
= 17.16.
15.14
1.2.3.4
= Ex.
2.
Here
r
2x3)17.
therefore
,«
38080 ai%i2.
Find the fourteenth term of (3
+
1
=
,,
i„ X 16 ai%i2
-
a) is.
14, therefore
= ^^Cis(Sy\ - a)^^ = 15(72 x(-9ai3) = - 945 ai3.
the required term
[-Art. 395.]
411. Simplest Form of the Binomial Theorem. The most convenient form of the binomial tlieorem is tlie expansion of This is obtained from the general formula of Art. (1 + xy. 408, by writing 1 in the place of a, and x in the place of b.
Thus (1
+ xy = 1 + "Cio; +
=1 ,
4-
nx
+
the general term being
"Cx^
1) ^
^^^^
—
^ .^'=^
+
^-^
^^
+ hy=
a(l
+
"= a\l
^\
I
+
h "C>''
+ x^,
—
^
'-x"".
may always term
first
+ c)'
is
be made to thus
unity
where
c
;
=—
I
Ex. Find the coefficient of
ici^ ij^
- 2 x) i'^ (x-^-2x)i'^ (x'^
We have
•••
r^
412. The expansion of a binomial
depend upon the case in which the (a
''CX
h
H
—
71(71
the expansion of (x^
—
2xy^.
= x2Vl-?j x^ V 1 - ? y^
(2\ x) 1
^'^ 2X1*^ ]
have in tains
thi.s
,
w(
expansion to seek the coefficient of the term which con-
— X4
Hence the required
coefficient
=
i'^C'4(— 2)'^
^ 10.0.8.7 1.2.3.4
X 16
=
3360.
BINOMIAL THEOREM.
343
PROOF BY MATHEMATICAL INDUCTION.
By
413.
identities
actual multiplication
we
obtain the following
:
+ by = o? + 3a-& -^3ah- + W, (a -\-by = a +4 a% + 6 a-b' + 4 air + b\ (a
Selecting any one of these, and rewriting so as to exhibit the laws of formation of exponents and coefficients, we have (a
+ by = a' + ^a'-'b + ^ci'-'^' + ^4^^^'"'^'
+ ^;^;3;^ ^^"^^(Art.216). If these laws of formation hold for (a
+
by, n being any
positive integer, then (a
+ by = a" +
na^^-'b
+ ^^-^a"-^^^
Multiplying each side of the assumed identity by (a
+ b)
and combining terms, we obtain (a
+ by+' = a'^+i + (n + l)«"^-> + ''^^' \^\e'-'b'
^ .(» + m»^-l) ,„-.,3^.... It will be seen that
n
in (1)
is,
.
(2).
in every instance, replaced
{n H- 1) in (2). Hence if the theorem be true for value of 71, it will be true for the next higher value.
by
any
We
have shown by multiplication that the theorem is true when 71 successively equals 2, 3, and 4; hence it is true when n = 5, and so on indefinitely. The theorem is therefore true for
all
positive integral values of
7i.
ALGEBRA.
344
414. lu the expansion (a
+ h}" = a" + nce'-'b
-f
'^~^a»-'b'+ ••.
we observe that in any term (1) The exponent of b, the second term of the binomial, is one less than the number of the term from the first. (2) The sum of the exponents is n. (3) The last factor of the denominator of the coefficient the same as the exponent of the second term of the
is
binomiaL (4)
The
last
factor of the numerator of the coefficient
the exponent of the
is
byl. Hence the
(r
+
Jii'st
term of the binomial increased
i)th or general term of (a
7i(n-l)...(/^-r+l)
^'
''
1.2.3...r
Ex. Find the Gth term in the expansion of (2 a Here n = 10, and r + 1 = 6,
We
have
^^-^-^'^'^ (2
= ^.2^7.6
ayb^
+ by is
+
.^
= 252(2ya^b^ = Note. the same
EXAMPLES XXXVIL (a:
+ +
2.
(x
3.
(a.+
4.
(a
the following binomials
.
11. 12.
13.
a
;
3)5. x)"^.
Write in simplest form 10.
8064a^b\
2)4.
- xy. (\-2yy.
5.
y^
The student should observe that the coefficient contains number of factors in both numerator and denominator.
Expand 1.
by^.
(-1)
- («-^)
•
•
:
The 4th term of (1 + x^^. The Gth term of (2 - yy. — 5 &) ^ The 5th term of (a The 15th term of (2x - ly^ :
^^^^ ^^
/^_ iy\
^^
^^^^
^^^^
^^
,^,j^^
^^^^ ^^^.^^ of f
3x
+ ^V.
;
BINOMIAL THEOREM.
17.
Expand ( Vl -
18.
Find the
- V^)* +(x+ V'^)^x^ + 1)5 -(Vl - x-^ - 1)5. coefficient of x^- in (x^ + 2xy^.
19.
Find the
coefficient of
16.
Find the value
345
of (x
x in
f
x^
—
20.
Find the term independent
21.
Find the
—
.
|
2x)
V
of x in
coefficient of x--^ in
|
2 x^
x2
2
6
X--
—
. j
\ 15
Li the expansion of (1 + x)'' the Coefficients. of terms equidistant from the beginning and end
415. Equal coefficients
are equal.
The
coefficient of the (r
+
l)th term from the beginning
is "C,.
The (r + l)th term from the end has n + 1 — (r + 1), or n — r terms before it therefore counting from the beginning it is the (u — r -\-l)th. term, and its coefficient is "C^ r^ which has been shown to be equal to "(7^ [Art. 395]. Hence ;
the proposition follows.
To find the greatest coefficient in 416. Greatest Coefficient. expansion of (1 -f xy\ The coefficient of the general term of (1 -f xy is "(7^ and we have only to find for what value of r this is greatest.
the
;
By "(7„;
Art. 401,
when n
is
when n odd,
it
is
even, the greatest coefficient is or "C„+i; these coefficients
is "(7„_i,
being equal. 417. Greatest Term. + hy\
To find
We
hy
the greatest term in the expan-
sion of (a
have
(a
-}-
=
a'^
fl
therefore, since a" multiplies every
+ - Y' term in
[
1
+-
j
,
it
will
be sufficient to find the greatest term in this latter expansion.
—
b
ALGEBRA.
346
Let the rth and (r -f- l)th be any two consecutive terms. (r -|- l)th term is obtained by multiplying the ?*th term n-r + l b, ^^^^ ^ /n + 1 _ ^^^ r a Ja \ r
The
A
.
_
-"
The
— ?i.
1
-I-
^!^
factor
1 decreases as r increases
;
hence the
r (/•
+ l)th
term
is
not always greater than the rth term, but
only until f^^—t
becomes equal to
1,
or less than
1.
^!i±i-l^^>l, solongas^-^ii-l>^^
Now
Ja
r
\
r
^i±i>^4-l,
thatis,
If
1 )_
^^
——^ +b
a
b
&^±Al^>r
or
denote be an integer, ^
it
by p ^ "^
; '
.
.
then
(1).
.
if r
= ^n
1, and the (p + l)th term and these are greater than any other
the multiplying factor becomes equal to the pth term. is
y^
If
"*"
/
;
be not an integer, denote
its
integral part
by
q; then the greatest value of r consistent with (1) is g; hence the (q + l)th term is the greatest. Since we are only concerned with the numerically greatest
same for (a — by thereany numerical example it is unnecessary to consider the sign of the second term of the binomial. Also it will be found best to work each example independently of the term, the investigation will be the
;
fore in
general formula. Ex. Find the greatest term in the expansion of (1 + 4a;)^, when x has the value \. Denote the ?'th and (r+ l)th terms by 7^ and Tr+\ respectively then ;
r
hence that
is,
Tr+i
> 30
Tr, so long as
-
3
r
4 r
> 3 r,
9
—
r
x
4
-> 1
r
3
or 30
> 7 r.
;
BINOMIAL THEOREM. The term
greatest value of r consistent with this
the sixth, and
is
its
Sum
418.
5
hence the greatest
;
value
= 8C5X
=
(1)5
8(73
X (1)5
=
AI34A
To find
of the Coefficients.
expansion of(l
cients in the
is
347
sum of the
the
coeffi-
+ xy\
In the identity -j-
(1
put
a;
=1
xy =1
+
C\x
Cox'
= sum
.
•
.
+
0,0;"
of the coefficients.
Ci+a + C34--- + a = 2«-l; the total number of combinations of n things taking
is,
some or
all
of them at a time
is
2"
— 1.
equcd
is
+
.!•)",
to the
the
sum of
sum of the
[See Art. 400.]
To
419. Sums of Coefficients equal. exjxinsion of (1
terms
+
thus
;
Cor. that
+ C^ +
that
jrrove
the coefficients
coefficients
in
the
of the odd
of the even terms.
In the identity
+ xy = 1 +
(1
put x
= — l; .-.
1
C,x
+
C,x'
+
Cx'^
+
•
•
•
+
C„x%
thus
+ a + c,+ - = c\ +
c,
+
c,
+
....
420. Expansion of Multinomials. The Binomial Theorem also be applied to expand expressions which contain more than two terms.
may
Ex. Find the expansion oi {x^ +2x — ly. Regarding 2 x — 1 as a single term, the expansion
= =
(a;2)3
x6
+
+
- 1) + 3 x%2 x - 4 x3 - 9 x2 + 6 X -
3 (a:2)2(2 ^
6 x5
+9
X-*
l)"^
1
,
+ (2 a;
1)^
on reduction.
421. Binomial Theorem for Negative or Fractional Index. full discussion of the Binomial Theorem when the index is not restricted to positive integral values the student
For a
ALGEBRA.
348 is
referred to Chapter xly.
is less
n(:)i-l)
(1 is
shown that when
It is there
it-
than unity, the formula
nhi-l)(n-2) ^
+ xy = 1 + nx + —^^T^-^'-v- + .,
-^
-^J^3
,
of^
+-
true for any value of n.
When
n
is
negative or fractional the
number
of terms in
the expansion is unlimited, but in any particular case we may write down as many terms as we please, or we may find the coefficient of any assigned term. Ex,
Expand
1.
+
(1
x)~^ to four terms.
1.2- 3
1-2
= l-Sx + 6x^Ex.
2.
oTi
Expand
,
3
3x
+
(4
,
3
10x3
3 x)
9x2
3 -
+
...
to four terms.
27x3
1
"1 ,
422. In finding the general term we must now use the formula 7i(7i
- l)(n - 2)
'•'
- r -\-l)
(n
^'^
\r
written in full; for the 71
is
symbol "0^ cannot be employed when
fractional or negative.
Ex.
1.
Tlie (r
Find the general term
+
l)th term
zr
in the
expansion of (1
K^ - 1X2 - 2)
••
(J:
-
r
+
+
x)^.
1) ^,
\r
^ l(- l)(-3)( -5) .-.(-2r + 2r[r
3) ^,
:
BINOMIAL THEOREM.
numerator is r, and r — 1 of these by taking — 1 out of each of these negative may write the above expression
The number are negative factors, Y{Q
349
of factors in the
therefore,
;
1.3.5...(2r-3)^,^ 2r\r_
Ex.
The
(,•
2.
+
Find the general term 1
)th
expansion of
in the
(- 3)(- 4)(term =
...
5)
(-
-
3
,
^(_iy3-4-5.-(r + 2)
(1
—
x)~^.
+
1)
^_,^y
,^,
\r_
^
1.2.3...r
^
1-2 by removing
like factors
from the numerator and denominator.
423. The following expansions should be remembered
- x)-^ = l-\-x-\-a? + Q(?+ + (l-x)-'' = l^-2x^3x^-\-4.a? + ...
(1
(1
_ x)-^ ^ 1 424.
.t""
+ l^x'-"
-^ 3a; H- 6a;^
The following example
H
.
+ (r + 1) iV + + ^ ^^%'^'^^ ^' + ...
...
^''
I
.
illustrates a useful
cation of the Binomial Theorem. Ex. Find the cube root of 126 to 5 places of decimals.
+
(126)^ =(53
1)^:. 5^1
3
V
=
+iy
9
53
56
59
81
/
5+1.1-1.1 + ^.1-... 3
55^81
9
52
57
= 5+1. 2i_l._2^,J_.^ 3
= = =
r
,
'
.04
i02
'+T 5
+
9
'
.00032
.013333
105 ,
81
'
107
.0000128
g-^-si ^.000035
•.•
+
•••.
Zi
•.•
5.01329, to five places of decimals.
appli-
.
ALGEBRA.
350
EXAMPLES XXXVII. In the following expansions find which
yy' when x = 4, - ?/)-« when x = 9, + xy when = f
1.
(x-{-
)j
2.
(x
^/
3.
(1
= =
is
-
b.
the greatest term 4 6)i5 ^vhen a
S.
4.
(«
4.
5.
{1 x-\-2}/)^
6.
(2
a;
a;
+ 3)'*
=
:
12,
when x=8, when x = |,
In the expansion of (1 + o:y^ the coefficients of the (2 r and (r + 5)th terms are eqnal find r, 7.
= 2.
1>
y= ?i
U.
=
+
15.
l)th
:
8.
(1
Find
+ xy 9.
of (r
when
7i
the coefficients of the 16th and 2Gth terms of
are equal.
Find the relation between r and n in order that the coefficients + 3)th and (2 r — 3)th terms of (1 + x)3» may be equal.
10.
Find the
11.
Find the middle term
12.
Find the sum
of the coefficients of (x
13.
Find the sum
of the coefficients of (3 x
coefficient of x"* in the
of (1
+
expansion of
x~
+
xj
simplest form.
.x)2« in its
+
(
yy^.
+
yy.
Find the rth term from the beginning and the rth term from the end of (a + 2x)". 14.
15.
Expand
Expand 16.
(l
(a^
+
+
2 «
1)^
and
(x2
-
4x
+
2)3.
to four terms the following expressions
+ x)i
19.
(l
+
:
3x)-2.
22.
(2+x)-3.
3
17.
(1+x)*.
20.
(l-x2)-3.
23.
(l+2x)"ri
18.
(l+x)i
21.
(l+3x)-».
24.
(a-2x)"2.
Write in simplest form
:
+
25.
The 5th term and the 10th term
26.
The 3d term and
27.
The 4th term and
the (r
+
l)th term of (1
28.
The 7th term and
the (r
+
l)th term of
29.
The
Find 30.
(r
+
of (1
the 11th term of (1
l)th term of (a
-
6x)-i,
+
and
x)"^. 2x)"2".
+ x)-^. (1 - x)^.
of (1
-
nx)"".
to four places of decimals the value of
\/l22.
31.
\/620.
32.
v/31.
33.
1--a/99.
BINOMIAL THEOREM. Find the value
of
+ V2)*+(a^-V2)*-
34.
(x
35.
(\/.r.2_rt2^a:)5-(\/x2-a^-a-)5.
38.
Find the middle term of fl-\-^y^.
39.
Find the middle term of
40.
Find the
coefficient of x^^ in
41.
Find the
coefficient of x^^ in {ax*^
42.
Find the
coefficients of x^-
43.
Find the two middle terms Write
in simplest
44. Tlie 8th
351
(V2 + l)^-(V2-l)'• (2- \/l-a;)6+(2+ VT^:)".
36.
j
37.
-^
1
[
.
j
+ '—- \
x^
—
bxy.
and
x~i"
of
3 «
i
.
"
1
I
\i^
(-i) '
-—
form
term of
(l
+ 2x)"^.
45.
The 11th term
of
(l-2x3)"2
46.
The 10th term
of
(l+3a2)"^A
47.
The 5th term
48.
The (r+l)thtermof {X-xy-.
of (3 a
-
.
2 ?>)-i.
49.
The (r+l)th term
of
50.
The (r+l)th term
of (1
51.
The (r+l)th termof
52.
The 14th term
53.
The 7th termof
(l-x)-*.
(l
+ x)^.
+ x)'^'.
of (2i'>-2'a:)^^. (38
+ 64x)
?^\
CHAPTER XXXVIII. Logarithms. 425. Definition. The logarithm of any number to a given base is the index of the power to which the base must be raised in order to equal the given number. Thus if a' = N, X is called the logarithm of JSf to the base a. Examples.
(1) Since
3"^
= 81, the logarithm of 81 to base = 10, 102 ^ iqo, lO^ = 1000,
(2) Since 10^
the natural 100, 1000,
numbers
•••
1, 2, 3,
•••
3
is 4,
•••
are respectively the logarithms of 10,
to base 10.
426. The logarithm of log^iV, so that the
N to
same meaning
base a is usually written is expressed by the two
equations a^
= ^;
x
= log« N.
Ex. Find the logarithm of 32^4 to base 2^2. Let X be the required logarithm then, by definition, ;
(2v2)-
= 32^4;
;
LOGARITHMS.
353
PROPERTIES OF LOGARITHMS. The logarithm of 1 is 0. 428. Logarithm of Unity. For a" = 1 for all values of a therefore log 1 = 0, whatever the base may be. ;
429. Logarithm
The logarithm of
the Base.
of
the base
itself is 1.
For
a^
=a
;
therefore
430. Logarithm tvhose base
is
log,,
a
The logarithm ofO, in any system
of Zero.
greater than unity,
For
a-="
=
= co,
431. Logarithm
of a Product.
sum of
is 7}iinus infinity.
^
Also, since «+'"
is the
= l.
= 0.
the logarithm
of
+
cc is
+
The logarithm of a product
the logarithms of its factors.
Let 3IN be the product let a be the base of the system, and suppose X = log„ 3f, y = loga N; ;
so that
a^
Thus the product
= M,
= N. a^ = a'+^
a^
MN= ce x log„ MN =x-\- y = log„ M-{- log„ N.
whence, by definition, Similarly,
log,
IINF = log, 31
and so on for any number of Ex.
log 42
=
432. Logarithm is the
log (2
-\-
log, JSf+ log, P;
factors.
X 3 X 7)= log 2
+
log3
+
log7.
The logarithm of a quotient
of a Quotient.
logarithm of the cUvidend minus the logarithm of the
divisor.
Let
M be the fraction, and suppose — = log, 3/, a^ = M, a.'
so that
2a
y cfT
= \og^N'^ = JS'.
; ;
354
ALGEBRA.
Thus the
— = — ^a"" N a"
fraction
''•''.
M = x — y = log„ 3f— log„ N.
whence, by definition, log„— Ex.
log (21)
= =
log y-
= log 15 -
log 7
-
=
log (3 X 5)
log 7
log 3
- log 7.
log 5
-I-
Logarithm of a Power. The logarithm of a number any power, integral or fractional, is the logarithm of number multiplied by the index of the power.
433.
raised to the
Let loga (3/^) be required, and suppose
=
X
log„ 31, so that
a""
= 31]
3P={ay = a^'\
then
whence, by definition, \o%^{M^) =x)x; that
is,
Similarly,
log„
{M^)
= p log„ 31.
log„
{3P)
= - log" 31. r
Express the logarithm of -^^^ in terras of log
Ex.
a, log b,
and
logc,
3
log
-^ = log =
^=
I log a
From
434.
—
log
J - log
(log c^
4-
log
the equation
common logarithms
(C562) b'^)
10"=
=
I log
=
a
N,
—
it
5 log c
is
—
2 log
b.
evident that and that
will not in general be integral,
they will not always be positive.
For instance,
> 10^ and < 10^ = 3 + a fraction. .06 > 10"- and < 10-^ log .06 = — 2 + a fraction.
3154 .-.
Again, .-.
log 3154
Negative numbers have no common logarithms. 435.
D?:fixitiox.
The
called the characteristic,
written that
The
it is
integral part of a logarithm is
and the decimal
part,
when
it is
so
positive, is called the mantissa.
characteristic of the logarithm of
base ID can be written by inspection, as
any number
we
to the
shall nQ^Y show.
:
LOGARITHMS. The
436.
;;
355
Characteristic of the Logarithm of
Any Number
It is clear that a number with two Greater than Unity. digits ill its integral part lies between 10^ and 10^ a number with three digits in its integral part lies between 10^ ;
and
10'^;
Hence a number with n digits in its between 10""^ and 10'\ be a number whose integral part contains n digits and so
on.
integral part lies
Let then
^
JCr -— i()(n-l)+ a fraction
.-.
Hence the
log
-^=
(?i
characteristic
— is
1)
-f-
—
?i
tic of the logarithm of a number one than the number of digits
.
a fraction.
1 that is, the characterisgreater than unity is less by
in
;
its
integral part,
and
is
positive.
437. The Characteristic of the Logarithm of a Decimal FracA decimal with one cipher immediately after the decimal point, such as .0324, being greater than .01 and less than .1, lies between 10""^ and 10~^; a number with two tion.
ciphers after the decimal point lies between 10~^ and 10~^ and so on. Hence a decimal fraction with n ciphers immedi-
the decimal point lies between 10"^'*+^' and 10~". be a decimal beginning with n ciphers then
atel}^ after
Let
D
;
jT)
.-.
Hence the
log
-— J^Q-(n+l)+ a
D == —
fraction
.
+ a fraction. — {n + 1) that is,
{n 4- 1)
characteristic is
;
the charac-
of the logarithm of a decimcd fraction is greater by unity than the number of ciphers immediately after the decimcd point and is negative. teristic
438. Advantages
of
Common
Logarithms.
Common
loga-
rithms, because of the two great advantages of the base 10, These two advantages are as follows are in common use. already proved it is evident that the the results From (1) characteristics can be written
by inspection, so that only the
mantissse have to be registered in the Tables.
ALGEBRA.
356 (2)
The mantissse
numbers which have
same for the logarithms of all same signijicaiit digits; so that it is
are the the
the logarithms of
tabulate the mantissse of
sufficient to integers.
This proposition we proceed to prove.
N
be any number, then since multiplying or 439. Let dividing by a power of 10 merely alters the position of the decimal point without changing the sequence of figures, it follows that integers, are
as those of
Now
^x
10^, and ^-i-10^, where j) and q are any numbers whose significant digits are the same
JSf.
log(iV^x 10'')=:logiV^
Again, log
(N -^
10')
+ i:)loglO = logi\r-f = log J^-q 7>
=log]Sr-q log 10
•
(1).
.
(2).
an integer is added to log N, and in (2) an integer is subtracted from logiV; that is, the mantissa or decimal portioyi of the logarithm remains unaltered.
In
(1)
In this and the three preceding articles the mantissse have been supposed positive. In order to secure the advantages of Briggs' system,
we arrange our
Avork so as always to
keep the mantissa positive, so that when the mantissa of any logarithm has been taken from the Tables the characteristic is xn-efixed
with
its
appropriate sign, according to the rules
already given.
440. In the case of a negative logarithm the minus sign written over the characteristic, and not before it, to indicate that the characteristic alone is negative, and not the
is
Thus 4.30103, the logarithm of .0002, is and must be distinguished from an expression in which both the integer and the
whole expression. equivalent to
— 4.30103,
— 4 + .30103,
decimal are negative. In working with negative logarithms an arithmetical artifice will sometimes be necessary in order For instance, a result such to make the mantissa positive. as — 3.G9897, in which the whole expression is negative, may be transformed by subtracting 1 from the characteristic and adding 1 to the mantissa. Thus,
- 3.69897 = - 4 +(1 - .69897)= 4.30103.
LOGARITHMS.
857
Ex. 1. Required the logarithm of .0002432. In Seven-Place Tables we find that 3859636 is the mantissa of log 2432 (the decimal point as well as the characteristic being omitted) and, by Art. 437, the characteristic of the logarithm of the given number is ;
-4; .
may
This
Ex.
•
log 0002432
.
.
be written 6.3859636
-
4.
3859636.
Find the value of V. 00000165, given
2.
log 165
= 2.2174839,
log 697424
Let X denote the value required log
= 10.
x
= log (.00000165)^ =
=
5.8434968.
then
;
^ log
(.00000165)=
i-(^-
2 174839);
the mantissa of log. 00000165 being the same as that of log 165, and the characteristic being prefixed by the rule.
K6.2174839)=K10 + 4.2174839) = 2.8434968 hence ic is a number conis the mantissa of log 697424 these same digits, but with one cipher after the decimal
Now
and .8434968 sisting of
point.
;
[Art. 437.]
Thus
X
=
.0097424.
441. Logarithms transformed from Base a to Base 6. Suppose that the logarithms of all numbers to base a are known
and tabulated. Let JV be any number whose logarithm
to
base b
is
required.
Let y that
= logft^,
so that
¥ = JSF;
ylos,b
is,
•
y^-.
= \og,N', 7Xlog,iV,
log« b
logj,N=-
or
-xlog„i\^
(1).
^ogab
N
N
Now since and b are given, log„ and log„ b are knoAvn from the Tables, and thus log^^may be found. Hence
to
transform logarithms from base a
midtiply them all by
;
to
base b
this is a constant quantity,
log„/->_
is
given by the Tables
;
it is
known
as the modulus.
ive
and
;
ALGEBRA.
358 Cor. If in equation
(1)
we put
It
a for N,
-xlog„« ""
1
logj,a=-
=
logjj'
\og,b
logja X log„6
.-.
we obtain
1
= l.
The fol442. Logarithms in Arithmetical Calculation. lowing examples illustrate the utility of logarithms in facilitating arithmetical calculation.
Ex.
1.
=
Given log 3
.4771213, find log {(2.7)3 x (.81)^
-
(90)^.
= 3(log - 1) + I (log 34 - 2) - J (log 32 + 1) = (9 + ¥-i)log3-(3 + | + |) = ^i\og3 = 4.6280766 - 5.85 = 2.7780766. 3-3
m
The student should
notice that the logarithm of 5 and
can always be obtained from log 2 log 5
Ex.
2.
;
= log i# = log 10 - log 2 =
Find the number log 2
log (87516)
hence the number of
= = = =
log7
16 log (7 X 125)
1
-
log 2.
= .84.50080. = 16 (log 7 +
16 X
digits is 48.
[Art. 436.]
of
V^^ and
a.
.03125 to base ^2, and 100
2.
Findthe value
3.
Write the numbers whose logarithms to bases
of log4 512, logs .0016, ^ogsio\, log49 34.3.
2,
-
3,
Simplify the expressions
6.
Find by
625.7, 3.502,
3 log 5)
+ 3-31og2) 2.9420080 = 47.072128
16(log7
1. Find the logarithms and .00001 to base .01.
-
powers
of digits in S~b^^, given
..3010.300,
EXAMPLES XXXVIII.
1.7, 1000, are i,
its
thus
5,
-1,2, -
25, 5, .02,
1,
— 4,
f respectively.
•
in.spection the characteristics of the logarithms of 3174,
.4,
..374,
.000135, 23.22065.
.
LOGARITHMS. 7.
The mantissa
of log 37203
359
.5705780: write the logarithms of
is
37.203, .000037203, 372030000. 8.
The logarithm
of 7623
3.8821259: wi'ite the numbers whose
is
logarithms are .8821259, 6.8821259, 7.8821259.
=
Given log 2
=
.3010300, log 3
.4771213, log 7
=
.8450980, find the
value of log 729.
10.
log 8400.
11.
log. 256.
12.
log
13.
logv'392.
14.
log. 3048.
15.
Show
16.
Find to
9.
5.
832.
+ log f f
that log \l six
- 2 log ^ =
log 2.
decimal places the value of
_ 2 \o(T JL.Q_
Incr 2 2 5
17.
2-
i
Iqct
5_1_2
Simplify log {(10.8)^ x (.24)^ --(90)--}, and find
its
numerical
value. 18.
Find the value of log (\/i26
Find
20.
Find the number
21.
Show
Vm
--
\/I008
.
v/162).
value of log J 588 x 768 ^686 X 972
19.
tlie
.
/HI
that
I
—
of digits in
42-*-.
\ 1000
is
greater than 100000.
J
22.
How many
ciphers are there between the decimal point /
le first
significant digit in
•>
-
\ 1000
?
)
\3; 23.
Find the value of
\/. 01008,
log 398742 24.
5.6006921.
Find the seventh root of .00792, having given log 11
25.
having given
=
=
1.0413927 and log 500.977
Find the value
of 2
log||
+ log
J^s^
-
= 2.6998179.
3 log f f
and
360
N 10
ALGEBRA.
LOGARITHMS.
N 55
361
ALGEBRA.
362
USE OF THE TABLE. 443.
On
pages 360-361 Ave give a four-place table concommon logarithms of all
taining the mantissae of the integers from 100 to 1000.
444. To
find the logarithm of a
number.
(a) Suppose the number consists of three figures, as 56.7. In the column headed JV find the first two significant On a line with these and in the column having at figures. the top the third figure will be found the mantissa. Thus on a line with 56 and in the column headed 7 we find 7536.
To
this, Avhich is the
decimal part of the logarithm, prefix
the characteristic [Art. 436], and log 56.7 (b)
Since in
common
we have
= 1.7536.
logarithms the mantissa remains
unchanged when the number is multiplied by an integral power of 10, we change one or two-figure numbers into three-figure numbers by addition of ciphers before looking for the mantissas.
The mantissa
of log
5ij
will be that of
560, the only change in the logarithm being in the characteristic.
Thus
log 560 log
56
= 2.7482, = 1.7482.
In the same manner log 7 has for mantissa that of log 700. log 700 log (c)
7
= 2.8451, = 0.8451.
Suppose the logarithm of a number of more than
three figures, as 62543,
is
required.
Since the
number
lies
between 62500 and 62600, its logarithm lies between their logarithms. In the column headed we find the first two figures, 62 on a line with these and in the columns headed Prefixing 5, and 6, Ave find the mantissee .7959 and .7966. the characteristic [Art. 436], Ave have
N
;
log 62000
log 62500
= 4.7966, = 4.7959.
;
LOGARITHMS.
863
Therefore while the number increases from 62500 to Now our number is hence if to the logarithm of 62500 we add y\% of .0007, a nearly correct logarithm of 62543 is obtained.
62600, the logarithm increases .0007. j%\ of the way from 62500 to 62600
Thus
log 62543
;
= 4.7959 .0003 correction
= 4.7962 Suppose the logarithm of a decimal, as .0005243, is The number lies between .0005240 and .0005250. we find the first two significant In the column headed on a line with these and in the columns headed figures, 52 Prefixing 4, and 5, we find the mantissse .7193 and .7202. the characteristic [Art. 437], we have (f?)
required.
N
;
log .0005240
= 4.7202 = 4.7193
differences .0000010
.0009
log .0005250
Now
.0005243
is
.0000003 greater than .0005240; hence
log .0005240 plus log c> i ^ .0005243 equals 1
-M^^ or -^ of .0009 .0000010
10
(the difference of logarithms)
that
log .0005243
is,
= 4.7193 .0003 (nearly)
= 4.7196 In practice negative characteristics are usually avoided by adding them to 10 and writing — 10 after the logarithm.
Thus
in the above
example 4.7196
= 6.7196 - 10.
445. The increase in the logarithms on the same line, as pass from column to column, is called the tabular differIn finding the logarithm of 62543, we assumed that ence. the differences of logarithms are proportional to the differences of their corresponding numbers, which gives us results that are approximately correct. For greater accuracy we
we
must use tables
of
more
places.
ALGEBRA.
364
find the number corresponding to a logarithm. Suppose a logarithm, as 1.74G(), is given to find the corresponding number. Look in the table for the mantissa .7466. It is found in the column headed 8 and on the line with 55 in the column headed N. Therefore we take the figures ooS, and, as the characteristic is 1, x^oint off two places, obtaining the num-
446. To (a)
ber 55.8. (b) Suppose a logarithm, as 3.7531, is given to find the corresponding number. The exact mantissa, .7531, is not found in the table, therefore take out the next larger, .7536, and the next smaller, .7528, and retain the characteristic in arranging the work.
Thus, the number corresponding to 3.7536
and the number corresponding
3.7528
to
is is
.0008
differences
5670 5660 10
Now the logarithm 3.7531 is .0003 greater than the logarithm 3.7528, and a difference in logarithms of .0008 corresponds to a difference in numbers of 10 therefore we should increase the number corresponding to the logarithm 3.7528 by .0003 3 p^,, ;
or
.0008
Thus
-
01 10.
8 the
number corresponding 3.7531
=
to the logarithm
5660 3.7 correction
= 5663.7 (c)
Suppose a logarithm, as 8.8225 number.
— 10
or 2.8225,
is
given
to find the corresponding
Take out the mantissie
as in the previous example.
The number corresponding The number corresponding
to 2.8228 is .0665 [Art. 437.] to 2.8222 is .0664
differences
.0006
.0001
Now tlie logarithm 2.8225 is .0003 greater than the logarithm 2.8222, and a difference in logarithms of .0006 corresponds to a difference in numbers of .0001 therefore we ;
LOGAKITHMS.
365
should increase the number corresponding to the logaritlim 2.8222 by •^
^^^^
or - of .0001.
.0006
6
Thus the
number corresponding
to the logarithm 2.8222
the number corresponding to the logarithm 2.8225
= .0664 = .0664 .00005
Correction,
= EXAMPLES XXXVIII. Find the common logarithms
.06645
b.
of the following
:
1.
50.
4.
.341.
7.
12345.
2.
203.
5.
0.045.
8.
0.010203.
3.
6.73.
6.
5265.
9.
354.076.
Find the numbers corresponding to the following common logarithms
:
10.
1.8156.
12.
4.0022.
14.
3.8441.
11.
2.1439.
13.
1.9131.
15.
7.4879-10.
447. Cologarithms. The logarithm of the reciprocal of a is called the cologarithm of that number.
number
Thus colog 210 Since log 1
= log 2io = log 1 -
= 0,
we
write
it
in the
log 210.
form 10
—
10 and then
subtract log 210, which gives
colog 210
=
(10
- 2.3222)-
10
=
7.6778
- 10.
Hence Rule. To find the cologarithm of a mnnher, subtract the 10 after the logarithm of the number from 10 and ivrite
—
result.
448. The advantage gained by the use of cologarithms the substitution of addition for subtraction.
is
^
ALGEBRA.
366
4.26
Ex. Find by use of logarithms the value of
^^
log ° 7.42 X .058
=
log 4.20
+
—
7.42 X .058
+
log ='7.42
log
-^ .058
= log 4.26 + colog7.42 + colog .058 = .6294 +(9.1296 - 10)+ 1.2366 = 10.9956 - 10. The number corresponding to this logarithm In finding colog .058 we proceed as follows
is 9.9.
:
colog .058
= log^— =
10
-
[log. 058]
-
10 (Art. 447),
.0o8
= =
10
-
[8.7634
10
-
8.7634
- 10] - 10 + 10 - 10 =
(Art. 437), 1.2366.
449. Exponential Equations. Equations in which the unquantity occurs as an exponent are called exponential equations, and are readily solved by the aid of logarithms.
known
Ex. Find the value of x in 15== = 28, Taking the logarithms of both sides of the equation, we have log 15^ .-.
X log 15 loo- 28
= log 28; = log 28.
log 15
1.4472
EXAMPLES Find by use of logarithms 24.051 X .02456 1. .006705 X .0203 145.206 X (-7.564) ^ •
448.1
x(-.2406)(-
c.
8.
.00010101 x (7117.1)6.
x (5.672)8
(285.42)1.^
^
^^ ^ ^-^2
x ^=T24:89
47.85) 12.876 x V.068x (.005157)^ ^ ^/
^^
3^+^
^c^
\{)^
13.
123^-*
\/9.8149 X 80.80008
14.
2'=
^8283 X (.0006412)4
18.
2-+y
845692.1 x .845856.
16.
3i
r742 8024)3 (742.8024).^
4.
(-.0012045)_^.
5.
V^^
X -y.002 X -^442.6
(18)2 X
7.
XXXVIII.
:
10.
3 3.
^
1.2305 +.
1.1761
.73
X (3.4562)^
^
29.029 X (52.8iyx (.4)9
,
= 405. = 9i
^^
x ^
6^
= y
•2x
18" 2x
x
=
-
6^ iz:
4
5-^
3y,
=
1458.
x
1^-'.
=
* Treat negative quantities occurring in logarithmic Wlien the numerical result is obtained, determine
tive.
ordinary rules of multiplication and division.
3 x 2y+\
2--2*-i
= 33^'-*.
work its
as posi-
sign by the
;
CHAPTER XXXIX. Interest and Annuities. 450. Questions involving Simple Interest are easily solved but in Compound Interest the calculations are often very laborious. We shall now show how these arithmetical calculations may be simplified by Instead of taking as the rate of the aid of logarithms. interest the interest on $ 100 for one year, it will be found more convenient to take the interest on $ 1 for one year. If this be denoted by $ r, and the amount of $ 1 for 1 year
by the rules of Arithmetic
by $ E, we have
Ii
;
= l-{-r.
451. To find the interest and amount of a given sum in a given time at compound interest. denote the principal, Ji the amount of $ 1 in one Let
P
year, n
number
the
of years,
I the
interest,
and 31 the
amount.
The amount since this at the
is
P
of at the end of the first year is PR-, and, the principal for the second year, the amount
end of the second year
is
PR
x
R
the amount at the end of the third year
hence the amount in n years
and therefore
is
or is
PR"; that
PR\
Similarly
PR^, and so on is,
M=PR"; /= P(R' - 1).
Ex. Find the amount of $ 100 in a hundred years, allowing comhaving interest at the rate of 5 per cent, payable quarterly
pound
;
given
log2
=
.3010300,
logs
=
.4771213,
367
log 14..3906
=
1.15808.
ALGEBRA.
368 The number
of
payments
is
400.
If
M be the amount, we have
^3/ =100(fi)400; .-.
logil/
= log 100 + 400 (log 81 - log 80) = 2 + 400 (4 logs - 1 - 3 log2) = 2 + 400 (.0053952)= 4.15808
;
3/= 14390. G. whence Thus the amount is $ 14390.60. Note. At simple interest the amount
is
$ 600.
452. To find the present value and discount of a given sum due in a given time, allowing compound interest. Let be the given sum, the present value, the disthe amount of $ 1 for one year, n the number of count,
F R
D
V
years.
Since Fis the sum which, put out to interest at the present \N\\\ in n years amount to P, ^ve have
time,
P= .-.
and
FP";
V=PE% I)
= P-V=P(1- R-'^y
ANNUITIES.
An
a fixed sum paid periodically under the payment may be made either once a year or at more frequent intervals. Unless it is otherwise stated, we shall suppose the payments annual.
453.
annuity
is
certain stated conditions
;
454. To find the amount of an annuity left unpaid for a given number of years, allowing compound interest.
Let A be the annuity, R the amount of $ 1 for one year, n the number of years, Jf the amount. At the end of the first year A is due, and the amount of this sum in the remaining n — 1 years is AR''~^ at the end of the second year another A is due, and the amount of this sum in the remaining ?t — 2 years is AR"'^; and so on. ;
.-.
M= AR' =zA(l
'
+ AR'^-'-^'"-\-AR'
+ R-{- R^-\-"'to
-\-
AR + A
n terms)
= ^:'^"~
R
;
INTEREST AND ANNUITIES. 455. To a given
an annuity
find the present value of
number
of years, allowing
compound
369 to continue for
interest.
A
Let be tlie annuity, R the amount of $ 1 in one year, n the number of years, V the required present vahie.
The present vahie
of
the present value of tlie present value of
A A
A due
AR~^
in 1 year is
due in 2 years due in 3 years
is
AR-''
is
AE~^; and
:
so on.
[Art. 452.]
Now V
the
is
sum
of the present values of the different
payments
V= AR-^ + AR-" + AR--^ H
.-.
ton terms
= AR-^^-:^^=A^-^-'\
R-1
1-R-^
Note. i)/,
This result
may
also be obtained
given in Art. 454, by E'K
by dividing the value
of
[Art. 451.]
CoR. If we make n infinite value a perpetual annuity
R-1
we
obtain for the present
r
EXAMPLES XXXIX. year 1600 a sura of f 1000 had been left to accumulate its amount in the year 1000, reckoning compound Given interest at 4 per cent per annum. 1.
If in the
for 300 years, find
log 104
=
2.0170333 and log 12885.5
=
4.10999.
2. Find in how many years a sum of money will amount to one hundred times its value at 5^ per cent per annum compound interest. Given log 1055 = 3.023. 3.
pound
Find the present value log2
4.
$6000 due in 20 years, allowing comannum. Given .47712, and log 12875 = 4.10975.
of
interest at 8 per cent per
=
.30103, log3
Find the amount
compound
of
=
an annuity of $ 100 in 15 years, allowing annum. Given
interest at 4 per cent per
log 1.04
=
.01703,
and log 180075
=
5.25545.
the present value of an annuity of $1000 due in 30 years, allowing compound interest at 5 per cent per annum ? 5.
AVhat
is
2b
CHAPTER
XL.
LoiiTiNG Values and Vanishing Fkactions. 456. It will be convenient here to introduce a phraseology will frequently meet with
and notation which the student in his mathematical reading.
457. Functions. tity, as X,
An
expression which involves any quanis dependent on that of x, is Functions of x are usually denoted
and whose value
called a function of x.
of the form f(x), f'{x), F(x), cf>(x), and read function of x,'' " the /' function of a;,'' etc. the equation y =f(x) may be considered equivalent
by symbols " the
/
Thus
any change made in the value of x will produce a consequent change in y, and vice versd. The quantities x aud y are called variables, and are further distinguished as the independent variable and the dependent to a statement that
variable.
An
independent variable
a quantity which may have to it, and the corresponding value determined as soon as the is
any value we choose to assign
dependent variable has its value of the independent variable
is
known.
458. Definition. If?/ =f(x), and if when x approaches a value a, the function f(x) can be made to differ by as little as we please from a fixed quantity b, then b is called the limit of y when x = a. For instance, if S denote the sum of n terms of the series l
+ l + i-|-i+
'222
...-
then
S= 370
2-~ 2"~
— AND VANISHING FRACTIONS.
LIMITING VALUES
S
Here
we when
is
a function of n, and
-^-—^
371
can be made as small
please by increasing n\ that
as
n
8 = 2. The
00
words
'^
=
sign
approaches as a
We
459.
may
This
infinite.
is
is
is, the limit of /S' is 2 be expressed by writing
sometimes used instead of the
limit.-
have occasion
shall often
to deal
with expres-
sions consisting of a series of terms arranged according to
powers of some common
where the
letter,
such as
«3'"are finite quantities and the number of terms may be limited
coefficients a^, a^, ^9,
independent of
x,
or unlimited. It will therefore be convenient to discuss
some proposi-
tions connected with the limiting values of such expressions
under certain conditions. Limiting Value.
460.
tto
when X (i.)
The
+ a^x -f
is indefinitely
limit of the series
ag-'^^
diminished
Suppose that the
+ <:h^ + is
• •
•
o-q.
series consists of
an
infinite
number
of terms.
Let let
h be the greatest of the coefficients a^, %, %, us denote the given series by «o ^ then
+
S
if
x
< 1,
S<
we have
• • •
;
and
'>
...;
—— X ^
1
Thus when x is indefinitely diminished, S can be made we please hence the limit of the given series
as small as
;
is ao. (ii.)
is less is
If the series consists of a finite
number
of terms,
than in the case we have considered, hence
the proposition true,
still
S
more
ALGEBRA.
372 Value of
461.
Any Term. ctQ
In the series
+ a^x + a^^
-f-
a^x^
+
• •
,
may make any term as large as tve please compared ivith the sum of all that folloiv it; and by taking x large enough tve may make any term as large as we by taking x small enough we
please compared tvith the
The
(i.)
follow
ratio of
sum of
all that jyrecede
any term, as
a„a;",
to tlie
it.
sum
of all that
it is
a„«"
When X large as
we
we
made made as
indefinitely small, the denominator can be
is
as small as
a„
^^
please
that
;
is,
the fraction can be
please.
Again, the ratio of the term that precede it is (ii.)
a,^x''
to the
sum
of all
a„x" -,
where y
or
=X
When
X
is
indefinitely
large,
y
is
indefinitely small;
hence, as in the previous case, the fraction can be large as
462.
we
made
as
please.
The following
particular
form of
the foregoing
very useful. In the expression
proposition
is
+ Oo?
+
a„_ia;"~^
+
consisting of a finite
number
of terms in descending powers
a^ic"
•••
-j-aiX
by taking x small enough the last term ccq can be made we please compared with the sum of all the terms that precede it, and by taking x large enough the first term a,^x^ can be made as large as we please compared witli of X,
as large as
the
sum
Ex. 7i*
—
5
of all that follow
it.
1.
By
taking n large enough
?i3
—
w
7
+
as large as
we
we can make
please
the
first
term of
compared with the sum
of all
;
.
AND VANISHING FRACTIONS.
LIMITING VALUES
373
the other terms that is, we may take the first term ii'^ as tlie equivalent of the whole expression, with an error as small as we please pro;
vided n be taken large enough.
Ex. IS
Find the limit
2.
'^
of
^^
5 x3
zero.
~ ^ ~ ^ when —4X+ 8 ^^^
(1) x
is
numerator and denominator we may disregard
(1) In the
;
(2) x
all
terms
infinite
but the first ; hence
- 2 x2 - 4 ^ 3
limit 3 x3 a:
When
(2)
x
is
hence the
last:'
=
5x3-4ic + 8
CO
indefinitely small
—
4
liniit is
^3 5*
we may disregard
all
terms hut the
1
or
,
a;3
5x3
2
8
VANISHING FRACTIONS. Suppose
463.
it is
required to find the limit of x^
-\-
ax
x^
x=
when
—2
ci?
a-
a.
= a-\- h,
we put x
If
—
then h will approach the value zero
as x approaches the value a.
Substituting a x'
-\-
h for
+ ax - 2 x'-a""
and when h is
ft"
x,
^ 3 ah + h^ ^ 3a + h 2ah + h' 2 a + h' ^
indefinitely small the limit of this expression
is
f
There
however, another way of regarding the question
is,
for x^
-i-
and
if
we
— 2 a^ _(x — a) (x + 2 a) _ x-\-2a X+ {x — {x -f put x — a the value of the expression
ax
—
x^
non^
ft)
ft-
ft
ft)
is
|,
as before. Tp it
m •
J.1
•
•
the given expression
— a;^
H-—fta;
y?
—
—2
ft^
we put x ,
found that it assumes the indeterminate [Art. 183] also has this form in consequence of the factor
before simplification, it will be
form
we
^,
the value of which
see that
it
—a
cv-
is
;
,
.
ALGEBRA.
374 X
— a appearing
numerator and denominator.
in both
Now
we cannot
divide by a zero factor, but as long as x is not absolutely equal to a, the factor x.— a may be removed, and we then find that the nearer x approaches to the value a,
the nearer does the value of the fraction approximate to f or in accordance with the definition of Art. 458,
when X =
x^
.. n .1 the T limit oi
-,
a,
— 2 a^ — a-
-^ax ^
x^
.
3
is -•
2
If /(^') and f(x) are two funcwhich becomes equal to zero for some
464. Vanishing Fractions. tions of X, each of
particular value a of
and
§-,
called a vanishing fraction.
is
Ex.
the fraction ^-.^ takes the form
x,
li
1.
x
= 3,
find the limit of
x-^
When
=
—
X"^
— ox
the expression reduces to the indeterminate form g but by removing the factor x — 3 from numerator and denominator,
x
3,
;
^x +
1
+
1
x2 4- 2
which
is
Ex.
2.
X
therefore the required limit.
The
fraction
—Vx +
y^ X
To
find its limit,
X
-
To
The
Thus the
1-(1 +
Now
h
=
Q
;
^^^^^^
^^^
/.)^
when
the fraction then becomes 2
-{-
=
fraction
find its limit,
Theorem.
a
a)
whence by putting x 3.
a ^y^^Q^^Q^
a
multiply numerator and denominator by the surd
V3 x —
(?,
(x
—
— Vx + a - (X a) — a) ( V3 X — a + Vx +
conjugate to
Ex.
^j^g^^ ^
«
we
\/3 x
—
find that the limit
^^
becomes -
1-^x
put x
a)
=
1
-\-
?i,
+ Vx +
a
is
-
V2a when x = 1.
and expand by the Binomial
fraction
l-Cl+l/^-^A^ + .-V -.-r.. x = l hence the required limit is f ;
a
:
:
AND VANISHING FRACTIONS.
LIMITING VALUES
We
465.
shall
now
discuss
some
peculiarities
875
which may
arise in the solution of a quadratic equation.
Let the equation be ax^
If c := 0, then
-{-
bx
whence
.t
= 0,
c
-\-
=
ao^ -\-bx
= 0. 0',
or
a that
one of the roots
is,
If b
= 0,
is
zero
;
and the other
is finite.
the roots are equal in magnitude and opposite
in sign.
If a = 0, the equation reduces to bx -\- c = 0; and it appears that in this case the quadratic furnishes only one root,
But every quadratic equation has two
namely,
and in order to discuss the value of the other root proceed as follows roots,
Write - for x in the original equation and clear of y tions
;
thus,
Noiv put a
cy^ -\-by
= 0,
a
-\-
we
frac-
— 0.
and we have
= 0;
cif -\-by
the solution of which
is
y
= 0,
or
;
that
is,
07=00, or
b
Hence,
in
any quadratic equation one root
finite if the coefficient
ivill
become
in-
of x^ becomes zero.
This is the form in which the result will be most frequently met with in other branches of higher Mathematics, but the student should notice that it is merely a convenient abbreviation of the following fuller statement In the equation ax^ -f 6a? + c = 0, if a is very small, one root is very large, and as a is indefinitely diminished this root becomes indefinitely great. In this case the finite root approximates to
—b
as its limit.
ALGEBRA.
376
EXAMPLES Find the (1)
limits of the following expressions
when x=
7x-^-6x + 4 '
3
x*
+
9
(x
^
:
when x =
(2)
(2a;-3)(3-5x)
.
XL.
-
3)(2
'
*
+ 2x3)(x-5) (4x3-U)(l + x)"
e
2 x^
10.
+
1)
-
1
1.
8.
2 x2
(3-x)(x + 5)(2-7x) '
(7x-l)(x+l)3
Find the limits of
whenx-
5a;)(3x
(2x-l)3
'
(3
^'+1,
-
0.
V^-^2«+Vx-
CHAPTER
XLI.
CONVERGENCY AND DIVERGENCY OF
SERIES.
466. We ha\'e, in Chapter xxxiv., defined a series as an expression in wliich the successive terms are formed by some regular law if the series terminates at some assigned term, it is called a finite series if the number of terms is unlimited, it is called an infinite series. In the present chapter, we shall usually denote a series by ;
;
an expression of the form Ui
+ no-^i's H
h
w„H
Suppose that we have a series conn terms. The sum of the series will be a function of n if n increases indefinitely, the sum either tends to become equal to a certain finite Ihnit, or else it becomes 467. Definitions.
sisting of ;
infinitely great.
An
infinite series is said to
be convergent
when
the
n terms cannot numerically exceed some quantity, however great n may be.
the
first
An the
infinite series is said to
be divergent
when
the
sum
of
finite
sum
of
n terms can be made numerically greater than any quantity by taking 7i sufficiently great.
first
finite
TESTS FOR CONVERGENCY. 468. is
When
Known.
Sum of the First n Terms of a Given Series we can find the sura of the first n terms of a we may ascertain whether it is convergent or the
If
given series, divergent by examining whether the series remains becomes infinite, when n is made indefinitely great 377
finite,
or
.
ALGEBRA.
378
For example, the sum of the
first
l-\-x-\-x--{-x^-]
n terms of the is
If X is numerically less than 1, the finite limit
1
—X
and the
,
is
X
be
and by taking n
,
—
1 greater than
made
any
1-x sum approaches
to the
series is therefore convergent.
If X is numerically greater than
terms
series
finite
1,
sum
the
of the first
sufficiently great, this
quantity
;
n
can
thus the series
is
divergent.
If X
= 1,
the series If x
the
sum
of the first
n terms
is n,
and therefore
is divergent.
= — 1,
the series becomes 1
-1 + 1-1 4-1-1 +
•••.
The sum of an even number of terms is zero, while the sum of an odd number of terms is 1 and thus the sum ;
oscillates
to a class
between the values and 1. This series belongs which may be called oscillating or periodic conver-
gent series.
When the Sum of the First n Terms of a Given Series Unknown. There are many cases in which we have no method of finding the sum of the first n terms of a series. We proceed therefore to investigate rules by which we can 469.
is
test tlie
convergency or divergency of a given series without
effecting its summation.
An
470. First Test. alternately positive
numerically
less
infinite series in ivhicli the
and negative
is
than the 2yreceding term.
Let the series be denoted by Ui
where
— U2 + ^3 — + — Uc + > u^ > > n^ > ?(4
if j
'^^3
terms are
convergent if each term
V.r,
ii^^
•
.
.
• • •
is
:
CONVERGENCY AND DIVERGENCY OF The given
may
series
379
SERIES.
be written in each of the following
forms {u^-U2)-^(us-u^)-\-(ih-nQ)-{iii—(u.2
....
— Us)-(u4 — ih) — (uQ — Uj)—
.
.
.
(1), (2).
From (1) we see that the sum of any number of terms is a positive quantity and from (2) that the sum of any number of terms is less than u^ hence the series is convergent. ;
;
For example, in the
series
1—1-1-1—1-4-1—1-1-... the terms are alternately positive and negative, and each term is numerically less than the preceding one hence the ;
series is convergent.
An
471. Second Test.
infinite series in ivhich all the
divergent if each term some finite quantity, hoivever small. are of the
For the
each term
if
sum
same sign
of the first
by taking
is
terms
greater than
greater than some finite quantity
is
n terms
is
sufficiently great,
n,
is
greater than
na
;
and
a,
this,
can be made to exceed any
finite quantity.
472. Before proceeding to investigate further tests of convergency and divergency, we shall lay down two important principles, which may almost be regarded as axioms.
divergent
mij finite is
it will remain convergent, and remain divergent, when we add or remove number of its terms; for the sum of these terms
If a series
I.
if
a
it
is
convergent
will
finite quantity.
II.
If a series in
which
vergent, then the series
is
the terms are positive is conconvergent when some or all of
all
the terms are negative for the sum all the terms have the same sign. ;
We
sliall
suppose that
the contrary
is
stated.
all
is
clearly greatest
when
the terms are positive unless
;
ALGEBRA.
380
An
473. Third Test.
and
;
some fixed term
after
ing term
numerically
is
numerically
mfinite series is convergent if
from
of each term to the x>recedthan some quantity which is itself
the rcUio
less
than unity.
less
Let the series beginning from the fixed term be denoted by i
+
r
+
?<4
H
—
....
n^
< 1.
Then
u^
-j-
^
^
,
-\-
u^
-{-
u^
-\-
• • •
?fo
Uo
U.
Uo
Uo
U2
Ui
U^
U2
Ui
since r
< 1.
?6o
<
u^
+ Ui
1
is,
^«3
no
Ui
that
+
—
cand let
where
^«2
Hence the given
)
series is convergent.
474. In the enunciation of the preceding article the student should notice the significance of the words "from and after a fixed term." Consider the series
l-\-2x-\-3x^+4:X^-\
f-7ia?"-^-|-....
J^ = ^^^ A + .^-V.; n — 1 \ n — lj
Here
^«-i
and by taking ?i sufficiently large we can make this ratio approximate to x as nearly as we please, and the ratio of each term to the i)receding term will ultimately be x.
Hence
if
But the
x
the series
^^
is,
until n
convergent.
will not be less than 1, until
rin-i
that
is
~ —< n-
1
1
> — X 1
Here we have a case of a convergent series in which the may increase up to a certain point, and then begin to
terms
CONVERGENCY AND DIVERGENCY OF For example,
decrease.
if
x
— ^^^,
then 1
381
SERIES.
—X
= 100,
and
the terms do not begin to decrease until after the 100th term.
475. Fourth Test. A71 infinite series in ivhich all the terms are of the same sign is divergent if from and after some fixed term the ratio of each term to the xyreceding term is greater than unity, or equal to unity.
Let the fixed term be denoted by
Wj.
If the ratio
is
the
sum
of
n terms
equal to
is
7iUi',
equal
and hence the series is
to unity, each of the succeeding terms is equal to
?^i,
divergent. If the ratio is greater than unity, each of the terms after
the fixed term
u^, and the sum of n terms hence the series is divergent.
greater than
is
greater than nui
;
is
476. In the practical application of these tests, to avoid having to ascertain the particular term after which each term is greater or less than the preceding term, it is con-
when
venient to find the limit of increased If
/<
If
Z
;
1,
> 1, = 1,
?i
indefinitely
is
is
be denoted by I. [Art. 473.] convergent.
is
divergent.
let this limit
the series the series the series
[Art. 475.]
may
be either convergent or divergent, and a further test will be required for it may happen that If
Z
;
u
—— <
1, Jnit
continually approaching to 1 as
its limit
ivhen n
In this case we cannot name any increased. quantity r which is itself less than 1 and yet greater Hence the test of Art. 473 fails. If, however, than I.
is indefinitely
finite
—u^>1, but continually approaching
its limit,
the
by Art. 475. u ^" as an abbreviation of the words We shall use "Lim u^. u ^ when n is infinite.'' the limit of
series is divergent
—
*'
to 1 as
—
382
ALGEBRA.
Ex. 1, Find whether the series whose nth term convergent or divergent.
is
-^
—
—^
,
;
CONVERGENCY AND DIVERGENCY OF between the greatest and
lies
— and
least of the fractions •••
-^,
,
therefore ^finite quantity,
is
+ W2 + «3 H
Uy
.'.
Hence
if
one series
h
one series
—
L
say
= ^K +
^
'^^2
+
-^3
h ^n)-
H
in value, so is the other
is finite
in value, so is the other
is infinite
383
SERIES.
;
if
;
which proves
the proposition.
The
478. Auxiliary Series.
application of the principle
of the preceding article is very important, for
by means
of
a given series with an auxilianj series whose convergency or divergency has been already estabThe series discussed in the next article will frelished. quently be found useful as an auxiliary series.
we can compare
it
479. The
infinite
divergent except
Case
The than
I.
first
2
—
;
when p
Let p
term
series
>
is
1
j;
is
+
97,
+ ^ + j^ H
always
positive and greater than 1.
1.
the next two terms together are less
;
the following four terms together are less than
the following eight terms together are less than
Hence the
on.
^^
—
;
—4
and
.
so
than 4 8_ ,. ^"^2^ "^4^"^ 8^"^"
series is less 1
-
2.
,
'
that
less
is,
than a geometrical "progression whose
9
ratio
—
is
less
than
1,
since
i^
>1
;
common
hence the series
is
convergent.
Case
The The J
;
II.
series
Let p = 1. now becomes
l
+ Y + i + i + i+**'-
third and fourth terms together are greater than J or the following four terms together are greater than | or
1
ALGEBKA.
384 i i
;
;
the following eight terms together are greater than j\ or and so on. Hence the series is greater than
and
is
[Art. 475.]
therefore divergent.
Case III. Let ^>< 1, or negative. Each term is now greater than the corresponding term Case
II.,
therefore the series
Hence the
when })
is
always divergent except in the case and greater than unity.
series is
positive
+-+^+ 14 9
Ex. Prove that the series -
Compare the given
series with !
+
-
•••
^-^-~
+
+
•••
is
divergent.
n-
+ -+•••
H
\-
'•••
Thus if Un and r„ denote the «th terms of the given series auxiliary series respectively, we have
Lim— =
1.
both divergent. the given series
is
hence
in
divergent.
is
Mn
_ +
Vn
n^
?i
1
!_ » +
.
.
n
n
and therefore the two
But the auxiliaiy
!
and the
series are
series
is
both convergent or
divergent, therefore also
divergent.
This completes the solution of Ex.
1.
[Art. 476.]
To show that 480. Convergency of the Binomial Series. .r/' by the Binonikd TJieorem is con-
the expansion of (1
vergent ivhen
a:
+
< 1.
Let u^, I'r^x represent the rth and expansion; then Vr^i y,
_n—r •
(r
+ l)th
terms of the
-{-
r
+
that is, from this AYhen r> n 1, this ratio is negative point the terms are alternately positive and negative when x is positive, and always of the same sign when x is negative.
Now when since x
<
r is infinite,
;
Lim -?^ =
1 the series is
x numerically
convergent
if all
;
therefore
the terms are of
more is it convergent when some of the terms are positive and some negative. [Art. 472.]
the same sign
;
and therefore
still
:
CONVERGENCY AND DIVERGENCY OF
EXAMPLES Find whether the following 1.
385
XLI. convergent or divergent
series are
1—1- + —
SERIES.
L^+...,
i
X X -\r a X -{- 2 a x + 3a X and a being positive quantities.
3.
1-
,
.A
X and y being positive 4.
+
..
(juantities.
^^ + -J^ + ^+^+.... 1-2
2-3
3-4
4.5
5.
-^ + _^ + -^ + ^!_+.... 1-2 3. 45. 67-8
6
H-lL%§! + f4""••
7-
10.
"^[2
8-
l+3x + 5x2+7x3 + 9:c4 +
9
—2 + —3 + —4 + —5 +•.
"^13^14
Vi+Vf + Vf + V5 + -1+? + ?+^+...+
'
1^
2i'
3p
4p
'''
10
M-2
+
1
Note. For further information on the subject of Convergency and Divergency of Series the reader may consult Hall & Knight's Higher Algebra, Chapter xxi.
CHAPTER
XLII.
Undetermined Coefficients. FUNCTIONS OF FINITE DIMENSIONS. 481. In Art. 105
it
was proved that
when x
gral function of x vanishes
X
any rational
if
= a,
inte-
divisible
is
it
by
— a. Let
jh^""
+ jpia;"-^ + 2^2^'"'^ +
+Pn
• • •
be a rational integral function of x of n dimensions, which vanishes when x is equal to each of the unequal quantities tt],
f%? •••
02?
Denote the function by /(x) by X — cii, we have
;
f'/i-
then since f(x)
= (x-a,)(p,x^'-'+ "•), — 1 dimensions. we Similarly, since f(x) is divisible by x — + = — «2)(poa^""- +•••)? the quotient being of u — 2 dimensions and _ _ a3)(poa?"-3 ^ Po^"-2 4_
is
divisible
f(x)
the quotient being of
?i
a.,,
^)o.i-"-^
have
•••
G^'
;
...
Proceeding in this
...
(.y
way, we shall
)_
finally obtain after
n
divisions /(•^')
482.
=i>oO^'
- (h)(x -
^to)(.c
- «,)
If a rational integral function of
.••(.-«- o„).
n dimensions vanishes coefficient of each
more than n values of the variable, the power of the variable must be zero. for
Let the function be denoted by f(x), where f(x)
= pox^' + p,x'^ + i)o.«"-2 + '
386
• • •
+ Pn
;
:
UNDETERMINED COEFFICIENTS. and suppose that
Let
c
vanishes
f(^x)
the unequal values
cto,
«!,
O3,
• • •
when x
«„
;
is
387
equal to each of
then
f(x) = 2h{^ — cti) (x — a2)(x — a,) -'-(x- a„). be another value of x which makes f{x) vanish; then
= 0, we have - (c - as) (c - as) (c - a,) = = 0? since, by hypothesis, none of the and therefore since /(c)
Po(c
•••
cii)
;
2>o
Hence f{x) reduces
factors is equal to zero.
By
other
to
hypothesis this expression vanishes for more than n
values of
and therefore
x,
In a similar manner cients 2h, Ihr This result
• '
2^1
= 0-
we may show
that each of the Pn niust be equal to zero. may also be enunciated as follows
coeffi-
'
If a rational integral function of n dimensions vanishes for n values of the variable, it must vanish for every value_ of the variable.
ynore than
If the function f{x) vanishes for more than n values the equation f(x)= has more than n roots.
Cor. of
.1",
Hence n roots
also, if
it is
an
an equation of n dimensions has more than
identity.
Ex. Prove that (x
(a
— -
b)(x
—
&)(«
-cy
c)
—
c) (x
—
—
a) (x
—
{b-
c)(&
-a)'^ (c-
a){c
-b)~
(x
(x
a)
?>)_-, '
This equation is of two dimensions, and it is evidently satisfied by each of the three values a, b, c hence it is an identity. ;
483.
If
two
rational integral functions of n dimensions are
equal for more than n values of the variable, they are equal for every value of the variable.
Suppose that the two functions 2)oX''
goo^"
are equal for (i^o
- Qo)
^'*
+Pia;"-i +P2X''-" H
+ gi^"-' +
g2^"-'
+
more than n values of x
4- (Pi
hp„, •••
;
+ qn,
then the expression
- qi) x^'-' + (2h - q2) x^-' +
•
• •
+ (p„ - qn)
ALGEBRA.
388
vanishes for more than n values of x; and therefore, by the j)recedmg article,
Po-go = that
= = Pi
Pi-qi
(^,
2h = qo,
is,
(^,
(Ji,
Hence the two expressions
= ^,'-'i\-Qn = 0; = 92,'" Pn = q^ i>2
P2-q2
and therefore Thus
are identical,
equal for every value of the variable.
If
is
rational intec/ral functions are identically eqncd,
tii'O
may
we
equate the coefficients of the like powers of the variable.
CoR. This proposition still holds if one of the functions For instance, if of lower dimensions than the other. Po-c" +pi.x'*-i +p2*'«"~^
= we have only q^
are
==0, gi
^2-'^""^
+ gs^""^ H
\-qn,
to suppose that in the above investigation
= 0, and then we obtain P^ = 0, 2h = 0, P2 = qo, Ih = qs, •"Pn =
The theorem
484.
yPn
+Pz^''~" H
q,r
established in the preceding article
for functions of finite dimensions is usually referred to as
the Principle of Undetermined Coefficients. The application of this principle is illustrated in the following examples. Ex.
1.
Find the sum of the
Assume that 1.2 + 2. 3 + 3.
4
series
1.2
+ 2.3 + 3.4
+
+ n(n+l)=^ + ^«+CH2 + l>«H^H*+-,
...
4---+ n(n
+
l).
where A, B, C, D, E, '• are quantities independent of have to be determined. Change n into n + 1 then
n,
(1)
whose values
;
1.2 + 2.3
= By (n
A-\-
+ «(?i + l) + (n + l)(n + 2) Bin + 1) + C(n + + />(n + 1)3 + EOi + +...
1)^ +....
l)--^
(2)
subtracting (1) from (2),
+
l)(n
+
2)
=
J5
+
(7(2
n
+
1)
+ +
2>(3
n'^
£'(47i3
+ +
3
7i
Gn2
+ 1) + 4n +
1)
+ ....
This equation being true for all integral values of w, the coefficients of the respective powers of n on each side must be equal thus and all succeeding coefficients must be equal to zero, and ;
32)=1; whence
+ 2C = 3; D -h C = h C=l, -B = |.
3Z) I>
-\-
B=2
;
E
UNDETERMINED COEFFICIENTS.
^
=A+2
Hence the sum
To
A, put
find
71
=
I
1
Note.
-2
+
2
•
3
+
1
+ -lA
n-
-\-
o
o
the series then reduces to
;
= ^ + 2,
2
Hence
»
3
•
+
4
its first
term, and
^ = 0.
or
+
...
389
+ 1)= ^ 'K« +
?K«
be seen from this example that
1)(h
+
2).
when
the nth term is a rational integral function of /?, it is sufficient to assume for the sum a function of n which is of one dimension higher than the nth term of the series.
Ex.
2.
It will
Find the conditions that
+ px^
x^
-\-
qx
+
r
may
be divisible
by
+
x^
The quotient pendent of
X.
x^
Equating the
+ px"^ +
qx
+
b.
= (x +
r
+
ct
=
the last equation
^•
+
ak
j>,
=-
;
b
r
is,
=
b
(p
—
k)
(.x^
^
ax
powers of
=
b
=
kb
g,
=
q\
b(q
—
'^-\-b
-{-
x,
b).
we have
r.
b
and ar
a),
namely, x and a term inde-
;
hence by substitution we obtain
T-^a-p, and that
+
coefficients of the like ^•
From
ax
two terms Hence, we assume will contain
=
b);
which are the conditions required.
EXAMPLES Find by the method
of
XLII.
a.
Undetermined Coefficients the sum
of
+ + 52 + 72 + to n terms. 2. 1 2 3 + 2 3 4 + 3 4 5 + ••• to M terms. 1 22 + 2 32 + 3 42 + 4 52 + 3. to terms. 13 + 33 + 53 + 73 + terms. 4. to to n terms. 5. + + 3^ + 44 + Find the condition that x^ — Spx + 2q may be divisible by a 6. factor of the form x2 + 2 «x + a^. Find the conditions that ax^ + bx^ + ex 7. d may be a perfect 1.
3^
1-
.
...
.
.
.
.
.
.
.
2-1
•••
.
...
l-i
.
?i
71
...
-\-
cube.
;
ALGEBRA.
390
Find the conditions that
8.
cfix^
+
+
h'jfl
cx^
-\-
dx
-\-
p
may
be a
perfect square.
Prove the identity
9.
- b)(x - c) {a-b){a-c)
a^(x
-c)(x- a) ib-c)(b-a)
b'^jx "^
- a)(x - b) _ ^ ~^' {c-a)(c-b)
c'^jx "^
FUNCTIONS OF INFINITE DIMENSIONS. 485.
+
the infinite series Aq
If
is
+
Oi-*'
equal to zero for every finite value of
x
^2-*'"
for
+ f^^x^ +
which the
must be equal
convergent, then each coefiicient
• • •
is
series
to zero identi-
cally.
Let the series be denoted by S, and let S^ stand for the ••• expression a, + a^x -\- cigX^ then /S = oto 3,nd ^'-^i? therefore, by hypothesis, clq -\- xSi for all finite values of
+
+
;
=
S is convergent, Si cannot exceed some finite therefore by taking x small enough, xSi may be made
But since
X.
limit
;
we please. In this case the limit of S is a^ but ahuays zero, therefore a^ must be equal to zero identi-
as small as
S
is
;
cally.
Eemoving the term of X
that
;
values of
is,
Similarly,
we have xSi -\-
a^x^
we may prove
If
two
a.^,
+
powers
Suppose
a^,
• • •
=
finite
all
finite
in succession that each of the
convergent and equal to one
value of the variable, the coefficients of
of the variable in the
two
that the tAvo series are
series are equal.
denoted by
+ ctiX + a^x' + a^x^ + Aq -f AiX + AzX^ + A^o^ + tto
and
for all finite values
vanishes for
•••is equal to zero identically.
infinite series are
another for every like
ciq,
+ a^x
x.
coefficients ai,
486.
tti
• • •
• • •
then the expression tto
- A +(«! - Ai)x-\-(a.2 - A.?}x^-^(as - A^jx^ +
•••
;
UNDETERMINED COEFFICIENTS. vanishes for
391
values of x within the assigned limits;
all
therefore by the last article
— A^ = 0, «! — ^li = 0, = Aq, a^ = Ai, that is, cIq
ciq
ao a.2
— A2 = 0, % — ^3 = 0,
= A^,
a^
hence the coefficients of like poioers of which proves the proposition.
= A^,
• •
•,
• • •
the variable are equal,
EXPANSION OF FRACTIONS INTO SERIES. 487.
'-^
Expand 1
of
-\-
—
^
X— X
in a series of ascending
powers
X.
2
Let
+ x = A-^Bx^Cx^ + Dx^ X—X -^
1
-[-
where A, B, C, D, determined then
•••,
are constants whose values are to be
;
2-{-x'
= A(l + x- x')+ Bx(l + ^ - a^2^^+ -hB
+ x- x^) + x-x')-j-...
Cx\l
-hDx\l
ALGEBRA.
392 Ex. Expand 2 x2
-
in a series of ascending
3
powers of
x.
5C3
term of the numerator, by cc^, of the first term of the denominator, we obtain x-'^ for the first term of the expansion therefore we assume Dividing
cc",
of the first
;
2a;2_3x3 then
Equating the
^=
1;
coefficients of like
2^-3^ = 0,
powers of
x,
20-3 5 = 0,
we
liave
2Z>-3C = 0,
-
.
UNDETERMINED COEFFICIENTS. Equating the
A=
...
we have
B'-{-2AC=0, ...
•
Vl + I-
thus
x,
2AD + 2BC = 0, B = i; D = ^\', C=-i', C' + 2BD-h2AE = 0;
2AB=1,
l',
powers of
coefficients of like
393
...
^—
•
128
= 1+--- + — X
x^
2
8
a^
,
16
5
5x* 128
.
,
Note, The expansion can be readily effected by the use of the Binomial Theorem [Art. 421 J.
EXAMPLES Expand
XLII.
c.
the following expressions to four terms
1.
Vl -X.
3.
Va2 -
2.
Va -
4.
v'2
x.
+
:
x^.
x.
6.
(1
6.
(1
+ x')^ + x + x^)^.
REVERSION OF SERIES. 490.
To
revert a series y
= ax + bx-
cx^
-{-
press X in a series of ascending powers of
...
-\-
to ex-
is
y.
Revert the series y
Assume
x
= x-2x^-\-3x^-4.x'-\-"= Ay-{- By^ + Cy^ + Dy* +
.
• •
.
.
.
(1).
•
Substituting in this equation the value of y as given in (1),
we have
x=A(x-2x~+3x^-4:x'+
...)
=A{x
-2x'-^3x^-Ax'-^'-.)
-{-Blx-2x'+3a^-4.x'-\- '"y=B(x''-{-4.x'-4:X^-i-6x'-\-'.')
-{•C{x-2x'^3a^-4.x'+ -y=C(:x'-6x'
+ Dlx-2x^-\-3x^-4tx'-{Equating the
...y=D(:x'-^
coefficients of like
)
...
powers of
)
x,
A = l; B-2A = 0, C-AB + 3A = 0, .'.0 = 5; B = 2; D-6C-\-10B-4,A = 0, B = U. x=y +2 -^ 5 y' U y' .... .'.
...
Hence
y'-
-\-
-\-
;
ALGEBRA.
394
491. If the series he y = 1 + 2x -^3x^ + put ?/ - 1 = then z = 2x-^3x^ + 4.of-{-'-'.
4.x^
-\-
...
2;
Assume as in Art.
x= Az
-{-
Bz^
Cz^
-\-
+
-"
490 after which replace
EXAMPLES
z
and complete the work by its value y — 1-
XLII.
d.
Revert each of the following aeries to four terms
=
+
+
+
+
....
1.
,^
2.
y=x + 'ix^-\-bx^-\-l:id^+
3
^/
—
^,
a:2
/y2
a:3
/).3
;;C— —-4-—
4
2
a;4
••'
4.
2/
5.
y
= l+a:+ 1+|+|^+
.
=
x
——^
^3
^5
n,^
3
5
7
/V.4
—+
•'•
*
8
6.
:
\-
....
•"•
y=ax-^hx:- + cx^ + dx'^+
—
.
PARTIAL FRACTIONS. 492. A group of fractions connected by the signs of addition and subtraction is reduced to a more simple form by being collected into one single fraction whose denominator is the lowest common denominator of the given frac-
But the converse process of separating a fraction into a group of simpler, or partial, fractions is often required.
tions.
For example,
if
we wish
to
expand
^
— 4i« + 3af 1—
-
—-
in a series
x, we might use the method of Art. and so obtain as many terms as we i)lease. But if we wish to find the general term of the series, this method is inapplicable, and it is simpler to express the given fraction
of ascending powers of
487,
in the equivalent form
1
—2 —
^ — 1 — 3 — Zx)~^ can (1 \-
1
.'K
-
Each
of
the ex-
.T
— x)~^ and now be expanded by the Binomial Theorem, and the general term obtained. pressions (1
493. We shall now give some examples illustrating the decomposition of a rational fraction into partial fractions. For a fuller discussion of the subject the reader is referred to treatises on Higher Algebra, or the Integral Calculus.
.
UNDETERMINED COEFFICIENTS.
395
it is proved that any rational fraction be resolved into a series of partial fractions and that
In these works
may
;
(1) To any factor of the first degree, as x — a, in the denominator there corresponds a partial fraction of the
A
.
lorm X
—a
factor of the first degree, as x — b, occurring times in the denominator there corresponds a series of n partial fractions of the form (2)
To any
71
x
B —b
C (;^•
R
.
— by
- by x^ + j^x + q, (x
in the de(3) To any quadratic factors, as nominator there corresponds a partial fraction of the form
Ax-{-B X-
+2^^+
Q
quadratic factor, as x"^ + px + q, occurring 7i denominator there corresponds a series of partial fractions of the form (4)
To any
times
in the
71
Ax + B
+
(x^ -\-px
+D + P^ + gy
q)
Here the quantities A, B, pendent of
We
shall
Ex + S
Cx (^"
C,
(.'«"
D,
•••
+
l^'-c
+ qy
R, S, are
all
inde-
x.
make use
of these results in the examples that
follow.
Ex.
1.
Separate
5x —
11
Since the denominator 2ocr
into partial fractions. -\-
x
—
5x-ll ^ A ~X+ 2 x- + X - 6 where
A and B are
Q
=
{x
-\-
'2)
(2 x
—
3),
we assume
B 2
2X
-
3'
quantities independent of x
whose values have to
be determined. Clearing of fractions, 5 X - 11 = ^1 (2 X - 3) + ^(x + 2) Since this equation is identically true, we may equate coefficients of like powers of x thus ;
whence
2^ + J3=5, -3^ + 2i? = -ll; ^ = 3, B = — \. 5X-11 ^ 3 "2x2 + x-6"x + 2
1__^
2x-3*
•
ALGEBRA.
396 Ex.
''^^^'
Kesolve
2.
^^
"^
into partial fractions.
— a){x + h) B A mx + n x + h X — a {x — a){x -\-h) mx+ n = A(x +h)+ B{x- a) (x
Assume
,
.-.
.
.
.
.
(1).
A and
We might now equate coefficients and find the values of but it is simpler to proceed in the following manner
B,
:
Since
we
A
and
B are
independent of
we may
x,
give to x any value
please.
In (1) put x
—
a =0, ov x
=
a\ then .
puttmg X
+
b
=
3.
b,
71
Resolve
mx + n — a){x +
-^ h
1
ma +
/
mb —
^i
n\
I
a
b)
X— — (2x- 1)(9-
11x2 ^^^
23 ^^-^
.
'
—
B= a
(x
Ex
=-
or X
0,
_ ma + a + 6
-\-
b\ X
—
x
a
+
h
/
into partial fractions. x2)
B C A 23x-llx2 %2x-l)(3 + x)(3-x) = 273T + 3T^'^33^ = ^ (3 + x)(3 - X) + B(2x - 1)(3 - x) 23x - 11 + 0(2x-l)(3+x). .
,,>..
.
,
^^^^"^ .-.
*
^'^
x'-^
equating the coefficients of like powers of x, or putting in - 1 = 0, 3 + x = 0, 3 - x = 0, we find that
By
succession 2 x
^=1, ^ = 4, C = -l. "'
Ex
4.
23x-llx2
1
^4
(2x-l)(9-x--2)
2x-l
3
Resolve
—
+
x
(x-2)'-^(l
Assume .-.
X
1
3-x
x
partial fractions.
.^^^.^
+ -^+ ^^,^ l-2x x-2
^ (x-2)'-2
+ x-2 =.l(x-2)2 + J3(l-2x)(x-2) + 0(l-2x). - 2 X = 0, then A = - I; - 2 = 0, then C = -4. 3x2
Let
To
—
-2 -2x)
+
^^'^t^"^ -2x)
(x-2)--2(l
let
'
1
find B, equate the coefficients of
3
3x2 (X
-
+
a:
2)2(1
x'^
;
thus
=
-2 -
2 X)
3(1
-
2 X)
3(x
- 2)
(x
-
2)2
UNDETERMINED COEFFICIENTS. Ex.
5.
Resolve
—+ — 42
19
'ic
into partial fractions.
'-
•
(x2
397
-
1) (X
4)
42-19x ^Ax±B^ + 1) (X - 4) ~ a;2 + 1 x - 4 42-19x=(^x + ^)(ic-4)+ (7(^2+1). Let X = 4, then O=—2 = A + C, and A = 2; equating coefficients of 42 = — 4 i> + (7, and i? = — equating the absolute terras, 42 - 19 _ 2 x - 11 _ 2 " (x^ + 1) (x - 4) ~ x^ + 1 X - 4" Assume'
'
(x2
.-.
;
x'^,
11,
a;
494. In all the preceding examples the numerator has been of lower dimensions than the denominator; if this is not the case, we divide the numerator by the denominator until a remainder is obtained which is of lower dimensions than the denominator. Ex. Resolve
By
6 x^
4-
—
5 x^
7
3x2-2x-l
into partial fractions. ^
division,
6x3+5x2-7^2x + 3+ 8X-4 3x2_2x-l 3x2-2x-l 1 -— 8x-4 = 5 3x2-2x-l 3x + l h x-1 6x3+5x2-7^,^^3^ 5 3x2_2x-l 3x + l x-1 .
,
,
and ^._
We
now explain how resobe used to facilitate the expansion of a rational fraction in ascending powers of a;. 495. The General Term.
lution into partial fractions
Ex.
a
1.
series of
By
Ex.
—
——-
Find the general term of "" ascending powers of Art. 493, we have
shall
may
l-3x + 2x2
when expanded
x.
1,
= __I
1-tA^
^ = 7(1-2
l-3x + 2x2l-2xl-x
x)-i ^
-
6(1 ^
-
^
= 7[l+(2x) + (2x)2+... + (2x)'-+...] +...). -6(1 +X + X2+...+ X'-
Hence the
(r
+
l)th, or general
term of the expansion,
7(2x)'--6x'- or
[7(2)'-
-
6]x'-.
is
x)-i ^
in
ALGEBRA.
398 Ex.
Find the general term
2.
—-
of
(x
powers Art. 493, we have
in a series of ascending
By 3a;2
(X
-
Ex.
4,
+ x-2
2)2(1
_ 3(1
-
a;)
3(x
2)
(x
-
a;)
(2
-
4
-
15 1
3(1
2x)
5 2
when expanded
-
of x.
1
-2x)
-^-2)"^(1
-2 a;)
3(l-2x)
3(2
-
2)2
^_
5
a:)2
4
4(l-^V
Gll-fi
-|(—)-^-^(-2)-^-(^-^)'^
.
= -i[l+(2x) + (2x)2+.-. + (2x)'-+...]
-[l0(|) + 3g.... + (,..X)(^-)%...]. Hence the
(r
+
l)th or general term of the expansion
V
3
6
2r
2'-
is
I
496. The following example sufficiently illustrates the to be pursued when the denominator contains a quadratic factor.
method
Ex.
Expand
i-^
(1+X)(1+X2)
in ascending
powers of x and
the general term.
7+x
Bx + O _ A + x2)-l + x+ l+x2 7 + x = ^(l + x2) + (5x+ C)(l + x). 1 + X = 0, then ^ = 3 Let C, whence (7 = 4; equating the absolute terms, 1 = A = ^ + J?, whence B = — S. equating the coefficients of x2, 4-3x 7+x _ 3 ~ 1 + X 1 + x2 (1 + X) (1 + a;2) = 3(1 + x)-i +(4 - 3x)(l + x2)"i = 3{1 - X + x2 + (- l)pxp +•••} + (4 -3x){l-x2 + x4-... + (-l)i'x2i' +...}. Assume ^^^"""^^
(l
+
,
'
x)(l
.-.
;
-\-
find
;
UNDETERMINED COEFFICIENTS. To
find the coefficient of
(1) If
4(_
r
is
— 3(—
1)
2
is
x'"
even, the coefficient of
1)2; therefore in the
(2) If r r-l
odd,
399
x*"
the second series
in
expansion the coefficient of
the coefficient of
x*"
in
x*"
is
3+4( — 1)2.
second series
the
r+l ,
and the required
coefficient is
EXAMPLES Resolve into partial fractions 46
XLII.
3(—
—
2
1)
3.
e.
:
+
13X
1
3
r2x2-llx-15*
(1
26 x2
+
+
3X
+
2 x2
-2x)(l -x2)"
208 X
is
is
;
CHAPTER
XLIII.
CoxTixrED Fractions. 497.
An
expression of the form a H
is
^
c-f-
,
e -r
here the letters a, b. o, any quantities whatever, but for the present a continued fraction
;
consider the simpler form
cii
may
we
where
. :j
denote
shall only a^,
a.2,
This will be usually written in
are positive integers. the more compact form a^,. •••
«i
H
• • •
called
•••
+
-••••
498. When the number of quotients a.i, a^ a^, ••• is finite the continued fraction is said to be terminating ; if the number of quotients is unlimited the fraction is called an infinite
continued fraction.
reduce every terminating continued fracby simplif\-ing the fractions in succession beginning from the lowest. It is possible to
tion to an ordinary fraction
499.
Let
To convert a given
—
fraction into a continued fraction.
be the given fraction
quotient and
p
the remainder
m= n
«i
;
;
p -
divide
thus
+ n= 400
m
1
«!
+ n-
by
n
;
let a^
be the
;;
CONTINUED FRACTIONS. divide n by
j^i
1^^
^-2
401
the quotient and q the remainder
t'e
thus
^
-=
divide
p by
5. let
and so
on.
Thus
/7l
Or,
//i
=
;
than
n,
the
—
It will
first
if
1 '-'i
in
the remainder
1
-r
quotient
is zero,
and
Tve
put
1
as before.
be observed that the above process
and
a
are
is
the same as
common measure of m and n commensurable, we shall at length
that of finding the greatest
hence
r
•••
m
and proceed
p
be the quotient and
= Oi H
is less
p
-
o^
If
,1. ^ 4--; + -=a2 .
a.,
p
;
a stage where the division is exact and the Thus every fraction whose numerator and denominator are positive integers can be converted into a terminatiner continued fraction. arrive
at
process terminates.
Ex.
Keduce
:
ALGEBRA.
402
The
500. Convergents.
fractions obtained
by
stopj^ing
quotients of a continued fracconvergents, because, tion are called the first, second, third, as will be shown in Art. 506, each successive convergent is at the first, second, third,
• • •
• • •
a nearer approximation to the true value of the continued any of the preceding convergents.
fraction than
501. To show that the convergents are alternately less and greater than the continued fraction.
Let the continued fraction be
cii -\
The
is
convergent
first
is
cii,
too small because the part
too great because the denominator a^
is
The
third convergent
a^H
is
and
,
is
is
is
%
—
-\
too small.
too small because
<^2~r ^3
-j
a.i-\
and
The second convergent
omitted.
and
—
is
.
is
too great;
and
so on.
«3
When
if the given fraction is a proper fraction, aj = we agree to consider zero as the first convermay enunciate the above results as follows ;
in this case gent,
we
of an odd order are all less, and the conan even order are all greater, than the continued
Tlie convergents
vergents of fraction.
502. To establish the law of formation of the successive convergents.
Let the continued fraction be denoted by
,111 a2+
then the
first
%+
0^4+
three convergents are aiCig
tti
1
'
+1
a^
a3(aia2 -f 1) ^'
ag
•
cig
-h 1
oti ^
'
;
CONTINUED FRACTIONS.
403
and we see that the numerator of the third convergent may be formed by nmltiplying the numerator of the second convergent by the third quotient, and adding the numerator of the first convergent; also that the denominator may be formed in a similar manner. Suppose that the successive convergents are formed in a similar way let the numerators be denoted by 2^1, Ihy 2h) •*•? and the denominators by q^, q^, qr„ •••• Assume that the law of formation holds for the 71th con;
vergent
;
that
2\
The
(n
+
is,
suppose
= Clnlh-l +Pn-2,
having the quotient a„
+ l)th
(n
= ^^Jln-l + qn-2'
Qn
l)th convergent differs from the nth only in H
in the place of a„
;
hence the
convergent
«H+
)Pn-l-\-Pn-2
,
X
+ Pn -2) + Pn-l ,
^^n+lj
(^n+l{<^nPn-l
a
.
by supposition.
If therefore
we put
see that the numerator and denominator of the (n + l)th convergent follow the law which was supposed to hold in the But the law does hold in the case of the case of the nth. third convergent, hence it holds for the fourth, and so on therefore it holds universally.
we
Ex. Reduce f |^| to a continued fraction and calculate the successive convergents.
ByArt.499,^4 = 313
2+^^^^-l. 6+1 + 1 +
The successive quotients are
The
2,
successive convergents are f ,
11
6, ^f-,
+
2
1,
1,
-y-,
ih
11,
2.
iff? fit*
:
ALGEBRA.
404
Explanation. With the first and second quotients take the first and second convergents, which are readily determined. Thus, in this example, 2 is the first convergent, and 2 -f a or J/ the second convergent. Tlie numerator of the third convergent, 15, equals the numerator of the preceding convergent, 13, multiplied by 1, the third quotient, The plus 2, the numerator of the convergent next preceding but one. denominator is formed in a similar manner thus 7 = 1x6 + 1. :
rru
The
, «f.u fifth convergent
=
11(28)+ 15
323
= ^• ^^^r^ (13) +
503. If the fraction
is
a proper fraction, we may consider and proceed as follows
zero as the first convergent,
Reduce
-^ to a continued
fraction,
and calculate the
suc-
cessive convergents.
Proceeding as in Art. 499, 227)84(0
00 84)227(2 •
168 59)84(1
59 25)59(...
We
obtain
-f
--
— l+-2+-2+-l-f3-}-2 — -— — — — -
-
-
:^
-•
2-f
The
successive quotients are
0, 2, 1, 2, 2, 1, 3, 2.
Writing ^ for the first convergent and arranging the work as show in the example of the preceding article, we have Quotients
Convergents
021221 f,
i i
f, -^,
^,
3
2.
j\\,
^.
504. It will be convenient to call a„ the ?ith partial quothe complete quotient at this stage being
tient
We by
;
shall usually denote the complete quotient at
fc.
any stage
;
CONTINUED FRACTIONS.
We
have seen that
qn
^K^ln-l
+
qn-2
Let the continued fraction be denoted by
from
405
cc
f then x differs
— only in taking the complete quotient k instead of the
partial quotient a„
thus
;
^
505. To show that
^ ^Pn-l-hPn-2 Mnl + qn-2
if
,
— be the /ith convergent to a continued
fraction, then
Let the continued fraction be denoted by tti
^111 + + H
«2
then
p,, q^^i -Pn-l Qn
But hence
When
• • •
«3 4- «4
= i^nPn +Pn^2)qn-1 -Pn~l((^nqn-1 + ^n-2) = (- l)(l>n-ign-2 -Pn-2qn-l) = (- iy2)n-2qn-s~Pn-3qn-2), similarly, 1
= (-iy-\2hqi-i>iq2)+ 1) — «! ^2 = 1 = ( — 1)^ 2hqi —Piq2= Pn^»-i - Pn-iqn =(•
(«'i'^2
I)''-
the continued fraction
is less
will still hold if Ave suppose that a^
convergent
is zero.
When we
Note,
than unity, this result and that the first
= 0,
are calculating the numerical value of the sucabove theorem furnishes an easy test of the
cessive convergents, the
accuracy of the work. 1. Each convergent is in its loivest terms; for if 2)n had a common divisor it would divide Pnqa~\ ~Pn~\q,o unity which is impossible.
Cor.
and or
Qn
;
;
ALGEBRA.
406
is is
The difference between two successive convergents Cor. 2. a fraction tvhose numerator is unity, and whose denominator the product of the denominators of these convergents ; for
qn^qn^l~
"quQu-l'
qnqn~l
506. Each convergent is nearer to the continued fraction than any of the preceding convergents.
—
Let X denote the continued fraction, and three consecutive convergents
;
then x
?
qn
'
differs
-^?
-^
qn+i
qn+2
from —
^ only
qn+2
in taking the complete f^n+2
j
denote this by k
;
(n
+ 2)th
Mn+l Pn
^ qn Pn+l
and
^n+1
Now
A;
is
^
quotient in the place of
thus
+ qn ^
KPn+'iqn^Pnqn+l) qn{kqn+i
+ qn)
1
Pn+iqn^PnQn+l qn+l(^qn+l
+ ^n)
greater than unity, and Qn
+ qn)'
qn{Ht+i
+ ^n)
qn+l(Hn+l is less
on both accounts the difference between
than
Pn+l
-^
q^^i
and x
;
hence is
less
qn+i
than the difference between
—
and x; that
is,
every con-
qn
vergent is nearer to the continued fraction than the next preceding convergent, and therefore nearer than any preceding convergent. Combining the result of this article with that of Art. 501, it follows that
The convergents of an odd order continually
increase, hut
are always less than the continued fraction The convergents of an even order continucdly decrease, but
are ahcays greater thari the contimied fraction. ^ The sign
~
ineani? " difference
between."
—
;
CONTINUED FRACTIONS.
407
507. To find limits to the error made in taking any convergent for the continued fraction.
—
Let
?
-^? -^^
be three consecutive convergents, and
k denote the complete (w
let
-|-
2)tli
^^^iWL±P«
then
Now
k
Pn
^
qn
qn{Mn+l
greater than
is
the continued fraction x,
1 ,
a7id
1,
1
+ ^n)
and any
is less
than
> q„,
q,^+i
— 1
-
From P
taking
—
1
^" ,
or
—
and greater than
,
is less
than
^^"
+ qn)
—
the error in taking
instead of x
^"
1
-—
2q\^i
qn
508.
Pn
convergent,
1 qn(qn+l
Again, since
^n(^«+l+f
therefore the difference between
greater than
qn^n+l
qnqn+i
quotient
the last article
it
appears that the error in
instead of the continued fraction
—
1 ;
that
is,
less
1
than
less
is
-;
than hence
a^+iqJ
qn{a'n+iqn-hqn-i) ^^
the larger a^+i
is,
the nearer does
—
approximate to the
continued fraction therefore, any convergent which immediately precedes a large quotient is a near apijroximation to ;
the continued fraction.
Again, since the error
is
less
than
—
^,
it
follows that in
qn-
order to find a convergent which will differ from the con-
tinued fraction by less than a given quantity
-,
Ave
have
a only to calculate the successive convergents up to
q^
is
greater than
a.
— ^'*
,
where
ALGEBRA.
408
The properties
509.
of continued fractions enable us to
whose ratio closely approximates to that of two incommensurable quantities, or to that of two quantities whose exact ratio can only be expressed by large tind tAvo small integers
Ex. Find a
approximating to 3.14159.
series of fractions
process of finding the greatest common measure of 14159 and 100000, the successive quotients are 7, 15, 1, 25, 1, 7, 4. Thus
In
tlie
o-in-Q o.l41o9
The
=
..
11 7+15+1+25+1-7+4 1
o T ,
1
1
1
1
successive convergents are
1 last
3.13
2_2.
1'
This
7
'
106'
Aii 13'
...
1
convergent which precedes the large quotient 25
near approximation, the error beini! less than
is
a very
and there-
,
25 X (113)-^ fore less than
,
or .000004.
25 X (^100)-
Any
510.
convergent
is
nearer to the continued fraction than is less than that of the
any other fraction whose denominator convergent.
Let
./:
be the continued fraction.
tive convergents, - a fraction
than
1-^',
^ "~^
q..
qn-i
whose denominator
s is less
5„.
If possible, let - be nearer to
.r
than
s
nearer to x than -?^
two consecu-
and
-^^"~\ it
h^
1^,
then - must be s
q^
[Art. 506]
;
follows that i must
and since x
lies
between
lie
1-^
between
and
h^.
Hence
r^^^^<^«^^, thatis<^-; s
qn-i
qn
qn-1
7./y«-i
qn
that
is,
an integer less than a fraction
Therefore
;
which
is
impossible.
^ must be nearer to the continued fraction than
-•
CONTIXUED FRACTIONS.
EXAMPLES
XLIII.
409
a.
Calculate the successive convergents to
'>+J_
1
J__L A_ J_
1
2"
1^ 3+ 5- 1+ 1+
1111111 4+ 2+
2
2+ 2+ o
3
,
6'
1+
34-
J_ J_ J_ _i_JL 3+ 1+
2-f-
2+ 1+
1 9*
Express the following quantities as continued fractions and find the fourth convergent to each also determine the limits to the error made by taking the third convergent for the fraction. :
4.
fff.
6.
iifl.
8.
.37.
10.
..3020.
5.
ffi
7.
^VtV
9-
1-139.
11.
4.316.
12.
Find
13.
Find an approximation to
limits to the error in taking ff f yards as equivalent to a metre, given that a metre is equal to 1.0936 j'ards.
i
which
differs
+ X_LJ.^^... 3+ 5+ 7+ 9+ 11 +
from the true value by
less
than .0001.
Show by
the theory of continued fractions that f^ difiers from 1.41421 by a quantity less than xriso' 14.
RECUERING COXTIXTED FRACTIONS. 511. We have seen that a terminating continued fraction with rational quotients can be reduced to an ordinary fraction with integral numerator and denominator, and therefore cannot be equal to a surd but we shall prove that a quadratic surd can be expressed as an infinite continued fraction whose quotients recur. TTe shall first consider a numerical example. ;
Ex. Express ^^19 as a continued fraction, and find a series of fractions approximating to
its
Vl9 = V19 + 3
4
^
value.
4
^
+(V19 - 4)= ^
V19 3
2
^
^
4
-^-^|-^; 5
.
^
V19 +
2
ALGEBRA.
410 V19 +
2 ^-^
V^Q-3 ^
I
5
5
^19 +
^'
2
2
Vl^ +
S'
V19 +
2'
Vl9 +
3 _-.
V19-2 = 1 +
,
5
5
Vl9 +
Vl9-4 ^^
^o
2
3
1
V19 +
3
^19 +
4'
4=8+(Vl9-4) = 8+...
after this the quotients 2, 1, 3, 1, 2, 8 recur
^
hence
;
2+ 1+ 3+ 1+ 2+
8+
It will be noticed that the quotients recur as soon as quotient which is double the first.
Explanation. and we write ^19
We =
4
first find
+(^^19
;
We
(Vl9-4)(Vl9 +
.
to a
the greatest integer in -^19 this is 4, then express ^19 — 4 as an
— 4).
Thus
equivalent fraction with a rational numerator.
^,na
we come
4)
_
3
-^19 +
V19+4
4"
The work now stands
V19 =
We
3
4
.
V19 +
begin the second line with
1
,
V19 +
4
v/l9
——4 4-
,
4
the denominator of this
o
complex nator.
fraction,
The
which
is itself
a fraction with a rational denomi-
greatest integer in this fraction
Vl9 +
4
is 2,
and we write
V19-2
^o I
3
We then to a/19 ^
—
3
multiply numerator and denominator by the surd conjugate 2,
so that after inverting ° the result
V19 +
,
2
we
again begin ° ^
a line with a rational denominator. The same series of operations performed in each of the following lines. The first seven convergents formed as explained in Art. 502 are
is
I
CONTINUED FEACTIONS. The
error in taking the last of these
411
than
is less
and
,
is
(320)-^' 1
therefore less than and still less than .00001. Thus or 102400 (320)2 the seventh convergent gives the value to at least four places of deci,
mals.
512. Every periodic continued fraction
equal to one of
is
the roots of a quadratic equation of which the coefficients are rational.
Let X denote tlie continued fraction, and y and suppose that
tlie
periodic
part,
,11111 c+ k+ ,1111 u+ r+ /<+
h-\-
I
yz=m-\
and1
y
•
,
7i+
where
a, h,
c, •••
h, k,
m,
n,
y
"-u, v are positive integers.
-,
p' —,
be the convergents to x corresponding to the
quotients
h,
k respectively; then since y
P
Let
quotient, ^
'
T
Let
-,
s
—+
we have x = —f-
q'y
t' -7 s
—•.
q'
whence ^y
is
= ^,
the complete
—^•
q'x-p<
be the convergents to y corresponding to the I
quotients u, v respectively
;
then y
,
= --—-—
Substituting for y in terms of x and simplifying, Ave obtain a quadratic of which the coefficients are rational. The equation s'y^-{-(s—r')y—r 0, which gives the value of y, has its roots real and of opposite signs if the positive
=
;
value of y be substituted in x
= —i^ly
denominator the value of x A, B,
C are
is real.
integers,
B
is
+
of the
—. on rationalizing the 'l A-\- /B
form
—^^^^—
,
where
being positive since the value of y
:
ALGEBRA.
412 Ex. Express iH
—
as a surd.
2+ 3+ 2+ 3+
Let X be the value of the continued fraction
.-1=^ 2+ whence 2a;2 + 2x-7 = 0. The continued fraction is equal and
is
+ (x-l)
; '
to the positive root of this equation,
—
v
therefore equal to
\
3
then
;
EXAMPLES
XLIII.
b.
Express the following surds as continued fractions, and find the sixth convergent to each 1.
V3-
2.
V^-
3.
V6-
4.
V8Vl^-
5.
:
6.
17.
Find limits of the error when
18.
Find
19.
Find the
limits of the error
when
^-^^- is
taken for ^^17.
f Jf
taken for ^23.
is
first
convergent to ^\0\ that
is
correct to five places
first
convergent to y^l5 that
is
correct to five places
of decimals. 20.
Find the
of decimals.
Express as a continued fraction the positive root of each of the following equations 21.
x2
+ 2x-l=0.
22.
x2-4a;-3=0.
23. 7 a:'^-8ft:-3=0.
CHAPTER
XLIV.
Summation of
Series.
513. Examples of the summation of certain series (Arithmetic and Geometric) have occurred in previous chapters. We will now consider methods for summing other series.
A
+
+
series ?
in
In the series
515. Scale of Relatione
each term after the second
is
and
— x^
;
5x' is,
by the
of the tAvo
constants 2 x
these quantities being called constants because
they are the same for
that
sum
equal to the
X)receding terms multiplied respectively
all
values of
Thus
n.
= 2x 4ar^+(-^')u^ = 2 xu^ — xhu .
3i«';
;
and generally, when n is greater than 1, each term is connected with the two that immediately precede it by the equation U,, = 2 Xlln_i — X^Un-2, or,
21^
— 2 xit,,_i + x\,_2 = 0.
In this equation the coefficients of ivitJi
their x>^'oper signs,
form what
relation.
413
?t„, u.^_i,
is
and
?^„_2? tnJcen
called the
scale
of
:
ALGEBRA.
414 Thus the
series
l-^2x-\is
3.t2
+ 4ar^ + 5x'+
...
a recurring series in which the scale of relation
is
- 2 + x\
1
a;
516. To find any term when the scale of relation
is
given.
any term can be found when a sufficient number of the preceding terms are known. As the method of procedure is the same, however many terms the scale of relation may consist of, the If the scale of relation of a recurring series is given,
following illustration will be sufficient 1
If is
—
p.x*
—
qx'
— rx^
the scale of relation of the series ciiX
«o -h
+ a^ 4- ctg,^ +
• • •
we have a^^x""
=2)x
'
+ qx^ a„_2.T"~- + rx^ = pa„ + qa^ + ran^^
a„
•
.
a,^__lX''~'^
or
2
i
a,^_^x'"~^,
I
thus any coefficient can be found when the coefficients of the three preceding terms are known. 517. To
If a sufficient
find the scale of relation.
number
of the terms of a series be given, the scale of relation
may
be found. Ex. Find the scale of relation of the recnrring series 2
This
is
+
5a:
+
13 a;2
+
35
cc^
+
plainly not a series of the
order, to obtain
p and
q
we have
+
275 x^
first
order.
97 r/
+
=
-f
.•-.
be of the second
the equations ;
and obtain the fifth and sixth 5,
scale of relation
If it
= 5p + 2(/, and 35 = 13|) + 5g By using these values of p q = — 6.
13
whence p
793 x^
coefficients
;
and q, we can hence they are correct, and the
is
1
-bx + 6x^
could not have pbtained the remaining coefficients with these values of p and q, we would have assumed the series to be of the third order, and formed the equations If
we
35
= 13p + 5g + 2r,
97 =SBp-\- 13 g
275
=
97j;
+
+ 35^ +
5
r,
13»';
;
SUMMATION OF
SERIES.
415
whence values for p, g, and r would have been obtained, and trial with the seventh and following coefficients would have shown whether they were
correct.
518. If the scale of relation consists of 3 terms it involves p and g and we must have 2 equations to determine 2> and q. To obtain the first of these we must know at least 3 terms of the series, and to obtain the second we must have one more term given. Thus to obtain a scale of relation involving two constants we must have at least 4 terms given. If the scale of relation be 1 — p.r — qx^ — rr^, to find the 3 constants Ave must have 3 equations. To obtain the hrst of 2 constants,
;
these w^e must
know
obtain the other two
at least 4 terms of the series,
and
to
we must have two more terms given
hence to find a scale of relation involving 3 constants, at least 6 terms of the series must be given. Generally, to find a scale of relation involving 7n constants, at least 2 consecutive terms. Conversely, if 2 consecutive terms are given, we may
m
we must know
m
assume for the 1
519. The
method
scale of relation
— 2h^ — Ih^"^ — Ih^^ —
Sum
...
— p^x"^.
n Terms of a Recurring Series. The sum is the same whatever be the for simplicity we shall suppose it to conof
of finding the
scale of relation
tain only
;
two constants.
Let the series be ciq
and
let
the
-f a-^x -f acfc^
sum be S
let
;
+ a^x^^ +
•
• •
the scale of relation be 1 —px—qx^ n greater than 1, we have
so that for every value of
Now^
S=
a-o
+ «i^' +
— qgi^S = — qaQSc^ —
«:2'^^
...
+
•••
+ (f'n^i^"~^)
— qa,^_^x"~^ — qa,^_^'' —
qa,^^^x'^^^.
;
+
—
ALGEBRA.
416
Hence (1
—px — qx^) S=a^^+ (oi —pao) ^ — (p(^n-
qO'n-2)^'*
1
for the coefficient of every other power of x
is
zero in con-
sequence of the relation
— pa„_i — ^a,i_2 = ^• jX(,,_i + (y^n-s)-^" + ^a,_ia;"+^ ^ ^ + (ai —iyao)x ( — 1 — qx^ 1 —2)X a„
ctp
j^.-c
Thns the sum nominator
is
(/a^^
of a recurring series is a fraction
whose
de-
the scale of relation.
520. If the second fraction in the result of the last article decreases indefinitely as n increases indefinitely, the formula for the sum of an injinite number of terms of a recurring series of the second order reduces to
^^ ao+(fti-jX(o)a; 1 — px — q^x?
If
we develop
this fraction in ascending
powers of x as
explained in Art. 487, we shall obtain as many terms of the original series as we please for this reason the expression ;
«o
1 is
(«i - PCIq)x —px — qx^
+
called the generating, function * of the series.
The summa-
tion of the series is the finding of this generating function.
If the series is of the third order,
Oo
521.
=
From
«o 4- {ch
— i^«o) ^ + («2 — P(^\ — 7«'o) ^ 1 —px — qar — rx^ 5
ij
the result of Art. 519,
ao-{-(ai-pao)x
_
,...,,. {pan -i
we
°
obtain
o
+1
+ gan-2)^" + qa'n~ix'''^\ l—px — qx"
* Sometimes called the generating fraction..
SUMMATION OF
417
SERIES.
from whicli we see that althougli the generating function «o+K-j)ffo)^^•
— px — qx^
1
be used to obtain as many terms of the series as we please, it can be regarded as the true equivalent to the
may
infinite series «o 4-
only
4- «2a^^ H
«ia^*
,
the remainder
if
— px — qx^
1
vanishes when n is indefinitely increased only when the series is convergent.
;
in other
words
522. The General Term. When the generating function can be expressed as a group of ]3artial fractions the general term of a recurring series may be easily found. Ex. Find the generating function, and reciuTiug series 1
tlie
-7a;-a:2-43x3
Let the scale of relation be
— px —
1
qocr
;
then
-\ + lp-q = 0, - 43 + p + whence j9
Let
=
1,
g
8 denote
=
6
the
and the scale of relation
;
sum
of the series
S=\-lx-xS= -
x
-Qx^S= ...
general term, of the
;
7
g
=
;
is
then
z2-43x3
+ 7x2+ x3+-.. -6x2 + 42x3+...
(I_x-6x2)^'=l -8x, 1
S.
-8x 6x2
which If
is
the generating function.
we
separate 1
1
—
-
X
8r
-
—
2 l
2 E
into
partial
6 x2
+ 2x
1_^
l-dx
fractions,
we
obtain
:
ALGEBRA.
418 By
actual division, or
=
2 1
+2a;
1
—
2 [1
-
by the Binomial Theorem. 2X
+ i2xy
+ (-
-•..
1)'-(2
xy+
...]
1
Whence
3a:
the (r
+
l)th, or general term,
[2(2'-)(-
!)'•
-
3'-] X'-
= {(-
EXAMPLES
is
l)'-2'-+i
XLIV.
-
a.
Find the generating functions of the following 1.
1
2.
2
3.
3
4.
2
5.
4
6.
1
7.
1
8.
1
3'-}x'-.
series.
+ 24 x2 + 84 x3 + •••• 2 X - 2 x2 + 6 x3 - 14 + 16 x + 42x2-94x3+ 5 X + 4 x2 + 7 x3 - 26 x* + .-. + 5x + 7x2 + 11x3+ + x + 2x2 + 2x3 + 3x'i + 3x5 + 4x6 + 4x7+ + 3x + 7x2+ 13x3 + 21x4 + 31x5+ -3x + 5x2- 7x3 + 9x*- 11x5 + .... + + -
6X
x"*
-
•••
.-..
....
••..
....
Find the generating function and the general term in each
of the
following series 9.
10. 13.
11. 2 + 3x + + 5x + 9x2+ 13x3+ .... .... 7 -6x + 12. 5x2-7x3+ 2-x + 3 + 6x + 14 x2 + 36 x3 + 98 x^ + 276 x^ + .... 1
5x2
9x2
+ 9x3+ .... + 27x4 + ....
THE METHOD OF DIFFERENCES. denote some rational integral function of n, u^y--- denote the values of i(„ Avlien for n the values 1, 2, 3, 4, ••• are written successively. From the series Ui^ ti2, y^, Xa^ ^s, ••• obtain a second series b}^ subtracting each term from the term which immediately 523. Let
and
follows
The is
u,^
let Ui, u^,
%,
it.
series
?<2
—
'^hf
^'s
—
'^h^
'^h
—
"^'s?
^'s
—
'^^y
'"
called the series of the first order of differences,
^^^^^s found and may be
conveniently denoted by Duy, Du^, Dii^, Du^, •••• By subtracting each term of this series from the term that immediately follows it, we have Du^ — Dui, Du^ — Du2,
SUMMATION OF SERIES.
Du^—
Du^,
...
From
may
wliicli
order of differences, this series
419
be called the series of the second
and denoted by
D2U2, B2IIS,
D.^rti,
we may proceed
to
form the
....
series
of the
orders of differences, the general terms of these series being D^u^, D^n^, D^u^, ••• respectively.
third, fourth, fifth,
• • •
From
524. Any Required Term of the Series. formation of the series U2,
Ul,
llr^,
Uq,
W5,
W4,
Dui, Du2, Duq, Du^, Du^,
A^i,
D2112,
D.^t^,
A^'sj A^*4j
D3U2, D3U3,
•
•
the law of
'
•••
•••
•••
appears that any term in any series is eqnal to the term immediately preceding it added to the term below it on the it
left.
Thus
By
U2
= Ui + Dili, and Du2 = Dui + D2U1. + Vu2 = u^, we have = u^ -f 2 Bill + ^2^^i-
addition, since
712
1(3
In an exactly similar manner by using the second, third, and fourth series in place of the first, second, and third, we obtain Dwg
By
= Dui
-\-
2
W4
+ Ds^h. + Dn^ = we
l>2^^i
addition, since u^
"^^4?
have
= Ui-\-S Dill + 3 A^'i + A"i-
So far as we have proceeded, the numerical coefficients follow the same law as those of the Binomial Theorem. We shall now prove by induction that this will always be the case. For suppose that Un+i
= ui + nDui+ ''^^~^'> D2U1 +
then by using the second to the (n of the first to the (n
Du^^i '
+ l)th
series
...
-f
+ "CA^^i +
...
+ Djii',
2)th series in the place
we have
= Dai + nD2Ui + ''^''^~^^^ D,Ui +
...
+ "CUiA'^i -\-
'•'
+ D^+i^i.
;
;
.
ALGEBRA.
420 "By addition, since
y^„+i
+ D?f„+i =
?i„+2?
we
obtain
But
_ Hence for ?<5,
(;,
+ l),,(,,-l)
(n^-r-\-l) _
...
the law of formation holds for n^^^ it also holds trne in the case of u^, therefore it holds for
if
?(„^2j b^it it is
and
u,,
Hence
tlierefore universally.
= u, + (n - l)Du, + ^''~f-^~^'^ D,u, +
•••
+ i>n-i^^i-
we take a
as the first term of a given series, di, d^, d^, terms of the successive orders of differences, any term of the given series is obtained from the formula If
as the
.
.
^first
(,,_iy(,,-2)(..-3)
Sum
525. The
series Ui, U2, Us,
n Terms of the Series. Suppose the the first order of differences of the
of
•••
is
series ^IJ
then
y^+i
=
(Vn+l-'^n)
'i^S,
'^'2?
+
(i'n
^'-IJ
•••>
- 1'n-l) +
•••
+
(^'2
"
^'l)
+ ^1,
identically .'.
Hence
V„+l
=
U,,
+ Un-l
h Ih
-\
+
^^1
+ ^1.
in the series V2,
0,
?
the law of formation .-.
v^^,
= +
is
v„
V3,
'?%
?/2,
V,'?«4"-
the same as in the preceding article;
nu,
+ ^^^f^ Du, -f JL
•
ij
.. .
+ i>„"i
SUMMATION OF that
is,
111
,
+ n
+ + — 1) ?'3
'^2
()i
+
•••
j^.
,
421
SERIES.
^U
— l)(n — 2)
'}i(n
^,
,
r,
,
a is the first term of a given terms of the successive orders of of n terms of the given series is obtained
as in the x^receding article,
If,
series, di, d^, d^,
differences, the
•
the first
• •
sum
from the formula e
nCn
,
Ex.
7
— V)(n — 2)
nCn
,
Find the 7th term and the sum
1.
the series
The
— 1)
,
of the first seven
terms of
4, 14, 30, 52, 80, .....
successive orders of differences are
6,
6,
Here n
—
22,
16,
10,
0.
0, 7,
28,
6,
= 4.
and a
Hence, using formula, Art. 524, the 7th term
=
+
4
+^.6 = 154.
6. 10
Using formula. Art. 525, the sum of the
= Ex.
7
,
+
—
40,
280,
168,
90,
50,
22,
34, 6,
0,
Hence the (,
o«
?ith
(,,
112,
78,
28, 6,
19
.
1.2.3
6
=
448.
432,
series
....
successive orders of difference are 28,
-
+ llAll
10
.
1.2
Find the general term and the sum of n terms of the
2.
12,
The
4
seven terms
first
152,
40,
6,
...
...
...
0,...
term [Art. 524] 1)
= ^3 + 5 ^2 + 6 n.
22 {n I
-
l)(n
-
2)
6 (n
-
l)0^
-
2)Cn
-
3)
:
ALGEBRA.
422
Using the formula for the sum of n terms we obtain 2871(71-1) _io,, - + ^„_iZ7i+ ,
227i(n-l)(n-2)
,
(s'
+
6 n(n
- l)(n-2)(n-3)
^
j3
= — (3 w2 + 26 w + 69 + 46) 91
=
tV n(n
+
1)(3
7i2
+
23
7i
+
46).
526. It will be seen that this method of summation when the series is such that in forming the orders of differences we eventuallj come to a series in which all the terms are equal. This will always be the case if the nth term of the series is a rational integral will only succeed
function of
n.
AND
PILES OF SHOT
SHELLS.
To find the number of sliot arranged 527. Square Pile. a complete pyramid on a square base.
in
The top layer consists of a single shot the next contains 4 the next 9, and so on to n^, n being the number of layers hence the form of the series is ;
;
Series
1%
2%
32,
^\-",n\
1,
4,
9,
16, •••,?il
1st order of differences
2d order of
differences
3d order of
differences
3,
n{n-l)
,
•'^+
528. Triangular in
angle.
we
obtain
n(n-l)(n-2) '^- n{n+l){2n+l) 6 1.2.3
Pile.
a complete pyramid
7,
2, 0.
Substituting in Art. 525,
o_,, ^-''+ 1.2
5,
2,
To find
the base
the
member of shot arranged is an equilateral tri-
of ivhich
SUMMATION OF The top 3
;
the next contains the next 10, and so on, giving a series of the
layer consists of a single shot
the next 6
;
423
SERIES. ;
form Series
1,
1+2,
1+2 + 3,
1,
3,
6,
1st order of differences 2,
2d order of differences
1+2 + 3+4,... 10,
3,
4,
1,
1,
3d order of differences
0.
Hence ,
n(n-l)
"^"''"^ ^
1.2
^ '^'^ ,
n(7i-l)(n-2) 1.2.3 To find
529. Rectangular Pile. in
a complete
The top this
row
2(m
+ 1)
is
6
number of shot arranged a rectangle.
row of shot. Snppose then the next layer contains
layer consists of a single
to contain ;
the
pile the base ofivhich
_ n(n -\-l)(n-{-2)
the next
m
shot
3(m
;
+ 2),
and so
on, giving a series of
the form
2m + 2, 3m + 6, 4m + 12, m + 2, +4, + 6,
m, 1st order of differences
7^1
2d order of differences
7>i
2
3d order of differences
••.
2, 0.
Now let I and iv be the number of shot in the length and width, respectively, of the base then 7n 1 — w -j-1.
=
;
Making
these substitutions,
we have
n(n-{-l)(3l-iv-\-l) 6
EXAMPLES 1.
XLIV.
b.
Find the eighth term and the sum of the
first
eight terms of the
series 1, 8, 27, 64, 125,.... 2.
Find the tenth term and the sum
series 4, 11, 28, 55, 92,
....
of the first ten
terms of the
:
ALGEBRA.
424 Find the number of shot
in
:
A square pile, having 15 shot in each side of the base. A triangular pile, having 18 shot in each side of the base. A rectangular pile, the length and the breadth of the base
3. 4. 5.
con-
taining 50 and 28 shot respectively.
An
6.
and a
incomplete triangular
a side of the base having 25 shot,
pile,
side of the top 14.
7.
An
incomplete square pile of 27 courses, having 40 shot in each
side of the base. 8.
Find the ninth term and the sum
series 1, 3
+
5, 7
The numbers
+
9
1,
+
11,
2,
of the first nine
terms of the
....
3,
are often referred to as the natural
••.
numbers. 9.
Find the sum
of the squares of the first
10.
Find the sum of the cubes
11.
The number
of the first
of shot in a complete rectangular pile is 24395
there are 34 shot in the breadth of the base,
length 12.
Find the number
if
its
;
how many
Find the number
is
and
169,
shot does the pile contain
?
of shot in a complete rectangular pile of 15 its
base.
an incomplete rectangular pile, the the sides of its upper course being 11 and 18, and of shot in
number of shot in the number in the shorter Find the
nth.
side of
its
term and the sum
30, 52, 80, 114,
lowest course being 30.
of
15.
4, 14, 8,
26, 54, 92, 140, 198,
17.
2,
12, 36, 80, 150, 252,
series
16, 0,
••..
....
-64, -200, -432,....
18.
8,
19.
30, 144, 420, 960, 1890, 3360,
20.
What
is
n terms of the
....
16.
pile
;
are there in
of shot in the top layer of a square pile
courses, having 20 shot in the longer side of 14.
how many
?
The number
in the lowest layer is 1089 13.
n natural numbers.
n natural numbers.
the
number
•••.
of shot required to complete a rectangular
having 15 and 6 shot in the longer and shorter upper course ?
side, respectively,
of its
The number of shot in a triangular pile is greater by 150 than number of shot in a square pile, the number of layers in each being the same find the number of shot in the lowest layer of the 21.
half the
:
triangular
pile.
SUMMATION OF Find the number of shot
22.
courses
when
the
number
425
SERIES.
an incomplete square pile of 10 upper course is 1005 less than
in
of shot in the
in the lowest course. 23. Show that the number of shot in a square pile is one-fourth the number of shot in a triangular pile of double the number of courses.
INTERPOLATION. 530. The process of introducing between the terms of a conforming to the law of the series is called interpolation. An important application is in finding numbers intermediate between those given in logarithmic and other mathematical tables. For this purpose we may employ the formula used in finding the ?ith term series intermediate values
by the Differential Method, giving fractional values Ex.
=
Given log 40
=
=
1.6021, log 41
=
1.6128, log 42
to n.
1.6232, log 43
1.6335, •••find log 40. 7.
Series
.0104,
.0107,
-
2d order of differences, 3d order of differences,
+
=
1.6021
we have
+ 1(.0107)+
1(-1)(^3^
7
1.6021
+
.00749
+
/
.000031
.0001,
.0002.
+ 10'A/'__3_V_1^\ 10 jv lojl =
1.6335, .0103,
-
.0003,
Substituting in formula of Art. 524, log 40.7
1.6232,
1.6128,
1.6021,
1st order of differences,
+
.000009
.0002 \ j
[3
=
1.6096 +.
Here log 40 is the first term (n = l); log 41 is the second term = 2) hence in introducing the intermediate term log 40.7 we give to n a value 1.7. (n
;
1.
=
0.7782, 2.
=
Given
...
Given
V^, 4.
;
a/tTI,
Given
^5^5L18.
;
=:
EXAMPLES XLIV. C. log 4 = 0.6021, log
5
=
0.6990,
log 6
log 53
=
1.7243,
log 54
0.4771,
find log 4.4.
Given log 51
1.7324, 3.
...
log 3
=
1.7070,
=
1.7160,
2.449,
^7 =
log 52
find log 51.9.
V^ =
2.236,
^Q =
2.645,
^S =
2.828
;
find
and VtTtI.
^51 =
3.7084,
^52 =
3.7325,
^53 =
3.7563,
...
;
find
; .
CHAPTER XLV. Any
Binomial Theorem. 531. In Chapter xxxvii.
we
Index.
investigated the Binomial
Theorem when the index was any
positive integer
;
we
sliall
now
consider whether the formulae there obtained hold in the case of negative and fractional values of the index. Since,
common
by Art. 411, every binomial may be reduced type,
it
binomials of the form By actual evolution (1
+
0.')^
and by actual (1
-
to one
will be sufficient to confine our attention to (
1
+ xy.
we have
= Vr+^-
1 H-io.'- ia;2
+ J^a;3_
...
+ 40.^ +
...
division,
.^•)-'
= 1 + 20^ + = 7T-^. (1
xy
3a.-2
and in each of these series the number of terms is unlimited. In these cases we have by independent processes obtained an expansion for each of
+
We
the expressions
(l+a;)"' and
they are only particular cases of the general formula for the expansion of (1 -f xy, where n is any rational quantity. This formula was discovered by Newton. (1
.^•)~^.
shall presently prove that
532. Suppose we have two ascending powers of x, such as
and
expressions
arranged
l+nx + »<" " ^> x' + "(" ' ^X" ,' 2) ^. + 426
...
in
(o).
BINOMIAL THEOREM. The product
of these
ascending powers of x
ANY INDEX.
two expressions denote it by
427
will be a series in
;
l-\-Ax-^ Bx^ -^Cx^
+ Dx'-{-
'•-',
it is clear that A, B, (7, ••• are functions of m and n, and therefore the actual values of A, B, (7, ••• in any ^2iYticular case will depend upon the values of m and n in that But the way in which the coefficients of the powers case. of X in (1) and (2) combine to give A, B, C, is quite independent of 7)1 and n in other words, whatever values m and n may ha,ve A, B, G, "'2^reserve the same invariahle form. If therefore we can determine the form of A, B, C, -•- for any value of m and n, we conclude that A, B, C, ••• will have the same form for all values of m and n. The principle here explained is often referred to as an example of "the permanence of equivalent forms"; in the present case we have only to recognize the fact that in any algebraic prodtict the form of the result will be the same whether the quantities involved are whole numbers, or frac-
then
• • •
;
tions
;
We
positive, or negative.
shall
make use of this principle in the general proof Theorem for any index. The proof which
of the Binomial
we
give
is
due to Euler.
533. To prove the Binomial Theorem when the index
is
a
positive fraction.
Whatever he
-,
of m, positive or negative, integral or symbol f(m) stand for the series
the value
fractional, let the
m(m —
, ,
1)
9
m(m — l)(m — 2)
,
3
,
'
1.2.3
1.2 then/(?i) will stand for the series -,
,
,
n(n
— 1)
9
,
n(n
— l)(n — 2)
3
If we multiply these two series together the product will bo another series in ascending powers of x, whose coefficients
ivill
be unaltered in
form
tvhatever
m
and n may
be.
;
ALGEBRA.
428
To determine
this invariable
form of
the
product
we may
m
and n any valnes that are most convenient for this jDurpose suppose that m and n are positive integers. In this case f{m) is the expanded form of (1 + x^, and fiii) and therefore is the expanded form of (1 + xy give to
;
;
X fill) = (1
fiin)
when m + a')'"+" is
but (1
and
+ xy^
X
(1
+ xy = (1 + xY^%
n are positive integers, the expansion of
This then is the form of the product of /(m) x/(?i) in all whatever the values of m and n may be; and in agreement Avith our previous notation, it may be denoted by cases,
f(^n
-I-
n)
;
therefore for
all
values
ofm and n
f(:m)xf(n)=f(m-{-n). f{m) X f(n) X f(p) = f(m
+ n) x f(p) = f{m + n + similarly.
Also
_2J),
Proceeding in this way we
/(m) x/(>0 x/(p)
•••
to
^'
may show
that
factors=/(m + ?i+i9-|---. to k terms).
Let each of -these quantities, m, where h and k are positive integers
n, p,
•••,
be equal to
{/©['=;(«>: but since h
but /( -
)
is
a positive integer,
f(Ji)
stands for the series
-(--A
= (1 +
a;)^
-,
;
BINOMIAL THEOREM.
.
ANY INDEX.
429
-("--A (l
.-.
+
^)'
= l+|^ + ^^^a^+...,
which proves the Binomial Theorem
for
any positive
frac-
tional index.
534. To prove the Binomial Theorem when the index any negative quantity.
is
been proved that
It has
/(m)x/(n) = /(m4-^0 for all values of 7n is
and
n.
Replacing
m
—
by
n (where n
we have
positive),
/(-n)x/00 = /(-« + n) = /(O)=l, since all terms of the series except the first vanish
but/()i)
= (l + xy
fkr for
''-''''
any positive value of
n
;
(l_|-a;)-«=/(-90.
or
But/(— ?i)
stands for the series
l+(-n>-+ (-")(7^"-^) x^+-; .-.
(1
+
x)-»
= 1 +(- n)x + (-n)(-n-l) ^ ^^,
...
which proves the Binomial Theorem for any negative index.
when x <1, each of the and /(m + n) is the true arithmetical equivalent of /0>0 X f(n). But when X > 1, all these series are divergent, and we can only assert that if we multiply the series denoted by f(m), by the 535. It should be noticed that
f(m + ?i)
is convergent,
series
f(m), f(n),
series
denoted by f(n), the first r terms of the product will first r terms of /(m + n), whatever finite
agree with the
value r
may
have.
—
3
CHAPTER XLVI. Exponential and Logarithmic Series. 536. The advantages of common logarithms have been explained in Art. 438, and in practice no other system is But in the first place these logarithms are calculated used. to another base and then transformed to base 10. In the present chapter we shall prove certain formula?, known as the Exponential and Logarithmic Series, and give a brief explanation of the way in which they are used in constructing a table of logarithms.
537. To expand a^ in ascending powers of
By
the Binomial Theorem,
^
,
1
,
if
nx(nx—l)
1
\2
n^
n
xlx
)
putting x
= l,
we
,
> 1.
nx(nx—V)(nx—2)
)
obtain
430
,
1
]{x
1_1
1 n"^
— ^ -+
x(x
=1+^'+-^^+-^ By
n
jr.
fl-lYl-2\
w-
EXPONENTIAL AND LOGARITHMIC SERIES. hence the series x( X 1
the
x{ X V
V^+
+ .+
and
(1) is
)
x^ower of the series (2)
o^th
](
that
is,
X n,
''A
^
however great n may we have
this is true
;
481
be.
If,
therefore, n be
indefinitely increased,
The is
series
1
+ 1+1 + 1 + 1+...
usually denoted by
e
;
hence /yi^
Write ex for
.t,
then
e-
=1+
ex-
+
/yvO
—+^+ [2
Now we
let e"
= a,
so that c
/yt^
••••
|3
= log^ a
;
by substituting
for c
obtain
a'= This
is
1
+ »log.a + ^i!fl|i^Vi?^(^V
....
the Exponential Theorem.
538. The series
which we have denoted by e, is very important, as it is the ba^e to which logarithms are first calculated. Logarithms to this base are
known
as the Napierian system, so
after Napier, the inventor of logarithms.
They
named
are also
—
6
ALGEBRA.
432
called natural logarithms
from the fact that they are the
hrst logarithms which naturally
come
into consideration in
algebraic investigations.
When logarithms are used in theoretical work it is to be remembered that the base e is always understood, just as in arithmetical work the base 10 is invariably employed. From the series the approximate value of e can be determined to any required degree of accuracy to 10 places of decimals it is found to be 2.7182818284. ;
Ex.
1.
Find the sum
of the infinite series l
We have
^
and by putting
a;
...e
hence the Ex.
2.
sum
=
^
=—
+
^
1 in
+ e- =
+l+i+i+ [2
[4
+|i +
|"i
'"
'
we
e^,
obtain
+i+i+i+
of the series is \{e +e"i).
Find the coeihcient of
^?-ii^
+ |i +
the series for
2(l
....
1
x''
in the expansion of ^
~
^ .
= (a-&x)e-^
= («-6x){l-x + |-| + ... + (^H-...}. The
coefficient required
—
( = ^^—— IV ^
•
a
— (— -^-
iy~i ^
\r-\
\r
h
|r
539. To expand
From
log, (1
+ x)
in ascending
powers
of x.
Art. 537,
if(\o^,af
a'-=l+ylog,a+-^^^|-
+
vYlog^rt)^
,
"^'"' |3
— EXPONENTIAL AND LOGARITHMIC SERIES. In this series write 1
-{-
x for a
;
thus
(1
we have
in (2) the coefficient of y is
(-1)^.
^ ^+1.2'' I
that
xy
<1
Also by the Binomial Theorem, when y
Now
-\-
433
(-!)(- 2) ^. 1.2.3
+ I
is,
ic
Equate
,
+
"-
—
——
4-^?^
2
3
(-l)(-2)(-3) *• 1.2.3.4
+
.
'
^4-..., 4
this to the coefficient of y in (1)
;
thus
we have
log,(l+x)=..;-J + |--|+.... This
known
is
as the Logarithmic Series.
540. Except when x is very small the series for log^ (1 + x) of little use for numerical calculations. We can, however, deduce from it other series by the aid of which Tables of Logarithms may be constructed. is
541. In Art. 539 log,(l
changing x into
—
log,(l 6eV
By
we have proved
+ a;)=a;-- + ^-
x,
•
we have
- x)=-x-~--J
3
2
subtraction, log.
^ =2 L± X
Put
that
x^
x^
^
1
—X
= ^^—
,,
2f
-\ '
so that X
3
'
= 2 91
71
log,(?i+l)-W ""' ?i=2i ^'^ ^
X \^'
(271+1
1
5
h
^
+1
we thus
:
1
obtain
h
3(27i+l)3^5(2n+l)^^
'
ALGEBRA.
434
From this formula by putting n = l we can obtain log, 2. whence Again by putting n = 2 we obtain log, 3 — log, 2 logc 3 is found, and therefore also log, 9 is known. Now by putting n = d we obtain log, 10 — log, 9 thus the value of log, 10 is found to be 2.30258509 .... ;
;
into logarithms to base
To convert Napierian logarithms 10
by we multiply ^
441] of the
which
,
the
is
modulus
[Art.
^
—
log, 10'
common
system, and
its
value
is
qnororAa
we shall denote this modulus by M. we obtain multiplying the last series throughout by logarithms. of common calculation the to adapted formula a
or .43429448
.
;
. •
M
By
Thus 31 log, (n o ,^
+ 1)— M log, n =
is,
5(2 n
M
M
M
(271
+1
+ l/
3(2n
,
+ 1)^ "^
"*"
2n
^(
1
,
+ 1 3(27^ + 1)^ logio(n + 1)— logio?i = (
that
1
1
(
'*
"t.
)
,
+ 1)^
5(271
'
j
>
the logarithm of one of two consecutive numbers be known, the logarithm of the other may be found, and thus a table of logarithms can be constructed.
Hence
if
EXAMPLES 1.
Show
XLVI.
that
(2^
e'-l-l+l + i + l +
Vl + x
...
ascending powers of
Z.
Expand
3.
Prove that
loge 2
4.
Show
,og„(^J = j^^(x + f + f +...).
5.
Prove that log
log
that
Show
that
if
in
i
+
yV
1±^ = — 1
6.
=
+
4:X
_i_
+
+
^l
4x^
3X
x>l, logVx2-l =
+
x.
....
+ ^^^ x^ +
20 x^
+ -.
o
logx-i-i-^.
CHAPTER
XLVII.
Determinants. 542. Consider two homogeneous linear equations ctiX
a^x
+ biy = 0, -^-boi/ =
0',
multiplying the first equation by b.,, the second by tracting and dividing by x, we obtain a/^s
This result
is
.....
— 02^i =
bi,
,
sub-
(1).
sometimes written
-0. and the expression on the left is called a determinant. It consists of two rows and two columns, and in its expanded form or development, as seen in the first member of (1), each term is the product of tAvo quantities; it is therefore said to be of the second order. cipal diagonal,
The
and the
letters a^,
b^,
a,,
The
line ci^2 is called the prin-
line bia2, the secondary diagonal. ?>2
determinant, and the terms
are called the constituents of the cti^g,
a2&i ^^'^ called
the elements.
THE VALUE OF THE DETERMINANT AFTER CERTAIN CHANGES. 543. Since
= ai^o —
fto^i
=
follows that the value of the determinant is not altered by changing the roivs into columns, and the columns into roivs.
it
Again,
it is
^1
easily seen that
ALGEBRA.
436
is, if loe interchange two 7'ows or two columns of the determincmt, we obtain a determinant ivhich differs from it onbj in
that
sign.
544. Let us
now
consider the homogeneous linear equa-
tions
a^x
+
bii/
a-^x -h 622/
a^x
By
eliminating
«1
or
(^2<^3
—
^3^2)
b
Co
+
x, y, Zy
+
^1
63?/
we
(^2%
+ CiZ = 0, H- ^2^ = Oj 4- C3Z = 0.
obtain
— C3«2) +
^1 («'2^3
"
«3^2)
= 0,
DETERMINANTS. 546. Minors. ((,
From
the preceding article,
437
ALGEBRA.
438 Thus
appears that if two adjacent columns, or rows, of of the determinant changed, hut its value remains unaltered. If for the sake of brevity we denote tlie determinant it
the determinant are interchanyed, the sign is
by
{a^boC^,
cti
hi
Ci
a.,
62
C2
a.
&3
C3
then the result we have just obtained
may
be
written {Hh(^z)
Similarly
= - (ctibiCs).
we may show that (r-i(/,6,)
= - (aicjj^) = 4- (chhCs).
548. Vanishing of a Determinant.
umns of the determinant
If two rows or
tivo col-
are identical the determinant vanishes.
For let D be the value of the determinant, then by interchanging two rows or two columns we obtain a determinant whose value is — 7J; but the determinant is unaltered; = — D, that is D = 0. Thus we have the followhence
D
ing equations,
- a^A^ -f a^A.^ = D, h^Ai - M2 + hA = 0, CiAi — C2^2 4- Cg^S = 0.
aiAi
549. Multiplication in
any row, or
in
any
then the determinant
For
of a
Determinant.
columii, is midtiplied
is
If each constituent bij
the
multiplied by that factor. mcii
same
factor,
DETERMINANTS.
439
550. A Determinant expressed as the Sum of Two Other Determinants. If each constituent in any row, or column, consists of two terms, then the determinant can he expressed as the
sum of two
other deter minayits.
Thus we have «1
+
f^l
^1
Cl
ALGEBRA.
440
Of these four determinants the first three vanish, Art. 548; thus the expression reduces to the last of the four determinants hence ;
its
value
= _ {c(c2 - ah) - h{ac = '6abc- «3 - 63 - c3. Ex.
We 67
2.
Find the value of
have
+
b-^)
67
19
21
39
13
14
81
24
20
.
a(a^
-
be)]
:
DETERMINANTS.
441
and what has been here proved with reference to the first colamn is equally true for any of the columns or rows; appears that in reducing a determinant we may by a new row or column formed in the following Avay
hence
it
replace any one of the rows or columns
Tal:e the constituents
and
of the row or column
to he replaced,
any equimultiples of the of one or more of the other rows or
increase or diminish them by
corresponding constituents columns.
After a
may
little
practice
it
will be
found that determinants
often be quickly simplified by replacing two or more
rows or columns simultaneously see that ai-\-pb,
bi-qci
c.
:
for example,
it is
easy to
ALGEBRA.
442 The given determinant 3
:
DETERMINANTS.
EXAMPLES
XLVII.
Calculate the values of the determinants 1.
1
1
1
35
37
34
23
26
25
443
a.
ALGEBRA.
444
APPLICATION TO THE SOLUTION OF SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. 552.
The
of
properties
may
determinants
employed in solving simultaneous Let the equations be
be usefully
linear equations.
+ biy + c^z + d^ = 0, 4+ c^z d2 = 0, a^x + + + = by A-^, — A2, A.^ respectively aix
a>;fc
-\-
622/
C02;
^32/
multiply them results, Ai, A^,
c?3
A^ being minors of
;
c/j,
cu,
and add the
a^ in the deter-
minant
D h
^3
^3
The coefficients of y and z vanish in virtue of the relations proved in Art. 548, and we obtain (a^Ai
— 662^2 + ^3^3)^ +
Similarly
we may show
(t^i^i
—
f?2^2
+ ^^3^3) = 0.
that
- h,B, + h,B.^y + {d,B, - d.rB, + d,B.^ = 0, and {c,C\ - cA + c,C,)z + (d,C, - d,a + dA) = 0. Now tti^i — ^2^2 + cisAs = — (biBi — b2B.2 + 63B3) = CjOi — C2O2 ~r ^3^3 ^^ Q),B,
J-^i
hence the solution
may
be written
-y
z
-1
DETERMINANTS. Ex. Solve
a;
have
2
?/
y
+
4?/
3a;
We
+
2x-\-
+ 3 ^ - 13 r= 0, + z- 7=0, + 3S-21 = 0.
445
ALGEBRA.
446 This
may
be more concisely written in the form cLi
hi
Ci
d|
«2
^2
<-'2
^2
Og
h^
Cg
^3
a4
^4
C4
f?4
= 0;
the expression on the left being a determinant of the fourth order. f?! taken Also we see that the coefficients of with their proper signs are the minors obtained by omitting the row and column which respectively contain these con-
stituents.
554. More generally, equations
a^i
if
we have n homogeneous
+ 62^2 + ^2% +
• • •
linear
+ ^2^'n = ^?
unknown quantities x^, x.,, x^, ••• x,^, these quancan be eliminated and the result expressed in the form
involving n tities
The
left-hand
CU
"62
«„
h„
member
Co'--
IC2
c.
A",
• • •
of this
equation
is
a determiuant is called a
which consists of n rows and n columns, and determinant of the
The
ni\\ order.
more general form of determinant beyond the scope of the present work it will be sufficient here to remark that the properties which have been established in the case of determinants of the second and third orders are quite general, and are capable of being extended to determinants of any order.
is
discussion of this
;
;
DETERMINANTS.
.
;;
447
Although we may always de555. Signs of the Terms. velop a determinant by means of the process described above, it is not always the simplest method, especially when our object is not so much to find the value of the whole determinant, as to find the signs of its several elements. 556.
The expanded form
of the determinant
«2
= c^hp. —
chhoji^i
^2
^2
% h
Co
4- a2^3Ci
— (-h^h^z +
<^h^h<'2
—
^3^2^!
appears that each element is the product of three factors, one taken from each row, and one from each column also the signs of half the terms are + and of the other half — When written as above the signs of the several elements may be obtained as follows. The first element a^lKf^, in which
and
it
.
the suffixes follow the arithmetical order, is positive; we shall call this the leading element every other element may be obtained from it by suitably interchanging the suffixes. is to be prefixed to any element according The sign -f or ;
—
as the
number
of inversions of order in the line of suffixes
even or odd; for instance in the element O362C1, 2 and 1 are out of their natural order, or inverted with respect to 3 1 is inverted with respect to 2 hence there are three inver-
is
;
sions fta^iCa
and the sign of the element is negative in the element there are two inversions, hence the sign is positive. ;
557. The determinant whose leading element thus be expressed by the notation
is a^b^t^d^-"
may
the S ± placed before the leading element indicating the aggregate of all the elements which can be obtained from it by suitable interchanges of suffixes and adjustment of signs. Sometimes the determinant is still more simply expressed by enclosing the leading element within brackets; thus ••• (aib2C,fili " •) is used as an abbreviation of 2 ± a/JaCgd^ * The Greek letter Sigma.
ALGEBRA.
448 Ex.
In the determinant {(iiJ^Csche^) what sign
is
to be prefixed to
the element aih^r^che2 ? Here 3, 1, and 2 are inverted with respect to 4 1 and 2 are inverted with respect to 3, and 2 is inverted with respect to 5 hence there are six inversions and the sign of the element is positive. ;
;
If in Art. 554, each 558. Determinant of Lower Order. h^, Cj, ••• A'l is equal to zero, the determinant reduces to -aiA-^ in other words it is equal to the l)th order, and product of tti and a determinant of the {ii we easily infer the following general theorem. of the constituents
;
—
If each of the constituents of the first row or column of a determinant is zero except the first, and if this constituent is times that deterequal to m, the determinant is equal to minant of lower order which is obtained by o^nitting the first
m
column and
first roiv.
Also since by suitable interchange of rows and columns any constituent can be brought into the first place, it follow^s that if any row or column has all its constituents except one equal to zero, the determinant can be immediately expressed as a determinant of lower order. This is sometimes useful in the reduction and simplification of determinants. Ex.
Find the value of 30
1
20
6 11
19
Diminish each constituent of the first column by twice the corresponding constituent in the second column, and each constituent of the fourth colunm by three times the corresponding constituent in the second column, and we obtain 8
11
20
5
3
15
-2
36
9
7
6
17
4
and since the second row has three zero constituents,
this
determinant
DETERMINANTS. =
3
8
20
5
449
CHAPTER
XLVIII.
Theory of Equations. 559. General Form of an Equation 2)oX''-\-2JiX''~'^-^P2X^~^-\
nth Degree. Let be a rational integral us denote it by f(x)
of the
+i\-i^+Pn
function of x of w dimensions, and let is the general type of a rational integral equathen f(x) = ;
Dividing throughout by tion of the nth. degree. that Avithout any loss of generality we may ta.ke
as the general
form
i\,
we
of a rational integral equation of
see
any
degree.
Unless otherwise stated the coefficients always be supposed rational. If
tion
any of the is
coefficients p^, p^, p^,
• • •
said to be incomplete, otherwise
560.
Any
^^i,
Ps?
=
P» will
are zero, the equa-
p,^
it is
called complete.
value of x which makes f{x) vanish
root of the equation fix)
•••
is
called a
0.
We
shall assume that every equation of the form has a root, real or imaginary. The proof of this proposition will be found in treatises on the Theory of Equations; it is beyond the range of the present work.
561.
/'(.x*)=0
562. Divisibility
of Equations.
If a
is
a
tion f(;x)= 0, then is f(x) exactly divisible by
Divide the
first
member by x
—a
root of the equa-
x
—
a.
until the remainder no
longer contains x. Denote the quotient by Q, and the remainder, if there be one, by E. Then we have
f(x)=Q(x-a)-{-Ii 450
= 0.
THEORY OF EQUATIONS.
Now
since a
is
a root of the equation x
Q(a--a)-\-R
R=
hence that
is,
the
first
hj x
divisible
—
member
=
=
a,
451 therefore
0,
0',
of the given equation
is
exactly
a.
563. Conversely, if the first member of f(x) = x — a, then a is a root of the equation.
exactly
is
divisible by
For, the division being exact,
Q(x
and the substitution of a a is a root.
DIVISION BY
— a) =
0,
for x satisfies the equation
;
hence
DETACHED COEFFICIENTS.
564. The work of dividing one multinomial by another be abridged by writing only the coefficients of the The following is an illustration. terms.
may
Ex. Divide 3x5^8a;4-5x3 + 26x2-33a: + 26 by x3-2a^2_4x+8. 1
+
2
+ 4-8)3-8- 5 + 26-33 + 3 + 6 + 12-24 -2+ 7 +
-2- 4-
26(3
-2 + 3
:
ALGEBRA.
452 Let
lis
take the example of the preceding article. work is as follows
arransreinent of the 1
The
'
THEORY OF EQUATIONS.
453
second term of the quotient, which multiplied by — 2 gives 14 for the second term of the second horizontal line the addition of 14 and gives 14 for the third term of the quotient, which multiplied by — 2 gives — 28 for the third term of the second line, and so on. ;
Every equation of the n\h degree 567. Number of Roots. has n roots, and no more. Denote the given equation hjf(x)= 0, where
The equation /(.7;)= has a root, real or imaginary let this be denoted by aj; then /(a*) is di^^sible hjx Gi, so that ;
—
/(..)= (,.-a,)/,(x), n — 1 dimenAgain, the equation f(x) = has a root, real or imaginary let this be denoted by a., then fi(x) is divisible by
where
fi(x) is a rational integral function of
sions.
;
X
— Qv,
;
so that
A(x) = (x-a,)Mx),
where f2(x) is a rational integral function of
Thus
n
— 2 dimensions.
f(^^ = (x - a,)(x - a,)Mx).
Proceeding in this way, we obtain /(.v.)=.(.^._a,)(.r-«,)-(-^--«n).
Hence the equation f(x) =
has n roots, since f(x) vanishes w^hen X has any of the values aj, a^, czs, ••• ««• Also the equation cannot have more than n roots for if x ;
has any value different from any of the quantities a^, a^, an, «„, all the factors on the right are different from zero, and therefore f(x) cannot vanish for that value of x. In the above investigation some of the quantities rtj, a.,, ttn, ... a^ may be equal in this case, however, we shall suppose that the equation has still n roots, although these are not all different. .
.
;
568. Depression of Equations.
If one root of an equation
known it is evident from the preceding paragraph that we may by division reduce or dej^ress the equation to one of the is
next lower degree containing the remaining roots.
So
if
k
ALGEBRA.
454 roots are (n
known we may
— k)t\i
depress the equation to one of the All the roots but two being known, the
degree.
depressed equation
a quadratic from which the remaining
is
roots are readily obtained.
=
569. Formation of Equations. Since f(x) (x—a^(x—o,^ — (7„j [Art. '307], we see that an equation may be formed
••(u.-
by mtbtracting each the continued
root
from
the
unknovm quantity and
j^larAng
product of the binomial factors thus formed equal
to 0.
Form
Ex.
1,-2, and
the equation whose roots are
2x3
.-.
+
^2
-5s:
EXAMPLES
+
=
2
0.
XLVIII.
-
|.
a.
_
7.
2 j: + 24 = 0. Show that 4 is a root of r^ 5 Show that + 3 is a root of r^ + 7 x^ + 7 x + 15 = 0. Show that - i is a root of 6 x^ + 17 x^ _ 4 x - 3 = 0. _ 9 ^ + 4 = 0. Show that f is a root of 10 x'' - 3 One root of x^ + 6 x2 - 6 x - 63 = is 3 find the others. One root of x^ - 23 x^ + 166 x - 378 = is 7 find the others. what are the others ? One root of x^ - 2 x- + 6 x - 9,^^ = is
8.
Two
1.
2. 3.
4. 5. 6.
3r2
a;'^
;
;
t^
roots of x^
-
15x-
+
24
=
43 x
4-
+
lOx
;
and 3
are 2
;
find the
others. 9.
x*
-
root of x^
+
Two roots of
3
x''
- 21 x- +
60
=
are 3 and 5
;
find
the others. 10.
One
2 axr
+
5 a-x
+
4 a'
—
is
—
a
;
what are the
others ? 11. 12.
Form Form
the equation
whose roots are
the equation whose roots are
— —
1,
—
2,
2,
—
3, 4- i,
and
—
5.
and
—
570. Relations between the Roots and the Coefficients. us denote the equati(jn l)y
and the roots by of
a, b, c,'-'k',
then we have identically
+pix"-^ +i>2^~^
= (x — a)(x —
H
b)(x
\-
Pn-i^
-\-P'"
— c)--'(x — k)
;
\.
Let
:
:
THEORY OF EQUATIONS. hence, by multiplication,
=
X"*
455
we have
- ^i.i-"-^ + S..V--'
+ (- iy''S„_,x + (-
Equating the coefficients of we have
like
1)''*^„.
powers of x in
this
identit}^,
Ih
=
^2)
which Si stands for the sum of the roots a, b, c -" k; S2 stands for the sum of the products of the roots taken two at a time, and so on to S„, which equals the continued product of all the roots. That is in
(1)
The
equals the (2)
The
coefficient of the
sum
second term
icith its sign
changed
of the roots.
coefficient of the third
term equals the sum of
the products of the roots taken two at a time. (3) The coefficient of the fourth term icith its sign changed equals the sum of all the products of the roots taken three
all
at a time, (4)
The
and
so on.
last
roots, the sign
term equals the continued product of all the -\- or — according as n is even or odd.
being
571. It follows that (1)
The sum
if
the equation
of the roots
is
zero
is
if
in the general
form
the second term
is
wanting. (2)
One
572.
root, at least, is zero if the last
The student might suppose
term
is
wanting.
that the relations estab-
would enable him to solve the number of the relations is
lished in the preceding article
any proposed equation
;
for
A
little reflection will equal to the number of the roots. show that this is not the case; for suppose we eliminate any n—l of the quantities a, h,c, -•• k, and so obtain an equation to determine the remaining one; then since these quan-
tities are
involved symmetrically in each of the equations,
;
.
ALGEBRA.
456
clear that we shall always obtain an equation having the same coefficients this equation is therefore the original equation with some one of the roots a, h, c, "- k substituted it is
;
for X.
Let us take for example the equation
and
let a, 6, c
be the roots
then
;
= P2, = —Pz. equations by a-, — a, 1 o? = — PiCr — pM — p^, a^ 4-i>i«' +iV^ + pz = 0, ab
ac
-\-
-\-
be
-\-
ahc
^Multiply these
add that
;
thus is,
which
is
the original equation with a in the place of
The above process is
respectively
and
x.
of elimination is quite general,
and
applicable to equations of any degree.
573. If two or more of the roots of an equation are connected by an assigned relation, the j)roperties proved in Art. 570 ^^-ill sometimes enable us to obtain the complete solution. Ex. 1. Solve the equation Ax^ — I^ x- + 23 z + 18 = 0, having given that the roots are in arithmetical progression. Denote the roots hj a — h, a, a -\- h then the sum of the roots is 3 n the sum of the products of the roots two at a time is 3 a- — h- ; and the product of the roots is a (a-— 6^); hence we have the equations ;
;
3a =
6,
3a2
- 62 = ;5^,
a{(fi
-
62)
=_
|
from the first equation we find a = 2, and from the second 6 = -t ^, and since these values satisfy the third, the three equations are conThus the roots are — \, 2, f sistent. Ex. 2. Solve the equation 2A.t^ - \\x- - Q^x being double another. Denote the roots by a, 2 a, 6 then we have
+
\b
= Q,
;
.3
From
the
a
+
first
6
= f.,
2 a2
+
3 a6
= - ^J-,
2
a% = - V-
two equations, we obtain
8a2-2a-3 = 0; .-.
a
=
I
or
—
\,
and h
=—
^
or f |.
one root
:
.
THEORY OF EQUATIONS. be found on
will
It
=
(7
=— ^
=— ^'
h
|,
Thus the
—
roots are |, |,
= — i, 6 = f f do not hence we are restricted to the
that the values a
txial
satisfy the third equation 2 a-b
values
457
;
f
we may not be able to find the roots we can make use of the relations proved
574. Although
of
in Art. 570 to determine the values of sj^mmetrical * functions
equation,
ail
of the roots. Ex. Find the sum of the squares and of the cubes of the roots of the .r^ — px- + qx — r = 0. Denote the roots by a, b, c then a + b -\- c =p, be + c
equation
;
i,-2
;
^3
_|.
a3
.-.
1$ 4.
+
c^-p(a^
?>s_j-
c^
^
i'2
^ c^^^
+ c)- Sr = S r = p^ - Spq + Sr.
q(^a -r b
= p{p^ - 2 q)- pq
-{-
EXAMPLES XL VIII. Form
the equation whose roots are
1.
I,
3.
2, 2,
—
-
2,
6.
2, 0,
b.
:
2.
0, 0,
4.
a
iv'3.
I,
;
+
2, b.
2,
a
-
-3, -3. b,
-
a
+
b.
a
—
and
7.
-
b.
Solve the equations 5.
.r*
6.
4
—
.(-3
16
.r^
1(5
_|_
+
X-
86 x- — 176 + — 9x — 06 = 0, jc
105
=
the
sum
0,
two roots being of
two
1
of the roots being
zero. 7.
4 x^
8.
3 x^
+ 20 x'^ — 23 -f 6 =0, two of the — 26 x"^ + 52 X — 24 = 0, the roots JC
roots being equal.
being in geometrical pro-
gression. 9.
2x^
—
X-
—
22
.r
—
24
= 0,
9^
—
9
two
of the roots being in the ratio of
3:4. 10. 24 .r^ the roots. 11.
S.r^
+ 46 J*-
—
2.r^
—
-t-
27.r-
+
= 0,
6.r
-I-
9
one root being double another of
=
0,
two
of the roots being equal
but opposite in sign.
*
A
function is said to be symmetrical with respect to its variables value is unaltered by the interchange of any jmir of them thus X + y + z, be + ca + ab/x^ + y^ + 2^ - xy:: are symmetrical func[See An. tions of the first, second, and third de^ees respectively.
when
319.]
its
:
ALGEBRA.
458
54 x^-S0x'^-26x+ 10
12.
=
0,
the roots being in geometrical pro-
gression.
—
32 x^
13.
48
x"2
—
3
40x2 -
7
+
22
ic
=
0, tlie
roots being in arithmetical pro-
gression,
6x4
14.
-
roots being 15.
29x3
+
- 2 x^ - 21 x2 +
X*
a:
-
12
=
0,
the product of two of the
2.
+
22 X
=
0,
the roots being in arithmetical
520 x
+
192
40
progression. 16. 27 x4 - 195 x3 + 494 x^ geometrical progression.
+
18x3
17.
81 x2
+
121 X
+
_ GO
=
0,
=
0,
the
roots
being in
one root being half the sum of
the other two.
Find the sum
18.
x*
+
qx^
+
rx
of the squares
and
of the cubes of the roots of
+ s = 0.
An equation ivhose coefficients are of the first term being unity, cannot have a rational fraction as a root. 575. Fractional Roots.
integers,
that
If possible suppose the equation
lias for
a root a rational fraction in
sented by
-.
its
lowest terms, repre-
Substituting this value for x and multiplying
h
through by 6"^\ we have h
Transposing,
This result is impossible, since it makes a fraction in its lowest terms equal to an integer. Hence a rational fraction cannot be a root of the given equation. In an equation 576. Imaginary Roots. imaginary roots occur in pairs.
icith real
coeffi-
cients
Suppose that f{x) = is an equation with real coefficients, and suppose that it lias an imaginary root a -f ih we shall show that a — ib is also a root. ;
;
THEORY OF EQUATIONS. The
factor of f(;x) corresponding to these (x
—a—
—a
ib)(x
-|-
or (x
ib),
(x — ay +
Let fix) be divided by by Q, and the remainder, if any, by
In this identity put x
-ay +
i;x
b''
h^
two roots
— af
is
61
-\-
denote the quotient
;
Rx + W then q\ix - af + h''\-\- Rx + R'.
fix) =^
also
459
=
= a-{-
ib,
hence
0',
;
then f(x) =
R (a +
ib)
by hypothesis
-\-R'
= 0.
Equating to zero the real and imaginary parts,
Ra + R' = and
by hypothesis
b
.-.
Hence
= a — ib
0;
not zero,
is
R=
and R'
fix) is exactly divisible
hence x
Rb =
0,
(x
— a—
is
also a root.
by
=
0.
i^x
— a)^ +
— a-^ ib)
ib) ix
6^,
that
is,
by
;
577. In the preceding article we have seen that if the has a pair of imaginary roots a ± ib, then equation f(x) = (x — ay + 6^ is a factor of the expression /(.»). are the imagiSuppose that a ± ib, c ± id, e ± ig, nary roots of the equation fix) = 0, and that * ^ i;x) is the product of the quadratic factors corresponding to these imaginary roots then • • •
;
^ix)={ix- af
Now X
;
+ bmix - cy + dm(x -
ey
+ f\^-.
each of these factors is positive for every real value of
hence 578.
cf)
As
ix) is
always positive for real values of
in Art. 576
we may show
x.
that in an equation with
rational coefficients, surd roots enter in pairs; that
a
+ -^b
is
a root then a
— ^b
Ex. 1. Solve the equation 6 given that one root is 2 — -^3. *
a;*
is
-
The Greek
is,
if
also a root.
13 x^
-
35
letter Phi.
a:^
-
x
+
3
=
0,
having
:
ALGEBRA.
460
Since 2 — y'S is a root, we know that 2 + y'S is also a root, and corresponding to this pair of roots we have the quadratic factor
x2-4x +
1.
Also 6a;4
-
13
-
a;3
-X+
35x2
= (x2 - 4x +
3
1)(6
a:2
+
11 X
+
3)
;
hence the other roots are obtained from
+
6x-^
Ex.
Form
2.
+
llx
-
thus the roots are
3=:
0,
-h
i,
+
or
(3x
2
+ V^?
+ 3)=0; - ^d.
l)(2x 2
the equation of the fourth degree with rational coeffi-
V—
3. one of whose roots is ^2 + Here we must have y'2 + v^— 3, ^2— V — 3 as one pair 3 as another pair. and — y'2 + 3, — y'2 —
cients,
x'^
of roots,
V—
V—
Corresponding to the first pair we have the quadratic factor 2y/2x + 6, and corresponding to the second pair we have the
—
quadratic factor
+2^2x4-5.
x2
Thus the required equation (x2
+
2
or
is
V2 X + 5) (x2 - 2 V2 x + (x2 + 5)2 -8x2 = 0, x4
or
+
2 x2
EXAMPLES
+
25
=
5)
=0,
0.
XLVIII.
C
Solve the equations 1.
3 x*
2.
x*
3.
x*
4.
X*
+ +
-
10 x3
+
4 x2
-
X
-
6
= 0,
one root being
^
"^^~^
+ 72 X - 30 = 0, one root being 3 - ^3. + 5 x2 + 2 X - 2 = 0, one root being - 1 + V+ 6 x2 + 4 X + 5 = 0, one root being V— 1.
-
36 x2 4 x^ 4 x^
1.
TRANSFORMATION OF EQUATIONS. 579. The discussion of an equation is sometimes simpliby transforming it into another equation whose roots bear some assigned relation to those of the one proposed. Such transformations are especially useful in the solution of cubic equations. fied
THEORY OF EQUATIONS.
461
580. To transform an equation into another whose roots are those of the original equation with their signs changed.
Let /(x) =
be the equation.
is satisfied Put — y for x\ then the equation /(— y)= with its sign changed tlius the by every root of f(x) = ;
required equation is /(— y)= If the given equation is
then
it is
0.
evident that the required equation will be
therefore the transformed equation original equation
by changing
the sign
is obtained from the of every alternate term
beginning with the second. Note. If any term of the given equation supplied with zero as a coefficient.
is
missing
it
must be
into another Ex. Transform the equation cc*-17x2-20x-6 = shall have the same roots numerically with contrary signs. We may write the equation thus
which
:
x4
By
the rule,
+ 0x3- 17a:2-20a:-6 =
0.
we have a;4
_
or
-
17 x2
x4-
17x'^
x3
+ 20 x - 6 = + 20x-6 =
0, 0.
581. To transform an equation into another whose roots are equal to those of the original equation multiplied by a given factor.
Let fix) quantity.
=
be the equation, and
Put y
= qx,
so that
let q
value, y is q times as large; then x
equation
denote the given
when x has any y
= -,
particular
and the required
is
?)"--(i)"«(r Multiplying by
fy",
we have
—©"=»
:
ALGEBRA.
462
Therefore the transformed equation is obtained from the original equation by multiplying the second term by the given factor, the third term by the square of this factor, and so on. 582. The chief use of this transformation equation of fractional
Remove
Ex.
Put X
=-y
we
fractional coefficients
g
=
4
all
to clear
an
from the equation
and multiply each term by
By putting 2,
is
coefficients.
the terms
cf
become
;
thus
and on dividing by
integral,
obtain
^Sy- -y +
ys
Q
= 0.
583. To transform an equation into another whose roots exceed those of the original equation by a given quantity.
=
be the equation, and let h be the given quan= x -\- h, so that for any particular value of X, the value of y is greater by h thus x = y — h, and the required equation is f(;y — h) = 0. Similarly if the roots are to be less by h, we assume y — X — h, from which we obtain x = y -\- h, and the required
Let
f(;x)
Assume y
tity.
;
equation
is
584. If
-f h)
f(y 71
is
= 0.
small, this
method of transformation is For equations of a higher
effected with but little trouble.
degree the following method
Let put x
is
to be preferred
f(x) = poO;" + p^x''-^ + p2X^-'- +
=y
-\-
h,
• • •
+ i>.-i^' + Pn
J
and suppose that f(x) then becomes
+ ^i2/"~' 4- ^2^"' + + gn-iV + g„. — Now y = X h; hence we have the identity ^02/"
=
ryo(x-
• • •
- hy + q,(x - hf-' +
• • •
+
fy„_
,{x
- h) + q„
;
therefore q„ is the remainder found by dividing f(x) x h; also the quotient arising from the division is
—
q,(x
- hy + qi(:x - hy + '
-
••.
+
^,._i.
by
THEORY OF EQUATIONS. Similarly g„_i expression by x sion
the remainder found by dividing the last and the quotient arising from the divi-
is
— h,
is
qo^x
and
468
so on.
- hf + q,(:x - h)-^ + -'
Thus
g„, g^„i, g„_2?
The
Synthetic Division. equal to
Hence
•••
• • •
+ qn-^2
;
ma^y be readily found by go? ^-ncl is obviously
last quotient is
p^.
to obtain the transformed equation,
Divide f{x) by x±h according as the roots are to be greater or by h than those of the original equation, and the remainder ivill be the last term of the required equation. Divide the quotient thus found by x ± h, and the remainder will be the coefficient of the last term but one of the required equation; and less
so on.
Ex. Find the equation whose roots exceed by 2 the roots of the equation
4x4
+
32x3
+ 83x2+76x4-21 =0.
be obtained by substituting x — 2 for x hence in Horner's process we employ x + 2 as divisor, and the calculation is performed as follows
The required equation
in the
proposed equation
will
;
:
4
+2
ALGEBRA.
464
Let the given equation be V Pn-i ^ + Pn = + PiX""-^ + V^""'^^ H y = x — h, we obtain the new equation
_p„a^"
then
if
+ hy +2h(y + ^0'"' -^I'^iv + ^O""' +
Po(y
when arranged
which,
If the
;
+k =
•••
o,
in descending powers of y, becomes
term to be removed is the second, we put n2)Qh-\-2)i=0,
so that h
—
=
:
if
the term to be removed
is
the third,
we put
'l^^pji'-^{n -
l)pji
+p, =
0,
and so obtain a quadratic to find h and, similarly, we may remove any other assigned term. Sometimes it will be more convenient to proceed as in the ;
following example. Ex.
Remove
the second term from the equation
px^
Let
a,
&,
c
qoiP
+
rx
-{-
be the roots, so that
increase each of the roots
sum
+
by —-,
of the roots will be equal to
the second term will be zero.
—
^
=
q h
^
0.
a-\- b -r c
q -
q = — --
Th6n
if
we
transformed equation the
in the
Hence the required transformation X
s
;
that
is,
the coefficient of
^ will
be effected by substituting
for X in the given equation.
op As the
general type of a cubic equation can be reduced to a more simple form by removing the second term, the student should carefully notice that the transformation is effected by substituting x minus the coefficient of the second term divided by the degree of the equation, for x in the given equation.
THEORY OF EQUATIONS.
465
586. To transform an equation into another whose roots are the reciprocals of the roots of the proposed equation.
Let x
/(.«)
1
= -\
=
be the proposed equation •
;
/1\
•
then the required equation
is
y
f{-] \yJ
put y
= -,
so that
^
= 0.
One
of the cliief uses of this transformation is to obtain values of expressions whicli involve symmetrical functions of negative powers of the roots. tlie
the equation
If a, 6, c are the roots of
Ex.
x^
— poc?
find the vakie of
Write - for
multiply by
— r — 0,
+ 7^ + -o"
-^ a-
x,
qx
-{
c2
h-
?/3,
and change
all
the signs
;
then the
y resulting equation r>f
has for
its
-
qff
+ i^y -
1
=
0,
111.
roots
a *
hence
u
c
V— =^-
S1 = 2 a
ah
r
into
-, it is
an equation
If
587. Reciprocal Equations.
by changing x
r*
is
unaltered
called a reciprocal equation.
X If the given equation is X''
-\-p^X''-^
+P2^''~~
-\
V Pn-2^^
the equation obtained by writing
-
+ Pn-V^ + Pn = 0,
for x,
and clearing of
fractions, is j^x"" +p„_ia5""^
+ Pn-2^''~^ H
+P2^^ +Pl^ + 1 = we must have
0.
If these two equations are the same,
Pn-2
Pn-l
*
S1
stands for the
a
2h
•
sum
Ih
of all the terms of
1
Pi
which a
is
the type.
ALGEBRA.
466
from the last result we have j>„ = two classes of reciprocal equations, If
(i.)
= 1,
p,,
±
1,
and thus we have
then
2h=2\-l, P2=l\-2, lh=Pn-3, •••; that is, the coefficients of terms equidistant from the beginning and end are equal. (ii.)
= — 1) then = - l\-l, 2h = - Pn-2, the equation is of 2 m
If 2\
Pl
Ps
= - Pn-A,
• • •
;
dimensions p^ = — i)„„ or In this case the coefficients of terms equidistant 2j^ = 0. from the beginning and end are equal in magnitude and opposite in sign, and if the equation is of an even degree the middle term is wanting.
hence
if
588. Standard Form of Reciprocal Equations. that f(x) is a reciprocal equation.
= If f(x) = is has a root — 1
Suppose
and of an odd degree
of the first class
it
If (x) so that f(x) is divisible by x-\-l. is the quotient, then cf>(x) is a reciprocal equation of the ;
=
and of an even degree.
first class
If /(.y) it
=
has a root
as before class
;
=
<^(.t)
=
has a root
ble
1
in this case/(.T) is
is
and of an odd degree, by x — 1, and
divisible
the reciprocal equation of the
first
and of an even degree.
If f(x) it
of the second class
is -+-
by
a^
is
of the second class
and a root — 1 in and as before (f>(x) =
+1
— 1,
;
and of an even degree, this case f(x) is divisiis
a reciprocal equa-
tion of the first class of an even degree.
Hence
ariy reciprocal equation is
last ter7n positive,
of an even degree
or can he reduced to this form ;
icith its
which may
therefore be considered as the standard form of reciprocal equations.
to
589. A reciprocal equation of the standard form can be reduced an equation of half its dimensions. Let the equation be
aa^"*
+ 6a;-"*-^
-f- c^ji^""-"^ -\
+
kx'''
H
\-
ex"
+ hx -\-a = 0\
;
THEORY OF EQUATIONS. dividing by
.«"'
and rearranging the terms, we have
z for
xPj\
^
af+^
hence writing
;«
+ -,
xj
and giving
3
Ij 2,
•••
we
xF-^J
\
p
to
X
values
467
in succession the
obtain
+ x^ \ = z'-2',
x'
x^'^
- = z(z' - 2) - z = z' - 3z: x^
x'-^-^
= z{f - 3z) -
(z^
- 2) = z' -
^z' -{-2
X*
and
so on
;
and generally
x"^
is
-\
m
of
dimensions in
z,
X
and therefore the equation in 590. To
find
z is of
m
dimensions.
the equation whose roots are the squares of
those of a proposed equation.
Let f(x) =
by putting y = be the given equation and therefore the required equation ;
= ^y,
we have x
X',
is
/(V2/)=0Ex. Find the equation whose roots are the squares of those of the equation x3
Putting X
= y/]i^
+ pix^ + P2X + p3 = 0.
and transposing, we have (2/+jP2)V2/=-(Pi2/+i'3);
whence or
(1/ -\-2p2y -V IH^) y 1/3
+(2i92 -i?i2)?/2
= Pi^v'^ +
+ (p22
-
2 pip^y
+ ps^,
2pipz)y -ps^
=
0,
:
ALGEBRA.
468
EXAMPLES XLVIIL 1.
d.
Transform the equation rc^ — Sx* + Oa:^ — 9.x2 -f ox — 1 which shall have the same roots with contrary signs.
=
into another
2. Transform the equation 2x^ — 4:X^ + 7x — 3 = whose roots shall be those of the first multiplied by 3. 3.
x^ — 7x — 6 multiplied by
Transform the equation
roots shall be those of the
first
=
into another
into another
—
whose
|.
into another 4. Transform the equation x^ — 4^^ + ix — ^ = with integral coefficients, and unity for the coefficient of the first term. 5.
Transform the equation 3 x* — 5x3 + first term is unity.
—x+1 =
x^
into another
the coefficient of whose
6. Transform the equation x^ + 10 x^ -f 39 x^ another whose roots shall be greater by 4. 7. Transform the equation x* — 12 x^ another whose roots shall be less by 3. 8.
Diminish by
+
+
x'^
17
76 x
+
— 9x +
65
7
= =
into
into
the roots of the equation
1
2x4-
13x2
+ 10x-
19
= 0.
Find the equation whose roots are greater by 4 than the corresponding roots of x4 + 16 x3 + 72 x^ + 64 x - 129 = 0. 9.
10.
Solve the equation
3 x^
-
22 x^
+
48 x
-
36
=
are in harmonical pro-
32
=
0,
the roots of
which are in harmonical progression. 11.
The
gression
roots of x^
—
11
x'^
+ 36x —
find them.
;
Remove
the second term from the equations
12.
x3-6x2 + 10x-3 =
13.
x4
14.
Transform the equation
+
4 x3
+
2 X-
-
0.
-
4X
2
=
0.
x^
-
X
— -3 =
into
one whose roots
4
exceed by
|
4 the corresponding roots of the given equation.
15.
Diminish by 3 the roots of the equation
16.
Find the equation each of whose roots
x5
a root of the equation 17.
x''
—
5 x^
-f-
6x
4X
—
3
+6=
0.
is
greater by unity than
= 0.
Find the equation whose roots are the squares x4
18.
+ 3 x2 -
4 x4
Form the
+
x3
+ 2 x2
-f
X
+
1
=
of the roots of
0.
equation whose roots are the cubes of the roots of x3
+
3x2
+2=
0,
:
THEORY OF EQUATIONS. If «, h, c are the roots of x^
qx
-{-
-{ r
=
469
form the equation whose
0,
roots are 19.
ka-\ kb-\ kc-K
21.
?^+^, ^:^+^, a^
20.
6%^ c%^
a-^6-^.
22.
6c
«^^.
b^
+
^,
ca
c-
+ ^,
a6
+ ^.
Solve the equations
+ x3-^6x2 + + 2 = 0. _ iqjcS 26x'^ - lOx + 1 = 0. - 5 x* + 9 ic^ - 9 + 5 - 1 = 0. _ 24 + 57 x* - 73 + 57 x^ -
23.
2x4
24.
a:4
25.
x^
26.
4
a:
_|.
a;2
a;5
a:6
a:
a:^
24
.r
+
4
= 0.
DESCARTES' RULE OF SIGNS.
When
each term of a series has one of the signs 4a continuation or permayience occurs when the signs of two successive terms are the same : and a change or variation occurs when the signs of two successive terms 591.
and
—
before
it,
are opposite.
592. Descartes* Rule. Li any equation, the number of poscannot exceed the number of variations of sign, and in any complete equation the number of negative roots cannot exceed the number of X)ermanences of sign. itive roots
Suppose that the signs of the terms in a multinomial are
+ + — — + — —
|-_-|-_5we
this multinomial is multiplied
shall show that if by a binomial whose signs one more change of sign in
are -1- — there will be at least the product than in the original multinomial. Writing only the signs of the terms in the multiplica,
tion,
we have
;
;
ALGEBRA.
470
a double sign^ spoken of as an ambiguity, being placed wher-
ever there
is
a doubt as to whether the sign of a term
is
positive or negative.
Examining the product we
see that
An
ambiguity replaces each continuation of sign in the original multinomial (ii.) The signs before and after an ambiguity or set of (i.)
ambiguities are unlike (iii.) change of sign
A
is
introduced at the end.
Let us take the most unfavorable case and suppose that from (ii.) all the ambiguities are replaced by continuations we see that the number of changes of sign will be the same whether we take the upper or the lower signs let us take the upper thus the number of changes of sign cannot be less than in ;
;
;
+ + -- +
and
+ - + -+,
this series of signs is the
same
as in the original multi-
nomial with an additional change of sign at the end. If then we suppose the factors corresponding to the negative and imaginary roots to be already multiplied together, each factor x—a corresponding to a positive root introduces at least one change of sign therefore no equation can have more positive roots than it has changes of sign. To prove the second part of Descartes' Rule, let us suppose the equation complete and substitute — y for x then the permanences of sign in the original equation become Now the variations of sign in the transformed equation. transformed equation cannot have more 2)ositive roots than it has variations of sign, hence the original equation cannot have more negative roots than it has permanences of sign. be complete or incomplete Whether the equation f{x) = its roots are equal to those of /(— a;) but opposite to them ;
;
in sign; therefore the negative roots of /(^)=0 are the positive roots of /(— x)=0-, but the number of these positive roots cannot exceed the number of variations of sign in
f{—x)]
that
is,
the
number
of negative roots of
cannot exceed the number of variations of sign in
/(a;)=0
/(— x).
THEORY OF EQUATIONS.
We may
471
therefore enimciate Descartes'' Rule as follows:
=
An
equation f(x) cannot have more positive roots than there are variations of sign in f(;x), and cannot have more negative roots than there are variations of sign inf(—x). Ex. Consider the equation cc^ + 5 x^ — x^ + 7 x + 2 = 0. Here there are two changes of sign, tlierefore there are at most two positive roots.
Again f(—x) = — x^-\-5x^ + x^ — lx-\-2, and here there are three changes of sign, tlierefore the given equation has at most three negative roots, and tlierefore it must have at least four imaginary roots.
593. It is very evident that the following results are included in the preceding article. (i.) If the coefficients are all positive, the equation has no positive root thus the equation x^ + o;^ + 2 x + 1 = cannot have a positive root. If the coefficients of the even powers of x are all of (ii.) one sign, and the coefficients of the odd powers are all of the contrary sign, the equation has no negative rootj thus the equation ;
x'
-j-x'-2x'-\-x'
-3x^ + 7 X-
5
=
cannot have a negative root.
EXAMPLES XL VIII. Find the nature of the roots 1.
x4
+
2.
x4
-
3.
3x*
4.
Show
imaginary 5.
2x3
10 x3
+
+
12x2
35 x2
14 X
-
of the following equations
+
+
50 X
+ 5x-4 =
24
:
= 0.
24
= 0.
0.
—
x*
+
4 x^
-
5
=
has at least four
x5
be inferred respecting the roots of the equation
x^-2x-3 = 0?
Find the
-
-
that the equation 2 x^
What may
tion x^
13 x2
roots.
xio_4x6 + 6.
-
e.
+
least possible x4
+
x"-^
+
1
=
number 0.
of imaginary roots of the equa-
ALGEBRA.
472
594. Derived Functions. f(x)
is
a rational
Let f(x) = PqX""
fix
4-
To find
the value off(x-\-h), ivhen
integral function of x.
+ i\x''- + p^''-" H h Pn-i^ + Pn then = p^i:x + hf -^p^ix hy-^ +po^{x + hy-^' +'"
li)
1
;
-f-
Expanding each term and arranging the ing powers of
h,
- l>i.T"-^+(vi - 2)p.pf-^-\-
••.
+2\-i\
- l)p^x--'+{n - l)(^n - 2)p,x-'-\-
...
+ 2p,_,\
4- h\np,x^^-^
-\-^~\n{n
+~
.
result in ascend-
we have
.
+
(^n
.
•
+ ^5n(n-l)(H-2)...2.1poS. 1
71
This result
fix
is
usually written in the form
+ /o=/w+ ¥'(^o+|V'(^*)+||V"(^)+ -
and the functions f{x), fix), f\x), second, third,
...
are called the
first,
derived functions of f{x).
Examining the f'(x)
•••
+|V''(^x
coefficients of h, ,—
•••,
we
see that to obtain
from fix) tve nudtiply each term in fix) by the index of x and then diminish the index by unity.
in that term,
Similarly
we
obtain fix), f"'(x),
595. Equal Roots. to a,
....
If the equation f{x) = has r roots equal = will have r—1 roots equal to a.
then the equation /'(x)
Let
(f>{x)
be the quotient
when
then flx) = (x-aycf>(x). Write X -{- h in the place of x
;
fi;x) is
divided by (x — aY;
thus
fi^x + h) = (x - a + hycf>
i^x
+ h)
;
;
THEORY OF EQUATIONS. +
.•./(x)+/i/(x)
={(x
- ay +
- ay-^h +
r(x
|/'(a;)+...
...}{0(x)+
In this identity, by equating the
f(x)=
r{x
/i0'(a;)
+ ^0"(x)+
coeflficients of
- aY-'
(^x
-
/<,
.••}.
we have
aycj>\x).
—
Thus /' (x) contains the
—
factor x 1 times a repeated r the equation f\x) has r 1 roots equal to a. Similarly we may show that if the equation f(^x)== has roots equal to b, the equation /'(it*) has s 1 roots equal
that s
473
—
=
is,
—
=
to b
;
and so
on.
From
the foregoing proof we see that if f(x) contains a ay, then /'(a?) contains a factor (x — ay~'^ and thus f(;x) and /'(a?) have a common factor (a; ay~^. Therefactor (x
—
;
—
fore if f(x) and f'(x) have no common factor, no factor in hence the equation f(x) has or f(^x) will be repeated
=
;
has not equal roots, according as f(x) and f'{x) have or have not a
common
factor involviyig
x.
596. It follows that in order to obtain the equal roots of the equation /(x) 0, ive must first find the highest common factor of f{x) and f\x), and then placing it equal to zero,
=
solve the resulting equation.
Ex. Solve the equation equal roots.
Here
f{x)
=
x^
x*—
-
11
\\x^
^3+ 44 cc2_
+
/(a:)=4Gc3_33a^2
^ix'^
+
7(5
^
^ 48 =
- IQx +
o,
which has
48,
88a;- 76;
and by the ordinary rule we find that the highest common factor fix) and f{x) is a; — 2 hence {x — 2)2 is a factor of f{x)\ and ;
/(x)
= (x-2)2(a;2-7x+12)
= (x-2y\x-^){x-4)]
of
ALGEBRA.
474
LOCATION OF THE ROOTS. 597.
If the variable
x
changes continuously from a to 6 the
function f(x) will change continuously from f(a) to f{b).
Let and b.
and
c
We
/(o
c
h be any two values of x lying between a
-\-
have
+ h) -/(c) = //'(c) +|/"(c) +
...
+^r(c);
and by taking h small enough, the difference between f(c-\-h) and /(c) can be made as small as we please hence to a small change in the variable x there corresponds a small change in the function f(x), and therefore as x changes gradually from a to h, the function /(a;) changes gradually from /(a) to f(b). ;
598. It
is
important to notice that we have not proved
that f(x) always increases from f(a) to f(b), or decreases from f(a) to f(b), but that it passes from one value to the
other without any sudden change creasing and at other times
599.
If
it
;
may
sometimes
it
may
be in-
be decreasing.
f(a) and f(b) are of contrary signs then one root of
the equation f(x)
=
must
lie
between a and
b.
As x changes gradually from a to b, the function f(x) changes gradually from f(a) to f(b), and therefore must but since /(a) and pass through all intermediate values f(b) have contrary signs the value zero must lie between them that is, f(x) = for some value of x between a and b. has only one root between It does not follow that/(a;) = a and b neither does it follow that if /(a) and f(b) have has no root between a and b. the same sign f(;x) = ;
;
;
600. Every equation of an odd degree has at least one whose sign is opposite to that of its last term.
real
root
In the function f(x) substitute for x the values successively, then
— 00,
/(+^)= + ^,
/(0)=p,.,
f(-oo)
= -c^.
+ ^,
0,
;
THEORY OF EQUATIONS. If
2^n
is
and — between
then f(x)
positive, 00,
and if p^ and + go.
is
=
475
has a root lying between = has a root lying
negative f(x)
601. Every equation which is of an even degree and has its term negative has at least two real roots, one positive and
last
one negative.
For
in this case
/(+«))= + CO, but 2\
and
is
f(0)=p,, f(-c^)= + c^hence f(x) = has a root lying between and a root lying between and — go.
negative
+ CO,
;
602. If the expressions f(a) and f(b) have contrary signs, will lie between a and 6 an odd number of roots of f(x) and if f(a) and f(b) have the same sign, either no root or an
=
even number of roots will
Suppose that a represent
is
lie
b.
b, and that c, d, e, ••• A; which lie between a and when f(x) is divided by the
greater than
the roots of f(x) (x) be the quotient
all
Let cfi b. product (x
between a and
= 0,
— c)(x — d) (x — e) — k) then — — — = (x — 'k)4> (^)(x c) (x d)(x e) f(x) /(a) = (a — c){a — d) (a — e) (a — k)cf) (a). = e) c) d) (b). (& (6 (5 (6 - k) jXb) • •
•
(a;
;
• • •
Hence
• • •
...
Now
must be of the same sign, for otherwise a root of the equation <^ (x) = 0, and therefore of f(x) = 0, would lie between a and b [Art. 599], which is contrary to the hypothesis. Hence if /(«) and f(b) have (^
(o)
and
<^
(b)
contrary signs, the expressions
— c)(a — d) (a — e)
...
(a
(b-c){b-d){b-e)
...
ib-k)
(a
must have contrary
— k),
Also the factors in the first expressions are all positive, and the factors in the second are all negative hence the number of factors must be odd, that is, the number of roots c, d, e, " k must be odd. Similarly if /(a) and /(6) have the same sign the numsigns.
;
•
ALGEBRA.
476
ber of factors must be even. dition
is
satisfied if
c,
d, e,
--•
In this case the given conk are all greater than a, or
less than 6; thus it does not necessarily follow that /(a;) = has a root between a and b.
EXAMPLES 1.
XLVIII.
f.
Find the successive derived functions of 2x'^—x^—2x^-\-5x — l.
Solve the following equations which have equal roots
:
S = 0. 3. X* - 6x^ + I2x- - lOx 2. x'^ - 9x2 + 4 x + 12 = 0. 4. x^ - 13 + 07 x^ - 171 x^ + 216x- 108 = 0. 5. x^ -x^ + 4:x^ -?>x + 2 = 0. 4 a;3 _ 18 6. Sec* + 11 _ 2 = 0. 7. Show that the equation lOx^ — 17^2 + + 6 = has a root between and — 1. Sx- + S5x — 10 = has a 8. Show that the equation x* — 5x^ root between 2 and 3, and one between — 2 and — 3, has a root 9. Show that the equation x* — 12x2 + 12x — 3 = between — 3 and — 4, and another between 2 and 3. has a root between 10. Show that x^ + 5x* — 20x2 — 19x — 2 = 2 and 3, and a root between — 4 and — 5. -\-
ic*
_|.
;^2
a;
a:
-\-
STUEM'S THEOREM AND METHOD. 603. In 1829, Sturm, a Swiss mathematician, discovered a method of determining completely the number and situation of the real roots of an equation. 604. Let f(x) be an equation from which the equal roots have been removed, and let fi(x) be the first derived funcNow divide f(x) by fi(x), and denote the remainder tion. Di\'ide /^ (x) by fo (x) and conivith its signs changed by /o (x). tinue the operation, which is that of finding the H.C.F. of f{x) and f\{x), except that the signs in every remainder are changed before it is used as a divisor, until a remainder is obtained independent of x the signs in this remainder must No other changes of sign are allowed. also be changed. ;
:
:
THEORY OF EQUATIONS.
477
The expressions /(.i'), fi(^x), f2(x), -•• f^(x) are called Sturm^s Functions. Let Qi, Qo,'" Qn-i denote the successive quotients obtained; then the steps in the operation may be represented as follows
=Qi /i(.^') -f2(x), =Q, Mx) -f,(x\ Mx) =Q, fsi^) -M^\ f-2{^) /(.^•)
From (1)
these equalities
Two
we
obtain the following
consecutive functions cannot vanish for the same
value of X.
For
if
they could,
all
the succeeding functions would is impossible, as it is inde-
vanish, including /„('^)j which pendent of X. (2)
When any
ticular value of
function except the
jr,
first
vanishes for a par-
the two adjacent functions have opposite
signs.
Thus
in
A(x) = Qfs(x)-f,(x)
if
Mx)= 0,
we have
A(x)=-Mx).
We may now
state Sturm's Theorem.
If in Stimn^s Functions ice substitute for
x any
2^(i'''ticuJcir
value a and note the number of variations of si^n ; then assign to X a greater value b, and again note the number of variations of sign; the number of variations lost is equcd to the number
of real roots of f(x) ichich
lie
between a and
b.
be a value of x which makes some function for example, fix), so that /^ (<;•)= 0. = Xow when x c, f-iix), and f+ilx) have contrary signs, and thus just before x = c and also just after x = c, the three functions f-i^x), f(x), fr^i(x) have one permanence of sign and one variation of sign, hence no change occurs in the (1)
Let
c
except the first vanish
;
ALGEBRA.
478
number of variations when x passes through a value which makes a function except J\x) vanish. (2) Let c be a root of the equation /(a.*) = Let h be any positive quantity.
Now
/(c
+
so that /(c)
h)=f(c)+ //^(c)+|y;(c) + ...,
= 0.
[Art. 594.]
\A
and as
c is
a root of the equation /(c
/(.y)
= 0,
/(c)
= 0,
hence
+ /o=/t/;(c)+|/,(c)+-.
we may disregard the terms conhigher powers and obtain
If h be taken very small,
taining
its
/(c
+ /0=//i(c),
and as h is a positive quantity, /(c + h) and / (c) have the same sign. That is, the function just after x passes a root has the same sign as fi(x) at a root. In a like manner we may show that /(c — li)= — hfi(c), or that the function just before x passes a root has a sign opposite to fi(x) at a root. Thus as x increases, Sturm's Functions lose one variation of sign only when x passes through a root of the equation /(;r) = 0. There is at no time a gain in the number of variations of sign, hence the theorem is established. 605. In determining the whole number of real roots of an equation f(x) = we first substitute — oo and then oo for X in Sturm's Functions: the difference in the number of variations of sign in the two cases gives the whole number
+
of real roots.
By
substituting
number and
— oo
for x gives the
606.
and
for x
of negative real roots,
When
number
-h co or
— oo
is
we may determine
and the substitution of
the
+ co
of positive real roots.
substituted for
x,
the sign of
any function will be that of the highest power of x in that function.
THEORY OF EQUATIONS.
479
607. Let us determine the number and situation of the real roots of x"
-^3x'
-9x - = 0. 4.
Kevef,(x)=Sx' + 6x-9. Now any positive factor may be introduced finding
by
or
removed
in
the sign of the result is not affected hence multiplying the original equation by 3,
/2(a;),/3(x), etc., for
so doing
;
we have 3aj2
_l_
6a;
- 9)3
o;-*^
+
9.^•^
- 21 x - 12(j; + 1
f.lx)=^x 2(3x-
+5
+ l.
ALGEBRA.
480 /(^O
= — CO x=— x = -4. x = -3 x = -2 x = -l x= x=l x = 2 X= 3
/3C^•)
-
+
a;
X==
3 3 2 2 2 2
variations. variations. variations. variations. variations. variations. 1 variation. 1 variation. 1 variation, no variation, no variation.
+
T)
+
-
±
+ + + + +
+
CO
Thus one root between and —
+ + + + +
+
and 5 a second lies between 2 and 3.
between and tlie third
lies 1,
:
I:
EXAMPLES XL VIII. Determine the number and situation 1.
2.
- 4 - 6 X + 8 = 0. 2x^-nx^ + Sx-16 = 0. _ 7 ^ + 7 ^ 0. c4 _4^3 + 6;;c2- 12x + 2 = x3
a:2
g-.
of the real roots of
5.
x*
6.
x^
-
+
4 x^
x^
+
6
a:
:
+ 2 = 0.
X - 1 = 0. + 23x- 10 + x^-2x^ + 2x x^
+
ix'-^
.,:3
0.
8.
x^
lies
= 0. 1
=0.
GRAPHICAL REPRESENTATION OF FUNCTIONS. Coordinates. 608. Fig. 1
Two
lines
drawn
at right angles to each other as in
form a simple system of
lines of reference.
intersection,
XX' L X'r-
the along
are called abscissas; dis-
XX' on a line paral-
tances from lel to
Their
called
Distances from
origin.
P
0, is
YY'
are called ordinates.
609. Abscissas measured to
o
-6
-h
the right of the origin are con-
and to the left, Ordinates measured above XX' are considered pos/tive, and when taken below XX' sidered
P'
Fig.
1.
]iositive,
negative.
are negative.
THEORY OF EQUATIONS.
481
610. The abscissa and ordinate of a point are called the and the lines XX' and are called Co-ordinate Axes, Axis of X, and Axis of Y, or Axis of Abscissas, and Axis of Ordinates.
YV
co-ordinates of that point,
Any
point in the plane can be given by means of thns the point of Fig. 1 is located by measuring the distance a to the right of O on the axis of abscissas, and then taking a distance b vertically upwards.
611.
these co-ordinates
P
:
Since a and b can be either positive or negative, a point is found by taking a positive and b negative; F" is found by taking a negative and b negative and P'" is found by
P'
;
taking a negative and b positive. Abscissas and ordinates are generally represented by x and y respectively. Thus for the point F, x = a, and y = b; for P', X = a, and y = — b, etc. 612. Instead of writing "the point whose co-ordinates and 3," a more concise form is used thus the point (5, 3) means that the point will be found by taking an abscissa of 5 units and an ordinate of 3. Locate the points (3, — U) (5, 8) (— 4, 4) (—8,-3).
are 5
:
;
;
;
482
ALGEBRA.
Ex. 1. Construct the graph of 2 x — 1. Let 2x — \ =y. Giving to x successive values, we obtain the corresponding values of y as follows:
x
= -2,
THEORY OF EQUATIONS. 614.
In giving values to x
y=0,
tion of a root gives
488
evident that the substituthe ordinate, or distance 0; hence where the graph
it is
that
is,
from the Axis of Abscissas, is we have the location of a real root, and cuts the Axis of as many times as the the graph will cross the Axis of equation has real and unequal roots. If the roots of the equation be imaginary, the curve will not touch the Axis of X.
X
X
EXAMPLES XL VIII.
h.
Construct the graphs of the following functions 1.
2 a; -3.
2.
x^'
+ 2x^l.
:
3.
oc^-5.
5.
x^-2.
7.
4.
x-HiC-1.
6.
x3-5x + 3.
8.
- 3 »:2 + 3. xH3x2 + 5x-12. ./•;>
SOLUTION OF HIGHER NUMERICAL EQUATIONS.
Commensurable Eoots. 615. is
A
real root
which
is
either an integer or a fraction
said to be commensurable.
By
Art. 582
we can transform an equation with
fractional
another which has all of its coethcients integers, that of the first term being unity hence we need Such equations canconsider only equations of this form. not have for a root a rational fraction in its lowest terms [Art. 575], therefore we have only to find the integral roots. By Art. 570 the last term oi f(x) is divisible by every integral root, therefore to find the commensurable roots of f{x) it is only necessary to find the integral divisors of the last term and determine by trial which of them are roots.
coefficients
into
:
numbers by actual substitution whether they are roots. In other cases we may use the method of Arts. 562 and 566 or the Method of Divisors, sometimes 616. Newton's Method.
we may
If the divisors are small
readily ascertain
called Newton'' s Method.
Suppose a to be an integral root of the equation ic"
+Pi^""^
-{-
p^x''-^
-\
\-Pn^iX
+ 'Pn = 0,
:
ALGEBRA.
484
By
substitution ^ye have
Transposing and dividing throughout by
^=in
which
ing
it is
Ihci"''
i>«-i
evident that
Q + i^-i =
-Ih^C"-
Pn — must be
— by Q and transposing —
a,
we obtain
- a"-\
an integer.
Denot-
Pn-\,
pocC'^ —picC"'"
— cC~^.
Dividing again by a gives
Q + Prx-\
...
„_2
,,_o
Again, as before, the first member of the equation must be an integer. Denoting it by Q., and proceeding as before, we must after n divisions obtain a result Qn-l-\'Pl
a
_ ~
^
Hence last
if a represents one of the integral divisors of the term we have the following rule
Divide the
last
term by a and add the
coefficient
of x
to the
quotient.
Divide to
it
this
sum by
the coefficient
Proceed
in this
each quotient
icill
be
The advantage
and
a,
^if the
quotient
of or. manner, and if a
an
integer
and
is
is
an
integer
add
a root of the equation
the last quotient
icill
be
—1.
Newton's method is that the obtaining any point of the division shows
of
of a fractional quotient at
at once that the divisor is not a root of the equation.
Ex. Find the integral roots of x*
-f
4
Descartes' Rule the equation cannot have nor more than three negative roots.
The
integral divisors of
stitution
shows that
—
1
is
s-^
-
:c^
-IGx-
more than one
- 12 are ± 1, a root, and that
±
+
2,
±.3,
1 is
12
=
0.
By
positive root,
±4, ±6.
not a root.
Sub-
:
:
THEORY OF EQUATIONS. To
ascertain
if
2
a root, arrange the
is
+
1
4
-
1
-
-1-6-11Hence
2
is
2
-
12
-
16
work
-
486
as follows
12 [2
6
22
a root.
Explanation. The first line contains the coefficients of the original equation, and the divisor 2. Dividing the last term, — 12, by 2 gives a quotient — 6 adding — 16, the coefficient of x, gives — 22. Dividing — 22 by 2 gives — 11 adding — 1, the coefficient of x^, gives —12. Dividing — 12 by 2 gives — 6 adding + 4, the coefficient of x^, gives — 2, which divided by 2 gives a final quotient of — 1, hence 2 is a ;
;
;
root.
Since the equation can have no more than one positive root, only make trial of the remaining negative divisors, thus
will
1
+
4-1-16+ -
—
Hence 1
6
is
+ Hence
+ 3 + 3-12 — 3 is a root.
Hence
—
4
is
not a root.
1+4-1- 16- 12
+4_1 _ 16-121 -3 4
121-4
3
-13
14
not a root.
-1-1+4+
Solve
1+4-1-16
12|-6
2
-
+
+5+ 6 2 + 4-10 Hence — 2 is a root. 1
2
i^
we
;
ALGEBRA.
486 617. The Cube Roots
Suppose that .-.
X
is
Unity.
of
= 1, or — 1 = l)(aj2 + (x + 1) = 0. — 1 = 0, or + + 1 =
= ^1
then x^
;
:)t^
;
ic
either
ic
^-^
x^l,
Avhence It
;
.t
=
or x
may be shown by actual when cubed is equal to
values
;
=involution that each of these unity. Thus unity has three
cube roots,
-14-V33 -1-V33 '
'
2
'
2
two of which are imaginary expressions. Let us denote these by a and b then since they are the ;
roots of the equation x^
+ +1 = x-
0,
their product is equal to unity
that
ct?b
.'.
that
= 1\ = «^; 6 = a^, since that a =
ab
is,
is,
Similarly
we may show
oj^
= 1.
b^.
Since each of the imaginary roots is the square of the usual to denote the three cube roots of unity
618.
other, it is
by 1, CO, Also
0)1* (X)
satisfies
the equation x^ .-.
that
is,
the
Again
sum of
+ +
1
0)
-\-
(0^
x-\-l
the three cube roots
w
•
w^
= 0-,
= 0; of unity
is zero.
= w^ = 1
therefore (1) the product of the
tivo
(2) every integral poiver
imaginary roots
of
is unity.
* The Greek letter Omega.
is
unity;
THEORY OF EQUATIONS.
487
CARDAN'S METHOD FOR THE SOLUTION OF CUBIC EQUATIONS. 619.
The
general type of a cubic equation x"
P^^
-f-
+
Qx
is
+ R = 0,
but as explained in Art. 5S5 this equation can be reduced
form
to the simpler
which we 620.
Let
a^
qx
-{-
-\- 7'
= 0,
shall take as the standard form of a cubic equation.
We proceed to solve the equation x^ qx z; then x=y = y^ + z^ Syz(y z) = f + z^ + Syzx,
-\-
-f-
r
= 0.
-\-
-\-
ix^
-\-
and the given equation becomes y^-^z'
+ (3yz + q)x-\-r = 0.
two quantities subject to the conequal to one of the roots of the given equation if we further suppose that they satisfy the equaWe tion 3yz + q = 0, they are completely determinate.
At present
y, z
dition that their
are any
sum
is
;
thus obtain f-{-z'
hence
cf
- ^^^ = -
= -r
r,
or y^
(1),
+ ry^ = ^
•
Solving this equation,
Substituting in
We
(1),
^^=-^Vi3- •••••• = -|-^^ + ^
2;»
obtain the value of x from the relation x
-I 2+V4^27> +
i,
2
= y -\-Z',
\4 + 27)
(^)-
(4).
thus
^'
— ALGEBKA.
488
The above
solution
generally
is
known
as Cardan^ s Solu-
was first published by him in the Ars Magna, in Cardan obtained the solution from Tartaglia; but
tion, as it
1545. the solution of the cubic seems to have been due originally to Scipio Ferreo, about 1505.
= y -\-z,
In this solution we assume x 2
=
—^,
hence
a cubic equation of
to solve
32/
ive substitute
y
——
for
—
—
(
3y
V
or
)
1/
+
/
2/3+ 15^
- l^x =
y^
y^
+ i^=126,
-
126 y^
= - 125. = 125, y = 5.
y^
.-.
=
- = 6. y Dividing the given equation x^ obtain the depressed equation v
126,
y
./
.-.
But x
126.
- for x, then y y
whence
form
x.
+ ^ + 1^- 15y-^ =
or
the
CI) find
+ qx-\- r =
a^
Ex. Solve the equation x^
Put y
and from
+
— \bx —
126
=
by
a;
+ 6a; + 21 = 0, + 2V— 3, and — 3 — 2\/— 3. - 15x = 126 are 6, - 3 + 2V^^,
— 6, we
a;2
the roots of which are
Thus the
—
roots of x^
3
and
-
3
BIQUADRATIC EQUATIONS. 621.
We
shall
now
give a brief discussion of some of the
methods which are employed of a biquadratic equation.
the methods tion;
we have
and thus
it
to obtain the general solution
It will be
first to
found that in each of
solve an auxiliary cubic equa-
will be seen that as in the case of the
cubic, the general solution is not
adapted for writing down
the solution of a given numerical equation.
\
;
;
:
THEORY OF EQUATIONS.
489
622. The solution of a biquadratic equation was obtained by Ferrari, a pupil of Cardan, as follows
first
Denote the equation by
+ 2px'^ + gx^ + 2 + s = side {ax + 6)^, the quantities x''
add to each
?-aj
make
determined so as to then a?*
;
a and h being side a perfect square
the left
+ '^n^ + (g + «') ^^ + 2 (r + a6) X + s 4-
=
?>'
(aa.-
+ h)\
Suppose that the left side of the equation is equal to {q^- -^px -\- Tif then by comparing the coefficients, we have ;
= q-\- a?,
p^ -\-2k
by eliminating a and
=
pA:
r
-\-
ah,
Ic-
=s
-\-
h'^
from these equations, we obtain
h
(pA; - rf = (2k +p' - q) (k' - s), = 0. 2k^ - qk^ + 2 (pr - s) k +ph - qs -
or
o""
From
this cubic equation one real value of k can always
be found [Art. 600]
+ 2^^ +ky = (ax + bf', px k = ± (ax + b)
(x'
x^
.'.
thus a and b are known.
;
-{-
-\-
and the values of x are
Also
;
be obtained from the two
to
quadratics
+
x'
and
(k -b) = 0, + (k -\-b) = 0.
-a)x +
(p
x^-\-(p-^ a) x
,
Ex. Solve the equation
Add «%2
_|.
xi-2x^-\-
x4
_
+
b^ to
2 abx (a2
then by equating a2
-
5)
^
2
+
6,
(2^•
+
A:
...
trial,
But from
we
+
a;2
find that
ab
lOx
- 3 = 0.
= -k-b,
6)(A:2
;
it
+
= A;2 + 3 + 5)2;
5^
3) zz(^^
5A;2-4A--7=0.
= -\;
the assumption,
+
we have
2F + k
5a;2
each side of the equation, and assume 2 (a6 + 5) x + b^ - S = (x'^ - x + k)'^
coefficients,
.-.
By
-
2x3
hence a^
follows that
- 4^ 62-4^
ab
=- 4.
ALGEBRA.
490
Substituting the values of k, a and
that
is
ic2
-
3
1
= 0,
—
^^^—,
+
a;
whence the roots are
The following
623.
and
b,
x^
we have -\-
-
x
S
the two equations
=
;
~— was given by Descartes
solution
in
1637.
Suppose that the biquadratic equation form
+ qx^ + rx -f s = 7'x s = (x^ + kx + x'^
assume
x'^
+ qx^
then by equating I
From
the
-\-
-\-
-\-m first
reduced to the
;
I) (oi^
— kx + m)
;
we have
coefficients,
— k'^= q,
is
k(m
— l)= r,
Im
=
s.
two of these equations, we obtain
2m = k'-^q +
'y,
= k' + q-l',
2l
k
k,
hence substituting in the third equation,
+ r)(k^ + qk - = 4 sk% - r^ = 0. k' + 2qk'-h {(f - 4
Qi^ -\-qk
or
?')
s)k''
This
is
a cubic in
k'^
which always has one
real positive
solution [Art. 600] thus when k'^ is known the values of I are determined, and the solution of the biquadratic is and ;
m
obtained by solving the two quadratics x^ -\-kx-\-l
Ex.
=
0,
x^
— kx + m = 0.
Solve the equation
a;4_ 2x2
Assume
x*
—
then by equating Z
2
a;^
+
8x
—
coefficients,
+w-
whence we obtain or
and
k'
{k^
3
+ 8:^-3 = = (x^ +
0.
A'»; -|-
{x-
—
kx
+
we have
= - 2, k{m -1)='^, Im = - 3 - 2 + 8) {k^ - 2 ^ - 8) = - 12 k% l^ - 4 + 1(3 - 04 = 0. ;
/j
A;*
k:^
iii)
;
;
:
THEORY OF EQUATIONS. This equation
is
clearly satisfied
when
k"^
—
4:
491
=
one of the values of k
will he sufficient to consider
;
0, or k = ± putting k =
2.
2,
It
we
have
+ = 2, tn — = 4; that is, =— 1, m = 3. - 2 x2 + 8x - 3 = (x2 + 2 X - 1) (x2 - 2 X + 3) — 1 = 0, and cc^ — 2x + 3=0; ic^ + 2 hence and therefore the roots are — 1 ± y/2, 1 ± V— 2. tn
Thus
I
I
I
a:*
a:
624. The general algebraic solution of equations of a degree higher than the fourth has not been obtained, and Abel's demonstration of the impossibility of such a solution If, however, the is generally accepted by mathematicians. coefficients of an equation are numerical, the value of any real root may be found to any required degree of accuracy by the method of Art. 626.
EXAMPLES
XLVIII.
k.
Solve the following equations 1.
x^-lSx =
S5.
2.
x3
3.
x3
+ 72 X+ 63x- 316
= 0. = 0.
4.
x3
+ 21x +
=0.
5.
28x3-9x2 + 1=0.
1720
-
6.
x3
7.
2x3
15x2
+
3x2
342
-33x +
+ 3x +
14.
x*
x3-6a;2 8^3
10. 11. 12.
847 1
13.
+ 3x-18 = 0. - 36x + 27 = 0. x3-15x-4 = 0. x4 + 8x3 + 9x2- 8x- 10 = 0. x4 + 2x3-7x2-8x + 12=0. x^ - 3x2 - 6x - 2 = 0.
8.
9.
=
= 0. 0.
- 2x3 -
12x2
+
lOx
+ 3 = 0.
INCOMMENSURABLE ROOTS. 625. The incommensurable roots of an equation cannot be found exactly. If, however, a sufficient number of the initial figures of the root have been found to distinguish it from the other roots we may carry the approximation to the exact value to any required degree of accuracy by a method first
published in 1819 by
W.
G. Horner.
;
ALGEBRA.
492
HORNER'S METHOD OF APPROXIMATION. 626. Let
it
be required to solve the equation
o^-3x'-2x-\-5
=
(1).
By
Sturm's Theorem there are 3 real roots and one of them lies between 1 and 2 we will find its value to four places of decimals, which will sufficiently illustrate the method. Diminishing the roots of the equation by 1 [Arts. 583, ;
584]
,
we have 1
-3 1
-2 1
-1
_2 -2 -4 -1 -5
+5 -4
|1
~~
1
1
The transformed equation
is
f-5y-i-l =
(2).
The roots of (1) has a root between 1 and 2. equation (2) are each less by 1 than those of equation (1) This root hence equation (2) has a root between and 1 being less than unity the higher powers of y are each less than y. Neglecting them, we obtain an approximate value Equation
.
from — 5y -\-l =0, or y = .2. Diminishing the roots of (2), the tion, by .2, we have
of y
±0 .2
first
transformed equa-
;
THEORY OF EQUATIONS. The transformed equation ;33
Equation
(2) lias
+
.
493
is
+ .008
6 2^-4.88 2
a root between
and
....
(3).
the roots of than those of equation (2) .2
.3
;
equation (3) are less by .2 hence equation (3) has a root between and .1. Neglecting in equation (3) the terms involving the higher powers, as was done in the case of the first transformed equation, we have
- 4.882; -h .008 = 0, Diminishing the roots of by .001, we have
(3),
or z
= .001.
the second transformed equr-
tion,
+ .6 .001
.001
494
-3
THEORY OF EQUATIONS.
495
627. It sometimes happens that the division of the last term of the first transformed equation by the coefficient of x In that in that equation gives a quotient greater than unity. case, as where the signs of these terms are alike, we obtain another figure of the root by the method used to obtain the integral part of the root.
628. If in any transformed equation after the first the two terms are the same, the figure of the root used in making the transformation is too large and must be diminished until these terms have unlike signs. signs of the last
629. If in any transformed equation the coefficient of the power of the unknown quantity is zero, we may obtain the next figure of the root by using the coefficient of the second power of the unknown quantity as a divisor and taking the square root of the result. first
630. Negative incommensurable roots may be found by transforming the equation into one whose roots shall be posiThis tive [Art. 580], and finding the corresponding root. result with its sign changed will be the root required.
By Horner's Method we 631. Any Root of Any Number. can find approximately any root of any number for placing ^a equal to x we have for solution the equation x" = a, or ;
a;«
-a=
0.
EXAMPLES Compute
the root which
following equations 1. 2.
3. 4. 5. 6. 7.
is
XLVIII.
1.
situated between the given limits in the
:
root between 2 and 3. + 10 x2 + 6 X - 120 = root between 2 and 3. x^ — 2x — 6 =0 root between 1 and 2. x* - 2 ^3 + 21 X - 23 = root between 9 and 10. x^ + X - 1000 = x^ + x^ 4- X — 100 = root between 4 and 5. 2 x3 + 3 x2 - 4 X - 10 = root between 1 and 2. x3 - 46 x2 - 36 X + 18 = root between aijd 1. ic3
;
;
;
;
;
;
;
:
ALGEBRA.
496 8.
x^
X
—
9.
x3
2
r.
+ + x3 +
11.
;
root between
=
13.
Find the
10 x2
and
2.
between 2 and
3.
root between
-
;
root between 4
;
3.
-
3 and
and
4.
5.
:
- 8 x3 + 12 x2 + 4 X - 8 = 0. x4 + x3 + x2 + 3 X - 100 = 0.
x4
16.
17.
0.
3.
root between 2 and
;
real roots of the following equations
- 3 X - 1 = 0. x3 - 22 X - 24 =
x3
Find
1
root between 2 and
;
;
12.
15.
= 20
+ 8 X - 120 = root 3 x3 + 5 X — 40 = X* - 12 x2 + 12 X - 3 = x^ — 4 X* + 7 x^ — 863 =
10.
14.
S
-
four decimals, by Horner's Method, the value of the
to
following 18.
^11.
^13.
19.
^5.
20.
MISCELLANEOUS EXAMPLES -
1.
Simplify b
2.
Find the sum of
a
+
b
—
{b
2{c-\- d), 6
3.
4.
It
5.
Find the square
6.
Solve
7.
Find the H. C.F.
8.
Simplify
=
+c—
3
=
y
Q,
4i,
=
z
of 2
3x
+
-{b - a a)
and
c
^7.
VII. b)]
+
d
-\-
2
a}.
- 4 (a +
5).
1?/.
S, find
—
[b
(c?
ix + f?/byx—
Multiply
X
-{a + b)-
21.
the value of
+
v2 x +
Sy
-\-
z.
x^.
^^^ + ^^ = 2. X
—
X
1
—
6
of a^
-
2 a
-
4
and
a^
-
a^
-
4.
^+-^-4+i^. — a
-\-b
a
b
a^
b^
x+y=v6] 9.
10.
added is
44
:
11.
5
Solve
Two
which form a number, change places when 18 is number, and the sum of the two numbers thus formed
digits,
to the
find the digits. If
a
=
1,
6
=-
2,
c
= 3,
d
= - 4,
10a-(c+&)2
find the value of
,
MISCELLANEOUS EXAMPLES -
+
12.
Subtract
13.
Write the cube of
14.
Simphfy
x^
^
+
a;
from the sum
-^ x
-
-
-•
=^
16.
Find the H.C.F. and L.C.M. of
17.
Find the square root of 4
18.
Solve
Simplify
When
^
"^
2
3J
-\-
and
«*
+
x^
9(1
4.
+ 3 x2 - 4. _ 2 a) + 3 a'^(l -4a).
.
x_\^^j^ X
a
added
1 is
8)
+ x^-\-2x-4
(-^ \x
20.
f (x
+^+
^'
Solve f (2 X
7)
of
8 y.
15.
x^
tain fraction the result its
^2
; x
,
-
_
?/2
497
VII.
—
a)
+
X"
ax
numerator and denominator of a cerequal to f and when 1 is subtracted from
to the
is
;
numerator and denominator, the
result
is
equal to 2:
find the
fraction.
Show
21. is
six
-11«
that the sum of \2a-\-Qh—c, times the sum of 25 a + 13 & — 8 + 6 + 10 c.
22.
Divide x^
23.
Add
—
xy
+
j% y^
together 18 |
hj x
—-
-
—
—7 a — b + c and
-
c,
13 a
-
13 &
+ 6 + 6c, - c, and
ffl
^y.
[^ + sVv
KT-^^)-"<^^«{lf-5(— >} 24.
Find the factors of
2x-l
25.
Solve
26.
Find the value of
5
(5
when 27.
28.
(i.)
=—
Find the H.C.F. of
7
0,
/>
Simplify
2k
79 x
1,
-
8.
(ii.)
729 x^
-
y^.
11
a-Sh){a-b)-b{3a|.
=
+
x"^
17
=
a
10
5a:4-3^3_4.x-118^
c
r(4
!/2
?/2
x^ x'-^
b^(a
+ 10 and - 6xy + iy^
ic3-10x2-7
x^ - 7 xy + 12 x^ + 5 xy + 6
a-b)-
+
a:
X2/
—
2
?/2
+
c)},
2 x^-x'^-2 x-\-l.
ALGEBRA.
498 Solve
29.
of
^«^^+
^= ^n. =
4abx + Sy
17 6
J
30. Find the two times between 7 and 8 o'clock when the hands a watch are separated by 15 minutes.
31.
If
=
a
= - 2, c = 3, = - 4, - 4 5 + a^ - Vc^ ^
&
1,
find the value of
fZ
y/cP
h'^ -\-
a
+
Multiply the product of Ix'^-lxy + y^ and
33.
Simplify by removing brackets a*
_[_ 2 - (a* - (- 4 «3 —
x^-Sy-^ i)|
^^ _1_
- 4 a)} - (8 a - 1)]. when 6x'^ — 7x^-\-Sx'^-x + S is
Find the remainder
34.
by x
d.
^x+y by — (4 a^ _ (6 a^ _ 4
32.
6 «2
divided
4.
35.
Simplify
36.
Solve
2
+
^
=18
2/
[•
„
2x + y^^^ = 29\ 4
J
37.
Fmd the
38.
Solve .006 x
39.
Find the L.C.M.
40.
A
41.
Simplify
square root of 4 x^
-
+
.491
x
.723
- 12 + = - .005. :»•*
28 x^
+
9 x^
-
42
a:
+
49.
Sx^+2xy-y'^, and x^-x-y+xy^-
of x^+t/,
paid with quarters and half-dollars, and twice the number of half-dollars exceeds three times that of the how many of each are used ? quarters by 10 bill of
$12.50
is
:
42.
by
^2 43.
(a+6 + c)2-(«-6-(-c)2-{-(a-f 6-c)2-(-a + + c)2. /)
Find the remainder when - «6 + 2 62. If
a
=
0,
=
6
c
1,
(3 abc 44.
— 3x—
Simplify
+
x
-^-^. a
&2
46.
Find the cube root of 8 x^
47.
Solve
9x+82/ = 43xn. 9y
= 42 xy
i
—
&"*
b
-
XlP"
2 x-y
+ -^
216
is
divided
of
will divide both 4 ^2 4- 3
1_1
ab
—^p^ -I-
a'^b'^
fZ
I
a2
8X
2
15 without remainder.
^
45.
— 3 a% +
= - 2, = 3, find the value - 2 bed) \/a^bc - c%d + 3-
Find an expression which
4a:3-f-7x2
a*
x
—
10 and
MISCELLANEOUS EXAMPLES
—^
Simplify
49.
Find the L. C. M.
+
8 x3
38 x2
A boy spent
^^^
?
48.
+
499
VII.
of
59 X
+
30 and 6 x^
money
-
13
.r^
-
13 x
+
30.
one shop, one-third of the remainder in a second, and one-fifth of what he had left in a third. He had 20 cents at last how much had he at first ? 50.
lialf
of his
in
:
Find the remainder when divided by x^ — 5 x + 4. 51.
is
52.
|« -|(5 ^^
Simplify 4
«
=
x'^
f|, 5
=
c 1= I,
53.
If
54.
Find the L. C. M. of
1,
}
—
10
a;^
1 1 (2 «
8 x^
-H
- &)+
- 7 x + 12,
Find the sum of the squares of ax ax and express the result in factors.
—
+
3x
-
c)
}
2 (5
—
11
•
—
c2
3 x2 - 6 x - 9, and 2 x2 - 6 x - 8.
55.
by
7 r^
prove that
y/a x^
—
-f-
—
bx
hij^
ay, ay
-{-
bx,
;
56.
57.
+
Solve 6
4
Simplify
a 58.
Solve
69.
Add
x-f3x- ^^;+^ ^ = ^(2x + 67) + ^[l+-y
together the following fractions
—
2 x2
-f-
x^
+
i/2
x3
4X
-
:
—
x^
y^
?/2(x
-
?/)2
x'^y
x2
-
y^
A man agreed to work for 30 days, on condition that for every work he should receive $2.50, and that for every day's absence from work he should forfeit $ 1.50 at the end of the time he received $ 51 how many days did he work ? 60.
day's
;
:
61.
Divicle-2^+27-iM^-4x* + Il^'-5i^by5f+S-x.
62.
Find the value
4
4
when X = —
\
and y
of
=
2,
8
4
2
—
'
ALGEBRA.
500 10
Simplify
63.
- 11 - 1)
:/:
3 (x2 64.
Find the cube root
65.
Solve
4.-17 X
66.
_
^
4
3
-
+
X
^
of
ofi
2
-
.X
Solve
68.
Simplify ^
(x3
+5 + 1)
.r
(a:
--x^
5^, x-1
xH5x2+:*- + 5.
(i.)
2 1)
-{-^-^x'^
2x-l
3
-
-
x2-2 x?/-323
(ii-)
?/2.
= 2l]
+ 2/)+2^ -^ z)=Q5\
Ix-y
^
x^
,
1)
-^-^x^
Sx-^(ij
67.
1
+
^ 8..-30
10.x-13
Find the factors of iG>-
10 X (x-2
— ——
—
;
2
^
+
x-^
3 xy
69.
Find the square root of -(3
70.
The united ages
6
man and Two years
of a
•
-
„_
.,
-
3o y^
-2c-
2
ay\2
{a
+
c)
-
3
h}.
his wife are six times the united
ago their united ages were ten ages of their children. times the united ages of their children, and six years hence their united ages will be three times the united ages of the children how many children have they ? :
71. .x2
Find the sum of
-^xy -
72.
From
73.
If
a
=
3 x3
+
^2^
xy -^y'^-\-
- («-6)(6-x)}
y^,
and
subtract
Solve
+
6 x2
-
189
X.
«2
(ii.)
+ q)x-{q-p)y
+
2 «6
+
= QU\. r
)
Simplify
Solve
2
x?/
-
^x. +
^L^ + +
-V.4
Reduce
—L_ +
2(x
7
*"
_|_
:-
2 x3
-
X
-
^ 2x-i5.
2x-6
7)
O y
y2
4(x3-|)
f
1
2
to its lowest terms.
i
if.
(a-hby-2bx.
6
+
c)3.
of
PX
x 78.
?/3
b = i, c = 3, find the value of + (6 + c)3 + (c + ay + (a + 6)3 - (« +
2x2
77.
f
5,
{p
76.
-
?/2
Find the factors (i.)
75.
2
{(« + 6)(a-x)
v/6 abc 74.
?/2,
I
6^ _^
a
+
b.
MISCELLANEOUS EXAMPLES 79.
Add
together the fractions 1
2 x2 80.
zero.
A
501
VII.
-
1
4X
+
2 '
2
.x^
+
4
'
+
.>;
and
1
2
-
1
x2
number
consists of three digits, the right-hand one being If the left-hand and middle digits be interchanged, the number
diminished by 180 if the left-hand digit be halved, and the middle and right-hand digit be interchanged, the number is diminished by is
;
336
:
number.
find the
81.
Divide
82.
If
p=
1 1,
-
5x
^
=
+ W- ^^ -
J,
84.
^-
-
-^cf
x^
by
1
-
x
-
{i
+ q-{p-{q-p)}
'ip
83. Multiply
llf ^*
find the value of
5 x^
+^+
9
- - x + 3.
by
Find the L. C. M. of {(fih
-
2
ah^)'^,
2a^-Zah-2
85.
Solve
86.
Reduce
87.
Find the square root of
—^^
^
—
3x3-2x2 + 16x-48 ^a'^
Solve
and 2(2
Ifi,
cfi
+
a6)2.
2iL±i = i^±Ll5^3x + 3. 4x + 4 3x+l x + 1
+ ^iffi + —\ +
= ^+^ 36
2a
to
a
+
b
its
lowest terms.
\2 a{(fi
-M) +
18.
x^.
ALGEBRA.
502 93.
Find the L.
15(p3 94.
of
_pq +
Resolve into factors (i.)
95.
+
M.
C.
5(^2
^3),
_
a3
8
_
(ii.)
+ ^+^^^ +h x+a+
+
a;2
35 65
(fih^c:^
rt^^c
-
-
49 6%3
91
x*.
8?/2
-
12?/
2y2-2ij -(50
7x-9^ + 4. =
Solve
16
+ y_x
]
-^
y
-{-
z
2
Simplify
99.
Find the square root 4 d^
An
^-^).
of
9 6^
12 ah
New York
express leaves If
x
- 6 /)c + 4 <7r + 4 «- + 9 c- — 12 ac
-
4- c^
and reaches Albany and arrives at New
at 3 p.m.
Albany
the ordinary train leaves at 6.
J
_^^. (-.-^^^
98.
York
+
+
?/^-7?/
....
d^b-^c^'
3
;
1
2X
^
2x-3// + 4^-5 =
100.
-
at 1.30 p.m.
both trains travel uniformly, find the time
when they
will meet.
101.
Solve
(i.)
.6x
+ .75x37
(ii.)
—
x'^
102.
Simplify
(i.)
^ (fi
(ii.)
.16
5X
+
^
+
+
X
=
.73-
2
3
.583x4-5.
4
,
X
6
7
—
—
X
— + -V^— —^+ + +
(1 4- X)2
-
ax^
-
I
ax
cfi
1
x"^
q,'^
4-
-x +
+
X
I
+
x^
Find the square root of «'
+
-^
- ef «4 + -\ +
15
id'-
+ -,) -20
also the cube root of the result.
Divide
l-2x
by l-t-3x
a-x^
"—^
+
1
104.
'^^
-''
1
103.
^^2).
6
X
at 6
6(792
Simplify .J. ^^'^
97.
_
and
?.a
X 96.
^2)^
:
615.
x
Solve
^pq +
4(p2
^^2)^
to 4 terms.
;
.
-\-x^
MISCELLANEOUS EXAMPLES
503
VII.
105. I bought a horse and carriage for $ 450 I sold the horse at a gahi of 5 per cent, and the carriage at a gain of 20 per cent, making on the whole a gain of 10 per cent : find the original cost of the ;
horse. 106.
8
(rt
Find the divisor when (4a2 + 7fl6 + 5&'')2 is the dividend, the quotient, and b'^ (9 « + 11 b)'^ the remainder.
+ 2 &)^
107.
Solve
5x
(i.)
-
(x
=
3)
-
2 (x
7).
x-2
(x-l)(x-2) 108.
If
x
=
prove that (x 109.
a
+ 5+
—
—
a)'^
C^-^^)' ,
(y
—
=
b)'^
Find the square root
b'^.
of
105]^_14xf_6x
49^
25 110.
Solve
^
ax
+
^^
(fi
111.
+
+
+ x'^
rt^
C. F.
^^
x (a*
+
«2^2
^4^
.)_
^
x'^
fir"^
and divide the difference by Find the H.
5
^ + - X - 12 2 ^ — 12^ ^. x2 + 7 X + 12
^-±-^ from X 12
x'^
112.
5
''-'' — ax + x^
—+ -—
Subtract
x-1
= ^L+-^ + -i^,
and2/
1 H
and
L. C.
M.
of
2x2+ (6a-10?>)x-30«6 and 3x'^- (9a + 156) x + 45a&. 113.
- abx + 2 abd = 4 cdx. 8x-l ^____2j\= (ii.) 2(x + 3)_ -" x2-9 4(x-3) = 1, b=2, c = 3, = 4, find the value of
Solve
2
(i.)
cx'^
^""
114.
6«
If
«
+
c''
fZ
+
d«
+(« +
&) (6
+
c)
U
^
^
&
cy
rode one-third of a journey at 10 miles an hour, one-third more at 9, and the rest at 8 miles an hour if I had ridden half the journey at 10, and the other half at 8 miles per hour, I should have been half a minute longer on the way what distance did I ride ? 115.
I
;
:
116.
The product
one of the factors a
+
=
is
two factors
of
x
—
y
:
If
118.
Resolve into factors
(i.)
x^
+
ij^
b
+ Sxy
l,
(x
+
y).
(3x
+
2?/)3
-
(2x
+
3?/)3,
and
find the other factor.
prove that (d^
117.
is
-
b'^y^
—
n^
=
a^
+
b'^
-
ab.
7i^)
+
:
(ii.)
m^
— m (m^ —
ii
(m — ny\
:
ALGEBRA.
504 119.
.
Solve
x^-y^ = 28\ + xy + y^ = 7 i'
(i.) x:^
120.
Find the square root of
121.
Simplify the fractions
(a- by -
(i-)
+
2 (a2
-6)2 +
62) (a
«
+
-
Find the H. C. F.
123.
A village had two-thirds of
—
h'^c
124.
Solve
^
of
abc
—
ah"^
and ax2
+
a6
—
«2
_
53^2,
voters Republicans
its
and 60 w^ent over to the Democrats How many voters were there altogether
Simplify
X
y
(i.)
U
^
+
(a
-
2x1
\y
(X
'^
l)x3
cc2_3ic-l-a
+ yl±^.=J^\ ^ 2yz
(x
+ 1)3 + \y -
(X
t+Jt^.
-
2xy
)
1)3 1)4"
-
+
1) x2
24
?/2
(2a
+
(a2
+ 4a -
5)
a;
+
3a
+6
+ 2.
x2
+
5
a:?/
-
+
128.
Find the square root of p2
129.
Solve
130.
Find the H.
+ (4a -
X
-
4 3
_ 3g
(ii.)
y.
x^
to three terms.
= ^-^. (1.)^-*^ a:-6 x-1 x-2 x-Z (ii.)
131.
-
(a:
Ix \
I
Resolve into factors (i.)
3 xi
?
Divide X*
127.
elec-
a + 6 6(i-±) = = 2. +
,...
by
an
the voters
?
\
126.
;
+
(ii.)5
125.
in
:
2«^
_^+(«-6)6 «
(i.)
lf'
^
tion 25 refused to vote,
were now equal.
= 9\
¥).
^lLA
l
122.
+
+
2 (a*
A
(ii.)
+
«
lly"^
x-3y =
^^-1
«2-
a^h
-6xy +
(ii.) x'^
2 6)x
ax+\ =
hy
-\-\
=ay +
hx.
C. F. of
- 2 a6 +
a2
and
a;3
+ (2 a -
h)x''-
-(2ab- a'^)x - a^b.
Simplify
M! ^ W! ^ M!:.
(ii.)
xM My - 4-
MISCELLANEOUS EXAMPLES VIL
505
At a cricket match the contractor provided dinner for 27 perand fixed the price so as to gain 12^ per cent upon liis outlay.
132.
sons,
Six of the cricketers being absent, the remaining 21 paid the fixed and the contractor lost $ 3 what was the charge for the dinner ?
price for their dinner,
133.
Prove that
:
+
a;(y
2)
+
-
+-
y
'X'
= -^l—
X
equal to
is
= ^^-^.
and y
2/+1 134.
Find the cube root
2
of
l^-i? + l.
x3-12x2 + 54x-112 +
.
a, if
\^j
135.
Find the H. C. F. and L. C. M.
136.
Simplify
+
x^
(i.)
^
"^
2
rtx^
|
36
+
138.
Solve
(i.)
(ii.)
is
Sx-^y
^
+
2 aofi
a'^x
3x-2y
^g|
— 2 a^.
2x-3y }
d^-^b'^'
18
9 52^3) i^to four factors.
ab'^)
^^ =
_^l(_:^70, x + y x^- z
—
^
.
X
a
—
^
.
X
b
the same whatever value x 140.
—
x^
- (32 a^ + SVS^ - l=V75x-29. +
cfi{x^
*K,
_^i-
84,
y
=
140.
z
-h
that the difference between
^ X
+2 a^ and
a -'6b
«
Resolve 4
Show
aP-x,
4x-3y
42
137.
139.
+
%Aj
of
Multiply x^
+
2
y'^
+
—c may 3 z^
and
-A_ + _^+ X
—
a
—
x
c
x
b
—
c
have.
by x^
-
2 y^
-
3 z^.
141. Walking 4| miles an hour, I start 1| hours after a friend whose pace is 3 miles an hour how long shall I be in overtaking him ? ;
142.
Express in the simplest form
/ 9« 2
(i.)
143.
(83
_3
3
+
42) X 16
I
^.
(ii.)
X
yy
32
X
—
3-n
^;^-^
Find the square root of
y
.
yx
1 )
- 27«
ALGEBRA.
506 144.
Simplify
^'[x-l '
145.
\a'
-2ay -^y-
Find the value
x'^
a-y
'
\
i
+
oy^
+
a^
+
y^
Solve
'
147.
Show
+ bY - c3 + b)— c 2(« +
+ c)3 - a^ +c—a + c)^ + «2 +
equal to
b
5
Divide
a*
+
149.
Find the square root
of
150.
(« How much
by
1)4
+
2(a4
2
VSS + U^'G.
b)
c
+
&2
+ ft)^ - b^ +a—6 ^2.
+
a^ _ x^. - 2(a2 +
1)
are pears a gross
the price five cents a dozen
when
1) (a
12
-
more
1)2.
for a dollar lowers
?
Show that if a number of two digits is six times the sum of its number formed by interchanging the digits is five times their
digits, the
sum. 152.
Find the value of 1
(a 153.
1
1
-b)ib-
r)
(b
-
c)(«
-
c)
(c
-
a){b
-
^O'
Multiply
3^, ^_12 + 4103 + 86.2 ^y^_^^^26.r-8x2_u — 4
154.
If
ic
-i=
1
,
+
7
prove
X
o
tliat
a;^
+ — = 3,
x 155.
y^
-x + 4 a^x^ — 4 a^x^
148.
a
(c
(?)
(a
151.
+
that
(g
is
ay
h)
-\-
(ii.)
-
cfi
(ii.)
x-b_x,::jc_^ 2(a X — a X — b X —{a 2x + Zy = n
(i.)
1
of
VS + V50 - VT8 + \/48.
(i.)
146.
x-^lj' x^+l'
Solve
and
a:^
x^
x'^
(i.)
(ii.)
^^ + ^^^
= l(x + =
2a-2
-3?/2
2xy
-Sy^=
4
23-)
3
J
b).
;*:
-—=
4.
— MISCELLANEOUS EXAMPLES
507
VII.
liV20-3v5-Vi- (ii.)^(^)^^.
156.
Simplify
157.
Find the H.C.F. of (p^
(i.)
y
p{p + 168.
Reduce
to
+(3p _ i)x - p{p -(p2 _ 2p - l)x-{p - 1).
1)0:2
_
^_i
-A iW-i)'
5x2
y^
y
Find the square root of
A
-
1
(i.)
22H+1
+
42"-
on the same day X
=
161.
If
162.
Solve
2
morning
+ V2,
(i.)
V^ + y/x -
= V^ ^x
« h
Find the product
of
Resolve 9 x^if-
i/?«
o-
"^
-
576 y-
(x
:
vT+ic -
..'2
7/2
(c- ?>)(«-
+
(a-c)(6-c)
6)
,
3
V5-7 5+V7
1
-
i x^
+
256 x2 into six factors.
(i.)
7(r+s) ^
((1^^^)^"+^^^/
21
I
^f
x?/2
^4(m2-«2)/J
•
' («~i)--p (ii.)
168.
VU - V132.
Find the H. C. F. and L. C. M. 20x4
+ x2-
3^
vr^^
(x + a)^ x(x + 2a) ^—^^' ^t(x - ^-^7-^ ^^rr? (x2 - a^) (x + a)2 a) + a)
^"^w + n^L Simplify
VM^ + vT:^ ^
Hi.)
—
/• X
(1.) ^ ^
+ i..
iV^, i^2, ^80, ^5, and divide
V5 + 165.
^^
+
^^
8-4V5
167.
What time should it indicate it may be right at 7.15 p.m.
find the value of x^
Simplify
V* Smiplify ^
+ 4".
6"
.
?
«2
166.
2
in order that
(h-a^ic-a) 164.
-
9'»
(ii.)
clock gains 4 minutes a day.
at 6 o'clock in the
163.
and
x^+ — y\^ Wij-iy^
+y_^
160.
1)
simplest form
its
ax
159.
1)^2
1,
25xi
•
of
+ 5x3^x-
1,
25x4-
10x2
+
1.
+
ALGEBRA.
508 169.
Solve
a
(i.)
(ii.)
x
-\-
+
+ V2 ax +
x
x^
=
b.
1
.11
9f
8
7
170. The price of photographs is raised $ 3 per dozen, and customers consequently receive ten less than before for $ 5 what were the prices charged ? :
+
=
-^
prove that ^3
171.
If f
172.
Find the value of
rt
3,
_|.
1 _ q.
x-{-2a_^x-2a_^ 4ab ^^en 2b - X 2b -j-x x:- -4 bReduce
173.
^ "^
^
174.
^
,'^\
ij
z]
f^
x-^-4:
\
\x^
x
+
ij-
+
+ sl\
-
z-^
-
xy
-
zx
x
+
y
\
+
z)
x-^6j
X-4:
+
^2+2-
(16)^
^L_ + _iL_.
Solve
(i. )
(4)-^
(ii.)
-
56 V3.
^^^ + ^-^^ - ^+^ + ^-±^. — — — — x
(ii.)
Sa
X
4a
x
4a
a;
3a
3x2+ xy + 3y'' = Sl 8x2-3x^ + 8^2^171
Find the square root of
^^^'^
a2w 178.
1
yz
Simplify (i.)
177.
b
Express as a whole number
(8)-3
176.
+
X-\-2J -^ z
( '
(27)3
175.
«^ a
to fractions in their lowest terms
.'^
U
=
x
+ 2 aW + + 2 a'»x" +
54^4^
x2«
Simplify
what 179. A boat's crew can row 8 miles an hour in still water the speed of a river's current if it take them 2 hours and 40 minutes to row 8 miles up and 8 miles down ? :
is
MISCELLANEOUS EXAMPLES 180.
If
=
a
—
x'^
—
yz, h
—
a^
— zx^ c = z^ — xy, = x{ax + by cz).
]p-
be
509
VII.
prove that
-{-
Find a quantity such that when it is subtracted from each c, the remainders are in continued proportion.
181.
of the quantities a, b,
Simplify
182.
(i.)
+
y
2(7
a;
fx
^^^^
^x
1
x-\-y (ii)
-4)
x
^^6x2-7x + 2
729
-
+
2916 x^
4860
-
x'^
4320 x^
—
Simplify
'-^= +
(i.)
+ Vx^ —
X
Solve
(ii.)
-
+
64 x^'^
1
7
^9.
= x.
cc2?/2
6-x
+
i92
= 28a:?/1 =S i
Simplify b
—
c
Solve
G
,
(i.)
(ii.)
xy
=
ab{a
cc
-
2(x
+
15|
+
If
189.
Find the H. C. F. of
_
3«
6)
(4a2 Multiply
+
_
+
——
X
a
,
x^
-
+ (a^ +
V2^ + \/2(2a;-
= 6. +
?/2
=
= ^. aS
+
^3^
prove that
a)
7
a
1)
^
\/2x
__£_4. V2(2a:- 1)-V2a;.
V2x
b
15|
x?/
b)\b
—
c2-(a-6)2
+ 2)a; - a2 _ + i)a;2_(4(u2 + 2a):K + a2,
2)x2
4rt
a
i/--^)=S(x-^-y)
and
188.
(2 a2
—
52_(c-«)-
Va
by
576 x^^
^-^—
a2_(^,_c)2
190.
x^
+ ^192 -
512
x-\-y
and
2160
5
(i.)
6
187.
'^x^-l'
'
6
186.
+
— Vx^ —
x
1
^16 + ^81 - V-
(ii.)
185.
2(4a;-l)
10
Find the sixth root of
183.
184.
-
(Syfi-x-2,
2 a
510
ALGEBRA.
191.
Divide
+
ci^h^
by
d^b 192.
ft^c^
+
c^a~
4- 6'^c
+
c^a
- a^h^ - h^-c^ - cH'^ — ah- — hc^ - caP:
Simplify 1
lOic-1
6(a:-l)
3(xHx+l)
r
(i.)
2(x+l) Hi
+ Vx — v'-'«
)
)
«
Vx +
_
Va;
a
ft
«
Vx^ -
^
^
\/(x
>
+
a)'^
g^
—
ax
lip be the difference between any quantity and its reciprocal, q the difference between the square of the same quantity and the square 193.
of its reciprocal,
show
that
194. A man started for a walk when the hands of his watch were When he finished, the coincident between three and four o'clock. hands were again coincident between five and six o'clock. AVhat was the time when he started, and how long did he walk ?
n be an integer, show that
195.
If
196.
Simplify
V
'^ ~i)
+
^ 197.
Find
7-"+i
V~
+
198.
+
If
«
c.ay^
8.
^r(-^)' :;
the value of
7 -3V5 7 + 3V5 ^^ 7-3v5 7 + 3V5' ^1+^ + Vl - X _i (ii.) Vl +ic - VI -X
4(«6
always divisible by
1 is
+ 6 + c + = 2 s, prove - (rt^ + 52 - a^ - d-y^ = (Z
2 6 ^'^
+
1
a) (s
-
6) (s
that
10(s
-
-
c) (s
- d).
A man buys a number of articles for $5, and sells for $5.40 but two at 5 cents apiece more than they cost how many did he
199. all
;
buy? 200.
Find the square root of 2(810:'*
201.
{he 202.
If
X
+
ca
If
a
:
a
+
::
+ y
?/*)- 2(9x2 :
h
ahy\x^
man
he saved ^ 20 the to % 1700 ?
z
::
+
:
c,
?/
+
y2^(^Sx
- yY + (Sx-
yy.
prove that
+
z^)
= {hz +
ex
+
ayY^a^
+
^2
+
c2).
save $ 10 more than he did the previous year, and if year, in how many years will his savings amount
first
MISCELLANEOUS EXAMPLES
—
203. Given that 4 is a root of the quadratic x^ the vaUie of q and the other root.
204.
A
person having
an hour after the
511
VII. 5x
+
(/
=
miles to walk increases his speed one mile and finds that he is half an hour less on
7
mile,
first
the road than he would have been had he not altered his rate. long did he take ? 205.
If {a
0, find
+h + c)x = {-a + + J>
1111 —=
show that
- H
1
Z
IJ
={a-h + c)z = {a + h-c)w,
i')ii
W
How
•
X
206. Find a Geometrical Progression of which the sum two terms is 2|, and the sum to infinity 4i.
of the first
xV 1
207.
Simplify
^
'^" v)
^
^
A
man has a stable containing 10 stalls 208. could he stable 5 horses ? 209.
In boring a well 400 feet deep the cost
is
27 cents for the first is the cost
and an additional cent for each subsequent foot what boring the last foot, and also of boring the entire well ?
foot of
how many ways
in
;
;
210.
If a, h
are the roots of
x'^ -\-
px
+
q
= 0, show
that p^ q are the
roots of the equation
+ (a +
x2
-
6
Extract the square root of
212.
If
213.
^-^i^ y
=- = X
^— z-y
If a, &, c are in
f3 \a
,
H. P.
7
,
show
c J \c
h
Find the number the letters of the words
0.
determine the ratios x
h
a
_9
&2
I
of permutations
Consequences,
:
y.
z.
that
+ 3_2W3^3_2N
214.
(i.)
- ab{a + 6) = — 30 V— 2.
ah)x
211.
^25^ rtc
which can be made from
(ii.)
all
Acarnania.
and find the (2 a — ?yx^^ numerically greatest term in the expansion of (1 -f x)'^^ if x = |, and 215.
?i
=
Expand by
the Binomial
Theorem
7.
216.
When
x
=
—
,
find the value of
4
l+2x 1+Vl + 2x
1-2 X l-Vl-2ic
;
ALGEBRA.
512 Simplify
217.
(«
+
——-^ +
vyv
-r {h
—
c){b
+ 2)2 =
a:^
-
c)
a)
{c
—
a){c
—
h)
Solve the equations
218.
(i.)
219.
r— — 6)(a — {xP-
-
5X
5
a:
+
22.
Prove that {y
-
2)3
+ (x -
2/)3
+
-y)(x-
3(0:
z)(u
-z) = (x-
zy.
and 5 vowels, how many words can be formed each containing 4 consonants and 2 vowels ? 220.
221.
Out
of 16 consonants
If h
—
—
that d 222.
c is
a is a harmonic mean between c - a and d a harmonic mean between a — c and b — c.
how many ways may
In
2 red balls, 3 black,
223.
The sum
series is 36,
tively
:
of a certain
and the
find the
and
first
number
number
of terras of
last of these
terms are
1
and
;
in
how
an arithmetical and 11 respec-
and the common difference
of terms,
show
a,
white, 2 blue
1
be selected from 4 red, 6 black, 2 white, and 5 blue many ways may they be arranged ?
—
of the
series.
224.
Expand by (i.)
226.
V5 227.
-— y.
Solve
225.
3
[2
the Binomial
Simplify
;,
given
6
By
-^ x
Theorem (ii.)
(1
-
f
x^
—
=
35.
xy
-y^ -^y =
15.
xy
15\/21
-i-
x
^5Vg
x)^ to
five terms.
^^^ ^^^ ^^^ ^^^^^
^^
4 Vis 3V27 7V48 that V^ = 2.2-36.
the Binomial
Theorem
find the cube root of
128 to six
places of decimals.
There are 9 books, of which 4 are Greek, 3 are Latin, and in how many ways could a selection be made so as to include at least one of each language ? 228.
2 are English
229.
;
Simplify
^45^
(i.)
- VSO x^ + a — X
.... x^ + x'^ ^"'Mx^'-X+l j
-x~^ \ [ x^ + 2 X + X + 1)"( X3-1
x^ X2
VSS .
^
- 2X X^+l
x^
^
\ ^
MISCELLANEOUS EXAMPLES 230.
Form
(i.)
the quadratic equation whose roots are 5
the roots of prove that p2 _ 4 ^ _ ] If
(ii.)
231.
Solve x3
232.
Find
518
VII.
+
=
1
— j>x
x'^
—
q
-\-
=
y/6.
0.
810/^
+
x^
?/);
+
=
x
9(y^
+
1).
Vl28, log2 ^ and having given .3010300 and log3 = .4771213,
logie 128, log4
=
log2
±
are two consecutive integers,
;
find the logarithm of .00001728.
A
233.
A
B
and
from the same point,
start
B
five
days after
A
;
day, 2 miles the second, 3 miles the third, and so on B travels 12 miles a day. When will they be together ? Explain the double answer. travels
mile the
1
first
;
Solve the equations
234.
sum
term
=
(ii.)
s^
= =
:
8j'+i, ?/2^,
= S^-K = 2 X 4^,
9y
2^
common
+
x
y
+
z
=
16.
an arithmetical
of the first 10 terms of
of the first 5 terms as 13 is to 4
to the
series is to
find the ratio of the first
;
difference.
Find the greatest term
236.
X
2^
The sum
235.
the
(i.)
in the expansion of {1
—
x)''^
when
if.
237. Five gentlemen and one lady wish to enter an omnibus in which there are only three vacant places in how many ways can these places be occupied (1) when there is no restriction, (2) when one of the places is to be occupied by the lady ? ;
238.
(i.)
Given log 2 = 301030, log3 = 477121, and log7 = 845098, .
.
find the logarithms of .005, 6.3, (ii.)
Find
239.
If
.'K
from the equation
P and
Q
.
and (2^)^. 188-4x ,= (54^2)3^-2.
vary respectively as
1 y"^
i
and y^ when z
is
constant,
and as z'^ and z'^ when y is constant, and if x = P + ^, find the equabetween x, y, z it being known that when y z= z =64, x = 12; and that when y = 4 z = 16, x = 2. tion
;
240.
Simplify log.y_3
number
+
2
logi^
-
log-VV-
+
logiVV
permutations of n things 4 at a time is to the number of combinations of 2 n things 3 at a time as 22 to 3, find n. the
241.
If
242.
If -
a metic 9,nd
mean
+-=
of
——
—
prove that 2 6 is either the aritha 2b —c between 2 a and 2 c, or the harmonic mean between a
c.
2l
c
2b
1
,
:
ALGEBRA.
iS14
243.
;
"Cr denote the number of combinations of n things taken
If
r together, prove that
244.
Find
the characteristic of log 54 to base
(i.)
logio (.0125)-^".
(ii.)
=
Given logio2 245. its
.30103, logio3
Write the
+
(r
(iii.)
=
the
3.
number
of digits in
3*5.
and express
it
.47712.
l)th term of (2 ax^
—
x^)^^
in
simplest form.
At a meeting
were 9 speakers and 4 for the negative. In how many ways could the speeches have been made, if a member of the affirmative always spoke first, and the speeches were alternately for the affirmative and the negative ? 246.
of a debating society there
5 spoke for the affirmative,
247.
Form
the quadratic equation whose roots are
a
+
6
+ Va- +
b-'
^
and «
248.
A
+
6
point moves with a speed which
same
?>2
different in different
is
speed in any mile varies inversely as the number of miles travelled before it commences this mile. If the second mile be described in 2 hours, find the time taken miles, but invariable in the
mile,
and
-'^^
+ y/ci^ +
its
to describe the 7ith mile.
249.
Solve the equations (i.)
x\h -
(ii.)
250.
{x^
c) + ax{c - a)+a\a -b)=0. — px + p") (qx 4- pq + p^) = qx^ + p'^q^ +
VS
Prove by the Binomial Theorem that
infinity, of
3
3^
3.5.7
is
p^.
the value, to
._
44.84.8.12 251.
and B run a mile race. In the first heat A gives B a start and beats him by 57 seconds in the second heat A gives of 81 seconds and is beaten by 88 yards in what time could
A
of 11 yards
B
a start each run a mile
A
;
:
?
meets with an accident which proceeds at three-fifths of its former rate and arrives 3 hours after time but had the accident happened 50 miles farther on the line, it would have arrived 1 ^ hours sooner firicl the length of the journey. 252. detains
it
train, an hour after an hour, after which
starting,
it
;
:
MTSCELLANEOirS EXAMPLES Expand
253.
for 4
515
VTI.
terms by the method of Undetermined Co-
2
efficients
2x3
A body
men were formed
into a hollow square, three deep, that with the addition of 25 to their number a solid square might be formed, of which the number of men in each
254.
when
of
was observed
it
would be greater by 22 than the square root of the number of each side of the hollow square required the number of men.
side in
men
:
255.
Expand
256.
Solve the equation
257.
Separate
into a series
7 -^ r2 4- 99 r
to
<
y/a'^
V2x -
+ 1
h'^.
+V3a:-2 = \/4x-3+ Vbx-4:.
^
-I-
partial fractions. ^
i»^to
+ 3)(x2-l) equation x* — 5 x^ — 6 x — (x
= 0.
258.
Solve the
259.
Find the generating function of 1 + 5x+ 7x2 + 17x3 + 31x4
260.
ox-x - 4 (X2+ l)(x2 -X-2)
Separate
+
5
+
...^
.^^^ partial fractions.
=
+
261.
Solve the equation x^
262.
Express |f | as a continued fraction and find the fourth con-
3 x^
16 x
60.
vergent. 263.
"What
264.
The sum
the
is
to 65 times the
sum
of n terms of the series
of 6 terms of the series
sum
to infinity
;
Convert 2y'5 into a continued fraction.
266.
Find
limits of the error
267.
Sum
to infinity the series
-
268.
Find value
269.
Solve,
X
-
of
2x2
-
when ^if
16x3
-
-^ J- J-
is
taken for v'23.
28x4
-
676x5
-i-
J-
x3
-
Solve the equation x* is
2
?
equal
+
....
J--...
by Cardan's Method, the equation .30
X
+
133
= 0.
Solve the equation x3 — 13x2 + 15x that one root exceeds another root by 2. 271.
•••
3+2+1+3+2+1 +
270.
one root
64,
find x.
265.
3
8, 27,
1,
1— xV— 1— x2+...is
+ V—
3.
— 4 x^ +
8x
+
+
35
189
=
0,
= 0,
having given
having given that
-
272.
Sum
273.
Solve the equation x*
274.
Solve the equation
to infinity the series
4
-
9
a-
+
16
-
.r'^
25 x3
275.
Solve the equation 2 x^
276.
Given log2
=
.30103,
=
Lo
_
6^
Separate (x-
49 x^
2
9a:
x^
and
+
+
36
=
8.X
+
1
=
0.
12 x 16a:
3x-8
(i.)
(ii.)
is
2_9,
in the
of 1
Solve the equations
2
V^ + x/5^ =
Find the general term when x2— 4x— ing powers of x. 280.
+
+ 2 = 12 x^ + 12x2. = .47712, solve the equations
(ii.)
279.
expanded
:
— Vx — y y/x-\- y + Vx-y \/2 xMH; + V2 x2 -
Vx'-\- y
X-
1
=
-f
y-
x^
—
4
X-*
10x3
/
+
40x2
form
of a
partial fractions,
=
65.
^
\/3-2.x2 (iii. )
0.
x
log 3
6-x.
•• ••
X
3
12 X
1
+
+
+
72
47 x2
+ ;^ ^ L _1 3+ 2+ 3+ 2+ x3 + 7 x2 - X - 8 into + X + l)(x-- 3x-l)
Find the value
(luadratic surd.
-
36 x*
+
6x4-2
4x
+ x+
(i.)
+
12 x^
.r
278.
:
ALGEBRA.
516
277.
4
+ 9x-
36
= 0.
in ascend-
Date Due
mM Sme'-' V
a>9e*^-
'
AUTrH^l
^ iCnight ^^
TITLE
aementarv- Alr;ebra
QA HI 67
^
