A college algebra

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m^M,^M.^M^M^

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Eflgin. Lifcrary

1903

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Eflgin. Lifcrary

1903

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COLLEGE ALGEBRA

BY

G. A. WENT WORTH

Author of a SicAncs of Tbxt-Books in Mathematics

REVISED EDITION

BOSTON, U.S.A.
GINN & COMPANY, PUBLISHERS

1903

4

CJOPYRIGHT, 1888, 1902, by

,^.

G. A. WENTWOKTH

ALL KIGIITS BESBRVED »

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PREFACE

This work, as the name implies, is interide'^ for colleges and
scientific schools. The first part is simply a re 'iew of the prin-
ce ciples of Algebra preceding Quadratic Equatio^8, with just enough
^ examples to illustrate and enforce these princip^ •^. By this brief
treatment of the first chapters sufficient space allowed, without
^ making the book cumbersome, for a full discussion of Quadratic
Equations, The Binomial Theorem, Choice, Chanr^e, Series, Deter-
minants, and The General Properties of Equatioiis. Every effort
has been made to present in the clearest light each "ubject discussed,
and to give in matter and methods the best traiuing in algebraic
analysis at present attainable. Many problems and sections can be
omitted at the discretion of the instructor.

The author is under great obligation to J. C. l shan, LL.D.,
Ottawa, Canada, to Professor J. J. Hardy, Ph.D., Laf.' '*^tte College^
Easton, Pa., and to W. H. Butts, A.M., Michigan Uni ersity, Ann
Arbor, Mich., who have read the proofs and given valuable sugges-
tions on the subject-matter.

Answers to the problems are bound separately in paper covers,
and will be furnished free to' pupils when teachers apply to the
publishers for them. ....

Any corrections or suggestions relating to the work will be
thankfully received.

G. A. WENTWORTH.

ExBTBB, N.H., May, 1902.

1 ■

iii

139399

TABLE OF CONTENTS

CHAPTER I

SBCTIONS PAGES

1-33. FUNDAMBNTAL IdEAS 1-13

CHAPTER II
34-72. The Elbmbntart Operations . , 14-37

CHAPTER III
73-98. Factors 38-65

CHAPTER IV
09-106. Symmetry 66-63

CHAPTER V
106-116. Fractions 64-70

CHAPTER VI
117-130. Simple Equations 71-79

CHAPTER VII
131-136. Simultaneous Simple Equations 80-88

CHAPTER VIII
137-rl63. Involution and Evolution 89-99

CHAPTER IX ^

164-170. Exponents 100-llC

CHAPTER X
171-183. Quadratic Equations 111-1'

IV

TABLE OF CONTENTS V

CHAPTER XI

IXOTIOVI PAOM

184-186. SlMULTANBOUB QUAPRATIO EQUATIONS . . . . . 134-146 ^

CHAPTER XII
187-100. Equations solved as Quadratics 147-164

CHAPTER Xin
191-200. Properties of Quadratic Equations .... 155-163

CHAPTER XIV
201-220. Surds and Imaoinaribs ......... 164-176

CHAPTER XV
221-223. Simple Indeterminate Equations . . ^ . . . 176-181

CHAPTER XVI
224-228. Inequalities . 182-183

CHAPTER XVII
220-264. Ratio, Proportion, and Variation . . . . . 184-201

CHAPTER XVIII
266-284. Progressions . 202-218

CHAPTER XIX
286>295. Binomial Theorem ; Positive Integral Exponent 210-225

CHAPTER XX
206-320. Logarithms 226-243

CHAPTER XXI
821-335. Interest and Annuities 244-252

CHAPTER XXII
336-363. Choice 263-275

vi TABLE OP CONTENTS

CfiAPTE^R XXin
S64-87i. Chancb ............... 276-298

CHAPTER XXIV
37^-387. . Variables, and Limits 299-307

CHAPTER XXV
388-439. Series . . 308-366

CHAPTER XXVI
440-462. CoNTiNDEix Fractions. ... . » 367-377

CHAPTER XXVII
463-467. Scales OP Notation. . . . . . . . . . . 378-383

CHAPTER XXVIII
468-476. Theory of Numbers .384-390

CHAPTER XXIX
476-506. Determinants . . . . ... . . ... . . 891-414

CHAPTER XXX
.606-663. -Genbral Properties of Equations 416-466

CHAPTER XXXI
664-684. NuMERiOAL Equations » 467-492

CHAPTER XXXn
686-698. Gbnbra-l Solution, of Equations . . . ... 493-609

CHAPTER XXXIII
6^9-617. Complex Numbbrs . . .... 610-630

COLLEGE ALGEBRA

CHAPTER I

FUNDAMENTAL IDEAS

1. Magnitude, Quantity, and Number. Whatever admits of
increase or decrease is called a magnitude. Every magnitude
must therefore admit of comparison with another magnitude of
the same kind in such a way as to determine whether the first
is greater than, less than, or equal to the other.

A measurable magnitude is a magnitude that admits of being
considered as made up of parts all equal to one another.

To measure any given measurable magnitude, we take as
standard of reference a definite magnitude of the same kind
as the magnitude to be measured and determine how many
magnitudes, each equal to the standard of reference, will
together constitute the given magnitude.

A quantity is a measurable magnitude expressed as a mag-
nitvde actually measured. Hence, the expression of a quan-
tity consists of two components. One of these components
is the name of the magnitude that has been selected as the
standard of reference or measurement. The other component
expresses how many magnitudes, each equal to the standard
of reference, must be taken to make up the quantity. The
standard magnitude is termed a unit, and the other component
of the expression is termed the numerical value of the expres-
sion. Hence,

A unit is the standard magnitude employed in counting any
collection of objects or in measuring any magnitude.

1

2 COLLEGE ALGEBRA

A number is that which is applied to a unit to express how
many parts, each equal to the unit, there are in the magnitude
measured.

The endless succession of numbers one, two, three, four, etc.,
employed in counting is called the natural series of numbers.

2. In the statement James walked 12 miles, the number of
miles is actually stated, and the 12 is therefore called a knoixm
number, or it is said to be explicitly assigned.

In the statement If from five times the number of miles
James walked, ten is subtracted, the remainder will be fifty,
the number of miles, though not directly given, may be found
from the data to be twelve and is therefore said to be implied
in the statement, or it is called an implicitly assigned number,
or more commonly, an unknown number.

In the statement If from the double of a number six is
subtracted, the result will be the same as if three had been
subtracted from that number and the remainder doubled, the
number to be doubled is assigned neither explicitly nor implic-
itly, since the statement is true for any number whatever.
A number of this kind, which may have any value whatever, is
called an arbitrary number. Arbitrary numbers are frequently
called known numbers, as they are often assumed to be known,
though not definitely assigned.

3. Numbers explicitly assigned are represented in Algebra,
as tliey are in Arithmetic, by the numerals or figures 0, 1, 2,
3, 4, 5, 6, 7, 8, 9, and combinations of these. Each figure or
combination of figures represents one and but one particular
number. Numbers implicitly assigned and arbitrary numbers
are usually represented by the letters of the alphabet. The
first letters of the alphabet, as a, b, c, etc., are generally used
to represent arbitrary numbers, while z, y, x, w, etc., commonly
represent unknown numbers.

4. When any letter, as x, is used in the course of a calcu-
lation it denotes the same number throughout. We may also

FUNDAMENTAL IDEAS 3

represent different numbers by the same letter with marks

afi^xed.

Thus, instead of writing a, 6, c for three different numbers, we may
represent these numbers by the symbols ai, o^, as (read a sub-one, a aub-
two, etc.), or by a\ a^', a'*^ (read a prime, a second, etc.)-

5. In Arithmetic the figures that represent numbers are
generally themselves called numbers ; and, similarly, in Algebra
the symbols that stand for numbers are themselves called
numbers. Letter-symbols are called literal expressionsy and
figure-symbols numerical expressions.

The number which a letter represents is called its value,
and if represented arithmetically , its numerical value.

6. In elementary Algebra we consider all quantities as
expressed numerically in terms of some unit, and the symbols
represent only the purely numerical parts of such quantities.
In other words, the symbols denote what are called in Arith-
metic abstract numbers.

7. An algebraic expression is the expression of a number in
algebraic symbols.

8. Certain words and phrases occur so often in Algebra that it
is found convenient to represent them by easily made symbols.

Symbols of Relation.

= , read equals, is equal to, will be equal to, etc.

=^, read is not equal to, etc.

>, read is greater than, thus 9 > 4.

<, read is less than, thus 4 < 9.

:, ::, the signs of proportion, as in Arithmetic.

Thus, a:b ::c:d, ot a:b = c:d, is read a is to h as c is to d.

Symbols for Words.

.'., read therefore, consequently, hence,

'.', read because, since.

Thus, •.• a = 6, and 6 = c ; .*. a = c, is read since a eqtioUs b, and b
equals c; ther^ore a equals c.

4 COLLEGE ALGEBRA

•••, the symbol of continuation, is read continued by the
same law.

Thus, 1, 2, 3, 4, . . . means that we are to continue the numbers by the
same law ; Xi, Xs, Xs? * " i ^n means Xi, a^, Xs, 24, Z5. <and so on to x^.

9. Signs of Operation. The principal operations of Algebra
are Addition, Subtraction, Multiplication, Division, Involu-
tion, Evolution, and Logarithmation. A mark used to denote
that one of these operations is to be performed on a number
is called a sign of operation. These signs of operation will
now be explained.

10. The sign of addition is + (vQdAplus). As in Arithmetic,
it denotes that the number before which it stands is an
addend.

Thus, a + h means that & is to be added to a ; so that if a represents
6 and h represents 4, a + 6 represents 6 + 4, which is 10. a-\-b ■\- c
denotes that 5 is to be added to a, and then c added to their sum.

The sum of two or more numbers is expressed by writing
them in a row with the sign -\- before each of them except the
first number,

11. The sign of subtraction is — (read minus). As in Arith-
metic, placed before a number it denotes that that number is
a subtrahend, *

Thus, a — h (read a minus 6), indicates that the numbei 'represented
by 6 is to be subtracted from the number represented by a ; so that, if a
represents 6 and h represents 4, a — 6 is equivalent to 6 — 4, which is 2.

Hence, to indicate that a number is to be subtracted from
another number, as a from a?, write the subtrahend after the
minuend with the sign — between them.

The expression a -}-b — c denotes that b is to be added to a,
and then c subtracted from the sum. a — b — c denotes that
b is to be subtracted from a, and then c subtracted from the
remainder.

FUNDAMENTAL IDEAS 5

12. Numbers to be multiplied together are called factors^
and the resulting number is called the prodiict of these factors.
Multiplication is indicated in two ways :

1. By a sign of operation. 2. By position.

The signs of multiplication are x and • (read intoy times, or
mtUtiplied by).

Thus, 3*4*5, or3x4x5 indicates the continued product of the three
factors 3, 4, and 6. In like manner, a • &, or a x 6, indicates the product
of the factors a and h.

If all the factors or all but one are represented by letters,
the signs of operation, x and • , are generally omitted ; this
method is called indicating multiplication by position.

Thus, five times a is vnritten 5 a (read^ve a), and f of the product of
m and z is written f mz.

A number which multiplies another number is called a
coefficient of that number. A coefficient (literally, co-factor) is
therefore simply a multiplier, numerical or literal.

Thus, in the expression 5 amXf

6 is the numerical coefficient of amx,
6 a " literal *» »* ma;,

6 am »* *» " ** X.

If no numerical coefficient is tvritten, unity is understood as
the actual numerical coefficient.

13. The sign of division is -f- (read divided by), and denotes
that the number immediately following it is a divisor.

Thus, a-^h (read a divided by h) means that a is to be divided by h.
If a represents 12 and h represents 4, a -^ & represents 12-^4, or 3.

Division is also indicated by arranging the numbers in the
form of a fraction with the dividend for numerator and the

divisor ioi denominator.

a (XX

Thus, a -*- 6 may be written -; az-^by may be written -~»

b by

This method is called indicating division by position.

6 COLLEGE ALGEBRA

14. In an expression such as 7aX'\-5ci/ — Sdz (read seven
ax plus five cy minus three dz) the multipliecUions are to be

performed before the additions and subtractions,

ax by cz

In an expression such as h -^ the multiplications

m, n q

and divisions are to be performed before the additions and
subtra^ctions, so that in this expression the quotient of ax bj
m is to be increased by the quotient of by by n, and the sum
diminished by the quotient of cz by q,

15. A power of a number is the product obtained by using
that number a certain number of times as a multiplier, starting
with unity as first m^ultiplicand. . The operation of forming a
power is called inyolation ; the number used as a multiplier is
called the base of the power ; the number of successive multi-
plications by the base is called the degree of the power ; and
the number indicating the degree of the power is called the
exponent or index of the power and is written in small char-
acters to the right and a little above the line of the base.

ThuB, 1 X a X a is represented by a* (read a »quare)'y here a is the
base^ 2 is the exponent (or index), and a* is the second pou>er of a.

I'C'C'C \s represented by c> (read c cube); here c is the base, 3 is the
exponent, and the number c> is the third power of c.

In x^ (read x to the fifth), z is the base, 5 is the exponent, and the
namber x^ is the fifth power of z.

Since the exponent denotes how many multiplications by the
l)ase are to be made, the first to be performed on unity, it fol-
lows that a^y the first power of a, represents 1 x a, or simply a.

Hence, also, a^, the zero power of a, denotes that no multi-
plication by a is to be made, or, in other words, that the unit-
multiplicand is not to be multiplied by a. Therefore a^ = 1 for
any value of a whatsoever.

16. In writing a power at full length as a product it is
usual to omit the imit-multiplicand, just as it is usual to omit
a unit-coefficient where such occurs.

Thus, instead of writing x« = 1 x x x x x «, we write sfi = x xx x x.

FUNDAMENTAL IDEAS 7

In this method of expressing the value of a power the expo-
nent denotes the number of times the base is taken as a factor.

17. Comparing powers, the second power is said to be higher
than the first, the third higher than the second, etc.

18. In an expression such as 4 a^b* -^ c* (read 4 a square
b euhe divided by c square) the involutions are to be performed

• before the multiplications and divisions,

19. Inyolution is the operation of forming a power by taking
the same number several times as a factor.

Evolution is the inverse of Involution, or the operation of
finding one of the equal factors of a number. A root is one of
the equal factors. If the number is resolved into two equal
factors, each factor is called the square root; if into three
equal factors, each factor is called the cube root ; if into four
equal factors, each factor is called the fourth root ; and so on.

The root sign is V* Except for the square root, a number-
symbol is written over the root sign to show into how many
equal factors the given number is to be resolved. This number-
symbol is called the index of the root

Thus, V64 means the square root of 64 ; V64 means the cube root of 64.

20. Logarithmation is the operation of determining the index
or exponent which the given base must have in order that the
resulting root or power may be equal to a given number. The
index or exponent is called the logarithm of the given niimber
to the given base.

Thus, if a and b are given numbers and a** = &, n is called the loga-
rithm of 6 to the base a.

21. PositiYe and Negatiye Numbers. There are quantities
which stand to each other in such an opposite relation that,
when combined, they cancel each other entirely or in part.

Thus, six dollars gain and six dollars loss just cancel each other ; but
ten dollars gain and six dollars loss cancel each other only in part. For
the six dollars loss will cancel six dollars of the gain and leave four dollars
gain.

8 COLLEGE ALGEBRA

An opposition of this kind exists in assets and d^ts, in
motion forwards and motion backwards, in motion to the right
and motion to the left, in the rise above zero and the fall below
zero of the mercury of a thermometer.

From thiis relation of quantities a question often arises
which is not considered in Arithmetic ; namely, the subtract-
ing of a greater number from a smaller. This cannot be done
in Arithmetic, for the real nature of subtraction consists in
counting backwards along the natural series of numbers. If
we wish to subtract 4 from 6, we start at 6 in the natural
series, count four units backwards, and arrive at two, the
difference sought. If we subtract 6 from 6, we start at 6 in
the natural series, coimt six units backwards, and arrive at
zero. If we try to subtract nine from six, we cannot do it,
because, when we have counted backwards ias far as zero, the
natural series of numbers has come to an end,

22. In order to subtract a greater number from a smaller, it
is necessary to assume a new series of numbers, beginning at
zero and extending backwards. If the natural series advances
from zero to the right, by repetitions of the unit, the new
series must recede from zero to the left, by repetitions of the
unit; and the opposition between the right-hand series and
the left-hand series must be clearly, marked. This oppositiou
is indicated by calling every number in the right-hand series
2^ positive number, and prefixing to it, when written, the sign 4- ;
and by calling every number in the left-hand series a negative
number, and prefixing to it the sign — . The two series of
numbers will be written thus :

4,-3,-2,-1, 0, + 1, + 2, + 3, 4- 4, . . .

I I I .. I I 1 1 \ \

and may be considered as forming but a single series consist-
ing of a positive portion or branch, a negative portion or
branch, and zero. The complete series thus formed is called
the scalar series.

FUNDAMENTAL IDEAS 9

If, now, we wish to subtract 9 from 6, we begin at 6 in
the positive branch, count nine units in the negative direction
(to the left), and arrive at — 3 in the negative branch. That
is, 6 - 9 = - 3.

The result obtained by subtracting a greater number from a
less, when both are positive, is always a negative number.

If a and b represent any two numbers of the positive branch,
the expression a — b will denote a positive nulhber when a is
greater than b ; will be equal to zero when a is equal to b ; will
denote a negative number when a is less than b.

If we wish to add 9 to — 6, we begin at — 6 in the negative
series, count nine units in the positive direction (to the right),
and arrive at -h 3 in the positive branch.

We may illustrate the use of positive and negative numbers

as follows :

-5 0 8 20

-1 1 1 1

DA C

Suppose a person starting at A walks 20 feet to the right of A,
and then returns 12 feet, where will he be ? Answer : at C, a point
8 feet to the right of A, That is, 20 feet - 12 feet = 8 feet; or,
20 ~ 12 = 8.

Again, suppose he walks from A to the right 20 feet, and then returns
26 feet, where will he now be ? Answer : at D, a point 6 feet to the
left of A. That is, if we consider distance measured in feet to the left of
A as forming a negative series of numbers, beginning at ^, 20 — 26 = — 6.
Hence, the phrase, 6 feet to the left of ^, is now expressed by the nega-
tive number — 5.

23. Numbers with the sign -|- or — are called scalar numbers.
They are unknown in elementary Arithmetic, but play a very
important part in Algebra. Numbers regarded without refer-
ence to the signs + or — are called absolute numbers.

Every algebraic number, as -h 4 or — 4, consists of a sign -|-
or — and the absolute value of the number ; in this case 4.
The sign shows whether the nimiber belongs to the positive
or the negative series of numbers ; the absolute value shows

10 COLLEGE ALGEBRA

what place the number has in the positive or the negative
series.

When no sign stands before a number the sign -h is always
understood.

Thus, 4 means the same as + 4, a means the same sua + a.

But the sign — is never omitted.

Two numbers which have, one the sign -\- and the other the
sign ^f are said to have unlike signs.

Two numbers which have the same absolute values, but
unlike signs, always cancel each other when combined.

Thus, +4-4 = 0, +a-a = 0.

24. Meaning of the Signs. The use of the signs + a^id — ,
to indicate addition and subtraction, must be carefully distin-
guished from their use to indicate in which series, the positive
or the negative, a given number belongs. In the first sense
they are signs of operations and are common to both Arith-
metic and Algebra. In the second sense they are signs of
opposition and are employed in Algebra alone.

25. When an expression is made up of several parts con-
nected by the signs -h, —, each of these parts taken with the
sign immediately preceding it (-h being understood if no
wi'itten sign precedes) is called a term.

Thus, a + 6 — c + d + e consists of the five terms +a, +6, — c, +(i, -fe.

A term whose sign is -\- is called a positive term ; a term
whose sign is — is called a negative term.

An expression which consists of but one term is called a
monomial or simple expression.

An expression which consists of two or more terms is called
a polynomial or compound expression.

A polynomial of two terms is called a binomial, A poly-
nomial of three terms is called a trinomial. Polynomials of
three or more terms are sometimes called multinomials.

PtlNDAMENTAL IDEAS 11

26. If two terms differ only in one having the sign + and
the other the sign — , they are called complementary terms.

Thus, +6 and —b are complementary terms in the expression a+b—b;
so — c and + c are complementary terms in a — c + 6 + c.

27. The degree of a term is the number of literal factors it
contains, and each literal factor is called a dimension of the term.

Thus, 3 a^c^ is of seven dimensions.

This term, a^b^c^, is said also to be of two dimensions in a,
of two dimensions in b, and of three dimensions in c.

The dimensions of a polynomial are determined by the
dimensions of its highest term.

Thus, 1 + a2 + 3 abc is of three dimensions because its highest term,
3 abc, is of three dimensions.

A polynomial is said to be homogeneous when all its terms
have the same dimensions.

Thus, x^ + S x2y 4- 3 zy^ + y^ is homogeneous.

28. Like terms are terms that have the same letters, and
the corresponding letters have the same exponents.

Thus, 6a2&8, 3a268, - 7 a^b^ are like terms; but Sa^b and Sal^ are
unlike terms because, though they contain the same letters, the corre-
sponding letters do not have the same exponents.

29. If an expression contains any like terms, these may be
united, and the expression is said to be simplified.

Thus, as in Arithmetic, 2 dozen + 3 dozen = 5 dozen ; 2 times 8
+ 3 times 8 = 5 times 8 ; so in Algebra, 2a6 + 3a6 = 5a6;2 a^b^+S a^b^=z
6 a^¥.

Similarly, in the case of negative terms ; 6a5~3a6 = 2a6; 5 aVi^ —
3 a268 = 2 a^fts. Hence,

To reduce two or more like terms to a single equiyalent term,

Form the sum of the numerical coefficients of the positive
terms and also of the negative terms, then take the difference of
these sums, affix the literal parts and prefix to the result the sign
of those terms whose numerical coefficients give the greater sum.

12 COLLEGE ALGEBRA

Thus, in the expression 6a^b-7 ad^+^a^b-Qac'^-ia'^b-Qa^b+l^ac'^,
the sum of the coefficients of the positive terms in a^b is 8, and the sum
of the coefficients of the negative terms is 10 ; the difference of these is 2,
to which we affix the literal part a^b, getting 2 a^b ; and as the sum (10)
of the coefficients of the negative terms is the greater, we prefix the
sign — , getting ^2a^b; similarly, combining the terms in oc^, we get
+ 2 ad^j and the whole expression is simplified to — 2 d?h + 2 ac^^ or
2 ac2 - 2 aV).

30. The reciprocal of a number is 1 divided by that number.

Thus, the reciprocal of a is - ; the reciprocal of a^fts ig

a ' '^ a262

The product of any number and its reciprocal is unity.
Thus, 6 X - = 1.

0

Hence, a divisor may be replaced by its fractionally expressed
reciprocal as a multiplier. If, for example, the product of a
and ft is to be divided by m, and the quotient divided by w,
this may be represented by

oo -^ m -^ w, or by a6 X — X — , or by

•^ m n mn

31. Compound Expressions. Every algebraic expression^ how-
ever complex, represents a number and may be treated in any
operation as a single symboL If an expression is to be so
treated, it is generally enclosed in brackets ; or a line called a
vinculum is drawn over it.

Thus, 7 + (8 — 3) denotes that 3 is to be subtracted from 8 and the
remainder added to 7.

7 — (8 — 3) denotes that 3 is to be subtracted from 8 and the remainder
subtracted from 7.

7 • (8 — 3) or 7 • 8 — 3 means that 3 is to be subtracted from 8 and the
remainder multiplied by 7.

Similarly, suppose a4-6 — cistobe operated on as a single symbol ;
tben,

X -\- {a -\- b — c) denotes that the number is to be added to a;,

SB — (a + 6 — c) " ** " " subtracted from as,

z{a + b-c) ** ** ** ** multiplied by X,

FUNDAMENTAL IDEAS 13

(a + 6 — c) -7- X denotes that the number is to be divided by ic,
(a + b-c)^ " " '* *» cubed,

V(a + b — c) ** " cube root of the number is to be

extracted.

32. An expression that has a part enclosed in brackets may
itself be enclosed in brackets to form part of a longer expres-
sion ; this again may be enclosed in brackets to form part of
a still longer expression; and so on to any extent. When
several paii's of brackets are thus employed it is usual to make
each pair different from the others in size or shape.

1. If it were required to multiply a -\- b into the sum of a (x + y) and
b{x — y), the result would be expressed thus,

{a + h) {a{x -{- y) + b{x - y)},

2. 10 a — [5 6 — {4 c + 2 (3 6 — a)}] denotes that a is to be subtracted
from 3 h, that the remainder is to be doubled, that the product is to be
added to 4 c, that the sum is to be subtracted from 5 &, that the remainder
is to be subtracted from 10 a.

Hence, if a = 15, 6 = 6, and c = 1, we have

10a -[66 -{4c + 2(36 -a)}]

= 150 - [30 - {4 + 2 (18 - 15)}]
= 150 - [30 - {4 + 2 X 3}]
= 160 - [30 - {4 + 6}]
= 150 - [30 - 10]
= 160-20
= 130.

33. Substitution. Two quantities, two numbers, or two oper-
ations are equal if either can be substituted for the other in
algebraic expressions without changing the values of the
expressions. From this it follows at once that

Numbers that are equal to the same number are equal to
one another.

In symbols : If a = c and & = c, then a = ft.

CHAPTER II

THE ELEMENTARY OPERATIONS

34. The introduction of negative numbers requires the mean-
ings of addition, subtraction, multiplication, and division to be
made wider and more comprehensive in Algebra than they are
in Arithmetic, but these enlarged meanings must be consistent
with the older arithmetical meanings, and the elementary
operations when thus generalized must still conform to the
fundamental laws which govern these operations in Arithmetic.
We now proceed to state these fundamental laws and to explain
these wider meanings.

ADDITION

35. In Algebra, as in Arithmetic, numbers which are to be
added ai'e called addends, and the result of the addition is
termed the sum of the addends ; but it must be borne in mind
that in Algebra under the term numbers are included not only
the numbers indicated by single letters but also those whicli
are the arithmetical values of compound algebraic expressions,
just as in Arithmetic numbers are expressed either by single
digits or by combinations of digits.

Addition is the operation of combining two or more numbers
or algebraic expressions into a single nimiber or expression
according to the following laws :

I. If equal numbers are added to equal numbers the sums
are equal.

If the sum of one pair of addends is equal to the sum, of a
second pair, and either addend in the first pair is equal to the

14

THE ELEMENTARY OPERATIONS 15

corresponding addend in the second pair, the remaining addend
in the first pair is equal to the remaining addend in the
second pair,

II. The sum of two addends is the same, whether the
second addend is added to the first, or the first addend is added
to the second.

ITT. The sum of three addends is the same, whether the
sum of the second and third addends is added to the first, or
the third addend is added to the stem of the first and second.

IV. Adding zero to any number leaves the number unchanged.

36. These laws expressed in algebraic symbols are :
Suppose a, b, c, 2Lnd d have each one and only one value,

zero being a possible value for any one or more of them.

I. If a = c and b = d, then a -\-b = c -\- d.
li a = c and a -\- b = c -}- d, then b = d,
lfb = d and a -\- b = c + d, then a = c.

Hence^ Addition is completely uniform.

II. a -\- b = b -i- a.

Proposition II is expressed by,
Addition is commutative.

III. a-\-(b-\-c) = (a-\-b)-\-c.
Proposition III is expressed by,
Addition is associative.

IV. a 4- 0 = a.
The modulus of addition is zero.

37. Cor. 1. (a + c)-\-b = a -\-(c + b) (III)

^a+(b + c) (II)

= (a + b) + c. (Ill)

Hence, adding any number to an addend adds an equal
number to the sum.

16 COLLEGE ALGEBRA

Cor. 2. If a + & = a, then & = 0. (IV and I)

Zero is the only addend whose addition to a number leaves
the number unchanged,

38. An algebraic number which is to be added or subtracted
is often enclosed in brackets, in order that the signs + and —
which are used to distinguish positive and negative numbers
may not be confounded with the -|- and — signs that denote
the operations of addition and subtraction.

Thus, -f 4 -f (— 3) expresses the sum of the numbers + 4 and — 3 ;
and + 4 — ( — 3) expresses that — 3 is to be subtracted from + 4.

39. Monomials. In order to add two algebraic numbers, we
begin at the place in the scalar series which the first number
occupies and count, in the direction indicated by the sign of
the second number, as many units as there are units in the
absolute value of the second number.

4,-3,-2,-1, 0, +1, +2, +3, +4,...

I \ \ \ \ \ \ I I ^

Thus, the sum of + 4 + (+ 3) is found by counting from + 4 three
units in the positive direction and is, therefore, + 7 ; the sum of
+ 4 + (— 3) is found by counting from + 4 three units in the negative
direction and is, therefore, + 1.

In like manner, the sum of — 4 + (+ 3) is — 1, and the sum of
_4 + (_3) is -7.

1. To add two numbers with like signs, find the sum of
their absolute values, and prefix the common sign to the sum.

2. To add two numbers with unlike signs, find the differ-
ence between their absolute values, and prefix to the difference
the sign of the number that is the greater in absolute value.

Thus, (1) + a + (+ 6) = a + 6 ; (3) - a + (+ 6) = - a + 6 ;
(2) + a + (- 6) = a - 6 ; (4) - a + (-&) = - a - 6.

By successive application of the above rules we readily obtain
rules for adding any number of terms.

THE ELEMENTARY OPERATIONS 17

Thus^ 4a + 5a + 3a + 2a=14a;

-3a-16a-7a + 14a-2a£=14a-27a=-13a;
4a-36-9a + 76 = -6a + 46.

40. Polynomials. Two or more polynomials "are added by
adding their separate terms.

It is convenient to arrange the terms in columns, so that
like terms shall stand in the same column.
Thus, 2 a* - 3 a^ft + 4 062 + 53

a8 + 4a26-7a62-268

-3a8+ a26-3a62_468

2a8 + 2a26-i-6a62-358

2a8 + 4a26 -868

Addition in Algebra does not necessarily imply augmenta-
twriy as it does in Arithmetic.
Thus, 7 + (-5) = 2.

The word sum, however, is used to denote the result.

Such a result is called the algebraic sum, when it is necessary
to distinguish it from the arithmetical sum, which would be
obtained by adding the absolute values of the numbers.

Exercise 1

Add:

1. 9a^ + 3a + 45, 2a2-4a + 55, 5a-2b-6a\

2. 7x^-2xy-{-y% 4:xy-2y% Sx^ - 9xy -{- 12y\

3. 7 a^b + 9 a5« - 13 ftS Sa^-\-2aP-7 b\ ab^ -a%-Q a»,
5^,8 _ 7^8 - ab\ 4:b^- 2a» + a%.

4. 5x^ -{-2x^ — 7, 4a;« + a; - 9, 1 + a; — x^,

a.6 + a;* - a;» - a;2 - 7, 9 x^ + 9 x"^ - 12 x - 4:X^ -^ 10.

5. 3m*4-2m»7i + 5mV-9< 7 n^ - 3 mn^ - S m^n^,

11 m/i' — 4 m^n^ + 6 wi*n, 5 m* -{- 2 m^n — 15 mn* — 7 n*.

6. 2a;«4-3a;V-4a;V, 2 y« - 3 a;?/* + 4 a;y - 10 a;y,
5xY -\-^xY - 9y^ Sa;^ -7 x^ + 6a;Y - 8a;y.

18 COLLEGE ALGEBRA

SUBTRACTION

41. Subtraction is the operation by which, when the sum of
two addends and one of the addends are given, the other
addend is determined. In symbols : Subtraction is the opera-
tion symbolized hj a — b, such that

(a — b)-{' b = a-,

and hy b — a, such that

a -{-(b — a) =b.

With reference to this operation, the sum is called the
minuend, the given addend is called the subtrahend, and the
required addend is called the remainder.

42. The laws of subtraction are not fundamental but are
derived from this definition combined with the laws of
addition. They are;

i. If equals are subtracted from equals, the remainders are
equal,

ii. Subtracting any number from an addend subtracts an
equal number from the sum,

iii. Adding any number to the minuend adds an equal mim-
ber to the remainder,

iv. Subt7*acthig any nuTnber from the minuend subtracts an
equal number from the reinainder,

V. Adding any number to the subtraheiid subtracts an equal
number from the remainder,

vi. Subtracting any number from the subtrahend adds an
equal number to the remainder. ^

43. These laws expressed in algebraic notation are :
Suppose a, 5, c, and d liave each one and only one value, zero

included as a possible value for any one or more of them.

THE ELEMENTARY OPERATIONS 19

If a = c and b = dy

a — b = c — d, (i)
(a — c)-{- b =(a -\-b) — c,

and a -f (6 — c) = (a + 5) — c. (ii)

(a -{- c) — b = (a — b)-\- c. (iii)

(a — c) — ft = (a — 5) — c. (iv)

a — (ft + c) = (a — ft) — c. (v)

a — (ft — c) = (a — ft) + c. (vi)

44. By definition, (a — ft) + ft = a.
Therefore, if ft = 0,

(a - 0) + 0 = a.
That is, a - 0 = a. (IV, p. 16)

Conversely, if a — ft = a, then ft = 0,

for in this case (a — ft) + ft = (a — ft), .

and therefore ft = 0. (Cor. 2, p. 16)

45. Monomials. In order to find the difference between two
algebraic numbers, we begin at the place in the scalar series
which the minuend occupies and count in the direction opposite
to that indicated by the sign of the subtrahend as many units
as there are units in the absolute value of the subtrahend.

Thus, when we subtract + 3 from + 4 we count from + 4 three units
in the negative direction, and arrive at + 1 ; when we subtract — 3 from
+ 4 we count from + 4 three units in the positive direction, and arrive
at + 7. In likt manner, + 3 from — 4 is — 7 ; — 3 from — 4 is — 1.

Hence,

1. Subtracting a positive number is equivalent to adding
an equal negative number.

2. Subtracting a negative number is equivalent to adding
an equal positive number.

20 COLLEGE ALGEBRA

To subtract one algebraic number from another,

Change the sign of the subtrahend and then add the suhtron
hend to the minuend.

Thus, (1) +a-(+&) = a-&; (3) - a-(+h) = -a-b:

(2) + a - (- 6) = a + & ; (4) - a - (- 6) = - a + 6.

46. Polynomials. When one polynomial is to be subtracted
from another place its terms under the like terms of the other,
change the signs of the subtrahend, and add.

From 4 as* — 3 x^y — xy^ + 2 y*

take 2x^ — x^y + 5 xy^ — Sy^.

Change the signs of the subtrahend and add :

4x8-3x2y- xy2^2y8
-2g8+ x%-5gy2 + 3y»

2X8-2X22/ - 6xy2 4. 5y8

In practice, instead of actually changing the signs of the
subtrahend we only conceive them to be changed.

47. Parentheses. Propositions III, p. 15, and ii, v, and vi,
p. 19, may be written

a-{-{-\-h — c)=^a-{-h — Cf
a — (+^H-o) = a — 5 — c,
a — (+6 — c) = a — ft-f-c,

and, therefore, by § 43, p. 19, and IV, p. 16, and Cor. 2, p. 16,

a+(— ^ + c) = a — 5 + c,
a— (— 5 + c) = a4-^ — c.

Hence, when the parenthesis enclosing a polynomial is pre-
ceded by a plus sign the parenthesis and plus sign may be
removed or omitted without making any change in the signs
of the terms of the enclosed polynomial other than inserting
the sign + before the first term if that term has no sign
expressed.

THE ELEMENTARY OPERATIONS 21

When a parenthesis enclosing a polynomial is preceded by
a minus sign the parenthesis and minus sign may be removed
if the signs of the terms of the enclosed polynomial are all
changed,

48. Expressions often occur with more than one parenthesis.
These parentheses may be removed in succession by removing
first the innermost parenthesis; next, the innermost of all
that remain, and so on.

Thus, a-[h-{c + {d-e -/)}]

= a-[b-{c + (d-e+f)}]

= a-[&-{c-f d-c-f/}]
=za — [h — c — d-\-e — /]
= a — 6 + c + d — c + /.

49. The rules for introducing parentheses follow directly
from the rules for removing them :

1. Any number of terms of an expression may be put within
a parenthesis, and the sign + placed before the whole.

2. Any number of terms of an expression may be put within
a parenthesis, and the sign — placed before the whole ; if the
sign of every term vnthin the parenthesis is changed.

Thus, a + &-c-d = (a-f&)-(c + d)

= a + (6 — c) — d
= a 4- (ft — c — d).

50. By II, p. 16, and ii and iv, p. 19,

a -\-h = h -{- a,
a — c-{-h = a-{-h — Cj
a — c — h = a — h — c.

Hence, the terms of any polynomial may be combined in any
order whatever.

Thus, a + 6 — c — d = a — d + (6 — c)

= a — c — (d — 6)
= — (c — 6) — (d — a), etc.

22 COLLEGE ALGEBRA

ExerciBO 2

1. From 4a4-55 — 3c take 2a-f-95 — 8c.

2. From 7x^-x^ + 4:X-2 take 2a;» -f- Sa;^ - 9a; + 8.

3. From 3 a» + 3 a»6 - 9a6* + 35«
take 2a«-5a»Z»-h7a«>2-9R

4. From J aJ + 4 a^ - f 52 + J a take a* - ^V ^^ + i a.

5. From 4 cc' — 6 cc^ 4- 8'a; — 7 take the sum of

8a;« + 7-8a;2 + 7a; and - 9a;« - 8a;2 + 4a; + 4.

Simplify :

6. 2-3a;-(4-6a;)-|7-(9-2a;)|.

7. Sa-(a-b-c)-2\a + c-2(b-c)\.

8. 4a-[3a-|2a-(a-^)| + 55].

9. [8 a - 3 I a - (J - a) I ] - 4 [a - 2 I a - 2 (a - 5) I 4- 5].

10. aj(y 4- «) + y [a; — (y + «)] — «[y — a;(« — a;)].

11. 2a;»(a;-3a)-2[2a;*-a2(a;2-a^]
- 3 a[a;« - 2a; |a* + a; (a - a;) ^ + a'*].

MULTIPLICATION

51. In Algebra, as in Arithmetic, numbers which are to be
multiplied together are called /actors, and the result of the
multiplication is termed the product of the factors. Under
the term numbers we include not only the numbers symbolized
by single letters but also those which are the arithmetical
values of compound algebraic expressions. In the case of two
factors, the factor which is to be multiplied by the other is
called the multiplicayid, and the factor by which the multi-
plicand is to be multiplied is called the multiplier.

THE ELEMENTARY OPERATIONS 23

Multiplication is the operation of combining two or more
numbers or algebraic expressions into a single number or
expression according to. the following laws:

I. If equal numbers are multiplied by equal numbers,
the prodiccts are equal.

If the product of one pair of factors is equal to the product
of a second pair, and if either factor in the first pair is equal
to the corresponding factor in the second pair and is not zero,
the remaining factor in the first pg,ir is equal to the remaining
factor in the second pair.

II. The product of two factors is the same whether the first
factor is multiplied by the second or the second factor is multi-
plied by the first,

III. The product of three factors is the same whether the
first factor is multiplied by the product of the second and third
or the product of the first and second foAstors is multiplied by
the third.

IV. Multiplying by unity leaves the multiplicand unchanged,
V. If the multiplier is zero, the product is zero.

52. These laws expressed in algebraic symbols are :
Suppose a, b, c, and d have each one, and only one, value,

zero included as a possible value for any one or more of them

except where noted.

I. K a = c and b = d, then axb = ex d.
If a = c^O, and axb = cxd, then b = d;

and if b =d=^0, and axb = cxd, then a = c.

These propositions are condensed into the single statement,

Multiplication is completely uniform for actual or non-zero
factors.

II. axb =ibx a.

Proposition II is expressed by.

Multiplication is commutative.

24 COLLEGE ALGEBRA

IIL a X (b X c) = (a X b)x c.

Proposition III is expressed by,
Multiplication is associative.

IV. axl = a.

The modulus of multiplication is unity.

V. a X 0 = 0.

The annihilator of multiplication is zero.

53. Cor. 1. Multiplying a factor by any number multiplies
the product by that number.

Cor. 2. If the product of two factors is equal to one of the
factors, the other factor is unity, the case of the product and
its equal factor both being zero excepted.

Cor. 3. If the product of two or more factors is zero, one at
least of the factors is zero.

Proofs of these corollaries are similar to the proofs in § 37,
p. 15.

54. The fundamental law connecting the operation of multi-
plication with the operations of addition and subtraction is :

VI. Multiplying the several terms of a polynomial by any
number multiplies the polynomial by that number.

In symbols : ad -^ bd — cd = (a -{- b ^ c)d.

Proposition VI is expressed by,

Multiplication is distributive, relative to addition and sub-
traction.

Hence, (a -^ b) (m + 7i) = a (m + n) -\- b (m -{- n) (VI, p. 24)

= (m + n)a -{- (m -{- n)b (II, p. 23)

= ma -{- na -{- mb '\- nb (VI, p. 24)

= ain -{- an + bm -{- bn. (II, p. 23)

THE ELEMENTARY OPERATIONS 25

Also, (a — i) (m — ?i) = a (m — 7i) — i (m — w) (VI, p. 24)

= (m — 7i) a — (m — w) 5 (II, p. 23)

= ma -- na — (mb — ri5) (VI, p. 24)

= ma — wa — mh + ^^ (vi, p. 19)

= am — an — bm -|- bn. (II, p. 23)

55. Law of Signs. Let (+ a) and (+ c) denote positive
scalar numbers whose product is -f ac, and (— b) and (— d)
denote negative scalar numbers, a, b, c, and d being the abso-
lute values of the numbers without reference to the relation
positive-negative, then

(+a) + (-J) = (+a)-(+J),
and (+ e) + (- d) = (+ c) - {+ d).

= \(+a)-i+b)n(+c)-(+d)\. (I,p.23)
By §64, l(+a) + (-b)\\(+c) + (-d)i

= (+aX+c) + (+a)(-d)

+ i-hX+e) + (-bX-a). [1]

Since (+ a) (.+ c) = (+ ac),

K+«)-(+*)IK+<')-(+'^l

= (+ ac) - (+ ad) - (+ Jc) + (+ 6<0

= (+ac)+(-a«f)+(-6c)+(+J««). [2]

Compare the right-hand members of [1] and [2], term by
term.

Since (+ a) x (+ c) = (+ ac),

then (+ a) X (— <^ = (— ad),

(-b)x(+c) = (-bc),
(-b)xi-d) = (+b<l).

Hence, the Urw of signs in multiplication,

Like sigaa give pltis ; unlike signs give minu».

26 COLLEGE ALGEBRA

The product of more than two scalar factors, each preceded
by the sign — , will be positive or negative, according as the
number of such factors is even or odd.

56. Index Law. The product of two or more powers of any
number is that number with an exponent equal to the sum of the
exponents of the several factors,

Eor, a"* X a" = (aaa • • • to m factors) (aaa • • • to n factors)

= aaaaaa • • • to (m + ^) factors

Similarly for more than two factors.

57. Monomials. The product of numerical factors is a new
number in which no trace of the original factors is found.

Thus, 4 X 9 = 36.

But the product of literal factors is expressed by writing
them one after the other.

Thus, the product of db and cd is expressed by dbcd^ and generally
the product of ae^* and be'* is a6c"» + '», for ae"* x6e» = ax6xe"»xc"
by the commutative law, =ax&xe"» + » by the index law.

Hence, to find the product of two monomials,

Multiply the coefficients ; affix all the literal parts, each with
an exponent which is the sum of its exponents in the separate
foAitors ; prefix the sign -f if the signs of the monomials are
alike, the sign — if they are unlike,

58. Polynomials. To multiply a polynomial by a monomial,
the distributive law, § 54, p. 24, may be applied, giving as rule :

Multiply every term of the polynomial by the monomial mul-
tiplier, observing the law of signs, § bb, p. 25.

To multiply a polynomial by a polynomial, we apply the
distributive law, as in § 54, p. 24, and obtain as rule :

Multiply every term in the multiplicand by every term in the
multiplier, observing the law of signs, § 55, p. 25.

THE ELEMENTARY OPERATIONS 27

59. In multiplying polynomials it is a convenient arrange-
ment to write the multiplier under the multiplicand, and place
like terms of the partial products in columns.

(1) Multiply 5 a - 6 ft by 3a -4ft.

6a - 66

3a - 46
15a2-18a6

-2006 + 246^
15a2- 3806 + 2462

(2) Multiply a* + ft^ + c* — aft — ftc — flwj by a -f- ft -f- c.

Arrange according to descending powers of a.
a2-a6-ac+62- be + c^

a + 6 -f c

a^ — a26 — a^c + a62 — a6c + ac*

H-a26 -a62-a6c H-6«-62c + 6c2

+ g^ - dbc-ac'^ + 62c - 6c2 -f c»

a8 -3a6c +68 +c«

Observe that, with a view to bringing like terms of the partial products
into columns, the terms of the multiplicand and of the multiplier are
arranged in the same order,

60. Detached Coefficients. In multiplying two polynomials
that involve but one letter, or are homogeneous (§ 23, p. 10)
and involve but two letters, we shall save much labor if we
write only the coefficients.

(1) Multiply 2a;« -h 4a; + 7 by a;» - 3a; + 4.

Since the z^ term in the first expression is missing, we supply a zero
coefficient. The work is as follows :

2 + 0+ 4+ 7
1-3+ 4
2+0+ 4+ 7
-6- 0-12-21

+ 8 + 0 + 16 + 28
2-6 + 12- 6- 6 + 28

Writing in the powers of x, the product is

2«» - 6a;* + 12x» - 6x2 - 6x + 28.

28 COLLEGE ALGEBRA

(2) Multiply a« -f- 2aaj« - 9a:« -f- 4a^ by x* - 2aaj — a\

Arranging by descending powers of x we have

-9x8H-2ax2 + 4a2x + a» and x«-2aaj~a?.

The work is as follows :

-9+ 2+4 + 1

1-2-1
-9+ 2 + 4 + 1
+ 18-4-8-2

+9-2-4-1

-9 + 20 + 9-9-6-1

Hence, the product is — 9 a^ + 20 ox* + 9 a^x' — 9 aW — 6 a*x — cfi.

61. Special Cases. The following products are of great
importance, and should be carefully remembered:

(a - b)» = a» - 2 ab 4- b* ;

(a + b)» = a^ + 2ab + b»;

(a4-b)(a-b) = a»-b«;

(a + b + c)^ = a* + b* + c* 4- 2 ab 4- 2 ac + 2 be.

The square of any polynomial may be immediately written
by the following rule :

Add together the squares of the several terms and ttviee the
product of each term into ea^h of the terms that follow it.

Also, {a ± ^»)8 = «» ± 3 a% + Zab^± h\

The double sign ± is read plus or minus, and signifies the sum
or the difference of the numbers between which it is placed.

62. Again, consider the product

(x + a) (aj + 6) = aj^ + (a -f h)x -|- ah.

The coefficient of x is the algebraic sum of a and h ; the
third term is the product of a and h.

Thus, (x + 3) (X + 7) = x2 + lOx + 21 ;

(x - 3) (X + 7) = x2 + 4x - 21 ;
(X + 3) (X - 7) = x2 - 4x - 21 ;
(X - 3) (X - 7) = x2 - lOx + 21.

\- ^! ■■'■ V-g

THE ELEMENTARY OPERATIONS 29

Exercise 3

Find the product of :

1. Sx'\-2y and 4 a — 5 y.

2. 2x^ — 5 and 4aj + 3.

3. 2x^4- 4a; -3 and 2 a;^ -h 3 a; - 4.

4. a;* + 2a;* + 4 and a;* - 2a;2 + 4.

5. a;* + 2 a;?/ — 3 y* and a;* — 6 a;?/ + 4 y\

6. 9a;* + 3a;2/-f-y^-6a; + 2y + 4 and 3a;-y + 2.

7. lla» + 46»-4a*(a-4ft) and a\b -{-Sa)-U^(a-{-b),

8. (a + by + (a- by and (a + by - (a - «>)2.

9. a; — 2y4-3« and x ^2y -^3z.

10. a;8 4-2a;2 — 4a; — 1 and a;» + 2a;2 — 4a; — 1.

11. 39<^^+''-^-54<^^-2y+i + 606Z^+3y a^jj(i 30(^2-^+2y

12. 24a;'»+2«-i__42a;2'»-8n + 2_^25a;2« + 3m-2 ^j^^ 25a;2-"'-2«.

13. a^-3a^-i + 4a^-2-6a^-3+5a^-''and2a8-a2 4.a.

14. a^n+\ _ ^n + i _ a» 4- a»-i and a» + 2 _ ^2 _ a + 1.

15. aP + 3aP-^ — 2aP-^ and 2aP + ^ + a^"*"^ — 3a^

DIVISION

63. Division is the operation by which, when a product and
one of its factors are given, the other factor is determined,
the given factor not being zero. In symbols : Division is the

operation symbolized by a -f- ft, or r-, or a : ft such that

(a -^ ft) X ft = a, or r- X ft = a, or (a:b)xb = a-,

and, as a consequence of law II, p. 23, such that

a
ft X (a -5- ft) = a, or ft X - = a, or ft X (a : ft) = a ;

in which a may have any value, and ft any value except zero.

80 COLLEGE ALGEBRA

In this operation the product is called the dividend; the
given factor the divisor ; and the required factor the quotient

64. The laws of division are not fundamental but are
derived from this definition combined with the laws of mul-
tiplication. They are :

i. If equals are divided by equals, the quotients are equaL

ii. Dividing a fcLctor by any number divides the product
by that number,

iii. Multiplying the dividend by any number multiplies the
quotient by that number, •-:

iv. Dividing the dividend by puy number divides the quotient
by that number,

V. Multiplying the divisor by any number divides the qtuh
tient by that number.

vi. Dividing the divisor by any number multiplies the quo-
tient by that number,

vii. If the quotient is equal to the dividend, the divisor is
unity,

viii. Dividing all the terms of a polynomial by any number
divides the polynomial by that number.

65. These laws expressed in algebraic notation are :

Suppose a, b, c, m, and n have each one and only one value,
zero included as a possible value for a, b, and c but not for
m and n.

If a = c and m = n,

then a -i- m = c-i-n, (i)

(a -5- m) X c = (a X c) -M?i,

and a X (c -5- m) = (a X c) -^ m. (ii)

(a X c) -V- m = (a -r- m) X c. (iii)

(a -5- n) -f- m = (a -^ m) -5- w. (iv)

THE ELEMENTARY OPERATIONS 31

a -i- (m X n) = (a -h m) -^ n. (v)

a -5- (m -5- n) = (a -T- m) X n. (vi)

If a -i-m = a, then m = 1. (vii)

a-hm + b-7-m — c-7-m=(a-{-b^c)-¥-m,

ah c a + h — 0 , ....

or 1 = (viii)

7n 7n ftn m ^ '

The fundamental law VI, p. 24, and law viii of this section
are both included in the single proposition :

Multiplication is compieteiT-.distributive relative to addition.

66. Since a x ft = + o^?

(— a) X ft = — aft,

ah

— ah

a X (— ft) = — aft,

(-a)x(-ft)= + aft,

— aft

-f-aft
.*. -^ — 7- = — a.
— 0

Consequently, the quotient is ^positive when the dividend
and divisor have like signs.

The quotient is negative when the dividend and divisor
have unlike signs.

67. Monomials. To divide one monomial by another,

Write the dividend, over the divisor with a line between them ;
if the expressions have common factors, remove the common
factors.

Thus,
Again,

26 ate
lO&cx

6a
"2c'

366c« 6«
30a6c 5a

o»

aaaaa
aa

= aaa = »•;

a«

aa

1 1

a*

aaaaa

aaa a*

32 COLLEGE ALGEBRA

_ - a"* aaa • • • to m factors

In general, — = — ■

a" aaa • • • to ti lactors

= aaa • • • to m — ti factors (if m > n),

or = 7 7-— — (if n > m).

aaa - "to n ^ m lactors ^

Hence, if a power of a number is divided by a lower power
of the same number,

The quotient is that power of the number of which the expo-
nent is the exponent of the dividend diminished by that of the
divisor.

If any power of a number is divided by a higher power of
tbe same number.

The quotient is expressed by 1 divided by that power of the
number of which the exponent is the exponent of the divisor
diminished by that of the dividend,

68. Polynomials by Monomials. When the divisor is a mono-
mial and the dividend a polynomial.

Divide each term, of the dividend by the monomial divisor ;
the required quotient is the sum of the partial quotients.

For since (a + & — c) X m = ma -\- mb — m^c,

,', (ma -f ^^ — "i^c) -i-m = a'{-b — c.

The signs are determined by § 66, p. 31. *

69. Division of Polynomials by Polynomials.

If the divisor (one factor) is a -{- b -{- c,

and the quotient (other factor) is n -^p -^ q,

C an -\-bn -\- en
then the dividend (product) is \ + ap + bp -\- cp

[ + aq -\-bq -\-cq.

The first term of the dividend is an, the product of a, the
first term of the divisor, by n, the first term of the quotient.

THE ELEMENTARY OPERATIONS 33

The first term n of the quotient is therefore found by dividing
an, the first term of the dividend, by a, the first term of the
divisor.

If the paxtial product formed by multiplying the entire
divisor by n is subtracted from the dividend, ap, the first
term of the remainder, is the product of a, the first term of
the divisor, by p, the second term of the quotient. Hence,
the second term of the quotient is obtained by dividing the
first term of the remainder by the first term of the divisor ;
and so on.

Therefore, to divide one polynomial by another.

Divide the first term of the dividend hy the first term of the
divisor.

Write the result as the first term of the quotient.

Multiply all the terms of the divisor by the first term of the
quotient.

Subtract the product from the dividend.

If there is a reniaindery consider it as a new dividend and
proceed a^ before.

It is of great importance to arrange both dividend and
divisor according to the ascending or the descending powers
of some common letter, and to keep this order throughout the
operation.

(1) Divide

22 a^ft* + 15 5* + 3 a* - 10 a»Z» - 22 a^« by a' -{- S b^ - 2 ab.

Arrange the dividend and divisor according to the descending powers
of a and divide.

3a* - lOa^h + 22a^l^ -22ab^ + 16¥\ a^ - 2 db -\- Sl^
3a*- 60864- Qg^b^ da^-^ab + blj^

- 4 a86 4- 13 a262 - 22 068

- 4a86 4- 8a262_i2a68

6a262_l0a68H-15&*
5a262_ 10068+ 15 &♦

84 COLLEGE ALGEBRA

The operation of division may be shortened in some cases
by the use of parentheses.

(2) Divide

x' -{■ (a -\- b -\- c) x^ -\- {ah -\- ac -\- he) x -\' abc by x -{- b.

x^ + {a + b + c)x^ + {ab + ac + bc)x + abc |a + &

x^ + { +b )x^ x^ + ia-^-c^z-hac

(a -{- c) x^ + (db + ac + be) x
(a +c)a^ + (fl6 +6c)g

acx + abc

OCX + abc

70. Detached Coefficients. In division, as in multiplication,
it is convenient to use only the coefficients when the dividend
and divisor are expressions involving but one letter, or homo-
geneous expressions involving but two letters.

Thus, the work of Example (1), § 69, may be arranged as followB :

3-10 + 22-22 + 1611 -2 + 3

3- 6+ 9 3-4 + 6

- 4 + 13-22

- 4+ 8-12

6-10 + 16
6-10 + 16

The quotient is 3 a^ - 4 a6 + 6 62.

71. Special Cases. There are some cases in division which
occur so often in algebraic operations that they should be
carefully noticed and remembered.

The student may easily verify the following results :

a* — h*

(1) ' ^ = a« + ab + b«.

^ ^ a— b

(2) ^^""f = a* + a^ + a^^ + a5» + 5*.
^ ^ a — h

In general, the difference of two like powers of any two
numbers is divisible by the difference of the numbers.

THE ELEMENTARY OPERATIONS 35

^ ^ a + b

(4) f5l±|! = a* - a^b + a'*' - <*«>• + b\

In general, the sum of two like odd powers of two numbers
is divisible by the sum of the numbers.
Compare (3) and (4) with (1) and (2).

(5) ^^ = x + y. (7) ^^* = a^« + a:V + a:y^ + y«.

(6) ^^=x«.y. (8) ^^ = x^^x^y-^xy^-y^

In general, the difference of two like even powers of two
numbers is divisible by the difference and also by the sum of
the numbers.

The sum of two like even powers of two numbers is not
divisible by either the sum or the difference of the numbers.

But when the exponent of each of the two like powers is
composed of an odd and an even factor, the sum of the given
powers is divisible by the sum of the powers expressed by the
even factor.

Thus, jc^ + ^ is not divisible by x + y, or by x — y, but is divisible by

The quotient may be found as in (3) and (4).
A factor of x** — ^ can always be found ; and a factor of
ajn _|_ yn Qg^j^ always be found unless n is a power of 2.

Thus, factors of x^ + y^, x* + y*, x^ 4- 2/®» etc., cannot be found.

Exercise 4
Divide :

1. (6 a^^c X 35 a^ftV) by (21 a'^b^c^ x 2 a^c").

2. 39 a^x^ + 24 a*a« + 42 a^x^ + 27 a*x^ by 6 a^«

3. 36aj« + 94aaj« + 52a^ + 8a* by 5aj + 2a.

36 COLLEGE ALGEBRA

4. a;* — 5 005* — a^x + 14 a* by x* — 3 oa; — 7 a\

5. 81a;* + 36a;V + 16^/* by Oa;* - 6a;y + 4y^

6. a;* + ft* - a^x* + 2 ftV by 0:2 + ^24. aaj.

7. a^-2b^ — Sc^ + ab + 2aC'{-7bc by a-b-\-Sc.

8. 4aj*-5a;V-8a;*-43^2_^4 + y*
by y^ + 2x^-2'-3xy.

9. 2a"»+i — 2a"+i-a"»+" + a*» by a" — 2a*

10. 625aj*~8l2/* by 5aj-3y.

11. a:8»4-2/*'* by x"" -^ y^,

125 64 ^ 5 4'

13. (a4-2ft)8 4-(ft-3c)» by a4-3(ft-c).

14. a"» — a"»+^4-37a"»+8 — 55a"»+* + 50a"»+*
by l-3a4-10al

15. 4 A^+i - 30 A^ + 19 hJ"^ + 5 A*-* + 9 A*-*
by A'-3 - 7 ^^-* + 2 A*-5 - 3 A*-«.

16. 6a;"'-'*+2 ^ a-m-n+i _ 22a;"*-'» + 19a;"»--'*--^ — 4a.m-»-8
by 3x^-'» — 4a:2-n^^i-«

72. Summary. The four elementary operations of Algebra
are performed subject to I, The Law of Uniformity; II, The
Associative Law; III, The Commutative Law; and IV, The
Distributive Law. The meanings of these laws have been
explained as occasion arose ; we here sum up the whole in
brief review.

1. From the number a and the number h there is deter-
mined by addition a definite number c which is expressed thus :

a -\-h -^c^ or c = a-\-h,

2. There is a determinate number which we name zero and
denote by 0, such that for every number a we have simultar

neously

a + 0 = a, and 0 + ^ = a.

THE ELEMENTARY OPERATIONS 37

3. There is a number which we name infinity and denote
by 00, such that for every number a we have simultaneously

a + 00 = 00, and oo -+- ^^ = oo .

4. If a and h denote given numbers, a not being infinity,
there always exists one and only one number x and also one
and only one number y, such that we have respectively

a-\- X = h, and y -\- a = h.

5. From the number a and the number h there is determined
by multiplication a definite number c which is expressed thus :

db = c, or c = ah,

6. There is a determinate number which we name unity and
denote by 1, such that for every number a we have simul-
taneously

a X 1 = a, and 1 x a = a.

7. For every finite number a we have simultaneously

a X 0 = 0, and 0 x a = 0.

8. If a and h denote given numbers, a not being zero and
b not being infinity, there always exists one and only one
number x and also one and only one number y, such that we
have respectively

ax — by and ya = b.

If a, by and c denote any numbers whatever, the following
laws of calculation always hold true :

9. a + (^ 4- c) = (a -f ^) 4- c.

10. a -f 5 = ft -f «.

11. a (be) = (ab) c.

12. ab = ba.

13. a (ft -h c) = aft -h «c-

CHAPTER III

FACTORS

73. Rational Integral Expressions. An expression is rational
if none of its terms contains indicated roots.

An expression is integral if none of its terms contains other
than positive integral powers.

74. Factors of Rational and Integral Expressions. By factors
of a rational and integral expression we mean rational and
integral expressions that will exactly divide the given expres-
sion.

•75. Factors of Monomials. The factors of a monomial may
be found by inspection.

76. Factors of Polynomials. The form of a polynomial that
can be resolved into factors often suggests the process of
finding the factors.

77. When the terms have a common monomial factor.

Resolve into factors 6 a" + 4 a^ + 8 a.
Since 2 and a are factors of each term, we have

Hence, the required factors are 2, a, and 3 a^ + 2 a + 4.

78. When the terms can be grouped so as to show a common
compound factor.

Resolve into factors ac — ad — be -\- hd,

ac - ad — be -f M = (ac — ad) — (6c — W)

= a(c — d) — h(c -- ^

=:{a-b){c- d).

Henoe, the required factors are a — b and c ~ d.

88

FACTORS 39

79. Square Roots. If an expression can be resolved into
two equal factors, one of the equal factors is called the square
root of the expression (§ 19, p. 7).

Thus, 16 x^y2 — 4 j-Sy x 4 x^y.

Hence, 4 xV is the square root of 16 afiy^.

The square root of a positive number may be either positive
or negative ; for

a^ = ax a, and a^ = (— a) x (— a).

Throughout this chapter the positive square root only will
be considered.

80. When a Trinomial is a Perfect Square. A trinomial is a
perfect square if the first and last terms are perfect squares
and positive, and the middle term is twice the product of the
square roots of the first and last terms (§ 61, p. 28).

Thus, 16 a* — 24 06 + 9 &2 is a perfect square.

To extract the square root of a trinomial that is a perfect square,

Extract the square root of the first term and of the last term
and connect these square roots by the sign of the middle term.

Resolve into factors aj^ — 18 aj + 81.

x2 - 18x + 81 = (X - 9) (X - 9) = (X - 9)2.
Hence, the required factors are x — 9 and x ~ 9.

81. When a Binomial is the Difference of Two Squares. The
difference of two squares is the product of two factors which
may be determined as follows :

Extract the square root of the first number and the square
root of the second number.

The sum of these roots will form the first factor.

The difference of these roots will form the second factor.

Thus, (1) a2 - &2 = (a + 6) (a - 6);

(2) (a - 6)2 - (c - d)2 = {(a - 6) + (c - (f)} {(a - 6) - (c - d)}

= {a - 6 -f c — d} {a — 6 — c + d}.

40 COLLEGE ALGEBRA

The terms of an expression may often be arranged so as to
form the difference of two squares, and the expression can
then be resolved into factors.

Thus, a^ + b^-c^-d^ + 2cU)-\-2cd

= a2 + 2a64-62-c2 + 2cd-€P

= (a2 + 2a6 + 62) _ (c2 _ 2c(i + cP)

= (a + 6)2 - (c - d)2

= {{a + 6) + (c - d)} {(a + 6) - (c - d)}

= {a + 6 + c - d} {a + 6 - c + d}.

A trinomial in the form a* + a%^ + b^ can be written as the
difference of two squares and resolved into factors.

Thus, X* + x2y2 + y4 = (x4 + 2aj2y2 + y4) - x2y2

= (X2 + 2/2)2 _ (ajy)2

= (x2 + 2/2 + X2^) (x2 + y2_x2^)
= (x2 + X2/ + 2/2) (x2 -xy + y2).

A binomial in the form cc* + 4 ?/* can be written as the dif-
ference of two squares and resolved into two factors.
Thus, 1 + 42/* = (1 +42/2 + 42/4) -4yi

= (1 + 2 2/2)2 - (2 2/)2

= (1 + 22/ + 22/2)(l - 22/ + 22/2).

Many expressions may be resolved into three or more factors.

Thus, x" - 2/^« = (x8 + 2/8) (x8 - 2/8)

= (x8 + 2/8) (X* + 2/*) (X* - 2/*)

= (X8 + 2/8) (X* + 2/4) (a;2 + y2) (3.2 _ y8)

= (x8 + 2/8) (x* + 2/4) (x2 + 2/2) (X + y)(z- y).

82. A Trinomial of the Form x^+ ax+ b, where a is the alge-
braic sum of two numbers and is either positive or negative,
and b is the product of these two numbers and is either posi-
tive or negative, can be resolved into factors.

Since (x -{- 5) (x -^ 3) = x^ -{- S x -\- 15,

the factors of x^ -\-Sx -^15 are a + 5 and x + 3.

Since (x + 5){x - 3) = x^ + 2x - 16,

the factors of cc^ + 2 a; — 16 are x -\- 5 and a? — 3.

FACTORS 41

Hence, if a trinomial of the form x^ -\- ax -{-b is such an
expression that it can be resolved into two binomial factors,
the first term of each factor will be x ; the second terms of the
factors will be two numbers whose product is b, the last term
of the trinomial, and whose algebraic sum is a, the coefficient
of X in the middle term of the trinomial.

(1) Eesolve into factors x^ -\-llx -\- 30.

We are required to find two numbers whose product is 30 and whose
sum is 11.

Two numbers whose product is 30 are 1 and 30, 2 and 15, 3 and 10,
6 and 6 ; and the sum. of the last two numbers is 11.

Hence, x2 + 11 jc -f 30 = (x + 6) (x + 6).

(2) Resolve into factors x^ — 7 x -{-12,

We are required to find two numbers whose product is 12 and whose
algebraic sum is — 7.

Since the product is -f 12, the two numbers are both positive or both
negative; and since their sum is — 7, they must both be negative.

Two negative numbers whose product is 12 are — 12 and —1,-6 and
-- 2, — 4 and — 3 ; and the sum of the last two numbers is — 7.

Hence, x^ - 7x + 12 = (x - 4) (x - 3).

(3) Resolve into factors a^ -f 2 a; — 24.

We are required to find two numbers whose product is —24 and
whose algebraic sum is 2.

Since the product is — 24, one of the numbers is positive and the
other negative ; and since their sum is + 2, the larger number is positive.

Two numbers whose product is — 24, and the larger number positive,
are 24 and — 1, 12 and — 2, 8 and — 3, 6 and — 4 ; and the sum of the
last two numbers is + 2.

Hence, x^ + 2x - 24 = (x + 6) (x - 4).

(4) Resolve into factors x^ ^Sx — 18.

We are required to find two numbers whose product is — 18 and
whose algebraic sum is — 3.

Since the product is — 18, one of the numbers is positive and the other
negative ; and since their sum is — 3, the larger number is negative.

42 COLLEGE ALGEBRA

Two numbers whose product is — 18, and the larger number n^ative,
are — 18 and 1,-9 and 2,-6 and 3 ; and the som of the last two
numbers is — 3.

Hence, x? - 3x - 18 = (x - 6) (x + 3).

Therefore, in general,

x^ + (a -^ b)x -\- ab = (x -{- a)(x + b)

whatever the values of a and b.

83. When a Trinomial has the Form ax^ + 2>x + c.

(1) Resolve into factors Sx^-22x-21.

Multiply by 8, the coefficient of x^, and write the result in the follow-
ing form :

(8x)a-22x8x-168.

Put z for 8x, sfl-22z- 168.

Resolve this expression into two binomial factors,

(z - 28) (z + 6). (§ 82, p. 40)

Since we have multiplied by 8, and put z for 8 x, we must reverse this
process. Hence, put 8 x f or z and divide by 8, and we have

(8x-28)(8x + 6)
8

As 4 is a factor of (8x — 28), and 2 is a factor of (8x + 6), we divide
by 8 by dividing the first factor by 4 and the second factor by 2.

(8x-28)(8x + 6)^

4x2 ^ '^ '

(2) Resolve into factors 2Ax^ — l^xy — 75 y*.

Multiply by 24, (24 x)* - 70 y x 24 x - 1800 y«.

Put z for 24 X, z2 _ 70 yz - 18OO yK

Resolve into factors, (2 - 90 y) (2 4- 20 y). (§ 82, p. 40)

Put 24x for 2, (24x - 90y) (24x -f 20y).

Divide by 6 x 4, (4x - 16y) (6x + 5y).

84. When a Binomial is the Sum or the Difference of Two Cabes.

From § 71, p. 34, ^' "^f = a^- o^ + ^^

a ■\-h

a* — b^

and — =^ a^ -\- ah ■\- b^,

a — b

FACTORS 48

/. a^ + b^=(a + h) (a^ - ah + ft«);
and a«-ft» = (a-ft)(a2 + a^-fJ2).

In like manner we can resolve into factors any expression
which can be written as the sum or the dijfference of two cubes.

(1) Resolve into factors 8 a« + 27 *«. V*i\^

8a« + 27 66 = (2a)8 + (362)8

= [2a + 362] [(2a)2 - (2 a) (3 62) + (352)2]
= (2a + 362) (4a2 - 6a62 + 96*).

85. When a Polynomial is the Product of Two Trinomials. The
following method is convenient for resolving a polynomial into
its trinomial factors :

Find the factors oi 2x^ — &xy -\-2y^ -\'l xz — byz + ^z\

1. Reject the terms that contain z,

2. Reject the terms that contain y.

3. Reject the terms that contain x.

Factor the expression that remains in each case.

1. 2x2-5xy +2y2_(aj_2y)(2a;-y).

2. 2x2+ 7xz-f322=:(x + 3z)(2x + z).

3. 22/2_6yz + 322 = (22/-32)(y -2).

Arrange these three pairs of factors in two rows of three factors each,
so that any two factors of each row may have a common term including
the sign.

Thus, 1. aJ-2y, x-f32, -2y + 32;

2. 2 X — y, 2 X + z, —y-\-z.

From the first row, select the terms common to two factors for one tri-
nomial factor :

X — 2 2/ + 3 z.

From the second row, select the terms common to two factors for the

other trinomial factor :

2 X — y + 2.

Then,
2x^ - 5xy + 2y2 + 7x2 - 6yz + 3z2 = (X - 2y + 32)(2x - y + «).

When a factor obtained from the first three terms is also a
factor of the remaining terms, the expression is easily factored.

Thus, x2 - 3xy + 2y2 _ 3x + 6y = (x - 2y) (x - y) - 3(x - 2y)

= (x-2y)(x-y-3).

44 COLLEGE ALGEBRA

THEORY OF DIVISORS

86. Theorem. The expression x — j is an exact divisor^ of
x° — y" when n is any positive integer,

Spce - x»-iy + x"-iy = 0, (5 23, p. 9)

Taking out x^-^ from the first two terms of the right side, and y from
the last two terms, we have

x» — y» = x*-i(x — y) + y(x»-i — y*-*).

Now X — y is au exact divisor of the right side, if it is an exact divigor
of X"-! — y»-i ; and if x — 2/ is an e?act divisor of the ri^t aide, it is
an exact divisor of the left side ; that is, x — y is an exact diyisor of
gjn _ yn ^it is an exact divisor of x*»— ^ — y^—\

Therefore, if x — y is an exact divisor of the difference of any ttoo like
• powers of x — y, it is an exact divisor of the difference of the next higher
powers of x — y.

But X — y 15 an exact divisor of x« — y* (§ 71), therefore it is an exact
divisor of x* — y* ; and since it is an exact divisor of x* — y^, it is an
exact divisor of x* — y^ ; and so on, indefinitely.

The method employed in proving this Theorem is called
Proof by Mathematical Induction.

87. The Factor Theorem. If a rational and integral expression
in X vanishes, that is, becomes equal to 0, when r is put far x,
then X — 1 is an exact divisor of the expression.

Given ax« -f 6x»-i -f • • • + ^ + A;. ' [1]

By supposition, ar^ + 6r»— ^ H [- hr -^k = 0, [2]

By subtracting [2] from [1], the given expression aasameB the form
a(x» — r^)^ 6(x»-i - r"-i) + \-h{x -r).

But X — r is an exact divisor of x" — r«, x«-i — r»-i, and so on. (§ 86)
Therefore, x — r is an exact divisor of the given expression.

Note. If x — r is an exact divisor of the given expression, r is an
exact divisor of k ; for A;, the last term of the dividend, la equal to r, the
last term of the divisor, multiplied by the last term of the quotient.

FACTORS 45

Therefore, in searching for numerical values of x that will make the given
expression vanish, only exact divisors of the last term of the expression
need he tried.

(1) Resolve into factors cc* + 3 cc^ — 13 aj — 15.

The exact divisors of — 16 are 1, — 1, 3, — 3, 5, — 6, 15, — 16.
If we put 1 for X in x^ -f 3 x^ — 13 x — 15, the expression does not
vanish. If we put — 1 for x, the expression vanishes.
Therefore, x — (— 1), that is, x + 1, is a factor.
Divide the expression by x + 1, and we have

x8 + 3x2 - 13x - 16 = (X -f 1) (x2 4- 2x - 15)

=z{x + l){x- 3) (X + 6).

(2) Resolve into factors x^ — 26x — 5.

By trial we find that the only exact divisor of — 5 that makes the
expression vanish is — 5.

Therefore, divide by x -f 5, and we have

x8 _ 26x - 6 = (x + 5) (x2 - 5x - 1).

As neither + 1 nor ~ 1, the exact divisors of — 1, will makex^ — 6x — 1
vanish, this expression cannot be resolved into factors.

Exercise 5

Resolve into factors :

1. 9ic* 4- 6a;8 4- 3a;2-f 2a;.

2. 2a^-Sa'^b'-14:a^ + 21ab.

3. 5x^-{-15x^y-4:xy^-12y\

4. d^x^ — b^xj/^ — ahx^ -h b'^cy\

5. aj3 + 8aj + 7. 12. a;^ - 14 a; - 176.

6. a;« - 17 a; + 60. 13. 81 a;* - 196 a^y.

7. x^-\-lx-l%, 14. 729a«-aj«.

8. a;2-2aj-24 15. ^4:x'^ ^ xy\

9. 9aj^4-30aj + 25. 16. (x^-^/V-y*-

10. 16 aj2 - 56 a; 4- 49. 17. (a« + 2 ft^)« - a^ft^.

11. a?« + aj-72. 18. (2»-3y)«-(aj-2y)^

46 COLLEGE ALGEBRA

19. 121 a;* - 286 a;^ + 169 y«.

20. a^-2ab + b^-x\

21. 49 a* - 15 a2j2 ^ 121ft*.

22. a^x^ + 14 abx + 33 b\

23. a;y + 23 a;z/« + 90 «2^

24. a^+a-132.

25. 8a2 4-14aft-15ft2

26. 6aj2 + 19icy-7y2.

27. 11 a»- 23 aft + 2*2.

28. a* + 64 ft*.

29. 2x^ — 5xy -{- 2y^ — xz — yz — z\

30. ex^-lSxy-^Qf-^-Uxz-lSyz + ez^

31. 2a;2^.5a;z/ — 3y2 — 4aj« + 22^«.

32. 4a;«-12a;2 + 9a;-l.

33. aj« + 9 a;2 + 16 aj + 4.

34. a;*H-5a;2 + 7a; + 2.

35. 2 a;^ — 3 xy + 4 aa; — 6 ay.

3p. aj V - 8 2/ V - 4 a;«7i2 4. 32 y«^« '

37. a;« - / - (x^ _ 2^2) _ ^3. _ ^^«^

38. aj* + 2x8-13a;2-38aj-24

39. x*-2(b^-c^)x^-{-b^-2b^c^ + c\

40. 15a;2-7a;-2.

41. lla;2-54a; + 63.

42. 21 ^2 _|_ 26 a; - 15.

43. 70x2-27aj-9.

44. a;* - 2 abx^ - a* - a^ft^ - ft*.

45. 5aj* + 4a;«-20a;-125.

46. 2x^-5x^^x^-2.

47. 6a;* - ax» - 2 aV -f- 3 a»x - 2a*.

FACTORS 47

HIGHEST COMMON FACTOR

88. A common factor of two or more integral and rational
expressions is an expression that divides each of them without

. a remainder.

Two expressions that have no common factor except 1 are
said to be prime to each other.

The highest common factor of two or more integral and
rational expressions is an integral and rational expression
of highest degree that will divide each of them without
remainder.

For brevity, H.C.F. will be used for highest common factor.

Find the H.C.F. of
8aV - 24a2a. -j. 16^2 and 12axhf - 12axy - 24 ay.

8aaa;2 __ 24a2x + lOa^ = 8a2(a;2 - 3aj + 2)

= 28a2(aj-l)(aj-2);
12 ax2y - 12 axy -2^ ay- 12 ay (x^-x- 2)

= 22x 3ay(a; + l)(a;-2).
.-. the H.C.F. = 22 a (X - 2) = 4 a (oj - 2).

Hence, to find the H.C.F. of two or more expressions,

Resolve each expression into its prime factors.

The product of all the common factors y each factor being taken
the least numher of times it occurs in any of the given expressions,
is the highest common factor required,

89. When it is required to find the H.C.F. of two or more
expressions that cannot readily be resolved into their factors,
the method to be employed is similar to that of the corre-
sponding case in Arithmetic. And as that method consists in
obtaining pairs of continually decreasing numbers which con-
tain as a factor the H.C.F. required, so in Algebra, pairs of
expressions of continually decreasing degrees are obtained,
which contain as a factor the H.C.F. required.

48 COLLEGE ALGEBRA

90. This method is needed only to determine the compound
factor of the H.C.F. Simple factors of the given expressions
should be taken out, and the highest common factor of these
factors reserved to be multiplied into the compound factor
obtained.

Modifications of this method are sometimes needed.

(1) Find the H.C.F. of 4 aj2 - 8 a; - 5 and 12x^-4:X-65.

4x2-8x-6)12a;2- 4aj-65(3

12a;2-24g-15
20X-50

The first division ends here, for 20 x is of lower degree than 4 x\ But
if 20 X — 50 is made the divisor, ix^ will not contain 20x an integral
number of times.

Now, it is to be remembered that the H.C.F. sought is contained in the
remainder 20 x — 60, and that it is a compound factor. Hence, If the
simple factor 10 is removed, the H.C.F. must still be contained in
2 X — 5, and therefore the process may be continued with 2 x — 6 for
a divisor.

2x-5)4x2- 8x-5(2x + l
4xg-10x

2x- 5
.-. the H.C.F. is 2 X - 5. 2x-5

(2) Find the H.C.F. of
21x^'-4.x^-15x-2 and 21 ic' - 32 «« - 54 oj - 7.

Writing only the coefficients (§ 70, p. S4), the work is as follows :

21-4-15-2)21-32-54-7(1

21 - 4-15-2
- 28 - 39 - 6

The difficulty here cannot be obviated by rem>oving a simple factor
from the remainder, for — 28x2 — 39x — 5 has no simple factor. In
this case, the expression 21x* — 4x2 — 15x — 2 must be mvUiplied by
the simple factor 4 to make its first term divisible by — 28 x^.

The introduction of such a factor can in no way affect the H.C.F.
sought; for the H.C.F. contains only factors com.mon to the remainder
and the last divisor, and 4 is not a factor of the remainder.

FACTORS

49

The signs of all the terms of the remainder may be changed ; for if an
expression A is divisible by — ^, it is divisible by + F.

The process then is continued by changing the signs of the remainder
and multiplying the divisor by 4.

28 + 39 + 6)84- 16- 60- 8(3
84 + 117 + 16

Multiply by — 4,

7 + :
+ 1.

_133_ 75 _ 8
- 4

Divide by - 63,
.-. the H.C.F. is 7 X

632 + 300 + 32 (19
632 + 741 + 96
-63) -441 -63

7+ 1

1)28 + 39 + 6(4 + 6
28+4

36 + 6
36 + 6

In practice the work is most conveniently arranged as follows

4-16-2

21-32-54-7

21 - 4-16-2

-1)- 28 -39 -6

28 + 39 + 6

28+4

36 + 5
35 + 6

3 + 19

4 + 5

21-

_4

84-16-60-8
84 + 117+ 15
-133- 76- 8

- 4

632 + 300 + 32
632 + 741 + 95

- 63) - 441 ^^

7+ 1

.-. theH.C.F. is7x + l.

91. In the exaanples worked out we have assumed that the
divisor which is contained in the corresponding dividend
without a remainder is the H.C.F. required.

The proof m2ij be given as follows :

Let A and B stand for two expressions which have no
monomial factors, and which are arranged according to the
descending powers of a common letter, the degree of B being
not higher than that of A in the common letter.

60 COLLEGE ALGEBRA

Let A be divided by JB, and let Q stand for the quotient,

and R for the remainder. Then, since the dividend is equal

to the product of the divisor and quotient plus the remainder,

we have

^ = J5Q + 72. [1]

Since the remainder is equal to the dividend minus the
product of the divisor and quotient, we have

R = A-BQ. [2]

Now, a factor of each of the terms of an expression is a
factor of the expression. Hence, any common factor of B
and 72 is a factor of JBQ -f 72, and by [1] a factor oiA. That
is, a common factor of B and 72 is also a common factor of
A and JB.

Also, any common factor of A and 5 is a factor of -4 — BQ,
and by [2] a factor of 72. That is, a common factor of A and
B is also a common factor of B and 72.

Therefore, the common factors of A and B are the same as
the common factors of B and 72 ; and consequently the H.C.F.
of A and B is the same as the H.C.F. of B and 72.

The proof for each succeeding step in the process is pre-
cisely the same; so that the H.C.F. of any divisor and the
corresponding dividend is the H.C.F. required.

If at any step there is no remainder, the divisor is a factor
of the corresponding dividend, and is therefore the H.C.F. of
itself and the corresponding dividend. Hence, this divisor is
the H.C.F. required.

92. The methods of resolving expressions into factors,
given in this chapter, often enable us to shorten the work of
finding the H.C.F. required.

(1) Find the H.C.F. of

aj* -f 3aj' + 12x - 16 and a' - 13aj + 12.

Both of these expressions vanish when 1 is put for x. Therefore, both
are divisible by a - 1 (§ 87).

FACTORS

The first quotient is x* + 4x2 + 4« + 16 = (x^ + 4)(x -\- 4).
The second quotient is x* + x — 12 = (x — 3) (x + 4).
Therefore, the H.C.F. is (x - 1) (x + 4).

(2) Find the H.C.F. of

2aj* + 9aj» + 14aj + 3 and 3aj* + 14aj» + 9aj + 2.

2x*-H9x«-Hl4x + 3

61

3x* + 14x8+ 9x
2

+ 2

6x* + 28x8-Hl8x
6x* + 27x8 + 42x

+ 4
+ 9

x8-24x-5

The remainder, x* — 24 x — 5, vanishes when 5 is put for x.

The quotient of x* — 24 x — 6 divided by x — 6 is x^ + 6 x -H 1.

Since 6 is not an exact divisor of 3, x >- 5 is not a factor of
2x* + 9x« -H 14x + 3 ; but x2 + 5x + 1 is found by trial to be a factor,
and is, therefore, the H.C.F. required.

(3) Find the H.C.F. of

28aj2 + 39aj + 5 and 84x« - 16aj«- 60a5 - 8.

By § 83, p. 42, the factors of 28x2 + 39x + 6 are 7x + 1 and 4x + 6.
The factor 7x + 1 is the H.C.F. required.

(4) Find the H.C.F. of

2x^ - 6aj» - aj2 + 15aj - 10; 4a;* + 6aj« - 4ic2 _ I5aj _ 16.

2x* - 6x» - x2 + 15x - 10

4x*+ 6x8-4x2-15x- 15
4x* - 12x8 - 2x2 _|_ 30a; - 20

18x8- 2x2 -46x+ 5

The remainder = 2x2(9x - 1) - 6(9x - 1) = (2x2 - 6) (9x - 1).
The factor 2x2 — 6 is the H.C.F. required.

LOWEST COMMON MULTIPLE

93. A common multiple of two or more integral and rational
expressions is an integral and rational expression that is
exactly divisible by each of the expressions.

The lowest common multiple of two or more integral and
rational expressions is an integral and rational expression of

52

COLLEGE ALGEBRA

lowest degree and of smallest numerical coefficient that is
exactly divisible by each of the given expressions.

For brevity, L.C.M. will be used for lowest common multiple.

Find the L.C.M. of 12 a% 14 hc\ 36 ab\

12 a^c = 2^ X Sa%
146c2 = 2 X 7&c2,
36 a62 = 22 X 32 062.

.-. the L.C.M. = 22 X 32 X 7 a^b^c^ = 252 a262c2.

HencC; to find the L.C.M. of two or more expressions^

Resolve each expression into its prime factors.
The product of all the different factors, each factor being
taken the greatest numJber of times it occurs in any of the given
expressions, is the lowest common multiple required,

94. When the expressions cannot be readily resolved into
their factors, the expressions may be resolved by finding their
H.C.F.

Find the L.C.M. of

6ic«-llajV + 22^ and 9x* - 22xy^ - Sy*.

6-11+0 + 2
6- 8-4

9+ 0-22- 8
2

- 3+4+2

- 8+4+2

18+ 0-44-16
18-33+ 0+ 6

11)33-44-22

3- 4- 2

2-1

Hence, 6x^ - Ux^ -\- 2y^ = {2x - y) {Sx^ - 4xy - 2y^,
and 9a;8 - 22a;y2 _ 82/8 = (3x + 4y) (3a;2 - 4ajy - 2y2).

.-. the L.C.M. =(2x-y)(3a; + 42/)(3aj2-4xy-2y2).

In this example we find the H.C.F. of the given expressions
and divide each of them by the H.C.F.

Instead of dividing each expression by the H.C.F., we may
divide only one expression, and multiply the quotient by the
other expression.

FACTORS 68

95. The prodttct of the H, C,F, and the L, CM, of two expres-
sions is equal to the product of the given expressions.

Let A and B stand for any two expressions ; and let F stand

for their H.C.F. and M for their L.C.M.

Let a and b be the quotients when A and B respectively are

divided by F. Then,

A = aFy

and B = bF.

Therefore, AB = Fx obF, [1]

Since F stands for the H.C.F. of A and By F contains all
the common factors of A and B. Therefore, a and b have no
common factor, and obF is the L.C.M. of A and jB.

Put M for its equal, abF, in equation [1], and we have

AB = FM.

96. Since FM = AB,

^^ AB A „

F F '

or M=-— = — XA.

F F

That is : The lowest common multiple of two expressions
may be found by dividing their product by their highest com,-
mon factor, or by dividing either of them by their highest
common factor and multiplying the quotient by the other,

97. The H.C.F. of three or more expressions is obtained by
finding the H.C.F. of two of them ; then the H.C.F. of this
result and of the third expression ; and so on.

For, '\i A,B, and C stand for three expressions,

and D for the highest common factor of A and B,
and E for the highest common factor of D and C,
then B contains every factor common to A and JB,
and E contains every factor common to D and C ;
that is, E contains every factor common to -4, B, and C.

64 COLLEGE ALGEBKA

98. The L.C.M. of three or more expressions may be obtained
by finding the L.C.M. of two of them ; then the L.C.M. of this
result and of the third expression ; and so on.

For, \f Ay By and C stand for three expressions,

and L for the lowest common multiple of A and By
and M for the lowest common multiple of L and C,
then L is the expression of lowest degree that is exactly

divisible by A and JB,
and M is the expression of lowest degree that is exactly
divisible by L and C.
That is, M is the expression of lowest degree that is exactly
divisible by Ay By and C.

Ezercise 6

Find the H.C.F. of :

1. 12x^-17x + 6y 9x^-\-6x-S.

2. aj* — a*, ic^ -f 3 aa; — 4 a^, a;^ — 5 ace + 4 a\

3. aj*~6aj« + 13x2-12a; + 4, x* - 4a;» + 8aj«- 16aj + 16.

4. Sx^-x^'-2x''-{-2x-Sy 6a;* + 13a;«4-3aj2 + 20aj.
6. 96aj* + 8aj«-2ic, 32aj8-24x«-8aj + 3.

6. aj* + 5a;«-7x2-9x-10, 2aj* - 4x« + 4a; - 8.

7. 2aj« - 16a; + 6, 5a;« + 15aj« + 5aj + 15.

8. 2a* + 3a«a;-9aV, 6 a*a; - 3 ax* - 17 a'a;^ + 14 aV.

9. 2 a» - 4 a* + 8 a« - 12 a^ + 6 a,
3a« - 3 a« - 6 a* + 9 a« - 3a2

10. Sx^-7x^-'y^ + bxy\ x^y + Zxy'^ - 3a;» - y«,
3 a;' + 5 x^y + xy'^ — y^,

11. 36a;'-28a;« + 32a;* + 8a;»-16a;^
12a;» - 14a;* - 20a;« + lOa;^ + 4a;.

FACTORS 56

12. 15aj* + 2aj»-75a;2 + 5iB + 2,
35aj* + aj» - 175a^ 4- 30a; + 1.

13. 21aj*~4aj«-15aj2-2a;, 21x8 - 32x« - 64aj - 7.

14. 9icV-22ic2y»-3iC2/*4-10y^
9 a;V - 6 a^y + ay - 25 a;2/^

16. 6aj«-4aj*-llic«-3aj2-3aj-l,
4a;* + 2ic» - 18a;2 + 3a; - 5.

16. a;*-aa;«-a2x2-a«x-2a*, 3 a;' - 7 aa;* + 3 a*a; - 2 a«.

17. 12 (a;* - y*), 10 (a;« - 2/«), 8 (a;^ + a^/).

18. a;* + ^2^, a;V + y\ a;* + a;y + y*.

19. 2(a;*y — ajy^), 3(a;V-icy»), 4(a;V — «2^*), 5(a;*y — a;/).

Find the L.C.M. of :

20. a;^ — 3a; — 4, x2 - X - 12, a;" + 5a; + 4.

21. 6a^-13a; + 6, ^x^^^x-Q, 9a;2-4.

22. 3a;*-a;«-2a;2 + 2x-8, 6 a;« + 13 a;^ + 3 a; + 20.

23. 15 aV + 10 a*x« + 4 a^x^ + 6 a«a; - 3 a\
12 a;* + 38 ax^ + 16 a^a;^ - 10 a«x.

24. 2a;* + x« - 8a;2 - a; 4- 6, 4x* 4- 12 a;« - a;2 - 27a; - 18,

4a;* + 4a;« - 17a;2 - 9a; + 18.

26. a;* - 2 a;* + a;2^ 2 a;* - 4 a;^ — 4 a; - 4.

26. a;« - Bar* + 11 a; - 6, a;» - 9a;2 + 26a; - 24,
a.8_8a;» + 19a;-12.

27. 4a;' — oi^y — 3a;^, 3a;* — 3a;*y + xif — y*.

28. 4a;»-12x2 4-9x-l, a;*-2a;« + «^-8a; + 8.

29. 2x*-8x* + 12x'-8a;2^2x, 3x«-6x« + 3a;.

80. x«-6ar» + 5x + 12, x«-5x2 4-2x4-8,

X* — 4 X* 4- ic 4- 6.

CHAPTER IV
SYMMETRY

99. S3rnimetrical Expressions. An algebraic expression that
involves two or more letters is symmetrical with respect to any
two of them if these letters can be interchanged without alter-
ing the expression in value or in form-
Thus, x^ 4- OL^ + a6 + 62x is symmetrical with respect to a and h ; for
if a is substituted for h and h for a, the expression becomes

which differs from x^ + a^x + a6 + ft^x only in the order of its terms and
the order of their factors.

Again, x^ + a^x + a6 + ft^x is not symmetrical with respect to a and x ;
for if a is substituted for a and a for a, the expression becomes

a2 + x^a + x6 + 6%,

which differs in form from x^ 4- a^ + a6 + Wc.

In like manner it may be shown that x^ + a^x + a6 -H 6%c is not sym-
metrical with respect to x and 6.

An expression is symmetrical with respect to three or more
of its letters if it is symmetrical with respect to each and
every pair of these letters that can be selected.

Thus, x^ + y* + «8 — 3 xyz + a6 is symmetrical with respect to x, y,
and z ; for it remains the same if x and y are interchanged, or if y and z
or X and z are interchanged.

An expression is completely symmetrical if it is symmetrical
with respect to each and every pair of its letters that can be
selected.

Thus, X* + y* -H «' + 3 xyz is completely symmetrical ; for it remains
the same If x is interchanged with y, y with z, or z with x ; and these three
pairs are all the pairs that can be selected from x, y, and z,

50

SYMMETRY 67

100. Cyclo-S3rnimetrical Expressions. An algebraic expression
is cyclo-symmetrical with respect to certain letters in a given
order when the value and form of the expression is not altered
by substituting the second letter for the first, the third for the
second, and so on, and the first for the last.

Thus, the expression ah -\- he •{■ cd -\- da is cyclo-symmetrical with
respect to the cycle (aJbcd) but is not completely symmetrical with respect
to a, 6, c, and d.

Every expression that is symmetrical with respect to a set of
letters is also cyclo-symmetrical with respect to these letters ;
but as is seen by the last illustration an expression may be
cyclo-symmetrical with respect to a set of letters without
being symmetrical with respect to the letters.

101. Principle of Symmetry. An expression which in any
one form is symmetrical or cyclo-symmetrical with respect to
any set of letters will in every other form be symmetrical or
cyclo-symmetrical, as the case may be, with respect to these
letters.

Thus, a' + 6^ 4- c' — 3 ahc is symmetrical with respect to a, 6, and c.
Hence, it is symmetrical with respect to a, 6, and c when written in any
other form, as i {a + & + c) [(6 - cf + (c - a)2 -f (a - 6)2].

Again, (a — 6)* + (6 — c)^ + (c — a)' is cyclo-symmetrical with respect
to (a, 6, c), but not completely symmetrical. Hence, it remains cyclo-
symmetrical with respect to (a, 6, c), but not completely symmetrical
when written in any other form, as 3 (a — h) (6 — c)(c — a).

102. 2 Notation. A symmetrical expression is often written
by writing each type-term once, preceded by the Greek letter S,
where 2 stands for the words the sum of all the terms of the
same type as.

Thus, 'La=.a-\-h-\-c-\-d + "'

Soft = ab ■\- he -\- cd -\ -{■he-\-hd-\ f-cd-|-««-

If the three letters, a, 6, c, are involved,

Sa26 = aV) ■\- aJtf^ ■\- a^c -{- ac^ -\- l^c -\- ftc^.

68 COLLEGE ALGEBRA

Ezerciae 7

1. For a, b, c, write the following in full :

S[(a + cy - ft*]; Sa(ft + c)*; S(a + ft) (c - a) (e - ft).

2. For a, ft, c, «?, write the following in full :

Softc; Sa*ft; Sa^ftc; 2(a — ft)j
Sa*(a - ft) ; Sa^ft'c ; S(a + ft - c).

Show that the following expressions axe symmetrical :

3. (x + a)(a-\- ft) (ft 4- a) 4- aftaj, with respect to a and ft.

4. (a + ft)* -{-(a — by, with respect to a and ft, and also
with respect to a and — ft.

6. a* (ft — cy + ft* (c — a)* + c* (a — ft)*, with respect to a, ft, c.

6. (ac + bd)^ -h (ftc — a<£)*, with respect to a* and ft*, and
also with respect to c* and (^*.

Select the letters with respect to which the following
expressions are symmetrical :

7. (a* - c*)* + 4 ft*(a + c)* + (2 ac - 2 ft*)*.

8. ic«-/ + «*-3(x*-2/*)(y*-«*)(«* + a^.

9. a*ft* + ft V + c*a* - 2 aftc (a + ft - c).

10. (a + ft)*+(a — c)*+(ft + c)* + (a + c)*.

11. Show that (a - ft)* (ft - c) (c - e^ (a - c) (ft - dy(a - ^*'
is not symmetrical with respect to a, ft, c, and d. With respect
to which of its letters is the expression symmetrical ?

103. Rule of S3rmmetry. The applications of symmetry and
of cyclo-symmetry are numerous. From the definitions given
in §§ 99 and 100, we have the following :

T?ie sum, the difference, the product, or the quotient of two
aymmetrical or cyclo-symmetrical expressions is also a symmet^
rical or a cyclo-symmetrical expression.

SYMMETRY 69

104. In reducing a symmetrical expression from one form to
another, advantage may be taken of the principles of sym-
metry; for it is necessary to calculate only the type-terms.
The other terms may be written at once from these.

(1) Simplify (a-|-ft + c + <^ + e + -- 0^

The expression is symmetrical with respect to a, 6, c, • • • ; hence, the
expansion also is symmetrical, and as it Is a product of two factors, it can
contain only the squares a^, d^, c^, • • • , and the products a&, oc, od, • • • , 5c,
M, • • • ; so that the type-termB are a^ and ab.

Now (a + 6)* = a» -H 2 06 -H 6* ; and the addition of terms involving
c, d, e, • • • does not alter the terms a^ + 2 a6, but merely gives additional
terms of the same type. Hence, from symmetry,

(o + 6 + c + d + e-H. ••)* = «' + 2a6 + 2ac + 2ad + 2a€ +

-H 6a-H2 6c + 2W-H2 6e +

+ c» + 2c(l + 2ce +

+ d»-H2de-H

+ e» +
This equation may be compactly written,

(Sa)» = Sa2 + 2 2a6.

(2) SimpUfy (a + by.

The expression is of three dimensions, and is ssrmmetrical with respect
to a and h.

The type-terms are a«, aPh,

.'. (a + ft)** = a' + 6« + n(a?h + l^a), where n is numericcd.

To find the value of n, substitute for a and h any convenient values
that will not reduce either side of the equation to 0, as for instance
put a = 6 = 1.

Then, (1 + 1)« = !« + !« + n(12 x 1 + 1^ x 1).

Whence, n = 3.

.-. (a + 6)8 = Sa« + 3 Sa26.

(3) Simplify (a + y + «)».

The expression is of three dimensions and is symmetrical with respect
to X, y, and z. We have

(« + y + «)• = [{X + y) + «]« = (a; + y)8 4- . . . = jB« + Sx^y + . . .

The type-terms are x' and 3x^, and the only other possible type-term
iMxyz.

60 COLLEGE ALGEBRA

Now, since the expression contains Bz^, it must also contain 8a^ ;
that is, it must contain Sx^(y + z).

Hence, {x + y + z)^ = x^ + Zz^ (y -\- z)

H-n(xy«),

where n is numerical, and is found to be equal to 6 by putting x=:y = x=zl
in the last equation.

.-. (x + y + z)^ = 2x8 + 320% + 6xyz.

(4) Simplify (a + ft + c + • • •)'.

The type-terms are a', a^ft, abc.

Simplifying (a + 6 + c)^, we obtain a* + 3 a^ft + 6 aJbc + • • •

Hence, by symmetry, we have

{Sa)8 = Sa8 + 3 Za^b 4- 6 Sa6c.

(6) Simplify

(x -{- y + zy -{- (x — y — zy + (y — z — xy -}-(z — X — y)«.

The expression is symmetrical with respect to x, y, and 2.

The type-terms are x^, 3 x^y, 6 xy«.

x^ occurs in each of the first two cubes, and — x* in each of the second
two cubes. Therefore, in the result there are no terms of the type a*.

3 x^ occurs in the first and third cubes, and — 3 x^ in the second and
fourth. Therefore, in the result there are no terms of the type 3x^.

6 xyz occurs in each of the four cubes.

Therefore, the given expression simplifies to 24 xyz.

Ibcercise 8

Simplify :

1. (a -f- ft + c)» + (a -I- ft - c)^ + (^ + c - a)^ + (c + a — by.

2. (a-fft -{-cy — a(b -\- c — a) — b (a -{- c — b)— c(a-\-b — e).

B. (x + y -{-z-^-ny + ^x — y — z-i-ny
-{-(x — y + z — ny-i-(x-{-y-'Z'-ny.

4. (x - 2y — Szy -^-Q/ - 2z - Sxy + (z - 2x - Syy.

6. a(b -h c) (ft2 ■i'C^-a^'{-b(c + a) (c^ + a* - ft^

6. (aft -|- ftc -h cay — 2 abc (a -f ft -|- c).

SYMMETRY 61

Prove that :

7. (a 4- J + c)* +(^ + c - a)* -f (c + a - by+(a + b-ey
= 4 (a* + ^* + c*) + 24 (a^b^ + ^ V + c^a^.

8. (a 4- ^ + c)* = Sa* + 4 Sa'^ + 6 Sa^ft^ + 12 Sa^ftc.

9. (Sa)* = Sa* + 4 2a«6 + 6 Sa^^i^ + 12 Ja^Jc 4- 24 ^abcd.

10. (a - &)2(ft - cy + (b- cy(c - ay + (c - a)2(a - &)«
= (a2 4- J3 + c2 - aft _ ao - bey.

11. (ar^ 4- 2 brs 4- cs^) (ax^ + 2 ftxi/ 4- cy^

— [anc 4" ^ (^ + saj) + C5y]^ = (ac — b^ (ry — sxy.

12. (a^ 4- aft + ft^(c^4- cd + f?^) = (^c -\- ad -{- bd)^
4- (ac 4- <w? + bd) (be — ad) 4- (be — ady,

105. Factoring. The principles of symmetry can be used in
resolving expressions into factors.

(1) Find the factors of

(a 4- ^ + c) (aft 4- ftc 4- ca) — (a 4- ft) (ft + c) (e 4- a).

The expression is symmetrical with respect to a, 6, and c.

If there is a monomial factor, a must be one. If we put 0 for a, the
expression vanishes. Hence, a is a factor, § 87, p. 44, and by symmetry
h and c are also factors. Therefore, ahc is a factor.

There can be no other literal factor, for the given expression is of
only three dimensions and abc is of three dimensions.

There may be a numerical factor, however. Let m be a numerical
factor of the given expression.

Then (a + 6 + c) (aft + 6c + ca) — (a + ft) (ft 4 c) {c-\- a) = mdbc.

To find m, put a = 6 = c = lin this equation, and m = 1. •

Therefore, the given expression is equal to abc,

(2) Find the factors of

a* (ft _ c) 4- ft* (^ - «) + ^'(« - ^)-
If we put a = 0, the expression does not vanish. Hence, a is not a

foctor, and by symmetry neither ft nor c is a factor.

If we put a = ft in the expression, the expression vanishes. Hence,

at^PlBfk ^tor, § 87, p. 44, and by symmetry ft — c and c — a are factors.

62 COLLEGE ALGEBRA

Now the given expression is of four dimensions. Hence, in addition
to the three factors already found there must be one other factor of one
dimension ; and as this factor must be symmetrical with respect to a, 6,
and c, it must be a + & + c.

There may be a numerical factor.

Let m be a numerical factor of the given expression. Then,

a« (6 - c) + 6'(c - a) + c^(a - 6) = m(a - 6) (6 - c) (c - a) (a + 6 + c).
To find m, put a = 0, 6 = 1, c = 2.

Then, m = — 1.

Hence, the given expression = — (a — 6) (6 — c) (c — a) (a + 6 + c).

(3) Prove that a^ -\- b^ -\- c^ -}- 3 (a -\- b) (b + c) (c + a) is
exactly divisible hy a -\- b -\- Cy and find all the factors.

Let a + 6 + c = 0, or a = — (6 + c), and substitute this value of a.

Then the given expression becomes — (6 + c)^ + 6^ + c* + 3 6c (6 + c)
or - (6 + c)8 + (6 + c)8, or 0.

Hence, a + b + c is a, factor.

If we put a = 0, the expression does not vanish. Hence, a is not a
factor, and by symmetry h and c are not factors.

Since a ^b -]- c, the factor already obtained, is of one dimension, the
other factor must be of two dimensions, and since it must be symmetrical
with respect to a, 6, and c, it must be of the form

m(a^ + b^ + c2) + n{ab + bc + ca),

in which m and n are independent of each other, and of a, b, and c.
To determine the values of m and n, put c = 0 in the equation

a« -f &* + c8 + 3(a + 6) (6 + c) (c + a)

= {a + b + c)[m (a2 + b^ + c^) + n(db + be + ca)].

Then, a« + &' + Sab(a + b) = {a + b) [m(a^ + 6^) + nab].
But a« + 6* + 3 a6 (a + &) = (a + b)^.
Therefore, (a + b)^ = (a-\-b) [m (a^ + 62) + na6].

Therefore, (a + 6)2 = m {a^ + 62) + na6.

That is, (a2 + 62) + 2 a6 = m (a2 + 62) + nab.

Now, this equation is true for all values of a and 6.
Therefore, m = 1, and n = 2.

.-. a8 + 6» + c8 + 3(a + 6) (6 + c) (c + a)

= (a 4- 6 + c) [a2 + 62 + c2 + 2 (a6 + 6c + ca)]

= (a + 6 + c) (a + 6 + c)2

= (a + 6 + c)8.

SYMMETRY 63

9

(4) Show that aj" + 1 is a factor of a;*** + 2 a;«" + 3 aj" + 2.

Let a* + 1 = 0, or X" = — 1, and substitute.

TheU, a«»» + 2x2'» + 3x» + 2=-l + 2-3 + 2 = 0.

Therefore, x" + 1 is a factor of the given expression.

(5) Show that a^ +- ^Ms a factor of 2 a* +- a*b +- 2 a^b^ +- ab^

Let a» + 62 = 0, or a2 = - 62^ and substitute.

Then, 2a* + a^d + 2a262 + od^ = 26* - a68 - 26* + od^ = 0.

Therefore, a2 + 62 is a factor of the given expression.

Exercise 9
Resolve into factors :

1. (iB -f y + ;5J)» - (aj« + y' + «*).

2. be (b — c) — ca (a— c) — db (b — a).

3. (a« - by +- (b^ - cy +- (c« - ay.

4. aj(y+-«)^+-y(« + «)* + «(« + y)* — 4ajy«.

5. (a + by-(b-\-cy-\-(c-a)\

6. a(6 — c)»+-^>(c — a)«+-c(a — J)».

7. (a +- J +- c) (ab -^ be -\- ca) — aJc.

8. a*(c - b^) -^b\a- c^) +- c» (^> - a^) +- al)c{ahc - 1),

9. a^{b +- c) +- 62(c +. a) +- c2(a +- ^>) +- 2 abc,

10. ajy +- ajy +- x^z^ + ic2;sj* +- y^z'^ + y2<g;4 ^ 2 x^y^z\

11. (a - J)*^ ^-{b- cy +-((?- a)^

12. ab{a -{-b)-\- bc(b +- c) +- ca (c +- a) +- (a' +- 5» +- c»).

13. a* (c - ^«) +- b* (a - c^) +- c* (ft - a«) + a^c (a^^^^^ - 1).

14. x^ {y^ ^z^ + y* (z^ - x^) +- «* (x^ - yy

15. aj*+-2^+-«*-2ajy-2yV-2«V.

16. (a +- 5)» +- (6 + c)» +- (c +- a)«

+- 3 (a +- 2 6 +- c) (6 +- 2 c + a) (c + 2 a +- ft).

17. a*(ft - c) +- 5*(c - a) +- c*(a - ft).

18. Show that a* + aH^^ — aft^ — ft» has a* — ft for a factor.

CHAPTER V

FRACTIONS

106. An algebraic expression is integral when it consists of
a number of terms connected by -h and — signs, and each term
is the product of a coefficient into positive integral powers of
the letters involved.

In an integral algebraic expression the coefficients may be
fractional.

Thus, x^ — 1 0x2 -f I a is an integral algebraic expression.

107. An algebraic fraction is the indicated quotient of two
algebraic expressions, and is generally written in the form — •

The dividend, a, is called the numerator ; the divisor, b, the
denominator.

The numerator and denominator are called the terms of the
fraction.

108. Since the quotient is unchanged if the dividend and
divisor are both multiplied (or divided) by the satne factor,
the value of a fraction is unchanged if the numerator and
denominator are multiplied (or divided) by the same factor.

109. To reduce a fraction to lower terms.

Divide the numerator and the denominator by any common
factor.

A fraction is expressed in its lowest terms when both numer-
ator and denominator are divided by their H.C.F.

6x^ — 5x — 6

(1) Reduce to lowest terms

8a:2_2aj-15

By§83,p.42, ^^^ " 6^ - ^ ^ (2x - 3) (3x + 2) ^ 8x+j^
'8x2-2x-15 (2x-3)(4x + 5) 4x + 6

64

FRACTIONS 65

(2) Reduce to lowest terms ^ . — tz — 5 — 77;

^ ^ 3 a' — 14 a'* + 16 a

Since no common factor can be determined by inspection, it is neces-
sary to find the H.C.F. of the numerator and denominator by the method
of division.

We find the H.C.F. to be a - 2.

Now, if a8-7aa+16a-12 is divided by a-2, the result is aa-5a+6;
and if 3a« - l^a^ + 16a is divided by a - 2, the result is,3a« - 8a.

gg - 7 gg + 16 g - 12 _g2-5g4-6
" 3g8-14a2 + 16a ~ Sa^-Sg

110. Mixed Expressions. If the degree of the numerator
of a fraction equals or exceeds that of the denominator, the
fraction may be changed to the form of an integral or a mixed
expression by dividing the numerator by the denominator.

The quotient is the integral expression; the remainder (if
any) is the numerator, and the divisor the denominator, of the
fractional expression.

To reduce a mixed expression to a fractional form.

Multiply the integral expression by the denominator, to the
product add the numerator, and under the result write the
denominator.

The dividing line has the force of a vinculum or parenthesis
affecting the numerator; therefore, if a minus sign precedes
the dividing line, and this line is removed, the sign of every
term of the numerator must be changed,

_, g — 6 en — (a — b) en — a-\-b

Thus, n = ^ '- = —

c c c

111. To reduce fractions to equivalent fractions having the
lowest common denominator,

If'ind the L, CM. of the denom^inators.
Divide the L. CM. by the denominator of each fraction.
Multiply the first numerator by the first quotient, the second
numerator by the second quotient, and so on.

66 COLLEGE ALGEBRA

Ths products are the numerators of the equivalent fra^^ions.
The L. CM, of the given denominators is the denomincutor of
each of the equivalent fractions.

Thus, — -, -^, -— - are equal to — — -i ---^i — — -, respectively.
4a« 3a Oa* 12a« 12 a« 12a«

The multipliers 3 a, 4a^, and 2 are obtained by dividing 12 a*, the
L.C.M. of the denominators, by the respective denominators of the given
fractions.

112. To add fractionB,

Reduce the fractions to equivalent fractions having the lowest
common denominator.

Add the numerators of the equivalent fractions.
Write the result over the lowest common denominator.

To subtract one fraction from another we proceed as in
addition, except that we subtract the numerator of the subtra-
hend from that of the minuend.

(1) Simplify + j2

The L.C.D. is 84.

The multipliers are 12, 28, and 7 respectively.

36 a — 48 6 = 1st numerator,

- 56 a + 28 6 — 28 c = 2d numerator,

91a --28 c = 3d numerator.

71 a — 20 6 — 66 c = sum of numerators.

3a-46 2a-6-fc . 13a-4c 71a-206-66c
.% = .

7 3 12 84

Since the minus sign precedes the second fraction, the signs of all
the terms of the numerator of this fraction are changed after being
multiplied by 28.

The L.C.D. is (x + y)(x- y) (x^ + y%

FRACTIONS 67

Hie multipliers are

a^ + y*, (« - y) («* + y*), (x + y){x^y), (x + y)(«-y)(a^ + y«)

respectively.

«V + 2^ = 1st numerator,

- 05* + 2a^ - 2«2ya + 2x2^ - 2^ = 2d numerator,

2 «8y — 2 xy* = 3d numerator,

g* — y* = 4th numerator.

4«8y — x^2 — y* = sum of numerators.

.•. sum of fractions = ^ — *y —Ir

x*-2^

113. Since — = a, and — r- = a, it is evident that if the

o — 0

signs of both numerator and denominator are changed, the
value of the fraction is not altered.

Since changing the sign before the fraction is equivalent
to changing the sign before every term of the numerator or
denominator, therefore the sign before every term of the numer-
ator or denominator m,ay he changed, provided the sign before
the fraction is changed.

Since, also, the product of -h a multiplied by + ^ is aby and
the product of — a multiplied by — 6 is oft, the signs of two
factors, or of any even number of factors, of the numerator or
denominator of a fraction may be changed without altering
the value of the fraction.

By the application of these principles, fractions may often
be changed to a form more convenient for addition or sub-
traction.

Simplify ~ + -r—'

Change the signs before the terms of the denominator of the third

traction, and change the sign before the fraction.

2 3 2x 3

The result is » in which the several denomlna-

X 2x-l 4x2-1

tors are written in similar form.

68 COLLEGE ALGEBRA

TheL.C.D. isx(2a; - l)(2x + 1).

8 «* — 2 = 1st numerator,

— 6 «* — 3 X = 2d numerator,

— 2 g^ + 3 a = 3d numerator.

— 2 = sum of numerators.
-2

.*. sum of the fractions =

x(2x-l)(2x + l)

114. Multiplication of Fractions. Let it be required to find

CL G

the product of the two fractions - and -•

If we multiply the dividend a by c, we multiply the quotient
- by c J if we multiply the divisor h by c?, we divide the

quotient - by c?. Hence, the product of — and ;j is t;/*

Therefore, to find the product of two fractions,

Find, the product of the numerators for the numerator of tJie
product, and the product of the denominators for the denomi-
nator of the product,

115. Division of Fractions. Multiplying by the reciprocal of
a number is equivalent to dividing by the number.

The reciprocal of a fraction is the fraction with its terms
interchanged. Therefore, to divide by a fraction,

Interchange the terms of the fraction and multiply hy the
resulting fraction.

If the divisor is an integral expression, it may be changed
to the fractional form.

116. A complex fraction is a fraction which has a fraction in
the numerator, or in the denominator, or in both terms.

To simplify a complex fraction.

Divide the numerator hy the denominator.

Or we may multiply both terms of the fraction by the L.C.D.
of the fractions contained in the numerator and denominator.

FRACTIONS 69

Exercise 10
Reduce to lowest terms :

1.

42 a» - 30 a^ 6aV - 2a* + 18c^ - 6a^

S5ax^-25x*' 4a* + 2aV + 12^2 + Gc^'

2x*-{'5x^-12x x^-[-(2b'^^a^x^-i-b*

: 7x* + 25x^-12x' ^' x^-^2ax^-^a^x''-b^'

6x^ - 9x^ -\-llx''-\-6x^-10x
' 4-a;« + 10a;» + 10a;* + 4aj« + 60a;^*

Simplify :

3a;-2y 4y + 2a; 22y-9a;
3 5 ^ 16 '

^ 1 2a + 3 1 3a-2&

3a 2b 6a2 "^2aj2'^ 6ab

3 , 4a 5a2

X -— a (pc — <^y (x — a)'

a + b . b + c a — c

9. — -; - +

(b — c)(c — a) (c — a){a ^ b) (a — b)(b — c)

10. . L . + ..■ I, .+ '

a(a — b)(a — c) b(b — c) (b — a) c(c — a) (c — b)

16x^^l7x-i-12 27a;^ + 18a;-24 25x^-25x-\-6
' 12a:«-26a; + 12 12a;2 + 7a;-12 20x^-23x^-6

_ 2a^x'' 5a*b^ 15 bh^ 25a^x
12. -r-rr- X T—r^ X

3b^ 4cV 4a^a; * ISoZ^V
\x^ — y^ ' x^ — xyj ' \x — y ' xy — y^J

»• C-4^-)(^:)-G-0-

70 COLLEGE ALGEBRA

a;' - 7a; + 12 x^ + x^2 2a;^ + ga;-3

16.

17.

22.

23.

24.

26.

26.

6 gg - g - 2 Sa^-lOa + 3 12a^ + 17a + 6
y 2a; + y \ g^ - ^y ^ ^

a; a; + y

a

+

X -\- y X a -{-h a — h

1 + a; 1 + a;^ 64 a' - 96 a^ + 36 ga;'

1 + a;^ 1 + a;' 36 a« - 729 a;^

• l+a;« 1+:?' 486i«-27a;»

l + a;» 1+a;* S a^ - 72 ax -{- 162 x^

«* -- a;*y + x^y^ — a^y + a;y* — y'^ ^ x^ — 2xy-\'y^
x'^ + x^ + xY + ^Y + ^I/^ + y^ ar* + 2a;y + y*

1+x ^(l+a?)'

1 — a; \l — x/

\x 4- a/ \^ — ^/

\a; + a/ \^ ~ ^/

2 ^ 2 ^ 2 ^ (x-y)^ + (y-;.)«+(^-a;)«

« — y y — « « — ^ («--y)(y — »)(»-■«)

05 + 1 a;— 1 1 — 3a; i * , 1

2a;-l~2a; + l"a;(l-2a;) a;(4a;2-l)"^a;(16a;*-l)'

1 1 . y 5__

a — y a — x (a — i/)^ (a — xy

1 1

(a — y){a — x)^ {a — a;) (a — y)^

CHAPTER VI
SIMPLE EQUATIONS

117. Two different expressions that involve the same sym-
bols will generally have different values for different assumed
values of the several symbols ; for certain values of the symbols
involved the two expressions may have the same value.

118. An equation is a statement that two expressions have
the same value; that is^ a statement that two expressions
represent the same number.

Every equation consists of two expressions connected by
the sign of equality ; the two expressions are called the sides
or members of the equation.

An equation will, in general, not hold true for all values of
the symbols involved ; it will hold true for only those values
that give to the two members the same value.

Thus, the equation

4x2 - 3x + 5 = 3x2 + 4a; - 6

holds true when for x we put 2, since each member then has the value
16 ; also when for x we put 5, since each member then has the value 90.
If we give to x any other value, the two members will be found to have
different values, and the equation will not hold true.

119. An equation of condition is an equation that holds true
for only certain particular values of the symbols involved.

An identical equation, or an identity, is an equation that
holds true for all values of the symbols involved.

The two members of an identical equation are identical
repressions.

In identical equations it is customary to use the sign =,
called the sign of identity, instead of the sign of equality.

71

72 COLLEGE ALGEBRA

Thus, the two expressions {x + v)^ and x^ -f 2 xy + J^ have the same
value for all values of x and y^ and we accordingly write the identity,

This is read (x + v)^ is identically equal to x^ + 2 xy + y* ;
or (x + y)^ is identical with x^ + 2 xy + J^.

Wherever the term equation is used, it is to be understood
that an equation of condition is meant, unless the contrary is
expressly stated.

In any particular problem we have two kinds of numbers.

1. Numbers that are either given, or supposed to be given,
in the problem under consideration. Such numbers are called
known numbers; if given, they are generally represented by
figures ; if only supposed to be given, by the first letters of
the alphabet.

2. Numbers that are not given in the problem nnder
consideration, but are to be found from certain given rela-
tions to the given numbers. Such numbers are called unknown
numbers, and are generally represented by the last letters of
the alphabet.

The relations between the known and unknown numbers
are generally expressed by means of equations.

120. Simultaneous equations are equations in which the cor-
responding unknowns have the same values.

In order to find all the unknown numbers in a system of
simultaneous equations, we must have as many equations as
there are unknown numbers.

121. To solve an equation, or a system of simultaneous
equations, is to find the unknown numbers involved.

122. The degree of an equation is the sum of the exponents
of the several unknown numbers in that term in which the
sum of the exponents is greatest.

If the equation involves but one imknown number, the
degree is the same as the exponent of the highest power of
the unknown number involved in the equation.

SIMPLE EQUATIONS 73

Equations of the first, second, third, and fourth degrees are
called respectively simple eqiLations, qiiadratic equations, cubic
equations, and biquadratic equatiogts.

123. Literal equations are equations in which some or all of
the given numbers are represented by letters.

124. An equation that involves but one unknown number,
represented for example by x, will hold true for those values
of X which give to the two members the same value (§ 118),
and for no other values of x. The values of x for which the
equation holds true are called the roots of the equation.

Thus, the roots of the equation 4x2 — 3x + 5 = 3x2 + 4x — 5 a^e 2
and 5.

To solve an equation that involves one unknown number is,
therefore, to find the roots of the equation.

125. The various methods of solving equations are based
mainly upon the following general principle :

If similar operations are performed upon equal numbers, the
results are equal numbers.

Thus, the two members of a given equation are equal numbers. If
the two members are increased by, diminished by, multiplied by, or
divided by equal numbers, the results are equal numbers. Similarly, if
the two members are raised to like powers, or if like roots of the two
members are taken, the results are equal numbers.

126. Any term m,ay be transposed from one side of an equa-
tion to the other, provided its sign is changed.

Suppose X -{- a^b. Suppose x — a=^b.

Now, a=^ a, Now, a =• a.

Subtract, x =b — a. Add, x = a + ^.

To transpose a negative number we add that number to both
sides of the equation ; to transpose a positive number we sub-
tract that number from both sides.

127. The signs of all the terms on each side of an equation
may be changed ; for this is in effect transposing every term.

74 COLLEGE ALGEBRA

128. To solve a simple equation with one unknown number,

Transpose all the terms involving the unknoum number to
the left side, and all the othfr terms to the right side : combine
the like terms, and divide both sides by the coefficient of the
unknown number.

To verify the result, substitute the value of the unknown
number in the original equation.

Solve (x - 2) (a: -f 4) = (aj 4- 1) (« + 2).

Expand, «» + 2x-8 = x« + 3x + 2,

or 2x-8 = 3x + 2.

Transpose, 2x — 3x = 2 + 8.

Combine, — x = 10.

.-. X = - 10.

129. To clear an equation of fractions,

Multiply each term by the L, CM, of the denominators.

If a fraction is preceded by a minus sign, the sign of every
term of the numerator must be changed when the denominator
is removed.

(1) Solve I - ^^ = a: - 9.

Multiply by 83, the L.C.M. of the denominators.
Then, llx-3x + 3 = 33x- 297.

Transpose and combine, — 25 x = — 300.

.-. X = 12.

Since the minus sign precedes the second fraction, in removing the
denominator, the + (understood) before x, the first term of the numerator,
is changed to — ; and the — before 1, the second term of the numerator,
is changed to +.

If the denominators contain both simple and compamid
expressions, it is generally best to remove the simple expres-
sions first, and then each compound expression in turn.

SIMPLE EQUATIONS 76

/ox « 1 » 8a! + 5 7x-3 ix + 6
(2) Solve ^^ + ^_^ = ^^.

AQx 21

Multiply by 14, 8x + 5 + ^|^^ — =- = 8x + 12.

r«_ J ,-. 49X-21 ^
Transpose and combine, — = 7.

Multiply by Sx + 1, 49x - 21 = 21 x + 7.

Transpose and combine, 28 x = 28.

.•. X = 1.

Exercise 11

Solve :

1. 8(10-a;)=5(a; + 3).

2. 2x-S(2x-S)=l'-4:(x-2).

3. (x - 5) (aj + 6) = (cc - 1) (x - 2).

4. (2x + 3)(Sx-2) = x^-\-x(5x-\-S).

5. (x--3)(x + 5)=^(X'^l)(2x-3)-x^

6. (a; + 4) (aj - 2) = (cc 4- 3) (3 a; + 4) - (2 a; + 1) (x - 6).

7. (a;-3)(2aj + 5)=a;(a;4-4) + (aJ + l)(aJ4-3).

8. (a; + 2)2 + 3 a; = (aj - 2)^ + 5(16 - a;).

9. (aj-3)2 + (aj-4)« = (aj-2)2 + (aj4-3)2.

5a; — 6 3aj a; — 9

10. -:r- — t: = tt:- 14.

3 a;
5

X

~6~

26

16

a; -
3 a;-

-2
-5

6
19*

3a;

-5

2

2 a; + 10

3

3(5

x-3)

1 6

2(4

x + 3)

1 6

6 4 10

^ - ^ 12 - 3a; 3a;- 11 .

11. ^ . = ^. 16. — ^ 3— =^-

^«, ^ - 4a; + 17 . 3a;-10 ^

12. 5^ -^ = -- 16. 7-^ + J- = 7.

a; + 3 05 — 4

2x + 1 405 —3

76 COLLEGE ALGEBRA

ix-hS 3a;-4 7

18.

3a; + 4 4a;-3 12

6 0^ + 7 3 _o, . 1

20. — j-H = 5.

ax — b bx -{- c

21. = flwc.

c a

22. 777—^ — 77 +

3(a; + 5) 2(aj + a) 6

a; ~ 2 g lSa^-2x^
• a; + 3a aj^-9a^ ~ "^^

24.

a X a(x — a) x(x -{- a) _^ ax _

X a x(x -{- a) a(x — a) a^ — x^

130. Problems. In the statement of problems it is to be
remembered that the unit of the quantity sought is always
given, and it is only the number of such units that is to be
found. We have nothing to do with the quantities them-
selves ; it is only numbers with which we have to deal.

Thus, X must never be put for a distance, time, weight, etc., but for a
number of miles, days, pounds, etc.

(1) A and B had equal sums of money ; B gave A $5, and
then 3 times A's money was equal to 11 times B's money.
What had each at first ?

Let X = the number of dollars each had.

TUen 2 + 5 = the number of dollars A had after receiving f(,

and X — 6 = the number of dollars B had after giving A f6.

.-. 3(x + 5) = ll(x-5),

3x + 15 = lla;-55, ; -

-8aj = -70, w

X = Sf.
Therefore, each had at first $8.75.

SIMPLE EQUATIONS T7

(2) A can do a piece of work in 5 days, and B can do it
in 4 days. How long will it take A and B to do the work
together ?

Let X = the number of days it will take A and B together.

Then - = the part they can do together in one day.

Now, J = the part A can do in one day,

and J = the part B can do in one day.

.'. I + i = the part A and B can do together in one day^

1_^1 1

.*.- + - = -»
6 4 X

4a; + 5« = 20,

0x = 2O,

x = 2J.

Therefore, they can do the work together in 2} days.

Ezercise 12

1. The difference between two numbers is 3; and three
times the greater number exceeds twice the less by 18. Find
the numbers.

2. If a certain number is increased by 16, the result is
seven-thirds of the number. Find the given number.

3. A boy was asked how many marbles he had. He
replied, " If you take away 8 from twice the number I have,
and divide the remainder by 3, the result is just one-half the
number." How many marbles had he ?

4. The sum of the denominator and twice the numerator of
a certain fraction is 26. If 3 is added to both numerator and
denominator, the resulting fraction is f . Find the fraction.

6. A courier sent away with a despatch travels uniformly
at the rate of 12 miles per hour ; 2 hours after his departure
a second courier starts to overtake the first, traveling
uniformly at the rate of 13^ miles per hour. In how many
hours will the second courier overtake the first?

78 COLLEGE ALGEBRA

6. Solve Example 5 when the respective rates of the first
and second couriers are a and h miles per hour, and the inter-
val between their departures is c hours.

7. A certain railroad train travels at a uniform rate. If
the rate were 6 miles per hour faster, the distance traveled in
8 hours would exceed by 50 miles the distance traveled in
11 hours at a rate 7 miles per hour less than the actual rate.
Find the actual rate of the train.

8. A can do a piece of work in 10 days ; A and B together
can do it in 7 days. In how many days can B do it alone ?

9. A can do a piece of work in a days ; A and B together
can do it in ^ days. In how many days can B do it alone ?

10. If A can do a piece of work in 2 m days, B and A

together in n days, and A and C in m -f- ^ days, how long will
it take them to do the work together ?

11. A boatman moves 5 miles in f of an hour, rowing with
the tide ; to return it takes him 1^ hours, rowing against a tide
one-half as strong. What is the velocity of the stronger tide ?

12. A boatman, rowing with the tide, moves a miles in
h hours. Returning, it takes him c hours to accomplish the
same distance, rowing against a tide m times as strong as the
first. What is the velocity of the stronger tide ?

13. If A, who is traveling, makes ^ of a mile more per
hour, he will be on the road only J of the time ; but if he
makes ^ of a mile less per hour, he will be on the road 2^
hours more. Find the distance and the rate.

14. The circumference of a fore wheel of a carriage is
a feet; that of a hind wheel, h feet. What distance will
the carriage have passed over when a fore wheel has made
n more revolutions than a hind wheel ?

16. A is 72 years old, and B is two-thirds as old as A.
How many years ago was A five times as old as B ?

SIMPLE EQUATIONS 79

16. If three pipes can fill a cistern in a, 5, and c minutes
respectively, in how many minutes will the cistern be filled
by the three pipes together ?

17. Find the time between 2 and 3 o'clock when the hands
of a clock are together.

18. Find the time between 3 and 4 o'clock when the hands
of a clock make a right angle.

19. A merchant maintained himself for three years at an
expense of $1500 a year, and each year increased that part
of his stock, that was not so expended by one-third of it. At
the end of the third year his original stock was doubled.
What was his original stock ?

20. When a body of troops was formed into a solid square
there were 60 men over ; but when formed in a column with
5 men more in front than before and 3 men less in depth,
1 man more was needed to complete the column. Find the
number of troops.

21. A man engaged to work a days on these conditions :
for each day he worked he was to receive h cents, and for
each day he was idle he was to forfeit c cents. At the end
of a days he received d cents. How many days was he idle ?

22. A banker has two kinds of coins. It takes a pieces of the
first kind to make a dollar, and h pieces of the second to make
a dollar. A person wishes to obtain c pieces for a dollar.
How many pieces of each kind must the banker give him ?

23. A wine merchant has two kinds of wine which cost him,
one a dollars, and the other h dollars, per gallon. He wishes
to make a mixture of I gallons, which shall cost him on the
average m dollars a gallon. How many gallons must he take
of each?

Discuss the question (i) when a = ft ; (ii) when a or ft = m ;
(iii) when a = ft = w ; (iv) when a > ft and < w ; (v) when
a > ft and ft > w.

CHAPTER VII

SmULTANSOnS SIMPLS EQUATIONS

181. Equations that express different relations between the
unknown numbers are called independent equations.

Thus, 05 + y = 10 and « — y = 2 are independent equations ; they
express differerd relations between x and y. But x + y = 10 and
3 X + 3 y = 30 are not independent equations ; both express the same
relation between .the unknown numbers.

132. An equation is said to be satisfied hj a number, if we
can substitute that number for one of the unknowns in the
equation without destroying the equality.

133. Simultaneous equations are solved by combining the
equations so as to obtain a single equation with one unknown
number ; this process is called elimination.

There are three methods of elimination in general use :

I. By Addition or Subtraction.
II. By Substitution.
III. By Comparison.

We shall give one example of each method.

/ix « 1 2a;-3y= 4\ [1]

i^)^^^^ 3x + 2y = 32r [2]

Multiply [1] by 2, 4 x - 6 y = 8 [3]

Multiply [2] by 3, 9a + 6y= 96 [4]

Add [3] and [4], 13 x =104

.-. X = 8.
Substitute the value of x in [2], 24 + 2 y = 32.

.-. y = 4.

In this solution y is eliminated by addition.

80

SIMULTANEOUS SIMPLE EQUATIONS 81

(2) Solve ft^^ = !V ^1}

^^ 3x4-7^ = 7/ [2]

Transposes^ in [1], 2x=8 — 3y.

Divide by coefficient of x, x = =^. [3]

Substitute the value of x in [2],

24-9y „ ^
2~^H-7y = 7.

24 -9yH-14y = 14.
6y=-10.
.-. y = - 2.
Substitute the value of y in [3], .*. x = 7.

In this solution y is eliminated by substitution.

(3) Solve

2
3

x-9y = ll
x — 4:y= 7}

[1]
[2]

Transpose 9 y in [1],

2x = ll + 9y.

[8]

Transpose 4 y in [2],

3x = 7H-4y.

[4]

Divide [8] by 2,

^_ll+9y
2

[6]

Divide [4] by 3,

„i±iv.

[8]

Equate the values of x,

1

2 3

[7]

Reduce [7],

33 + 27y = 14 + 8y.

19y = -19.

••• y = — !•

Substitute the value of y in [5], .*. x = 1.

In this solution x is eliminated by comparison. ,

Each equation must be simplified, if necessary, before the
elimination is performed.

82

COLLEGE ALGEBRA

ix-l)(y + 2) = (x-3)(j,-l) + 8
(4) Solve 2a! -1 3(y-2)

5 4 ""

► •

[1]
[2]

Simplify [1], «yH-2x-y-2 = xy-x-3y + 3 + 8.

Transpose and combine, 3 x + 2 y = 13. [3]

Simplify [2], 8x - 4 - 16 y H- 30 = 20.

Transpose and combine, 8x — 16y— —6.

[4]

Multiply [3] by 8, 24 x -f 16 y = 104.

[6]

Multiply [4] by 3, 24 x - 45 y = - 18.

[6]

Subtract [6] from [6], 61 y = 122.

.-. y = 2.
Substitute the value of y in [3], 3 x -f 4 = 13.

.-. X = 3.

Fractional simultaneous equations, with denominators which
are simple expressions containing the unknown numbers, may
be solved as follows :

(5) Solve

^ + ^ =7
Sx 5y

7
6x

:r--T4- = 3

Multiply [2] by 4,
Add [1] and [3],
Divide by 19,

10 y
14 2

3x 6y
19

3x
J_

3x
.-. X = 1

= 12.
= 19.
= 1.

[1]
[2]

[3]

Substitute the value of x in [1],

6 + — = 7
by

Transpose,
Divide by 2,

2^

by

by
••• 2/ = i

= 2

= 1

SIMULTANEOUS SIMPLE EQUATIONS 83

134. Literal Simultaneous Equations. The method of solv-
ing literal simultaneous equations is as follows ;

ax -{- by = m\ [1]

Solve ^ y Y' ^oT

ex -\-dy = n J [ JJ

Multiply [1] by c, . acx + hey = cm^ [3]

Multiply [2] by a, acx + ady = an. [4]

Subtract [4] from [3], (be — ad)y =cm — an.

Divide by coeflBcient of y, y = . •

he — ad

Multiply [1] by d, ajdx + hdy = dm. [6]

Multiply [2] by b, bcx + bdy = bn. [6]

Subtract [6] from [5], (ad — bc)x = dm— bn.

Divide by coefficient of x, x =

ad — be

135. If three simultaneous equations are given, involving
three unknown numbers, one of the unknown numbers must
be eliminated between two pairs of the equations; then a
second between the resulting equations.

2x — 3i/'\-4:z= 4
Solve Sx-\-5y-7z = 12

5x — y — Sz = 5

Eliminate z between two pairs of these equations.

Multiply [1] by 2, ix-Qy + Sz= 8.

[3] is bx — y — Sz= 5.

Add, dx-7y =13. [4]

Multiply [1] by 7,

Multiply [2] by 4,

Add, 2Qx- y =76. [5]'

Multiply [6] by 7, 182 x - 7 y = 632. [6]

[4] is

Subtract [4] from [6], 173 x = 519.

.-. X = 3.
Substitute the value of x in [6], 78 — y = 76.

.-. y = 2.
Substitute the values of x and y in [1], 6 — 6 + 42 = 4.

/. 2 = 1.

> .

[1]

[2]
[3]

9x- 7y

= 13.

14

x-21y + 282

= 28.

12

X + 20 y -

-282

= 48.

26

X- y

= 76.

182x-

-72/

= 632.

9x-

-72/

= 13.

84

COLLEGE ALGEBRA

Likewise, if four or more equations are given, involving
four or more unknown numbers, we must eliminate one of the
unknown numbers from three or more pairs of the equations,
using every equation at least once; then a second unknown
number from pairs of the resulting equations ] and so on.

Exercise 13

Solve the following sets of equations ;

1.

2.

3.

4.

6.

6.

7.

8.

9.

6a; -f- 5y = 46

2x
3x

-52/ = 16j

4a;-f 9y = 79
7a;--17y = 40

2x-7y= 8
4t^-9a; = 19

X =

y

= 16-42/1
= 34-4a;J '

5x = 2y-h7S)
3y = a;-f 104 J

^ + ^ = 10

7/ __ 5x — 7
4" 19

4 + 2/ =

Sx^

x-S

_4y
5

>'

a^ -\- ax -\- y = 0
b^ -{-bx-{-y = 0

10.

11.

12.

13.

14.

16.

3

+ a; = 15

^ + .-6

i..

8 "^ 6

2a;

+

= 2
2y-5_

21

= 3

>•

5 + 5 = 3

— -- = 4
X y

>

j4^ 6_^861
5 a; 6y 15

4a;

5y

11
20

r-

a;-2 lO-g^y — 10
5 3 "" 4

2y4-44a; + yH-13
3 ~ 8

a -f c)a; -f (a — c)y = 2a5 1
a -f- ft) y — (a — ^) « = 2ac J

SIMULTANEOUS SIMPLE EQUATIONS

86

17.

18.

19.

•20.

21.

16.

X —

y-

2y-

-X

23-

y-

-X

-3

a; -18

= 20 +

= 30-

2a; -591
2

7S-Sy

2x — 3y = 5a — b
3x-^2p = 5a-{'b

ad c

a

^•

a

g + y_

x — y b ^ c

X -\- c _ a -\- b
y+b" a-hc

X — a a — b^
y — a a -\-b

X ^a^ — b^
y""aM^

8aj-|-4y — 3« = 6

x + 3y — z = 7
4a; — 5y + 4^ = 8

i^.

I

x+^--=a
b c

it if
22. y + --- = d }.
<? a

« i T = <^

a b

(a^h)x+(b + c)y-\'(c
27. (a + c)a;+(a + ^;)y+(^>

(* + c)a;+(a4-c)?/4-(a

3x-^-hz
4

23. 2x-?^

o

y

= 7i

= 5i^

24.

26.

2a;-§ + 4« = ll

3y-2x 1
~2

= 1

3«-7

5« — a;
2y-3«

y-2^ _.

3y-2a;

^ + ^^ = 3
x y z

26. 5 + ^-^ = 1

x y z

2a _^ , c
X y z

^ = a

x + y

xz

X -\- z

= n

^ = c

2^ + «

4- a) « = ad + ftc -f ca
+ c) « = ad + ac + dc
+ d)« = a" + d2 + c"

r

86 COLLEGE ALGEBRA

136. Problems. It is often necessary in the solution of
problems to employ two or more letters to represent the num-
bers to be found. In all cases the conditions must be sufficient
to give just as many equations as there are unknown numbers.

If there are more equations than unknown numbers, some
are superfluous or inconsistent ; if there are fewer equations
than unknown numbers, the problem is indeterminate.

If A gives B $10, B will have three times as much money
as A. If B gives A $10, A will have twice as much money
as B. How much has each ?

Let X = the number of dollars A has,

and y = the number of dollars B has.

Then y -f 10 = the number of dollars B has after receiving $10,

X — 10 = the number of dollars A has after giving f 10,
X + 10 = the number of dollars A has after receiving $10,
and y — 10 = the number of dollars B has after giving $10.

2^ + 10 = 3 (X - 10),
and X + 10 = 2 (y - 10).

From the solution of these equations, x = 22 and y = 26.
Therefore, A has $22 and B has $26. *

Exercise 14

1. Three times the greater of two numbers exceeds twice
the less by 27 ; and the sum of twice the greater and five times
the less is 94. Find the numbers.

2. A fraction is such that if 3 is added to each of its terms,
the resulting fraction is equal to | ; and if 3 is subtracted
from each of its terms, the result is equal to ^. Find the
fraction.

3. Two women buy velvet and silk. One buys 3^- yards of
velvet and 12f yards of silk ; the other buys 4^ yards of vel-
vet and 5 yards of silk. Each woman pays $63.80. Find the
price per yard of the velvet and of the silk.

SIMULTANEOUS SIMPLE EQUATIONS 87

4. Each of two persons owes $1200. The first said to the
seoofnd, "If you give me f of what you have, I shall have
enough to pay my debt." The second replied, " If you give
me f of what your purse contains, I can pay my debt." How
much does each have ?

. 6; Two passengers have together 400 pounds of baggage.
One pays $1.20, the other $1.80, for excess above the weight
allowed. If all the baggage had belonged to one person he
would have paid $4.50. How much baggage is allowed free ?

6. A number is formed by two digits. The sum of the
digits is 6 times their difference. The number itself exceeds
6 times the sum of its digits by 3. Find the number.

7. A number is formed by two digits of which the sum
is 8. If the digits are interchanged, 4 times the new number
exceeds the original number by 2 more than 5 times the sum
of the digits. Find the original number.

8. Three brothers. A, B, C, have together bought a house
for $32,0.00. A could pay the whole sum if B would give him
f of what he has ; B could pay it if C would give him f of
what he has ; and C could pay the whole sum if he had i of
what A has together with ^^ of what B has. How much
does each have?

9. A and B entered into partnership with a joint capital
of $3400. A put in his money for 12 months ; B put in his
money for 16 months. In closing the business, B's share of
the profits was greater than A's by 3*^ of the total profit. Find
the sum put in by each.

10. A capitalist makes two investments ; the first at 3 per
cent, the second at 3^ per cent. His total income from the
two investments is $427. If $1400 was taken from the sec-
ond investment and added to the first, the incomes from the
two investments would be equal. Find the amount of each
investment.

88 COLLEGE ALGEBRA

11. A cask contains 12 gallons of wine and 18 gallons of
water ; a second cask contains 9 gallons of wine and 3 gallons
of water. How many gallons must be taken from each cask,
so that, when mixed, there may be 14 gallons consisting half
of water and half of wine ?

12. A and B ran a race to a post and back. A returning
meets B 30 yards from the post and beats him by 1 minute.
If on arriving at the starting place A had immediately
returned to meet B, he would have run ^ the distance to the
post before meeting him. Find the distance run, and the time
A and B each makes.

13. A and B together can do a piece of work in 16 days.
After working together for 6 days, A leaves off and B finishes
the work in 30 days more. In how many days can each do
the work ?

14. A and B together can do a piece of work in 12 days.
After working together 9 days, however, they call in C to aid
them, and the three finish the work in 2 days. C finds that
he can do as much work in 5 days as A does in 6 days. In
how many days can each do the work ?

15. A pedestrian has a certain distance to walk. After
having passed over 20 miles, he increases his speed by 1 mile
per hour. If he had walked the entire journey with this
speed, he would have accomplished his walk in 40 minutes
less time; but, by keeping his first place, he would have
arrived 20 minutes later than he did. What distance had he
to walk ?

16. A person invests $10,000 in three per cent bonds,
$16,500 in three and one-half per cents, and has an income
from both investments of $1056.25. If his investments had
been $2750 more in the three per cents, and less in the three
and one-half per cents, his income would have been 62^ cents
greater. Find the price of each kind of bonds.

CHAPTER yill

mVOLUTION AND SVOLUTION

137. Involution is the operation of raising an expression to
any leqaired power. (See § 15.)

Every case of involution is merely an example of multipli-
cation, in which the factors are equal.

138. Index Law. If w is a positive integer, by definition

a^ssaxaxa-'tow factors. (§ 16)

- Consequently, if m and n are both positive integers,

(a*)** = a* X a" X a" • • • to w factors

B= (a X a • • • to ri factors) (a x a • • • to ri factors)
. . . taken m times

= a X a X a • • • to mn factors

The above is the index law for involution.
Similarly,

(a*)* = a*" = (a")**.

Also, (ahy ^ abxdb" 'to n factors

= (a X a • • * to ri factors) (bxh-to n factors)

139. If the exponent of the required power is a composite
nmnbei^.the exponent may be resolved into prime factors, the
power denoted by one of these factors found, and the result
raised to a power denoted by another factor ; and so on.

Thus, the fourth power may be obtained by taking the second power
vf ttsB second power ; the sixtli by taking the second power of the third
power ; and so on.

89

90 COLLEGE ALGEBRA

140. From the Law of Signs in multiplication it is evident
that all even powers of a scalar number are positive ; all odd
powers of a scalar number have the same sign as the number
itself. (§ 55)

The even powers of two compound expressions which have
the same terms with opposite signs are identical.

Thus, (6 - a)2 = { - (a - b)}^ = (a - b)^

141. Binomials. By actual multiplication we obtain

(a-{-by = a^-{-2ab -f ^^

(a + i)8 = a« 4- 3 a% -f- 3 a5« + ^»;

(a 4- by = a* 4- 4a8^ 4- 6 a^b^ 4- 4a*» 4- b\

In these results it will be observed that :

1. The number of terms is greater by one than the expo-
nent of the power to which the binomial is raised.

2. In the first term the exponent of a is the same as the
exponent of the power to which the binomial is raised, and
it decreases by one in each succeeding term.

3. b appears in the second term with 1 for an exponent,
and its exponent increases by 1 in each succeeding term.

4. The coefficient of the first term is 1.

5. The coefficient of the second term is the same as the
exponent of the power to which the binomial is raised.

6. The coefficient of each succeeding term is found from
the next preceding term by multiplying the coefficient of
that term by the exponent of a, and dividing the product by
a number greater by 1 than the exponent of b.

If b is negative, the terms in which the odd powers of b
occur are negative.

Thus, (a - 6)* = a* - 4 a86 4 6 a262 _ 4 a68 4 &*.

By the above rules any power of a binomial of the form
a -{- b or a — b may be written at once.

INVOLUTION AND EVOLUTION 91

142. The same method may be employed when the terms
of a binomial have coefficients or exponents,

(1) (a - 6)« = a* - 3a26 H- 3a62 _ 68

(2) (6 JB^ - 2 y8)8 = (6 x2)8 - 3 (5 a;2)2(2 yS) + 3 (6 a;2) (2 y«)^ - (2 y*)'

= 125x5 - 150x*y8 + 60 xV - 8y».

In like manner, a polynomial of three or more terms may be
raised to any power by enclosing its terms in parentheses, so
as to give the expression the form of a binomial.

(3) {x»- 2x2 + 3x4-4)2

= {(x8-2x2) + (3x + 4)}2

= (x8 - 2x2)2 4. 2(x8 - 2x2)(3x H- 4) + (3x + 4)2

= x« ~ 4x6 + 4x4 H- 6x4 - 4x3 - iqx^ + 9x2 + 24x + 16

= x« - 4x6 + lOx* - 4x8 - 7x2 + 24x H- 16.

Exercise 15

Perform the indicated operations :

1. (2 ay. 4. {-4.h^c)\ 7. (-5a«ftV)«.

2. (3aV)«. 5. {-a^'^cy, 8. (Ba^Z^V)*.

/2a^Y 6 (3 a*Z>^)^ (- 3 a'a;^^

V3 c»a;V • (9 a^^y * (6 a^Z^a;^)^ '

(66V)''(a^6)'' (9a;y)* ■ (2^)" '

12. (a; + 3)". 16. (l-4a;)«. 18. (^ - ^)

13. (1 - 2xy. 16. Tl - ^ j . 19. (1 + 3 a;)'.

92 COLLEGE ALGEBRA

2a%^- — y 23. (l+3a;-a;y.

143. Evolution is the operation of resolving a number into
factors all equal to one another. If a number is resolved into
two equal factors, either factor is called a square root of the
number ; if into three equal factors, each factor is called a cube
root ; if into four equal factors, each factor is called a fourth
root ; and generally if a number is resolved into n equal
factors, n being any positive integer, each of these factors
is called an nth root of the number.

Under the term number is included any algebraic expression
(§7), whether monomial or polynomial, integral or fractionaL

The symbol which denotes that a square root is to be
extracted is V 5 ^^^ ^ or other roots the same symbol is used,
but with a figure written above to indicate the root; thus,
V^ V> 6tc., signify the third root, fourth root, etc.

144, If k, m, and n are positive integers, we have

(a"*)** = a*^. (§ 138)

Therefore, a"* is an Tith root of a*^.
That is, a"* = one value of Va"~.

Also, § 138, {a^^^y = a^h^.

Consequently, a^"^ = one value of Va**ft"^.

Hence, a root of a monomial is found by dividing the expo-
nent of each factor by the index of the root and taking the
product of the resulting factors, first expressing the jiumerical
coeflBicient, if other than 1, as a product of its prime factors.
The root thus obtained is called the principal root of the mono-
mial for the given index.

Thus, 3 a8 is the principal fourth root of 81 a^^ (= Z^a^,

INVOLUTION AND EVOLUTION 93

By the Law of Signs for Multiplication, § 55,

(+a)x(+a) = +a2,
and (— a) X (— a) = + a^.

Therefore, V+ a^ may be either + a or — a,
but V— a^ can be neither + a nor — a.

Hence,

I. Every positive number has two square roots, equal in
absolute value but opposite in sign, one being positive, the
other negative. This is indicated by writing the double sign
± before the root, which sign is read jplus or minus.

Hence, also, any even-indexed root of a positive number will
have the double sign ±.

II. No scalar number can be the square root of a negative
number.

III. An odd-indexed root of a scalar number has the same
sign as the number itself.

145. The indicated even root of a negative number is called
an imaginary or orthotomic number.

146. Square Roots of Compound Expressions. Since the square
of a-\-h is a^ -\-2db -\- b% the square root of a^ -\-2ah -\- h^
is a + ft.

It is required to devise a method for extracting the square
rpot a-\-b when a^ -f 2 aft -f ft^ is given.

The first term a of the root is obviously the square root of the first

term a^ of the expression.
a^ ■}■ 2 ab ■}■ y^(a -{■ h Ifa^is subtracted from the given expres-

a^ sion, the remainder is 2 a6 + 62. There-

2a4-ft 2a64-62 fore, the second term 6 of the root is

2a6 -f ft^ obtained by dividing the first term of this

remainder by 2 a, that is, by double the
part of the root already fownd. Also, since 2 a6 + 62 = (2 a -f 6) 6, the
divisor is com/pldedi, by adding to the trial-divisor the new term of the root.

The same method applies to longer expressions, if care is
taken to obtain the trial-divisor at each stage of the process

94 COLLEGE ALGEBRA

by dovhling the part of the root already foundj and to obtain
the complete divisor by annexing the new term of the root to the
trial-divisor.

Find the square root of

1 + 10 aj2 + 25 a:* + 16a:« - 24aj^ - 20a;« - 4a:.

Arrange according to ascending or descending powers of x. Thus,

16x6 - 24x5 4- 25 X* - 20x8 + lOx^ -4x4-1 (4x8 - Sx^ 4-2x - 1
16x6

8x8-3x2

-24x5 + 25x*
-24x6+ 9x*

8x3-6x2 + 2x

16x*- 20x8 + 10x2
16x*-12x8+ 4x2

8x8-6x2 + 4x- 1

- 8x8+ 6x2-4x+l

- 8x3+ 6x2-4x+l

It will be noticed that each successive trial-divisor may be obtained
by taking the preceding complete divisor with its last term doubled.

147. Square Roots of Arithmetical Numbers. In extracting
the square root of a number expressed by figures, the first step
is to separate the figures into groups.

Since 1 = 12, 100 = 102, io,000 = 1002, and so on, it is evident that the
square root of any integral square number between 1 and 100 lies between
1 and 10 ; the square root of any integral square number between 100 and
10,000 lies between 10 and 100 ; and so on. In other words, the square
root of any integral square number expressed by one or two figures is a
number of one figure ; the square root of any integral square number
expressed by three or four figures is a number of two figures ; and so on.

If, therefore, an integral square number is divided into groups of two
figures each, from the right to the left, the number of figures in the root
is equal to the number of groups of figures. The last group to the left
may consist of only one figure.

Find the square root of 3249.

32 49 ( 57 In this case, a in the typical form a2 + 2 a6 + 6^ represents
25 5 tens, that is, 50, and h represents 7. The 25 subtracted is

107)7 49 really 2500, that is, a2, and the complete divisor 2a + 6 is
7 49 2 X 50 + 7 = 107.

INVOLUTION AND EVOLUTION 95

The same method applies to numbers of more than two
groups by considering that a in the typical form represents
at each step the part of the root already found, and that a rep-
resents tens with reference to the next figure of the root.

148. If the square root of a number has decimal places, the
number itself has twice as many.

Thus, if 0.21 is the square root of some number, this number is

(0.21)2 = 0.21 X 0.21 = 0.0441.

Hence, if the given sq: ^ number contains a decimal, we divide it

into groups of two figures each, by beginning at the decimal point and
proceeding toward the left for the integral number and toward the right
for the decimal. We must be careful to have the last group on the right
of the decimal point contain two figures, annexing a cipher when neces-
sary.

Find the square root of 41.2164, and of 965.9664.

41.21 64(6.42 9 66.96 64(31.08

36 _9_

124J621 61)66

496 61

1282 )25 64 * 6208 ) 496 64

26 64 496 64

149. If a number contains an odd number of decimal places,
or gives a remainder when as many figures in the root have
been obtained as the given number has groups, then its exact
square root cannot be foimd. We may, however, approximate
to the root as near as we please by annexing ciphers and con-
tinuing the operation.

Find the square root of 3, and of 357.357.

3(1.732 ... 3 67.36 70(18.903- • •

1 1

27)2 00 28) 2 67

189 2 24

343)1100 369)33 36

10 29 33 21

3462)7100 37803)14 70 00

69 24 11 34 09

96 COLLEGE ALGEBRA

Ezerciee 16

Simplify :

1. Vl6 a^b\ 3. V'Sl a^h'^ 5. •V^1024 a^^'bK

„ ■V^27 a%^ ^J^625^ •v^216 «•«»

^. — » 4. - • g, •

Extract the square root of :

7. l+4a; + 10aj2 + 12aj8 + 9a;*.

8. 9-24a;-68a;2 4-112a;» + 196a;*.

9. 4-12a; + 5a:2 + 26a;»-29a;*-10a:^ + 25aj«.

10. 36 ic2 - 120 a'x - 12 a^aj + 100 a* + 20 «• + a*.

11. 4 + 9y2_20aj + 25ar2 + 30a;?/-12y.

12. 4x* + 9?/6-12a;y + 16ic2 + 16-24y«.

sc* y^ x^ xh/ 1 y*
^^•'4"^9""^T~"3""^16~'6* •

Extract to four places of decimals the square root of :

14. 326. 15. 1020. 16. 3.666. 17. 1.^1213.

150. Cube Roots of Compound Expressions. Since the cube of

a -h ^ is a' + 3 a^^ + 3 «Z»^ + h^, the cube root of

tt^ 4- 3 tt^Z, + 3 aft* + ft« is a + ^.

It is required to devise a method for extracting the cube
root a -\-h when a^ + 3 a^ft 4- 3 ah^ -j- h^ is given.

Find the cube root of a^ + 3 a^h + 3 aP + h\
3a2 a3

+ 3a6 + 62

3a2 + 3a6 + 62

3a26 + 3a62 4.&8
3a26 + 30624. 68

The first term of the root is a the cube root of a*.

INVOLUTION AND EVOLUTION 97

II a^ is subtracted, the remainder is 3 a^ft + 3 06^ 4. 5? j therefore, the
second term h of the root is obtained by dividing the first term of this
remainder by three times the square of a.

Also, since 3 a^d + 3 aft^ 4. &» = (3 0^2 ^ 3 gb + 62)5^ the complete divisor

is obtained by adding 3 06 4- 6^ to the trial-divisor 3 a^.

The same method may be applied to longer expressions by
considering a in the typical form S a^ -\- S ab -\- b^ to represent
at each stage of the process the part of the root already found.

Find the cube root of aj** — 3 a^ + 5 x* — 3 aj — 1.

|X2 - X - 1

3fi -3x6 4.6x8-3x-l
3x* x6

(3X2 _ X) (- X) = -3x8-f X2

3x*-3x8+ x2

-3x6 4-5x»

-3x6 +3x*- x8

3(x2 - x)2 = 3x* - 6x8 4. 3x2
(3x2- 3x _ 1) (_ 1) = -3x2 4-3x4-1

3x*-6xs 4-3x4-1

-3x*4-6x3-3x-l

-3x*4-6x8-3x-l

The first trial-divisor is 3x*, and the first complete divisor is
3x* - 3x8 4- x2. The second trial-divisor is 3(x2-x)2, or 3x*-6x84-3x2.
The second term of the root is found by dividing - 3 x^, the first term
of the remainder, by 3x*, the first term of the fu\st trial-divisor. The
second complete divisor is 3x* — 6x8 4-3x4-1.

151. Cube Roots of Arithmetical Numbers. In extracting the
cube root of a number expressed by figures, the first step is to
separate the figures into groups.

Since 1 = 18, 1000 = 108, 1,000,000 = 1008, and so on, it follows that
the cube root of any integral cube number between 1 and 1000, that is,
of any integral cuTse number that has one, two, or three figures, is a num-
ber of one figure ; and that the cube root of any integral cube number
between 1000 and 1,000,000, that is, of any integral cube number that
has four, five, or six figures, is a number of two figures ; and so on.

If, therefore, an integral cube number is divided into groups of three
figures each, from right to left, the number of figures in the root is equal
to the number of groups. The last group to the left may consist of one,
two, or three figures.

98

COLLEGE ALGEBRA

If the cube root of a number has decimal places, the number

itself contains three times as many.

Hence, if a given number contains a decimal, we divide the figures
into groups of three figures each, beginning at the decimal point and pro-
ceeding toward the left for the integral number, and toward the right for
the decimal. We must annex ciphers if necessary, so that the last group
on the right shall contain </iree figures.

If the given number is not a perfect cube, zeros may be
annexed and an approximate value of the root found.
152. In the typical form, the first complete divisor is,

3 a2 + 3 a5 + h\

and the second trial-divisor is 3 (a + 6)^, that is,

3a^-\-6ab-{-3b%

which may be obtained from the preceding complete divisor
by adding to it its second term and twice its third term.

Extract the cube root of 5 to five places of decimals.

5.000(1.70997

1

3 X 102 = 300

4000

3(10 X 7) = 210

72= 49^

559 y

3913

259 J

87000000

3 X 17002 = 8670000

3 (1700 X 9) = 46900

92 = 81 ^

8716981

>

78443829

46981 >

86661710

3 X 17092 = 8762043

78868387

67033230

61334301

After the first two figures of the root are found, the next trial-divisor
is obtained by bringing down the sum of the 210 and 49 obtained in com-
pleting the preceding divisor ; then adding the three lines connected by
the brace, and annexing two ciphers to the result.

INVOLUTION AND EVOLUTION 99

The last two figures of the root are found by division. The rule in
such cases is, that two less than the number of figures already obtained
may be found without error by division, the divisor to be employed being
three times the square of the part of the root already found.

153. Since the foiu'tli power is the square of the square, and
the sixth power the square of the cube, the fourth root is the
square root of the square root, and the sixth root is the cube
root of the square root. In similar manner, the eighth, ninth,
twelfth, • • • roots may be found.

Exercise 17

Extract the cube root of :

1. 27 - 108 a; + 144 a;2 - 64 «».

2. a;»-3ic'* + 5aj«-3a;-l.

3. «._-«.j + ___.

4. 1 - 6a; + 21a;2-44aj8 + 63a;* -54a;'^ + 27 a?«.

5. 27 + 296a;« - 125a;« - 108a; + Oa;^ - 15a;* - 300a;^

6. 64a;« + 192 a;*^ + 144a;* - 32 a;^ - 36 a;*-^ + 12a; - 1.

7. 1 - 3a; + 6a;2 - 10a;« + 12a;* - 12 a;^ + 10 a;« - 6a;^

+ 3a;8-a;».

8. a«-12a^* + 60a*62_i5o^8^8_^240a^6*-192a*« + 64ft«.

9. 8a«-f-48a«^> + 60a*^>2-80a868-90a25*-hl08a6«-276«.

10. 12a;2~i^-54a;-59 + — + 8a;8 4-^-

a;" X X*

11. 8a;»-36aa;2 + -V5^ + 66a2a;-^'-63a».

X^ X X*

Extract to three places of decimals the cube root of :

12. 517. 13. 1637. 14. 3.25. 15. 20.911.

> .

CHAPTER IX

EXPONENTS

154. Positive Integral Exponents. If a is any definite num-
ber or any algebraic expression having one and only one value^
and m and n are positive integers, we have, by the definitions
of involution and evolution, §§15 and 19,

a'' = axaxa--ton factors,
and ("va)" = a.

We also know that a» = a"~^ X a, (§ 56)

and a^ = 1. (§ 15)

We now easily deduce the following Laws of Calculation :
If a is any definite number or a single-valued algebraic
expression and m and n are positive integers,

I. a" -5- a" = a"*~", if n<m, or if n = m;
II. a"^a" = — —-> if n>m;

III. (a»)" = a"" ;

IV. (•v^)» = a»

155. To obtain an interpretation of negative exponents we
extend law I to include the case n>m\ that is, we assume
that law I holds true for all integral values of w — n, negative
as well as positive, and interpret the result so that it shall be
consistent with law II.

To obtain an interpretation of fractional exponents we
extend law III to include all cases in which mn is int^pral ;
that is, we assume that law III holds true for all integral

100

EXPONENTS 101

▼alnes of n and mn, negative as well as positive, and so inter-
pret the results that they shall be consistent with laws II
and IV.

156. Negative Integral Exponents. If we divide a* succes-
sively by a in the ordinary manner, we have the series

If we divide again by a by subtracting 1 from the exponent
of the dividend, we have, since law II holds true, the series

a', a^, a}, a°, a~^, a~^, a~^ [2]

If we compare [1] and [2], we see that

aP = l, a~^ = -> a~^ = — > a~* = ~'

From the preceding we see at once that we may interpret
a"" as equivalent to — consistently with law II.

Hence, aJ* = a x a x a -- -to n factors ;

and or"" = -x-X to?i factors.

a a a

157. Positive Fractional Exponents. If t^ is a positive integer,

- is a positive fraction.
n

We have, by the extended interpretation of law III,

1 1

(««)**= a«''" = a^ = a.

Taking the nth root of each side, we obtain

1

a^ = >y^; (§144)

1

that is, a" may be taken as denoting any number which when
raised to the ?ith power produces a, and this is exactly what

Va denotes. For example, 4* = Vi = ± 2.

102 COLLEGE ALGEBRA

Again, if m and n are both positive integers, by the extended
interpretation of law III,

m m

, — . — X •

(ary = a* = a" ;
but ( Va*")* = a"*. .'. a" = Va".

Hence, in a fractional exponent, the numerator indicates a
power, and the denominator a root.

158. Negative Fractional Exponents. If n is a positive integer,

is a negative fraction, and we have, by the extended inter-

n

pretations of laws I and III,

1 I 1

/ — -\« — x« , JL

(a ») =a » =a-i = — .
^ ^ a

Taking the nth root of each side, we obtain

a-i = ^ = \- (S 144)

Again, if m and n are both positive integers, by the. extended
interpretations of laws I and III,

(a ») =a » =a "• = —

Taking the nth root of each side, we obtain

_!= 1 1

a " = = — •

Vo^ a-
Hence, whether the exponent is integral or fractional^ we
have always a~* = — •

It is worthy of notice that while we have by definition

1

it does not necessarily follow that

{a*y = a.

EXPONENTS 103

An example will make this plain.

(4*)2 = (± 2)2 = 4 ;
but (42)* = 16* = ± 4.

Hence, if a" = ^"j it does not necessarily follow that a = b;
all we are entitled to say is that if b takes in succession all
possible values, one of these values must be a.

159. Index Laws. We shall reserve further discussion of
this subject and the full and complete statement of the Index
Laws to Chapter XXXIII. Meanwhile, if we take into consid-
eration only the principal values of all roots indicated, we may
enunciate the Index Laws as follows :

If a and b are single-valued expressions or numbers and m, n,
r, s are any scalar integers, excluding zero values of m and n,

I. a"xa"=a^" ""^

r ■ Ts

II. (a»)" = a"".

III. (ab)» = a»b".

160. Compound ezpressions are multiplied and divided as
follows :

(1) Multiply aj* -f- x^y^ + y* by x* — x^y^ + y*.

X* 4- x^y^ + 2/*
X* — x^y^ + y*
X + x^y^ 4- x*2/*

— x^y^ — x*2/* — xV*

+ x*y* + x^y^ + y

X 4- xV* 4- y

(2) Divide -J^ 4- Vx - 12 by -J^ - 3.

x'4- x*-12|x*-3

x^-3x* x*4-4

4-4x*-12
4-4x*-12

104 COLLEGE ALGEBRA

Exercise 18

1. Express with radical signs and positive exponents

2. Express with fractional exponents :

V^; ^; 4-5 \/i; 4='

3. Express with positive exponents :

— 2

4. Express without denominators ;

(4^)^' V5^' 33^-^' 3^ •
Simplify :

6. a* X a* X Vo^ ; i Vc -^ (cx)^ ; (a* Voa)*.

.. (3.)'Vpir. (^)-, (jf^.)-, (^)-

Multiply :

8. x^ — x^-\-l by aj^ + 1.

9. x^P -\- xPyP + y^P by oj^i' — a''?/'' + 2^^

10. 8a* + 4a^^>-^4-5a*^>"* + 96~* by 2a*-r*.

Divide :

11. aj^^ + y""* by a;" + 2/*.

12. x — y-^ by aj* — x^"^ + ic*y"^ — 3/^'.

13. a* H-^^ + c"*- 3 a*^>*c~* by «♦ + ** + <?"*

EXPONENTS ' 106

RADICAL EXPRESSIONS

161. An indicated root that can be obtained approximately
but not exactly is called a surd.

The index of the required root shows the order of a surd ;
and surds are named quadratic, cubic, biquadratic, according
as the second, third, or fourth roots are required.

The product of a rational factor and a surd factor is called
a mixed surd ; as, 3 V2, b Va.

When there is no rational factor outside of the radical sign,
the surd is said to be entire ; as, V2, Va.

162. Since Va x V^ x Vc = Vaic, the product of two or
more surds of the same order will be a radical expression of
the same order, the number under the radical sign being the
product of the numbers under the several radical signs.

In like manner, Va^ = Va^ x "V^ = a V^. That is,

A factor under the radical sign the root of which can be taken
may, by having the root taken, be removed from under the radical
sign.

Conversely, since a V^ = Vo^,

A foMor outside the radical sign may be raised to the corre-
sponding power and placed under the radical sign.

By "va, where a is positive, is meant hereafter in this
chapter the positive number which taken n times as a factor
gives a for the product.

163. A surd is in its simplest form when the expression
under the radical sign is integral and as small as possible.

Surds which, when reduced to the simplest form, have the
same surd factor are said to be similar.

Simplify -v^lOS; "v^T^.

106 COLLEGE ALGEBRA

164. The product or quotient of two surds of the same oi
may be obtained by taking the product or quotient of
rational factors and of the surd factors separately.

Thus, 2 Vs X 5 V7 = 10 V36.

Surds of the same order may be compared by express
them as entire surds.

Compare f Vl and f VlO.

|VlO = Vi^.
V-^= v^, and V^ = V^.
As V^ is greater than V^, | Vio is greater than f V?.

165. The order of a surd may be changed by changing
power of the expression under the radical sign.

Thus, V5 = V2b ; Vc = V^.

Conversely, V26 = Vs ; Vc^ = Vc.

In this way, surds of different orders may be reduced to
same order and may then be compared, multiplied, or divic

(1) Compare V2 and V3.

v^ = 2* = 28 = v^= -Vs;
\/3 =*3* = 3* = -v/p = "Vo.
.*. V3 is greater than V2.

(2) Multiply -vTa by V6a.

\/4a = (4a)* = (4a)' = \/(4 a)2 = \^16 a« ;
Vex = (6 x)* = (6 x)* = ■V'(6 x)8 = \^216«».
.-. \/4a X Vex = Viea2 x 216x8 = 2 V64a^.

(3) Divide "v^ by Vg^.

VSa = (3 a)* = (3 a)' = V(3 a)2 = V9^ ;
V66 = (6 6)i = (e 6)8 = ■N^(6 6)8 = V'2166«.

.-. \/3a -- V66 = \/-^ = — "^1944 a268.

\2i668 66

EXPONENTS 107

Exercise 19

Express as entire surds :

1. 3V5; 5V32; a%y/hc\ Sy^v^; a«-v^a«P.

2. 6ahc^ahc-^\ f^^J (aJ + y)\-

aj2^

Express as mixed surds :

3. vTeOoV; •V^54 ajy ; -v^eioy; •V'1372 a"6".

Simplify :

4. 2 ^80 a^hh^ ; 7V396aj; VTh; ^5 \X^'

•■ (f )fe)' (f )(?)' <--)>«'^'-

6. Show that V20, V45, V| are similar surds.

7. Show that 2 Va*P, VS^, ;^ \T ^^® similar surds

8. Arrange in order of magnitude 9 V3, 6 V7, 6 VlO.

9. Arrange in order of magnitude 4 vi, 3 V6, 6 "v3.

10. Multiply 3 V2 by 4V6; J "^ by 2^2.

11. Divide 2V5 by 3Vl5; | V2I by ^^ V^.

^. ,.^ 2VT0 7V48 4V15

12. Simplify 7= X 7= H 7=«

3V27 6V14 I5V2I
Arrange in order of magnitude :

13. 2 V^, 3 V2, j ^. 14. 3 Vi9, 5 -v^, 3 ^.
Simplify :

16. -v^a^xVa^; 3 "V^Io^ ^ V2a^.

16. Va«)'xV(f|)^; (</My X ^/(an^\

108 COLLEGE ALGEBRA

166. In the addition or the subtraction of surds each surd
must be reduced to its simplest form; then, if the restdting
surds are similar.

Add the rational factors, and to their sum annex the common
surd factor.

If the resulting surds are not similar,

Connect them with their proper signs.

167. Operations with surds will be more easily performed
if the arithmetical numbers contained in the surds are expressed
in their prime factors, and if fractional exponents are tised
instead of radical signs.

(1) Simplify V27 + ViS -f VHt. ''

y/21 = (38)* = 3 X 3* = 3 VS ;
Vis = (2* X 3)* = 22 X 3* = 4 X 3* = 4 VS;
Vii? = (72 X 3)i = 7 X 3* = 7 Vs.'
.-. V^ + Vis + Vii7 = (3 + 4 + 7) V3 = 14 V3.

(2) Simplify 2 ^^320 - 3 "v^.

2-^320 = 2{2« x5)* = 2x2«X*6* = 8'V6;
3 \/iiQ = 3(28 X 6)* = 3 X 2 X 5* = 6-^6.
.-. 2 -^320 - 3 Vlio = 8-v/6-6'v/6 = 2'v/6.

168. If we wish to find the approximate value of -—pj it

will save labor if we multiply both numerator and denomi-
nator by a factor that will render the denominator rational ;
in this case by V2.
Thus 3 _ 3\^ _8V^

* VS" V2 X V^~ 2

169. It is easy to rationalize the denominator of a fraction
when that denominator is a binomial involving only quadratic
surds. The factor required will consist of the terms of the
given denominator, connected by a different sign.

EXPONENTS 109

Thus, 7= will have its denominator rationalized if we multiply

6 + 2V6 .

both terms of the fraction by 6 — 2 v6. ^

7 - 3 VS (7 - 3 Vs) (6 - 2 Vs) _ 72 - 32 Vs _ 9 . /r
For, 7= = 7= 7=- — — J Vo.

6 + 2 V6 (6 + 2 V6) (6 - 2 V6) 16 ' 2

170. By two operations the denominator of a fraction may
be rationalized when that denominator consists of three quad-
ratic surds.

Thus, if the denominator is V6 + Vs - >^, both terms of the frac-
tion may be multiplied by VS — VS + V^. The resulting denominator
will be 6 — 6 + 2 Vo = 1-1-2 V6 ; and if both terms of the resulting frac-
tion are multiplied by 1— 2 V6, the denominator becomes 1—24 or —23.

Ezerciae 20

Simplify :

1. V27 + 2V48 + 3Vi08; 7 "v^ H- 3 "V^ H- ■v^432 .

2. 2V3H-3Vij-V5j; 2 V| -fV60 - Vi5 - Vj.

i^c /^ /^^ ^ /2 . ^ /T , fr

4. 2-v^ + 3-v'T08 + ^J^500---v^320-2-^^
6. (-v^)*; (V^)*; (-v^)'; (^)'.

6. (a-V^)-«; (aj-V^)"*; (i?'V^)*; (a-«V^^)-*.

Extract the square root of :

7. a;*"'4- 6a*"'2/»H-lla^"'2^"4-6aj"'2^''H-y*".

8. 1 H- 4a;"* ^^x"^ _ 4a;-i H- 25a;"* - 24aj"* H- 16a-«.

9. 9aj-*-18aj-«2^*H-15aj-«y — 6a;-iy* + y2^

10. Extract the cube root of

8a;« -I- 12ar^ - 30x - 35 -I- 45 oj-^ + 21 yr^ - 27aj-»

no COLLEGE ALGEBRA

Simplify ;

11,

■ ("Mr- "• (i^)^<-^>

aj«~p\p— «

17. 3 (a* + ft*)2 - 4 (a* + 5*) (a* - ft*) + (a* - 2 ft*)«.

Find equivalent fractions with rational denominators for
the following, and find their approximate values:

7V5

18. — = =• 24.

19. = =• 26.

20. =• 26.

V7 + V6

7

2V5-V6

4-V2

I + V2

6

5-2V6

2

V3

1

21. =• 27.

22. — :=• 28.

V7 + V3'

7-2V3H-3V2
34-3V3-2V2

3V5-4V2
2V5 + 3V2'

V5-V6
2V5-V6*

1

V5 + V3 + V7'

V5-V2 V5-3V2 + VT

30. Extract the cube root of

a-f _ 6a-Jft-i + 15 a-^ft-i - 20a-tft-l + 16ar-l6-i
-6a-ift-J + ft-i

CHAPTER X

QUADRATIC EQUATIONS

We now resume the subject of equations where we left it
at the end of Chapter VII. Having considered equations of
the first degree with one or more unknowns, we come next to
the consideration of quadratic equations.

171. A quadratic equation that involves but one unknown
niunber can contain only :

1. Terms involving the square of the unknown number.

2. Terms involving the first power of the unknown number.

3. Terms which do not involve the unknown number.

If the similar terms are combined, ej5?;ery quadratic equation
can be made to assume the form

ojx^ -+- fta; -i- c = 0,

where a, ft, and c are known numbers, and x the imknown
number.

If a, ft, c are given numbers, the equation is a numerical
quadratic If a, ft, c are numbers represented wholly or in
part by letters, the equation is a literal quadratic.

Thus, x^ — 6x + 5 = 0isa numerical quadratic,

and 0x2 + 26x + 3c — a6 = 0isa literal quadratic.

172. In the equation ax^ + ftaj -j- c = 0, the numbers a, ft,
and c are called the coefficients of the equation. The third
term c is called the constant term.

If the first power of x is wanting, the equation is a pure
quadratic ; in this case, ft = 0.

If the first power of x is present, the equation is an affected
or complete quadratic.

Ill

112 COLLEGE ALGEBRA

173. Solution of Pore Quadratic Equationa.

(1) Solve the equation 6 x* — 48 = 2 05*.

We have 5x*-48 = 2x^.

Collect the terms, 3 z* = 48.

Divide by 3, x^ = 16.

Extract the root, x = :t 4.

Observe that the roots are numerically equal, but one is positive and
the other negative. There are but two roots, since there are but two
square roots of any number.

It may seem as though we ought to write the sign ± before the x as
well as before the 4. If we do this, we have

+ x=+4, -x=-4, +« = -4, — « = + 4.

From the first and second, x = 4 ; from the third and fourth, a; = — 4 ;
these values of x are both given by x = ± 4. Hence, it is wtneeessory,
although perfectly correct^ to write the ± sign on hoik sides of the reduced
equation.

(2) Solve the equation 3 a* — 15 = 0.

We have 3x« = 16,

or X* = 5.

Extract the root, x = ± Vs.

The roots cannot be found exactly, since the square root of 6 oannot
be found exactly ; it can, however, be found as accurately as we please ;
for example, it lies between 2.23(306 and 2.23607.

(3) Solve the equation 3 ar^ + 15 = 0.

We have 3 x= = - 16,

or x^ = — 5.

Extract the root, x = ± V— 6.

There is no scalar square root of a negative number, sinoe any scalar
number, positive or negative, multiplied by itself, gives a positive reault.

174. A root that can l>e found exactly is called an
root or rational root. Such roots are either whole numbers
or fractions.

QUADRATIC EQUATIONS 113

A root that is indicated but can be found only approxi-
mately is called a surd root. Such roots involve the roots
of imperfect powers.

Exact and surd roots are together called real roots.

A root that is indicated but cannot be found as a number
in the arithmetical scale, either positive or negative, is called
an imaginary root. Such roots involve the even roots of nega-
tive numbers.

Exercise 21

Solve :

, a;«-5 , 2aj3-f 1 1 ^3 1 7

1. n r r» = rt' 3.

6 2 4:x^ ex^ 3

2. -A- 4. ^^__ = 8. 4. 6aj2- 9 = 2a;2 -I- 24.

1 +x 1 —X

5 15 26

3a;^-27 90 + 4a;^
aj2 + 3 ■*" x^ + 9 "■

7.

8.

4a;^ + 5 ___ 2x^-5 _ 7x^-25
10 15 "" 20 '

10x2-M7 12a;2 + 2 5a;2-4

18 11 a;2 - 8

9. x^ -]- bx '\- a = bx(l — bx),

10. oic* -f- ^ = c.

11. x^ — ax -{- b = ax(x ^1).
ab — X b — ex

12.

b — ax be — X

10 3(a; + a) 2 a; -fa ^

10. . — 7: ; =s 1,

4a; — a 2 a -\- X

114 COLLEGE ALGEBRA

14. — •+ ^

X — 5 a X -{- 3 a (x — 5 a) (a; + 3 a)
2(a'\-2b) a-2x _ h^

16. V r. '^ 4-

a'\-2x a '\-b (a -\- b) (a -^ 2 x)

175. Solution of Affected Quadratic Equations.

Since (« ± 6)^ is identical with x^±2bx -\- b^, it is evident
that the expression x^±2bx lacks only the third term b^ of
being a perfect square.

This third term is the square of half the coeflBicient of x.

Every affected quadratic may be made to assume the form
x^±2bx = c by dividing the equation through by the coeffi-
cient of x^ (§ 171).

To solve such an equation :

The first step is to add to both members the sqtiare of half
the coefficient of x. This is called completing the square.

The second step is to extract the square root of each member
of the resulting equation.

The third step is to solve the two resulting simple equsr
tions.

(1) Solve the equation a;* — 8 « = 20.

We have x^-Sx = 20.

Complete the square, x^ — Sx -\-lQ = SQ,

Extract the root, x — 4 = ± 6.

Solve, X = 4 + 6 = 10,

or x = 4 — 6 = — 2.

The roots are 10 and — 2.

We write the ± sign on only one side of the equation, for the reason
given after the first example of § 173.

Verify by putting these numbers for x in the given equation :

x = 10.

102 _ 8 (10) = 20,

100 - 80 = 20.

x = -2.

(_2)2-8(-2) = 20,
4 + 16 = 20.

1

1 '■

(2) Solve the equation

QUADRATIC EQUATIONS 116

aj + l 4.x — S

^ y / ' -^

— ■. >

x — 1 a; + 9

Free from fractions, (x + 1) (x + 9) = (x - 1) (4» - 3).

Simplify, 3x2-17» = 6.

Divide by 3, x2-J^x = 2.

Complete the square, x^-^x-^- ( V^)« = ^.

Extract the root, x — J^ = ± J^.

Solve, x = V + V = V = «,

or x = V-V = -i = -f

The roots are 6 and — J.

Verify by putting these numbers for x in the original equation :
x = 6.
6 + 1 _ 24 - 3
6-l~ 6 + 9*

* =

-i-

-i+l_

-1-

3

-i-1

-t+

0

* _
-*

-¥
¥'

-i=

-H-

176. When the coeflBicient of x^ is not unity, we may pro-
ceed as in the preceding section, or we may complete the
square by another method.

Since {ax±.hy is identical with a^^ ± 2 aftaj .+ ft^, it is
evident that the expression a^x^ ± 2 ahx lacks only the third
term h^ of being a perfect square.

This third term is the square of the quotient obtained by
dividing the second term by twice the square root of the first
term.

Every aifected quadratic may be made to assume the form
aV ±2abx=^c{% 171).

To solve such an equation.:

The first step is to complete the square ; to do this, we divide
the second term by twice the square root of the first term, square
the quotient, and add the result to each member of the eqtuitian.

The second step is to extract the square root of each member
of the resulting equation.

The third step is to reduce the two resulting simple equations.

116 COLLEGE ALGEBRA

177. Numerical quadratics are solved as follows :

(1) Solve the equation 16 a;^ -f 6 a; — 3 = 7 aj^ — a; + 45.

16x« + 5x-3 = 7x2-x + 46.
Simplify, 9 x* + 6 x = 48.

Complete the square, 9x^ + 6x + i = 49.
Extract the root, 3 x + 1 = ± 7.

Solve, 3x = -l + 7or-l-7.

.'. 3x = 6 or -8.
.'. X = 2 or — 2f .

Verify by substituting 2 for x in the equation

16x» + 5x - 3 = 7x* - X + 46.

16(2)2 + 6(2) - 8 = 7 (2)2 - (2) + 46,
64 + 10 - 8 = 28 - 2 + 46,
71 = 71.

Verify by substituting — 2f for x in the equation

16x2 + 6x - 8 = 7x2 - X + 45.

16(-})2 + 6(-f)-3 = 7(-f)2-(-J) + 46,
iPj^ _ _^ _ 3 = Afi + f + 46,
1024 - 120 - 27 = 448 + 24 + 406,
877 = 877.

(2) Solve the equation 3 «^ — 4 x = 32.

Since the exact root of 3, the coefficient of x*, cannot be found, it is
necessary to multiply or divide each term of the equation by 3 to make
the coefficient of x2 a sqiuire number.

Multiply by 3, 9 x2 - 12 x = 96.

Complete the square, 9x2 — 12x + 4 = 100.
Extract the root, 3 x — 2 = i 10.

Solve, 3x = 2 + 10 or 2 - 10.

.-. 3x = 12 or -8. .
.-. X = 4 or - 2).

Or, divide by 8, x2 - -^ = -— .

o o

o 4x 4 32 . 4 100
Complete the square, x2 — -. + - = — + - = ---.

o 9 o V 8

^ ^ 2 10

Extract the root, x - - = ± — •

QUADRATIC EQUATIONS 117

2 ±10

= 4 or - 2J.
Verify by substituting 4 for x in the original equation,

48 - 16 = 32,
32 = 32.

Verify by substituting — 2f for x in the original equation,

2H + lOf = 32,
32 = 32.

(3) Solve the equation — 3a;*4-5aj = — 2.

Since the even root of a negative number is impossible, It is necessary
to change the sign of each term. The resulting equation is

3x2 -fix = 2.
Multiply by 3, 9x2-16x = 6.

Complete the square, Ox^ — 15x + ^ = ^.
Extract the root, 3 x — f = i: }.

Solve, 3x= »

2

.-. 3x = 6 or — 1.

.-. X = 2 or — J.

Or, divide by 3, x^- — = -'

3 3

^ , , « fix 25 49

Complete the square, x* 1 — = — .

3 36 36

Extract the root, x — J = ± J.

...x = i^ = 2or-f
6 ^

If the equation 3x2 — fix = 2is multiplied by /our times the coefficient
of x^, fractions will be avoided.

36x2-60x = 24.
Complete the square, 36x2 - 60 x + 25 = 49.
Extract the root, 6 x — 6 = ± 7.

Solve, 6x = 6±7.

.-. 6 X = 12 or - 2.
.*. X = 2 or — J.

It will be observed that the number added to complete the square by
this last method is the square of the co^cient of x in the original equa-
tion 3 x2 — 6 X = 2.

118

COLLEGE ALGEBRA

(4) Solve the equation

5 — X 2x — 5

= 2.

Simplify, 4x2-23x = -30.

Multiply by four times the coefficient of x^, and add to each side the
square of the coefficient of x,

^«* - { ) + (23)« = 529 - 480 = 49.
Extract the root, 8x-23 = ±7.

Solve, 8x = 23±7.

.-. 8x = 30 or 16.
.-. X = 3} or 2.

If a trinomial is a perfect square, its root is found by taking the roots
of the first and third terms and connecting them by the sign of the middle
term. It is not necessary, therefore, in completing the square, to write
the middle term, but its place may be indicated as In this example.

(5) Solve the equation 72 x* — 30 a; = — 7.

Since 72 = 2* x 3>, if the equation is multiplied by 2, the coefficient of
z* in the resulting equation, 144 x^ — 60x = — 14, is a square number,
and the term required to complete the square is (}})* = (})* = ^.

Hence, if the original equation is multiplied by 4 x 2, the coefficient
of X* in the result is a square number, and fractions are avoided in the
work.

Multiply the given equation by 8,

676xa-240x = -6«.

Complete the square, 676x» - ( ) + 25 = - 3L

Extract the root, 24x- 5 = j: V-31.

Solve, 24x = 5±V~31.

.•.x = ^{5±V-31).

Note. In solving the following equaticxis care must be taken to select
the method best adapted to the examine under considention.

Szerdie 22

Solve :

1. jr*-2jr = 15,

%. x«-14jr = -48.

S. x«-x = 12.

4. x*-8x = 28.

6. jr«-13x-f42 = 0.

6. or* -21x + 108 = 0.

7. 2x*-fx = 6.

8. 4x* + 7x = 15.

QUADRATIC EQUATIONS 119

9. 3aj2- 19a; + 28 = 0. 11. 6aj^-a; = 12.
10. 4a;2 4-l7a. — 16 = 0. 12. 5x^ — 3^x -{■ 4^ = 0.
13. 6aj2_7aj + | = 0.

M4. ^^+(a; + l)(a; + 2)=0.

16. (oj - 6)2 + a;2 - 6 == 16(aj + 3).

aV 3aj-19 11+aj ^^ 6 .a; 8

16. -^ + 5 = — ^' 21. ;r - + - = -.

6 3 3 2aj — 6 3 — aj x

2aj2-ll aj-fl «« aj + 2 4 - aj 7
17 = — 1 — . 22. — ■ = -.

2a; + 3 2 aj - 1 2aj 3

a; -f 1 a; __ 11 a; — 6, ^ + 5.,

^®- "^"^6"2^' ^^' ^:r2"^2^Ti"

a;2-4 . 2aj .l-2a; ^ aj - 3 . aj - 4 1

19. — r f--=- = icH z 24. 7 +

3aj '6 5 aj + 4 ' 2(a;-l) 2

20. , + ^ = 2(a;~2). 25. ■^ + '-" '

a._6 ^ ^ a;2-4 ' a; + 2 5(x-2)

x-5 x-S_ 80 1
a; + 3"^a;-3""a;2-9"^2*

1.7 14 a; -4

27. =

aj-3 aj + 3 x^-9 x-^S

3 a; + 5 a;H-3 _ a; -- 1
^^' a;4-3 aj-3~aj2-9*

^ aj + l.ajH-2 2ajH-13
a; — 1 x — 2 X'\-l

2aj-l . 3a;-l 7-x ,

30. — TT" + — rir + — ? = '*•

aj-hl a; + 2 x — 1
1 — oaj 1-foa; 25 a;^ — 1

120 COLLEGE ALGEBRA

X'\-7 1—x

32.

9-4a;2 2a; + 3 2«-3

^^- a: + 3 ^ x + 2 "" ^^-

178. Literal quadratics are solved as follows :

(1) Solve the equation ax^ + bx -{- c = 0.

Transpose, ax^ + 6x = — c.

Multiply the equation by 4 a and add the square of 6,

4agxg + () + 5^ = &^-4ac.
Extract the root, 2 ox + 6 = ± V6^ -4ac.

Solve, 2 ox = - 6^± V6a-4ac.

-'6± V6a-4ac

.*. X = •

2a

<

(2) Solve the equation adx — ctcx^ = 6ca: — bd.

Transpose hex and change the signs,

acx^ + 6cx — adx = bd.

Express the left member in two terms,

acz^ + (6c — ad) x = bd.
Multiply by 4 ac,

4a2cax2 + 4ac(6c - ad)x = 4abcd.

Complete the square,

4a2cax2 + ( ) + (6c - ad)2 = ftac^ + 2a6cd + a«d».

Extract the root, 2 acx + (6c — ad) = db (6c + ad).

Solve, 2 acx = — (6c — ad) ± (6c + ad).

.*. 2 acx = 2 ad or — 2 6c.

d 6

.*. X = - or .

c a

CO'

(3) Solve the equation ^a;^ — ^aj + g'x* -{- qx = -^ —
Express the left member in two terma,

Multiply by 4 times the coefficient of x^,

4(p + q)^* - 4(p2 _- g2)x = 4pq,

QUADRATIC EQUATIONS 121

Complete the square,

4(i) + g)2x2 - ( ) + (p - g)2 = 1)2 + 2pg + g2.

Extract the root, 2 (p +g) x— (p — g) = ± (p + ?)•

Solve, 2(p + g)x = (p - q) ± {p + g).

.'.2(p-{'q)x = 2poT—2q.

p q

.-. X = —4-— or — ^

Observe t^a^ f Ae left member of the simplified equation must he expressed
in two terms, simple or compound, the first term involving x^, ike second
involving x.

Exerciae 23
Solve :

11. 2a^ + — = (a + 6)aj.

2. x^'\-7a^ = Sax. ^

3. 4a;(a;-a)H-a2 = J«. 12. (a; + w)2 + (a;-m)«=5mx.
.4. ---j = 2a(x + 2a). is. aa!» + 5 «*« + ^ = 0.

6. a^ = oa; + J. 14 J (a - «)» = (6 - 1) «».

lo. h

6.

(x + a)* (x - a)2

7.

, X 3

a 4a^

8.

05* — (a + ^) aj = — a^.

16.

a—x X a—x

a^ — ab _^x-^ a
x^b ¥~"

mTi x — 2a a a

2x(a — x) a a/x a'\-x _ Ba-^-x

• Sa^2x ~"4' • V^~^ 2b

19. 7— = a -f- 5 — (a — &)«.

ax — bx ^ ^

Bah — ^b^ — ax 2a-\-x

20. = ^ —

2a^x 3

122 COLLEGE ALGEBRA

It 4

3a 2a 4a , a
22. — : 1 — jr- = V

23.

a54-a x-\-2a x x-\-Sa

a — b-\-x a-\-b _ ^ ^^ 4:(x-\-a) __ 3(^4-^) _ ^

a + b + x x-\-b ' ' a-\-b x + a

a; + 2ft fl;-25 a 4a« + 95« " '

27. (3a« + 5«)(aj«-aj + l) = (a« + 3ft«)(i»« + aj + l).
4a« 5* 4a«-ft«

28.

29.

30.

a; 4- 2 a; -2 a;(4-a;2)

a-{-2b_ a^ Ab^

a-2b'' (a-2b)x x^

x-\-l 2 x-\'2
c ex ax — bx

a — c x — a Sb(x — e)

3J^^ . =5 i i — .

x—a a — c (a — c)(x — a)
32. x(X'\'b^-b):=ax(a + l)^(a + by (a - b).
X . C4 m* — n*) mn 4 m* -|- n*

84. -4--fl + — V + - + - = 0.
w -f n \ mw/ m n

2ab (3x-l)b' ^(2x-^l)a*
^^' 3a; + l 2a; + l " 3a; + l

x-\-2a — 4:b 8b — 7a x — 4:a

^^' 2bx ax-2bx^ 2(aJ}-2i^)'^

1 X . ar-65 ar + 195-2a

a4-25 a*-45«^(a + 25)a; 25aj-aa?

QUADRATIC EQUATIONS 123

a; + 25"^ a;-6a + 3ft

a; + 35 . 35 a-|-35

39. 77-= 777-7 +

8a«-12aft 4a2-9i2 (2a + 3i)(a;-35)

1 1 a 2hx^h

2aj« + a;-l"*"2a^-3aj + l~2to-ir"^ a-aaj«"

1 4aa^ + 35(2--a;)
«■*" 2ax' + 2a + Sh

x — a , 2(db-ax-\'2m 1

2 5 (« 4- a) a (a; -h a)* a

ax-\'h ax — b a^x^ — 6*

44. ^i^±» + li£±£l = 2.
05 — 3a-|-o oj-l-o + c

179. Solutions by a Formula. Every affected quadratic may
be reduced to the form x^ -\- px -i- q = 0, in. which p and q rep-
resent numbers, positive or negative, integral or fractional.

Solve X* -^px -f 2' = 0.

Complete the square, 4x2 + ()+i)2=p2 — 4g.

Extract the root, 2z+p = ± -y/p^ — 4 g. 0 ( ( j,

.••« = -|±2Vp^-4g. ^

By this formula the values of aj in an equation of the form
x^ -\- px -\- q = 0 may be written at once.

Thus, take the equation

3x2-5x4-2 = 0.

Divide by 3, «2-Jx + f = 0.

Here, p = — }, and q — \.

= 1 or f

124 COLLEGE ALGEBRA

180. Solutions by Factoring. A quadratic whioh has been
reduced to its simplest form, and has all its terms written
on one side, may often have that side resolved into factors by
inspection.

In this case the roots are seen at once without completing
the square.

(1) Solve «« -I- 7 a; - 60 = 0.

Since x^ + 7a. _ 60 = (at + 12) (« - 6),

the equation x^ + 7 a: - 60 = 0

may be written (x 4- 12) (x — 5) = 0.

It will be observed that if either of the factors x + 13 or s — 5 is 0,
the product of the two factors is 0, and the equation is satisfied. '

Hence, x 4- 12 = 0, or x — 6 = 0.

.-. X = — 12, pr X = 6.

(2) Solve a;2 -f 7 a; = 0.

The equation x^ + 7 x = 0

becomes x (x -f 7) = 0,

and is satiafied if x = 0, or if x + 7 = 0.

Therefore, the roots are 0 and — 7.

It will be observed that this method is easily applied to an equation
all the terms of which contain x.

(3) Solve 2a;8-a;«- 6a: = 0.

The equation 2x8-xa-6x = 0

becomes x (2 x^ — x — 6) = 0,

and is satisfied if x = 0, or if 2 x^ — x — 6 = 0.

By solving 2 x^ — x — 6 = 0 the two roots 2 and — | are foond.
Therefore, the equation has three roots, 0, 2, — {.

(4) Solve a;«-fa;^-4:a;-4 = 0.

The equation x^ + x^ — 4x — 4 = 0

becomes x^ (x + 1) — 4 (x + 1) = 0,

or («^ - 4) (X + 1) = 0.

Therefore, the roots of the equation are — 1, 2, — 2.

(5) Solve «« - 2 a;2 _ 11 x -f 12 = 0.

By trial we find that x — 1 is a factor of the left member (§ 87).

QUADRATIC EQUATIONS 125

The given equation may be written

(X - 1) (x2 - X - 12) - 0,
or (X - 1) (x + 3) (X - 4) =z^

Therefore, the roots are 1, 4, ~ 3.

(6) Solve the equation a? (a^ — 9) = a (a} — 9).

If we put a for x, the equation is satisfied ; therefore a is a root (§ 87).
Transpose all the terms to the left member and divide by x — a.
The given equation may be written

(X - a) (x2 + ox + a^ - 9) = 0,

and is satisfied if x — a = 0, or if x^ + ax 4- a^ — 9 = 0.
The roots are found to be

- a + V36 - 3 a« -a - V36 - 8 a^
a, — , .

Exercise 24

»
Find all the roots of :

1. (a;-l)(a;-2)(x«-4a;-h8) = 0.
3. a« + 27==0.

6. a:8-27 4-4:(aj2-9) = 0.

6. a^4-9ar^-16(x»-|-9) = 0.

7. 2a;«-f 3ic2-2aj-3 = 0.

9. «* — aj — 6 =» 0.

10. a^-6iB«-|-llaj-6=xO.

11. a;*-3a;«-8a^ + 6a;-|-4»:0.

12. a»-faj*- 14a; -24 51=0.

18. a5*--6a» + 9»*H^4aj-.18»0,

126 COLLEGE ALGEBRA

14. aj(a;-3)(aj + l) = a(a-3)(a + l).

15. x(x-S)(x-\-l) = 20.

16. (a;-l)(a;-2)(aj-3) = 24.

17. (aj + 2) (x - 3) (x + 4) =' 240.

18. (a; + l)(« + 5)(a;-6) = 96.

181. Character of the Roots. Every quadratic equation can
be made to assume the form

ax^ -{- bx •}- c = 0,

Solving this equation, § 178, Example (1), we obtain for its
two roots

_j-|-V62-4ac -J-V^^34

ac

J

2a 2a

There are two roots, and only two roots, since there are two,
and only two, square roots of the expression ft* — 4 ac.

As regards the character of the two roots, there are three
cases to be distinguished :

1. P — Aac positive. In this case the roots are real and
different That the roots are different appears by writing
them as follows :

b . y/b' '-4:ac b V^^TI

-7r- + ^ » -

ae

2a 2a 2a 2a '

these expressions cannot be equal since ft* — 4 ac is not zero.

If ft* — 4 ac is a perfect square, the roots are rational If
ft* — 4 oc is not a perfect square, the roots are surds.

2. ft* — 4 ac = 0. In this case the two roots are real and

eqtial, since they both become — ;r— •

^ a

3. ft* — 4 ac negative. In this case both roots have a real
part and an imaginary part and are called imaginary roots.

QUADRATIC EQUATIONS 127

If we write them in the form

_h_ -slW-^ac h ■y/b^-4: ac

2a 2a 2a 2a

we see that two imaginary roots of a quadratic cannot be equal,
since 6* — 4 ac is not zero. They have the same real part,

— ;r— f and the same imaginary parts with opposite signs.
^ a

Such expressions are called conjugate expressions.

The above cases may also be distinguished as follows :

1. b* — 4 ac > 0, roots real and different.

2. b^ — 4 ac = 0, roots real and equal

3. b* — 4 ac < 0, roots imaginary.

182. By calculating the value of i^ — 4 ew we can determine
the character of the roots of a given equation without solving
the equation.

(1) a;«-5a;-f 6 = 0.

Here a = 1, 6 = — 6, c = 6.

Therefore, 62 _ 4 ^c = 25 - 24 = 1.

The roots are real and different, and rational.

(2) 3ar^-|-7a;-l = 0.

Here a = 3, 6 = 7, c = — 1.

Therefore, 62 _ 4 ^c = 49 + 12 = 61.

The roots are real and different, and are both surds.

(3) 4ar»-12a;-f 9 = 0.

Here a = 4, 6 = — 12, c = 9.

Therefore, 62 _ 4 ^c = 144 - 144 = 0.

The roots are real and equal.

(4) 2a;«-3aj-f 4 = 0.

Here a = 2, 6 = — 3, c = 4.

Therefore, 62 _ 4 ac = 9 - 32 = - 23.

The roots are both imaginary.

128 COLLEGE ALGEBRA

(5) Find the values of m for which the equation
277ta^ -f (5m + 2)a; + (4m -f 1) = 0
has its two roots equal.

Here a = 2m, &=:5m + 2, cs4m + l*

If the roots are to be equal, we most have 6^ — 4 oc =: 0, or

(5 m + 2)2 - 8 m (4 m + 1) = 0.
The solution of this equation gives m = 2 or ~ |.
For these values of m the equation becomes

4x* 4- 122 + 9 = 0, and 4a;« -4a; + 1 = 0,
each of which has its roots equal.

Bzaroifle 25

Determine^ without solving, the character of the roots of
each of the following equations :

1. ar^- 6a; 4- 8 = 0. 6. IGac*- 66a; + 49 — 0.

2. a:2-4a; + 2 = 0. 7. Sac*- 2a; + 12 = 0.

3. a;* -I- 6a; -1-13 = 0. 8. 2a;*- 19 a; -f 17 = 0.

4. 4a;2-12a;-|-7 = 0. 9. 9a;* + 30aj + 25 = 0.

5. 5a^-9a; + 6 = 0. 10. 17a;*- 12aj + ff = 0.

Determine the values of m for which the two roots of each
of the following equations are equal :

11. (3m + l)a^ + (2m + 2)a; + m = 0.

12. (m-2)a^ + (m-6)a; + 2m-5 = 0.

13. 2ma;* + a;* — 67na; — 6a; + 6m + l=0.

14. ma;* + 2x* + 2m = 3?7ia;-9a; + 10.

183. Problems involving Quadratics. Problems that involve
quadratic equations apparently have two solutions, as a quad-
ratic equation has two roots. When both roots axe positive
integers they will give two solutions.

QUADRATIC EQUATIONS 129

Fractional and negative roots will in some problems give
solutions ; in other problems they will not give solutions.

No difficulty will be found in selecting the result which
belongs to the particular problem we are solving.

Sometimes, by a change in the statement of the problem,
we may form a new problem which corresponds to the result
that was inapplicable to the original problem.

Imaginary roots will in some problems give solutions. Their
interpretation in such cases will be given in Chapter XXXIII.

(1) The sum of the squares of two consecutive numbers is
481. Find the numbers.

Let

X = one number,

and

X + 1 = the other.

Then

x2 + (« + 1)2 = 481,

or

2x2 + 2x +1 = 481.

The solution of which gives x = 16 or — 16.

The positive root 16 gives for the numbers, 16 and 16.

The negative root — 16 is inapplicable to the problem, as consecviive
numbers are understood to be integers which follow one another in the
common scale, 1, 2, 3, 4 • • •

(2) What is the price of eggs per dozen when 2 more in a
shilling's worth lowers the price 1 penny per dozen ?

Let X = the number of eggs for a shilling.

Then - = the cost of 1 egg in shillings,

X

12
and — = the cost of 1 dozen in Bhillings.

X

But if X + 2 = the number of eggs for a shilling,

12

= the cost of 1 dozen in shillings.

then X + 2

12 12 1

... _ = _ (1 penny being ^ of a shilling).

X X "T ii 1«

The solution of which gives x = 16 or — 18.

And, if 16 eggs cost a shilling, 1 dozen will cost |} of a shilling, or
9 pence.

Therefore, the price of the eggs is 9 pe^ce per dozen.

ISO COLLEGE ALGEBRA

If the problem is changed so as to read : What is the price
of eggs per dozen when 2 less in a shilling's worth raises the
price 1 penny per dozen ? the algebraic statement is

12 12 _ 1
x-2 « ~12'

The solution of this equation gives a; = 18 or — 16.

Hence, the number 18, which had a negative sign and was inapplicable
in the original problem, is here the true result, while the — 16 is inappli-
cable in this problem.

Exercise 26 -^

1. The product of two consecutive numbers exceeds their
sum by 181. Find the numbers.

2. The square of the sum of two consecutive numbers ex-
ceeds the sum of their squares by 220. Find the numbers.

3. The difference of the cubes of two consecutive numbers
is 817. Find the numbers.

4. The difference of two numbers is 5 times the less, and
the square of the less is twice the greater. Find ^ttie numbers.

5. The numerator of a certain fraction exceeds the denomi-
nator by 1. If the numerator and denominator are inter-
changed, the sum of the resulting fraction and the original
fraction is 2^^^^. Find the original fraction.

6. The denominator of a certain fraction exceeds twice the
numerator by 3. If 3^;^ is added to the fraction, the result-
ing fraction is the reciprocal of the original fraction. Find
the original fraction.

7. A farmer bought a number of geese for $24. Had he
bought 2 more geese for the same money, he would have paid
J of a dollar less for each. How many geese did he buy, and
what did he pay for each ?

State the problem to which the negative solution applies.

QUADRATIC EQUATIONS 131

8. A laborer worked a number of days and received for
his labor $36. Had his wages been 20 cents more per day,
he would have received the same amount for 2 days' less
labor. What were his daily wages, and how many days did
he work ?

State the problem to which the negative solution applies.

9. For a journey of 336 miles, 4 days less would have
sufficed had I traveled 2 miles more per day. How many
days did the journey take ?

State the problem to which the negative solution applies.

10. A farmer hires a number of acres for $420. He lets
all but 4 acres for $420, and receives for each acre $2.50 more
than he pays for it. How many acres does he hire ?

11. A broker sells a number of railway shares for $3240.
A few days later, the price having fallen $9 a share, he buys,
for the same sum, 5 more shares than he had sold. Find the
number of shares transferred on each day, and the price paid.

12. A man bought a number of sheep for $300. He kept
15 and sold the remainder for $270, gaining half a dollar on
each sheep sold. How many sheep did he buy, and what did
he pay for each ?

13. The length of a rectangular lot exceeds its breadth by
20 yards. If each dimension is increased by 20 yards, the

^^rea of the lot will be doubled. Find the dimensions of the
lot.

14. Twice the breadth of a rectangular lot exceeds the
length by 2 yards ; the area of the lot is 1200 square yards.
Find the length and the breadth.

15. Three times the breadth of a rectangular field, the area
of which is 2 acres, exceeds twice the length by 8 rods. At $5
per rod, what will it cost to fence the field ?

182 COLLEGE ALGEBRA

16. Two pipes running together fill a cistern in lOf hoars ;
the larger pipe will fill the cistern in 6 hours less time than
the smaller pipe. How long will it take each pipe, running
alone, to fill the cistern ?

17. Three workmen, A, B, and C, dig a ditch. A can dig
it alone in 6 days more time, B in 30 days more time, than the
time it takes the three to dig the ditch together ; G can dig
the ditch in 3 times the time the three dig it in. How many
days does it take the three, working together, to dig the ditch ?

18. A cistern with a capacity of 900 gallons can be filled
by two pipes running together in as many hours as the larger
pipe brings in gallons per minute ; the smaller pipe brings in
per minute 1 gallon less than the larger pipe. How long will
it take each pipe by itself to fill the cistern ?

19. A number is formed by two digits, the second being
less by 3 than one-half the square of the first If 9 is added
to the niuuber, the order of the digits is reversed. Find the
number.

20. A number is formed by two digits ; 5 times the second
digit exceeds the si^uare of the first digit by 4. If 3 times
the first digit is added to the number, the order of the digits
is reversed. Find the number.

21. A boat's crew row 3 miles down a river and back again
in 1 hour and 15 minutes. Their rate in still water is 3 miles
per hour faster than twice the rate of the current. Find the
rate of the crew and the rate of the current.

22. A jeweller sold a watch for S22.76 and lost on the cost
of the watch as many per cent as the watc*h cost dollars. What
was the cost of the watch ?

23. A farmer sold a horse for $ 138 and gained on the cost
^ as many per cent as the horse cost dollars. Find the cost
of the horse.

QUADRATIC EQUATIONS 133

24. A broker bought a number of $100 shares, when they
were a certain per cent below par, for $8500. He afterwards
sold all but 20, when they were the same per cent above par,
for $9200. How many shares did he buy, and what did he
pay for each share ?

25. A drover bought a number of sheep for $110 ; 4 having
died,^ he sold the remainder for $7. 33 J a head and made on
hii^ investment 4 times as many per cent as he paid dollars
for each sheep bought. How many^sheep did he buy, and how
many dollars did he make ?

26. A certain train leaves A for B, distant 216 miles;
3 hours later another train leaves A to travel over the same
route ; the second train travels 8 miles per horn* faster than
the first, and arrives at B 45 minutes behind the first. Find
the time each train takes to travel over the route.

27. A coach, due at B 12 houi's after it leaves A, after
traveling from A as many hours as it travels miles per hour,
breaks down ; it then proceeds at a rate 1 mile per hour less
than half its former rate and arrives at B 3 hours late. Find
the distance from A to B.

28. Several boys spent each the same sum of money. If
there had been 5 boys more and each boy had spent 25 cents
less, the amount spent by the boys would have been $37.50.
If there had been 5 boys less and each boy had spent 25 cents
more, the amount spent would have been $30. Find the
number of boys and the amount each boy spent.

29. A detachment from an army was marching in regular
column with 5 men more in depth than in front. On approach-
ing the enemy the front was increased by 845 men, and the
whole detachment was thus drawn up in 5 lines. Find the
number of men.

CHAPTEE XI

SIMULTANSOUS QUADRATIC EQUATIONS

Quadratic equations that involve two unknown numbers
require different methods for their solution according to the
form of the equations.

184, Case I. When from one of the equations the value
of one of the unknown numbers can be found in terms of the
other; and this value substituted in the other equation.

solve 3.»-2., = 51 [1]

X'-y = 2j [2]

Transpose x in [2], ^ = x — 2.

Substitute in [1], 3aj2 _2x(x - 2) = 6.
The solution of which gives x = 1 or — 6.

.*. y = — 1 or — 7.

Special methods often give more elegant solutions than the
general method by substitution.

1, When equations have the form x ± y = a, and xj = h;
x^ ± y* = a,, and xy = b ; or, x ± y = a, and x* -f y * = b.

(1) Solve

x-\-y= 40'
xy = 300

>>.

[1]

[2]

Square [1],

x2 + 2x2/4-2^ = 1000.

[3]

Multiply [2] by 4,

4x2/ = 1200.

M

Subtract [4] from [3],

x2-2x2/ + y^ = 400.

[6]

Extract the root,

X - 2^ = ± 20.

[«]

Add [6] and [1],

2x = 60 or 20.

Subtract [6] from [1],

22/ = 20 or 60.

x = 30^ x = 101
••2/=10|^'2/=30;-

134

SIMULTANEOUS QUADRATIC EQUATIONS 135

(2) Solve

«-y= 41 [1]

a;« + y« = 40j* [2]

Square [1], ««- 2xy + 2^ = 16. [3]

Subtract [2] from [3], - 2 xy = - 24. [4]

Subtract [4] from [2], x^ + 2 xy + y* = 64.

Extract the root, ' x + y = ± 8. [6]

Combine [5] and [1], y = 2} ^' y = - 1}

(3) Solve

11^ 1.^
X y 20

1^ 1 ^ 41

x' 7/' 400

[1]
[2]

1 Q 1 Q1

Square [1], _!_ + jL+J_ = ^. [31

^ ^ ^' x2 xy y2 400 ^ ^

Subtract [2] from [3], ^ = ^- W

xy 400

Subtract [4] from [2], 1 _ i. 4. i = J- .
•" -■ •" ■" x2 xy y2 400

Extract the root, = ± — . [61

X y 20 ■■ •■

Combine [1] and [6], IZb} ^^IZl}'

2. When one equation may he simplified by dividing it by
the other,

aj»-|-y» = 91\ [1]

(4) Solve

;»-hy» = 91|

[2]

Divide [1] by [2], x^ - xy + y^ = 13. [3]

Square [2], x* 4- 2 xy + y* = 49. [4]

Subtract [3] from [4], 3 xy = 36.

Divide by - 3, - xy = - 12. [6]

Add [5] and [3], x* - 2xy + y^ = 1.

Extract the root, x — y = ± 1. [6]

Combine [6] and [2], J = JJ or ^^ J}-

\

136

COLLEGE ALGEBRA

185. Case II. When each of the two equations is homo-
geneous and of the second degree.

Solve

2y2-4xy-f 3a;«=17
y* — ic^ = 16

Let y = »x, and substitute twj for y in each equation.
From [1], 2 v^ - 4 raj* + 3 x^ = 17.

• 3*2 —

17

From [2],

Equate the values of x^^

t^2 _ ajs = 16.
16

.-. a* =

17

r2-l
16

2u2-4t?-f3 1)2-1
32 r2 - 64« + 48 = 17«2 - 17,
15r2_64?y = -66.

The solution gives
t) = ¥»

Substitute in [2],

r = V- or f

13 X

26

X2 = V-,

13x 13

6x

Substitute in [2],

9

x2 = 9,
x = ±3,

[1]

[2]

186. Case III. When the two equations are symmetrical

with respect to x and y.

In this case the general rule is to combine the equationi in such a
manner as to remove the highest powers of x and y.

(1) Solve

a:+?/ = 12 J

Divide [1] by [2], 7^ - xy + y^ =

3xy

[1]

[2]
P]

SIMULTANEOUS QUADRATIC EQUATIONS 137

To remove x'^ and y^, square [2],

a;2 + 2a;y + y2 = i44. [4]

Subtract [4] from [3], - 3 xy = ^ - 144,

which gives xy — 32.

We now have

a; + y = 12
xy

= 32/

/p _ g"j jc = 41
Solving as in Case I, we find, _ 4 r or _ o r •

To remove x* and y*, raise [2] to the fourth power,

at* + 4a^ 4- 6x2y2 + 4x2^ + y* = 2401. [3]

Subtract [1] from [3], 4 xV + 6 x^y^ + 4 xy^ = 2064.
Divide by 2, 2 xV + 3 xV + 2 xy* = 1032. [4]

Square [2] and multiply the result by 2 xy,

2x8y + 4x2y2 + 2xy8 = 98xy. [5]

Subtract [6] from [4], - xV = 1032 - 98 xy,

or xV - 98 xy = - 1032.

This is a quadratic equation, with xy for the unknown number.

Solving, we find xy = 12 or 86.

We now have to solve the two pairs of equations,

X

+ y= 7^ x + y= 71

xy = 12j' xy = 86j

From the first, "~ „ l- or ~ . L

y = 3J y = 4J

7±V-296l

x =

From the second,

y =

2
7TV-296

► «

The preceding cases are geifijeroX met?iod8 for the solution of equations
that belong to the kinds referred to; often, however, in the solution
of these and other kinds of simultaneous equations, involving quadratics,
a little ingenuity will suggest some step by which the roots may be found
more easily than by the general method.

138

COLLEGE ALGEBRA

Exercise 27

1.

2.

8.

4.

5.

6.

7.

8.

9.

10.

11.

12.

18.

xy = 15j

x + y = 6']
xy + 27 = 0}

= 24/-

VI}

X'-y =
xy

x — y =
xy + 60

x + 2y =
xy

2x-\-3y =
xy + 15

X

2

= 9-3aj 1
= 10-a;yJ

aj-f 2y = 12
ay + y* = 35

aj-3y-f 9 = 0
ay — J^^ + 4 = 0

a' + y' =

= 1001
= 14/

aj2 -f y2 ^ 17
4aj + y = 15

2a;»-
3a;

a;
2a;

-2^2 + 8 = 0\
-y-2=0/

« + a;y = 40 T
^-3y= I/-

14.

aj2 _ y2 = 13 1
;a;-2y= 9j

15.

a; y 18

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

r

a;y = 54,

a; y 36
a; -2^ + 15= 0

a;« + 4y + ll = 0'

3a; + 2y+ 1 = 0J*

a+3y+l=0

a;' + y» = 106 1
a;y = 45 J

a;» + y» = 52 1
a;y-f24= OJ

x^ — xy = 31
y» + a;y = 10J

a;^ + a;y + y*= 371
a* + ay + y* = 481 J

a;» + 3a;y + y«= 11
3a;2H-a;y + 32r' = 13j ■

3a;y + 2a; + y = 485l
3a;-2y= OJ

x«-y« = 0l
3a;«-4a;y + 5y« = 9j

SIMULTANEOUS QUADRATIC liQUATIONS 139

26.

27.

28.

29.

38.

39.

40.

41.

42.

xy -\- y^ = 4
2a;2-y2^17

y^ + ccy = 40 J

a;^ + 2 icy — 2/^
3a;2 + 2a;2/+22/^

28

72

30.

31.

32.

. 33.

cc* — 4 icy =

y^ — xy

= 451
= 6/

= 18 1

03^ + 3 icy =
2y^H-ajy

x^ — xy -\' y^ —

a;2 4- aj?/ + 2 2/^ =
2 a;* — a-y -f

= 37\
2/2 = 44 1

34.

8 a;2 - 3 a;?/ - 2/2 = 40
9aj2-f a;2/ + 2/ = 60

35.

3a;2 + 3a;y + 2/^= 52
5aj2 + 7a;2/ + 4y2 = i40

36.

4aj2-f 3a;2/H-52/2 = 27
7x2 + 5a;2/ + 92/2 = 47

= 521
= 140/

= 271
= 47 I'

37.

5a;2-f3a;2/ + 22/2 = 188
x^ — xy -{-y^ =. 19

= 1881
= 19/

aj8 + 2/' = 65
a; + 2/= ^

a;8 - 2/^ = 98 ^
x-y= 2

43.

44.

aj' + 2/* -
aj-f 2/

= 2791
= 3/

45.

aj8 - 2/" = 218 1
aJ-y= 2j'

a;8 + 2/» = 152 1
a;2 — ajy + y2 — 19

aj8 - 2/8 = 1304 1

a^ + 252/ H- 2/" = 163 J

a;8 -f 2/^ = 91
^2/(« + 2/) =84

cc8 - 2/^ = 98 ^

30

X — y — —
^ a;2/

}

X

2^_27^

46.

2/ aj 2
a; 2/ 2

140

COLLEGE ALGEBRA

47.

48.

49.

50.

51.

52.

53.

54.

55.

56.

y a

X y

X y

x^ y^

191
6

1
6

..

n

2
36

a;' — y* = 7 ajy

a^ +2^ =

aj + y =

2

bxy

>

6

a;y- 16x^ + 60 = 0

ajy = 4 ajy + 12

ccy = X -f y + 1

«-.^=2^i

a + y =

5aj?/
"6^

a^ -f y2 = 67 — xy
aj H- 2/ = a?y — 5

aj' + y* = 1 — 3 a?y
aj* + 2/* = ajy + 37

a:* + y=706\
a; + 2^== 2

58.

59.

aj

;6 — ?/ =

a;

y» = 211\

60.

a;6 -f 2/' = 3368 1

y

61

ar^ + 2/« = a;y + 19 \
xH-?/ = ar2^ — 7 J

a? + y a; — 7/ __ 10
62. aj — y aj -h y 3
ar» + y* = 45

g^ a:* + x22^ + y* = 133l
^ — xy -^ if =■ 19 j

ar + ajy + ^ = 49 J

a:' + a;2^ + 2/* = 84 I
aj-h V^4-2^= 6 J *

a;» + 2/> = 819-a;// |

a; + 2^ = 21+V^J

= 97 1

= 49-aj»2^J

65.

66

+ 2^ =

a;

ar*H"2r

2a^H-3aJ2^
^®- 3a: + 5

r + 12 = 32^\

y + l = 0 /•

SIMULTANEOUS QUADRATIC EQUATIONS 141

69.

76.

82.

83.

a 0
X y

\

«* = aa; + fry 1

72. ^-«^2/ = «' + ^'l
ajy — ^ = 2 a* J

4xy

73.

x' + 2/' +

X -f y = 18 1

x -f y = ay -f 1

76. «^2^ + ^«2^=*l

77.

aj y a + y

i + i

x^ y

a'

78.

a^ "^ ^3 -«

79.

a;" +

x^ + y^ = 2(a^-{-b^)^

y2-8a^ = 6(a» + ^^1
xy-5a^^ = 2(a2 + ^^J

80. ^' + 2^' = «^yl.
a; -f y = ^a;y J

2(x^-\-y^)=5xy-9ab\
' 2(a -\- b)(x -h y) = S(xy ^ ab) j'

x

''' + y* + «" = 49

« + y + » = 11

4;s;= 6

2x + 3y

*y 4- y« + «« = 40 1

:(jcy

2ajy + a; + y = 22
'^ =68

= 32

4aj = 3y = 2«H-4

X'

4-y»-f i5« = 84^
84. a-f y + « = 14 ^

y* = ««.

85. 2 y» -\- y '\- z
2xz -\- X'\' z

ix? '\' xy + xz i=i a? i

86. y^H-y»H-a;y = 2a^ l-

87

«^ H- «« -f y« s= i^* J

a» + y» = 24 + 5(x-y)1
xy = 15 J

142 COLLEGE ALGEBRA

Exercise 28*

1. If the length and breadth of a rectangle were each
increased 1 foot, the area would be 48 square feet; if the
length and breadth were each diminished 1 foot, the area
would be 24 square feet. Find the length and the breadth of
the rectangle.

2. A farmer laid out a rectangular lot containing 1200
square yards. He afterwaids increased the width 1^ yards
and diminished the length 3 yards, thereby increasing the
area by 60 square yards. Find the dimensions of the original
lot.

3. The diagonal of a rectangle is 89 inches; if each side
were 3 inches less, the diagonal would be 85 inches. Find
the area of tlie rectangle.

4. The diagonal of a rectangle is 65 inches; if the rect-
angle were 3 inches shorter and 9 inches wider, the diagonal
would still be 65 inches. Find the area of the rectangle.

5. The difference of two numbers is | of the greater, and
the sum of their squares is 356. Find the numbers.

6. The sum, the product, and the difference of the squares
of two numbers are all equal. Find tlie numbers.

Hint. Represent the numbers by x -f y and x —y.

7. The sum of two numbers is 5, and the sum of their cubes
is 65. Find the nmnbers.

8. The sum of two numbers i^ll, and the cube of their
sum exceeds the sum of their cubes by 792. Find the numbers.

9.* A number is formed by two digits. The second digit is
less by 8 than the square of tlie first digit ; if 9 times the
first digit is ailded to the nmnber, the order of the digits is
reversed. Find tlie number.

SIMULTANEOUS QUADRATIC EQUATIONS 143

10. A number is formed by three digits, the third digit
being the sum of the other two ; the product of the first and
third digits exceeds the square of the second by 5. If 396 is
added to the number, the order of the digits is reversed. Find
the number.

11. The numerator and denominator of a certain fraction are
each greater by 1 than those of a second fraction ; the sum of
the two fractions is J^ J. If the numerators were interchanged,
the sum of the fractions would be |. Find the fractions.

12. There are two fractions. The numerator of the first
is the square of the denominator of the second, and the
numerator of the second is the square of the denominator
of the first ; the sum of the fractions is ^/, and the sum of
their denominators 5. Find the fractions.

13. If the product of two numbers is increased by their
sum, the result is 79. If their product is diminished by their
sum, the result is 47. Find the numbers.

14. The sum of two numbers which are formed by the same
two digits is |{ of their difference ; the difference of the squares
of the numbers is 3960. Find the numbers.

15. The fore wheel of a carriage turns in a mile 132 times
more than the hind wheel ; if the circumference of each were
increased 2 feet, the fore wheel would turn only 88 times more.
Find the circumference of each wheel.

16. Two travelers, A and B, set out at the same moment
from two distant towns, A^to go from the first town to the
second, and B from the second town to the first, and both
travel at uniform rates. When they meet, A has traveled
30 miles farther than B. A finishes his journey 4 days, and
B 9 days, after they meet. Find the distance between the
towns, and the number of miles A and B each travel per day.

144 COLLEGE ALGEBRA

17. Two boys run in opposite directions around a rectan-
gular field, the area of which is 1 acre; they start from
one comer, and meet 13 yards from the opposite comer. One
boy runs only | as fast as the other. Find the length and
breadth of the field.

18. A man walks from the base of a mountain to the summit,
reaching the summit in 5^ hours ; during the last half of the
distance he walks J mile less per hour than during the first
half. He descends in 3 J hours, walking 1 mile per hour faster
than during the first half of the ascent. Find the distance
from the base to the summit and the rates of walking.

19. A garrison had bread for 11 days. If there had been
400 more men, each man's daily share would have been 2 ounces
less ; if there had been 600 less men, each man's daily share
could have been increased by 2 ounces, and the bread would
then have lasted 12 days. How many pounds of bread did
the garrison have, and what was each man's daily share ?

20. Throe students. A, B, and 0, agree to work out a set of
problems in preparation for an examination ; each is to do all
the problems. A solves 9 problems per day and finishes the
set 4 days before B ; B solves 2 more problems per day than
C, and finishes the set 6 days before C. . Find the number of
problems in the set.

21. A cistern can be filled by two pipes ; one of these pipes
can fill the cistern in 2 hours less time than the other; the
cistern can be filled by both pipes running together in Ij
hours. Find the time in which each pipe will fill the cistern.

22. A and B have a certain manuscript to copy between
them. At A's rate of work lie would copy the whole manu-
script in 18 hours; B copies 9 pages per hour. A finishes
his portion in as many hours as he copies pages per hour.

SIMULTANEOUS QUADRATIC EQUATIONS 145

B is occupied with his portion 2 hours longer than A is with
his. Find the number of pages copied by each.

23. A and B have 4800 circulars to stamp and intend to
finish them in two days, 2400 each day. The first day A,
working alone, stamps 800, and then A and B stamp the
remaining 1600, A working in all 3 hours. The second day
A works 3 hours and B 1 hour, and they accomplish only
^^ of their task for that day. Find the number of circulars
each stamps per minute and the number of hours B works on
the first day.

24. A, in running a race with B to a post and back, meets
him 10 yards from the post. To come in even with A, B
must increase his pace from this point 41^ yards per minute.
If, without changing his pace, he turns back on meeting A, he
will come in 4 seconds behind A. Find the distance to the
post.

25. A boat's crew, rowing at half their usual speed, row
3 miles down stream and back again, accomplishing the dis-
tance in 2 hours and 40 minutes. At full speed they can go
over the same course in 1 hour and 4 minutes. Find the rate
of the crew and of the current.

26. A farmer sold a number of sheep for $286. He received
for each sheep $2 more than he paid for it, and gained thereby
on the cost of the sheep ^ as many per cent as each sheep cost
dollars. Find the number of sheep.

27. A person has $1300, which he divides into two parts
and loans at different rates of interest in such a manner that
the two portions produce equal returns. If the first portion
had been loaned at the second rate of interest, it would have
yielded annually $36 ; if the second portion had been loaned
at the first rate of interest, it would have yielded annually $49.
Find the two rates of interest.

146 COLLEGE ALGEBRA

28. A person has $5000, which he divides into two portions
and loans at different rates of interest in such a manner that
the return from the first portion is double the return from the
second portion. If the first portion had been loaned at the
second rate of interest, it would have yielded annually $245 ;
if the second portion had been loaned at the first rate of
interest, it would have yielded annually $90. Find the two
amounts and the two rates of interest.

29. A number is formed by three digits ; 10 times the
middle digit exceeds the square of half the sum of the three
digits by 21 ; if 99 is added to the number, the digits are in
reverse order ; the number is 11 times the number formed by
the first and third digits. Find the number.

30. A number is formed by three digits ; the sum of the
last two digits is the square of the first digit ; the last digit
is greater by 2 than the sum of the first and second ; if 396
is added to the number, the digits are in reverse order. Find
the number.

31. There are two numbers formed of the same two digits
in reverse order. The sum of the numbers is 33 times the
difference between the two digits, and the difference between
the squares of the two numbers is 4752. Find the numbers.

32. A boat's crew, rowing at half their usual rate, row
2 miles down a river and back in 1 hour and 40 minutes.
At their usual rate they would have gone over the same course
in 40 minutes. Find the usual rate of the crew and the rate
of the current.

33. A railroad train, after traveling 1 hour from A, meets
with an accident which delays it 1 hour ; it then proceeds at
a rate 8 miles per hour less than its former rate and arrives
at B 5 hours late. If the accident had happened 50 miles
farther on, the train would have been only 3^ hours late.
Find the distance from A to K

CHAPTER XII

EQUATIONS SOLVED AS QUADRATICS

187. An equation is in the quadratic form if it contains but
two powers of the unknown, and if the exponent of one power
is twice the exponent of the other power.

(1) Solve 8 «« -f 63 a:« = 8.

This equation is in the quadratic form in x^.
We have 8x6 + 63x8 = 8.

Multiply by 32 and complete the square,

256 x6 H- ( ) + (63)2 = 4225.
Extract the square root, 16 x^ + 63 = ± 66.
Hence, x^ = J or — 8.

Extracting the cube root, we find two values of x to be i and — 2.
To find the remaining roots, solve completely the two equations

x8=J, x3 = -8.

8x8-1=^0, We have

We have

or (2x-l)(4x2 + 2x + l) = 0.

.-. 2x- 1 = 0,
or 4x2 + 2x+l=0.

Solving these, we find for three
values of x,
1 -i+VTs _ 1 -VITs
2' 4 ' 1

x8 H- 8 = 0,
or (x-f 2)(x2-2x + 4) = 0.

.-. X + 2 = 0,
or x2-2xH-4 = 0.

Solving these, wo find for three
values of x,

- 2, 1 + V^, 1 _ V33.

These six values of x are the six roots of the given equation.

(2) Solve Va:8 - 3 Va:« = 40.

Using fractional exponents, we have x^ — 3 x^ = 40.
This equation is in the quadratic form in x^, if we regard x^ as the
unknown number.

147

148

COLLEGE ALGEBRA

Complete the square, 4 x* — 12 x^ + 9 = 169.
Extract the root, 2 x* - 3 = ± 13.

.-. 2x^ = 16 or -10.
.-. x^ - 8 = 0,
or X* H- 6 = 0.

.-. (X* - 2) (X* + 2x* + 4) = 0 ; or (x* + 5^) (x* - 6*x* + 5') = 0.

.-. X* = 2 or - 1 ± vCTs ; or X* = - -^6 or J -v^Cl i: V^).

.-. X =10 or 8(-l±V^); or x= 6V5 or fV5(-lTV^).

Exeroiae 29

Solve :

1. ar« + 7a:» = 8.

2. j-*-oj-^ + 4 = 0.

3. a-«-f4x* = 9G.

4. 370-^-9 = 4j-*.

5. 10x« = 17x*-l.

6. 32u'^^^ = 33x^^-1.

7. j-«-f 14j-* + 24=:0.

8. 19 J"* + -10 X- = J-.

9. j-*-22j-*h-21 = 0.
10. j^"--3.r" = 4.

12. y^* -^ ox^'= 10.

13. dr**-i- 2tix* = Sci*.

14. x-*~4x-* = ll\

1^, x-^-S-r-*- ir>4 = 0.

17. 9aj-*-f4aj-« = 5.

18. 4ic*-.3aj* = 10.

19. 2ar*-3ar* = 9.

20. Vx* = ^J^ -f 12.

21. x = 9Vx + 22.

22. V^-4vGr = 32.

23. 2Vx*-3V^x» = 35.

1.1 3

24. --+-^ = 7-

Vx vGr

25. jr"*H-jr""* = |.

26. 3jr~*-h4ar~* = 20.
37. 2jr~'^x~* = 4o.

28. 4 vx^-f 3^3^ = 27.

«.

29. \2x-f V4x*=72.
50. \^^4x = l.

EQUATIONS SOLVED AS QUADRATICS 149

188. Equivaleirt Equations. Two equations that inyolve the
same unknown number are called equivalent equations, if the
solutions of either include all the solutions of the other.

Thus, 7x — 35 = 6x + & and 4 x = 8 6 are equivalent equations, for
the solution of each is x = 2 6.

A single equation is often equivalent to two or more equa-
tions.

Thus, the equation x' + 1 = 0 may be written

(X + 1) (x2 - X + 1) = 0 ;
and this equation is equivalent to the two equations

X + 1 = 0 and x^ - x + 1 = 0.

In solving x* + 1 = 0, we sliould write it as x + 1 = 0 and x^ — x + 1 = 0,
and solve each of these equations.

If each TneTriber of an equation is multiplied by the same
factor and this factor involves an unknown number of the
equation, new solutions are in general introduced.

Thus, if we multiply x — 3 = 0 by x — 5, we get (x — 3) (x — 6) = 0,
and introduce the solution of x — 6 = 0.

But if the multiplying factor is a denominator of a fraction
of the equation, new solutions are in general not introduced.

Thus, = 3 + X becomes, when multiplied by x — 1,

X — 1

5 = (X - 1) (3 + x), or x2 + 2x - 8 = 0 ;

that is, (x + 4) (x - 2) = 0. Whence, x = - 4 or 2.

Therefore, the solution x = 1 is not introduced, and this solution is tlie
only solution that could be introduced by the factor x — 1.

In general, new solutions are not Introduced in clearing an
equation of fractions if we proceed as follows :

1. Combine fractions that have a common denominator.

2. Reduce fractions to their lowest terms.

3. Use the L.C.M. of the denominators for the multiplier.

If ea^h member of an equation is raised to the same power,
neiv solutions are, in general, introduced.

150 COLLEGE ALGEBRA

Thus, if, we square each member of the equation x = 2, we have «* = 4,
or x2 - 4 = 0 ; that is, (x + 2) (x - 2) = 0.

Therefore, the solution of x + 2 = 0 was introduced by squaring both
members of x = 2.

In solving an equation^ if we raise each member to any power^
we must reject the solutions of the resulting equation that do not
satisfy the given equation.

Solve by clearing of radicals

Vx -f 4 -f V2a;-f 6 = V7 a; -f 14.

Square, x + 4 + 2 V(x -f 4)(2x + 6) + 2x -f 6 = 7x + 14.
Transpose and combine, 2 V(x -f 4) (2 x + 6) = 4 x + 4.
Divide by 2 and square, (x + 4) (2 x + 6) = (2 x + 2)2.

Reduce, x2-3x = 10.

Therefore, x = 6 or — 2.

Of these two values only 5 will satisfy the given equation.

Squaring both numbers of the original equation is equivalent to
transposing V? x + 14 to the left member, and then multiplying by the
rationalizing factor

Vx + 4 + V2x + 6 -f V7 X + 14.
The result reduces to

V(x-f 4)(2x + 6) - (2x + 2) = 0.
Transposing and squaruig again is equivalent to multiplying by
(Vx + 4 - V2X + 0 - V7 X + 14) (Vx + 4 - V2x + 6 + V7 x + 14).
Tlierefore, the equation x^ — 3 x — 10 = 0 is really obtained from

(Vx + 4 + V2 X + (> - V7x+ 14)

X (Vx + 4 + V2X + 6 + V7 X + 14)
X (Vx + 4 - V2X + 6 -V7x + 14)
X (Vx + 4 - V2 X + n + V7 X + 14) = 0.

This equation is satisfied by any value that will make any one of the
four factors of its left member equal to zero. The firist factor is 0 for
X = 6, and the last factor is 0 for x = — 2, while no value can be found
to make the second or third factor vanish.

Since — 2 does not satisfy the given equation but is introduced by
multiply nig' by another equation, it is called an extraneous nalue of x.

EQUATIONS SOLVED AS QUADRATICS 161

189. Some radical equations may be solved as follows :

Solve 7x2 - 5a; -f 8 VTa^-Sx-f 1 = - 8.
Add 1 to each side,

7x2-5x + l + 8V7x2-5x+l=-7.
Solving for 7 x2 — 6 X + 1, we have

7x2-5x + l = 1, or 7x2-6x + l=49.

Solving these, we find 0, ^, 3, — -*/ for the values of x.
All these values are extraneous values, and the given equation has no
solution.

190. Various other equations may be solved by methods
similar to that of the last section.

(1) Solve x* - 4a;« + 5x2 - 2x - 20 = 0.
Begin by attempting to extract the square root.

X* -4x8 + 5x2-2x-20(x2-2x

X*

2x2-2x

-4x8 + 6x2
-4x8 + 4x2

x2-2x-20
We see from the above that the equation may be written

(x2 - 2x)2 + (x2 - 2x) - 20 = 0.
Solving, x2 — 2 X = — 6, or x2 — 2 X = 4.

Solving these two equations, we find for the four values of x,
1 + 2V^, I-2V-T, I + V5, I-V5.

(2) Solve a;«-f-^4-a; + - = 4

Add 2 to each member,

^ + 2 + ;i + x + i^6,

X X

or

(a= + l)V(x + l) = 6.

Extract the root, x + - = 2, orx + -= — 3.

XX

Solving these two equations, we find for the four values of x,

-3 + V5 : -3-V6 ;
1» 1» :: » ;; •

162 COLLEGE ALGEBRA

Exercise 30

Solve :

1. Vaj + 4 + V2ic — 1 = 6. 3. V« -f V4 -f « = 3.

2. Vl3a;-1-V2ic-1 = 6. 4. Vaj« - 9 -f 21 = x\

5. Vaj + 1 + Va + 16 = Vx -f 25.

6. V27TT-V^T4 = ^^^^.

7. Va5 + 3 + Vic + 8 = 5 VJ.

8. Vx + 7 + Vx - 5 + V3 05 + 9 = 0.

9. Vaj + 5 4- VS — 2a5 -f- V9 — 4a; = 0.

10. V7 - a; + V3a; + 10 -h VaJ + 3 = 0.

11. V2aj2 + 3a; + 7 = 2a;2 + 3aj-6.

12. aj^-3a; + 2 = 6 Va;^ _ 3aj - 3.

13. 6 aj2 - 3aj - 2 = V2a;2 - 35.

14. 16a; - 3a;2 - 16 = 4 Va;^ _ 6a; + 5.

15. 6 a;« - 21 a; + 20 = V4 a;* - 14 a; + 16"

16. V36a;2 + 12a; + 33 = 41 - 8a; - 24x*

17. 4a;* - 12a;» + 6a;* + 6a; -15 = 0.

18. a;*-10a;» + 35a;«-50a; + 24 = 0.

19. a;*-4a;»-10a;2 + 28a;-15 = 0.

20. 18a;* + 24a;«-7a;2- 10a; -88 = 0.

21. 4a;*-12a;» + 17a;2-12a;-12 = 0.

22. ■N/S + Va; + 3= __=>

Va;-f-3

23- 6 + Va^^-1= /"^^ •■ '

Va;*-i

EQUATIONS SOLVED AS QUADRATICS 153

24. — _!_ ^ —- .^

Vflj + 1 Va;-1 Va;^-1

25 "V^^ + 2 — Va; — 2 _ a?

Vaj + 2 -f Vaj - 2 "" 2

3 aj + V4 X — x^
26. = = 2.

3 aj — V4 a; — a?^

27.

28.

29

V3aj« + 4 + V2^«Tl "" "^ "

V7a;^ + 4 + 2V3a;-l_
V7a;2 + 4-2V3a;-.l"'

V5aj — 4 + V5^^ 2 Vi + 1

I ■ I I ■ CS ill III! .

V5 aj — 4 — V5 — X 2Va — 1

L V(aj + a)^ + 2 aft + Z»* + a; + a = ft.

31. V3 _ ^_

V2ar-1- Vaj-2 Va; - 1

32. ^/f7^+V^^=^/f .

33. \/l + --Vl-- = l-

^ a ^ a;

34. Vaj« + a^ ^ 3aa; + Va;^ + a^- 3aaj = -s/2a^-\-2

35. 4aj* - 3(aj* + 1) (ar* - 2) = x*(10 - 3 aj*).

36. (a;^ - 2) {x^ - 4) = x^ (a;* - 1)*^- 12.

37. 3 Va;» + 17 + Va;» + 1-2 V6a;» + 41 = 0.

2 x 4: x ^ X

38

2 2

39. , H , = X.

x-\-^2-x^ x-^2-x^

154

COLLEGE ALGEBRA

40.

41.

42.

+

1 + Vl - aj l-VT^

X

2x
9

■y/ax~+b -f- Va^ __ 1 -f- Vaa; — b
y/ax + b — Vaoj 1 — Vaa; — 5

Va — X + V^ — ^ _ V^+ "V^
Va — X — Vi — X V« — Vft

43

. V^ + V a — Vaa; + aj^ = Va.

44 x2-hy^4-a; + y = 48^

46.

47.

45. / , >"

X

a^ -\-xy + y^ =

X + Wxy + 1/ =

3Vx + 2Vi/
4 Vi — 2^/y

= b I

6

a;2 4- 1 y2 _ (54

16

aj^

• (x-\-7jy = 2{x-yy }

49. V:

3x

x + y

+

^ oa;

X -\- y = xy — 54:

CHAPTER XIII

PROPERTIES OF QUADRATIC EQUATIONS

191. If we represent the roots of the quadratic equation

ax^ -f 5a5 + c = 0
by a and )3, we have (§ 181)

a -

-h^

V52.

— 4ac

2a

-b-

Vz»2.

— 4ac

2a

Adding,

a

Multiplying,

a

If we divide the equation ax^ -\- bx -\- c = 0 through by a,

b c
we have the equation x^ + -x + - = 0; this ma^ be written

b 6

x^ + px -\- a = 0, where p = -9 o' = - •
^ a a

It appears, then, that if any quadratic equation is made to
assume the form x^ -\- px -{- q = 0, the following relations hold
between the coefficients and roots of the equation :

1. The sum of the two roots is equal to the coefficient of
x with its sign changed.

2. The product of the two roots is equal to the constant
term.

Thus, the sam of the two roots of the equation x^ — 7x + 8 = 0 is 7,
and the product of the roots 8.

166

156 COLLEGE ALGEBRA

192. The expressions a -i- fi, afi are examples of symmetric
functions of the roots. Any expression that involves both root^
and remains unchanged when the roots are interchanged, is a
symmetric function of the roots.

From the relations oc -\- p = —p, al3 = q, the value of any
symmetric function of the roots of a given quadratic may be
found in terms of the coefficients.

Given that a and /3 are the roots of the quadratic «* — 7x + 8 = 0, we
may find the values of symmetric functions of the roots as follows :

(1) a2 + /32.

We have a + /3 = 7,

and ap = 8.

Square the first, a^ + 2ap + ^ = 49

Subtract, 2ap =16

and we have a^ + jgs = 33

(2) 0:8 + /38.

^ ' ^ a8 + 3 a2/3 + 3 a/32 + /38 = 343

3 a)3(a + ^3) or Sa^^ + Sa^ = 168

Subtract, «» + /38 = 176

a2 i32

(3) ^ + ^.

a^ + fls 176

This is —1 which is

ap 8

193. Resolution into Factors. By § 191, if a and fi are the

roots of the equation x^ -\- px -}- q = 0, the equation may be

written 2 / , m , o a

x^ — (a -{- P)x -{- afi = 0,

The left member is the product of a; — a and x — fi, so that
the equation may be also written

(x - a) (a; - /8) = 0.

It appears, then, that the factors of the quadratic expression
x^ -{- px -{- q are x — a and x — fi, where a and p are the roots
of the quadratic equation x^ -h px -^ q = 0.

The factors are real and different, real and alike, or imagi-
nary, according as a and /3 are real and unequal, real and equal,
or imaginary.

PROPERTIES OF QUADRATIC EQUATIONS 157

li fi =: a, the equation becomes

(x — a)(x — a)=:i 0, or (x — a)^ = 0.

If, then, the two roots of a quadratic equation are equal, the
left member, when all the terms are transposed to that member,
is a perfect square.

If the equation is in the form ax^ -\- bx -{- c = 0, the left
member may be written

afx''{'-x-\--\ or a{x-a)(x-P). (§191)

194. If the roots of a quadratic equation are given, we can
form the equation.

Form the equation of which the roots are 3 and — J.

The equation is (x — 3) (x + f ) = 0,

or (x - 3) (2 X + 5) = 0,

or 2 x2 - X - 15 = 0.

195. Quadratic expressions may be factored by the principles
of § 193.

(1) Resolve into two factors x^ — 5x -\-S.
Write the equation x^ — 5 x + 3 = 0.

The roots are found to be and

2 2

The factors of x^ — 5 x + 3 are

5 + Vi3 ^ 5- Vl3

X and X

2 2

(2) Resolve into factors 3 a;^ — 4 a; -f- 5.

Write the equation 3x2 — 4x-f5 = 0.

2 j_ \/ w 2 ">/ 11

The roots are found to be and

3 3

Therefore, the expression 3 x^ — 4 x + 5 may be written (§ 193)

3 2 + vrn^^_ !-vTiI

(-^-i^')(-^^^)

158 COLLEGE ALGEBRA

Exercise 31

Form the equations of which the roots are :

1. 3, 2. 6. a + 35, a — 3b.

2. 4,-5. 7 a + 2b 2a + b

^- 3 ' 3
^' -^y -^' 8. 2 + V3, 2 - V3.

^' h i' 9. _ 1 + V5, - 1 - V5.

5. -h - f • 10. 1 + Vf , 1 - Vf .

Resolve into factors, real or imaginary :

11. 3x2-15x-42. 15. x^'-Sx + A.

12. 9x^-27x-70. 16. x^ + x-i-l.

13. 49 aj2 ^- 49 a; -h 6. 17. 4 ic^ - 28 aj + 49.

14. 169x2-52ir + 4. 18. 4x2 + 12a; + 13.

In Examples 19-27, a and p are to be taken as the roots of
the equation a;^ — 7 a; + 8 = 0.

Find the value of :

19. (a-py. ^^ ^ + ^

20. a^p + ap\

a + P

11 1.1

22. ^ + ^. 26. (a^-P^y.
P a

23. •^ + 5- 27. -^H-^-
P^ o^ P? o?

In Examples 28-33, a and /? are to be taken as the roots of
the equation a;^ + ^x + g' = 0.

PROPERTIES OF QUADRATIC EQUATIONS 159

Find in terms of p and q the value of :

28. 1 + 1. ^^- -'^ + -^-

"^ ^ 32. a* + )8*.

29. a«)3 + a/?. ^ ^

30. a« + )3«. ^^' i^"^^*

34. When will the roots of the equation ax^ -f 5a5 -f- c = 0
fee both positive ? both negative ? one positive and one nega-
tive?

196. The Roots in Special Cases. The values of the roots of
the equation ax^ -\-bx -\- c=^0 are (§ 191)

_ ^, + VZ>2 _ 4 ac -b--y/b^-4.

ac

9

2a 2a

[1]

Multiplying both numerator and denominator of the first
expression by — b — V^^ — 4 ac, and both numerator and
denominator of the second expression hy — b -{- V5^ — 4 ac,
we obtain these new forms for the values of the roots:

'^ '' :• [2]

_ 5 _ V62 - 4 ac - 5 + Vi2__4

ac

We proceed to consider the following special cases :

1. Suppose a to be very small compared with b and c. In
this case P — Aac differs but little from b^, and its square root
but little from b. The denominator of the first root in [2]

will be very nearly —2 b, and the root itself very nearly — - ;

the denominator of the second root in [2] will be very small,

and the root itself numerically very large.

The smaller a is, the larger will the second root be, and the

/*

less will the first root differ from — 7 •

o

160 COLLEGE ALGEBRA

The first root may be found approximately by neglecting
the 01? term and solving the simple equation ftas + c = 0. In
fact, the quadratic equation itself approximates the form

2. Suppose both a and h to be very small compared with c.

In this case the first root, which differs but little from — - >
also becomes very large, so that both roots are very large.

The smaller a and h are, the larger will the roots be. The
quadratic equation in this case approximates the form

Oar^-hOaj + c = 0.

3. Suppose c = 0 while a and h are not zero. In this case

the first root in [1] becomes zero, the second root becomes

The quadratic equation becomes

ao? + ^aJ = 0, or X (ax -f- 5) = 0 ;

one root is 0, the other is

a

4. Suppose ^ = 0 and c = 0 while a is not zero. In this
case the equation reduces to ax^ = 0, of which both roots
are zero.

5. Suppose b = 0 while a and c are not zero. In this case

the two roots become -\-\ and —\j

^ a ^ a

The equation becomes the pure quadratic aa? -\- c = 0.

197. Collecting results, we have the following :

1. If a is very small compared with h and c ; one root is
very large.

2. If a and b are both very small compared with c ; both
roots are very large.

3. If c = 0, a and b not zero ; one root is zero.

4. If b = 0, c = 0, a not zero ; both roots are zero.

5. If ^ = 0, a and c not zero ; the equation is a pure quad-
ratio with roots numericallv equal but opposite in sign.

PROPERTIES OF QUADRATIC EQUATIONS 161

198. Variable Coefficients. When the coefficients of an equa-
tion involve an undetermined number the character of the
roots may depend on the value given to the unknown number.

For what values of m will the equation

27n^ + (5m + 2)a; +(4m + 1) = 0

have its roots real and equal, real and unequal, imaginary ?

We find 62_4ac = (5m + 2)2-8m(4m + l)

= 4 + 12m-7m5»
= (2-m)(2 + 7m).

Roots equal. In this case b^ — 4 ac is zero. (§ 181)

.-. 2 - m = 0, or 2 4- 7 m = 0.
.-. m = 2, or m = — ^.

Roots real and unequal. In this case 6^ _ 4 ac is positive. (§ 181)

The factors 2 — m, 2 + 7 m, are to be both positive or both negative.
If m lies between 2 and — f , both factors are positive ; both factors
cannot be negative.

Roots imaginary. In this case b^ — 4 oc is negative. (§ 181)

Of the two factors 2 — m, 2 + 7 m, one is positive, the other negative.
If m is greater than 2, 2 — m is negative and 2 + 7 m positive ; if m is
less than — ^, 2 + 7 m is negative and 2 — m positive.

199. By a method similar to that of § 198 we can often
obtain the maximum or the minimum value of a quadratic
expression for real values of x.

(1) Find the maximum or the minimum value oil -{- x — x^
for real values of x.

Let 1 + X — X* = m.

a , 1 ±V5-4m

Solve, X = •

2

Since x is real, we must have

5>4m or 6 = 4 m.
Therefore, 4 m is not greater than 5.

That is, m is not greater than J.

The maximum value of 1 + x — x^ is | ; for this value x = J.

162 COLLEGE ALGEBRA

(2) Find the minimum value of au^ -f- 3 05 + 4 f or real values
of X,

Let x2 H- 8 X H- 4 = m.

Then, x2 + 3 x = w - 4. i

o 1 -3 ± V4m - 7
Solve, X = = .

2

Since x is real, we must have

4m>7or4m = 7.
Therefore, 4 m is not less than 7.

That is, m is not less than J.

The minimum value of x^ + 3 x + 4 is { ; for this value x = — f .

Note. Instead of solving for x, we might have used the condition for
real roots, viz. , 6^ — 4 ac greater than or equal to zero.

200. Tlie existence of a maximum or a minimum value may
also be shown as follows :

Take the first expression of the last article,

1 + X - x2.
This is f - (t - « + a;2),

or J - (aJ - S)'-

(x — \Y is positive for all real values of x ; its least value is zero, and
in this case the given expression has its greatest value, J.
Similarly for any other expression.

Exercise 32

For what values of m are the two roots of each of the fol-
lowing equations equal, real and unequal," imaginary ?

1. (3m-f l)a;2 + 2(m + l)aj + m = 0.

2. (m-2)x2 + (77i-5)x-f 2m-5 = 0.

3. 2 mx^ + x^ — 6 mx — 6aj-|-6m + l = 0.

4. mx'^ -\-2x^ + 2m- 3mx + 9x — 10 = 0.

5. 6mx^-\-S7rix + 2m = 2x -x^ — :i.

PROPERTIES OF QUADRATIC EQUATIONS 163

For real values of a, find the maximum or the minimum
value of each of the following expressions :

6. x^-6x-\-13. x^-x — 1

7. 4a;2-12aj + 16. ' x^ - x + 1

8. 3 + 12a;-9aj2. ^^ x^-h2x-3

9. aj» + 8a; + 20. x^-2x-\r3

10. 4aj2-12aj + 25. 17,

11. 25 a;2 - 40 a; - 16.

1 1

2 + x 2 — 05

X — O 18 ! ! .

(a: + 12)(a.-3), ^^ f*' + ^)' •

^^* a-a a;2 — x 4- 1

4aj 2x^-2x + 5

(ic + 2)^ aj^-2aj4-3

21. Divide a line 2 a inches long into two parts such that
the rectangle of these parts shall be the greatest possible.

22. Divide a line 20 inches long into two parts such that
the hypotenuse of the right triangle of which the two parts
are the legs shall be the least possible.

23. Divide 2 a into two parts such that the sum of their
square roots shall be a maximum.

24. Find the greatest rectangle that can be inscribed in a
given triangle.

25. Find the greatest rectangle that can be inscribed in a
given circle.

26. Find the rectangle of greatest perimeter that can be
inscribed in a given circle.

CHAPTER XIV

SURDS AND IMAGINARIES

201. Quadratic Surds. The product or the quotient of two
dissimilar quadratic surds is a quadratic surd.

For every quadratic surd, when simplified, has undiBr the
radical sign one or more factors raised only to the first power ;
and two surds which are dissimilar cannot have all these
factors alike.

202. The sum or the difference of two dissimilar quadratic
surds cannot be a rational number, nor can it be expressed as
a single surd.

For, if Va ± '\/b could be equal to a rational number c, then
squaring and transposing,

± 2 -^/ab = <^-a-b.

Now, as the right side of this equation is rational, the left
side should be rational ; but Va5 cannot be rational (§ 201).
Therefore, Va ± V^ cannot be rational.

In like manner it may be shown that Va ± V^ cannot be
expressed as a single surd Vc.

203. A quadratic surd cannot be equal to the sum of a

rational number and a surd.

For, if Va could be equal to c + V^, then squaring and
transposing,

2c V^ = a — 6 — c*;

that is, a surd would be equal to a rational number ; but this
is impossible.

104

SURDS AND IMAGINARIES 165

204. ijr a -f- Vb = X + Vy, then a is equal to x, and b is
equal to j.

For, transposing, "Vb — V^ = a; — a ; and if b were not equal
to y, the difference of two unequal surds would be rational,
which is impossible. (§ 202)

.*. b = y, and a = x.

In like manner, if a — V^ = x — Vy, a is equal to x, and b
is equal to y.

An expression of the form a + V^, where Vi is a sui-d, is
called a binomial surd.

205. Square Root of a Binomial Surd.

(1) Extract the square root of a + V^.

Let Va + V6 = Vx + Vy.

Square, a + V6 = a + 2 Vxy + y.

.'. X -{■]/ = a, and 2 Vxy = V6. (§ 204)

From these two equations the vahies of x and y may be found.
Or, since a = x -{- y and Vb = 2 Vxy,

a — Vb = x — 2 Vxy + y.
Extract the root, v a — Vb = Vx — Vy.

... (Va + Vb) ( Va - Vft) = ( VS + v^) (Vi - V^).

.*. Va2 — b = x — y.
And, as a = x-\-y^

the values of x and y may bo found by addition and subtraction.

(2) Extract the square root of 7 + 4 V3.

T^t -v^ + V^ = V? + 4 Vs. . [1]

Then, V^-V^- V? - 4 Vs. [2]

Multiply [1] by [2], x - y = V49 - 48.

.-. X — y — \.
But X + y = 7.

.-. X = 4, and y = 3.

.-. Vi + Vy = 2 + Vs.
... Vy-f 4 Vs = 2 + Vs.

166 COLLEGE ALGEBRA

A root may often be obtained by inspection. For this pur-
pose, write the given expression in the form a -f 2 V^, and
determine the two numbers that have their sum equal to a,
and their product equal to b.

(3) Find by inspection the squgjjpoot of 18 + 2 V77.

The two numbers whose sum is 18 and product 77 are 11 and 7.
Then, 18 + 2 Vfj = 11 + 7 + 2 Vll x 7

= (Vll +V7)2.
That is, VTT + V7 = the square root of 18 + 2 V77.

(4) Find by inspection the square root of 75 — 12 V2i.

It is necessary that the coefficient of the surd be 2 ; therefolre,

75 - 12 V2T must be put hi the form 75-2 V766.
The two numbers whose sum is 75 and product 756 are 63 and 12.
Tlien, 75-2 V756 = 63 + 12-2 V63 x 12

= (V63-Vl2)2.
That is, V63 - VT2 = the square root of 75 - 12 V2T ;

or 3 V7 - 2 V3 = the square root of 75 - 12 v^.

Exercise 33

Extract the square root of :

1. I4+-6V5. 6. 20-8V6. 11. 14-4V6.

2. I7 + 4V15. 7. 9-6V2. 12. 38-12ViO.

3. IO+-2V2I. 8. 94-42V5. 13. 103-12ViT.

4. 16+-2V65. 9. I3-2V3O. 14. 57-12Vi5.

5. 9-2Vi4. 10. II-6V2. 15. 3^-ViO.

16. 2a + 2^a^-b\ 18. 87 - 12 V42.

17. a^-2b-y/a^-b\ 19. (« + Z,)^ _ 4 (« _ ft) Vo^.

SURDS AND IMAGINARIES 167

206. Orthotomic Numbers. The squares of all scalar num-
bers are positive scalar numbers ; hence, a negative scalar
number cannot be the square of a scalar number, and conse-
quently the square root of a negative scalar number cannot be
a scalar number (§ 144). For the complete treatment of
evolution and of equations of the second and higher degrees,
account must be taken of the square roots of negative scalar
numbers, and as these roots are not scalar numbers it is neces-
sary to assume a new series of numbers distinct from the scalar
series, but such that the square of each and every number in
the new series is a number in the negative branch of the scalar
series. These new numbers being distinct from the scalar
numbers require a distinguishing name, and accordingly they
have been named orthotomic numbers or imaginary numbers.
Hence,

An orthotomic numJ)er is any indicated square root of a nega-
tive scalar number or any scalar multiple thereof.

The complete series of orthotomic numbers includes a posi-
tive branch and a negative branch with zero as common origin.

207. Complex Numbers. The sum of any two scalar numbers
is a scalar number, and it will presently be shown that the
sum of any two orthotomic numbers is an orthotomic number,
but the sum of a scalar number and an orthotomic number is
evidently neither a scalar number nor an orthotomic number
and therefore requires a distinctive name. The name gener-
ally given is complex number. Hence,

A complex number is the indicated sum or difference of a
scalar number and an orthotomic number.

Thus, if g and h are scalar numbers either positive or negative but not
zero, and p is a positive scalar number, but not zero, ^ V(— P) is an
orthotomic number and fl' + ^ V(— 1^) is a complex number. If g and h
may take any scalar values, zero included^ the form fir + ^ V(— p) includes
the wliole assemblage of the scalar, the orthotomic, and the complex
numbers.

'^

168 COLLEGE ALGEBRA

Such assemblage is named the uniplanar or coplanar assem-
blage of algebraic numbers. It will be shown hereafter that
this uniplanar assemblage includes all the numbers necessary
to be considered in ordinary algebra ; that is, the algebra of
the four elementary operations, Addition, Subtraction, Multi-
plication, and Division, performed subject to the Laws of
Uniformity, Association, Commutation, and Distribution as
given in § 72.

208. The introduction of orthotomic numbers requires the
meanings of the four elementary operations to be made more
general in the algebra of complex numbers than they are in the
algebra of scalar numbers, but these enlarged meanings must
be consistent with the older meanings of scalar algebra and
include them as special cases ; and the elementary operations,
when thus generalized, must be performed subject to the
four fundamental laws which govern or define them in scalar
algebra. (§ 34)

A full statement of these wider meanings with ' illustrative
applications of them will be given in Chapter XXXIII.

209. It is necessary, however, to notice here the generali-
zation of the Law of Signs which results from the action of
the Associative and Commutative Laws of multiplication with
the Law of Distribution of the square root operation over the
factors of a product.

If a and b are both positive scalar numbers, the distribu-
tion of the square root operation over the factors of a product
gives

+ Vab = (+ V^) (+ V^) = (- Va) (- V5), (i)

_ V^ =(- Va)(+ Vb) = (+ Va)(- V5). (ii)

In extending this law to orthotomic numbers it is assumed
that the law still holds when either factor (or both factors)
under the radical sign is negative, provided the distribution is
made over the factors taken with their signs unchanged. Thus,

SURDS AND IMAGINARIES 169

+ Va (- b) = (+ V^) (+ V^)

= (- V^) (- V^)» (iii)

- Va(- J) = (- Va)(+ V^)

and + V(- a) (- 6) = (+ V^) (+ V^)

= (- V^) (- V^). C^)

- V(- a) (- 6) = (- V^) (+ V^)

= (+ V^) (- V^).* (vi)

Hence, if J = 1, we have as special cases of (iii) aiid (iv)

+ V:^=+V^7=ri)=:(+V^)(+V^), (Vii)

and - V^ = - Va (- 1) = (- Va) (+ V^). (viii)

Now, by the Associative Law of multiplication, we have

+ V(-a)(-&) = + V(-l)^a( 6)^
which, by (vii), = (+ V^) I + Va(-J)^,

which, by the Associative and Commutative laws.

= (+V=l)^+V(-l)(a^)|,
which, by (vii), = (+ V^) (+ V^) (+ Vab)

But, by (v), + V(- a) (- b) = (+ V^T^) (+ V^).
/. (+ V^) (+ V^) = - V^.t

* Notice that (iii), (iv), (v), and (vi) are all included in the forms (i)
and (ii), if a and b are not restricted to be positive scalar numbers but
may be any scalar numbers whatever. This generalization of (i) and (ii)
is the proper statement of the distributive law of the square root.

t Notice that from this we have

(H- V^) (+ V36) = - {(+ V^) (+ Vft)}.

170 COLLEGE ALGEBRA

Similarly, it may be shown that

(_ V^) (- V^) = - Va^,
(_ -vCT^) (+ V^) = + Va^,

The generalized Law of Signs in multiplication may now
be enunciated as follows :

I. Two scalar factors with like signs give a positive scalar
product ; two scalar factors with unlike signs give a negative
scalar product,

II. Two orthotomic factors with like signs give a negative
scalar product ; two orthotomic factors with unlike signs give a
positive scalar product.

III. Two factors^ the one scalar the other orthotomicj give a
positive orthotomic product if the factors have like signs, a negor
tive orthotomic product if the factors have unlike signs.

210. The successive powers of V— 1 are ;
( V^)2 = - 1.

( V- 1)« = ( V- 1)2 V^ = (- 1) V^ = - V^ ;

(Vri)4 = (V3i)2(V3i)2 = (-i)(-i) = + !;__

(•vCTi)'^ = ( V^)* V^ = (+ 1) V^ = + V^^.

It appears that the successive powers of V— 1 form the
repeating series + V— 1, — 1, — V— 1, + 1 ; and so on.

211. Every orthotomic number is of the form ± m V— j9,
wherein m and p are positive scalar numbers. Now,

± m V— p = ±m -y/p \ V— 1\,

and in this the factor ± m V^ is a scalar number ; hence,
every orthotomic number may be written in the fonn a V— 1,
in which a is a scalar number ; and, conversely, if a is a scalar

SURDS AND IM AGIN ARIES 171

number^ a V— 1 will be an orthotomic number. Hence, the
sum of two orthotomic numbers a V— 1 and h V— 1 is an
orthotomic number or is zero, for

a V^ 4- h V^ =(a + h) V^,

and (a + 6) is a scalar niunber, a and h being scalar numbers,
or is zero if ft = — a.

212. Every imaginary number may be made to assume the
form a-\-h V— 1, where a and h are scalar numbers, and may
be integers, fractions, or surds.

The form a + ft V— 1 is the typical form of complex numbers.

Reduce to the typical form 6 + V— 8.

This may be written 6 -f Vs V- 1, or 6 -f 2 V2 V^ ; here a = 6,
and 6 = 2 V^.

213. The algebraic sum of two complex numbers is in general
a complex number.

Add a + ft V— 1 and c -\- d V— 1.

a + ftv-Ti
c + dV^

The sum is (a + c) + (& + d) V^

This is a complex number unless ft + e^ = 0, in which case
the number is scalar, or a + c = 0, in which case the number
is orthotomic.

214. The product of two complex numbers is in general a
complex number.

Multiply a + ft V^ by c + c^ V^.

The product is (oc — 6d) + (6c + ctd) V— 1,
which is a complex number unless ftc -|- ew? = 0 or ac — bd = 0.

Then,

172 COLLEGE ALGEBRA

215. The quotient of two complex numbers is in genieral a
complex number.

Divide a + b V— 1 hj c -\-d V— 1.

^ , , a 4- 6 V— 1

The quotient is ■== •

c + dvCi

Multiply both numerator and denominator by c — d V — 1.

{a + h V^) (c-d V^)
(c + d V^) (c-d V3T)

_ (oc + ^) + (&c - od) V^

_ ac 4- ?^ , 6c — ad / — r
~ c2 + d2 "^ ca + d2

This is a complex number in the typical form.
If he — ad^ 0, the quotient is scalar.

216. Two expressions of the form a-\-h V— 1, a — b V— 1
are called conjugate numbers.

Add a-^-h V— 1 and a — & V— 1.
The sum is 2 a.

Multiply a-\-h V— 1 by a — 5 V— 1.

a + 6vCi;

The product is a^ + 6^

From the above it appears that the sum and the product of

two conjugate numbers are both scalar.

The roots of a quadratic equation, if they are not scalar
numbers, are conjugate numbers. (§ 181)

217. A complex number cannot be equal to a scaZar number.

For, if possible, let a + & V— 1 = c. •

Then, h V- 1 z=c — a^

and -h^ = (c- a)2.

Since h^ and (c — a)^ are both positive, we have a negative number
equal to a positive number ; but this is impossible.

SURDS AND IMAGINARIES 173

218. If two complex mtmbers are equal, the scalar parts are
equal and the orthotomic parts are equal.

For, let a + & V^ = c + d V^.

Then, (&-d)V^ = c-a.

Square, - (h - df = (c - a)\

This equation is impossible unless h= d and a = c.

219. If X and j are scalar a7id x + y V— 1 = 0, then x = 0
and y = 0.

For, y V— 1 = — X,

Square, — y^ = x2.

Transpose, x^ + y^ = o,

This equation is tnie only when x = 0 and y = 0.

220. If the roots of ax^ -{- bx -\- c = 0 ai-e not scalar, then
ax^ -\- bx -i- c is positive for all scalar values of x, if a is posi-
tive; and negative for all scalar values of x, if a is negative.

Let the two roots be y + 8 V— 1 and y — 8 V — 1, where y
and 8 are scalar.

Then, by § 193, the expression ax^ -{-bx-{-c is identical with

a(x-y-8 V^) (a; - y 4- 8 V^).

This product reduces to a [(a; — y)^ -h 8^].

For all scalar values of x, (x — y)^ + 8^ is a positive scalar
number. Hence, ax^ -\- bx + c is positive if a is positive, and
negative if a is negative.

Examples :

(1) The roots of the equation x^ - 6x + 13 = 0 are 3 + 2 V- 1 and
3 _ 2 V— 1. The expression x^ — 6x + 13 may bo written (x — 3)^ -f 4,
which is positive for all scalar values of x. ^^__

(2) The roots of the equation 12x - 13 - 4x2 = 0 are -^

3 — 2 V— 1
and . The expression 12 x — 13 — 4 x^ may he written

- (4aJ« - I2x + 9 + 4J or - [(^x - 3)2 + 4],
which iA negative for all scalar values of x.

174 COLLEGE ALGEBRA

The expressions (aj- 3)^+4 and -[(2aj - 3)^ -h 4] of
Examples (1) and (2) cannot become zero for any scalar
values of x ; they accordingly have either a rninimum value
below which they cannot fall, or a maximum value above
which they cannot rise. (§ 199)

Ezercise 34
1. Multiply

V^Ts by V^; 2V^ by 4 V- 27 ; 3V^ by ^

V27

2. Divide

V7 by V^; V^ by V^; 3V^ by V2V^^.

3. Reduce to the typical form

4 + V- 81 ; 5 + 2 V^; (3 -|- Vi:^)2.

Multiply :

4. 4 -f V^ by 4 - V^.

5. V3 - 2 V^ by V3 + 2 V^.

6. 7 -h V-27 by 4 + V^.

7. 5 4. 2 V^ by 3 - 5 V^.

8. 2 V3 - 6 V^ by 4 V3 - V^.

9. Va -h ^> V— c by Vc 4- a V— ^►.
Divide:

10. 26 by 3 -h V^; 86 by 6 - V^.

11. 3 + V^ by 4 4- 3 V^^.

12. - 9 + 19 V^ by 3 4- V^.
Extract the square root of :

13. l4.4Vir3. 15. _17 + 4V^n6.

14. 10-8V-6. 16. _38-15V^^^.

r

SURDS AND IMAGINARIES 175

17. Show that 4: x^ — 12 x -\- 25 is positive for all scalar
values of x, and find its minimum value.

18. Show that 6 a; — 4 — 9 a;^ is negative for all scalar
values of x, and find its maximum value.

19. Show that each of the two complex roots of the equa-
tion ic* = 1 is the square of the other complex root.

20. Show that, if CO is a complex root of x^ = 1,
x^ -\- y^ -\- z^ — S xyz

= (aj + y -h «) (aj -h coy 4- w^«) (« + wV + w«).

21. Find all the fourth roots of — 1.

22. Find all the sixth roots of -h 1.

23. Find all the eighth roots of + 1.

24. Reduce to the typical form

(2-3 V^)(3 + 4 V^)
(6 + 4 Viri)(16 - 8 V^) '

Simplify :

25. (1 + V^)* 4- (1 - V^)*.

26. (1 + V^)8 - (1 - V^)8.

27. V(3 + 4V-1) + V(3-4V^).

28. V(3 + 4V^)-V(3-4V^).

29. V(5 + 2V^) + V(5-2V^).

30. V( V3 4- V- 106) - V( V3 - v^nios).

CHAPTER XV

SIMPLE INDETBRMHrATE EQUATIONS

221. If a single equation involving two unknown numbers
is given, and no other condition is imposed, the number of
solutions of the equation is unlimited; for if one of the
unknown numbers is assumed to have ar^y particular value,
a corresponding value of the other may be f oimd.

Such an equation is called an indeterminate equation.

Although the number of solutions of an indeterminate equa-
tion is unlimited, the values of the unknown numbers are
confined to a particular range; this range may be further
limited by requiring that the unknown numbers shall be posi-
tive integers.

222. Every indeterminate equation of the first degree, in
which X and y are the unknown numbers, may be made to
assume the form

ax ± ^2^ = ± c,

where a, b, and c are positive integers and have no common
factor.

223. The method of solving an indeterminate equation in
positive integers is as follows :

(1) Solve 3 X + 4 y = 22, in positive integers.
Transpose, 3 x = 22 — 4 y.

the quotient being written as a mixed (expression.

^ ^ -y

170

SIMPLE INDETERMINATE EQUATIONS 177

Since the values of x and y are to be integral, x + 2/ — 7 will be integral,
\ —y
and hence — - — will be integral, though written in the form of a fraction.

o

1-2/

Let '~~^ ~ ^*» ^^ integer.

Then, 1 -y = Sm.

.-. 2/ = 1 — 3 m.

Substitute this value of y in the original equation,

3« + 4- 12 m = 22.

.*. X = 6 + 4m.

The equation y = 1 — Sm shows tliat m may be 0, or have any nega-
tive integral value, but cannot have a positive integral value.

The equation x = 6 + 4 m further shows that m may be 0, but cannot
have a negative Integral value greater in absolute value than 1.

.-. m may be 0 or — 1,

and then ^ m ^^ aY-

(2) Solve 6 aj — 14 y = 11, in positive integers.

Transpose, 6 x = 11 + 14 2/,

1 + 4?/

x = 2 + 22/ + ^?-Y^. [1]

5
Since x and y are to be integral, x — 2 y — 2 will be integral, and hence
will be integral.

1 + 4i/
Let = 771, an integer.

__ 6m — 1

Then, 2/ = -. >

4

or y = m-\ — • [2]

4

Now, must be integraL

4

■r X m — 1

Let — - — = n, an mteger.

4

Then, m = 4 n + 1^

Substitute value of wi in [2], y = 6 n -h 1.
Substitute value of y in [1], x = 14 n + 5.

178 COLLEGE ALGEBRA

Obviously x and y will both be positive integers if n has any positive
integral value.

Hence, x = 6, 19, 33, 47, • • • ,

2/ = 1, 6, 11, 16, ...

Another method of solution is the following :

From the given equation we have x = •

5

Here y must be so taken that 11 + 14 y is a multiple of 6; take y = 1,
then X = 5, and we have one solution.

Now, 6 X — 14^ = 11,

and 6(5) -14(1) = 11.

Subtract, 6 (x - 5) - 14 (y - 1) = 0,

x-5 14

or = — •

y-l 5

Since x — 6 and y — 1 are integers, x — 6 must be the same multiple
of 14 that 2/ — 1 is of 6.

Hence, if x — 6 = 14 m, then y — 1 = 6m.

.-. X = 14 m + 6, and y = 6 m + 1.
Tlierefore, x = 6, 19, 33, 47, • . • ,

and 2/ = 1, 6, 11, 16, ...

It will be seen from [1] and [2] that when only positive integers are
required the number of solutions will be limited or unlimited according
as the sign connecting x and y is positive or negative.

(3) Find the least number that vvrhen divided by 14 and 6
will give remainders 1 and 3 respectively.

If JV represents the number, then

JV^-1 ^ N-S

= X, and = y.

14 ' 6

,'.N= 14x + l, and JV=5y + 3.
.-. 14x + l = 62/ + 3.
52/ = 14x-2,
6y = 15x — 2 — X.

2 + x

... 2/ = 3 X —

5

2 •+ X
Let = ?n, an integer.

.*. X = 5 m — 2.

SIMPLE INDETERMINATE EQUATIONS 179

y = l(l^x — 2), from original equation.
.*. y = 14 m — 6.
If wi = 1, X = 3, and y = 8,

.-. jy^= 14x + 1 = 6y + 3 = 43.

(4) Solve 5x -\-6y = S0, so that x may be a multiple of y,

and both aj and y positive.

Let X = my.

Then, (6m + 6)y = 30.

30

.'.y =
and X =

6m4-6
30 m

6m + 6

If m = 2, X = 3f , y = If

If m = 3, x = 4f, y = If ;

and so on.

(6) Solve 14a5 + 22y = 71, in positive integers.

1 -8y
« = 5-y + — — 5^.

14

If we multiply the fraction by 7 and reduce, the result is — 4 y + J, a
form which shows that there can be no integral solution.

There can be no integral solution ofax±6y=±cifa and b have a
common factor not common also to c ; for, if d is a factor of a and also
of &, but not of c, the equation may be written

mdz ± ndy = ± c, or nx ± ny = ± - ;

a

which is impossible, since - is a fraction, and mx d: ny is an integer, if x
and y are integers.

Exercise 35

Solve in positive integers ;

1. aj + y = 12. 5. Bx + Sy = 105.

2. 2aj + lly = 83. 6. faj4-5y = 92.

3. 4aj + 92/ = 53. 7. |aj + ^y = 27.

4. Sx-\-5y = 7L 8. fa; + Jy = 63.

180 COLLEGE ALGEBRA

Solve in least possible intiiegers :

9. 7x-2i/ = 12, 12. llaj — 52/ = 73.

10. 9a; -5?/ = 21. 13. 16 « - 47 ^^ = 11.

11. 7i»-4 7/ = 45. 14. 23«-14y = 99.

15. Find two numbers which, multiplied respectively by 7
and 17, have for the sum of their products 1135.

16. If two numbers are multiplied respectively by 8 and
17, the difference of their products is 10. What are the
numbers ?

17. If two numbers are multiplied respectively by 7 and
15, the first product is greater by 12 than the second. Find
the numbers.

1 8. Divide 89 in two parts, one of which is divisible by 3,
and the other by 8.

19. Divide 314 in two parts, one of which is a multiple of
11, and the other a multiple of 13.

20. What is the smallest number which, divided by 6 and

by 7, gives each time 4 for a remainder ?

21. The diiference between two numbers is 151. . The first
divided by 8 has 5 for a remainder, and 4 must be added
to the second to make it divisible by 11. What are the
numbers ?

22. Find pairs of fractions whose denominators are 24
and IG, and whose sum is ^J.

23. How can one pay a sum of $87, giving only bills of $5
and $2 ?

24. A man buys calves at $5 apiece, and pigs at $rS
apiece. He spends in all $114. How many did he buy of
eacli ?

SIMPLE INDETERMINATE EQUATIONS 181

25. A person bought 40 animals, consisting of pigs, geese,
and chickens, for $40. The pigs cost $5 apiece, the geese
$1, and the chickens 25 cents each. Eind the niunber he
bought of each.

26. Solve 18 aj — 5y = 70 so that y may be a multiple of x,
and both positive.

27. Solve 8 a; H- 12 y = 23 so that x and y may be positive,
and their sum an integer.

28. Divide 70 into three parts which shall give integral
quotients when divided by 6, 7, 8 respectively and the sum
of the quotients shall be 10.

29. In how many ways can $3.60 be paid with dollars and
twenty-cent pieces ?

30. In how many ways can 300 pounds be weighed with 7
and 9 pound weights ?

31. Find the general form of the numbers that, divided by
2, 3, 7, have for remainders 1, 2, 5 respectively.

32. Find the general form of the numbers that, divided by
7, 8, 9, have for remainders 6, 7, 8 respectively.

33. A farmer buys oxen, sheep, and hens. The whole
number bought is 100, and the total cost £100. If the oxen
cost £6, the sheep £1, and the hens I5. each, how many of
each does he buy ?

34. A farmer sells 15 calves, 14 lambs, and 13 pigs, and
receives $200. Some days after, at the same price, he sells
7 calves, 11 lambs, and 16 pigs, for which he receives $141.
What is the price of each ?

CHAPTEK XVI

INEQUALITIBS

2SJ4. An inequality is a statement that two expressions do
not have the same value ; that is, a statement that two expres-
sions do not represent the same number.

Every inequality consists of two expressions connected by
a sign of inequality ; the two expressions are called the sides
or members of the inequality.

225. We say that a'>h when a — ft is positive ; that a <. b
when a — ft is negative,

226. The symbols <t and > are used for the words not less
than and not greater than respectively.

227. In working with inequalities the following principles
are easily shown to be true :

The sign of an inequality remains unchanged if both ment'
hers are increased or diminished by the same number ; if both
members are multiplied or divided by the same positive numh&r ;
if both members are raised to any odd power, or to any power
when both members are positive.

The sign of an inequality is reversed if both members are mul-
tiplied or divided by the same negative number ; if both m,embers
are raised to the same even power when both memhers are negative.

228. Fundamental Theorem. If a and b are unequal scalar

numbers, a^ -|- b^ > 2 ab.

For {a — by must be positive.
That is, a^ - 2 aft -h ft2 > 0.

r.a'' + b''> 2 aft. (§227)

182

INEQUALITIES 183

(1) If a and h are unequal positive scalar numbers, show
that a^-\-h^> a^h -\- ah\

We shall have a^ + &* > a^ft + a62, - C "^ '"

if (dividing by a + 6) a^ - ab -{- y^ > ab, [\/'^
if a2 + &2>2a6.

But a2 + 62>2a6. (§228)

(2) Show that a^ + 5^ -h c^ > a5 + ac + he.

Now, a2-f62>2^,

a2 + c2>2ac, (§228)

&2 + c2>26c.
Add, 2a2 + 2&2 + 2c2>2a6 + 2ac + 26c.

.-. a2 + 62 _|_ c2 > a6 + oc + 6c.

Exercise 36

Show that, the letters being imequal positive scalar numbers :
1. a^ + Sh^>2h{a'\-h), 2. a'^b -^ ab^ > 2 a^l^.

3. (a* + 52) (a* + ^>*) > (a8 + ^>8)2.

4. a^^ + a^c -\- ah'^ -^ h^c '\- a(? -\- h(? > ^ ahc,

5. The sum of any fraction and its reciprocal > 2.

6. If a^ = a^ -h W, and ^ = c^ -f cF^, xy <fiac -\- bd, or ac? + 5c.

7. a5 -h ac -h 5c < (a + 5 — c)2+ (a + c — 5)* + (5 + c — a)l

8. Which is the greater, (a^ + 5^) (c^ + (P) or (ac + 5ef)» ?

9. Which is the greater, a* — 5* or 4 a^ (a — 5) when a>b?

10. Which is the greater, \j -\- \— or Va + V5 ?

11. Which is the greater, — jr — or — —7?

° 2 a -\- 0

12. Which is the greater, 75 + -^ or T + ~?

° (^ or b a ^

v3

« ■

/

CHAPTER XVII

RATIO, PROPORTION, AND VARIATION

229. Ratio of Numbers. The relative magnitude of two
numbers is called their ratio, when expressed by the indicated
quotient of the first by the second.

Thus, the ratio of a to b is -» or a -4- 6, or a: 6; the quotient is

0

generally written in the last form when it is intended to express a ratio.

The first term of a ratio is called the antecedent, and the
second term the consequent.

When the antecedent and consequent are interchanged the
resulting ratio is called the inverse of the given ratio.

Thus, the ratio 3 : G is the inverse of the ratio 6 : 3.

230. A ratio is not changed if both its terms are multiplied
by the same number.

Thus, the ratio a : 6 is represented by - « the ratio ma : mb is represented

^ ma , . ma a , " , ,

by — ; and smce — = - , we have 7na : too = a : 0.
mb mb b

A ratio is changed if its terms are multiplied by different
multipliers.

If m^Uy

then ma =^ na^

and

But

or

'IHCb

na

=^

rib^

nb

na

a

«"M •

nb

b

ma

a

nb

V

7na : rib =^

a:b*

184

RATIO, PROPORTION, AND VARIATION 185

231. Ratios are compounded by taking the product of the
fractions that represent them.

Thus, the ratio compounded of a : 6 and c:d\aac:bd.

The ratio compounded of a:b and a:b is called the duplicate
ratio a^ : b^.

The ratio compounded of a : bj a : b, and a : ^ is called the
triplicate ratio a^ib^; and so on.

232. Ratios are compared by comparing the fractions that
represent them.

Thus, a:6>or<c:d

according as - > or < - •

0 d

233. Proportion of Numbers. Four numbers, a, b, c, d, are
in proportion when the ratio a:b is equal to the ratio c:d.

We then write a:b = c:d (read, the ratio of 3. to h equals
the ratio of c to d, or a. is to h as a is to d).

A proportion is also written a:b ::c:d.

The four numbers, a, by c, d, are called proportionals ; a and
d are called the extremes, b and c the means.

234. The fourth proportional to three given numbers is the
fourth term of the proportion which has for its first three
terms the three given numbers taken in order.

Tiius, d is the fourth proportional to a, 6, and c in the proportion

a : 6 = c : d.

235. The numbers a, b, c, d, e are said to be in continued
proportion \i a-.b =^b:c =^ c:d=^ d-.e.

If three numbers are in continued proportion, the second is
called the mean proportional between the other two numbers,
and the third is called the third proportional to the other two
numbers.

Thus, h is the mean proportional between a and c in the proportion
a:h = h:c\ and c is the third proportional to a and 6.

186 COLLEGE ALGEBRA

236. If four numbers are in proportiony the prodtict of the
extremes is eqtcal to the prodtict of the means.

For, if

a:b = c :d,
a _c

Multiply by 6d,

ad=ihc.

The equation

ad :

= bc

gives

a :

be , ad

so that an extreme may be found by dividing the product of
the means by the other extreme ; and a mean may be found
by dividing the product of the extremes by the other mean.
If three terms of a proportion are given, it appears from the
above that the fourth term has one, and but one, value.

237. If the product of two numbers is equal to the product of
two others, either two may be made the extremes of a proportion
and the other two the means.

For, if ad = 6c,

divide by 6a, — = — ,

•^ ' 6d 6d

a c
or - = - .

6 d
.'. a : 6*= c : d.

238. If four numbers, a, b, c, d, are in proportion, they are
in proportion by inversion ; that is, h is to 2^, a^ d is to c.

For, if a : 6 = c : d,

*v a c

then - = - »

6 d

and 1 -H - = 1 -=- - ,

6 d

6 d
or - = -.

a c

.'. b : a = d : c.

RATIO, PROPORTION, AND VARIATION 187

239. If four nuwJberSy a, b, c, d, are in 'proportion^ they are
in proportion hy composition ; that w, a + b : b = c + d : d.
For, if aib = C',d,

n. t*.

then
and

or

5

■J'

V^'

■5-''

a-\-h

c + d

h

d

.*. a H- 6 : 6 =

.c-\-d:d.

240. If four numbers, a, b, c, d, are in proportion, they are
in proportion by division ; that is, a — b:b = c — d:d.
For, if a:b = c:d,

*u a c

then

and

or

b

d'

a
b

1

z=

C

d

1,

a —

b

c -

d

b

MIM

d

i^— •

,\a

-b:

b

^

c —

d:

d.

241. If four numbers, a, b, c, d, are in proportion, they are
in proportion by composition and division ; that is,

a + b:a — b = c + d:c — d.

a H- 6 c + d

For, from § 239,
and from § 240,
Divide,

b d

a—b_c—d

a + 6 c + d
a — 6 c — d
.-. aH-6:a — 6 = c + d:c — d.

242. If four numbers, a, b, c, d, are in proportion, they are
in proportion by alternation ; that is, a : c = b : d.
For, if a:b = e:d^

then 7 = ^ •

0 d

188 COLLEGE ALGEBRA

Multiply by -i

ab
be

he

a
c

b
d

.-. a : c

= b:d.

or

243. Like powers of the terms of a proportion are in pro-
portion.

For, if a:b = c:d,

4.U a c

then - = - .

b d
Raise both sides to the nth power,

a» _ c*
b*" d?^'
.-. a« : 6» = c« : d«.

244. If a:b = c:d, any ratio whose terms are two polyno-
mials in a and 6, homogeneous and both of the same degree,
is equal to the ratio whose terms are found from those of the
preceding ratio by substituting c for a and d for b.

To prove this in any particular case, it will be found suffi-
cient to substitute ra for b and re for d.

245. In a series of equal ratios, the sum of the antecedents

is to the sum of the consequents as any antecedent is to its

consequent,

T? t a c e g

wo may put r for each of these ratios.

n,u a c e g

Then, - = r, - = r, - = r, f = r.

b d f h

/. a = br^ c = dr, e =fry g = hr,

.'. a-\-c + e-\-g = {b-\-d +/+ h)r.

a-\- c -{■€-{• g _ _a

" 6 + <2+/ + ^~ b

.*. a + c -\-e-\- gib -^d -{•/-{• h = a:b.

In like manner, it may be shown that

ma ■}• nc -{■ pe + qg : mb -\- nd ■{■ pf -\- qh = a : b.

RATIO, PROPORTION, AND VARIATION 189

246. If four numbers^ a, b, c, d, are in continued proportion,
then a : c = a^ : b^ and a : d = a^ : b*.

For, if

6 c d

*u^ a h a a

then 7 X =7 X T»

0 c h h

.-. a : c = a2 : 62.

or

or

.i a h c a a a

-^^^» I^-^:i = lXTXT»

b c d h h h

a _a^

d"" 68'
.-. a : d = a* : 68.

247. The mean proportional between two numbers is equal to
the square root of their product.

For, if a : 6 = 6 : c,

*u ah

then - = -»

6 c
and 62 = (ic.

.'. 6 = Vaii.

248. TA-e products of the coi^esponding terms of two or more
proportions are in proportion.

For, if a : 6 = c : d,

e:/=gf:A,
A; : Z = m : n,

^, a c e gr A; m

then 7 = ^' T = 7' 7 = ~"*

0 a J h I n

Take the product of the left members, and also of the right members of

these equations,

aek __ cgm

.'. aek : bfl = cgm : dhn.

249. The laws that have been established for ratios should
be remembered when ratios are expressed in fractional form.

190 COLLEGE ALGEBRA

a;^ + a; -f 1 x^ — x -\- 2

By composition and division,

2x2 2x2

2(x+l) -2(x-2)

This equation is satisfied when x = 0 ;

2x2 1 1

or, dividing by — j when

2 X + 1 2 -x'

tliat is, when x = i.

(2) If a:b = c:d, show that

a^ -\- ab:b^ — ab = c^ -\- cd:<P — cd.

If

a c

b-r

then

a + 6 c -\- d
a — b c — d

and

a c
-b -d

a
" -b

a -f 6 c c + d
a — b — d c —d

that is.

a2 + a6 c2 + cd
62 _ ab~~ d^-cd'

or

a^ + ab

: 62 _ oft = c2 + cd : d2 - cd.

(§241)

(§248)

Exercise 37

1. Write the ratio compounded of 3 : 5 and 8 : 7. Which
of these ratios is increased, and which is diminished by the
compounding ?

2. Compound the duplicate ratio of 4 : 15 with the tripli-
cate of 5 : 2.

3. Arrange in order of magnitude the ratios 3:4, 23 : 25,
10 : 11.

Find the ratio compounded of :

4. 3:5, 10:21, 14:15. 5. 7:9, 102:105, 16:17.

RATIO, PROPORTION, AND VARIATION 1^1

0

6. a^ — x^:a^ -{- Sax -\-2x^ and a -{- x : a — x.

7. a^ — 4 : 2 a;^ — 5 X + 3 and a; — 1 : aj — 2.

8. Show that the ratio a : ft is the duplicate of the ratio

a -\- c :h -{- c M (^ = ah.

9. Two numbers are in the ratio 2 : 5, and if 6 is added to
each, the sums are in the ratio 4 : 7. Find the numbers.

10. What must be added to each of the terms of the ratio
m : n that it may become equal to the ratio p'^q'^

11. If a; and y are such that, when they are added to the
antecedent and consequent respectively of the ratio a; ft, its
value is unaltered, show that x:y = a:h.

Find x from the proportions :

12. 27: 90 = 45: a;. Sa 12a _Uc

13. lli:4i = 3i:a;. ' 5ft* 7c ~16ft*^*

Find the third proportional to :

^ , a2-ft2 ^ a-ft

15. fj and T^y. 16. and

c c

Find the mean proportional between :

17. 3 and 16*. 18. ^^!!^ and ^^^.

If a:b = c:dy prove that :

19. 2a-|-ft:ft = 2c-h(^:c?. 20. Sa — b:a =3c — die.

21. 4a -h 3ft:4a — 3ft = 4c-h3c?:4c — 3<^.

22. 2a«-|-3ft«:2a»-3ft« = 2c«-h3(£«:2c«-3d:».

If a : ft = ft : c, prove that :

23. a* -h aft : ft2 -I- ftc : : a : c. 24. a : c : : (a -|- ft)* : (ft + c)'.

25. If — r-^ = = 9 and X, y, z are unequal, show

that Z -h m -h w = 0.

192 COLLEGE ALGEBRA

Find X from the proportions :

26. x-\-l:x — l:=x-\-2:x — 2.

27. X'^a:2x — b = Sx-\-b:4:X — a.

28. x^-4.x-\-2:x^-2x-l = x^-4.x:x^-2x-2.

29. 3 -\- X : 4: -\- X = 9 -{- x lis -\- X.

30. a-{-x:b-\-x = c-}-x:d-\-x.

31. If a:b = c:df show that

a* + ^' : 7 = c2 -h ^ :

a -]- b c -{- d

32. When a, by c, d are proportionals and all unequal, show
that no number x can be found such that a -^ x, b -{- x, c -\- x,
d -\- X shall be proportionals.

RATIO OF MAGNITUDES

250. Commensurable Magnitudes. If two magnitudes of the
same kind are so related that a unit of measure can be found
which is contained in each of the magnitudes an integral num-
ber of times, this unit of measure is a common measure of the
two magnitudes, and the two magnitudes are commensurable.

Two magnitudes different in kind can have no ratio.
If two commensurable magnitudes are measured by the same
unit, their ratio is the ratio of their numerical measures.

Thus, \ of a foot is a common measure of 2^ feet and 3} feet, being
contained in the first 16 times and in the second 22 times.

Therefore, the ratio of 2 J feet to 3 J feet is the ratio of 16 : 22.

251. Incommensurable Magnitudes. Two magnitudes of the
same kind that cannot both be expressed in integers in terms
of a common unit are said to be incommensurable, and the eoRoct
value of their ratio cannot be fomid. But by taking the unit

RATIO, PROPORTION, AND VARIATION 193

sufficiently small, an approximate value can be found that
shall differ from the true value of the ratio by less than any
assigned value, however small.

Suppose a and ^ to be two incommensurable magnitudes of the
same Icind, Divide b into any integral number, n, of equal parts^
and suppose one of these parts is contained in a more than m

times and less than m -{- 1 times. Then, - lies between — and

4-1 ^1

and cannot differ from either of these by so much as - •

n -j^ n

But, by increasing n indefinitely, — can be made to decrease

indefinitely and to become less than any assigned value, how-
ever small, though it cannot be made absolutely equal to zero.
Hence, the ratio of two incommensurable magnitudes,
although it cannot be expressed exactly by numbers, may be
expressed approximately to any desired degree of accuracy.

Thus, if h represents the length of the side of a square, and a the

a r~
length of the diagonal, - = v2 .

Now, v^= 1.41421356 •••, a value greater than 1.414213, but less
than 1.414214.
If, then, a m\llwnlth part of 6 is taken as the unit, the value of the

ratio - lies between and , and therefore differs from

b 1000000 1000000

either of these fractions by less than

1000000

By carrying the decimal farther, a fraction may be found that will
differ from the true value of the ratio by less than a billionth^ a trillionthy
or by less than any other assigned value whatever.

Hence, the ratio - > while it cannot be expressed by numbers exactly,

b

may be expressed by numbers to any degree of accuracy we please,

252. The ratio of two incommensurable magnitudes is an
incommensurable ratio, and is a fixed value such that an
approximate value can be found which will differ from this
fixed value by a quantity whose absolute value shall be less
than that of any assigned constant, however small.

194 COLLEGE ALGEBRA

253. Equal Incommensurable Ratios. As the treatment of
Proportion in Algebra depends upon the assumption that it is
possible to find fractions which will represent ratios, and as
it appears that no fraction can be foimd to represent exactly
the value of an incommensurable ratio, it is necessary to show
that two incoinmensurahle ratios are equal if their approxi-
mate values remain equal when the unit of measure is indefi-
nitely diminished.

Thus, let a:b and a^ : ^ be two incommensorable ratios whose true

values lie between the approximate values — and , when the unit

n n

of measure is indefinitely diminished. Then they cannot difier from each

other by so much as - •

n

Let d denote the difference (if any) between a : h and a' :b' ; then

^ 1
d<-'
n

Now the true values of a : 6 and a' : b^ being fixed, their difference, d,
must be fixed, that is, d must be a constant.

By increasing n we can make the value of - less than. any assigned

1 ^

value, however small ; hence, - can be made less than d if d is not zero.

n

Therefore, d is 0, and there is no difference between the ratios a : b
and a' : b\ (Therefore, a : 6 = a' : b\

254. The laws which apply to the proportion of abstract
numbers apply to the proportion of concrete quantities, except
that alternation will apply only when the four quantities in
proportion are all of the same kind.

Exercise 38

1. A rectangular field contains 5270 acres, and its length is
to its breadth in the ratio of 31 : 17. Find its dimensions.

2. If five gold coins and four silver ones are worth as much
as three gold coins and twelve silver ones, find the ratio of the
value of a gold coin to that of a silver one.

RATIO, PROPORTION, AND VARIATION 195

3. The lengths of two rectangular fields are in the ratio of
2 : 3, and the breadths in the ratio of 5 : 6. Find the ratio of
their areas.

4. Two workmen are paid in proportion to the work they
do. A can do in 20 days the work that it takes B 24 days
to do. Compare their wages.

5. In a mile race between a bicycle and a tricycle their rates
were as 5 : 4. The tricycle had haK a minute start, but was
beaten by 176 yards. Find the rate of each.

6. A railway passenger observes that a train passes him,
moving in the opposite direction, in 2 seconds ; but moving
in the same direction with him, it passes him in 30 seconds.
Compare the rates of the two trains.

7. A vessel is half full of a mixture of wine and water.
If filled with wine, the ratio of the quantity of wine to that
of water is 10 times what it would be if the vessel were filled
with water. Find the ratio of the original quantity of wine
to that of water.

8. A quantity of milk is increased by watering in the ratio
4 : 5, and then 3 gallons are sold ; the remainder is increased
in the ratio 6 : 7 by mixing it with 3 quarts of water. How
many gallons of milk were there at first ?

9. Each of two vessels, A and B, contains a mixture of
wine and water ; A in the ratio of 7 : 3, and B in the ratio of
3:1. How many gallons from B must be put with 5 gallons
from A to give a mixture of wine and water in the ratio of
11 : 4 ?

10. The time which an express train takes to travel 180
miles is to that taken by an accommodation train as 9:14.
The accommodation train loses as much time from stopping
as it would take to travel 30 miles; the express train loses

196 COLLEGE ALGEBRA

only half as much time as the other by stopping, and travels
15 miles an houi* faster. What are their respective rates ?

11. A and B trade with different sums. A gains $200 and
B loses $50, and now A^s stock is to B's as 2 : ^. But if A
had gained $100 and B lost $85, their stocks would have
been as 15 : 3^. Find the original stock of each.

12. A line is divided into two parts in the ratio 2:3, and
into two parts in the ratio 3:4; the distance between the
points of section is 2. Find the length of the line.

13. A railway consists of two sections ; the annual expendi-
ture on one is increased this year 5 per cent, and on the other
4 per cent, producing on the whole an increase of 4^ per cent.
Compare the amounts expended on the two sections last year,
and also the amounts expended this year.

VARIATION

255. One quantity is said to vary as another when the two
quantities are so related that the ratio of any two values of
the one is equal to the ratio of the corresponding values of
the other.

Thus, if it is said that the weight of water varies as its volume, the
meaning is that one gallon of water is to any number of gallons as the
weight of one gallon is to the weight of the given number of gallons.

256. Function of a Variable. Two variables may be so related
that when a value of one is given the correspondiiig value of
the other can be found. In this case one variable is said to
be a function of the other.

Thus, if the rate at which a man walks is known, the distance he
walks can be found when the time is given ; the distance is in this case
^function of the time.

257. When two variable magnitudes X and F, not necessarily
of the same kind, are so related that when X is changed in

RATIO, PROPORTION, AND VARIATION 197

«

any ratio, Y is changed in the same ratio, Y is said to vary as
X, and the relation is denoted thus, Foe A'. The sign oc, called
the sign of variation, is read varies as.

Thus, the area of a triangle with a given base varies as its altitude ;
for, if the altitude is changed in any ratio, the area is changed in the
same ratio.

If Y ccX, and if when X has a definitely assigned value A,
Y takes the value B, then

B:Y=A:X, [1]

and therefore, by the theory of proportion, B has a value
definitely determined by the value of A.

Let the numerical measures of A, B, X,' and F be a, ^, a;,
y, respectively, so that

and b:y = B:Y:

Therefore, by [1], b:y = a:x.

.*. h : a = y : X. [2]

Since a and h are the numerical measures of the definitely
assigned magnitudes A and B^ they are themselves constant and
their ratio, h-.a, is constant. Also, x and y are the numerical
measures of the variable magnitudes A' and F; hence, by [2],

When two variable magnitudes X and Y are so related that
YccX, their numerical measures are so related that their ratio
is constant.

Hence, if yccx, y.x is constant, and if we represent this

constant by m,

-I y

y:x = m:l, or - = m. .'.y = 7nx.

Again, if y\ x' and y", x" are two sets of corresponding
values of y and x, then

y' :x' = y" : x",

or y' : y" = x' : x".

198 COLLEGE ALGEBRA

258. Inverse Variation. When x and y are so related that
the ratio of y to - is constant, y is said to vary inversely as x ;

thife is written y oc - •

Thus, the time required to do a certain amount of work varies inversely
as the number of workmen employed ; for, if the number of workmen is
doubled, hadved, or changed in any other ratio, the time required is halved,
doubled, or changed in the inverse ratio.

In this case, y : - = m.

X

m -

.*. y = —9 and xy = 7n;

that is, the product xy is constant.

1 1

As before, y':- = y":--,

X X

x'y' = x"y",
or y' : y" = a;" : a;'.

259. If the ratio of y:xz is constant, then y is said to vary
jointly as x and z.

In this case, y = mxz,

and y' : y" = x'z' : x"z",

X

260. If the ratio y : - is constant, then y varies directly as
X and inversely as z.

In this case.

mx

y= y
z

and

1 11 XX

y -y = ,' n

z' z"

261. Theorem I. li y cc x, and xccz, then yocz.
For y = niXy and x = nz.

/. y = tnnz,

,'. y ccz.

RATIO, PKOPORTION, AND VARIATION 199

262. Theorem II. If y oc x^ and « oc a;, then (y ± «) oc x.
For y = Ttvx^

and z = Tix.

«^

.'. (y ± «) ocaj.

263. Theorem in. If yccx when « is constant, and yccz
when aj is constant, and if x and « are independent of each
other, then yocxz when x and z ar« both variable.

Let x', y\ z' and aj", y", «" be two sets of corresponding
vahies of the variables.

Let X change from a;' to aj", while z remains constant, and
let the corresponding value of y be Y,

Then, y':Y = x': x'\ [1]

Now, let z change from «' to «'', while x remains constant.
Then, Y\y''^z'\ z\ [2]

From [1] and [2],

yy:y"F = a;V:a;"«", (§248)

or y' : y" = aj's' : aj"«",

or y' :x'z' = y" :x"z".

V
.'. the ratio -^ is constant, and y oc xz,

xz ^ ^

In like manner, it may be shown that if y vai'ies as each
one of any number of independent values x, z,tc,'", when the
rest are unchanged, then when they all change, y oc xzu • • •

Thus, the area of a rectangle varies ajs the base when the altitude is
constant, and as the altitude when the base is constant, but as the product
of the base and altitude when both vary.

The volume of a rectangular solid varies as the length when the width
and thickness remain constant ; as the width when the length and thick-
ness remain constant ; as the thickness when the length and width remain
constant ; but as the product of length, breadth, and thickness when all
three vary.

200 COLLEGK ALGEBRA

264. Examples. (1) It y varies inversely as x, and when
y =^ 2 the corresponding value of x is 36, lind the eorrespond-
iug value of x when y = 9.

Here, 2/ = — » or »w = xy.

.-. w = 2 X 36 = 72.
If 9 and 72 are substituted for y and m respectively in

m

!/ = —■>
X

72

the result is 9 = — , or 9« = 72.

X

.'. x = S.

(2) The weight of a sphere of given material varies as its
volume, and its volume varies as the cube of its diameter. If
a sphere 4 inches in diameter weighs 20 pounds, find the
weight of a sphere 5 inches in diameter.

Let W represent the weight,

V represent the volume,

D represent the diameter.
Then, WocV, and F«D8.

Put W = mX)8 ;

then, since 20 and 4 are corresponding values of W and D,

20 = m X 64.

.-. m = §J = r«g.

Therefore, when D = 6, W = ^^^ of 125 = Zdfg.

Exercise 39

1. If ?/ oc X, and y = 4 when x = 5, find y when x = 12.

2. If y oc X, and y = i when x = ^, find y when « = J.

3. If z varies jointly as x and y, and 3, 4, 5 are simulta-
neous values of x, y, z, find z when a; = y = 10.

RATIO, PROPORTION, AND VARIATION 201

4. If « oc-j and 05 = 4 and y = 3 when « = 6, find the
value of z when x = 5 and y = 7.

5. If the square of x varies inversely as the cube of y, and
x=:2 when y = 3, find the equation between x and y.

6. If z varies as x directly and y inversely, and if a; = 3
and y = 4: when z = 2, find z when x = 15 and y ^S,

7. The velocity acquired by a stone falling from rest varies
as the time of falling ; and the distance fallen varies as the
square of the time. If it is found that in 3 seconds a stone has
fallen 145 feet and acquired a velocity of 96f feet per second,
find the velocity and distance fallen at the end of 5 seconds.

8. If a heavier weight draws up a lighter one by means of
a string passing over a fixed wheel, the space described in a
given time varies directly as the difference between the weights,
and inversely as their sum. If 9 ounces draws 7 ounces
through 8 feet in 2 seconds, how high will 12 ounces draw 9
ounces in the same time ?

9. The space will also vary as the square of the time.
Find the space in Example 8, if the time is 3 seconds.

10. Equal volumes of iron and copper are found to weigh
77 and 89 ounces respectively. Find the weight of 10^ feet
of round copper rod when 9 inches of iron rod of the same
diameter weigh 31-^^ ounces.

11. The square of the time of a planet's revolution about
the sun varies as the cube of its distance from the sun. If
the distances of the Earth and Mercury from the sun are as
91 to 35, find in days the time of Mercury's revolution.

12. A spherical iron shell 1 foot in diameter weighs ^^
of what it would weigh if solid. Find the thickness of the
metal, it being known that the volume of a sphere varies as
the cube of its diameter.

CHAPTER XVIII

PROGRESSIONS

265. A successioD of numbers that proceed according to
some fixed law is called a series ; the successive numbers are
called the terms of the series.

A series that ends at some particular term is a finite series ;
a series that continues without end is an infinite series.

266. The number of different forms of series is unlimited ;
in this chapter we shall consider only arithme^iical series^
geometrical series, and harmonical series.

ARITHMETICAL PROGRESSION

267. A series is called an arithmetical series or an arithmet-
ical progression when each succeeding term is obtained by adding
to the preceding term a constant difference.

The general representative of such a series is

a, a -\- dy a -{-2d, a -{-Sd, • • • ,

in which a is the first term and d \he common difference ; the
series is increasing or decreasing according as e^ is positive
or negative.

268. The nth Term. Since each sticceeding term of the series
is obtained by adding d to the preceding term, the coefficient
of d is always one less than the number of the term, so that
the nth term is a -|- (^ — 1) d.

If the nth term is represented by /, we have

l = a^+(n-l)d. (I)

202

ARITHMETICAL PROGRESSION 203

269. Sum of the Series. If I denotes the nth term, a the
first term, n the number of terms, d the common difference,
and s the sum of n terms, it is evident that

5= a-\-{a-\-d)-\-(a-{-2d)-\ \-(l — d)-{-l

or 5= l-\-\l — d)-\-(l — 2d)-\ \-(a-\-d)-\-a

.-. 25 = (a + 0 + (^+ 0 + (^ + 0 H h(a4-0 + (« + 0

= w (a -h 0*
Therefore, s = 5 (a + 1). (II)

870. From the two equations,

pl = a + (n-l)d, (I)

^-

n

8 = ^ (a + 1), (11)

any two of the five numbers a, d, I, w, s may be foimd when
the other three are given.

(1) Find the sum of ten terms of the series 2, 5, 8, 11, • • •

Here, a = 2, d = 3, n = 10.

From (I), Z = 2 + 27 = 29.

Substitute in (II), s = V^ (2 + 29) = 156.

(2) The first term of an arithmetical series is 3, the last
term 31, and the sirni of the series 136. Find the series.

From (I), 31 = 3H-(n-l)d, [1]

From (II), 136 = - (3 + 31). [2]

From [2], n = 8.

Subfltitute in [1], d = 4.

Therefore, the series is 3, 7, 11, 16, 10, 23, 27, 31.

(3) How many terms of the series 5, 9, 13, ••. must be
taken in order that their sum may be 275?

From (I), Z = 6 + (n-l)4.

.-. i = 4n + l. [1]

204 COLLEGE ALGEBRA

From (II), 276 = ^(6 + 0. [2]

Substitute in [2] the value of I found in [1],

276 = J(4n + 6),

or 2n2 + 3n = 276.

We now have to solve this quadratic.
Complete the square,

16n2 + {) + 9 = 2209.

Extract the root, 4 n + 3 = ± 47.

.-. n = 11 or - 12i.

We use only the positive result.

(4) Find n when d^ I, s are given.

From (I),

a = l-(n-l)d.

From (H),

2s -In

a =

n

Therefore,

l-

-{n-

i)d-2»-'».

n

.'. In — dn^ + dn = 2 s — Zn.
.-. dn2-(2Z + d)n = -2«.

This is a quadratic with n for the unknown number.
Complete the square,

4d2n2 _ () ^. (2 Z + d)2 = {2Z + d)2 _ 8d«.

Extract the root,

2 dn - (2 Z + d) = ± V(2Z + d)2 - 8d«.

_2l + d ±y (2 f + d)2 - 8 d8
2d

Note. The table on the next page contains the results of the general
solution of all possible problems in arithmetical series, in which three of
the numbers a, Z, d, n, s are given and two required. The student is
advised to work these out, both for the results obtained and for the prac-
tice gained in solving literal equations in which the unknown quantities
are represented by letters other than sc, y, «.

AEITHMETICAI, •pRoGSESSION

No.

GiVKir

KKlHrTBKl.

Result

2
3
i

dns

■

l = a + {n-l)d.

[^-^di,^f2da + {a-^d}K

6
6

7

9
10
11
12

adn
ail
ani
dnl

•

a=in[2ai + (n-l)d].

l + a P-a^
'^ 2 ^ -Id '

s = Jn[2l-(n-l)d].

dnt
dns
dla

•

13
U
15
16
17
18
lit
20

anl

d

2 (» ^ an)
n{n-I>

ads

'

..t^+i.

d-2o±V{2fl-d)» + 8(b

" id

2i + d±V[2( + <H3-8d»
" 2d

206 COLLEGE ALGEBRA

271. The arithmetical mean between two numbers is the
number which, when placed between them, makes with them
an arithmetical series.

If a and h represent two numbers, and A their arithmetical
mean, then, by the definition of an arithmetical series^

A — a = h — A,
..A- ^

272. Sometimes it is required to insert several arithmetical
means between two numbers.

Insert six arithmetical means between 3 and 17.

Here the whole number of terms is 8 ; 3 is the first term and 17 the
eighth.

By (I), 17 = 3 + 7d.

.-. d = 2.

Therefore, the complete series is

3, [6, 7, 9, 11, 13, 16,] 17,
the terms within the brackets being the means required.

273. When the sum of a number of terms in arithmetical
progression is given it is convenient to represent the terms
as follows :

Three terms by ^ — y> ^9 « + y ;

four terms by a; — 3y, a — y, « + y; a5 + 3y;

and so on.

The sum of three numbers in arithmetical progression is 36,
and the square of the mean exceeds the product of the two
extremes by 49. Find the numbers.

Let X — y, X, X + y represent the numbers.
Then, adding, 3x = 36.

/. X = 12.
Putting for x its value, the numbers are 12 - y, 12, 12 + y.

ARITHMETICAL PROGRESSION 207

By the conditions of the problem,

(12)2 ^ (12 _ y) (12 + y) + 49,
144 = 144 - y2 4. 49^

Therefore, the numbers are 5, 12, 19 ; or 19, 12, 5.

ExerclBO 40

Find:

1. The tenth term of 3, 8, 13, • • •

2. The eighth term of 12, 9, 6, • • •

3. The twelfth term of - 4, - 9, - 14, .. .

4. The eleventh term of 2J, 1|, 1^, • • •

5. The fourteenth term of 1^, J, — |, • • •

Find the sum of :

6. Eight terms of 4, 7, 10, • • •

7. Ten terms of 8, 5, 2, • • •

8. Twelve terms of — 3, 1, 6, • • •

9. n terms of 2, 1 J, J, • • •

10. n terms of 2 J, 1|, 1^^, • • •

11. Given a = S, I = 55, n= 13. Find d and s.

12. Given a = 3J, Z = 64, « = 82. Find d and s.

13. Given a = 1, ti = 20, s = 305. Find d and I.

14. Given I = 105, w = 16, s = 840. Find a and d,

15. Given c? = 7, w = 12, s = 594. Find a and Z.

16. Given a = 9, c? = 4, s = 624. Find n and L

17. Given d = 5,l = 77,s = 623. Find a and n.

208 COLLEGE ALGEBRA

18. When a train arrives at the top of a long slope the last
car is detached and begins to desbend, passing over 3 feet in
the first second, 3 times 3 feet in the second second, 6 times
3 feet in the third second, and so on. At the end of 2 minutes
it reaches the bottom of the slope. What space did the car
pass over in the last second ?

19. Insert eleven arithmetical means between 1 and 12.

20. The first term of an arithmetical series is 3, and the
sum of 6 terms is 28. What term will be 9 ?

21. How many terms of the series — 5, — 2, +!,••• must
be taken in order that their sum may be 63 ?

22. The arithmetical mean between two numbers is 10, and
the mean between the double of the first and the triple of the
second is 27. Find the numbers.

23. The first term of an arithmetical progression is 3, the
third term is 11. Find the sum of seven terms.

24. Arithmetical means are inserted between 8 and 32, so
that the sum of the first two is to the sum of the last two as
7 is to 25. How many means are inserted ?

25. In an arithmetical series the common difFerence is 2,
and the square roots of the first, third, and sixth terms form
a new arithmetical series. Find the series.

26. Find three numbers in arithmetical progression of which
the sum is 21, and the sum of the first and second three-fourths
of the sum of the second and third.

27. The sum of three numbers in arithmetical progression
is 33, and the sum of their squares is 461. Find the numbers.

28. The sum of four numbers in arithmetical progression
is 12, and the sum of their squares 116. What are these
numbers ?

ARITHMETICAL PROGRESSION 209

29. How many terms of the series 1, 4, 7, • • • must be taken
in order that the sum of the first half may bear to the sum of
the second half the ratio 7 : 22 ?

30. The sum of the squares of the extremes of four num-
bers in arithmetical progression is 200, and the sum of the
squares of the means is 136. What are the numbers ?

31. A man wishes to have his horse shod. The blacksmith
asks him $2 a shoe, or 1 cent for the first nail, 3 for the
second, 5 for the third, and so on. Each shoe has 8 nails.
Ought the man to accept the second proposition ?

32. A number consists of three digits which are in arith-
metical progression, and this number divided by the sum of
its digits is equal to 26 ; if 198 is added to the number, the
digits in the units' and hundreds' places will be interchanged.
Required the> number.

33. There are placed in a straight line upon a lawn 50 eggs
3 feet distant from each other. A person is required to pick
them up one by one and carry them to a basket in the line of
the eggs and 3 feet from the first egg, while a runner, starting
from the basket, touches a goal and returns. At what distance
ought the goal to be placed that both men may have the same
distance to pass over ?

34. Starting from a box, there are placed upon a straight
line 40 stones, at the distances 1 foot, 3 feet, 5 feet, and so
on. A man placed at the box is required to take them and
carry them back to the box one by one. What is the total
distance that he has to accomplish ?

35. The sum of five numbers in arithmetical progression is
45, and the product of the first and fifth is five-eighths of the
product of the second and fourth. Find the numbers.

210 COLLEGE ALGEBRA

GEOMETRICAL PROGRESSION

274. A series is called a geometrical series or a geometrical
progression when each succeeding term is obtained by multi-
plying the preceding term by a constant multiplier.

The general representative of such a series is

a, ar, ar^, ar^, ar^, •••,

in which a is the first term and r the constant multiplier or
ratio.

The terms increase or decrease in numerical magnitude
according as r is numerically greater than or numerically less
than unity.

275. The nth Term. Since the exponent of r increases by
one for each succeeding term after the first, the exponent is
always one less than the number of the term, so that the »th
term is ar"""^

If the nth. term is represented by I, we have

l = ar»-\ ' (I)

276. Sum of the Series. If I represents the nth term, a the
first term, n the number of terms, r the common ratio, and s
the sum of n terms, then

s = a -{- ar -{- ar^ + • • • + a^~^ [1]

Multiply by r, rs = ar -{- ar^ -\- ar^ -\ h ar""^ + af^, [2]

Subtract the first equation from the second,

r5 — s = ar* — a.
Resolve each member into its factors,

{r-l)s = a{i--l). . )'■ \

Divide by r — 1. ^* ^ \ • *

Therefore, e = ^ _ ^ • (II)

GEOMETRICAL PROGRESSION 211

Since I = ar""^, rl = ar", and (II) may be written

rl — a

277. From the two formulas (I) and (II), or the two for-
mulas (I) and (III), any two of the five nimibers a, r, I, n, s
may be found when the other three are given.

(1) The first term of a geometrical series is 3, the last term
192, and the sum of the series 381. Find the number of terms
and the ratio.

From (1), 192 = 3 r«-i. [1]

From (in), 381 = ^^^ — ?. [2]

V — 1

From [2], r = 2.

Substitute in [1], 2'»-i = 64.

.-. n = 7.

Therefore, the series is 3, 6, 12, 24, 48, 96, 192.

(2) Find I when r, n, s are given.

I

From (1), a =

fti — 1

I

rl-

fti — 1

Substitute in (III), 8 = .

r — 1

' fn, — 1

(r — l)r'»— ^«
r« - 1

Note. The table on page 212 contains the results of all possible
problems in geometrical series in which three of the numbers a, r, 2, n, 8
are given and the other two required, with the exception of those in
which n is required ; these last require the use of logarithms with which
the student is supposed to be not yet acquainted.

\

212

COLLEGE ALGEBRA

No.

Given

Required

BEBUIiT

1
2
3
4

am

ar 8

an8

' r n 8

I

l = ar^-'K
^_a + (r-l)«

r
I (8 - 0~-^ - a (8 - a)«-i = 0.
^_{r-l)8fn-i

5
6

7
8

am
ar I
a n I
r n I

8

a(r«-l)

8 =: •

r-1

rl — a

8 =

r-X

n — l, — n — 1 ^—

Win — Va*

8 = 5 -. •

n — 1 ,- n — 1 /—

Vz- Va
Irn-l

8 = J-

r«-rn-i

9
10
11
12

13

14
16
16

r nl
r n 8
r I 8
n I 8

a

I

rn-i
a = rl — (r — 1)8.
a(8- a)»-i -1(8- 0»-i = 0.

a n I

a n 8
a I 8
n I 8

r

8 8 — a -

r» r + = 0.

a a

8 — a
*• — ■ .

s , I ^

rn — r»-i 4- : = 0.

S-l 8-1

278. The geometrical mean between two numbers is the
number which when placed between them makes with them
a geometrical series.

GEOMETRICAL PROGRESSION 213

If a and h denote two numbers, and G their geometrical
mean, then, by the definition of a geometrical series,

a " G*

.'. G = Vab.

279. Sometimes it is required to insert several geometrical
means between two numbers.

Insert three geometrical means between 3 and 48.

Here the whole number of terms is 5 ; 3 is the first term and 48 the
fifth term.

By (I), 48 = 3r*. •

.-. r* = 16,
and r = ± 2.

Therefore, the series is one of the following :

3, [ 6, 12, 24,] 48;
3, [-6, 12, -24,] 48.

The terms within the brackets are the means required.

280. Infinite Geometrical Series. When r is less than 1, the

successive terms become numerically smaller and smaller ; by

taking n large enough we can make the Tith term, ar^~^, as

small as we please, although we cannot make it absolutely

equal to zero.

The sum of n terms, — ' ^7 changing the signs of the

' ■*■ d fit?***

numerator and denominator, may be written -z > which

1 — r

CL CLT^ a

is equal to :; ; this sum differs from by

1 — rl — r 1 — r

the fraction ; by taking enough terms we can make

1 — r

an^^ and consequently this fraction, as small as we please;
the greater the number of terms taken the nearer is their

sum to Hence, -. — '■ — is called the sum of an infinite

1 — r 1 — r

number of terms of the series.

214 COLLEGE ALGEBRA

(1) Find the sum of the infinite series 1 — J-hi^ |4-*"

Here a = 1, r = — J.

1 2

The sum of the series is -, or -.

1 + i 3
Therefore, the sum of n terms is

*-f(-««, or J[l_(_i)n].
This sum evidently approaches | as n is increased.

(2) Find the value of the recurring decimal 0.12135135 • • •
Consider first the part that recurs ; this may be written

and the sum of this series is . ^ » or yj^. Adding 0. 12, the part
that does not recur, we obtain for the value of the decimal ^V^.

Ezercise 41

Find:

1. The eighth term of 3, 6, 12, • • •

2. The twelfth term of 2, - 4, 8, • • •

3. The twentieth term of 1, — J, ^, • • •

4. The eighteenth term of 3, 2, IJ, • • •

5. The nth term of 1, — Ij, 1{, • • •

Find the sum of :

6. Eleven terms of 4, 8, 16, • • •

7. Nineteen terms of 9, 3, 1, • • •

8. Twelve terms of 5, — 3, 1 J, • • •

9. n terms of 1|, |, 5^^, • • •

Sum to infinity :

10. 4-2 + 1 12. l-§4-^

11. j4_j + j4_... 13. j-f-^-f^ + ...

GEOMETRICAL PROGRESSION 215

Find the value of the recurring decimals :

14. 0.153153... 16. 3.17272...

15. 0.123535 ... 17. 4.2561561 . • •

18. Given a = 36, ^ = 2J, n = 5. Find r and s,

19. Given I = 128, r = 2,n = 7. Find a and s.

20. Given r = 2, ri = 7, s = 635. Find a and I.

21. Given I = 1296, r = 6,s = 1555. Find a and n.

22. Insert" three geometrical means between 14 and 224.

23. Insert five geometrical means. between 2 and 1458.

24. If the first term is 2 and the ratio 3, what term will be
162?

25. The fifth term of a geometrical series is 48, and the
ratio 2. Find the first and seventh terms.

26. Four numbers are in geometrical progression ; the sum
of the first and fourth is 195, and the sura of the second and
third is 60. Find the numbers.

27. The sum of four nimibers in geometrical progression is
105 ; the difference between the first and last is to the differ-
ence between the second and third in the ratio of 7 : 2. Find
the numbers.

28. The first term of an arithmetical progression is 2, and
the first, second, and fifth terms are in geometrical progression.
Find the sum of 11 terms of the arithmetical progression.

29. . The sum of three numbers in arithmetical progression
is 6. If 1, 2, 5 are added to the numbers, the three resulting
numbers are in geometrical progression. Find the numbers.

30. The sum of three numbers in arithmetical progression
is 15 ; if 1, 4, 19 are added to the numbers, the results are in
geometrical progression. Find the numbers.

216 COLLEGE ALGEBRA

31. There are four numbers of which the sum is 84; the
first three axe in geometrical progression and the last three in
arithmetical progression ; the sum of the second and third is
18. Find the numbers.

32. There are four numbers of which the sum is 13, the
fourth being 3 times the second; the first three are in geo-
metrical progression and the last three in arithmetical progres-
sion. Find the numbers.

33. The sum of the squares of two numbers exceeds twice
their product by 576 ; the arithmetical mean of the two num-
bers exceeds the geometrical by 6. Find the numbers.

34. A number consists of three digits in geometrical pro-
gression. The sum of the digits is 13 ; and if 792 is added
to the number, the digits in the units' and hundreds' places
will be interchanged. Find the number.

35. Find an infinite geometrical series in which each term
is 5 times the sum of all the terms that follow it.

36. If a, b, c, d are four numbers in geometrical progression,
show that

{a" -k-b^-k- c2) (^2 _^ ^2 ^ ^) ^ (^5 + bc + cd)\

HARMONICAL PROGRESSION

281. A series is called a harmonical series, or a harmonlcal
progression, when the reciprocals of its terms form an arith-
metical series.

The general representative of such a series is

111 1

a a -\- d a-^2d a-{-(n — l)d

Questions relating to harmonical series are best solyed by

writing the reciprocals of its terms, and thus forming au arith-

metical series.

4'

HARMONIC AL PROGRESSION 217

«

282. The harmonical mean between two nmnbers is the num-
ber which when placed between them makes with them a
harmonical series.

If a and h denote two numbers, and H their harmonical
mean, then, by the definition of a harmonical series,

H a" h h'

. 2 _1 l_a + b
' ' H a h ah

a-\-h

283. Sometimes it is required to insert several harmonical
means between two numbers.

Insert three harmonical means between 3 and 18.

Find the three arithmetical means between \ and ^.
These are found to be ^f , \\, -f^ ; therefore, the harmonical means are
ih \h V ; or 3|J, 5^. 8.

284. Since A = ^^-—-^ and G =Vab,

G'

H=—y or G=VaH,
A

That is, the geometrical mean between two numbers is also
the geometrical mean between the arithmetical and harmon-
ical means of the numbers, or

A:G = G:H,

Hence, G lies in numerical value between A and H.

Exercise 42

1. Insert four harmonical means between 2 and 12.

2. Find two numbers whose difference is 8 and the har-
monical mean between them 1}.

218 COLLEGE ALGEBRA

3. Find the seventh term of the harmonica! series 3, 3f ,
4, • • •

4. Continue to two terms each way the harmonical series
of which two consecutive terms are 15, 16.

5. The first two terms of a harmonical series are 6 and 6.
What term will be 30 ?

6. The fifth and ninth terms of a harmonical series are 8
and 12. Find the first four terms.

7. The difference between the arithmetical and harmonical
means between two numbers is If, and one of the numbers is
4 times the other. Find the numbers.

8. The arithmetical mean between two numbers exceeds the
geometrical by 13, and the geometrical exceeds the harmonical
by 12. What are the numbers ?

9. The sum of three terms of a harmonical series is 39,
and the third is the product of the other two. Find the terms.

10. When a, ft, c are in harmonical progression show that
a:c = a — h:h — c,

11. If a and h are positive, which is the greater, A ot H?

12. Show tha1[ a, ft, and c will be in arithmetical progres-
sion, in geometrical progression, or in harmonical progressioni
according as a — ft : ft — c is equal to a : a, to a : 6, or to a : e. •

CHAPTER XIX

BINOMIAL THEOREM; POSITIVE INTEGRAL EXPONENT

285. Binomial Theorem ; Positive Integral Exponent. By suc-
ce'fesive multiplications we obtain the following identities :

(a -h 5)8 = a8 -h 3 a% ^-^ah^-^r ^8;

(a -f hy = a^ -h 4a8Z^ -{-Qa%^ -h 4aZ^8 + h\

The expressions on the right may be written in a form
better adapted to show the law of their formation:

3-2 3-21

4'3 4-32 4. 321

(- + ^)* - -* + ^ «'* + 172 '^^^^ + itIts «*' + rlil **•

286. Let n represent the exponent of (a -f h) in any one of
these identities; then, in the expressions on the right, we
observe that the following laws hold true :

1. The number of terms is ti -h 1.

2. The first term is a"*, and the exponent of a decreases by
one in each succeeding term. The first power of h occurs in
the second term, the second power in the third term, and the
exponent of h increases by one in each succeeding term.

The sum of the exponents of a and h in any term is n.

3. The coefficient of the first term is 1; of the second
term, n\ of the third term, \ o '^ ^^^ ^^ ^^•

219

220 COLLEGE ALGEBRA

287. The Coefficient of Any Term. The number of factors
in the numerator of the coefficient of any term is the same as
the number of factors in the denominator of that term. The
number of factors in each numerator and denominator is the
same as the exponent of h in that term, and this exponent is
one less than the number of the term.

288. Proof of the Theorem. Show that the laws of § 286
hold true when the exponent is any positive integer.

We know that the laws hold for the fourth power ; suppose,
for the moment, that they hold for the A;th power, k being any
positive integer.

We shall then have

(a 4- hf = a* -h ka'^'^h + ^^^"I^^ a*-^^^

1 * ^

Multiply both members of [1] by a + J ; the result is
(a + J)*+> = a*+' +(k + l)a*i ^(^ + l)^^t-iy

In the right member of [1] for k put * + 1 ; this gives

«*+ 1 + (A, + 1) a*j + (^ + 1)(^ + 1-1) ^»-ij«

1 * z

^ 123 ^ u -r

This last expression, simplified, is seen to be identical with
the right member of [2], and this in turn by [2] is identical
with {a-{-hy-^\

BINOMIAL THEOREM 221

Hence, [1] holds when for k we put k -\-l\ that is, if the
laws of § 286 hold for the A;th power, they must hold for the
(A: -h l)th power.

But the laws hold for the fourth power; therefore, they
must hold for the fifth power.

Holding for the fifth power, they must hold for the sixth
power ; and so on for any positive integral power.

Therefore, they must hold for the nth. power if n is a posi-
tive integer ; and we have

(a + b)» = a» + na»-ib + ^^^-'^^ a»- »b»

,::^<^ 1-2

n(n — l)(n — 2) , «^« . ^at

+ ^ ^ g 3 a'-'b» + • • • [A]

289. This formula is known as the binomial theorem.

The expression on the right is known as the expansion of
(a -h hY\ this expansion is a finite series when ti is a positive
integer. That the series is finite may be seen as follows :

In writing the successive coefficients we shall finally arrive
at a coefficient which contains the factor n — n\ the corre-
sponding term will vgnish. The coefficients of the succeed-
ing terms likewise all contain the factor n — n, and all these
terms will vanish. ^

290. If a and b are interchanged, the identity [A] is written

(a -h hy = (6 -h a)" = h^ + nb^-^a ^ ^(^ - 1) ^n-2^2

^.(.-l)(.-2)^_,^,_^

This last expansion is the expansion of [A] written in
reverse order. Comparing the two expansions, we see that:
the coefficient of the last term is the same as the coefficient of
the first term ; the coefficient of the last term but one is the
same as the coefficient of the first term but one \ and so on.

222 COLLEGE ALGEBRA

In general, the coefficient of the rth term from the end is
the same as the coefficient of the rth term from the beginning.

In writing an expansion by the binomial theorem, after arriv-
ing at the middle term, we can shorten the work by observing
that the remaining coefficients are those already found, written
in reverse order.

291. If ^ is negative, the terms which involve even powers
of b are positive; and the terms which involve odd powers
of b are negative. Hence,

2

(a - b)» = a» - na—^b + ^ '_^' a»-«b»

_n(n-l)(n-2)^...^^ , [B]

If we put 1 for a and x for b in [A],

,^ V ^ ^i (^ — 1) o

(l+x)« = 1 -{- nx -{- -^Y-2

^Mn-mn-^^^_„ ^„j

If we put 1 for a and x for b in [B],
(1 - a;)" s 1 - nx + ^K» - 1) ^.2'

7i(n-l)(n-2) , ^^^

- 1-2.3 "+- t^J

292. Examples. (1) Expand (1 + 2 x^.
In [C] put 2 X for X and 5 for n. The result is

(1 + 2x)' = H- 5(2x) + ^(235)2 + ^^{2»)»

5.4.3.2 5-4-8-2-1

= 1 + lOx + 40x2 + 80x8 + 80 X* + 32 ac^.

BINOMIAL THEOREM 223

fl _ 2x^\

\x 3 ;

(2) Expand to three terms [ -

1 2x2

Put a for -» and b for

X 3

Then, by [B], (a - b)^ = a^ -Qa^b + 16a*62 _

Replacing a and b by their values,

~X6 X8 3 /**

293. Any Required Term. From [A] it is evident (§ 286)
that the (r -f l)th term in the expansion of (a -h by is

n(n — l)(n — 2)'" to r factors
1 •2-3 • • • r

The (r -h 1) th term in the expansion of (a — by is the same
as the above if r is even, and the negative of the above if r
is odd.

Find the eighth term of ( 4 — — ) •

Here, a = 4, 6 = — » n = 10, r = 7.

10.9.8.7-6-6-4 /x2\7
The eighth term is - t^-^-^-J-^^ WM ^ ) ' or - 60 x".

294. The Greatest Coefficient. Suppose that the coefficient of
the (r -f l)th term is the numerically greatest coefficient.

This coefficieiit and the preceding and following coefficients
are as follows :

rth term, ^ o *^

123

(r + l)thterm, -^^ ^

(r + 2)th term, ^(^""^>^

(n-r-^2)

.(r-1) '

(n — r + 2) (yi — r 4- 1) .

2.3-..(r-l)r '

(n — r + 2)(n — r -{- l)(n — r)

2.3..-(r-l)r(r-hl)

224 COLLEGE ALGEBRA

The coefficient of the rth term may be obtained by multi-
plying the coefficient of the (r + l)th by — r 5 the coeffi-
cient of the (r -h 2)th, by multiplying the coefficient of the
(r-hl)th by ^^=^. If the coefficient of the (r + l)th is
numerically the greatest,

< 1, and T < 1.

n — r -\-l r-f-1

Therefore, r < ti •— r -f 1, and r -h 1 > n — r.

Therefore, r< — - — > and r> — - — •

w

If n is even, r = -y and r + 1 = — - — > in this case there

is one middle term, and its coefficient is the greatest coefficient.

If n is odd, we can have only r — — - — j or r = — - — ; in

in this case there are two middle terms ; their coefficients are
alike and are the two greatest coefficients.

295. A trinomial may be expanded by the binomial theorem
as follows :

Expand {\-^2x-xy.

.-. (1 + 2a;-x2)8 = l +3(2x-x2) + 3(2x-aj2)« + (2«-a!^*

Ezercise 43
Expand :

1. (1+3 x)\ 4. (2 -f xy, 7. (3 a; - 2 y)«.

10. (1 + 4 a: + 3x2)*. ^^ (a« - aa; - 2 a^*.

BINOMIAL THEOREM 225

Find:

12. The fourth term

of (. + ^)'

13. The eighth term of r 2- — j •

14. The twelfth term of

/I _ v^Y*

15. The twentieth term oi I x —

(x - -^Y-

/ 8/— 1 \"

16. The fourteenth term of f Vx^ — — ;= ) •

V 2V^y

17. The (r + l)th term

"(^-€)'

18. The (r + l)th term of (\^ - -o^)''*

19. The (r + 3)th term of f-^ ^ ) •

\2y -y/SxJ

(3 £«\"

4 V 2" J '

(a /3ic\"

V2^ ^4a/

22. The rth term from the end of I -7-^ — \2" / '

23. In the expansion of (a + ^)'' show that the sum of the
coefficients is 2*.

24. In the expansion of (a — by show that the sum of the
positive coefficients equals the sum of the negative coefficients.

25. Expand

CHAPTER XX

LOGARITHMS

296. Definitions. Let any positive number except 1 be
selected as a base. Then, the index or exponent, which the
base must have to produce a given number is called the
logarithm of that number to the given base.

Any positive number except 1 may be selected as the base ;
and to each base corresponds a system of logarithms.

Thus, since 2^ = 8, the logarithm of 8 in the system of which 2 is the
base is 3.

That is, the logarithm of 8 to the base 2 is 3 ; this is abbreviated
log2 8 = 3.

In general, if a» = N^ then n = loga^.

Observe that a^ = N and n = logo^ are two different ways of express-
ing the same relation between n and N. The identity, ai<«a^=-y, is
sometimes useful.

The subscript which shows the base is usually omitted when there is no
uncertainty as to what number is being used as the base.

In this chapter only the positive scalar values of the root
will be considered ;* consequently, in a system with a positive
base, negative numbers cannot have scalar logarithms.

297. The logarithms of such numbers as are perfect powers
of the base selected are commensurable numbers; the logSr
rithms of all other numbers are incommensurable numbers.

Remark. By an incommensurable number is meant a number that
has no common measure with unity (§ 251).

Incommensurable logaiithms are expressed approximately to
any desired degree of accuracy by means of decimal fractions.

226

LOGARITHMS 227

298. A logarithm in general consists of two parts, an
integral part and a fractional part ; the integral part is called
the characteristic, and the fractional part the mantissa.

The calculation of logarithms to a given base will be con-
sidered in Chapter XXV.

299. Incommensurable Exponents. It will now be necessary
to prove that the laws which in Chapter IX were found to
apply to commensurable exponents apply also to incommen-
surable exponents.

Let a be any positive number except 1, and let m and n be
two positive incommensurable numbers.

To prove a"*a** = a"*•*■^

We can always find (§ 251) four positive integers, p, q, r,

s, such that m lies between — and , and n between -

and

5

p p + l r

Then, a™ lies between a^ and a « , and a" lies between a'

r-H

and a ' .

Therefore, a"*a** lies between a' a* and a "^ a ' .

But a'^a* = a« %

and a ^ a ' = a^ ' "f '.

Hence, a"*a** lies between a^ ' and a^ * '' *, and conse-
quently differs from a' ' by less than (a^ ' ^ ' — a^ ');
that is, by less than a' '(a*' ' — l)-

Also, since m lies between — and ^ > and n between -

q q 8

and ? a™"^** lies between a*^ ' and a« ' ' ', and con-

* £+: £+: i+1

sequently differs from a« ' by less than a* ' (a' ' — 1).

228 COLLEGE ALGEBRA

Therefore, the expressions a^al^ and a"'+* have the same
approximate value a« *, and each differs from this value by

p r 11

less than a^'^' (a^^' — 1).

Now let q and 5 be continually increased, p and r being

always so taken that m lies between ~ and ^- > and n

<1 9.

r r -h 1 1 1

between - and • Then, - and - continually decrease;

S S OS

a' ' approximates to a*^ or 1 ; and a*' ' (a« ' — 1) continually
decreases. „ .

Therefore, the difference between a"*a" and a« ' continually

decreases; the difference between a"*'^**and a' ' continually

decreases ; and each difference becomes as small as we please.

But, however great q and s may be, the expressions «"•«*

E+-
and a"*+** have the same approximate value, a' '.

Therefore, as in § 253, we must have

The foregoing proof is easily extended to the case in which
m and n are one or both negative.

Having proved for incommensui-able exponents that

it is easily proved that

a"* . . « — "•

— . = «»«-»» J (a"*)" = a"^ ; V a™ = a" ; a"*^"* = (aby.

a

300. Properties of Logarithms. Let a be the base, M and

N any positive numbers, m and n their logarithms to the

base a: so that „ ,^ « ,.x

' a™ = M, a" = N,

^ = logoiV/, n = log«JV.

Then, in any system of logarithms :
1. The logarithm of 1 is 0.
For, a«=l. .•.O = log„l.

LOGARITHMS 229

2. The logarithm of the base itself is 1.
For, a^ = a. .*. 1 = log^a.

3. The logarithm of the reciprocal of a positive number is
the negative of the logarithm of the number.

For, if a** = N, then ~: = — = a""*.

N a""

4. The logarithm, of the product of two or more positive
numbers is the sum, of the logarithm's of the several factors.

For, Mx N —a'^xa'' = a"*+^

.-. log«(M X iV) = m + 7i = log„ilf + logaiNT.

Similarly for the product of three or more factors.

5. The logarithm of the qitotient of two positive numbers is

the remainder found by subtracting the logarithm of the divisor

from the logarithm of the dividend.

^ Af a*"

For, — : = — = a"»-».

M
••• log„ ~= m — n=^ log^M — log^N.

6. The logarithm of a power of a positive number is the
product of the logarithm of the number by the exponent of the
power.

For, NP = (a^^y = a'^P.

.'. log^NP = np =p\og^N.

7. The logarithm of the real positive value of a root of a
positive number is the quotient found by dividing the logarithm
of the number by the index of the root.

For, -Vn = ■Va' = or.

r r

230 COLLEGE ALGEBRA

301. In a system with a base greater than 1 the logarithms
of all positive numbers greater than 1 are positive, and the
logarithms of all positive numbers less than 1 are negative.

Conversely, in a system with a positive base less than 1 the
logarithms of all positive numbers greater than 1 are negative,
and the logarithms of all positive numbers less than 1 are
positive.

302. Two Important Systems. Although the possible number
of different systems of logarithms is unlimited, there are but
two systems in common use. These are :

1. The common system, also called the Briggs, denary, or
decimal system, of which the base is 10.

2. The natural system, of which the base is the natural
base.

The natural base, generally represented by e, is the fixed
value which the sum of the series

^1^12^1. 2. 3^1- 2. 3. 4^

approaches as the number of terms is indefinitely increased.
The value of e, carried to seven places of decimals, is

2.7182818 . . .

The common system is the system used in actual calcula-
tion ; the natural system is used in the higher mathematics.

303. Common Logarithms. By logarithms in §§ 303-317 we

mean the common logarithms.

Since 10^= 1, 10-^ (=^^) =0.1,

10^= 10, 10-^(=t4^) =0.01,

10^ = 100, 10-»(=TTfc^)= 0.001,

therefore log 1 = 0, log 0.1 =—1,

log 10 = 1, log 0.01 =-2,

log 100 = 2, log 0.001 = - 3.

LOGARITHMS 231

Also, it is evident that the common logarithm of any num-
ber between

1 and 10 will be 0 -f a fraction,

10 and 100 will be 1 -f a fraction,

100 and 1000 will be 2 + a fraction,

1 and 0.1 will be — 1 -f a fraction,

0.1 and 0.01 will be — 2 -f a fraction,

0.01 and 0.001 will be - 3 -f a fraction.

304. With common logarithms the mantissa is always made
positive. Hence, in the case of numbers less than 1 whose
logarithms are negative, the logarithm is made to consist of a
negative characteristic and sl positive mantissa.

When a logarithm consists of a negative characteristic and
a positive mantissa it is usual to write the minus sign over the
characteristic, or to add 10 to the characteristic and to indicate
the subtraction of 10 from the resulting logarithm.

Thus, log 0.2 = 1.3010, and this may be written 9.3010 - 10.

305. The characteristic of the logarithm of an integral
number, or of a mixed number, is one less than the number
of integral digits in the number.

Thus, from § 303, log 1 = 0, log 10 = 1, log 100 = 2. Hence, the com-
mon logarithms of all numbers from 1 to 10 (that is, of all numbers
consisting of one integral digit) have 0 for characteristic ; and the common
logaritiims of all numbers from 10 to 100 (that is, of all numbers consist-
ing of two integral digits) have 1 for characteristic ; and so on, the char-
acteristic increasing by one for each increase of one in the number of
digits, and hence being always one leas than the number of integral digits,

306. The characteristic of the common logarithm of a deci-
mal fraction is negative, and is equal to the number of the
place occupied by the first significant figure of the decimal.

Thus, from § 303, log 0.1 =-1, log 0.01 =-2, log 0.001 =-3.
Hence, the common logarithms of all numbers from 0.1 to 1 have — 1

232 COLLEGE ALGEBRA

for characteristic (the mantissa being positive), the common logarithms of
all numbers from 0.01 to 0.1 have — 2 for characteristic, the common log-
arithms of all numbers from 0.001 to 0.01 have — 3 for characteristic, and
so on ; the characteristic always being negative and equal to the number of
the place occupied by the first significant figure of the decimal.

307. The mantissa of the common logarithm of any inte-
gral number, or decimal fraction, depends only upon the
sequence of the digits of the number, and is unchanged so
long as the sequence of the dibits remains the same.

For, changing the position of the decimal point in a number is equiva-
lent to multiplying or dividing the number by a power of 10. Its common
logarithm, therefore, is increased or diminished by the exponent of that
power of 10 ; and, since this exponent is integral, the mantissa, or deci-
mal part of the logarithm, is unaffected.

Thus, 27,196 = 104-4345^ 2.7196 = 100-««,

2719.6 = lOs-tsis, 0.27196 = 109««-io,

27.196 = lOi-is^, 0.0027196 = 107-««-io.

One advantage of using the number ten as the base of a
system of logarithms consists in the fact that the mantissa
depends only on the sequence of digits, and the characteristic
depends only on the position of the decimal point,

308. In simplifying the logarithm of a root the equal posi-
tive and negative numbers to be added to the logarithm should
be such that the resulting negative number, when divided by
the index of the root, gives a quotient of — 10.

Thus, log 0.002* = ^ of (7.3010 - 10).

The expression \ of (7.3010 - 10)

may be put in the form \ of (27.3010 - 30), which is 9.1003 — 10, isdnce
the addition of 20 to the 7, and of — 20 to the — 10, produces no ohange

in the value of the logarithm.

Agam^ log 0.0002* = ^ of (6.3010 - 10)

= i of (46.3010 - 60)
= 9.2602 - 10.

LOGARITHMS

233

Exercise 44

Given: log 2 = 0.3010; log 3 = 0.4771 ; log 6 = 0.6990;
log 7 = 0.8451.

Find the common logarithms of the following numbers by
resolving the numbers into factors and taking the sum of the
logarithms of the factors :

1. 6. 5. 25. 9. 0.021. 13. 2.1.

2. 15.

3. 21.

4. 14.

6. 30.

7. 42.

8. 420.

10. 0.35.

11. 0.0035.

12. 0.004.

Find the common logarithm of :

IT. 2«

18. 5»

Id. 7*

20. 5*

21. 2*

22. 5*

23. 5*

24. 7^.

25. 2^

26. 7*

27. 5'

28. 3^

29. 5*.

30. 2"^.

31. 5*.

32. f.

33. f.

34. f.

35. f

36. I

37. f

38. J.

39. J.

40.

41.

42.

43.

44.

45.

0.05

^^— ^^^ •

3

0.005
2

0.07

■ ■ ■ •

5

5

0.07'

0.05
0.003

0.007
0.02

14. 16.

15. 0.056.

16. 0.63.

46.

47.

48.

49.

50.

51.

0.02
0.007 *

0.005
0.07 *

0.02^
3»

3»
0.022*

7»
0.022 '

0.07«
0.003"

309. The remainder obtained by subtracting the logarithm
of a number from 10 is called the cologarithm of the number,
or arithmetical complement of the logarithm of the number.

The cologarithm is abbreviated colog, and is most easily
found by beginning with the characteristic of the logarithm

234 COLLEGE ALGEBRA

and subtracting each figure from 9 down to the last signifi-
cant figure, and subtracting that figure from 10.

Thus, log 7 = 0.8461 ; and colog 7 = 9.1649. We readily find colog 7
by subtracting, mentally, 0 from 9, 8 from 9, 4 from 9, 6 from 9, 1 from
10, and writing the resulting figure at each step.

310. If 10 is subtracted from the cologarithm of a number,
the result is the logarithm of the reciprocal of that number.

Thus, ' log ^ = log 1 - log JV

N

. =0-\ogN

= (10 - log N) - 10
= colog N — 10.

311. The addition of a (cologarithm — 10) is equivalent to
the subtraction of a logarithm.

Thus, colog JV - 10 = (10 - log JV) - 10 = - log N.

312. The logarithm of a quotient may be found by adding
the logarithm of the dividend and the cologarithm of the
divisor, and subtracting 10 from the result.

In finding a cologarithm when the characteristic of the logarithm is a
negative number, it must be observed that the suhtrajdion of a Tnjegatioe
number is equivalent to the addition of an equal positive number.

Thus, log = log 6 + colog 0.002 - 10

0.002 ^

= 0.6990 + 12.6990 - 10
= 3.3980.
Here, log 0.002 = 3.3010, and in subtracting — 3 from 9 the result is
the same as adding + 3 to 9.

2
Again, log — — = log 2 + colog 0.07 — 10

= 0.3010 + 11.1549-10
= 1.4569.

Also, log — = 8.8461 - 10 + 9.0970 - 10

= 17.9421 - 20

= 7.9421 - 10.
Here, log 2^ = 3 log 2 = 3 x 0.3010 = 0.9030.

Hence, colog 2^ = 10 - 0.9030 = 9.0970.

LOGARITHMS 235

313. Tables. A table of four-place common logarithms is
given at the end of this chapter, which contains the common
logarithms of all numbers under 1000, the decimal point and
characteristic being omitted. The logarithms of single digits
1, 8, etc., are found at 10, 80, etc.

Tables that co;ntain logarithms of more places can be pro-
cured, but this table will serve for many practical uses, and
will enable the student to use tables of five-place, seven-
place, and ten-place logarithms in work that requires greater
accuracy.

In working with a four-place table, the numbers correspond-
ing to the logarithms, that is, the antilogarithms, as they are
called, may be carried to four significant digits,

314. To find the Logarithm of a Number in this Table.

(1) Find the logarithm of 65.7.

In the column headed *'N" look for the first two significant figures,
and at the top of the table for the third significant figure. In the line
with 65, and in the column headed 7, is seen 8176. To this number
prefix the characteristic and insert the decimal point. Thus,

log 66. 7 =1.8176.

(2) Find the logarithm of 20,347.

r

In the line with 20, and in the column headed 3, is seen 3075 ; also in
the line with 20, and' in the 4 column, is seen 3096, and the difference
between these two is 21. The difference between 20,300 and 20,400 is 100,
and the difference between 20,300 and 20,347 is 47. Hence, ^ of 21 = 10,
nearly, must be added to 3075. Tliat is,

log 20,347 = 4.3085.

(3) Find the logarithm of 0.0005076.

In the line with 50, and in the 7 column, is seen 7050 ; in the 8 column,
7059 ; the difference is 9. The difference between 5070 and 6080 is 10,
and the difference between 5070 and 6076 is 6. Hence, ^^ of 9 = 5 must
be added to 7050. That is,

log 0.0005076 = 6.7055 - 10.

236 COLLEGE ALGEBRA

315. To find a Number when its Logarithm is given.

(1) Find the number of which the logarithm is 1.9736.

Look for 9736 in the table. In the column headed "N,'' and in the
line with 9736, is seen 94, and at the head of the column in which 9736
stands is seen 1. Therefore, write 941 and insert the decimal point as
the characteristic directs. That is, the number required is 9^.1.

(2) Find the number of which the logarithm is 3.7936.

Look for 7936 in the table. It cannot be found, but the two adjacent
mantissas between which it lies are 7931 and 7938 ; their difference is 7,
and the difference between 7931 and 7936 is 5. Therefore, f of the differ-
ence between the numbers corresponding to the mantissas, 7931 and 7938,
must be added to the number corresponding to the mantissa 793J.

The number corresponding to the mantissa 7938 is 6220.

The number corresponding to the mantissa 7981 is 6210.

The difference between these numbers is 10, and

6210 + ^ of 10 = 6217.
Therefore, the number required is 6217.

(3) Find the number of which the logarithm is 7.3882 — 10.

Look for 3882 in the table. It cannot be found, but the two adjacent
mantissas between which it lies are 3874 and 3892 ; their difference is 18,
and the difference between 3874 and 3882 is 8. Therefore, -^ of the dif-
ference between the numbers corresponding to the mantissas, 3874 and
3892, must be added to the number corresponding to the mantissa 8874.

The number corresponding to the mantissa 3892 is 2460.

The number corresponding to the mantissa 3874 is 2440.

The difference between these numbers is 10, and

2440 + j^j of 10 = 2444.
Therefore, the number required is 0.002444.

(4) Find the number of which the logarithm is 0.3664.

The two adjacent mantissas between which the given mantiflBa 8664
lies are 3655 and 3074 ; their difference is 19, and the difference between
3655 and 3664 is 0.

The number corresponding to the mantissa 3666 is 2320.

Therefore, the number required is 2.320 + A ©^ 1® = 2.826.

LOGARITHMS 237

Exercise 45

Find, from the table, the common logarithm of :

1. 60. 4. 3780. 7. 70,633. 10. 0.0004523.

2. 101. 5. 5432. 8. 12,028. 11. 0.01342.

3. 999. 6. 9081. 9. 0.00987. 12. 0.19873.

Find antilogarithms to the following common logarithms :

13. 4.2488. 16. 1.9730. 19. 9.0410-10.

14. 3.6330. 17. 0.1728. 20. 9.8420-10.

15. 4.7317. 18. 2.7635. 21. 7.7423-10.

Find the cologarithm of :

22. 428. 25. 4872. 28. 62,784. 31. 0.14964.

23. 567. 26. 9645. 29. 18,657. 32. 0.000762.

24. 841. 27. 0.478. 30. 0.00634. 33. 0.01783.

316. Computation by Logarithms.

(1) Find the product of 908.4 x 0.05392 x 2.117.

log 908.4 = 2.9583
log0.05392 = 8.7318 -10
log 2.117=0.3267

2.0168 = log 103.7.

Therefore, the requh-ed product is 103. 7.

When any factor is negative find its logarithm without regard to the
sign ; write n after the logarithm that corresponds to a negative number.
If the number of logarithms so marked is odd, the product is negative ;
if even, the product is positive,

(2) Find the product of 4.52 x (- 0.3721) x 0.912.

log 4.52 = 0.6561
log0.3721 = 9.6706- 10 n
log 0.912 = 9.9600 - 10

0.1857n = log -1.634.

Therefore, the required product is — 1.634.

238 COLLEGE ALGEBRA

(3) Find the cube of 0.0497.

log 0.0497 = 8.6964-10

3
6.0892 - 10 = log 0.0001228.

Therefore, the cube of 0.0497 is 0.0001228.

(4) Find the fourth root of 0.00862.

Iog0.00862 = 7.9365 -10

30. - 30

4)37.9356-40

9.4839- 10 = log 0.3047.

Therefore, the fourth root of 0.00862 is 0.3047.

(5) Find the value of ^3.1416 x 4771.2 x 2.718*

^30.13* X 0.4343* x 69.89*

log 3.1416 = 0.4971 = 0.4971

log 4771.2 = 3.0786 = 3.6786

J log 2.718 = J (0.4343) = 0.1448

4colog30.13 =4(8.6210-10)= 4.0840-10

I colog 0.4343 = I (0.3622) = 0. 1811

4 colog (39.89 = 4 (8. 1556 - 10) = 2.6224 - 10

11.2080 - 20

30. - 30

5)41.2080 - 60

8.2416 - 10 = logO.01744.

Therefore, the required value is 0.01744.

317. An exponential equation, that is, an equation in which
the exponent involves the unknown number, is easily solved
by logaritlims.

Find the value of x in 81^ = 10.

81^ = 10.
.-. log (81^) = log 10,
X log 81 = log 10,

x = '-^1^ = 1:520? = 0.524.
log 81 1.9085

LOGARITHMS

239

Exercise 46

Find by logarithms :

1. 948.76x0.043875.

2. 3.4097x0.0087634.

3. 830.75x0.0003769.

4. 8.4395x0.98274.
70654

9.

10.

5. 7564 X (- 0.003764).

6. 3.765 X (- 0.08345).

7. - 5.8404 X (- 0.00178).

8. -8945x73.85.
0.07654

54013
7.652

11.

83.947 X 0.8395

- 0.06875

13. 1.1768^

14. 1.3178^^

15. 11*.

16. a^y\

212x (-6.12) X (-2008)
365 X (- 531) X 2.576

17.
18.
19.
20.

906.80^.

(Mi)'-

25. VO.00476.

26. V- 325.

27.
28.
29.
30.
31.
32.

- 400)1
0.00065)1

- 0.0084)^.
O.00872)*.
0.8756)1

- 0.4762)^

36.
37.
38.
39.
40.

41.

21. 2.5637^.

22. (8j)2».

23. (5f|)«-8".

24. (9e)*.

- 4762)^.
4.861)1

- 0.00222)*.

- 0.03654)*.

- 0.00008) ^
-4)V

33. ^8462.

2«
42. (f)^.

•V^O.00052

43.

VO.0068125

34. VO.481.

35. (-286)^

44 ±(M?§^.
(- 257.14)*

240 COLLEGE ALGEBRA

4/0.008541^ X 8641 x 4.276* x 0.0084
^ 0.008548 y 182.63* x 82* x 487.27* *

W0.0075433^ X 78.343 x 8172.4* x 0.00052
^ 64285* X 154.27* x 0.001 x 586.79*

7l0.03271^ X 53.429 x 0.77542»
• ^ 32.769x0.000371*

jT 7.1206 X VO.13274 x 0.057389
VO.43468 X 17.385 x VO.0096372

46

47

48.

Find X from the equations :

49. 5^ = 12. 51. 7* = 25. 53. (0.4)-* =7.

50. 4^ = 40. 52. (1.3)^ = 7.2. 54. (0.9)** = (4.7)-*.

318. Change of System. Logarithms to any base a may be
converted into logarithms to any other base b as follows :

Let N be any number, and let

^ = ^^Ea^ 2Lnd m = log^iyT.

Then, N = a"" and N = b"".

.'. a" = ^>"».

Taking logarithms to any base r,

11 log^a = m log^Z>, (§ 300)

or, log^a X log^N = log^^> X logjiV,

from which log^iV may be found when log^a, log^ft, and log^iV
are given ; and conversely, log„iV may be found when log^a,
log^^, and logj,iV are given.

319. lia = 10,b = e,r = 10, and N = 10,

logiolO X logiolO = logioe X log,10. (§ 318)

.•.log,10 = , -•

LOGARITHMS 241

From tables, logio^ = 0.4342945.

.-. log, 10 = 2.3025851.

320. li a = 10, b = e, r = 10, and N is any number,

logiolO X logioN = logioe X log,N, (§ 318)

and logioiV = logio^ X log^N,

Hence, to convert common logarithms into natural loga-
rithms, multiply by 2.3025851 j and to convert natural loga-
rithms into common logarithms, multiply by 0.4342945.

Exercise 47

Find to four digits the natural logarithm of :

1. 2. 3. 100. 5. 7.89. 7. 2.001.

2. 3. 4. 32.5. 6. 1.23. 8. 0.0931.

Find to four digits :

9. log27. 11. log49. 13. log98. 15. log7l4.

10. log84. 12. log67. 14. log85. 16. logglO?.

17. Find the logarithm of 4 in the system of which J is
the base.

18. Find the logarithm of -^j in the system of which 0.5
is the base.

19. Find the base of the system in which the logarithm
of 8 is §.

20. Find the base of the system in which the logarithm
of 3 is - J.

COLLEGE ALGEBBA

N

0

1

2

3

4

5

6

7

a

9

lO

oooo

0043

0086

0128

0170

0212

0263

0294

0334

0374

11

0414

0463

0492

0531

0569

0607

0646

0682

0710

0755

12

0702

0828

0804

0899

0034

0009

1004

1038

1072

iioa

13

1139

1173

1206

1230

1271

1303

1367

ISiW

1430

14

1461

1492

1523

1553

1684

1614

1644

1673

1703

1732

15

1761

1790

1818

1847

1875

1903

1931

1959

1987

2014

16

2041

2068

2096

2122

2148

2176

2201

2227

2263

2279

17

2304

2330

2365

2380

2405

2430

2456

2480

2504

2629

18

2563

2577

2601

2625

2648

2672

2605

2718

2742

2765

19

2788

2810

2833

2866

2878

2900

2023

2045

2087

2989

SO

3010

3032

3054

3075

3096

3118

3130

3100

3181

3201

21

3222

3243

3203

3284

3304

3324

3345

3305

3386

3404

3124

3444

3404

3483

3502

3622

3641

3500

3579

3508

23

3ei7

3636

3655

3674

3711

3729

3747

3766

3784

24

3802

3830

3838

3856

3874

3892

3009

3927

3946

3062

25

3079

3907

4014

4031

4048

4066

4082

4009

4110

4133

26

4150

4160

4183

4200

4216

4232

4249

42«6

4281

4208

27

4314

4330

4346

4302

4378

4393

4409

4426

4440

4456

28

4472

4487

4602

4618

4633

4548

4664

4679

4594

4609

29

4024

4G39

4654

4060

4683

4698

4713

4728

4742

4767

30

4771

4786

4800

4814

4829

4843

4857

4871

4886

49D0

31

4914

4928

4043

4065,

40G9

4983

4907

5011

6024

5038

5051

5005

6070

6092

6105

6119

6132

5146

6169

5172

6185

5198

5211

5224

6250

5263

5278

5289

6302

34

5316

5328

5340

6306

5378

6391

6403

6416

6428

35

6441

5453

6466

5478

5400

5502

6614

5627

5639

6551

36

5563

55T6

5687

6609

6011

6623

6635

6847

6668

6670

37

5094

6705

6717

6729

6740

5752

6763

5776

6788

38

57il8

5800

6821

5832

5843

5855

6866

5877

5888

5899

39

5911

6922

6033

S044

6956

5960

6977

5988

6999

6010

40

6021

6031

6042

6063

0084

6076

6086

6096

0107

6117

41

6128

6138

6149

6100

6170

6180

6191

6201

0212

6222

42

0243

0253

6263

6274

6294

6304

6314

6325

43

6336

6:i4fi,

0356

6306

0376

0386

6396

6406

6416

642G

44

6435

9444
6642

0454

6404

8474

0484

6493

6503

6613

0622

45

6532

6561

6501

0571

6680

6590

6509

6600

6618

46

6628

6637

6646

6656

60fl5

6075

6684

6693

6702

6712

47

8721

6730

0739

0749

6758

6767

6776

6786

6794

6603

48

6812

0821

6830

6839

0848

6857

6806

8876

6884

68G3

49
50

6002

em

6011

0020

6928

8937

6946

6965

6964

6972

6981

6998

7007

7010

7024

7033

7042

7050

7059

7067

51

7076

7084

7093

7101

7110

7118

7126

7135

7143

7153

52

7100

7108

7177

7186

7103

7202

7210

7218

7226

7336

63

7243

7261

7269

7267

7276

7284

7292

7300

7308

7816

JJ

7324

7332

7340

7348

7356

7364

7372

7380

7388

7306

LOGARITHMS

243

N

55

56

57
58
59

60

61
62
63
64

65

66

67
68
69

70

71

72
73

74

75

76
77
78
79

80

81
82
83

84

85

86
87
88
89

OO

91
92
93
94

95

96
97
98
99

0

1

2

3

4

6

6

7

8

9

7404
7482
7659
7634
7709

7412
7490
7566
7642
7716

7419
7497
7574
7649
7723

7427
7605
7582
7057
7731

7436
7513
7589
7664
7738

7443
7520
7597
7672
7745

7451
7528
7604
7679
7752

7459
7536
7612
7686
7760

7466
7543
7619
7694
7767

7474
7551
7627
7701

7774

7782
7863
7924
7993
8062

7789
7860
7931
8000
8069

7796
7868
7938
8007
8076

7803
7876
7945
8014
8082

7810
7882
7962
8021
8089

7818
7889
7969
8028
8096

7825
7896
7966
8035
8102

7832
7903
7973
8041
8109

7839
7910
7980
8048
8116

7846
7917
7987
8066
8122

8129
8195
8261
8326
8388

8136
8202
8267
8331
8396

8142
8209
8274
8338
8401

8149
8215
8280
8344
8407

8166
8222
8287
8351
8414

8162
8228
8293
8367
8420

8169
8235
8299
8363
8426

8176
8241
8306
8370
8432

8182
8248
8312
8376
8439

8189
8254
8319
8382
8445

8461
8513
8573
8633
8692

8467
8519
8579
8639
8698

8463
8525

8585
8645
8704

8470
8531
8591
8051
8710

8476
8537
8597
8657
8716

8482
8643
8603
8663
8722

8488
8649
8609
8669
8727

8494
8666
8615
8676
8733

8600
8561
8621
8681
8739

8506
8567
8627
8686
8745

8761
8808
8866
8921
8976

8766
8814
8871
89*27
8982

8762

8820
8876
8932
8987

8768
8826
8882
8938
8993

8774
8831
8887
8943
8998

8779
8837
8893
8949
9004

8785
8842
8899
8964
9009

8791
8848
8904
8960
9016

8797
8864
8910
8965
9020

8802
8859
8915
8971
9025

9031
9086
9138
9191
9243

9036
9090
9143
9196
9248

9042
9096
9149
9201
9263

9047
QlOl
9164
9206
9268

9063
9106
9169
9212
9263

9058
9112
9165
9217
9269

9063
9117
9170
9222
9274

9069
9122
9175
9227
9279

9074
9128
9180
9232
9284

9079
9133
9186
9238
9289

9294
9346
9396
9446
9494

9299
9360
9400
9450
9499

9304
9366
9406
9466
9604

9309
9360
9410
9460
9509

9315
9365
9416
9465
9613

9320
9370
9420
9469
9518

9326
9375
9425
9474
9623

9330
9380
9430
9479
9628

9336
9385
9435
9484
9633.

9340
9390
9440
9489
9538

9642
9690
9638
9686
9731

9547
9695
9643
9689
9736

9562
9600
9647
9694
9741

9657
9605
9652
9699
9745

9662
9609
9657
9703
9750

9566
9614
9661
9708
9754

9571
9619
9666
9713
9759

9676
9624
9671
9717
9763

9581
9628
9675
9722
9768

9686
9633
9680
9727
9773

9777
9823
9868
9912
9956

9782
9827
9872
9917
9961

9786
9832
9877
9921
9965

9791
9836
9881
9926
9969

9795
9841
9886
9930
9974

9800
9846
9890
9934
9978

9805
9850
9894
9939
9983

9809
9854
9899
9943
9987

9814
9859
9903
9948
9991

9818
9863
9908
9952
9996

CHAPTER XXI

INTEREST AND ANNUITIES

321. Interest is money paid for the use of money.

322. Principal. The smn loaned is the principal,

323. Rate of Interest. The rate of interest is the interest
on $1 for one year.

324. Amount. The sum of the principal and interest is the
amount.

325. Compound Interest. Interest is compounded when it is
added to the principal and becomes a part of the principal at
specified intervals.

Compound interest is compounded annually, semiannually,
quarterly, or monthly according to agreement. Compound
interest is understood to be compounded annually unless
otherwise stated.

326. In interest problems four elements are considered :
principal, rate, time, and interest or amount. If three of
the elements are known, the fourth may be found.

327. Let r staiid for the interest on $1 for one year ; t for
the time in years between two successive conversions (com-
poundings) ; n the number of conversions ; A^ the original
amount, the principal ; A„ the amount after n conversions of
interest into principal; and /„ the total of the interest con-
verted in the n conversions. Then,

A^=^A^{l-\-rt),

A^ = A,{l-\-rt)=^A^(l + Hy,

A,=^A,{l^-rt)=A,{l + Hy, •
244

INTEREST AND ANNUITIES 246

A^ = A^(1 + rt) = ^„(1 + rty,

• •••••

• •••••

^n = ^,-1 (i+rt)= Ao (1 + rty,
and /„ = ^„ — ^0*

If R is written for (1 4- rt), these equations become

A. = AoR«, [1]

and I„ = Ao (R'^ - 1). [2]

Hence, also, log A^ = log A^-^n log(l + rt).
In the case of simple interest, n = 1.

328. If there should occur a broken P^^^^d whose time in
years is t', t' being less than t, the rate of increase for t' is
by commercial usage taken to be 1 4- rt'.

329. Sinking Funds.- If the sum set apart at the end of
each year to be put at compound interest is represented by S,

The sum at the end of the

first year = S,
second year = 5 4- SR,
third year = S -^ SR -^ SR^,

nth. year = S + SR -^ SR^ -\ h SR'^'K

That is, the amount A = S -}- SR -^ SR^ -\ h SR^'-K

,'. AR = SR, 4- SR'^ -}- SR^-\ h SR\

.-. AR - A = SR"" - S.

•*^" R-1 '

. S(R»-1)
or, A = — ^^ ^ .

r

(1) If $10^000 is set apart annually, and put at 6 per cent
compound interest for 10 years, what will be the amount ?

_S{R^-l) __ f 10,000 (1.0610 - 1)
~ r ~ 0.06

By four-place logarithmB the amount is $131,600.

246 COLLEGE ALGEBRA

(2) A county owes $60,000. What sum must be set apart
annually, as a sinking fund, to cancel the debt in 10 years,
provided money is worth 6 per. cent ?

Ar $60,000x0.06 ^,^,« ,^ ^ . . .
S = = ' ,^ = $4658 (by four-place logs.).

Note. The amount of tax required yearly is $3600 for the interest
and $4658 for the sinking fund ; that is, $8168.

330. Annuities. A sum of money that is payable yearly, or
in parts at fixed periods in the year, is called an annuity.

To find the amount of an unpaid annuity when the interest,
time, and rate per cent are given.

The sum due at the end of the

first year = S,
second year = .S 4- SR,
third year = S -\- SR -^ SR%
nth year = S -^ SR -it SR^ -\ h SB^-\

That is, A = ^(^''-^) . (§ 329)

An annuity of $1200 was unpaid for 6 years. What was
the amount due if interest is reckoned at 6 per cent ?

^ S{R^-\) $1200(1.066-1) ^^^^^ ,^ ^
A = -^ '- = ^-— '- = $8360 (by four-place logs.).

331. To find the present worth of an annuity when the time
it is to continue and the rate per cent are given.

Let P denote the present worth. Then, the amount of P for
n years is equal to A, the amount of the annuity for n years.
But the amount of P for n years

= 7^ (1 + ry = PR^, (§ 327)

SCR"" - 1)
and A = ^j^_^^' (§ 330)

INTEREST AND ANNUITIES 247

S R"^ - 1
P = — X

R** R — 1

This equation may be written

S

P =

R-

1^ R- R-l\ R-J

As n increases indefinitely, the expression 1 — — ap-
proximates to 1.

Therefore, if the annuity is perpetual,

S S

P =

R — 1 r

(1) Find the present worth of an annual pension of $105,
for 5 years, at 4 per cent interest.

„ S R^-1 3105 1.046 _ 1 «^ ,^ ,

P = — X = X = $466.20 (by logs.).

i2» R-1 1.046 1.04 - 1 "^ \ Jf 6 f

(2) Find the present worth of a perpetual scholarship that
pays $300 annually, at 6 per cent interest.

r 0.06

332. To find the present worth of an annuity that begins in
a given number of years, when the time it is to continue and the
rate per cent are given.

Let p denote the number of years before the annuity begins,
and q the number of years the annuity is to continue.

Then, the present worth of the annuity to the time it termir
nates i^ ijP+«_l

xV-r^' (§331)

RP-KI R—1

248 COLLEGE ALGEBRA

The present worth of the annuity to the time it begins is

^x-^-j-. (§331)

Hence p J ^-^^'^^ -'^\-( ^ ^^L:^\
' \i2'' + « R — 1 J \RP R — lJ

■"i2^ + ''L R-l J*

S R'l — 1
.•.P = X

RP + <1 R_l

If the annuity is to begin at the end of p years, and to be
perpetual, the formula

S R^-1

P = -:7-7- X

may be written P =

EP+9 R-1

S i?«-l

RP (R - 1) R^

R'^ — 1
Since — — — approaches 1 as g' increases indefinitely (§ 331),

S

P =

RP(R-1)

(1) Find the present worth of an annuity of $6000, to begin
in 6 yeais, and to continue 12 years, at 6 per cent interest.

S R^-l f!5000 1.0612-1 ^^^^^^ ,

P = X = X = $29,560 (by lofiB.).

Rp + 1 R-l 1.0618 0.06 ' V J^ "8»-;-

(2) Find the present worth of a perpetual annuity of $1000,
to begin in 3 years, at 4 per cent interest.

P = = ^^^^ = $22,226 (by logs.).

Rp(R-l) L048X0.04 * ' \ J iSP-f

333. To find the annuity when the present worthy the time,
and the rate per cent are (jiven.

INTEREST AND ANNUITIES 249

p^^i^lzLll. (§331)

22" — 1
.-. S = Pr X

What annuity for 5 years will $4675 yield when interest is
reckoned at 4 per cent ?

S = Prx —- — = 14676 x 0.04 x -^^^ — = $1053 (by logs.).

334. Life Insurance. In order that a certain sum may be
secured, to be payable at his death, a person pays yearly a
fixed premium,.

If P denotes the premium to be paid for n years to insure
an amount A, to be paid immediately after the last premium,
then Ty(T>n

. p A(R-l)^ At
R» — 1 R** — 1

If A is to be paid a year after the last premium, then

p A(R-l)^ Ar
R(R'*-1) R(R'*-1)

Note. In the calculation of life insurance it is necessary to employ
tables that show for every age the probable duration of life.

335. Bonds. If P denotes the price of a bond that has n
years to run, and bears r per cent interest, S the face of the
bond, and q the current rate of interest, what interest on his
investment will a purchaser of such a bond receive ?

Let X denote the rate of interest on the investment.
Then P(l+aj)" is the value of the purchase money at
the end of n years.

250 COLLEGE ALGEBRA

Sr(l + qy-^ + Sr(l + qY'^ H [- Sr -\- S is the amount

received on the bond if the interest received from the bond
is put immediately at compound interest at q per cent.

But Sr(l + qy-^ H- 5r(l + qy-^ H h 'S^^ is a geometrical

progression in which the first term is Sr, the ratio 1 + q, and
the number of terms n.

Therefore, Sr(l + qY'^ -\-Sr(l-\- qy-^ ^,.. + Sr + S

^s + ^rLQ^t^l^. (§276)
■•.P(i + .)- = 5 + ^-r(iY>'-lJ-

rSq + Sr(l+q)'-Sr-||

•••i+-=L — P^ — J-

(1) What interest will a purchaser receive on his investment
if he buys at 114 a 4 per cent bond that has 26 years to run,
money being worth 3^ per cent ?

/3.5-I-4 X 1.03628 -4\ A

1 + JC = I I •

V 114 X 0.036 /

By logarithms, 1 + x = 1.033.

That is, the purchaser will receive 3^ per cent for his money.

(2) At what price must 7 per cent bonds, running 12 years,
with the interest payable semiannually, be bought in order
that the purchaser may receive on his investment 6 per cent,
interest semiannual, which is the current rate of interest ?

In this case S = 100 ; and, as the interest is semiannual,

q = 0.025, r = 0.035, n = 24, x = 0.026.

„ „ 2.5 + 3.5(1.025)2* -3.6

Hence, P = ^ —- •

0.025(1.025)2*

By logarithms, P = 118.

INTEREST AND ANNUITIES 251

Exercise 48

1. In how many years will $100 amount to $1060 at 6 per
cent compound interest ?

2. In how many years will $^ amount to $B (1) at simple
interest, (2) at compound interest, r and R being used in their
usual sense ?

3. Find the difference (to five places of decimals) between
the amoimt of $1 in 2 years, at 6 per cent compound interest,
according as the interest is payable yearly or monthly.

4. At 6 per cent, find the amount of an annuity of $-4
which has been left unpaid for 4 years.

5. Find the present value of an annuity of $100 for 6
years, reckoning interest at 4 per cent.

6. A perpetual annuity of $1000 is to be purchased, to
begin at the end of 10 years. If interest is reckoned at 3^
per cent, what should be paid for the annuity ?

7. A debt of $1850 is discharged by two payments of
$1000 each, at the end of one and two years. Find the rate
of interest paid.

8. Reckoning interest at 4 per cent, what annual premium
should be paid for 30 years in order to secure $2000 to be
paid at the end of that time, the premium being due at the
beginning of each year ?

9. An annual premium of $150 is paid to a life-insurance
company for insuring $5000. If money is worth 4 per cent,
for how many years must the premium be paid in order that
the company may sustain no loss ?

10. What may be paid for bonds due in 10 years, and bear-
ing semiannual coupons of 4 per cent each, in order to realize
3 per cent semiannually, if money is worth 3 per cent semi-
annually ?

252 COLLEGE ALGEBRA

11. When money is worth 2 per cent semiannually, if bonds
having 12 years to run, and bearing semiannual coupons of 3J
per cent each, are bought at 114^, what per cent is realized
on the investment ?

12. If $126 is paid for bonds due in 12 years, and yield-
ing 3^ per cent semiannually, what per cent is realized on
the investment, provided money is worth 2 per cent semi-
annually ?

13. A person borrows $600.25. How much must he pay
annually that the whole debt may be discharged in 35 years,
allowing simple interest at 4 per cent ?

14. A perpetual annuity of $100 a year is sold for $2500.
At what rate is the interest reckoned ?

15. A perpetual annuity of $320, to begin 10 years hence,
is to be purchased. If interest is reckoned at 3^ per cent,
what should be paid for the annuity ?

16. A sum of $10,000 is loaned at 4 per cent. At the end
of the first year a payment of $400 is made, and at the end
of each following year a payment is made greater by 30 per
cent than the preceding payment. Find in how many years
the debt will be paid.

17. A man with a capital of $100,000 spends every year
$9000. If the current rate of interest is 5 per cent, in bow
many years will he be ruined ?

18. Find the amount of $365 at compound interest for 20
years, at 5 per cent.

19. A railroad company bought and paid for 860 freight
cars at $360 each. The company wishes to charge the cost
of the cars to operating expenses in six equal annual amounts,
the first charge to be made on the date of the purchase. If
money is worth 4%, what annual charge to operating expenses
should be made ?

CHAPTER XXII

CHOICE

336. Fundamental Principle. If one thing can he done in a
different ways, and, when it has been done, a second thing can
he done in b different ways, then the two things can he done
together in a X b different ways.

For, corresponding to the first way of doing the first
thing, there are h different ways pf doing the second thing;
corresponding to the second way of doing the first tiling,
there are h different ways of doing the second thing; and -so
on for each of the a different ways of doing the first thing.

Therefore, there are axh different ways of doing the two
things together.

(1) If a box contains four capital letters, A, B, C, D, and
three small letters, x, y, z, in how many different ways may
two letters, one a capital letter and one a small letter, be
selected ?

A capital letter may be selected in four different ways, since any
one of the letters -4, B, C, D may be selected. A small letter may be
selected in three different ways, since any one of the lettera jc, y, z may
be selected. Any small letter may be put with any capital letter.

Thus, with A we may put jr, or y, or z ;

with B we may put x, or y, or z ;
with C we may put x, or y, or z ;
with D we may put x, or y, or z.

Hence, the number of ways in which a selection may be made is 4 x 3,
or J 2, These ways are :

Az Bx Cx Dx ^

Ay By Cy By

Az Bz • Cz Dz

253

N.

264 COLLEGE ALGEBRA

(2) On a shelf are 7 English, 5 French, and 9 German
books. In how many different ways may two books, not in
the same language, be selected ?

An English book and a French book can be selected in 7 x 5, or 36,
ways. A French book and a German book in 5 x 9, or 46, ways. An
English book and a German book in 7 x 9, or 63, ways.

Hence, there is a choice of 36 + 46 + 63, or 143, ways.

(3) Out of the ten figures 0, 1, 2, 3, 4, 6, 6, 7, 8, 9 how
many different numbers of two figures each can be formed ?

Since 0 has no value in the left-hand place, the left-hand place can be
filled in 9 ways.

The right-hand place can be filled in 10 ways, since repetitiona of the
digits are allowed, as in 22, 33, etc.

Hence, the whole number of numbers is 9 x 10, or 90.

337. By successive application of the principle of § 336 it
may be shown that,

If one thing can he done in a different waySy and then a
second thing can he done in b different ways, then a third thing
in c different ways, then a fourth thing in d different vtays,
and so on, the number of different ways of doing all the things
together isaxbxcxdx--

For, the first and second things can be done together in
axh different ways (§ 336), and the third thing in c different
ways ; hence, by § 336, the first and second things and the
third thing can be done together in (a x ^) X <? different ways.
Therefore, the first three things can be done in axbx c dif-
ferent ways. And so on, for any number of things.

In how many different ways can four Christmas presents be
given to four boys, one to each boy ?

The first present may be given to any one of the boys ; hence, there
are 4 ways of disposing of it.

When the first present has been disposed of, the second present may
be given to any one of the other three boys ; hence, there are 3 ways of
disposing of it. .

CHOICE 255

When the first and second presents have been disposed of, the third
present may he given to either of the two other boys ; hence, there are 2
ways of disposing of it.

When the first, second, and third presents have been disposed of, the
fourth present must be given to the last boy ; hence, there is only 1 way
of disposing of it.

There are, then, 4 x 3 x 2 x 1, or 24, ways.

338. Combinations and Permutations. (1) In how many dif-
ferent ways can a vowel and a consonant be chosen, assuming
that the alphabet contains 6 vowels and 20 consonants ?

A vowel can be chosen in 6 ways and a consonant in 20 ways, and
both (§ 336) in 6 X 20, or 120, ways.

(2) In how many different ways can a two-lettered word be
made, containing one vowel and one consonant ?

The vowel can be chosen in 6 ways and the consonant in 20 ways ;
and then each combination of a vowel and a consonant can be written in
2 ways ; as, ac^ ca.

Hence, the whole number of ways is 6 x 20 x 2, or 240.

These two examples show the difference between a selection,
or combination, of different things and an arrangement, or
permutation, of the same things.

Thus, oc forms a selection of a vowel and a consonant, and ac and ca
form two different arrangements of this selection.

From (1) it is seen that 120 different selections can be made with
a vowel and a consonant; and from (2) it is seen that 240 different
arrangements can be made with these selections.

Again, a, 6, c is a selection of three letters from the alphabet. This
selection admits of 6 different arrangements, as follows :

abc bca cab

acb bac cba

A selection, or combination, of any number of things is a group
of that number of things put together without regard to theii
order.

An arrangement, or permutation, of any number of things is
a group of that number of things put together in a definite
order.

256 COLLEGE ALGEBRA

339. Permutations, Things all Different. The number of dif-
ferent arrangements or permutations of n different things
taken all together is

w(n-l)(7i-2)(7i-3)x---x3x2xl. ,

For, the first place can be filled in n ways, then the second
place in 71 — 1 ways, then the third place In ^ — 2 ways, and
so on, to the last place, which can be filled in only 1 way.

Hence (§ 337), the whole number of arrangements is the
continued product,

n(7i - 1) (71 - 2) (71 - 3) X • • • X 3 X 2 X 1.

For the sake of brevity this product is often written [n
or n\ (read factorial n).

Observe that 1 x 2 x • • • X (ti — 1) x 7i = [^.

How many different arrangements of nine letters each can
be formed with the letters in the word Cambridge ?

There are nine letters. In making any arrangement any one of the
letters can be put in the first place. Hence, the first place can be filled
in 9 ways.

Then, the second place can be filled with any one of the remaining eight
letters ; that is, in 8 ways.

In like manner, the third place can be filled in 7 ways, the fourth place
in G ways, and so on ; and, lastly, the ninth place in 1 way.

If the nine places are indicated by Roman numerals, the result is (§ 337)
as follows :

I II III IV V VI VII VIII IX

9x8x7x6x6x4x3x 2x1 = 362,880 ways.

Hence, there are 362,880 different arrangements possible.

340. The number of different j9crm?/fa^to7W of n different

things taken t at a time is

n(n — l)(n — 2)"- to r factors, -
that is, n(n — 1) {n — 2) " • \n — (r — 1)],

or n(n — V)(n — 2)" (n — r -\- 1).

CHOICE 257

For, the first place can be filled in n ways, then the second
place in ri — 1 ways, then the third place in n — 2 ways, and
so on^ and then the rth place in n — (r — 1) ways.

Let P„^ ^ represent the number of arrangements of n differ-
ent things taken r at a time. Then,

p^ ^ = n(n — 1) (71 — 2) '" to r factors
— n(ri — V)(n — 2)"'(n — r-^V),

How many different arrangements of four letters each can
be formed from the letters in the word Cambridge ?

m

There are nine letters and four places to be filled.

The first place can be filled in 9 ways. Then, the second place can be
filled in 8 ways ; then, the third place in 7 ways ; and then, the fourth
place in 6 ways.

If the places are indicated by I, II, III, IV, the result is (§ 337)

I II III IV

9x8x7x6 = 3024 ways.

Hence, there are 3024 different arrangements possible.

341. Combinations, Things all Different The member of dif-
ferent selections or combinations of n different things taken r at

a time is

n(n — l)(n — 2) " ' (n — r -\- 1)

To prove this, let C„ ,. represent the number of different
selections or combinations of n different things taken r at a
time.

Take one selection of r things ; from this selection [r
arrangements can* be made (§ 339).

Take a second selection ; from this selection [r arrangements
can be made. And so on, for each of the C„ ,. selections.

Hence, C^rX\r is the number of arrangements of n differ-
ent things taken r at a time.

That is, C„.,x[r = P,,,.

258 COLLEGE ALGEBRA

P

. ti(7i - l)(n - 2) • • ' (n - r ^1)

In how many different ways can three vowels be selected
from the five vowels a, e, i, o, u?

The number of different ways in which we can arrange 3 vowels out
of 5 is (§ 340) 5 X 4 X 3, or 60.

These 60 arrangements might be obtained* by first forming all the
possible selections of 3 vowels out of 5, and then arranging the 3 vowels
in each selection in as many ways as possible.

Since each selection can be arranged in |3, or 6, ways (§ 339), the
number of selections is Y-, or 10.

The formula applied to this problem gives

5x4x3

342. Combinations, Second Formula. Multiplying both numer-
ator and denominator of the expression for the number of
combinations in the last example by 2x1, we have

^5x4x3x2x1^ L?
'•' 1x2x3x2x1 |3|2*

In general, multiplying both numerator and denominator of
the expression for C„ ,. in § 341 by \n — r, we have

_ n{n — 1) ' " {n — r -\- 1) {n — r) • "1 _ [g

^n,r

n — r

\rx{n — r)-"l \

This second form is more compact than the first and is
more easily remembered.

Note. In the reduction of such a result \n — r cancels all the factors

112
of the numerator from 1 up to and including n — r. Thus, in r^rb* 1 7

l2ll
cancels all the factors of [12 from 1 up to and including 7 ; so that

li? ^ 12 X 11 X 10 X 9 X 8 _ .y^^o
[5^7 1x2x3x4x5

CHOICE 259

343. Theorem. The number of combinations of n things
taken r at a time is the same as the number of combinations
of n things taken n — t at a time.

\n \n

For, C^,_, = , r^=--. r = I *- ■ = C

n — r\n — (n — r)

n — r

n, r*

This is also evident from the fact that for every selection
of r things taken, a selection of n — r things is left.

Thus, out of 8 things, 3 things can be selected in the same number of
ways as 5 things ; namely,

[8 _8 X 7 X 6
[3|5 1x2x3

Out of 10 things, 7 things can be selected in the same number of ways
as 3 things ; namely,

[15 10 X 9 X 8

56 ways.

7 3 1x2x3

= 120.

344. Examples in Combinations and Permutations. Of the

permutations possible with the letters pi the word Cambridge,
taken all together :

(1) How many begin with a vowel ?

In filling the nine places of any arrangement the first place can be
filled in only 3 ways, the other places in [8 ways.

Hence, the answer is 3 x [8 = 120,960. (§ 337)

(2) How many both begin and end with a vowel ?

The first place can be filled in 3 ways, the last place in 2 ways (one
vowel having been used), and the remaining seven places in [7 ways.
Hence, the answer is 3 x 2 x [7 = 30,240. (§ 337)

(3) How many begin with Cam ?

The answer is evidently |6, since our only choice lies in arranging the
remaining six letters of the word.

(4) How many have the letters cam standing together ?

This may be resolved into arranging the group cam and t^ last
six letters, regarded as seven distinct elements, and then arranging the
letters cam.

260 COLLEGE ALGEBRA

The first can be done in [7 ways, and the second in [3 ways. Hence,
both can be done in [7 x [3 = 30,240 ways.

In how many ways can the letters of the word Cambridge
be written :

(5) Without changing the place of any vowel ?

The second, sixth, and ninth places can be filled each in only 1 way ;
the other places in [6 ways.

Therefore, the whole number of ways is [6 = 720.

(6) Without changing the order of the 3 vowels ?

The vowels in the different arrangements are to be kept in the order

a, z, e.

One of the 6 consonants can be placed in 4 ways : before a, between
a and i, between i and e, and after e. ,

Then, a second consonant can be placed in 5 ways, a third consonant
in 6 ways, a fourth consonant in 7 ways, a fifth consonant in 8 ways, and
the last consonant in 9 ways. Hence, the whole number of ways is

4x5x6x7x8x9, or 60,480.

(7) Out of 20 consonants, in how many ways can 18 be
selected ?

The number of ways in which the 18 can be selected is

|20 20 X 19

18

2

(8) In how many ways can the same choice be made so as
always to include the letter b ?

Takuig b first, we must then select 17 out of the remaining 19 conso-
nants. This can be done in

[19 19x18 ,^, ,. „,

^ = -^- = 171 ^ays. (5 342)

(9) In how many ways can the same choice be made so as
to include b and not include c ?

Taking b first, we have then to choose 17 out of 18, e being excluded.
This can be done in 18 ways.

"' \ \ c

CHOICE 261

(10) From 20 Republicans and 6 Democrats, in how many
ways can 5 different offices be filled, of which 3 particular
offices must be filled by Republicans, and the other 2 offices
by Democrats ?

The first 3 offices can be assigned to 3 Republicans in

20 X 19 X 18 = 6840 ways.
The other 2 offices can be assigned to 2 Democrats in

6 X 5 = 30 ways.
There is, then, a choice of 6840 x 30 = 205,200 different ways.

(11) Out of 20 consonants and 6 vowels, in how many ways
can we make a word consisting of 3 different consonants and
2 different vowels ?

Three consonants can be selected in = 1140 ways, and

6x6 1x2x3

two vowels in = 15 ways. Hence, the 6 letters can be selected in

1x2

1140 X 15 = 17,100 ways.

When 5 letters have been so selected they can be arranged in
[5 = 120 different orders. Hence, there are 17,100 x 120 = 2,062,000
different ways of making the word.

Observe that the letters are first selected and then arranged.

(12) A society consists of 50 members, 10 of whom are
physicians. In how many ways can a committee of 6 mem-
bers be selected so as to include at least 1 physician?

Six members can be selected from the whole society in

160

Six members can be selected from the whole society, so as to include
no physician, by choosing them all from the 40 members who are not
physicians, and this can be done in

140
. I ways.

[6 [34 ^ .

|60 140 . . , . .

Hence, r^=i= ■ , is the number of ways of selectmg the committee

SO as to include at least 1 physician.

262 COLLEGE ALGEBRA

345. Greatest Number of Combinations. To find for what

value of r the number of selections of n things, taken r at a

time, is the greatest.

The formula

_n{n — V)(n — T)" '(n — r -\-V)

*»•'■"' Ix2x3x-r

may be written

_n n — 1 _ n — 2 _ n — r -\-l

n,r

1 2

The numerators of the factors on the right side of this
equation begin with n, and form a descending series with
the common difference 1 ; and the denominators begin with 1,
and form an ascending series with the common difference 1.
Therefore, from some point in the series, these factors become
less than 1. Hence, the maximum product is reached when
that product includes all the factors greater than 1.

1. When n is an odd number the numerator and the denomi-
nator of each factor are alternately both odd and both even,
so that the factor greater than 1, but nearest to 1, is the factor
whose numerator exceeds the denominator by 2. Hence, in
this case, r must have such a value that

n — 1

n — r-\-l = r-\-2, or r = — - — •

It

2. When n is an even number the numerator of the first
factor is even and the denominator odd ; the numerator of the
second factor is odd and the denominator even ; and so on,
alternately, so that the factor greater than 1, but nearest to 1,
is the factor whose numerator exceeds the denominator by 1.
Hence, in this case, r must have such a value that

n — r-hl = ^-|-l, or r = -'

LI

(1) What value of r will give the greatest number of com-
binations out of 7 things ?

CHOICE

Here

n ifi

} odd,

and

r

n-1 7-
2 2

1

3.

• •

Gi,z

7x6x6
1x2x3

36.

If r =

= 4,

then

C^7,4

7 X 6 X 5 X
1 X 2 X 3 X

4_
4"

36.

263

When the number of things is odd there are two equal numbers of
combinations, namely, when the number of things taken together is just
under and just over one-half of the whole number of things.

(2) What value of r will give the greatest number of selec-
tions out of 8 things ?

Here n is even, and r = - = - = 4.
' 2 2

^ 8x7x6x6 «^

.-. Os. 4 = = 70,

'' 1x2x3x4

so that, when the number of things is even, the number of selections
will be greatest when one-JicUf of the whole are taken together.

346. Division into Two Groups. The number of different
ways in which p -\- q things, all different, can be divided into
two groups of p things and q things respectively is the same
as the number of ways in which p things can be selected from

p-^q thmgs, or -t — |

For, to each selection of p things taken corresponds a selec-
tion of q things left, and each selection therefore effects the
division into the required groups.

(1) In how many ways can 18 men be divided into 2 groups

of 6 and 12 each?

118
r^=^=- = 18,664 ways.
[6^[l2 ' ^

(2) A boat's crew consists of 6 men, of whom 2 can row
only on the stroke side of the boat, and 1 can row only on the
bow side. In how many ways can the crew be arranged ?

264 COLLEGE ALGEBRA

There are left 3 men who can row on either side ; 1 of these must row
on the stroke side, and 2 on the bow side.

The number of ways in which these 3 can be selected is

13
1-^— = 3 ways.

When the stroke side is completed the 3 men can be arranged in
[3 ways ; likewise, the 3 men of the bow side can be arranged iu |3
ways. Hence, the arrangement can be made in

3 X |3 X |3 = 108 ways.

347. Division into Three or More Groups. The number ol
different ways in which 7^ -\- q -{- r things, all different, can be
divided into three groups of p things, q things, and r things

,.- , . \P_±_1±1
res])ectively is ", . 1 — *

[Mr

For, J) -{- q -\- r things may be divided into two groups of p

ll^ -^ q -h r
things and q -^ r things in "1" \~ ways ; then, the group

p q -\-r

of q -{- r things may be divided into two groups of q things

\q -i-r
and r things in -, — 1 — ways. Hence, the division into three

groups may be effected in

\p -\- q -^r \q -\-r \ p -h q -\- r

1 — I — ; X ~T~] — or — T — r-i ways ;

P q + r \q\r [p\q\r

and so on, for any number of groups.

In how many ways can a company of 100 soldiers be
divided into 3 squads of 50, 30, and 20 respectively ?

1 100
The answer is r^jnoQ^ ways.

348. When the number of things is the same in two or more
groups, and there is 710 distinction to be made between these
groupsy the number of ways given by the preceding section is
too large.

CHOICE 265

(1) Divide the letters a, h, c, d into two groups of 2 letters
each.

The number of ways given by § 346 is r-=f- = 6 ; these ways are :

I. ab cd. III. ad be. V. bd ac.

II. ac bd. IV. be ad. VI. cd ab.

Since there is no distinction between the groups, IV is the same ae
III, V the same as II, and VI the same as I.

1 ' |4
Hence, the correct answer is - x i-=i^ , or 3.

2 [2[2

If, however, a distinction is to be made between the two groups in any
one division, the answer is 6.

In the case of three similar groups the result given by § 347
is to be divided by [3, the number of ways in which three
groups can be arranged among themselves ; in the case of four
groups, by [4 ; and so on, for any number of groups.

(2) In how many ways can 18 men be divided into two

groups of 9 each?

118
According to § 346, the answer would be p==- •

The two groups, considered as groups, have no distinction ; therefore,

permuting them gives no new arrangement, and the true result is obtained

118
by dividing the preceding by [2, and is . \ — . •

If any condition is added that will make the two groups different, — if,

for example, one group wear red badges and the other blue, — then the

118
answer will be r==- •

(3) In how many ways can a pack of 52 cards be divided
equally among 4 players. A, B, C, D ?

Here the assignment of a particular group to a different player makes

the division different, and there is, therefore, a distinction between the

\S2

groups ; the answer is

13|13 13 13

266 COLLEGE ALGEBRA

(4) In how many ways can 52 cards be divided into 4 piles
of 13 each ?

Here there is no distinction between the groups, and the answer is

52

[4 [13 [13 [13 [13 *

Ezerolse 49

1. How many numbers of 5 figures each can be formed
with the digits 1, 2, 3, 4, 5, no digit being repeated?

2. How many even numbers of 4 figures each can be formed
with the digits 1, 2, 3, 4, 5, 6, no digit being repeated ?

3. How many odd numbers between 1000 and 6000 can
be formed with the figures 1, 2, 3, 4, 6, 6, 7, 8, 9, 0, no
figure being repeated? How many of these numbers will
be divisible by 5?

4. How many three-lettered words can be made from the
alphabet, no letter being repeated in the same word ?

5. In how many ways can 4 persons, A, B, C, D, sit at a
round table ?

6. In how many ways can 6 persons form a ring ?

7. How many words can be made with 9 letters, 3 letters
remaining inseparable and keeping the same order ?

8. What will be the answer to the preceding problem if the
3 inseparable letters can be arranged in any order ?

9. A captain, having under his command 60 men^ wishes
to form a guard of 8 men. In how many different ways can
the guard be formed ?

10. A detachment of 30 men must furnish each night a
guard of 4 men. For how many nights can a different gaard
be formed, and how many times will each soldier serve ?

CHOICE 267

11. Out of 12 Democrats and 16 Republicans, how many
different committees can be formed, each committee consist-
ing of 3 Democrats and 4 Republicans ?

12. Out of 26 Republicans and 14 Democrats, how many
different committees can be formed, each committee consist-
ing of 10 Republicans and 8 Democrats ?

13. There are m different things of one kind and n different
things of another kind ; how many different sets can be made,
each set containing r things of the first kind and s of the
second ?

14. With 12 consonants and 6 vowels, how many different
words can be formed consisting of 3 different consonants and
2 different vowels, any arrangement of letters being consid-
ered a word ?

15. With 10 consonants and 6 vowels, how many words
can be formed, each word containing 5 consonants and 4
vowels ?

16. How many words can be formed with 20 consonants
and 6 vowels, each word containing 3 consonants and 2
vowels, the vowels occupying the second and fourth places ?

17. An assembly of stockholders, composed of 40 mer-
chants, 20 lawyers, and 10 physicians, wishes to elect a
commission of 4 merchants, 1 physician, and 2 lawyers. In
how many ways can the commission be formed ?

18. Of 8 men forming a boat's crew, 1 is selected as stroke.
How many arrangements of the rest are possible ? When the
4 men who row on each side are decided on, how many arrange-
ments are still possible ?

19. A boat's crew consists of 8 men. Either A or B must
row stroke. Either B or C must row bow. D can pull only

268 COLLEGE ALGEBRA

on the starboard side. In how many ways can the crew be
seated?

Note. Stroke and bow are on opposite sides of the boat

20. A boat's crew consists of 8 men. Of these, 3 can row
only on the port side, and 2 can row only on the starboard
side. In how many ways can the crew be seated ?

21. Of a base ball nine, either A or B must pitch ; either B
or C must catch ; D, E, and !F must play in the outfield. In
how many ways can the nine be arranged ?

22. How many signals may be made with 8 flags of differ-
ent colors, which can be hoisted either singly, or any number
at a time, one above another ?

23. Of 30 things, how many must be taken together in
order that, having that number for selection, there may be
the greatest possible variety of choice?

24. The number of combinations of w -f 2 objects, ta.ken 4
at a time, is to the number of combinations of n objects^ taken
2 at a time, as 11 is to 1. Find n.

25. The number of combinations of n things, taken r
together, is 3 times the number of combinations when r — 1
are taken together, and half the number of combinations when
r 4- 1 are taken together. Find n and r.

26. At a game of caixis, 3 being dealt to each person, any
one can have 425 times as many hands as there are cards in
the pack. How many cards are there in the pack ?

27. It is proposed to divide 15 objects into lots, each lot
containing 3 objects. In how many ways can the lota be
made ?

28. The number of combinations of 2 n things, taken n — 1
together, is to the number of combinations of 2 (» — 1) things,
taken n together, as 132 to 35. Find n.

CHOICE 269

349. Permutations, Repetitions allowed. Suppose we have
n letters, which axe all different, and that repetitions are
allowed.

Then, in making any arrangement, the first place can be
filled in n ways.

When the first place has been filled the second place can
be filled in n ways, since repetitions are allowed. Hence, the
first two places can be filled in nx n, or n% ways (§ 336).

Similarly, the first three places can be filled in nx nxn,
or w', ways (§ 337).

In general, r places can be filled in ti*" ways ; or, the number
of arrangements of n different things taken t at a time, when
repetitions are allowed, is n'.

(1) How many three-lettered words can be made from the
alphabet when repetitions are allowed.

Here the first place can be filled in 26 ways ; the second place in 26
ways ; and the third place in 26 ways. The number of words is, there-
fore, 268 = 17,576.

(2) In the common system of notation how many num-
bers can be formed, each number consisting of not more than
5 figures ?

Each of the possible numbers may be regarded as consisting of 5 fig-
ures, by prefixing zeros to the numbers consisting of less than 5 figures.
Thus, 247 may be written 00247.

Hence, every possible arrangement of 5 figures out of the 10 figures,
except 00000, will give one of the required numbers, and the answer is
10^ - 1 = 99,999, that ifi, all the nupabers between 0 and 100,000.

350. Permutations, Things Alike, All together. Consider the
number of arrangements of the letters a, a, h, h, h, e, d.

Suppose the a's to be different and the &'s to be different, and dis-
tinguish them by ai, Oj, &i, &2f h'

The 7 letters can now be arranged in [7 ways (§ 389).

Now supx>ose the two a's to become alike, and the three &*8 to become
alike. Then, where we before had |2 arrangements of the a's among

270 COLLEGE ALGEBRA

themselves, we now have but one arrangement, aa ; and where we before
had [3 arrangements of the 6's among themselves, we now have but one
arrangement, bbb. .,*

Hence, the number of arrangements is |-=j- = 420.

In generalj the number of arrangements of n things, of which
p are alike, q others are alike, and r others are alike, • • • , m

\n

(1) In how many ways can the letters of the word college
be arranged ?

If the two Ts were different and the two e's were different, the number
of ways would be [7. Instead of two arrangements of the two Ts, we have
but one arrangement, II ; and instead of two arrangements of the two e^s,
we have but one arrangement, ee. Hence, the number of ways is

rV = 1260.
|2[2

(2) In how many ways can the letters of the word Missis-
sippi be arranged ? . ^ .

rW=r- = 34,660.
IMl

(3) In how many different orders can a row of 4 white balls
and 3 black balls be arranged ?

I — I— = 36.

UH

351. Combinations, Repetitions allowed. We shall illustrate
by two examples the method of solving problems which come
under this head.

(1) In how many ways can a selection of 3 letters be made
from the letters a, b, c, d, e, if repetitions are allowed ?
The selections will be of three classes :

(a) All three letters alike.

(b) Two letters alike.

(c) The three letters all different.

CHOICE 271

(a) There will be 6 selections, since any one of the 5 letters may be
taken 3 times.

(b) Any one of the 5 letters may be taken twice, and with these may
be put any one of the other 4 letters. Hence, the number of selections
is 5 X 4, or 20.

5x4x3

(c) The number of selections (§ 341) is » or 10.

1 X i3 X o

Hence, the total number of selections is 6 + 20 + 10 = 36.

(2) How many different throws can be made with 4 dice ?
The throws may be divided into five classes :

(a) All four dice alike.

(b) Three dice alike.

(c) Two dice alike, and the other two alike.

(d) Two dice alike, and the other two different.

(e) The four dice different.

(a) There are 6 throws.

(b) Any of the 6 numbers may be taken 3 times, and with these may
be put any one of the other 5 remaining numbers. Hence, the number of
throws is 6 X 6, or 30.

(c) Any two of the 6 pairs of doublets may be selected. Hence, the

number of throws is » or 16.

1x2

(d) Any pair of doublets may be put with any selection of 2 different
numbers from the remaining 6. Hence, the number of throws is

6x^iii = 60.
1x2

(e) The number of throws is = 16.

^ ' 1x2x3x4

The answer is, then, 6 + 30 + 16 + 60 + 16 = 126.

352. Combinations and Permutations, Things Alike. We shall
illustrate by an example the method of solving problems
which come under this head.

How many selections of 4 letters each can be made from
the letters in the word proportion ? How many arrangements
of 4 letters each ?

272 COLLEGE ALGEBRA

There are 10 letters as follows :

0 p r t i n
0 p r

0

Selections : The selections may be divided into four claaseB :

(a) Three letters alike.

(b) Two letters alike, two others alike.

(c) Two letters alike, other two different.

(d) Four letters different.

(a) With the 3 o^s we may put any one of the 6 other letters, giving
6 selections.

(b) We may choose any 2 out of the 3 pairs, o, o; p, p; r, r.

= 3 selections.

1x2

(c) With any one of the 3 pairs we can put any two of the 6 remaining
letters in the first line.

6x4

3 X = 30 selections.

1x2

... 6x6x4x3,-,..

(d) = 16 selections.

^ ' 1x2x3x4

Hence, the total number of selections is 6 + 3 + 30 + 15 = 68.

Arrangements : (a) Each selection can be arranged in

6 X 4 = 20 arrangements.

li

(b) Each selection can be arranged in rr-r^ = 6 ways.

3 X 6 = 18 arrangements.

li

(c) Each selection can be arranged in r^ = 12 ways.

l4

■- =: 4 Mrays.

30 X 12 = 300 arrangements.

(d) Each selection can be arranged in [4 = 24 ways.

16 X 24 = 360 arrangements.

Hence, the total number of arrangements is

20 4- 18 4- 360 + 360 = 768.

CHOICE 278

353. Total Number of Combinations. I. The whole number of
ways in which a combination (of some, or all) can be m^de from
n different things is 2^ — 1.

For, each thing can be either taken or left ; that is, can be
disposed of in 2 ways.

There are n things ; hence (§ 337), they can all be disposed
of in 2" ways. But among these ways is included the case in
which all are rejected ; and this case is inadmissible.

Hence, the number of ways of making a selection is 2" — 1.

(1) In a shop window 20 different articles are exposed for
sale. What choice has a purchaser ?

The number of ways in which a purchaser may make a selection is

2» - 1 = 1,048,675.

(2) How many different amounts can be weighed with
1-pound, 2-pound, 4-pound, 8-pound, and 16-pound weights?

The number of different amounts that can be weighed is

26 - 1 = 31.

Note. Let the student write out the 31 weights.

II. The whole number of ways in which a selection can be
made from p + q + r + • • • things, of which p are alike, q are
alike, r are alike, ---, is |(p + 1) (q + 1) (r + 1) • • • | — 1.

For, the set of p things may be disposed of in ^ + 1 ways,
since none of them may be taken, or 1, 2, 3, • • •, or p, may be
taken.

In like manner, the q things may be disposed ot in q + 1
ways ; the r things in r + 1 ways ; and so on.

Hence (§ 337), all the things may be disposed of in

(^ + 1) (S' + 1) (r + 1) . • • ways.

But the case in which all the things are rejected is inad-
missible ; hence, the whole number of ways is

274 COLLEGE ALGEBRA

In how many ways can 2 boys divide between them 10
oranges all alike, 15 apples all alike, and 20 peaches all alike ?

Here the case in which the first boy takes none, and the case in which
the second boy takes none, must be rejected.

Therefore, the answer is one less than the result, according to IL

11 X 16 X 21 - 2 = 3694.

Exercise 50

1. How many three-lettered words can be made from the
6 vowels when repetitions are allowed ?

2. A railway signal has 3 arms, and each arm may take 4
different positions, including the position of rest. How many
signals in all can be made?

3. In how many different orders can a row of 7 white balls,
2 red balls, and 3 black balls be arranged ?

4. In how many ways can the letters of the word mathe-
matics j taken all together, be arranged ?

5. How many different signals can be made with 10 flags,
of which 3 are white, 2 red, and the rest blue, always hoisted
all together and one above another?

6. How many signals can be made with 7 flags, of which
2 are red, 1 white, 3 blue, and 1 yellow, always displayed all
together and one above another ?

7. In how many ways can 5 letters be selected from a, ft,
c, d, e, /, if each letter may be taken once, twice, three times^
four times, or five times, in making the selection ?

8. In how many ways can 6 rugs be selected at a shop
where two kinds of rugs are sold ?

9. How many dominos are there in a set numbered from
double blank to double ten ?

CHOICE 276

10. In how many ways can 3 letters be selected from n dif-
ferent letters, when repetitions are allowed ?

11. Five flags of different colors can be hoisted either
singly, or any number at a time, one above another. How
many different signals can be made with them?

12. If there are m kinds of things, and 1 thing of the first
kind, 2 of the second, 3 of the third, and so on, in how many
ways can a selection be made ?

13. How many selections of 6 letters each can be made
from the letters in the word democracy ? How many arrange-
ments of 6 letters each?

14. li oip -{- q-{-r things, p are alike, and q are alike, and
the rest different, show that the total number of selections is
Q> + 1)(^ 4- 1)2^-1.

15. Show that the total number of arrangements of 2n
letters, of which some are a's and the rest i's, is greatest
when the number of a's is equal to the number of ^'s.

16. If in a given number the prime factor a occurs m times,
the prime factor b, n times, the prime factor c, p times, and ^
these are all the factors, find the number of different divisors
of the given number.

17. If P„ represents tJie total number of permutations of n
different letters, %, a2> ^8> * ••>«„? and Q„ represents the number
of arrangements in which no letter occupies the place denoted
by its index (the complete disarrangement) y show that

Q, = P2-2Pi + Po, ^0 = 1,

Q* = ^4 - 4P8 + 6P2 - 4Pi + Po,

_ . , n ^ . ^ (^ — 1) ^

and, in general, Q^ = Pn - j ^„-i + \y^2 ""-^

n(n-l)(n-~2)

CHAPTER XXIII
CHANCE

354. Definition& If an event can happen in a ways and fail
in h ways, and all these a-{-b ways are equally likely to occur ;
if, also, one, and only one, of these a -{-b ways can occur, and
one must occur; then, the chance of the event happening is

7 > and the chance of the event failing is r •

a + b "^ ^ a + b

Thus, let the event be the throwing of an even number with a single
die.

The event can happen in 3 ways, by the die taming up a two, a four,
or a six ; and fail in 3 i??ays, by the die turning up a one, a three, or a
five ; and all these 6 ways are equally likely to occur.

Moreover, one, and only one, of these 6 ways can occur, and one mtut
occur (for it is assumed that the die is to be thrown).

Consequently, by the definition, the chance of throwing an even num-

3 1

ber is ? or - ; and the chance of throwing a number not even, that

3 + 32

3 1

is, odd, is » or - •

3 + 3 2

The above may be regarded as giving a definition of the
term chance as that term is used in mathematical works.
Instead of chance, probability is often used.

355. Odds. In the case of the event in § 354 the odds axe
said to be a to ft in favor of the event, if a is greater than h ;
and b to a against the event, if b is greater than a.

li a = b, the odds are said to be even on the event.

Thus, the odds are 5 to 1 against throwing a six in one throw with a
single die, since there are 5 unfavorable ways and 1 fovorable way, and
all these G ways are equally likely to occur.

276

CHANCE 277

356. Rules. From the definitions it is evident that^

The chance of an event happening is expressed hy the fraction
of which the numerator is the number of favorable ways, and the
denominator the whole number of ways favorable and unfavor-
aJfle.

For example, take the throwing of a six with a single die. The num-
ber of favorable ways is 1 ; the whole number of ways is 6. Hence, the
chance of throwing a six with a single die is }.

The chance of an event not happening is expressed by the
fraction of which the num,erator is the number of unfavorable
ways, and the denominator the whole number of ways favorable
and unfavorable.

Tor example, take the throwing of a six with a single die. The num-
ber of unfavorable ways is 5 ; the whole number of ways is 6. Hence,
the chance of not throwing a six with a single die is {.

357. Certainty. If the event is certain to happen, there are
no ways of failing, and ^ = 0. The chance of the event hap-
pening is then — —rr = 1. Hence, certainty is expressed by 1.

It is to be observed that the fraction which expresses a
chance or probability is less than 1, unless the event is cer-
tain to happen, in which case the chance of the event happen-
ing is 1.

358. Since — ^ + — ^ = 1,

a -{-b a -\-b

we have 7 = 1 —

a -\-b a -\- b

Hence, if p is the chance of an event happening, 1 — p is
the chance of the event failing.

359. Examples; Simple Event. (1) What is the chance of
throwing double sixes in one throw with 2 dice ?

Each die may fall in 6 ways, and all these ways are equally likely to
occur. Hence, the 2 dice may fall in 6 x 6, or 86, ways (§ 336), and

fA

278 COLLEGE ALGEBRA

these 36 ways are all equally likely to occur. Moreover, only one of the
36 ways can occur, and one must occur.

There is only one way which will give double sizes. Hence, the
chance of throwing double sixes is ^,

Remark. It may seem as though the number of ways in which the
dice can fall ought to be 21, the number of different throws that can be
made with two dice. These throws, however, are not all equally likely
to occur.

To obtain ways that are equally likely to occur we must go back to the
case of a single die. One die can fall in 6 ways, and from the construction
of the die it is evident that these 6 ways are all equally likely to occur.

Also the second die can fall in 6 ways, all equally likely to occur.
Hence, the 2 dice can fall in 36 ways, all equally likely to occur (§ 836).

In this case the throw, first die five, second die six, is considered a
different throw from first die six, second die five. Consequently, the
chance of throwing a five and a six is ^, or ^, while the chance of
throwing double sixes is only ■^^. This verifies the statement already
made, that the 21 different throws are not all equally likely to occur.

(2) What is the chance of throwing one, and only one, five
in one throw with two dice ?

The whole number of ways, all equally likely to occur, in which the
dice can fall is 86. In 5 of these 36 ways the first die will be a five, and
the second die not a five ; in five of these 36 ways the second die will be
a five, and the first not a five. Hence, in 10 of these ways one die, and
only one die, will be a five ; and the required chance is J{, or ^.

The odds are 13 to 6 against the event.

(3) In the same problem what is the chance of throwing at

least one five ?

Here we have to include also the way in which both dice fall fives, and

the required chance is ^J.

The odds are 25 to 11 against the event.

(4) What is the chance of throwing a total of 5 in one

throw with 2 dice ?

The whole number of ways, all equally likely to occur, in which the
dice can fall is 36. Of these ways 4 give a total of 5 ; viz.^ 1 and 4, 2
and 3, 3 and 2, 4 and 1. Hence, the required cliance is 3^, or |.

The odds are 8 to 1 against the event.

CHANCE 279

(5) From an um containing 5 black and 4 white balls,
3 balls are to be drawn at random. Find the chance that

2 balls will be black and 1 white.

There are 9 balls in the um. The whole number of ways in which

3 balls can be selected from 9 is , or 84.

1x2x3 g^4

From the 6 black balls 2 can be selected in , or 10, ways : from

1x2

the 4 white balls 1 can be selected in 4 ways ; hence, 2 black balls and

1 white ball can be selected in 10 x 4, or 40, ways.

The required chance is f f = iJ.

The odds are 11 to 10 against the event.

(6) From a bag containing 10 balls 4 are drawn and re-
placed ; then 6 are drawn. Find the chance that the 4 first
drawn are among the 6 "last drawn.

The second drawing could be made altogether in

110
i i^TT = 210 ways.

But the drawing can be made so as to include the 4 first drawn in

i-=p- = 15 ways,
\2\l ^ '

since the only choice consists in selecting 2 balls from the 6 not previously
drawn. Hence, the required chance is ^^^ = ^.

(7) If 4 coppers are tossed, what is the chance that exactly

2 will turn up heads ?

Since each coin may fall in 2 ways, the 4 coins may fall in 2* = 16

ways (§ 337). The 2 coins to turn up heads can be selected from the 4

4x3
coins in = 6 ways. Hence, the required chance is ^^ = |.

The odds are 5 to 3 against the event.

(8) In one throw with 2 dice, which sum is more likely to
be thrown, 9 or 12 ?

Out of the 36 possible ways of falling, four give the sum 9 (namely,
6 4- 3, 3 -f 6, 6 + 4, 4 + 5), and only one way gives 12 (namely, 6 + 6).
Hence, the chance of throwing 9 is four times that of throwing 12.

280 COLLEGE ALGEBRA

Note. It will be observed in the above examples that we sometimes
use arrangements and sometimes use selections. In some proUems the
former, in some problems the latter, will give the ways which are all
equally likely to occur.

In some problems we can use either selections or arrangements.

Exercise 51

1. The chance of an event happening is |. What are the
odds in favor of the event ?

2. If the odds are 10 to 1 against an event, what is the
chance of the event happening ?

3. The odds against an event are 3 to 1. What is the
chance of the event happening ?

4. The chance of an event happening is ^. Find the odds
against the event. ^

5. In one throw with a pair of dice what number is most
likely to be thrown? Find the odds against throwing that
number.

6. Find the chance of throwing doublets in one throw with
a pair of dice.

7. If 4 cards are drawn from a pack of 62 cards, what is
the chance that there will be 1 of each suit ?

8. If 4 cards are drawn from a pack of 62 cards, what is
the chance that they will all be hearts ?

9. If 10 persons stand in a line, what is the chance that 2

assigned persons will stand together ?

10. If 10 persons form a ring, what is the chaace that 2

assigned persons will stand together ?

11. Three balls are to be drawn from an urn contaixuxig
5 black, 8 red, and 2 white balls. What ii^ tbQ ^hfiOlQe of
drawing 1 red ball and 2 black balls?

CHANCE 281

12. In a bag are 5 white and 4 black balls. If 4 balls
are drawn, what is the chance that they will all be of the
same color ?

13. If 2 tickets are drawn from a package of 20 tickets
marked 1, 2, 3, •••, what is the chance that both will be
marked with odd numbers ?

14. A bag contains 3 white, 4 black, and 5 red balls ; 3 balls
are drawn. Find the odds against the 3 being of three dif-
ferent colors.

15. Show that the odds are 35 to 1 against throwing 16 in
a single throw with 3 dice.

16. There are 10 tickets numbered 1, 2, ••, 9, 0. Three
tickets are drawn at random. Find the chance of drawing a
total of 22.

17. Find the probability of throwing 15 in one throw with
3 dice.

18. With 3 dice, what are the relative chances of throwing
a doublet and a triplet ?

19. If 3 cards are drawn from a pack of 52 cards, what is
the chance that they will be king, queen, and knave ?

360. Dependent and Independent Events. Thus far we have
considered only single events. We proceed to cases in which
there are two or more events.

Two or more events are dependent or independent, according
as the happening (or failing) of one event does or does not
affect the happening (or failing) of the other events.

Thus, throwing a six and throwing a five in any particular throw with
one die are dependent events, since the happening of one excludes the
happening of the other.

But, with 2 dice, throwing a six with one die and throwing a five with
the other are independent events, since the happening of one has no effect
upon the happening of the other.

282 COLLEGE ALGEBRA

361. Events mutually Exclusive. When severskl dependent
events are so related that one, and only one, of the events can
happen, the events are said to be mutually exclusive.

Thus, let a single die be thrown, and regard its falling one up, two
up, three up, and so on, as six different events. Then, these six events
are evidently mutually exclusive.

362. If there are several events of which one, and only one,
can happen, the charice that one vnU happen is the sum of the
respective chances of happening.

To prove this, let a, a', a", •••be the number of ways
favorable to the first, second, third, ••• events respectively,
and m the number of ways unfavorable to all the events,

these a -\- a' -\- a" -\ -\- m ways being all equally likely to

occur, and such that one must occur.

Represent by 7i the sum a -\- a* -\- a" -{ \-m.

Of the n ways which may occur, a, a', a", ••• ways are
favorable to the first, second, third, ••• events respectively.

Hence, the respective chances of happening are

a a' a"

— • 5 >

n n n

Of the n ways which may occur, a + »' + a" H ways are

favorable to the happening of some one of the events. Hence,
the chance that some one of the events will happen is

<i + (/' + (i"H a a' a"

? or — I — + 1

n n n n

If, til en, jD, p\ p'\ • • • are the respective chances of happen-
ing of the first, second, thini, • • • of several mutually exclusive
events, the chance that some otie of the events will happen-
lSi>-f i> -rp -i

Thus, let the thn>\?iug of two. a four, and a &x^ with a aln^ die, be
three events. These three events are evidently mutually ezehiaive.

CHANCE 283

There are 6 ways, all equally likely to occur, in which the die can fall ;
of these 6 ways, one must occur, and only one can occur.

The chance of throwing a two is J ; of throwing a four, J ; of throwing
a six, I ; since there is but one favorable way in each case.

The chance of throwing an even number is |, since 3 out of the 6 ways
are favorable ways.

But } = ^ + 1 + ^ ; hence, | is the sum of the respective chances of
throwing a two, a four, a six. (Compare § 354.)

363. Compound Events. If there are two or more events,
the happening of them together, or in succession, may be
regarded as a compound event.

Thus, the throwing of double sixes with a pair of dice may be regarded
as a compound event compounded of the throwing of a six with the first
die and the throwing of a six with the second die.

364. Concurring Independent Events. The chance that two or
more independent events will happen together is the product of
the respective chances of happening.

To prove this, let a and a' be the number of ways favorable
to the first and second events respectively, and b and b' the
number of ways unfavorable to the first and second events
respectively ; the a-\-b ways being all equally likely to occur,
and such that one must occur, and only one can occur ; and
the a'-{-b' ways being all equally likely to occur, and such that
one must occur, and only one can occur.

Then, the respective chances of happenings are j and

— — 77 ; and the respective chances of failing are r and

a' -{-b' ^ a -\- b

b'
— — 7^- Represent the former by p and p'] then, the latter

will hel —p and 1 — p'.

Consider the compound event. It is evident, by § 336, that
there are (a -\- b) (a' + b') ways, all equally likely to occur. Of
these, one must occur, and only one can occur.

284 COLLEGE ALGEBRA

The number of ways in which both events can happen is
<w} \ hence, the chance that both events will happen is

Similarly, the chance that both events will fail is

the chance that the first will happen and the second fail is

(a + h) (a) + h') ""^^ ""-^'^'
the chance that the first will fail and the second happen is

Similarly for three or more events.

365. Successive Dependent Events. By a slight change in
the meaning of the symbols of § 364, we can find the chance
of the happening together of two or more dependent events.

For, suppose that, after the first event has happened^ the

second event can follow in a' ways and not follow in ft' ways.

aa
Then the two events can happen in , y. ways ; and

SO on, as in § 364. ^ ^^ ^

Hence, if p is the chance that the first event will happen,

and^' the chance that after the first event has happened the

second will follow, pp^ is the chance of both happening;

(1 —p)(l —p^, the chance of both failing; and so on.
Similarly for three or more events.

366. Examples. (1) What is the chance of throwing donble
sixes in one throw with 2 dice ?

Regard this as a compound event. Thd chance that the first die will
turn up a six is ^ ; the chance that the second die will torn up a six is | ;
the chance that each die will turn up a six is ^ x }, or ^.

The events are here independent. In Example (1), { 860, tha throwing
of double sixes is regarded as a 8imple event.

CHANCE 285

(2) What is the chance of throwing one, and only one, five
in a single throw with 2 dice ?

The chance that the first die will be a five, and the second not a five,
is i X j = ^ ; the chance that the first die will not be a five, and the
second die a five, is i x ^ = ^. These two events are dependent and
mutually exclusive, and the chance that one or the other of them will
happen is (§ 362) -f^ + y\ = ^. (Compare Example (2), § 369.)

(3) What is the chance of throwing a total of 5 in one
throw with 2 dice ?

There are four ways of throwing 5 : 1 and 4, 2 and 3, 3 and 2, 4 and 1.
The chance of each of these ways happening is ■^^. The events are mutu-
ally exclusive ; hence, the chance of some one happening is (§ 362)
A + tV + A + A = t- (Compare Example (4), § 369.)

(4) A bag contains 3 balls, 2 of which are white ; another
bag contains 6 balls, 5 of which are white. If a person is to
draw 1 ball from each bag, wliat is the chance that both balls
drawn will be white ?

The chance that the ball drawn from the first bag will be white is | ;
the chance that the ball drawn from the second bag will be white is {.
The events are independent; hence, the chance that both balls will be
white is I X f = f (§ 364).

(5) In the last example, if all the balls are in one bag, and
2 balls are to be drawn, what is the chance that both balls
will be white ?

The chance that the first ball will be white is | ; the chance that, after
1 white ball has been drawn, the second will be white is f ; the chance of
drawing 2 white balls is (§ 366) J X { = ^^j.

(6) The chance that A can solve this problem is §; the
chance that B can solve it is ^, If both try, what is the
chance (1) that both solve it ; (2) that A solves it, and B fails ;
(3) that A fails, and B solves it ; (4) that both fail ?

A*s chance of success is f ; A*8 chance of failure is \,
B*8 chance of success is ^ ; B's chance of failure is ^.

286 COLLEGE ALGEBRA

Therefore, the chance of (1) is f x y\ = J J

the chance of (2) is | x ^-^ = J|
the chance of (3) is J x ^^ = ^
the chance of (4) is J x ^^ = -J^.

The sum of these four chances is J§ + JJ + ^^ 4- ^^^ = 1, as it ought to
be, since one of the four results is certain to happen.

(7) In Example (6) what is the chance that the problem
will be solved ?

The chance that both fail is /^. Hence, the chance that both do not
fail, or that the problem will be solved, is 1 — ^5^ = || {§ 368).

(8) From an urn containing 5 black and 4 white balls,
3 balls are to be drawn at random. Find the chance that of
the 3 balls drawn 2 will be black and 1 white.

There are 9 balls in the urn. Suppose the balls to be drawn one at a
time. The white ball may be either the first, second, or third ball drawn.
In other words, 1 white ball and 2 black balls may be drawn in

r|=^ = 3 ways (§ 346).

The chance of the order, white black black, is J x f x ^ = JJ.
The chance of the order, black white black, is f x | x ^ = JJ.
The chance of the order, black black white, is f x | x ^ = J}.
Hence, the required chance is J? + i? + i? = M (§ ^62).
The method of Example (5), § 359, Is, however, recommended for
problems of this nature.

(9) When 6 coins are tossed what is the chance that one,
and onli/ one, will fall with the head up ?

The chance that the first alone falls with the head up is (§ 364)
Jxix^x^x^xi = ^^j; the chance that the second alone falls with
the head up is ^ ; and so on, for each of the 6 coins.

Hence, the chance that some one coin, and only one coin, falls with
the head up is ^\ + ^\ + ^{ + ^\ + ttt + ^V = /? = A-

(10) When 6 coins are tossed what is the chance that cU
least one will fall with the head up ?

The chance that all will fall heads down isjxjxjxjxjxjss^.
Hence, the chance that this will not happen is 1 — -^ = ft*

CHANCE 287

(11) A purse contains 9 silver dollars and 1 gold eagle, and

another contains 10 silver dollars. If 9 coins are taken out

of the first purse and put into the second, and then 9 coins are

taken out of the second and put into the first purse, which

purse now is the more likely to contain the gold coin ?

The gold eagle will not be in the second purse unless it (1) was among
the 9 coins taken out of the first and put into the second purse ; (2) and
not among the 9 coins taken out of the second and put into the first purse.
The chance of (1) is ^j^, and when (1) has happened the chance of (2)
is 1^. Hence, the chance of both happening is ^^^ x U = j*^. Therefore,
the chance that the eagle is in the second purse is -fg, and the chance that
it is in the fii*st purse is 1 — A = IS* Since \^ is greater than ^^y the
gold coin is more likely to be in the first purse than in the second.

Note. The expectation from an uncertain event is the product of tJie
chance that the event will happen by t?ie amount to be realized in case
the event happens.

(12) In a bag are 2 red and 3 white balls. A is to draw a
ball, then B, and so on alternately ; and whichever draws a
white ball first is to receive $10. Find their expectations.

A^s chance of drawing a white ball at the first trial is f . B's chance
of having/ a trial is equal to A's chance of drawing a red ball, or |. In
case A drew a red ball, there would be 1 red and 3 white balls left in the
bag, and B's chance of drawing a white ball would be J. Hence, B's
chance of having the trial and drawing a white ball is } X | = ^^^ ; and
B's chance of drawing a red ball is J x | = ^.

A's chance of having a second trial is equal to B's chance of drawing
a red ball, or -j^^. In case B drew a red ball, there would be 3 white balls
left, and A's chance of drawing a white ball wx)uld be certainty, or 1-

A's chance, therefore, is | + ^^ = t^ ; and B's chance is ^.

A's expectation, then, is $7, and B's $3.

367. Repeated Trials. Given the chance of an event hap-
pening in one trial, to find the chance of its happening exactly
once, twice, • • •, r times in n trials.

Let p be the chance of the event happening, and q the
chance of the event failing, in one trial ; so that q = 1 —p.

288 COLLEGE ALGEBRA

In n trials the event may happen exactly n times, ?i — 1
times, n — 2 times, • • • down to no times. The respective
chances of happening are as follows:

n times. The required chance, by § 364, is p^,

n — 1 times. The one failure may occur in any one of the
n trials; that is, in n ways. The chance of any particular way
occurring \^ p*~'^q\ the required chance is, therefore, np^^^q.

n — 2 times. The two failures may occur in any two of the

th (fh "■" 1^

n trials; that is, in — nr^ — - ways. The chance of any par-

ticular way occurring is ^"~y ; the required chance is, there-

n(n — V)
fore, —So — - P T-

r times. The n — r failures may occur in any n — r of the
n trials; that is, in i 1— ways. The chance of any particu-

n — r

lar way occurring is jp**^'**"''; the required chance is, therefore,
[n

71 — r

^'■2'"-"^

In
Similarly, the chance of exactly r failures is i p^~^q

E

n — r

The coefficients of the chance of r successes and of r failures
are the same, by § 343.

If, then, (p + qy is expanded by the binomial theorem, it
is evident that the successive terms are the chances that the
event will happen exactly n times, w — 1 times, • • • down to
no times.

The chances that the event will happen at least r times in

n trials is evidently jp** + np'^q -\ + 1 1- p^cj^

n

— rlr

\TeM — r

Note. Since p + g = 1, we have, whatever the value of n,

1 = p» + np»-ig 4- • • • + npg*— 1 + g«,

a somewhat remarkable equation inasmuch as there exists bat one
relation between p and g, mz.^ p + g = 1.

CHANCE 289

368. Examples. (1) What is the chance of throwing with
a single die a six exactly 3 times in 5 trials ? at least 3
times?

There are to be 2 failures. The 2 failures may occur in any 2 of

5x4
the 6 trials ; that is, in , or 10, ways. In any particular way there

will be 3 sixes and 2 failures, and the chance of this way occurring is
(i)' (i)^ J the chance of throwing exactly 3 sixes is, therefore,

\6/ \6/ 3888

The chance of throwing at least 3 sixes is found by adding the respec-
tive chances of throwing 5 sixes, 4 sixes, 3 sixes ; and is

(i)^ + 5(i)Mf) + io(i)8(j)2 = ^.

(2) A's skill at a game, which cannot be drawn, is to B's
skill as 3 to 4. If they play 3 games, what is the chance that
A will win more games than B ?

Their respective chances of winning a particular game are ^ and f .
For A to win more games than B, he must win all 3 games or 2 games.
The chance that A wins all 3 games is {^)^ = ■^. The chance that A
wins any particular set of 2 games out of the 3 games, and that B wins
the third game, is (^)2 x (^). As there are 3 ways of selecting a set of
2 games out of 3, the chance that A wins 2 games, and B wins 1 game,
is 3 X (})^ X ^ = Hi' Hence, the chance that A wins more than B is

0

(3) In the last example find B's chance of winning more
games than A.

B^s chance of winning all 3 games is (f)^ = ^. The chance that B
wins 2 games, and A wins 1 game, is 3 x {^)^ x ^ = iff. Hence, B's
chance of winning more games than A is ^ + JJJ = }J|.

Notice that A's chance added to B's chance, |f| + m, Is 1. Why
should this be so ?

(4) A and B throw with a single die alternately, A throw-
ing first ; and the one who throws an ace first is to receive a
prize of $110. What are their respective expectations ?

290 COLLEGE ALGEBRA

The chance of winning the prize at the first throw is ^ ; of winning at
the second throw, f X J ; of winning at the third throw, (f )2 x J ; of win-
ning at the fourth throw, (|)8 x J ; and so on.

Hence, A's chance is | + (i)^l + (f)*i + • • •, and B's chance is
(f)i + (l)*i + (i)^i + •• • Evidently B's chance is | of A's chance.
Since A's chance + B's chance = 1, A's chance is /j and B's /j. A's
expectation is ^ of $110, or $60; and B's t\ of $110, or $60.

Exercise 52

1. One of two events must happen. If the chance of the
first is § that of the other, find the odds against the first.

2. There are three events, A, B, C, of which one must hap-
pen, and only one can happen. The odds are 3 to 8 on A, and
2 to 5 on B. Find the odds on C.

3. In one bag are 9 balls and in another 6 ; and in each
bag the balls are marked 1, 2, 3, and so on. If one ball is
drawn from each bag, what is the chance that the two balls
will have the same number ?

-4. What is the chance of throwing at least one ace in 2
throws with one die?

5. Find the probability of throwing a number greater than
9 in a single throw with a pair of dice.

6. The chance that A can solve a certain problem is J, and
the chance that B can solve it is §. What is the chance that
the problem will be solved if both try ?

7. A, B, C have equal claims for a prize. A says to B :
" You and C draw lots, and the winner shall draw lots with
me for the prize." Is this fair ?

8. A bag contains 5 tickets numbered 1, 2, 3, 4, 5. Three
tickets are drawn at random, the tickets not being replaced
after drawing. Find the chance of drawing a total of 10.

CHANCE 291

9. A bag contains 10 tickets, 5 marked 1, 2, 3, 4, 5, and
5 blank. Three tickets are drawn at random, each being
replaced before the next is drawn. Find the probability of
drawing a total of 10.

10. Find the probability of drawing in Example 9 a total
of 10 when the tickets are not replaced.

11. A bag contains four $10 gold pieces and six silver
dollars. A person is entitled to draw 2 coins at random.
Find the value of his expectation.

12. Six $5 gold pieces, four $3 gold pieces, and 5 coins which
are either all gold dollars or all silver dimes are thrown together
into a bag. Assuming that the unknown coins are equally
likely to be dimes or dollars, what is a fair price to pay for
the privilege of drawing at random a single coin ?

13. A bag contains six $5 gold pieces, and 4 other coins
which all have the same value. The expectation of drawing
at random 2 coins is worth $8.40. Find the value of each
of the unknown coins.

14. Find the probability of throwing at least one ace in 4
throws with a single die.

15. A copper is tossed 3 times. Find the chance that it
will fall heads once and tails twice.

16. What is the chance of throwing double sixes at least
once in 3 throws with a pair of dice ?

17. Two bags contain each 4 black and 3 white balls. A
ball is drawn at random from the first bag, and if it is white,
it is put into the second bag, and a ball di-awn at random from
that bag. Find the odds against drawing two white balls.

18. A and B play at chess, and A wins on an average 2
games out of 3. Find the chance of A's winning exactly
4 games out of the first 6, drawn games being disregarded.

292 COLLEGE ALGEBRA

19. At tennis A on an average beats B 2 games out of 3.
If they play one set, find the chance that A will win by the
score of 6 to 2.

20. A and B, two players of equal skill, are playing tennis.
A needs 2 games to win the set, and B needs 3 games. Find
the chance that A will win the set.

21. If n coins are tossed up, what is the chance that one,
and only one, will turn up head ?

22. A bag contains n balls. A person takes out one ball,
and then replaces it. He does this n times. What is the
chance that he has had in his hand every ball in the bag ?

23. If on an average 9 ships out of 10 return safely to port,
what is the chance that out of 5 ships expected at least 3 will
safely return ?

24. At tennis A beats B on an average 2 games out of 3;
if the score is 4 games to 3 in B's favor, find the chance of
A's winning 6 games before B does.

25. The odds against a certain event are 5 to 4, and the odds
for another independent event are 6 to 5. Find the chance
that at least one of the events will happen.

26. A draws 5 times (replacing) from a bag containing 3
white and 7 black balls, drawing each time one ball; every
time he draws a white ball he is to receive $1, and every time
he draws a black ball he is to pay 50 cents. What is his
expectation ?

27. From a bag containing 2 eagles, 3 dollars, and 3 quarter-
dollars, A is to draw 1 coin and then B 3 coins ; and A, B, and
C are to divide equally the value of the remainder. What
are their expectations ?

28. What is the chance of throwing with a single die a five
at least twice in four throws ?

CHANCE 293

369. Existence of Causes. In the problems thus far con-
sidered we have been concerned only with future events ; we
now proceed to a different class of problems, problems of
which the following is the general type.

An event has happened. There are several possible causes,
of which one must have existed, and only one can have existed.
From the several possible causes a particular cause is selected ;
required the chance that this was the true cause.

Before proceeding to the general problem we shall consider
some examples.

(1) Ten has been thrown with 2 dice. Eequired the chance
that the throw was double fives.

Ten can be thrown in 3 ways : 6, 4 ; 4, 6 ; 6, 6. One of these three
ways must have occurred, and only one can have occurred.

Before the event the chances that these respective ways would occur
were all equal.

We shall assume that after the everU the chances that these respective
ways fiave occurred are all equal.

Then, precisely as in § 354, the chance that the throw was double fives
is i, and the chance that the throw was a six and a four is ^ + ^ = f .

(2) Fifteen has been thrown with 3 dice. Eequired the
chance that the throw was 3 fives.

Fifteen can be thrown in 10 ways :

654 646 456 663 366
645 564 465 636 665

One of these 10 ways must have occurred, and only one can have
occurred.

Before the event the chances that these respective ways would occur
were all equal.

We shall assume that after the event the chances that the respective
ways have occurred are all equal.

Then, precisely as in § 354, the chance that the throw was 3 fives is ^.

' (3) A box contains 4 white balls and 2 black balls. Two
balls are drawn at random and put into a second box. From

294 COLLEGE ALGEBRA

the second box 1 ball is then drawn and found to be white.
Required the chance that the two balls in the second box are
both white.

Before the event there were three cases which might exist. These
ca43es, with the respective chances of existence, were as follows:
The second box might contain :

(a) 2 white balls, of which the chance was |.

(b) 1 white and 1 black ball, of which the chance was ^,

(c) 2 black balls, of which the chance was ^.

Since 1 white ball has been drawn, (c) is impossible ; we have, there-
fore, only (a) and (b) to consider.

Supposing (a) to exist, the chance of drawing a white ball from the
second box was 1 ; supposing (b) to exist, the chance of drawing a white
ball from the second box was \.

Hence, the chance before the event that (a) exists, and we draw a white
ball, that is, the chance that we draw a white ball from 2 white balls,
was I X 1 = 1; the chance before the event that (b) exists, and we draw a
white ball, that is, the chance that we draw a white ball from a white
ball and a black ball, was i^? x ^ = ^.

Represent by Qi the chance after the event that (a) existed, and by Q2
the chance after the event that (b) existed.

We shall assume that Qi and Q2 are proportional to the chance brfore
the event that a white ball would be drawn from (a), and the chaDce
before the event that a white ball would be drawn from (b).

This assumption corresponds to the assumption in Examples (1) and (2),
in which the cases were equally likely to occur. We assume, then, that

. Qi ^ Qa ^ Q1+Q2
But Qi + Qa = 1, since either (a) or (b) miLSt exist ; also | + A = }.

" i A I*

.-. Qi = I, and Qa = f

The chance that both balls are white is f .

370. In general, let Pi, Pa? Ps, •• he the chance before the
event that the first, second, third, • • • cause exists ; and pi^ p^

CHANCE 295

* Pi9 ' ' ' t^G chance before the event that, when the first, second,
third, • • • cause exists, the event will follow. Let Qi, Q2^ Qj,

• • • be the chance after the event that the first, second, third,

• • • cause existed.

Then P^px is the chance before the event that the event
will happen from the first cause ; P^p^j the chance before the
event that the event will happen from the second cause ; and
so on.

We shall assume that Qi, Q2, Qzy • • • are, respectively, pro-
portional to Pii?i, PiP^^ PsPs) '"

rni J. • Ql Q2 Qi

That IS, — = — = -;; = . . .

PiPi P2P2 PzPz

Therefore, by § 245,

Ql Q2 Qz Q1 + Q2 + QZ + '"

PiPi P2P2 PzPz PiPi + P2P2 + PzPz -\

But Ql + Q2 + Qs H = 1> since some one of the causes

must exist. Hence,

_Qi_^_Q2_^_Q8_^ ^ 1 ^

PiPi P2P2 Pzpz PiPi + P2P2 -^ PzPz -\ '

from which Qi, Q2, Qj, • • • may readily be found.

Exercise 53

1. An even number greater than 6 has been thrown with

2 dice. What is the chance that doublets were thrown ?

2. A number divisible by 3 has been thrown with 2 dice.
What is the chance that the number was odd ?

3. Fourteen has been thrown with 3 dice. Find the chance
that one, and only one, of the dice turned up a six.

4. An even number greater than 10 has been thrown with

3 dice. Find the chance that the number was 14.

296 COLLEGE ALGEBRA

6. From a bag containing 6 white and 2 black balls a per-*
son draws 3 balls at random and places them in a second bag.
A second person then draws from the second bag 2 balls and
finds them to be both white. Find the chance that the third
ball in the second bag is white.

6. A bag contains 4 balls, each of which is equally likely to
be white or black. A person is to receive $12 if all four are
white. Find the value of his expectation.

Suppose he draws 2 balls and finds them to be both white.
What is now the value of his expectation ?

7. A and B obtain the same answer to a certain problem.
It is found that A obtains a correct answer 11 times out of
12, and B 9 times out of 10. If it is 100 to 1 against their
making the same mistake, find the chance that the answer
they both obtain is correct.

8. From a pack of 52 cards one has been lost; from the
imperfect pack 2 cards are drawn and found to be both spades.
Required the chance that the missing card is a spade.

371. Expectation of Life. The subjoined table gives the
mortality experience of thirty-five life insurance companies.
Columns A show the age-year ; columns D show the number .
of deaths during the corresponding age-years in columns A ;
and columns S show the number who survive at the end of
the year; that is, the number who attain the full age in
columns A,

Thus, out of 1000 healthy persons who attain the age of 10 years, 4 die
at that age, that is, during their 11th year, and 996 survive to attain the
fall age of 11 years. Again, looking opposite the 31 in column A^ we
find that of the 1000 persons arriving at the age of 10 years, 7 die during
their 31st year and 883 survive to attain the full age of 31. Hence, 890
out of the 1000 must have survived to the full age of 80, and 110 liad died
without attaining that age.

^

=

090

29

D

5

^

-

-■

-

^

-

^

«

S

11

9S7

47

10

750

66

20

492

83

17

87

12

992

30

SIX)

4S

10

740

63

21

471

81

16

72

13

«38

31

883

49

11

729

67

22

449

85

13

69

W

flS4

32

878

60

11

■J18

88

22

427

86

12

47

15

«80

33

869

51

11

707

69

22

405

87

10

37

16

fl78

34

862

52

12

695

70

23

382

88

9

28

17

fl72

35

8

854

63

12

883

71

23

359

89

7

21

18

mi

36

8

846

64

13

670

72

24

336

30

5.3

15.7

19

963

S7

8

838

65

13

657

73

25

310

91

4.4

11.3

20

fl50'

38

8

830

68

13

844

74

25

2S6

92

3.S

8.0

21

fl50

39

8

822

57

14

630

76

26

259

93

2.5

5.6

22

flU

40

g

814

58

15

615

76

26

234

04

1.8

3.7

23

flS8

41

0

806

66

15

800

77

25

209

95

1.3

2.4

24

932

42

9

798

00

10

684

78

23

1S8

M

0.9

1.5

25

1)25

43

9

787

81

17

667

79

23

163

97

0.6

0.9

28

fllS

44

fi

778

82

IS

549

SO

21

142

98

0.4

0.5

27

fill

45

9

769

83

18

631

81

20

122

99

0.3

0.2

28

^

O04

46

9

760

64

19

612

B2

18

104

100

0.2

0

(1) What is the chance that a person who has just completed

his 51st year dies before he is 52 ? ^ i

Out of every 707 healthy perBona who complete the 61st year ot their
lives, 12 die during their 52d year and 666 survive. Hence, the cliance
of the death daring hl£ 52d year of the person in queetloD is ^.

(2) What is the chance that a person aged 20 lives till he
is 50?

Out of every 966 persons wlio attain Uie age of 20 years, 718 survive to •
attain the full age of 50. Hence, the chance that the person in question f/^
lives till he is 60 is Ht V^

(3) What is the expe<;tation of life of a person who has just
completed his 90th year ?

The chance that he will illn during his Olit year la ^, during Iiia
92d year ff,, durlDg his »3d JtAT ffj, during bis 94tli year ^, and to
on u per table. Bat U be die* during hi* Slat year, he may die with

298 COLLEGE ALGEBRA

equal probability in any part of it ; hence, his expectation of life is \ year.
So if he dies during his 92d year, his expectation will be li years. If he
dies during his 93d year, his expectation will be 2i years, and so on.
Hence, his whole expectation will be

44x1+33x3+26x5+18x7 + 13x9+9x11+6x13+4x15+3x17+2x19

157x2
= MJ = 2Mi years of life.

(4) What is the expectation of life of a person who has just
completed his 80th year ?

The chance that he will die during his 81st year is ^o^, his 82d year is
^^, his 83d year is ^^, and so on. His expectation of life prior to the
completion of his 90th year is

20 X 1 + 18 X 3 + 17 X 6 + 15 X 7 + 13 X 9 + . . ■ + 6.3 X 19

142x2

= !fH = 3.5.

The chance that he will survive his 90th year is ^^^. Therefore, his
expectation of life subsequent to his 90th year is

the 10 years being added to the result of Example (3).
Hence, his whole expectation is 3.5 + 1.4 = 4.9 years.

Exercise 54

1. If B has just attained the age of 21, what is the chance of
his death within a year ? Within 5 years ? Within 10 years ?

2. If A is just 25 years old, what is the chance of his living
till he is 50 ? Till he is 60 ? TUl he is 75 ?

3. B and C are twins just 18 years old. What is the chance
that they will both attain the age of 50 ? That one, but not
both, will die before the age of 50 ?

4. A bridegroom of 24 marries a bride of 21. What is the
chance that they will live to celebrate their golden wedding ?

5. What is the expectation of life of a person who has
attained the age of 75? Of 70? Of 60?

CHAPTER XXIV

' , VARIABLBS AND LIMITS

372. Constants and Variables. A number that; under the
conditions of the problem into which it enters, may take
different values is called a variable.

A number that, under the conditions of the problem into
which it enters, has di, fixed value is called a constant.

Variables are generally represented by x, y, z, etc.; con-
stants, by the Arabic numerals, and by a, h, c, etc.

373. Functions. Two variables may be so related that a
change in the value of one produces a change in the value

of the other. In this case the second variable is said to be a

«

function of the first.

Thus, if a man walks on a road at a uniform rate of a miles per hour,
the number of miles he walks and the number of hours he walks are both
variables, and the first is a function of the second. If ^ is the number

of miles he has walked at the end of x hours, y and x are connected by

y
the relation ^ = ox, and ^ is a function of x. Also, x = - ; hence, x is

also a function of y.

When one of two variables is a function of the other, the
relation between them is generally expressed by an equation.
If any value of the variable is assumed, the corresponding
value or values of the function can be found from the given
equation.

The variable of which the value is assumed is generally
called the independent variable; and the function is called
the dependent variable.

299

300 COLLEGE ALGEBRA

In the last example we may assume values of x, and find the corre*
spending values of y from the relation y = ax\ or assume values of y,

y

and find the correspondmg values of x from the relation x = - • In the

first case x is the independent variable, and y the dependent; in the
second case y is the independent variable, and x the dependent.

374. Limits. As a variable changes its value, it may
approach, some constant; if the variable can be made to
approach the constant as near as we please, the variable is
said to approach the constant as a limit, and the constant is
called the limit of the variable.

Let X represent the sum of n terms of the infinite series

Then (§ 276), x = ^^^^ "" ^ = ^^-^ = 2 ^ .

^* ' i-1 2»-i 2»-i

Suppose n to increase ; then, — — - decreases, and x approaches 2.

Since we can take as many terms of the series as we please, n can be

made as large as we please ; therefore, — — - can be made as small as we

please, and x can be made to approach 2 as near as we please.

If we take any assigned positive constant, as xrr^ir* ^® ^^^ make the
difference between 2 and x less than this assigned constant ; for we have

only to take n so large that is less than ; that is, that 2*~* is

•^ ^ 2'»-i 10000

greater than 10,000: this is accomplished by taking n as large as 15.
Similarly, by taking n large enough, we can make the difference between
2 and x less than any assigned positive constant.

Since 2 —x can be made as small as we please, it follows that the sum

of n terms of the series 1 +i + J + jH , asnis constantly increased,

approaches 2 as a limit.

375. Test for a Limit. In order to prove that a variable
approaches a constant as a limit, it is necessary and sufficient
to prove that the difference in absolute value between the
variable and the constant can become and remain less than
any assigned constant, however small.

VARIABLES AND LIMITS 301

A variable may approach a constant without approaching it
as a limit.

Thus, in the last example x approaches 3, but not as a limit ; f or 3 — x
cannot be made as near to 0 as we please, since it cannot be made less
than 1.

376. Infinitesimals. As a variable changes its value, it may
constantly decrease in absolute value ; if the variable can
become and remain less in absolute value than any assigned
constant however small, the variable is said to decrease with-
out limit, or to decrease indefinitely. In this case the variable
approaches zero as a limit.

When a variable that approaches zero as a limit is conceived
to become and remain less in absolute value than any assigned
constant however small, the variable is said to become infini-
tesimal ; such a variable is called an infinitesimal number, or
simply an infinitesimal.

377. Infinites. As a variable changes its value, it may
constantly increase in absolute value ; if the variable can
become greater in absolute value than any assigned constant
however great, the variable is said to increase without limit,
or to increase indefinitely.

When a variable is conceived to become and remain greater
in absolute value than any assigned constant however great,
the variable is said to become infinite ; such a variable is called
an infinite number, or simply an infinite.

Infinites and infinitesimals are variables, not constants.
There is no idea of fixed value implied in either an infinite
or an infinitesimal.

A constant whose absolute value can be shown to be less

«

than the absolute value of any assigned constant however
small can have no other value than zero.

378. Finites. A number that cannot become an infinite
or an infinitesimal is said to be a finite number, or simply a
finite.

302 COLLEGE ALGEBRA

379. Relations between Infinites and Infinitesimals.

1. If X is infinitesimal and a is finite and not 0, then ax
is infinitesimal,

For^ ax can be made less in absolute value than any assigned
constant since x can be made less than any assigned constant.

II. If ILis infinite and a is finite and not 0, then aX is
infinite.

For, aX can be made larger in absolute value than any
assigned constant however large since X can be made larger
in absolute value than any assigned constant however large.

III. If X is infinitesim,al and a is finite and not 0, then -
is infinite.

For, - can be made larger in absolute value than any

assigned constant however large since x can be made less in
absolute value than any assigned constant however small.

IV. If a is infinite and a is finite and not 0, then — is
infinitesimal.

For, —: can be made less in absolute value than any assigned

constant however small since X can be made larger in abso-
lute value than any assigned constant however large.

In the above theorems a may be a constant or a variable ; the
only restriction on the value of 2l is that it shall not beeom,e
either infinite or zero,

380. From § 197 one root of the quadratic equation
ax^ -f Jaj -f c = 0 is infinite when a is infinitesimal, and
both roots are infinite when a and b are both infinitesimal.

381. Abbreviated Notation. An infinite is often represented
by 00. In § 379, III and IV are sometimes written

a ^ rv

Q = ^> 55 = ^-

VARIABLES AND LIMITS 303

The ezpresEdon - cannot be interpreted literally since we cannot diyide

by 0 ; neither can — = 0 be interpreted literally since we can find no

number such that the quotient obtained by dividing a by that number is
zero.

- = oo is simply an abbreviated way of writing : i/ - = X, and x
approacJies 0 as a limit, X increases withxmt limit,

ft 9^

— = 0 is simply an abbreviated way of writing : if — = Ti and X
increases without limit, x approaches 0 as a limit.

The symbol = is used for the phrase approaches as a limit
Thus, X == a means and is read as x approaxihes &as a limit,

382. Approach to a Limit. When a variable approaches a
limit it may approach its limit in one of three ways :

1. The variable may be always less than its limit.

2. The variable may be always greater than its limit.

3. The variable may be sometimes less and sometimes greater
than its limit.

If X represents the sum of n terms of the series 1 + i + i + iH ,

X is always less than its limit 2.

If X represents the sum of n terms of the series 3 — | — i^ — J^ — •••,
X is always greater than its limit 2.

If X represents the sum of n terms of the series 3 — } + | — JH ,we

have (§ 276)

1 + i ^ '

As n is indefinitely increased, x evidently approaches 2 as a limit.

If n is even, x is less than 2 ; if n is odd, x is greater than 2. Hence,
if 71 is increased by taking each time one more term, x is alternately less
than and greater than 2. If, for example,

n = 2, 3, 4, 6, 6, 7,

x=li, 2i, 1}, 2^, IJi, 2^j.

In whatever way a variable approaches a constant, the test
for a limit given in § 375 always applies.

304 COLLEGE ALGEBRA

383. Equal Variables. If two varicibles are always equal^
and ea^ch approaches a limit, then their limits are equal.

Let X and y be increasing variables, a and b their limits.

Kow, a = x -\- x' and b = y -{- y\ (§ 375)

where as' and y' are variables which approach 0 as a limit.
Then, since the equation x = y always holds true,

a — b = x' — y'.

But x' — y' can be made less than any assigned constant since
ic' and y' can each be made less than any assigned constant.

Since x' — y' is always equal to the constant a — b, x* — y'
must be a constant. But the only constant which is less than
any assigned constant is 0. (§ 337)

Therefore, x' — y' = 0,

Hence, a — b = 0, or a = b.

384. Limit of a Sum. The limit of the algebraic sum of
any finite number of variables is the algebraic sum of their
limits.

Let x,y,z,--- be variables, and a, b,c,"- their limits. Then,
a — Xf b — y, c — z, • • • are variables which can each be made less
than any assigned constant (§375). Then, (a — x)-^(b — y)
-{- (c — z) -{- " ' can be made less than any assigned constant.'

For, let V be the numerically greatest of the variables a — x, 6 — y,
c — z, • • •, and n the number of variables.

Then, (a — x) + (b — y) -^ (c — z) -\ < i? + u + u • • • to n terms.

But V -\- V -^V' " to n terms = nv.

Now nv can be made less than any assigned constant since n is finite,
and V can be made less than any assigned constant (§ 379, 1).

Therefore, (a — x) + (6 - y) -|- (c — «) H , which is less than n«, can

be made less than any assigned constant.

.'. (a + 5 + c H )—(x-\-y-^z-\ ) can be made less than

any assigned constant.

.'.a -\-b -{-c-\ is the limit of (x -\- y -{- z -\ ). (§ 376)

VARIABLES AND LIMITS 30S

385. Limit of a Product The limit of the prodv^ of two or
more variables is the product of their limits.

Let X and y be variables, a and b their limits.
To prove that ab is the limit of xy.

Put X =: a — x'y y = b — y' ; then x' and y' are variables that
can be made less than any assigned constant. (§ 375)

Now, xy = (a — x*) (b — y')

= ab— ay' — bx' -f x'y'.
.'. ab — xy = ay' -f bx' — x'y'.

Since every term on the right contains x' or y', the right
member can be made less than any assigned constant. (§ 384)
Hence, ab — xy can be made less than any assigned constant.
Therefore, ab is the limit of xy. (§ 375)

Similarly for three or more variables.

386. Limit of a Quotient. The limit of the quotient of two
variables is the quotient of their limits, if the divisor ^ zero.

Let X and y be variables, a and b their limits.
Put a — X = x', and b — y = y' ; then x' and y' are varia-
bles with limit 0. (§ 375)

X a — a?'

We have x = a — x', y = b — y', and - = ,

„ a aj__2 a — x' _ bx' — ay'

^""^^ b^y^l^T^'^bib-y')

The numerator of the last expression approaches 0 as a

limit, and the denominator approaches 6* as a limit; hence,

the expression approaches 0 as a limit. (§ 379, 1)

a X
Therefore, approaches 0 as a limit.

Therefore, ^ is the limit of -• (§ 375)

306 COLLEGE ALGEBRA

387. Vanishing Fractions. When variables are mvolved in
both numerator and denominator of a fraction it may happen
that for certain values of the variables the numerator and the
denominator both vanish. The fraction then assumes the

form jr7 a form without meaning; as even the interpretation

of f 381 fails, since the numerator is 0. If, however, there is
but one variable involved, we may obtain a value as follows :
Let X be the variable, aud a the value of x for which the

fraction assumes the form -• Give to a; a value a little greater

than a, B8 a -^ z; the fraction now has a definite value. Find
the limit of this last value as 2; is indefinitely decreased.
This limit is called the limiting value of the fraction.

x* — a'

(1) Find the limiting value of as a = a.

VThen x has the value a the fraction assomes the form -•
Put X = a + 2 ; the fraction becomes

(a-\-z)-a ~ z

Since z is not 0, we divide by z and obtain 2a-\- z.

As z is indefinitely decreased, this approaches 2 a as a limit.

Hence, 2 a is the limiting value of the fraction as x == a.

2x* 4a; -4-5

(2) Find the limiting value of o^s 1 o a^-i ^^®^ « = 00 .

2-1 + 1
^ , 2x^-4x4-5 x« x»

We have — — — — - — - = 7-*

3x» + 2x2-l „ 2 1

o -i

X X*

As X increases indefinitely, —»—»-» — approach 0 as a limit (§ 370,

X2 X' X X*

2
rV), and the fraction approaches - as a limit.

VARIABLES AND LIMITS 807

Ezerdae 55

Find the limiting value of :

1. ^ ^ , — ^ — ^ when X becomes infinitesimal.

7 aj' — 6 oj + 4

2. ^ . ^ c^^ when x becomes infinite.

X* + 36

3. ^—5 7- when x becomes infinitesimal.

X^ -\- 4:

x^ — S x -\- 15
*• ~2 — 7 To ^^®^ ^ approaches 3.

6. ^2 , 9a. . j^3 ^^®^ ^ approaches - 3.

x(x^-h4:X-}-S) , ,

6. . : n 1 , trr — T^ whcn X appioachcs — 1.
aj* + 3 aj^ + o aj + 3

a;» + aj2-2 ^ ^ ^

7. -jj — TT—i — t: 7 when x approaches 1.

«• -h 2 a;'* — 2 a; — 1

4 X -4- "v aj — 1

8. , when X approaches 1.

2a;-V^rri

9. , , when X approaches 1.

a;2_4

10. , , when x approaches 2.

V^rf2-V3a;-2

Vaj — a -f V^ — Va

11. . when a; approaches a.

Va;^ — a^

12. If a; approaches a as a limit^ and n is a positive integer,
show that the limit of a;* is a\

13. If a; approaches a as a limit, and a is not 0, show that
the limit of a;" is a% where n is a negative integer.

CHAPTER XXV
SERIES

CONVERGENCY OF SERIES

,'

388. Given Series. A series of numbers is said to be given

if a law is known by which any term of the series can be
calculated when its rank in the series is given.

389. An infinite series is a series in which the number of
terms may be made greater than any finite number.

Thus, if we divide the numerator of the fraction by the denom-

1 —X

inator, we obtain the series 1 + x + x* + x* + • • • Since we may carry
the division as far as we please, it is evident that we may make the
number of terms in the series greater than any finite number.
Hence, l + x-j-x^ + x'H isan infinite series,

390. Convergent Series. An infinite series is a convergent
series if the limit of the sum of the first n terms, when n
increases indefinitely, is a definite finite number.

Thus, if X < 1, the series l+x-|-x^ + x> + **>isan infinite decreasing
geometrical series and

. = -1-. (8280)

X — X V

That is, the limit of the sum of the first n terms of the series, when n

is made to increase indefinitely, is the definite finite number 1 and

the series is convergent, "

Every finite series is a convergent series.

391. Sum of Convergent Series. The limit of the sum of
ahe first n terms of an infinite convergent series, when n
increases indefinitely, is called the sum of the series,

308

CONVERGENCY OF SERIES 309

392. Divergent Series. An infinite series is a divergent series
if the sum of the first n terms may be made greater than any
assigned finite number if n is made large enough.

Thus, ifx = lorx>lm the infinite series 1 + x + jc^ _j. 358 _^ . . .^ it is
c vident that by making n large enough we can make the sum of the first
n terms greater than any assigned finite number.

Hence, if x = 1 or x > 1, the series l + x + x2 + x* + »'«is divergent.

393. Oscillating Series. An infinite series is an oscillating
series if the sum of the first n terms^ approaches different
finite numbers as n is increased.

Thus, if X = — 1 in the infinite series l + x-l-x2 + x' + '', the series

becomes 1 — 1 + 1 — IH If we take an even number of terms, their

sum is 0 ; if an odd number, their sum is 1. '

Hence, if x = — 1, the series l+x + x^ + x^+^is oscillating.

394. In general, we let Ui, u^, u^, • • •, w„, • • • represent any
infinite series each of whose terms is finite.

395. Residue of a Series. The difference between the sum
of an infinite series and the sum of the first n terms if n
increases indefinitely is called the residue of the series.

Let S represent the sum of a series,

5„ represent the sum of the first n terms,

and R^ represent the residue after the first n terms.

Then, by the definition of the residue,

S — S^ = R^.

396. If an infinite series is convergent, its residue is an
infinitesimal.

For S-S^ = R^. (§ 395)

Since by hypothesis the series is convergent,

S = the limit of 5„. (§ 390)

.'. 5 — 5^ is an infinitesimal. (§ 375)

.•./?„ is an infinitesimal. (§ 395)

^

810 COLLEGE ALGEBRA

397. If an infinite series is convergent, the nth term i «
an infinitesimal when n increases indefinitely, 'j

For S — S^_i is an infinitesimal, (§ 394|»)

also 5 — 5^ is an infinitesimal.

Hence, ^ — ^n-i—i^ — ^n) is an infinitesimal,

or, 'S^n "~ ^n-i is an infinitesimal.

' But S, - 5„_i = w„.

Therefore, u^ is an infinitesimal.

398. If an infinite series is convergent, m can he made so
large that the sum, of p consecutive terms beginning with the
(m 4- V)th will be an infinitesimal, however great p may be
made.

Let S = the sum of the series,

S^ = the sum of the first m terms,
and 5^^^ = the sum of the first m -\-jp terms.

Then, 5 — 5^ = the sum of all the terms after the mth,
and 5 — S^^^ = the sum of all the terms after the (m -\-p)ih..

Hence, 'S- 5^ -(5- 5^^^)= w^^i -f ^^^2 + ••* + ^m+p^
or, S^^p- S^ = w^+i + u^^2 4- • • • + u^+p' [1]

Let m-\-p = n, [2]

Then [1] becomes S^-S^ = u^^^ + w^+g H h t*n- [8]

Now let p increase indefinitely.

Then, by [2] n must also increase indefinitely,

and limit S^ = S,

Then [3] becomes

S-S^ = ^^l (u^^, + u^^, H- u^^, + ...). [4]

Now let m be made to increase indefinitely.

Then, 5 - 5^ is an infinitesimal. (§ 396)

Therefore, ^^^ (^m+i + ^m+2 H ) is an infinitesimal

CONVERGENCY OF SERIES 311

Therefore, by making m large enough we can make

an infinitesimal, however large p may be.

399. If in an infinite series the sum of p consecutive terms
beginning with the (m -{-l)th is an infinitesimal, however great
p may be made, then the series is convergent.

Let e represent any positive number taken as small as we
please.

Since by hypothesis

^m+l + W^ + a + Wm + 3 H h '^m+p [1]

is an infinitesimal, then

^m + l + ^m + 2 + ^m + 8 H + '^m+p < «• (§ 376)

But W^+1 -f U^+2 + Wm + 8 H \- Ki + p = 'S^m+p — ^m'

Let m H- j9 = n. [3]

Then [2] becomes S^ - s^<e. [4]

Now let p increase indefinitely.
By [3] n must also increase indefinitely,
and limit S„ = S,

Then [4] becomes S — S^<e. [6]

Since by [5] 5 — 5^ is an infinitesimal,

S = limit 5„. (§ 376)

Therefore, the series is convergent. (§ 390)

400. If an infinite series is convergent, the residue B^ is an
infinitesimal.

For, when p is made to increase indefinitely (§ 399),

Wm+i + Wm+2 + Wm+8 + • • • + '^m-s-p ^ au infinitesimal.

But W« + i + W„ + 2 -f Wm+8 + • • • + «^m+p = -»«•

Therefore, J?„ is an infinitesimal.

812 COLLEGE ALGEBRA

401. The theorems of §§ 399 and 400 may be stated,

If the residue E^ of an infinite series is an infinitesimal^ the
series is convergent ; andy conversely y if an infinite series is
convergent, B^ is an infinitesimal,

402. If an infinite series has positive terms only and S^
remains less than a known finite magnitude M, however great
n may he made, then the series is convergent.

For, if the series could be divergent,

limit iS„ _

and it would be possible to make n so great that S^ would be
greater than M, which is contrary to the hypothesis.
Therefore, the series is convergent.

403. If the infinite series Vj -|- Vg -f Vg H ha^ positive terms

only and is convergent, and if from a definite term onwards

Un s= Vn or Un < v^ then the series u^ -f Ug -f Ug H is also

convergent.

Let V^ = Vi + V2-\-v^-\ \-v^

and £/» = Wi 4- 1^2 4- i^s H h ^n-

Since the first series is convergent by hypothesis, we can
take m so great that

however great p may be made. (§ 398)

But u^+i > v^+i, w„+3 > t;„+2, • • •, u^+p > v^+p-

Hence, u„+i + u^+2-\ f- Wm+p < «•

Therefore, the series t^j -f ^2 + ^s H is convergent. (§ 399)

404. If the infinite series Vj -f V2 + Vg H ?ias positive

terms only and is divergent, and if from a definite term onwards
Un = Vn or Un > v^, then the series Uj + u, -f Uf + • • • w also
divergent

CONTERGENCY OF SERIES 318

For, if the u series could be convergent, then, since by
hypothesis v„ > u^, the v series by § 403 would be convergent,
which is contrary to the hypothesis.

Therefore, the series Ui -{- u^ -{- u^ + * " is divergent.

405. If the infinite series u^ + Uj -|- Ug H- • • • has positive
terms only, and if from a given term, onwards, say from the

nthf — ^^ > k < 1, where ^ is a constant independent of n, then

the series is convergent.

By hypothesis, after the nth term -^^ > k for every value
oin. ""^

Hence, Wn+i > ^^n^ ^^^ every value of n.
Therefore, we may make the following table :

i

^m + 3 > ^m + 2^ > ^m^*,

• • - • • •

We see that each term of the series

^m + ^m + 1 + W». + 2 + ^^m + 8 H [1]

is not greater than the corresponding term of the series

^m 4- ujc + ujc" + ujc"" + . . . [2]

But series [2] is a geometrical progression whose ratio is k,

and, since by hypothesis k<l, the progression is a decreasing

geometrical progression, and therefore convergent. (§ 390)

Hence, series [1] is convergent. (§ 403)

Therefore, the series u^ -\- v^ -{■ u^ -\ is also convergent.

406. If the infinite series Uj + Uj -f- Uj H has positive terms

only, and if from, a definite term onwards, say from the nth,

u

n-H

< 1, the series is divergent.

For all values of ,n less than m, u^^^ < u,

814 COLLEGE ALGEBRA

Therefore, we may make the following table :

««m+l < ^m,

^m + 2 < ^n. + l < ^m>

^m + 8 < ^m + 8 < ^mj

• • • •

Now the series u'^ + u^-\-u^ + u^ is divergent. (§ 392)

Hence, the series u^ -\- u^^^ + u^^^ H is divergent. (§ 404)

Hence, the series u^-\-U2-\ h w^ + ^V+i "^ ^m+aH is

also divergent.

407. -4n infinite series that contains both positive and nega^
tive terms is convergent if the series consisting of the absolute
values of its terms is convergent.

Let the given series be

t^i + i^a + ^8 H h Wn> [1]

and let S^^^ -S^ = u^+^ + u^+2 + • • • + u^+p- [2]

Let \u„\ represent the absolute value of u^, and let

Then, since series [3] is convergent by hypothesis, by § 398,
S«+p-S« = |w^+i| + |2^^+2|H----+|^m+p|<«. [4]

Let w^^.1 + w^+2 + i^^+3 H h u^+p [5]

contain all the positive terms in series [2],

and let t*,+i + t^,+2 + i^,+8H ^-^.+p [6]

contain the absolute values of all the negative terms in
series [2].

Since [5] contains only a part of the terms in [4],

Wr + l + Wr + 2 + ^r + 8 H + W^+p < 6. [7]

Since [6] contains only a part of the terms in [4],

^.+1 + W.+2 + ^.+8 H ^- ^.+p < «• [8]

CONVERGENCY OF SERIES 815

Hence^

Since [10] contains all the terms in [2],

Therefore, % + w^ + ^s H 1- ^*n is convergent. (§ 399)

406. If the absolute values of the terms of a given series
form a convergent series, the given series is said to be abso-
lutely convergent

409. Examples. (1) Per what values of x is the infinite

series a? — "ttH--^ 'i — :f--» convergent ?

Here, r = !^ = (-i^)x = f 1 ^-)x.

As n is indefinitely increased, r approaches x as a limit.
Hence, the series is convergent when z is numerically less than 1
(if 407, 406), and divergent when z is numerically greater than 1.
When X = 1 the series is convergent by § 403.
When X = — 1 the series becomes

and the series is divergent. (§ 392)

(2) Por what values of x is the infinite series

X X X iE**

z n + z — 3 ,+ o T ■) 1" ~~/ — T^TT convergent ?

1 X 2 , 2 x^J 3 X 4 «(» + !) ^

Here, ;}>^"'''\^^^ = (-^oh= f-^\ ^■

As n is indefinitely increased, r approaches x as a limit.

If X is nmnerically less than 1, the series is convergent. (§ 405)

If X is numerically greater than 1, the series is divergent. (§ 406)

If X = 1, the series is convergent. (§ 405)

If X == — 1, the series is convergent. ({ 407)

816 COLLEGE ALGEBRA

Ezerclse 56

Determine whether the following infinite series are conver-
gent or divergent :

'• *-[3 + [5-[7 + - ^- ^ + *+|J + |3 + -"

x^ 05* 05* 05* a;* cc^

^- ^~(2 + (i~[6 + "" ■*• *-3"^ 5"Y + '"

,1 «• 1-3 «», 1.3-6 a;'

6. aj H — • • • h • • •

FACTORIAL BINOMIAL THEOREM

410. Factorial Notation. The expression nV^ in which r is
a positive integer, denotes the product

1 X ^ X (w — 1) X (/I — 2) X • • • X (n — r -I- 2) X (n — r -f 1),

and is read factorial n o/ orc^ar r ; w is the primitive (factor),
and r is the index of the order. If the primitive and the index
of the order are equal, the latter is omitted.
Thus, n!" is written simply n!.

In writing out a factorial as a product, the initial imit-
multiplicand is usually omitted, so that the general practice
is to write

nl*' = 71(71" 1) (n — 2) • • • (n — r -f 1),
and Til = w(n — l)(n — 2)--*3'2-l.

However, inserting the initial unit-multiplicand gives at
once the interpretation of the zero index, n!® = l, and the
extension to negative indices,

^^"'^^ (7H-l)(7i-f 2)...(7i-f r)'
Thus, 6 1^ = 6x5x4 = 120; and5! = 6x4x3x2x 1 = ISli

FACTORIAL BIKOMIAL THEOREM

5T

Find the value of

•
•

1. 7!*.

2. 71*.

7.

fir

3. 7!.

8.

(- 2) '.*.

4 «■•.
** 3!

9.
10.

(- ") •'•

^- 2131'

11.
12.

my..

(3i):».

*• 4!4!'

13.
14.

SIT

15. Sl-\

16. 0!-^

17. (-5).^».

18. (i):-*.

19. n.^

20. (n-fl)(n^).

21. n(f^ + l)H«-l)^'^

22. (n-f l)(n + 2)(ii!).

Assuming that Formula [A], § 288, is true and that n is a
positive integer, show that :

23. (x + y)'' = x- + j^x— »y + -^x— y-l-...

r!
24. (. + ,).=„.{^ + ^-^ + ^-^ + ...

H ^ 1 y,

(n — r)lrl J

26. (x + yy = aj» + nl^OI-^x"- y + 7i!*0!-»x»-y -\

+ n!'"0!— 'a^-y H

Til

fr

411. Show that — 4-

nV

^'-' ^(7^4-1)!''
r! ' (r — 1)! "" r!

Now

n!*- (n-r-flX^!'""^

r!

and

nl—i

r!

(r-1)!"" r!

+

nr-^ ^ (yt ~ r 4- 1) Tn^-^) -f r(n\''-^)

r\ (r-1)! r!

(7^-hi)(nr-^) (n4- 1) r

rt rl

318 COLLEGE ALGEBRA

412. The Factorial Binomial Theorem. If r is a positiye
integer and m and n any numbers whatever,

1 1 • ^

By successive multiplications we obtain the following
identities :

(m + 7i)!« = m!8 + 3 ml^Ti !i + 3 ml^TiI^ 4- ^!' ;

(771 4- n)l^ = ml^ + 4m!»7i!i + 6 wI^ti!^ 4- ^ml^n!* 4- t^!*.

The proper method of obtaining the expanded forms on the
right is as follows :

771 4 W = 771 4 71

7714 n — 1 = (771 — 1) 4 71 ; 771 4 (7i — 1)

771 (771 — 1) 4 77171

4 77171 4n(n — 1)

.•.(77i4n)!2 = w!2 +2mn +n\^ (i)

m4n-2 = (77t-2)4 7i; (tti - l)4(n- 1); 77t4(7i-2)

771 !2 (771 - 2 ) 4 2 77171 (771 - 1) 4 771 (7ll2) (ii)

4 7711^71 4 277in(7i - 1)4 n!2(n - 2) (ill)

.•.(77i4n)l' = 7Ml8 4 377i!2nli 4377ilin!a 4nl« (iv)

In the preceding multiplication the line (ii) is formed by
multiplying the first term of line (i) by (m — 2), the second
term by (m — 1), and the third term by m. Line (iii) is
formed by multiplying the first term of line (i) by n, the
second term by (n — 1), and the third term by (n — 2).
Hence, line (iv), which is the sum of lines (ii) and (iii),
contains the first term of line (i) multiplied by (m — 2) 4- n,
the second term multiplied by (m —1) 4- (^ — 1), and the third
term multiplied by m + (ri — 2). Therefore, line (iv) is equal
to line (i) multiplied by (m -\- n — 2).

FACTORIAL BINOMIAL THEOREM

319

Continuing this process to form a line (v) by multiplying
the first term of line (iv) by (m — 3), the second term by
(m — 2), the third term by (m — 1), and the fourth term by m ;
and a line (vi) by multiplying the first term of line (iv) by n,
the second term by (n — 1), the third term by (n — 2), and
the fourth term by (n — 3), we have

(m + n)\^ = m\^ -{-SmlHl^ +SfnlH\^ +n\^ (iv)

m + n - 3 =(m-3)+n; (m-2)+(n-l); (m-l) + (n--2); m+(n-3)

m!* +3m!«n!i +Sm\^\^ + m\H\^ (v)

4- mlHl^ +3m!2n!g +3m!in!»+n!* (vi)

.•.(m + n)I*=m!* +^nVM\^ -\-6m\H\^ +4w!in!8+n!* (vii)

These expansions may be written in a form better adapted
to show the formation of the coefficients of their terms :

3 3
(m-|-?i)!«=m !« + T m l^nl^ -|- -

4 4

-m!^

m!i7i!2 +

2
3

m!V!2 +

+

1
4

1
4

2
3

2
3

2

3
2

7il»;

3
21

3.4

Comparing these expansions with those of the powers of
(a -\' b) as given in § 285, we observe that corresponding
terms up to the fourth order and the fourth power have
the same coefficients and have the same indices of order as
exponents.

To prove that the corresponding coefficients and indices of
order are the same as the coefficients and exponents in the
expansion of the Binomial Theorem for all positive integral
values of index of order and exponent, we proceed exactly as in
§ 288. We assume that laws 1, 2, and 3 (§ 286) hold true up
to a given value, r, of the index of order, and prove that in such
case they hold true for the value r -|- 1 of the index of order.

820 COLLEGE ALGEBRA

Let it be granted tliat

(m + n)\^ = mr + -ml^htl^ -{■ -^ 'ml^-Hl^ + . . . + _ml'^nl«+-.. [1]

1 1*2 tl

Multiply the first term on the right of [1] by (m — r), the
second term by (m — r -\- 1), the third term by (m — r + 2),
and so on ; writing the partial products in order in a line.

Form a second line by multiplying the first term on the
right of [1] by n, the second term by (n — 1), the third term
by (71 — 2), and so on ; writing the first partial product of this
line under the second partial product of the first line.

Add the two lines of partial products, simplifying the coeffi-
cients of the sum by § 411, and we obtain the product of the
right member of [1] by (m + ti — r). Thus :

r r(r — 1\ r!'

ml'* + - ml'— in!i + -^^ ml'— %!2 + 1- — ml*— 'n!' H

1 1-2 tl

{m-r)+n; (m-r+l) + (n-l); (m-r+2) + (n-2) ; (m-r-\-t) + {n~-t)

m\r+i + -ml'-nli + ^^^ ~ ' ml^-H\^ + • • • + —m\^+^-*n\* + • • •

1 1 • 2 t\

r rl<— 1

+ ml'-nli + -mr-inl^ H + mK+i-'nl* -\

1 (t~l)l

m\r+i + !l:ti ml'-nli + ^^ "*" ^^^ml'-^nia + • • • + l^^-tHi'mlC'+D-'nl' + • • •
1 1-2 t\

= (m + n)l'-+i. [2]

Hence, if the expansion [1] is true for any given positive
integer r, it is true for (r + 1). Now expansion [1] is true for
r = 4, as shown on page 319. Therefore, it is true for r = 5 ;
and, being true for r = 5, it is true for r = 6 ; and so on. In
short. Formula [1] is true for all positive integral values of
the index of order.

Hence, for all positive integral values of r,

(m + n)l' = m!' + rml'-^nl* + ^^^"""^^ ml'-«nl* -f- • • .

+ L.mI'-*n!* + ...

Note. Theorem [1], § 412, is often named Vandermonde^s Factorial
Theorem, and Theorem [A], § 288, Newton^s Binomial Theorem.

BINOMIAL THEOREM ; ANY EXPONENT 821

BINOMIAL THEOREM

413. Binomial Theorem; Any Exponent. Let m and n be any
two scalar numbers and let the value of x be so taken as to
render convergent each of the three series

^ m m(m — 1) . ■ ^-^ « i

n . n (ti — 1) « . . n!'

^ + r+ 1.2

. m + n , (m 4- ^) (^ + ^ — 1)
"*""nr~^"*" 12 ^

g , (^ + n)!^.

^!

a*

(i)

(ii)

(iii)

Then, series (iii) is the product of series (i) and series (ii).

For, on forming the product of series (i) and series (ii) and
arranging it according to descending powers of x, the coeffi-
cient of »' is

f«-2ti!a

!«-8n.T8

ml* mV''^n mv~'nr iyii'~°nr

IT "^ (^-l)!l!i "*" (^-2)12! "*" (i5-3)I3! "^ * "

^! 1 X * ^

. t(t-l)(t-2) ,,_, „ ,

}

_ (m 4- ^) l^

(§ 412)

Hence, if a; is so taken as to make all three series con-
vergent, we may write

l+r:^ + ^(f7^)^»4..

I

12

= 1 +

►• *

t + ix + ^^^^^x' +

n

i

12

••■■}

m-^-n . (m + n) (m + ^ — 1) . , rn

— J— a^-f-^^ ^^72 ^»' + --- [1]

822 COLLEGE ALGEBRA

414. If m is any positive integer, then by § 291 series (i)
is equal to (1 + «)"•. If n = — m, so that ti + w = 0,
Formula [1], § 413, becomes

(1 + xY< 1 + -J- X + Y"-^ ^

12

— m{— m — 1) {— m — 2)
"*" 12. 3

aj«4.... 1=1.

Divide by (1 -h «)"•,
1 I —^ , — w(— m — 1) 1

— 7/1

Comparing this theorem with Formula [A], § 288, we see
that [2] merely extends [A] to all negative integral exponents.

415. Let m = n in [1], § 413. Then,

i'^r+ 172 "'+''•; =^+T"+ 1.2- ■^+'"

Multiply by 1 + ^a: + ^5^j^ a:^ + ...,

and reduce the resulting right-hand member by [1], § 413.
llien,

. , n ^ n(n-l) . I* ^ . 3/1 , 3n(3n-l) ."

1 ' 12

12

{

Mtiltiply again by 1 4- 7 aj H — \ o *' "I >

reduce by [1], § 413, and repeat to q factors, q being any
positive integer. Then,

{

BINOMIAL THEOREM; ANY EXPONENT 828

i . n ■ 71(71 — 1) „ , V ^ , S'» , qniqn — l) « ,

1 ' 1.2

1 ' 12

Let » = — > in which p is any integer, positive or negative.
Then, *

^ q

aj2 +

by Pormula [A], § 288, or Formula [2], § 414, according as p
is positive or negative.

Take the arithmetical qth root of each side. Then,

. I f(f-i) f(f-0(f-0

(±+x) ±-r^x-f j^ 2 '^ -»- 1.2.3 r3]

P

If p is prime to g, (1 + x) s has q different values, and series [3] gives
the arithmetical or principal value.

416. On comparing [A], § 2SS, [2], § 414, and [3], § 415,
it will be seen that [2] and [3] are in form included in [A],
Hence, for any rational scalar value of n,

(l+x)-=l+-x + -l-^x» + -i-^^-ii^x« + ...[A]

provided x is so taken as to render the series of the expansion
convergent.

For irrational scalar values of ti we may substitute approxi-
mate rational values, carrying the approximation to any
required degree; or we may carry the approximation closer
than any assigned difference, however small in absolute
value, and thus prove that Formula [A] holds true for all
scalar values of n.

824 COLLEGE ALGEBRA

417. If n is fractional or negatiye^ the expansion of (a 4- 6)"

/ A-
must be in the form a"(l-|--l ifa>5; and in the form

418. Convergency of the Binomial S^es. In the expansion
of (1 -h xy, the ratio of the (r + l)th term to the rth term is
(§ 294)

n -f- 1
If X is positive, and r greater than n + 1, 1 is

T

negative. Hence, the terms in which r is greater than n + 1
are alternately positive and negative.

If X is negative, the terms in which r is greater than » + 1
are all positive. In either case we have

n — r -h 1
X, or

'-r-i^-^y--

as r is indefinitely increased, this approaches the limit — x.
Hence (§ 405), the series is convergent if « is numerically less
than 1.

419. Examples. (1) Expand (1 -h x)K

2 2-5

= 1 + ix - -— xa + \ x»

' 3 6 3. 6-9

The above equation is true only for those values of x that make the
series convergent.

(2) Expand

4T3

-i-t^, -i— i-t

(1 - x)-i = 1 - (- J)x + y ^x« ^' J' ^x> +

VjZr^ ' ^ *' • 1.2 1.2.8

1-6 1-5.9

== 1 + IX + -— X2 + — — --X» + .

^* ^4.8 4. 8. 12 ^

if X is so taken that the series is convergent.

BINOMIAL THEOREM; ANY EXPONENT 326

A root may of tjpn be extracted by meaas of an expansion.
(3) Extract the cube root of 344 to six decimal places^

344 = 343(1 + ^h) = 7«(l + yiy).

••• ^^^ = 7 (1 + Tir)*

L 3\343/ 12 \343/ J

= 7 (1 + 0.000971817 - 0.000000944)
= 7.006796.

Exercise 58
Expand to four terms :

1. (l + «)^ 4. (1-aj)-*. 7. y/2-Sx,

2. (1 -h x)K 6. (1 4- «)*. 8. -^(2 - xy.

3. ^ 6. . 9.

■v^l-a; -y/a^ - x^ ^{1 + 2 a^)*

Find:

10. The eighth term of (1 - 2 a-)*.

11. The tenth term of (a — 3 x)"^.

12. The (r -|- l)th term of (a + ar)*.

13. The (r -|- l)th term of (a»— 4 ««)-•.

14. The square root of 65 to five decimal places.
16. The seventh root of 129 to six decimal places.

16. Expand (1 — 2 a; + 3 x^"^ to four terms.

(1 + 2 xY

17. Find the coefficient of x^ in the expansion of ^ tt^'

(1 + 3 a5)'

18. By means of the expansion of (1 -h x)* show that "y^
is the limit of the series

1 1 1.3 135

■^"^2 2.22'^2.3.2« 2. 3. 4. 2*"^*".

826 COLLEGE ALGEBBA

19. Find the first negative term in ^he expansion of
(1 + x)"^.

20. Expand \ in ascending powers of x to six terms.

^1 — a;

21. If n is a positive integer, show that the coefficient
of af"^ in the expansion of (1 — a)"* is always twice the
coefficient of af*.

22. If m and n are positive integers, show that the coeffi-
cient of «"• in the expansion of (1 — a;)~""^ is the same as the
coefficient of x* in the expansion of (1 — x)~"''~\

23. Find the coefficient of «•'* in the expansion of a/ "~ ^
in ascending powers oi x, "*~ *

24. Prove that the coefficient of xT in the expansion of

SERIES OF DIFFERENCES

420. Definitions. If, in any series, we subtract the first term
from the second, the second term from the third, and so on,
we obtain a first series of differences ; in like manner, from this
last series we may obtain a second series of differences ; and
so on. In an arithmetical series the second differences all
vanish.

There are series, allied to arithmetical series, in which not
the first, but the second, or third, etc., differences vanish.

Tlvus, take the series

1 6 12 24 43 71 110
First differences, 4 7 12 19 28 39 •••
Second differences, 8 6 7 9 11

Third differences, 2 2 2 2

Fourth differences, 0 0 0 • • •

^2

b.

^8

h

a^

K

««

h

ttg

^6

aj

«1

«2

Cb

c^

<?*

• • 1

SERIES OF DIFFERENCES 827

In general, if a^, a^^ ^39 * ** is such a series, we have

First differences, 5i

Second differences,

Third differences, d^ d^ d^ d^

Fourth differences, e^ 62 e^

and finally arrive at differences which all vanish.

421. Any Required Term. For simplicity let us take a
series in which the fifth series of differences vanishes. Any
other case can be treated in a manner precisely similar.
From the manner in which the successive series are formed
we shall have :

^2 = ^1 "f- ^1 ag = a2 + ^2 = ^1 4- 2 ^1 4- Ci

^2 = ^1 + ^1 ^3 = ^2 + ^2 = ^w + 2 Ci 4- d^

^2 = Ci 4- ^1 C3 = Cg + (:?2 = ^1 4- 2 e^i -H Ci

c?2 = ^1 4- ^1 (:?8 = c?2 4- ^2 = ^1 4- 2 61

62 = ^1 63 ^ 62 ^^ ^1

a^ = fltg -f- ^8 ^^ ^1 "f" ^ ^1 H~ 8 Cj -{- di
^4 = ^8 + <^3 = ^1 4- 3 Cj + 3 c?i 4- ^1
^4 == ^8 4- ^8 = ^^1 + 3 c?i 4- 3 61
^4 = ^3 4- e, = (^1 4- 3 61

a^ = a^ 4- ^^ = tti + 4 ^1 4- 6 Ci + 4 c^^ 4- ^1
5^ = 5^ 4. C4 = ^1 4- 4 Ci + 6 c?i 4- 4 ^1
C5 = C4 + c?^ = Cj 4- 4 c?i 4- 6 «!

ag = ^5 + ^5 = »! 4- 5 ^1 4- 10 c, 4- 10 c?i 4- 5 6x
^6 = ^5 + C5 = ^ + 5ci + lOe^i + 10^1

aj = ttg + 6g = a^ 4- 6 6j 4- 15 c^ 4- 20 d^ + 15 e^
and so on.

328 COLLEGE ALGEBRA

The student will observe that the coeflficients in tH^^xpres-
sion for a^ are those of the expansion of (x -f- y)*, and simi-
larly for ag and a^. Hence, in general, if we represent ai, 61,
Ci, etc., by a, by c, etc., and put a^^j for the (n + l)th term,
we obtain the formula

n (n — 1) n(n — 1) (n — 2) , ,
^ 1x2 1x2x3

rind the 11th term of 1, 5, 12, 24, 43, 71, 110, • • •

Here (§ 420), a = 1, 6 = 4, c = 3, d = 2, e = 0 ; and n = 10.

.-. an = a + 106 + 45c + 120d
= 1+ 40 + 135 + 240 = 416.

422. Sum of the Series. Form a new series of which the
first term is 0, and the first series of differences ai, a,, a^, • . .
This series is the following :

0, «!, ai + a2, «! -f ^2 -f ^s, fl^i + ^2 + ^8 H" ^4j • • •
The (n -f l)th term of this series is the sum of n terms of
the series ai, aj, (I9, • "

(1) Find the sum of 11 terms of the series 1, 5, 12, 24, 43,
71, . . .

The new series is 0 1 6 18 42 85 156
First differences, 1 5 12 24 43 71

Second differences, 4 7 12 19 28

Third differences, 3 5 7 9

Fourth differences, 2 2 2

Here, a = 0, 6=1, c = 4, ^ = 3, e = 2; and n = 11.

.:s = a+ 116 + 55c + 165d + 330e
= 11 + 220 + 495 + 660
= 1386.

If s is the sum of n terms of the series ai, a^, a^y • • •

n(n — l) n(n — l)(n — 2)
1x2 1x2x3

(2) Pind the sum of the squares of the first n natural
numbers, 1^, 2% 3^ 4^, ..., n\

SERIES OF DIFFERENCES 829

Given Beries, 1 4 9 16 26 • • • n^

First differences, 3 6 7 9 • • •

Second differences, 2 2 2 • • •

Third differences, 0 0

Tiierefore, a = 1, 6 = 3, c = 2, d = 0.

These values substituted in the general formula give

. n(n - 1) - . n(n - 1) (n - 2) ^
' = "+172 ^'+ 1x2x3 ^'

= 5{6 + 8n - 9 + 2n« - 6n + 4}

= g{2«« + 3n + l} = "<'*+^><^'* + ^>.
6 6

423. Piles of Spherical Shot. I. When the pile is in the
form of a triangular pyramid the summit consists of a single
shot resting on three below; and these three rest on a course
of six ; and these six on a course of ten ; and so on, so that
the courses form the series

1, 14-2, 14-2 + 3, 1 + 2 + 3 + 4, .., 1 + 2+.. + W.

Given series, 1 3 6 10 16 * * •

First differences, 2 3 4 6 .*•

Second differences, 1 1 1 • * *

Third differences, 0 0

Here, a = 1, 6 = 2, c = 1, d = 0.

These values substituted in the general formula give

, n(n-l) - , n(n-l)(ii-2)

8 = n + —^ X 2 4 — i^ — ^

2 2x3

_^n8 4 3n«4 2n

"" 6

_n(n4 1)(to4 2)

"" 1 X 2 X 3 '

in which n is the number of balls in the side of the bottom course, or ths
number of courses.

880 COLLEGE ALGEBRA

II. When the pile is in the form of a pyramid with a square

base the summit consists of one shot, the next course consists

of four balls, the next of nine, and so on. Therefore the

number of shot is

ia + 2« + 3« + 4« + --- + nl

mvi I n(n + 1) (2 n + 1)

This sum is -^-- — 9 — :r — ^» (§ ^2)

1x2x8 \» /

in which n is the number of balls in the side of the bottom course, or the
number of courses.

III. When the pile has a base which is rectangular, but
not square, the pile terminates with a single row. Suppose p
the number of shot in this row ; then the second course con-
sists of 2 (^ -h 1) shot ; the third course of S(p + 2); and
the nth course of n(p -^ n — 1). Hence, the series is

Py 2p + 2, 3^-1-6, •••, n(p + n — 1).

Given series, p 2jp + 2 Sp + 6 4p + 12

First differences, p + 2 l) + 4 p + 6

Second differences, 2 2 • • •

Third differences, 0 • • •

Here, a = p, 6 = jp + 2, c = 2, d = 0.

These yalues substituted in the general formula give

, Mn-l), . ox j^ ^i^ - 1) (n - 2) ^ ^

= ?{6i) + 3(n - l){jp + 2) + 2{n - l)(n - 2)}
o

- ?(6jp + 3np - 3i) + 6n - 6 + 2n« - 6n + 4)
6

= 5(3np + 8jp + 2n2-2)
6

= 5(n + l)(3i) + 2n-2).
o

If n' denotes the number in the longest row, then n' =p + n — 1, and

therefore p = n^ — n + 1* The formula may then be written

« = ?(n + l)(3n'-n + l),

in which n denotes the number of shot in the width, and n' in the length,
of the bottom course.

SERIES OF DIFFERENCES 331

When the pile is incomplete compute the number in the
pile as if complete, then the number in that paxt of the pile
which is lacking, and take the difference of the results.

ExerciBe 59

1. Find the fiftieth term of 1, 3, 8, 20, 43, • • •

2. Find the sum of the series 4, 12, 29, 66, ••• to 20 terms.

3. Find the twelfth term of 4, 11, 28, 55, 92, • • •

4. Find the sum of the series 43, 27, 14, 4, — 3, ••. to 12
terms.

6. Find the seventh term of 1, 1.235, 1.471, 1.708, • • •

6. Find the sum of the series 70, 66, 62.3, 68.9, • • • to 16
terms.

7. Find the eleventh t«rm of 343, 337, 326, 310, • • .

8. Find the sum of the series 7 x 13, 6 x 11, 5 X 9, • • • to
9 terms.

9. Find the sum of n terms of the series 3 x 8, 6 x 11,
9x14,12x17,...

10. Find the sum of n terms of the series 1, 6, 15, 28, 45, • • .

11. Show that the sum of the cubes of the first n natural
numbers is the square of the sum of the numbers.

12. Determine the number of shot in a side of the base of a
triangular pile which contains 286 shot.

13. The number of shot in the top course of a square pile
is 169, and in the lowest course 1089. How many shot are
there in the pile ?

14. Find the number of shot in a rectangular pile having
17 shot in one side of the base and 42 in the other.

16. Find the number of shot in the five lowest courses of a
triangular pile which has 16 in one side of the base.

882 COLLEGE ALGEBRA

16. The number of shot in a triangular pile is to the num-
ber in a square pile, of the same number of courses, as 22 to
41. Find the number of shot in each pile.

17. Find the number of shot required to complete a rectan-
gular pile that has 15 and 6 shot respectively in the sides of
its top course.

18. How many shot must there be in the lowest course of a
triangular pile that 10 courses of the pile, beginning at the
base, may contain 37,020 shot ?

19. Find the number of shot in a complete rectangular pile
of 16 courses which has 20 shot in the longest side of its base.

20. Find the number of shot in the bottom row of a square
pile that contains 2600 more shot than a triangular pile of the
same number of courses.

21. Find the number of shot in a complete square pile in
which the number of shot in the base and the number in the
fifth course above differ by 226.

22. Find the number of shot in a rectangular pile that has
600 in the lowest course and 11 in the top row.

COMPOUND SERIES

424. A compound series is a series in which the terms are
the sum or the difference of the terms of two other series.
(1) Find the sum of the series

1 1 1 1

1x2' 2x3' 3x4' ■' n(n + l)'

Each term of this series may evidently be expressed in two parts :

1111 11

12 2 3 n n-f 1

Hence, the sum is

(i-5)-a-i)-a-i)--(;-dn)

COMPOUND SERIES 333

in which the second part of each term except the last is canceled by
the first part of the next succeeding term.

Hence, the sum is 1

n -f 1

As n increases without limit, the sum approaches 1 as a limit.

(2) Find the sum of the series

111 1

3x5 4x6 5x7 ' w(»-f-2)

The terms may be written,

i(i_i), ici.i), .., 1(1—1-).

2\3 6/ 2\4 6/ 2V11 n + 2/

^1 1.1.1. .111 1 1 1\

2\3 4 5 6 n 6 6 n n -f 1 11 + 2/

=1(1+1— i 1-)

2V3 4 n-fl n + 2/

7

24 2(n-fl) 2(n + 2)
As n increases without limit, this sum approaches ^7 as a limit.

Exercise 60

Write the general term, and the sum to n terms, and to an
infinite number of terms, of the following series :

1x4^2x5^3x6^ 1x5^5x9^9x13^

« 1.1.1. 6.6.6.

1x3^2x4^3x5^ 2x7^7x12^12x17^

5 X 11 8 X 14 ^ 11 X 17 ^

3x86x12^9x16^

Write the series of which the general term is :

3n-f 1

7. —. rr^ —7:r' 8.

w(w + l)(n + 2) * (n -f 1) (n 4. 2) (n -f 3)

834 COLLEGE ALGEBRA

INDETERMINATE COEFFICIENTS

425. If two series which are arranged by powers of x are
equal for all values of x that make both series convergent, the
corresponding coefficients are equal ea^h to each.

Let the equation

a 4- Ja -f- ca^ -f ^«* H = ^ -f -Ba; -f- Cx* -f- Dx^ H [1]

hold true for all values of x that make both series con-
vergent.

Since this equation holds true for all values of x which
make both series convergent, it holds true when a; = 0.

For a; = 0, a = A. [2]

Subtract [2] from [1], and since for any value of x that is
not 0 we may divide by x, divide each member by x ; then

b-hcx'{'dx^'\ = B-{- Cx-]- Dx^ H [3]

Then for x = 0, b = B. [4]

In like manner, c = C ; and so on.

Hence, the corresponding coefficients are equal each to each.

426. Partial Fractions. To resolve a fraction into partial
fractions is to express it as the sum of a number of fractions
of which the respective denominators are the factors of the
denominator of the given fraction. This process is the
reverse of the process of adding fractions that have different
denominators.

Eesolution into partial fractions may be easily accomplished
by the use of indeterminate coefficients and the theorem of § 425.

In decomposing a given fraction into its simplest partial
fractions, it is important to determine what form the assumed
fractions must have.

Since the given fraction is the sum of the required partial
fractions, each assumed denominator must be a factor of the
given denominator.

INDETERMINATE COEFFICIENTS 336

1. All the factors of the given denominator may be real and
different.

In this case we take each factor of the given denominator
as a denominator of one of the assumed fractions.

Thus, — = h

(« - 2) (X - 8) « - 2 X - 3

2. All the factors of the given denominator may be equal.
In this case we assume as denominators every power of the
repeated factor from the given power down to the first.

Thus, - — -- = - — — -f - — — +

(X - 1)» (X - 1)» (X - 1)2 X - 1

3. All the factors may be real and some equaL

In this case we combine the methods of the first two cases.

^ 4x8 - 63x2 + 338X- 619 A , B , C , D
Thus, ; -— = — • + — + +

(x--6)8(x-7) (x-6)8 (x-6)2 x-6 x-7

4. All the factors may be imaginary.

The imaginary factors occur in pairs of conjugate imaginaries
80 that the product of each pair is a real quadratic factor.

For example, in the fraction -r-^ ; . ^ ox / o . ^ ;::7

the factor a;«-f-2aj+5 = (a+T-2V^)(aj-f H-2V^),

and the factor aj«-4aj + 13 = (a + 2-3V^)(a + 24-3V^).

In this case we assume a fraction of the form —z—. r

x^± ax -\- 0

for each quadratic factor in the given denominator.

,^ 7x»- 6x2 + 9x4-108 Az + B Cx + D

Thus, — = ■ 1-

(x2 - 4x -f 13) (x2 + 2x + 6) x2 - 4x + 13 x^ + 2x + 6

5. Some of the factors may be imaginary.
In this case we combine the method of the fourth case with
the method of one of the preceding cases.

_. 13x»-68x + 95 A . Bx + C
Thus, — = +

(X - 6) (x2 - 6x + 13) X - 6 x2 - 6x + 13

336 COLLEGE ALGEBRA

Sx — 7
(1) Resolve -z ^p-z rr into partial fractions.

Zx-7 A . B

Assume = •

(« - 2) (X - 3) X - 2 « - 3

Then, 3x -7 = A(x - 3) -f B(z - 2).

.-. J. + jB = 3 and 3^ + 2 jB = 7 ; (§ 426)

whence, -^ = 1 and jB = 2.

m^ * 3x-7 1.2
Therefore, — = h

(x - 2) (x - 3) X - 2 X - 3
This identity may be verified by actual multiplication.

3

(2) Resolve -g — r- into partial fractions.

The denominators will be x 4- 1 and x^ — x + 1.

3 A , Bx + C
Assume — = 1-

x8 + l X + 1 X2-X + 1

Then, 3 = ^ (x^ - x + 1) + (Bx + C) (x + 1)

= (A-{-B)x^-\-(B-\-C-A)x+{A+C).
Therefore, S = A + C, B-\-C -A=Oy A -f jB=0, (§ 426)

and ^ = 1, jB = - 1, C = 2.

3 1 x-2

Therefore,

x8-fl x + l X2-X + 1

(3) Resolve ^ _j_i\2 ^^^ partial fractions.

The denominators will be x, x^, x + 1, (x H- 1)*.

4x8-x2-3x-2 ^ B C D

Assume = — | 1 f-

x2(x + l)2 X x2 x + 1 (x + l)a

... 4x8 - x2 - 3x - 2 = ^x{x + 1)2 + Bix + 1)2 + Cx«(x + 1) + JDx«

= (A + C)x^ + {2A + B-\-C-\- 2))x2 + (-4 + 2 B)« + B.

Therefore, ^ + C = 4, (§ 426)

2A + B+C + D = -1,

A + 2B = -Z,

B = -2;

and ^ = 1, B = - 2, C = 3, D = - 4.

^ , 4x8-x2-3x-2 1 2 3 4

Therefore, = 1-

x2(x+l)2 X x2 x+1 (x + l)«

EXPANSION IN SERIES 837

Eacercise 61

Resolve into partial fractions :

(a; + 4)(a;-5) (2a; - l)(a;-5) * a;»-l

6 , x — 2 ^ 7? — x

2. -7 Tzz-. 7Z' 4. -T — 'rn* 6.

(a;-f-3)(a; + 4) * a;«-3aj-10 * a;(aj«-4)

^ 3a;«-4 ,, 13a; -f- 46

7. -TTz — — =r* 11.

a;2(a;4-5) 12a;2 - 11a; - 15

7a;2-a; _ 2a;2-lla; + 6

8. •: i-rz 7^' 12.

(a;-l)2(a; + 2) a;^ - a;^ - 11 a; -f 15

2a;2-7a;-f-l .« a;«-15a;-18

9. : T-^ 13.

a;»-l (a;-h3)(a;-3)(a;-l)

7a;-l 3a;«4-12a; + ll

10. -:r—i 3 -^- 14.

6a;«-5a;-f 1 ' (a; + 1) (a; + 2) (a; -f- 3)

EXPANSION IN SERIES

427. A series which is obtained from a given expression is
called the expansion of that expression. The given expression
is called the generating function of the series.

Thus (§ 389), the expression is the generating function of the

1 — X

infinite series l+x + a;2 + x' + «'.

If the series is finite, the generating function is equal to
the expansion for all values of the symbols involved.

Thus, (l±^Y^l. + ? + 12x + 8«..

\ X / x' X

If the series is infinite, the generating function is equal to
the expansion for only such values of the symbols involved
as make the expansion a convergent series.

Thus, is equal to the series 1 + x + x^ + x* + • • • when, and

1 —X
jonly when, x is numerically less than 1 (§ 390).

838 COLLEGE ALGEBRA

X

(1) Expand ^ , in ascending powers of x.

Divide « by 1 + x^ ; then,

X

= X ~ X* -f xP — •

1-f x2

provided x is so taken that the series is convergent. By §§ 407, 406, the
value of X must be numerically less than 1.

X

(2) Expand -j ^ in descending powers of x.

Divide x by x^ H- 1 ; then,

__!__!_ 1^_

1 + x2 X x8 x^

provided x is so taken that the series is convergent. By §§ 407, 405, the
value of X must be numerically greater than 1.

In the two preceding examples we have found an expansion of

for all values of x except ±1. l + a;

X

(3) Expand -t— — ^ in ascending powers of x by the

X -J" X

binomial theorem.

^ = (1 + x2)-i = l-xa + x*

X

= X — X* -f x^ — • •

provided x is so taken that the series is convergent.

(4) Expand -r— — — — ^ in ascending powers of sc.

X ~y" X ~y" X

2 4- 3x
Assume = ^ + 5x + Cx^ + Bgfi + . , .

1 + X + X2

Clear of fractions,

2 + 3x = ^ 4- 5x 4- Cx2 + Dx* + - •

+ ^x + 5x2 + 0x8 + . . .

+ ^x2 + J5x8 -f . . .

EXPANSION IN SERIES 339

Hence, A = 2, B + A = S, C + B + A = 0, D+C + B = 0. (§ 426)

I

Whence, -4 = 2, J5 = 1, C = ~ 3, Z) = 2 ; and so on.
2 + Zx

1 + X + ««

= 2 + X - 3aj2 + 2x8 + aj* - 3x6 + . • .

The series is of course equal to the fraction for only such values of x
as make the series convergent.

Remark. In employing the method of Indeterminate Coefficients the
form of the given expression must determine what powers of the variable
X must be assumed. It is necessary and sufficient that the assumed equa-
tion, when simplified, shall have in the right member all the powers of x
that are found in the left member.

If any powers of x occur in the right member that are not in the left
member, the coefficients of these powers in the right member will vanish,
so that in this case the method still applies ; but if any powers of x occur
in the left member that are not in the right member, then the coefficient
of these powers of x must be put equal to O'in equating the coefficients of
like powers of x; and this leads to absurd results. Thus, if it were
assumed in Exam^^le (4) that

2 + 3x

1 -f X + x2

= J.X + Bx^ + Cx8 +

there would be in the simplified equation no term on the right correspond-
ing to 2 on the left ; so that, in equating the coefficients of like powers of
X, 2, which is 2x0, would have to be put equal to Ox^ ; that is, 2 = 0, an
absurdity.

(6) Expand (a — aj)* in a series of ascending powers of x.

Assume (a - x)^ = A -\- Bx -\- Cx^ + Zte* + • • •

Square, a-x-A^-\- 2ABx +(2 J.C + B^)x^ -{-(2 AD + 2BC)7fl+ ...

Therefore, by § 426,

A^= a, 2 AB = - 1, 2 AC + & = 0,2 AD + 2BC = 0, etc.,

and ^ = o* 5 = --?-r, C = -— ,, D = - ^

2o*

Hence,

8 a* 16 a^

xs x»

.1 1 Jb Jj JC

(cL — X)* = a* . • .

2a* 8a^ 16a*

340

COLLEGE ALGEBRA

(6) Expand

7 + x

Assume

(l4-aj)(l+x»)

7-faj _ A

in ascending powers of x,
. Bx + C

(1 -f X) (1 -f x2) 1 + X 1 + aj»

.-. 7 + x = (A H- (7) + (jB+ C)x-f (-4 + B)z^

/. -4 + 0 = 7, B+C = l, -4 + J? = a

Whence, -4 = 3, JB = -3, 0 = 4.

7 + x 3 4-3x

— +

4-3x

(1 + X) (1 + x2) 1 + X 1 + xa
1

and

Now,? — ^ = (4 - Sx)!"— ^"j = (4 - 3x) (1 - xa + X* )

= 4-3x-4x» + 8x« + 4x* —

-^ = 3 C^— ") = 3 - 3x + 3aJ» - 8«» + 8x* -

1 + x Vl+x/

7 + x

Add,

(l + x)(l+x2)

7-6x-x« + 7»*-

REVERSION OF A SERIES

428. Reversion of a Series. If y is the sum of a convergent
series in Xy the writing of a? in terms of a convergent series in
y is called the retJersion of the series.

Given y = ax -{- bx^ + ex* -f dx^ -\ , where the series is

convergent, to find x in terms of a convergent series in y.

Assume x = Ay -f- By^ + Cy* 4- ^* + . . .

In this series for y put ax + bx^ 4-005*4- cfoj* 4 — • >

aj* 4- . . .

a; = aAx 4- ^-4

x^-^t-cA

x^-^-dA

4-a^5

+ 2abB

4-^*5

4-a'C

4-2ac5

+ 3a^bC

4-a*2>

REVERSION OF A SERIES 341

Equate coefficients (§ 425),

aA = l; bA-ha^B^O'j c^ + 2aJ5 + a»C = 0;

dA '\- b^B -{- 2 acB -{- S a^bC -{- a*D = 0.

. 1 „ b ^ 2b^-ae

a a' a^

^ 5 b* — 5 abc + €L^d

Z) = z — ■ 9 etc.

(1) Revert y = a; -f a;' + «■ H

Here, a = 1, 6 = 1, c = 1, d = 1, • • •

A = l, B = -l, C = l, 2) = -l, ...
Hence, x = y — y^ + y* — 2^+ •••

(2) Revert y = «--2+"3"""4"'

Here, a = 1, 6 = — i, c = ^, d = — i, • . •

Hence, x = y + ^+^+^ + ...

Ezercise 62

Expand to four terms in ascending powers of x :

1 1 4 1-a; a;(a; — 1)
l-2a;' l4.a;4.aj»* (a + l)(i»*4-l)

1 5- 2a; ^ x^'-x-\-l

2 . 5 . 8. ■ — •

2 -3a; *"• 1 + ^ - a;^ x^x^ - 1)

l-hx 4a; -6a;' 2a;' -1

2 + 3a;' ^' l-2a; + 3a;«' a;(a;» + l)*

342 COLLEGE ALGEBRA

Expand to four terms in descending powers of x :

,A ^ ,o 5-2a; ,^ 3x^2

10. 12. :: 14.

2-\-x l-f-3aj-a;* x(x-r-iy

2 — x gg — a; + 1 a;2 — aj-f-1

11. — • lo. z :r-"' 15.

3-f« x(x-2) " (a; - 1) (a* + 1)

Eevert :

16. y = x — 2x^'{'Sx^ — 4tX^-\

x^ . x^ x^

17. y = a,_- + --- +

X^ . X^ . x^

18. 3^ = 0^ + 5^ + 2:3 + 3:^ +

RECURRING SERIES

1 + a;
429. From the expression -z ^ 2 ^® obtain by actual

division, or by the method of indeterminate coefficients, the
infinite series

1 + 3 a; + 7 ^2 -f 17 a« + 41 a* + 99 aj* + . . .

In this series any required term after the second is found
by multiplying the term before the required term by 2 a;, the
term before that by x^, and adding the products.

Thus, take the fifth term :

41x* = 2x(17x8)4-a:M7a').

In general, if w„ represents the nth. term,

^» = 2a;z^„_i + a;X-2-
A series in which a relation of this character exists is called
a recurring series. Recurring series are of the first^ second,
third, • • • order, according as each term is dependent upon one,
two, three, • • • preceding terms.

KECURRING SERIES 343

A recurring series of the first order is evidently an ordinary
geometrical series.

In an arithmetical or a geometrical series any required term
can be found when the term immediately preceding is given.
In a series of differences or in a recurring series several pre-
ceding terms must be given if any required term is to be found.

The relation which exists between the successive terms is
called the identical relation of the series ; the coefficients of this
relation, when all the terms are transposed to the left member,
is called the scale of relation of the series.

Thus, in the series

1 + 3x + 7 x2 + 17 x* + 41 X* + 99x5 + . . .
the identical relation is

aod the scale of relation is 1 ~ 2 x — x^.

430. If the identical relation of the series is given, any
required term can be found when a sufficient number of
preceding terms is given.

Conversely, the identical relation can be found when a
sufficient number of terms is given.

(1) Find the identical relation of the recurring series
1 + 4a; + Ua^ + 49a;» -f ITlx* + 597aj* + 2084a;« + ...

Try first a relation of the second order.

Assume Un = pxun-i + qxhin-a.

Put n = 3, and then n = 4.

14 = 4p + g,
49 = 14p + 4g;
whence, p = |, g = 0.

This gives a relation which does not hold true for the fifth and follow^
ing terms.

Try next a relation of the third order.

Assume w" = pxiLn - 1 + qxHbn-2 + rx»Un-8.

844 COLLEGE ALGEBRA

Put n = 4, then n = 6, then n = 6.

49= Up-h 4g+ r,
171 = 49i) + 14g4- 4r,
697 = 171p + 49g + 14r;
whence, 1> = 3, g = 2, r = — 1.

This gives the relation

which is found to hold true for the seventh term.
The accde of relation is 1 — 3 x — 2 x^ + x*.

(2) Find the eighth term of the above series.
Here, u^ = S xui + 2 x^ue — xhi^

= 3 X (2084 x«) -f 2 xa (697 afi) - x» (171 ap*)

= 7276x7.

SUMMATION OF SERIES

431. Infinite Series. By the sum of an infinite convergent
numerical series is meant the limit which the sum of n teims
of the series approaches as n is indefinitely increased. A
non-convergent numerical series has no true sum.

By the sum of an infinite series of which the successive
terms involve one or more variables is meant the generaHnt/
function of the series (§ 427), that is, the expression of whieh
the series is the expansion.

The generating function is a true sum when, and only when,
the series is convergent.

The process of finding the generating function is called
summation of the series.

432. Recurring Series. The sum of a recurring series can
be found by a method analogous to that by whicll tb^ sum of
a geometrical series is found (§ 276),

SUMMATION OP SERliiS 846

Take, for example, a recurring series of the second order in
which the identical relation is

or Ut -put^^ - quj,^^ = 0.

Let s represent the sum of the series; then,

-qs= -qu^ qu^^j^ - qu^_2 - qu^-i - qu^.

Now, by the identical relation,

1*8 —PU^ — qU^ = 0, U^ —PU^ — qU2 = 0, • • • , W„ —P^n^l — 2'^n-2 = ^•

Therefore, adding the above series,

Observe that the denominator is the scale of relation.
If the series is infinite and convergent, u^ and u^_i each
approaches 0 as a limit, and s approaches as a limit the frac-

1-p-q

If the series is infinite, whether convergent or not, this
fraction is the generating function of the series.

For a recurring series of the third order of which the iden-
tical relation is

% = P^k^i + qut^i 4- rui_s,

^efind ^ _ ^1-^(^2 -pui)-h(u, -pu^ - qu,)

1 —p — q — r

PUn -hqjUn-h u^-i) + r(u^ 4- t^n-i + u^ ->)

1 —JO — g — r

Similarly for a recurring series of higher order.

346 COLLEGE ALGEBRA

(1) Find the generating function of the infinite recurring
series

1 + 4a; + 13 a:^ -f- 43 »« -f 142a;* -f- •• •

By § 430 the identical relation is found to be

Hence, « = l + 4x+13x2 + 43x* + 142a;* -f • . •

-3x8= --3x- 12x2 -39x8 --129a;*

- x28= - xs- 4x8- 13x*

Add, (1 _ 3x - x2)« = 1 + X.

1 H-x
.-. s =

1 -3x-x»

(2) Find the generating function and the general term of
the infinite recurring series

1 _ 7a; _ a;2 - 43a;« - 49a;* - 307a;»

Here, Uk = xut - 1 + 6 xhik~~2.

« = l-7x- x2-43x8-49x*

- xs= - x + 7xa+ x8 + 43x*-f---
-6x28= -6x2 + 42x8+ 6x* + .»-

Add, (1 - X - 6x2) 8 = 1 - 8x.

_ 1 -8x _ 1 -8x

1 -x-6x2 " (1 + 2x) (1 - 3x)*
By § 426 we find '

l-8x _ 2 1__

(l+2x)(l-3x)"l + 2x l-8x' '

By the binomial theorem or by actual division,
1

1 +2x
1

= 1 - 2x + 22x2 - 28x8 + • . . + 2'-(- lyvr +
= 1 + 3x + 32x2 + 38x8 + . • • + 3'-x'- + . . .

• • • .

l-3x
Hence, the general term of the given series is

[2r+l(-l)r_3^jx^

SUMMATION OF SERIES 847

(3) Find the identical relation in the series

12 4- 22 4- 32 4- 42 4- 6^ 4- 6* 4- 72 4- •• •
The identical relation is found from the equations

16= 0p+ 4g+ r,
26 = 16p+ 0g + 4r,
36 = 26i) + 16g + 0r,
to be wjfe = 3 ujt-i — 3 Uk-2 + Uk—s*

Ezerdse 63

Find the identical relation and the generating function of :

1. l4-2a;4-7a;2 4-23a;«4-76a;*H

2. 3 4- 2a; 4- 3a;2 4- 7x« 4- 18a;* H

3. 3 4- 5 a; 4- 9 a;2 4- 15 a;« 4- 23 a;* 4- 33 a;^ 4- 45 a;« H

4. 1 4- 4a; 4- 11 a;2 4- 27a;« 4- 65a;* 4- 158a;5 4- 388 a;« 4- • • •

Find the generating function and the general term of :

6. 2 4- 3a; 4- 5a;2 4- 9a;« 4- 17 a;* 4- 33a;^ 4

6. 7 - 6a; 4- 9a;« 4- 27a;« 4- 54x* 4- 189a;^ 4- 567a;« 4- •••

7. l4-5a;4-9a;2 4-13a;«4-17a;*4-21a;*^4

8. 1 4- aJ - 7 a;8 4- 33 a;* - 130a;^ 4- 499 a;«

9. 3 4- 6a; 4- Ma;* 4- 36a;« 4- 98a;* 4- 276a;^ 4- 794a;« 4- •••

10. 1 4- 4a; 4- 9a;2 4- 16 a;« 4- 25 a;* 4- 36a;^ 4- 49 a;« H

Find the sum of n terms of :

11. 2 4- 5 4- 10 4- 17 4- 26 4- 37 4-50 4- •••

12. 1« 4- 2« 4- 3« 4- 4« 4- 5« 4- 6« 4- • • •

13. l4-2a;4-3a;2 4-4a;«4-5a;*H

14. l4-3a;4-6a;*4-10a;«4-15a;*4-21a;«4-28a;«4----
16. !« 4. 32 4- 5* 4- 7* 4- 9* 4- 11* 4- • • •

16. ia4.5«4-9»4-13*4-17*4-21«4-- •

848 COLLEGE ALGEBRA

INTERPOLATION

433. As the expansion of (a + by by tba bmoinial thaorem
has the same form for f ractioual as for integral values of w,
the formula (§ 421)

7i(n — V) , n(n — l)(n — 2)^ ,

may be extended to cases in which n is a fraction, and be
employed to insert or interpolate terms in a series between
given terms.

(1) The cube roots of 27, 28, 29, 30 are 3, 3.03669, 3.07232,
3.10723. Find the cube root of 27.9.

3.00000 3.03669 8,07232 3.10723

First differences, 0.03659 0.03673 0.03491

Second differences, - 0.00086 - 0.00082

Third differences, 0.00004

These values substituted in the general formula give

S,»„„..3»Hi(-i)(-i^).^(-i)(-ll)(«^)

= 3 + 0.082931 + 0.0000387 + 0.00000066
= 3.03297.

(2) Given log 127 = 2.1038, log 128 = 3.1072, log 129 =
2.1106. Find log 127.37.

2.1038 2.1072 2.1106

First differences, 0.0034 0.0034

Second differences, 0

Therefore, the differences of the second order Taiiish, and the required
logarithm is

2.1038 + ^oV of 0.0034 = 2.1038 + 0.001268

=x: 2.1061.

EXPONENTIAL AND LOGARITHMIC SERIES 849

(3) The latitude of the moon on a certain Monday at noon was
V 53' 18.9", at midnight 2* 27' 8.6"; on Tuesday at noon
2° 68' 55.2", at midnight 3* 28' 5.8"; on Wednesday at noon
3* 54' 8.8". Eind its latitude at 9 p.m. on Monday.

The series expressed in seconds and the successive differences are
6798.9 8828.e 10736.2 12485.8 14048.8

2020.7 1906.6 1750.6 1563.0

- 128.1 - 156.0 - 187.6

- 32.9 - 31.6

1.3

As 9 hours = | of 12 hours, n = |.

Also, a = 6798.9, b = 2029.7, c = - 123.1, d = - 32.9, e = 1.3.

These values substituted in the general formula

, . n(n-l) . n{n-l)(n-2) , , n(n- l)(n -2)(n - 3) ,

a-\- iiJb H — ^^ '^c-\ — ^ — -d H — ^ '— — ~-^ ^e H

1x2 1x2x3 1x2x3x4

give

(

„..,.!,^.„.?(-!)(-!fl).|(-l)(-?)(-!H)

-!(-l)(-D(-DO-

= 6798.9 + 1622.28 + 11.64 - 1.29 - 0.03 • • •

= 8331.4

= 2*^18' 51.4".

EXPONENTIAL AND LOGARITHMIC SERIES

434. EzponentUl Series. By the binomial theorem,

-• . iV'^ -i . V. 1 . nx(nx-l) 1
nj n 1x2 n^

nx (nx — 1) (nx — 2) 1
■^ 1x2x3 ^n«"^

/ 1\ / IV 2\

X\ X ) X\ X ][ X 1

=l+x + ^ + -^ + ... [1]

360

COLLEGE ALGEBRA

This equation is true for all real values of x. It is, how-
ever, true only for values of n numerically greater than 1,

since - must be numerically less than 1 (§ 418).

n

As [1] is true for all values of x, it is true when a; = 1. -

1 + 1 +

n ^ \ nj\ n)

^

13

- [(-0"]'=(-^

Hence, from [1] and [2],

. n \ nJ \ nj
l4-l + ^_ + ^ ^ -

+

)nx

[2]

(§ 299)

X

= l + a; +

(-4),4-0(-!)

[2

[3

+

This last equation is true for all values of n numerically
greater than 1. Take the limits of the two members as n
increases without limit. Then (§ 383),

(

1+1+^+1+

)■=

^ + ^"+1 + 1 +

[3]

and this is true for all values of x. It is easily seen by § 405
that each series is convergent for all values of x.

The sum of the infinite series in parenthesis is called the
natural base (§ 302), and is generally represented by e\
hence, by [3],

^=^ + ^+l2+[3 +

[A]

EXPONENTIAL AND LOGARITHMIC SERIES 361

To calculate the value of

e, we proceed as follows :

1.000000

2
3
4
6
6
7
8

\

1.000000

0.600000

0.166667

0.041667

0.008333

0.001388

0.000198

0.000026

V

0.000003

Add,

c = 2.71828.

To ten places,

e = 2.7182818284.

435. In [A] put

ex in place of x ; then,

e^^

= 1 4- caj

c^x^ c^x^

[2 ■ [3
Put e*" = a ; then c = log^a, and e*^ = a*'.

.-. a* = 1 4- a log,a + ^ .°' + ^ 13 '^ + • • •

Series [B] is known as the exponential series.

Series [B] reduces to [A] when we put e for a.

436. Logarithmic Series. In [A] put e* = 1 -f y ; then,

X = log, (1 -f y), and by [A],

.4

[B]

x^ . x}

x^

Revert the series (§ 428), and we obtain

= .y — TT + ir — r + '*-

x^y

But

a; = loge(l + y).
.Mog,(l+y) = y-|' + |'-^V-..

[C]

852 COLLEGE ALGEBRA

Similarly from [B],

The series in [D] is known a8 the logarithmic series ; [D]
reduces to [C] when we put e for a.

In [C] and [D], in order to have the series convergent, the
value of y must lie between — 1 and -f 1, or be equal to -f- 1,
by § 409, Example (1).

437. Modulus. Comparing [C] ^nd [D], we obtain

loga(l -hy)^ j^ log,(l 4- y) ;
or, putting iV f or 1 -|- y,

log„i\r«= logeiVT.

log, a

Hence, to change logarithms from the base e to the base a,

multiply by z = log^e ; and conversely (§ 318).

The number by which natural logarithms must be multi-
plied to obtain logarithms to the base a is called the modulus
of the system of logarithms of which a is the base.

Thus, the modulus of the common system is logjoe (8 320).

438. Calculation of Los^arithms, Since the series in [G] and
[D] are not convergent when x is numerically greater than 1,
they are not adapted to the calculation of logarithms in gen-
eral. We obtain a convenient series as follows :

The equation

log,(l+y) = y-|' + f-^ + -.- [1]

holds true for all values of y numerically less than 1 ; there-
fore, if it holds true for any particular value of y, it will hold
true when we put — y for y; this gives

log.(l-y) = -2/-f-f-^--.. [2]

EXPONENTIAL AND LOGAKITHMIC SERIES 858

Subtract [2] from [1]. Then, since

log, (1 + y) - log, (1 - y) = log, (jz:^)

Put y = -^—-', then t±i?t=5Jli,

and

log. ( ^"7~) = ^^' (» + !)- log.*

1 . 1

\2^ + l

+

-h

+

...j. [E]

3(2^4-1/ 5(2«-|-l/

This series is convergent for all positive values of z.
Logarithms to any base a can be calculated by the corre-
sponding series obtained from [D] ; viz.,

log„(« + l)-log««

1 1

log,a\2z + l

'^ S(2z + iy'^ 8(2z + iy

+

••)• m

(1) Calculate to six places of decimals log, 2, log, 3, log, 10,
logioe.

In [E] put z = 1 ; then 2z + 1 = 3, loggZ = 0,

and

3 2 8 2

^ 33x38 6x36 7x37

+

The work may be arranged as follows i

3
9
0
0
0
9
9

2.0000000

0.0666667+ 1 = 0.6666667

0.0740741 -i- 3 = 0.0246014
0.0082306+ 5 = 0.0016461

0.0000146 -i- 7 = 0.0001306
0.0001016+ 9=10.0000113

0.0000113 + 11 = 0.0000010

0.0000013 + 13 = p.OOOOOOl
log«a = 0.603147

864 COLLEGE ALGEBRA

lOgeS = 10ge2 + ? + -A^ + — ?— + • • •

6 3 X 68 6x6*
= 1.0986123.
loge9 = loge{32) = 2 loge3 = 2.1972246.

2 2 2

lOgelO ='l0ge9 + — + + + . . .

Be 5e ^ JQ ^ 3 >^ 198 ^ 5 X 19* ^

= 2.1972246 + 0.1063606
= 2.302686.

logioe = — i— = 0.434294.
loge 10

Hence, the modulus of the common system is 0.434294 (§ 320).
To ten places of decimals :

lo&.10 = 2.3026860928,

logioe = 0.4342944819.

For calculating common logarithms we use the series in fF]

logio(« + l)-logio«

= 0.8685889638 (^-^+.,,^ \ ix8 + ;t/o ViN5 + -'Y

\2«-hl 3(2«-fl)* 5 (2 « 4-1)* /

(2) Calculate to five places of decimals logioll.
Put z = 10 ; then 2 z + 1 = 21, logz = 1,

logll = 1 + 0.868688/'— + — —^ + — —^ + . ..V
^ \21 3 X 21* 6 X 21* '

)

21
441

0.868688

0.041361 -r- 1 = 0.041361

0.000094 - 3 = 0.000031

0.041392

1^

logioll = 1.04139

In calculating logarithms the accuracy of the work may be tested
every time we come to a composite number by adding the logarithms of
the several factors (§ 300). In fact, the logarithms of composite numbers
may be found by addition, and then only the logarithms of prime
numbers need be found by the series.

EXPONENTIAL AND LOGARITHMIC SERIES 365

439. Limit of f 1 -f - j . By the binomial theorem,

(. , a;\» . . X , n(n — l) x^

nj n 1x2 TT

n(n-l)(n-^^2) X'
"^ 1x2x3 "^n^^"'

. n „ \ nJ\ nJ ,

i^ [3 ^

This equation is true for all values of n greater than x
(§ 418). Take the limit as n increases without limit, x
remaining finite.

= „"f'«(l+^)" (§434)

Exercise 64

1. Show that the infinite series

1 +^-A.+

1x2 2x22'3x2« 4x2*
is convergent, and find its sum.

2. Eind the limit which Vl-h^wc approaches as n approaches
0 as a limit.

1/12 3 \

3. Showthat - = 2f r^+pg+ry + •••)•

4. Calculate to four places, logg4, log^S, logg6, log^T.

lOgef - j = .^..o^.o + O./a .. ^ + ^..L.r; + •• •

356 COLLEGE ALGEBRA

6. Find to four places the moduli of the systems of which
the bases are 2, 3, 4, 8, 6, 7.

6. Show that

5 7 9

X 2 X 3 "^ 3 X 4 X 5 "^ 6 X 6 X 7

7. Show that
log.a-log,^ = -^ + ^(^-^j-|-3(^-^j+...

8. Show that, if x is positive,

2» 3» 48

9. Show that l+rTj+nT+rrH = 5e.

L? E [4

X =^ 1 —'^ -^7- — 7--\ — Y = x —7- -k-7- — n-H —

[2^|I [6^ ' [3^[5 [7^

05^ X* X* . .. X' . X* X'

11. Expand ^ in ascending powers of x.

12. Expand -. in ascending powers of m.

13. Find the sum of n terms of the series

a a (a + x) a (a + a:) (a + 2 x)
^"^ Z>(Z> + «) ft(ft+i»)(^ + 2x) ^*"'

14. Show that, if n is any positive integer,
n

n(n — V) n{n — V){n — 2) ^

w + 1 "" (n + 1) (^ -h 2) (w + 1) (w + 2) (n + 3)

n(yi-l)(yi-2).»-3»2-l ^*1
(n + l)(n + 2)...(27i-l)(ail) 2'

CHAPTER XXVI
CONTINUED FRACTIONS

440. A fraction in the form

a

/+etc.
is called a continued fraction.

A continued fraction in which each of the numeratorf of
the component fractions is + 1 and each of the denominators
is a positive integer, as

1

.+ '

r + etc.
is called a simple continued fraction.

We shall consider in this chapter some of the elementary
properties of simple continued fractions.

441. Any proper fraction in its lowest terms may be con-
verted into a terminated simple continued fraction*

Let - be a fraction in its lowest terms.
a

Then, if p is the quotient and c the remainder of a -i-b,

^^1^_1

a a , e

357

858

COLLEGE ALGEBRA

If ^ is the quotient and d the lemainder of h

111

Hence,

p

H

e

P +

a

1

1

,d

P +

fif +

r-feto.

The successive steps of the process are the same as the steps
for finding the H.C.F. of a and b\ and since a and h are prime
to each other, a remainder, 1, will at length be reached, and
the fraction terminates.

Observe that ^, g^, r, • • • are all positive integers.

442. Convergents. The fractions formed by taking one, two,
three, • • • of the quotients Py q, r, " are

1

P

P + - p H

2

9 •

,+1

which simplified are

1

qr-{-l

p pq + '^ (pq + 1)^ -^p

and are called the first, second, third, • • • canvergents respec-
tively.

The value of the complete continued fraction is called
briefly the complete value.

443. The successive convergents are alternately greater than
and less than the complete value of the continued fraction.

CONTINUED FRACTIONS 869

Let X be the complete value of

1

1

i> +

,+ '

r + etc.
Then, since p, q, r, -" are positive integers,

P<P-^ T'

r + etc.

1 1

->

p-\

,^ '

r -t- etc.

That is, - > a.

P

1

Again, 9 <q-^

r -f etc.

1 1

->

9 1

^ 9 +

r -f etc.

^ <— 1

p + - p-{-

9 +

r + etc.

That is, T < X ; and so on.

p + -
9

Corollary. Hence, if — > — are consecutive convercrents to
X, then a > or < — according as — > or < — > and, there-
fore, a'* > or < — z according as — > or < — •

V2^ Vi V,

\

860 COLLEGE ALGEBRA

Therefore, xi^^ — u^ and wjVj — u^vi are gimiiltaiieouBly

positive or simultaneously negative. Therefore, ^

is always positive. * *

Note. Continued fractions are often written in a compact and con-
venient form ; thus, the fraction

1

a +

6 + -L

1

C +

..!

may be written in the form a + -

444. If — > — > — are any three consecutive convergentSy

and if m^, nij, nig are the quotients that produced them, then

Us ^ mgUg + Ui
Va mgVj -fvi

For, if the first three quotients are p, q, r, the first three
convergents are (§ 442)

1 q qr + 1

— )

p pq + 1 (pq -{'l)r-\-p

[1]

From § 442 it is seen that the second conyergent is

formed from the first by writing in it ^ 4- - for p ; and the

1 ^

third from the second by writing q -{-^ for q. In this way

r

any convergent may be formed from the preceding convergent.
Therefore, — is formed from — by writing i7u-\ for nu.

The numerator of the third convergent in [1] is equal to
r X (second numerator) -f (first numerator).

The denominator of the third convergent in [1] is equal to
r X (second denominator) -f (first denominfttof).

J

CONTINUEP FRACTIONS 881

Assume that this law holds true for tb9 thi^rd of tb^ tbi^Qe
consecutive convergents

Up Ui Wa

Vq Vi Vj

3othat t^^m^z^^+j^^

Then, since — is formed from — by using m^-i for m^,

Ij^ !■ a ■ * ' •

(^2 + — J^l + W

Substitute u^ and Vj for their values m^u^ + t«o 3^^^ '^h^i + ^o-
Then ^s ^ ^8^2 + ^i

Therefore, the law still holds true; and as it has been
shown to be true for the third convergent, the law is general
by mathematical induction.

Corollary. If -^r- > — ? — > • -i ~j • • • are the convergents to

1 Wa v, f^

U" , ^»-2 . 1

W«-i Wn-l ^n-l

. 1 1

= ?»„ H ^ , -^

W„_8
J. 1 1 1

w^-i + w«- 2 + 1V:J

862 COLLEGE ALGEBRA

and so on, until finally

u^ 1

In like manner, it may be shown that

V. .11 1

" =^n +

445. Examples. (1) Find the continued fraction equal to
f^f and also the successive convergents.

Following the process of finding the H.C.F. of 31 and 76, the succes-
sive quotients are found to be 2, 2, 2, 1, 1, 2. Hence, the equivalent
continued fraction is

1

2 + -^

2 +

2 +

l + -i

-I

or, in the compact form,

111111

2 + 2 + 2 + 1 + 1 + 2*

To find the successive convergents, write the successive quotients in
order in a line, and in the next line below write the initial convergents ^
and f to the left of the first quotient ; then, beginning with these initial
convergents, form the successive convergents as follows: Multiply the

i . I- of any known convergent by the quotient next on its

right and to the product add the -j , . i- of the convergent next

preceding. The sum is the i . . ^ I of the next succeeding con-
^ ® 1^ denominator J ^

vergent (§ 444). Write tbis convergent immediately below the quotient

producing it.

Thus, Quotients = 2, 2, 2, 1, 1, 2.

Convergents = i, f , i, f, A» tV. ih H-

#

CONTINUED FRACTIONS 868

If the given fraction is improper with an integral part n, the initial

In 1

convergents are - and ~ * Thus, the zeroth convergent is always - and

the first convergent is the integral part of the continued fraction, or is
zero if there is no integral part.

(2) Find the successive convergents to the continued fraction

•^^2 + 3 + 4 + 5
Quotients = 2, 3, 4, 6.

Convergents = J, {, f , V^, )i. f If

EzerclBO 65

Express the follovring numbers as simple continued fractions
and find the successive convergents :

1. f J. 4. J,^-. 7. 0.0498756.

3. §|f. 6. 0.43589. 9. 2.44949.

10. Find the value of

111. 111. 11111

4-1-3-1-2' 2-1-3 + 7' l-f-2-M-f 4-1-5*

11. Find a series of fractions converging to the continued
fraction that has as quotients 2, 1, 3, 1, 7, 2, 1, 2, 6, 4.

446. The difference between two consecutive convergents

Ui . Ua . 1
— and -^ is

Vi V2 V1V2

The difference between the first two convergents is

1 9

p pq + 1 p(pq + l)

Let the sign '^ stand for the words the difference between^
and assume the proposition true for — and — •

364 COLLEGE ALGEBRA.

Then !f2 ^ !^ = t^Qt^i - u^Vq ^ J^

^ Vo Vi VqVi VqVi

But

!f? 1^1 _ U2V1 ^ U1V2 _ (rHzUi + ^o) ^1 ^ ^1 (^a^i + ^0)

if we put for i^j and Vg their values, mgWi + Wo ^i^d mgVi 4- Vq.
Eeduce, — /^ — = = (by assumption).

Hence, if the proposition is true for one pair of consecutive
cbnvergents, it is true for the next pair; but it has been
shown to be true for the first pair ; therefore, it is true for
evert/ pair by mathematical induction.

Corollary. If — and ~ are two consecutive convergents,

Vi v^

U1V2 — u^Vi = -f- 1 or — 1 according as -^ > or < — •

Vi V2

447. Since by § 443 the complete value of x lies between
two consecutive convergents — and — > the convergent —

Vl Vj Vi

differs from a; by a number less than — '^ — > that is, by a

Vi V2

number less than ; so that the error in taking — for x is

ViV% Vi

less than > and therefore less than => as Va>m,Vij

since V2 = Wa^i -f- Vq,
Hence, the best convergents to select are those immediately

preceding large quotients.

ft

448. Any convergent — is in its lowest terms; for, if

Ui and Vi had any common factor, it would also be a factor
of i/^Va ^ U2V1, that is, a factor of 1.

449. The successive convergents approach more and more
nearly to the complete value of the continiced fra^stion.

CONTINUED FRACTIONS 366

Let — > — > — be consecutive convergent s.

Vq Vi V2

Now, — differs from x. the value of the complete fraction,

Va 1

only because mo is used instead of m- A

^ . ^ m, -f-etc.

Let this complete quotient, which is always greater than
unity, be represented by M,

Inen, since — = 9 x =

• » X

V2 m^Vi -f- Vq Mvi -f- Vo

Uy MUi + Wo Ui UqVi '^ ICiVq

/^»* — — — — ■ ■ ^^^ — — * ■ -■ —^

Vi Mvi + Vq Vi Vi (Mvi -f Vo) Vi (Mvi -f Vq)

- Uq Uq Mill + t^o M(uoVi '^ WiVo) ^

ana — *^ x = — ^^ —- = — —-- r-^ =

Vq Vq MVi-hVo Vo(MVi + Vo) Vo(MVi4-Vo)

Now, Km and Vi > t;©, and for both these reasons

Vi Vo
That is, — is nearer to x than is — •

Vi Vo

Corollary. Hence, the oc?rf-numbered convergents to the
continued fraction Ci H — "~ i ' * ' ^^^^^ ^^ increasing series

of rational fractions continually approaching to the value of
the complete continued fraction ; and the even-numbered con-
vergents form a decreasing series having the same property.

450. The /ration -^ is greater than or less than x^ accord-
ing as -^ is greater than or less than — •

For (§ 449), J/ > 1, Wj > -^i, and v^ > v^.

,\ MHL2V2 — UiVi >0.

366 COLLEGE ALGEBRA

Hence, (MSi^^^ — '^\^i){^i^2 — ^2^1) > 0 or < 0,
that is, MhbiU2V2 -\- U1U2V1 > or <ilf ^2*^iV2+^i^iV2,

and uiu^ (MV2 + Vi)^ > or < VjVa (Mu^ + t*i)*,

and, therefore, ^ > or < (^^^'

V1V2 \ Mv2 4- ^1/

according as — is > or is < — •

But ^ = ^^f^±^.

U1U2 ^ ^„ J. Ui ^ ^tu

.'. > or < JB^ according as — > or < — •

Corollary. w always positive,

U1V2-U2V1

451. Any convergent — is nearer the complete value x than
any other fraction with smaller denominator.

Let T be a fraction in which h < v^,

o \

If - is one of the convergents, x ^ -> — '^ x, (§ 449)

If — is not one of the convergents, and is nearer to x than

is — ) then, since x lies between — and — C§ 443), 7- must

^1 u. u, ^1 ^« ^

be nearer to — than is — •

V2 v^

rpv 4. • ^ ^2 ^ ^l ^2 V^a ^Uj) ^

That IS, _ ,^ -i5 <; _L ^ _* , or -^= -—^ <

b V2 Vi V2 v^ VjVj

Since h < v^, this would require that Vga '^ uj><l. But
Va^ '^ u^ cannot be less than 1, for a, ft, t^j, v, are all integers.

Hence, ~ is nearer to x than is -7 •

Vi o

CONTINUED FRACTIONS 367

Examples. Express VS in the form of a continued fraction. '
Let Vs = 1 + - (since 1 is the greatest integer in Vs).

X

Then, i = Vs - 1.

X

1 V3 + I
.'. X = — — =

V3.-I 2
Let = 1 + - ^ since 1 is the greatest integer in J •

_ 1 V3 + I , V3-I

Then, - = 1=

y 2 2

2 V3 + I

V3-I 1

Let = 2 + -( since 2 is the greatest integer in j .

Then, 1 = :1±1_2=V3-1.

z 1

1 V3 + 1

.'. z = — — = •

V3-I 2

This is the same as x above ; hence, the quotients 1, 2 will be con-
tinually repeated.

.-. V3 = l + ^ ,

1+ ^

2 + etc.
of which will be continually repeated, and the whole expression

may be written 1 i

1 + "■ ~'
1 + 2

• •

The convergents of the continued fraction - - are

.'. the convergents to Vs are 1, 2, J, |, ff , f f , JJi * ' '

This example shows how any pure quadratic surd may be
converted into a non-terminating simple continued fraction.

368 COLLEGE ALGEBRA

The following is another example with the work of con-
version exhibited in full in a convenient arrangement.

1 V7 + 2 «i

V7 + 2 , , V7-I , , 2 ,1

/. 2Ji = — = 1 + = 1 + -p = 1+ - •

3 3 V7 + 1 «i

V7 + I , , V7-I , , 3 ,1
... x^ = = 1 H = 1 -^ — ;= = IH

2 2 V7 + 1 X,

V7 + I , . V7-2 , , 1 ,1

.-. Xn = = 1 + — = 1 + -7= = 1 + - •

3 3 V7+2 »4

/. X4 = :; = 4 + = 4 + — = 4 + — .

1 1 V7 + 2 35i

...v7=2+i 1 1 i.
^1+1+1+4

QnoUents = 1, 1, 1, 4, 1, 1, 1, 4.

Convergente = I f , f , f, f , f J, f f, H» W» «*•

452. A non-terminating simple continued fraction in which
the denominators recur, and recur always in the same order^ is
called a simple periodic continued fraction.

453. Every qiuidratic surd may be converted into a simple
periodic continued fraction.

It is sufficient to consider the case of a pure quadratic surd,
as a mixed surd can always be reduced to a pure surd.

Let N be any given integer not a square, and let qi be the
integer next less than ViV, hence ViV^ — g'l < 1. Then, arrang-
ing as in the last example, we have

ViV = gr, + '^^ ^^ = q^-{'

ri

in which ki = qi and ri = N — q^*;

= 2^2 H = 2^2 +

if ki = rig'j — A^i and r^ = ^

CONTINUED FRACTIONS 869

■y/N + ka . VN-k

3 I ^8

^2

if k^ = r^q^ — k^ and rg =

' N^k,^

^2

)

r.

^n-l n.-l Vi\r4-A:„ ^_;^2

if A:, = r^-iqn - ^»- 1 and r„ =

K-i

Now, the numbers r^ Tj, Tg, • • • and kg, kg, • • • are positive
integers.

^n — 1 ^n ^n-f 1

For, let -^^-^ > -^ ) -^^^^ be the consecutive convergents cor-
responding to the partial quotients qn-n qro ^n+i- The com-

plete quotient next after q^ is -y and using this

7*

instead of q^ to form the next convergent, we obtain the
complete value ViV^ of the continued fraction, instead of

the convergent value -^^^ • (§ 445)

Equate rational and irrational parts of this equation.
Then, k^u^ + ^n^*„_i = v«^,

and k^v^ + V^n-i = "^n-

and r„ = — -^^^ = [2]

Hence, by § 450, Corollary, and § 443, Corollary, k^ and
r^ are both positive, and since w„-iv„ — '2*„v„_i =± 1, they

370 COLLEGE ALGEBRA

are also integral if t/._i, u^ v^-u and v. are int^^; that is,
\f ki,k^ k^ • • -f ^n-i an^ ^v ^» ^» * * '9 ^n-i are integers. Now,
ki = qi and rj = A' — q^^ and are, therefore, positiye int^^rs.
Therefore, A^j and r, are positive integers. Therefore, k^ and
r, are positive int^ers, and so on.

Therefore, ViV has been developed into the simple continned
fraction, ^ ^

gr, -f- — — • • •

The greatest value that any of the numbers k^ k^ k^ •••
can have is q^-^ for r,_ir, = A' — k^, and r,_i and r^ are posi-
tive integers; hence, k^ cannot be greater than the greatest
integer in ViV, that is, k^ cannot be greater than ^,.

The greatest value that any of the numbers r^r^r^ "• and
the greatest value that any of the numbers s^u ?» y» • • • can
have is 2 g'l ; for k^_^ -\-k^ = r^^^q^. Therefore, r^-ifl', cannot
be greater than 2 qi, and as neither r^_i nor q^ can be less than
1, therefore neither of them can be greater than 2 ^j.

As the numbers ki, k^ k^ "- cannot have more than q^ dif-
ferent values and the numbers r^, r^, r^ • • • cannot have more

V^4- k
than 2qi different values, the complete quotients -9

— > .•• cannot have more than 2gi* different valnes,

^2

and therefore one of these values must recur. If this happens,
all the quotients following the first recurring quotient must
recur in order, and consequently all the partial quotients recur
in a cycle which cannot have more than 2 qi^ terms.

Hence, Vn can he developed ifUo a simple periodic continued
fraction,

454. There cannot be more than one non-recurring partial
quotient in the conversion of ViV.

For, if possible, let there be t partial quotients that do not
recur, t being greater than 1, and let

CONTINUED FRACTIONS 871

/l^u^ll lil i

in which q^ is not equal to q^.

Let y = ^t+i + '- — , - — . ••• , r-^

and let — denote the nth. convergent to "wN.
Then, ViV^=^i4--^-^--.-^-

2^2 + 2^8+ -\-qt-\-y
,11 11 11

2^2 4- $'8 4- -\-qt-\-qt^i+ H-fi'm + y

[1]

^m— 1-1 — ]

Wm.

Eliminate y from these equations,

+ W|W«_ 1 - w^-i^m = 0. [2]

i^<-iv ^; 7. — J4-w<-iVm4-iv :n :: — J = ^-

I ^ _i ^< — 2 ^ ^m — 2 I

+ «.-.«„-. (.. + ^-,„-^;)=o. [3]

Now, since ^ > 1, '""^ is a positive proper fraction, and

^m — 1

is zero if ^ = 2, and is a positive proper fraction if ^ > 2.
Hence, -^^ ^^^ is a proper fraction, say ± /. So also

-^^ ^^^^ is a proper fraction, say ±/'. Hence, [3] may

be written

t*«-iv,_i (S'l -qm±f) + Ut _iv«-i (S'l - y« i/O = 0. [4]

V

t—

872 COLLEGE ALGEBRA

Now, / and /' both being proper fractions and y^^*^ q^ being
an integer, for ^^ was assumed unequal to ^„^ the numbers
qt — q,m^f ^^^ ^t — Qm^f are both positive or both negative^
and [4] becomes the sum of two positive numbers or of two
negative numbers is equal to zero; but this is impossible.
Therefore, t cannot be greater than 1.

If t = 1, then W|_i = 1, V|_i = 0, Wj = qi, v, = 1,
and equation [2] becomes

.-. v^_^N + qiu^_i - w« = 0, [5]

and u„-i-\-qiV„,^i-v^ = 0. [6]

4- 2^1 =

» , Ji ui -J- — — • * * ^^ Vm "i * " * ~~ *

(See § 444, Corollary.)

'''qm = ^qiy qm-i = q2y s'm-2 = s'» •••

,'.Vn = q, + ^ - •" - - jri-. [71

455. Eliminating 2'i from equations [5] and [6], we obtain

Now, w — 1 is the number of terms in the cycle in [7].
Therefore, u^_i^ — Nv„_{^ = -{- 1 or —1, according as the
number of terms is even or is odd, in the cycle of the simple
periodic continued fraction into which ViV" is convertible.

Let Ci = u^_i and Si = v^_i, that is, let — be the conver-

gent immediately preceding the partial quotient 22^1 in [7];
then equation [8] becomes

ci" - Nsi^ = 4-1 or - 1.

CONTINUED FRACTIONS 373

Consider the case c^ — Ns^ =4-1.

Let (ci 4- SiVNy = c, -f s^-y/N.

Then, (ci - ^iViV^)" = c„ - «„V^;

.-. c,» - NsJ" = (ci» - i\r«i^" = 1. [A]

Also (c^ 4- s^Vn) (c^ 4- «n ViV) = c^^^ 4- ««+« Viv". [B]

Multiply the factors on the left side of [B] and equate
rational and irrational parts ; then

^w + n = ^nfin 4" Vm J

456. These equations give a very easy and rapid method

of obtaining a close approximation to ViV^.

From the example on page 368, we find for Vt

. Cg 64 4- 7 X 3^ 127
'''s," 2x8x3 " 48 '

Ce 127« H- 3 X 7 X 127 X 48^ 8193151
56 - 3 X 1272 X 48 4- 7 X 48» "" 3096720 *

By § 447, the error of approximation is

4 X 30967202 " 10"

457. Compare equations [A], [B], and [C] with the trigo-
nometrical equations ;

cos^a — (— l)sin^a = 1,
(cos a 4- V— 1 sin a) (cos p 4- V— 1 sin fi)

= cos(a 4- /3)4- V^ sin(a 4- /3),
and cos (n- 4- ^) = cos a cos )8 4- (— 1) sin a sin )8,

sin (a 4- ^) = sin a cos )8 4- sin )8 cos a.

874 COLLEGE ALGEBRA

458. If Ci" ^ Nsi" = -- 1,
then Ca* — Ns2^ = + 1 ;
and, in general, Cg^" — NssJ^ = + 1,

459. Equation [2], § 453, gives
and therefore, if

then t*,+^« - Nv^^rJ = (- 1)" ^X+i-

460. If g is the H.C.F. of mg and a^ — N, to reduce the

mixed surd ^ a simple periodic continued fraction,

mg ^ ^—L '

it is sufficient to reduce 5 to a continued fraction

by the method of § 451. ^ ^

3 + V7 12 + V112 1

Thus,

8 32 12 -_ Vii2

__llillj_li

1+2 + 2 + 1 + 1+20 + 1 + 1*

461. The value of a simple periodic continued fraction can
be expressed as the root of a quadratic equation.

• •

Find the surd value of t t:-

1 + 2

Let X be the value of the continued fraction.

1 2 + x

Then, z =

1+ ' «+*

2 + x
.-. xa + 2x = 2.

.-. X = - 1 + Vs.

We take the + sign since x is evidently positive.

CONTINUED FRACTIONS

375

It is not true, however, that a value can be determined for
any periodic continued fraction ; for if we assume

we obtain

""^1-1-1-1-

1-x

a;^ - a; -f- 1 = 0.

• • •!/ "■"

i±V^

which is absurd. The continued fraction - - - •••is

1 — 1 — 1 —

not convergent, as may be seen on attempting to form the
principal convergents to it; these are

1, 00, 0, 1, 00, 0.

462. Exponential Equations. An exponential equation can
be solved by continued fractions.

Solve by continued fractions 10* = 2.

Let
Then,

or

Then,

y

1
10» = 2,

10 = 2,v.

,'. y = 3 + - (since 10 lies between 2' and 2*).
10 = 2 ^ = 2» X 2*.

.-. 2« = Y- = }.
.-. 2 = (f )«.

.'. 2 z= 3 + - (since 2 lies between (})« and (J)*).

Then,

2 = (J)'^« = (f)»x(})

The greatest integer in u is found to be 9.

376 COLLEGE ALGEBRA

Hence, x = 0 +

3 +

3 +

9 + etc.

The successive convergents are |, yV» ih ®^*
The last gives x = |f = 0.3010, approximately.

common logarithm of 2, is considerably less than 7;^^» that is, con-

NoTB. Observe that by the above process we have calculated the com-
mon logarithm of 2. By § 445, the error, when 0.3010 is taken for the

siderably less than 0.00011 ; so that 0.3010 is certainly correct to three
places of decimals, aiid probably correct to four places.

Logarithms are, however, much more easily calculated by the use of
series, as shown in Chapter XXV.

Ezercise 66

1. Find continued fractions for {^, ij^, W> tVt> "^9
Vn, 4 Vg ; and find the fourth convergent to each.

2. Find continued fractions for ^j, |^J, |f}|, W^J and
find the third convergent to each.

3. Find continued fractions for V2r, V22, V33, V55.

4. Obtain convergents, with only two figures in the denom-
inator, that approach nearest to the values of

V7, Vio, Vi5, Viz, Vis, V20, 3 - V5, 2 4- -v^.

5. If the pound troy is the weight of 22.8157 cubic inches
of water, and the pound avoirdupois of 27.7274 cubic inches of
water, find the fraction with denominator less than 100 which
shall differ from their ratio by less than 0.0001.

6. The ratio of the diagonal to the side of a square being
V2, find the fraction with denominator less than 100 which
shall differ from their ratio by less than 0.0001.

CONTINUED FRACTIONS 377

7. Find the next convergent when the two preceding con-
ViBrgents are ^^ and J|, and the next quotient is 5.

8. The ratio of the circumference of a circle to its diam-
eter is approximately 3.14159265 : 1. Find the first three con-
vergents to this ratio, and determine to how many decimal
places each agrees with the true value.

9. In two scales of which the zero points coincide the
distances between consecutive divisions of the one are to the
corresponding distances of the other as 1 : 1.06577. Find
what division points less than 100 most nearly coincide.

10. Find the surd values of

^44-2 ''^14-6 34-1 + 6 ^^24-34-4

11

. Show that (a + 'T -j(t -) = t

12. Show that the ratio of the diagonal of a cube to the
edge may be nearly expressed by 97 : 56. Find the greatest
possible value of the error made in taking this ratio for the
true ratio.

13. Find a series of fractions converging to the ratio of
5 hours 48 minutes 51 seconds to 24 hours.

14. Find a series of fractions converging to the ratio of a
cubic yard to a cubic meter, if a cubic yard is 0.76453 of a
cubic meter.

CHAPTER XXVII

8CAL£S OF NOTATION

463. Definitions. Let any positive integer be selected as a
radix or base ; then any number may be expressed as an alge-
braic expression of which the terms are multiples of powers
of the radix.

Any positive integer may be selected as the radix ; and to
each radix corresponds a scale of notation.

When we write numbers in any scale of notation, they are
arranged by descending powers of the radix, and the powers
of the radix are omitted, the place of each digit indicating of
what power of the radix it is the coefficient.

Thus, in the scale of ten, 2356 stands for

2xl0« + 3xl02 + 5xl0 + 6;
in the scale of seven for

2x7» + 3x 72 + 5x7 + 6;
in the scale of r for

2r» + 3r2 + 5r + 6.

464. Computation. Computations are made with numbers in
any scale, by observing that one unit of any order is equal
to the radix-number of units of the next lower order; and
that the radix-number of units of any order is equal to one
unit of the next higher order.

(1) Add 56,432 and 15,646 (scale of seven).

^^ The process differs from that in the decimal scale only in

i/vfUA that when a sum greater than seven is reached, we diyide by
seven (not ten), write the remainder, and carry the quotient to

106411 ^, ^ ,

the next column.

378

SCALES OF NOTATION 379

(2) Subtract 34,561 from 61,235 (scale of eight).

fii9<li> When the number of any order of units in the minuend is

oAtiA^ ^^^ *^*^ ^^® number of the corresponding order in the subtra-

TTTTT hend, we increase the number in the minuend by eight instead
of by ten, as in the common scale.

(3) Multiply 6732 by 428 (scale of nine).

6732

428 We multiply the number of units in each order in the multi-

61477 plicand by the number of units in each order in the multiplier,

12664 divide each time by nine^ set down the remainder, and carry

26238 the quotient.
2712127

(4) Divide 2,712,127 by 5732 (scale of nine).

428

6732 ) 2712127

25238 '^^ operations of multiplication and subtraction in-

■IYY22 volved in this problem are precisely the same as in the

12664 '*«"'''°** «=*1« 0' AOtation, with the exception that the
"TTTyy radix is 9 instead of 10.

51477

465. Integers in Any Scale. Ifxis any positive integer j any
positive integer N may he expressed in the form

N = ar** -f br"-^ H h pr* -f qr -f s,

in which the coefficients a, b, c, • • • are positive integers, each
less than r.

For, divide N by r*, the highest power of r contained in iV,
and let the quotient be a with the remainder N^.

Then, iV^ = ar* + N^.

In like maimer,

and so on.

By continuing this process a remainder s is at length reached
which is less than r. So that,

N = ar" -f h7^~^ -f . . . -f-^r* + qr -{- 8.

380 COLLEGE ALGEBRA

Some of the coefficients s, q, p, "• may vanisliy and every
coefficient is less than r ; that is^ the values of the coefficients
may range from zero to r — 1.

Hence, including zero, r digits are required to express num-
bers in the scale of r.

Express N in the form

ar^ + br^~^ -+-•••+ p^^ + q^ -\- s,

and show how the digits a^ b, -" may be found.

If N=a7^-\- br^-^ -\ \- pr^ -\- qr -^^ s,

N s

then — = ar""^ 4- b7^~^ + hpr-{-q-\

r T

That is, the remainder on dividing i\r by r is «, the last
digit.

Let N^ = ar»--i -f ^^""^ H hi»- + g.

N a

Then, — = ar^-* -f ^"^ H hi> + - •

T T

That is, the remainder is q^ the last but one of the digits.

466. Hence, to express an integral number in the scale of /*^

Divide the number by the radix, then the quotient by the radix,
and so on until a quotient less than the radix is reached. The
successive rernainders and t?ie la^t quotient are the successive
digits beginning with the units^ place.

(1) Express 42,897 (scale of ten) in the scale of six.

6)42897
6)7149

6)1191
67198
6ll

3
3
3
0
3

Therefore, 42,897 (scale of ten) is expressed in the scale of six 1^
630,333.

SCALES OF NOTATION 881

(2) Change 37,214 from the scale of eight to the scale of nine.

9)37214 The radix is 8. Hence, the two digits on the left, 37,

9)3363 • • • 1 do not mean thirtyseoen, but 3x8 + 7, or thirty-one,

9)305 • • • 6 which contains 9 three times, with remainder 4.

9)25 • • • 8 The next partial dividend is 4 x 8 + 2 = 34, which

2 • • • 3 contains 9 three times, with remainder 7 ; and so on.

Therefore, 37,214 (scale of eight) is expressed in the scale of nine by
23,861.

(3) In what scale is 140 (scale of ten) expressed by 352 ?

Let r be the radix ; then, in the scale of ten,

140 = 3r2 + 5r + 2, or Sr^ + 5r = 138.

Solving, we find r = 6.

The other value of r is negative and fractional, and therefore inad-
missible, since the radix is always a positive integer.

467. Radix-Fractions. As in the decimal scale decimal frac-
tions are used, so in any scale radix-fractions are used.

Thus, in the decimal scale, 0.2341 stands for

10 "^ 102 "^ 108 10* *
and in the scale of r it stands for

? + i + l + i.

f f.2 fA fi

(1) Express f|^ (scale of ten) by a radix-fraction in the

scale of eight.

245 a , b c , d ,

Assume — = — H — h***

256 8 82 88 8*

Multiply by 8, 7f J = a + g + ^ + | + •• •

Therefore, a = 7,

, 21 h c ^d ^

Multiply by 8, ^ = & + |+^-*-'**

32 8 82 88
c d
8 P

382 COLLEGE ALGEBRA

Therefore, 6 = 6,

l__c

4""8 ' 82

d
8

and T = « + — +

d
Multiply by 8, 2 = c + - +

Integral part : 6 )35

4 • • • 6.

Therefore, c = 2,

and 0 = (2, etc.

Therefore, }f{ (scale of ten) is expressed in the scale of eight by 0.752.

(2) Change 35.14 from the scale of eight to the scale

of six. 3

6
We take the integral part and the fractional part 16)18(1

separately. i^

"2

6
16)12^0

FractiorujU part : - H — = — = — . ^ ^ ,

8 82 64 16 I6J72X4

This is reduced to a radix-fraction in the scale of six ^

as in the margin.

Therefore, 36.14 (scale of eight) is expressed in the 48

scale of six by 45.1043.

Exercise 67

1. Add 435, 624, 737 (scale of eight).

2. From 32,413 subtract 15,542 (scale of six).

3. Multiply 6431 by 35 (scale of seven).

4. Multiply 4685 by 3483 (scale of nine).

5. Divide 102,432 by 36 (scale of seven).

6. Find H.C.F. of 2541 and 3102 (scale of seven).

7. Extract the square root of 33,224 (scale of six).

SCALES OF NOTATION 383

8. Extract the square root of 300,114 (scale of five).

9. Change 624 from the sdale of ten to the scale of five.

10. Change 3516 from the scale of seven to the scale of ten.

11. Change 3721 from the scale of eight to the scale of six.

12. Change 4535 from the scale of seven to the scale of nine.

13. Change 32.15 from the scale of six to the scale of nine.

14. Express ^y^ (scale of ten) by a radix-fraction in the
scale of four.

16. Express ^^ (scale of ten) by a radix-fraction in the
scale of six.

16. Multiply 31.24 by 0.31 (scale of five).

17. In what scale is 21 x 36 equal to 746 ?

18. In what scale is the square of 23 expressed by 540 ?

19. In what scale are 212, 1101, 1220 in arithmetical pro-
gression ?

20. Show that 1,234,321 is a perfect square in any scale
(radix greater than four).

21. Which of the weights 1, 2, 4, 8, • • • pounds must be
selected to weigh 345 pounds, only one weight of each kind
being used ?

22. Multiply 72,645 by 46,723 (scale of eight).

23. Divide 162,542 by 6522 (scale of seven).

24. A number of three digits in the scale of 7 is expressed
in the scale of 9 by the same digits in reverse order. Find
the number.

26. If two numbers are formed by the same digits in dif-
ferent orders, show that the difference between the numbers is
divisible by r — 1.

CHAPTER XXVUl

THEORY OF IIUMHKRS

468. Definitions. In the present chapter, by the term num^
her is meant positive integer. The terms prime, eampasite,
are nsed in the ordinary arithmetical sense.

A multiple of a is a number that contains the factor a, and
may be written mu.

An even number, since it contains the factor 2, may be writ-
ten 2 m ; an odd number may be written 2 m + 1, 2 w — 1,
2m-f 3, 2w — 3, etc.

A number a is said to divide another number b when — is
an integer.

469. Resolution into Prime Factors. A number can be resolved
into prime factors in only one way.

Let N be any number. Suppose N = ahe • • •, where ayb^e,---
are prime numbers; suppose also N=zaPy'y where a, fi,
y, ••• are prime numbers.

Then, abc • • . = afiy • • •

Hence, a must divide the product abc • • • ; but a, 6, c, • • • aie
all prime numbers ; hence, a must be equal to some one of
them, a suppose.

Divide by a, Jc • • • = j8y • • • ; and so on.

Hence, the factors in afiy • • • are equal to those in abe • • *,
and the theorem is proved.

470. Divisibility of a Product. \, If a number a divides a
product be, and is prime to b, it must divide c.

For, since a divides be, every prime factor of a mnst be
found in be ; but, since a is prime to b, no factor of a will be

384

THEORY OF NUMBERS 386

found in h ; hence, all the prime factors of a are found in c ;
that is, a divides c.

From this theorem it follows that :

II. If a prime number a divides a product bcde • • •, lY must
divide some factor of that product ; and conversely,

III. If a prime number divides b", it must divide b.

IV. If 2^ is prime to b and to c, it is prime to be.

'V. If 2^ is prime to b, every power of q, is prime to every
power of b.

a
471. If -y a fraction in its lowest terms ^ is equal to another

n

fraction -> then c and d are equimultiples ofz, and b.

r£ a c ^ ad

If T = ";^ then — - = c.

b d 0

Since h will not divide a, it must divide d\ hence, c? is a
multiple of h.

Let d = mhy m being an integer.

a c a G

Since t = j' and d = mh. - = — r ; therefore, c = ??ta.
0 a 0 mo

Hence, c and d are equimultiples of a and h.
From the above theorem it follows that :

Li the decimal scale of notation a common fraction in its
lowest terms will produce a non-terminating decimal if its
denominator contains any prime factor except 2 and 5.

For a terminating decimal is equivalent to a fraction with

a denominator 10". Therefore, a fraction - in its lowest

0

terms cannot be equal to such a fraction, unless 10" is a
multiple of b. But 10", that is, 2" x 5", contains no prime
factors besides 2 and 5, and hence cannot be a multiple of b,
if b contains any prime factors except 2 and 5.

386 COLLEGE ALGEBRA

472. Square Numbers. If a square number is resolved into
its prime factors, the exponent of each factor is even.

For, if N^a^xb^XC"'

Conversely: A number that has the exponents of all its
prime factors even is a perfect square; therefore,

To change any number to a perfect square,

Resolve the number into its prime factors, select the favors
which have odd exponents^ and multiply the given number by
the product of these factors.

Thus, to find the least number by which 250 must* be multiplied to
make It a perfect square.

250 = 2 X 5", in which 2 and 5 are the factors that have odd exponents.
Hence, the multiplier required is 2 x 5 = 10.

473. Divisibility of Numbers. I. If two numbers N and N'
when divided by a have the same remainder, their difference is
divisible by a.

For, if N when divided by a has a quotient q and a remain-
der r, then

N = qa -{- r.

And, if N' when divided by a has a quotient q* and a
remainder r, then

N' = q'a + r.

Therefore, N - N' =z(q - q^a.

II. If the difference between two numbers N and N' is divisi-
ble by a, then N and N' when divided by a have the same
remainder.

N — N^

Let = m, where m is an integer.

a

N r

Now, — = ^ 4- -> where r<a.

' a a

THEORY OF NUMBERS 387

and — = q' -\ — > where r^<a,

a a

N — N* r — r'
Subtract, = q — g' -\ •

N — N^

But is an integer by hypothesis.

a

Therefore, is an integer, or zero.

a

Now, r — T^ <r {r and r' being integers), and r < a.

Hence, a cannot divide r — r'. Therefore,

cannot be an integer, and hence must be zero.

r — r'

a

Therefore, r must equal r\

III. If two numbers N and N' when divided by 'a given
number a have remainders r and r', then NN' and rr' whsn
divided by a have the same remainder.

For, if N= qa -\-rj

and iV' = q'a + r',

then NN* = S'S''^^ + g'a^' + s^'a^ 4- ^^'

= (s'S^'a + g'r' + q^r) a -f- rr'.

Therefore, NN^ and it' when divided by a have the same

remainder.

Thus, 37 and 47 when divided by 7 have remainders 2 and 5.

Now, 37 X 47 = 1739, and 2 x 6 = 10.

The remainder when each of these two numbers is divided by 7 is 3.

From II it follows that, in the scale of ten,

1. A number is divisible by 2, 4, S, ' • - if the numbers denoted by its
last digits last two digits^ last three digits, • • • are divisible respectively by
2,4,8,...

2. A number is divisible by 5, 25, 125, . • • if the numbers denoted by its
last digit, last two digits, last three digits, • . • are divisible respectively by
5, 26, 125, ...

888 COLLEGE ALGEBRA

3. If from a number the sum of its digits is subtracted, the remainder
is divisible by 9.

For, if from a number expressed in the form

a + 106 + 102c + 108d + ...

a + b + 0 + d-\ is subtracted, the remainder is

(10 - 1)5 + (102 - l)c + (108 _ i)d + . . .
and 10 - 1, 102 - 1, lO^ - 1, . • . are each divisible by 10 - 1, or 9.
Therefore, the remainder is divisible by 9.

4. A number N may be expressed in the form 9 n + s (if s denotes the
sum of its digits); and N is divisible by S if 8 is divisible by 3 ; and by 9
if&is divisible by 9.

5. A number is divisible by 11 if t?ie difference between the sum of its
digits in the even places and the sum of its digits in the odd places isO or
a multiple of 11.

For, a number N expressed by digits (beginning from the right)
a, 6, c, d, • • • may be put in the form of

N=a-h 106 + 102C+ 10«d + ...

... jV _ a + 6 - c + d = (10 + 1) 6 + (102 - 1) c + (iqs + i) d + . . .

But 10 + 1 is a factor of 10 + 1, 102 - l, IQS 4. i^ . . .

Therefore, N — a + b — c -^d is divisible by 10 -f 1 = 11,

Hence, the number N may be expressed in the form

11 n + (a + c + • . •) - (6 + d + • • •)»

and is a multiple of 11, if (a + c H ) — (& + d H ) is 0 or a multiple

of 11.

474. Theorem. The product of r consecutive integers is

divisible hy Ir.

Eepresent by P^^j^ the product of k consecutive int^ers
beginning with n.

Then, p„^ = 7i(n + l)...(n + Aj - 1);

^n + 1, * + 1 = (^ + 1) (n + 2) . . . (71 + A;) (w + Aj + 1)

= n (n -\-l) (71 -{- 2) '"(n-^k)

-\-(k-\-l){n + 1) (71 4- 2) . . . (n + k).

• *• -^n + 1, Jt + 1 ^^ -^n, * + 1 "I" (^ "I" 1-) -^n + 1, A*

THEORY OF NUMBERS 389

Assume, for the moment, that the product of any k con-
secutive integers is divisible by \k.

Then, Pn+i.*+i = i^n,t+i +(^ + l)M\k',

or, ^n+i.*+i = ^n.*+i 4- M\k±l,

where ilf is an integer.

Hence, if Pn,*+i is divisible by \k + 1, ^n+i,*+i is also
divisible by \k + 1. But Pi.^+i is divisible by \k -\- 1 since
A.A+i = 1^ 4- 1« Therefore, ^2,*+! is divisible by [^ + 1;
hence, P^t+i is divisible by |^ + 1 ; and so on.

Hence, the product of any ^ + 1 consecutive integers is
divisible by |A: + 1, if the product of any k consecutive inte-
gers is divisible by \k. The product of any 2 consecutive
integers is divisible by [2 ; hence, the product of any 3 con-
secutive integers is divisible by [3 ; hence, the product of any 4
consecutive integers is divisible by [4 ; and so on. Therefore,
the product of any r consecutive integers is divisible by [r.

475. Examples. (1) Show that every square number is of
one of the forms 5 n, 5 ti — 1, 5 w 4- 1.
Every number is of one of the forms :

5n — 2, 5n — 1, 5n, 5n4-l» 5n + 2.
Now, (5n ± 2)2 = 25n2 ± 20n + 4 = 6(5 n^ ± 4 n + 1) - 1 ;
(5n ± 1)2 = 25 n2 ± iQn + 1 = 5(5n2 ± 2n) + 1 ;
and (5 n)2 = 25 n2 = 6 (5 n^).

Therefore, every square number is of one of the three forms :

5n, 5n — 1, 5n4- 1.

Hence, in the scale of ten, every square number must end in 0, 1, 4,
5, 6, or 9.

(2) Show that n^ — nis divisible by 30 if n is even.

n* - n = n(n - 1) (n + 1) (n^ + 1)

= n(n - 1) (n + 1) (n2 - 4 + 6)
= n (n - 1) (n + 1) [(n - 2) (n + 2) + 5].
Now, n (n - 1) (n + 1) is divisible by [3. (§ 474)

390 COLLEGE ALGEBRA

One of the five consecutive numbers n — 2, n — l, n, n + 1, n + 2 is
divisible by 6.

If n, (n— 1), or (n+1) is divisible by 6, the number on the right is divis-
ible by 5. If (n - 2) or (n + 2) is divisible by 5, then [(» - 2)(n + 2) + 5]
is divisible by 5, since 5 is divisible by 5.

Therefore, the number on the right is always divisible by 5.

Hence, n^ — n is divisible by 6 x [3, that is, by 30.

Ezercise 68

Find the least number by which each of the following must
be multiplied that the product may be a square number :

1. 2625. 2. 3675. 3. 4i374. 4. 74,088.

5. If m and n are positive integers, both odd or both even,
show that m" — n" is divisible by 4.

6. Show that n^ — n is always even.

7. Show that n^ — n iB divisible by 6 if w is even, and by
24 if 71 is odd.

8. Show that n^ — n ia divisible by 240 if n is odd.

9. Show that ri? — n is divisible by 42 if n is even, and
by 168 if n is odd.

10. Show that 7i(7i -f 1) (n + 5) is divisible by 6.

11. Show that every square number is of one of the forms
371, 3ri-f 1.

12. Show that every cube number is of one of the forms
On, 9n — 1, 97i-fl.

13. Show that every cube number is of one of the forms

771, 771 — 1, 771 + 1.

14. Show that every number which is both a square and
a cube is of the form 7 7i or 7 7i + 1.

15. Show that in the scale of ten every perfect fourth
power ends in one of the figures 0, 1, 5, 6.

CHAPTEE XXIX

DETERMINANTS

476. Origin. If we solve the two simultaneous equations
we obtain

cJ>2 — cjbi

X = — : —f

\

y =

Similarly, from the three simultaneous equations

we obtain

dih^s — dj)^2 + djl)jfi^ — djb^c^ + efg^iCg — djb^^

/J. ^ ' y

with similar expressions for y and ^.

The numerators and denominators of these fractions are
examples of-.expressions which often occur in algebraic work,
and for which it is therefore convenient to have a special
name. Such expressions are called determinants.

477. Definitions. Determinants are usually written in a
compact form, called the square form.

Thus, ai6j — 0261 is written

h. b,
and aibjpt — aibtCt + aJbtCi — Oj^iCj + at&i(!t — <hMi is written

.x^^•^*'^'r^-^>'

'^^.

at 0% og

Ci Cj c^

\

.K .

^®

w.

892 COLLEGE ALGEBRA

'IMiiH tK|uaru form is sometimes written in a still more abbreviated form,
'riiuri, tho liiHt two determinants are written |ai 6s| and \ai bt Ct\. This
lust notation hIiouUI, however, always suggest the square form. In any
pniliU^iii it gi^norally is advisable to write this abbreviated foim in the
roiiiploto Htpiaru form.

'I'lu^ individual symbols a-u a^, bi,b2, •" are called elements.

A hori/oi^t^il line of elements is called a row; a vertical line
a column.

'V\w two linos rfp hn, c^ and a^, h^, c^ are called diagonals ; the
iirst the principal diagonal, the second the secondary diagonal

'V\\i\ order of a determinant is the number of elements in a
row or colunin.

TliuH, tho last two determinants are of the second and third orders

nwiMH'.tively.

Hie expression of whicli the square form is an abbreviation

is calkul the expansion of the determinant.

The several terms of the expansion are called terms of the

determinant.

ai 02

Thus, the expansion of

hi 6j

LB 0162 — CI261.

Kemark. By some writers constituenJt is used where we use efement,

and elemevd where we use temi.

478. General Definition. In general, a determinant of the
nth. order is an expression involving n^ elements arranged in
n rows of n elements each.

479. Inversions of Order. In any arrangement of the letters
of .a determinant the occurrence of any one of them before
another which precedes it in the principal diagonal is called
an inversion of order.

Thus, if 1, 2, 3, 4, 5 is the order in the principal diag<»ial9 in the order
2, 3, 5, 1, 4 there are four inversions : 2 before 1, 3 before 1, 6 before 1,
6 before 4.

Similarly, if a, &, c, d is the order in the principal diagonal, in the
order &, d, a, c there are three inyersions ; & before a, d before a, d
before c.

DETERMINANTS 893

480. In any arrangement of integers (or letters) let two
adjacent integers (or letters) be interchanged ; then the num-
ber of inversions is either increased or diminished by one.

For example, in the arrangement 6 2 [6 IJ 4 3 7 Interchange 5 and 11

We now have 6 2 [1 5] 4 3 7.

The inversions of 6 and 1 with the integers before the group are the'
same in each arrangement.

The inversions of 6 and 1 with the integers after the group are the
same in each series.

In the first arrangement 5 1 is an inversion ; in the second series 1 5
is not an inversion.

Hence, the interchanging of 5 and 1 diminishes the number of inver-
sions by one.

Similarly for any other CBsbj\^

m

481. Signs of the Terms. The principal diagonal term
always has a + sign. _ ^ , .

To find the sign of any other term : u v> , ^ j

Add the number of inversions among the letters, and the ^
number of inversions among the subscripts. If the total
number is even, the sign of the term is -\-] if odd, — .

Thus, in the determinant \a\ &2 Cs d^\ consider the term c^%djl>i»
There are m c adh three inversions ; in 2 3 4 1 three inversions ; the
total is six, an even number, and the sign of the term is +•

482. Special Rules. In practice the sign of a term is easily
found by one of the following special rules :

I. Write the elements of the term in the natural order of
letters ; if the number of inversions am^ng the subscripts is
even, the sign of the term is -{-', if odd, —,

II. Write the elements in the natural order of subscripts;
if the number of inversions arnang the letters is even, the sign
of the term is -{-; if odd, — .

Thus, in the determinant |ai &a cg (24| consider the term C2as<2«2^*
Writing the elements in the order of letters, we have aj)ic^4. There
are twb inversions, viz,, 8 before 1, and 3 before 2, and the sign of the
term is +. Or, write the elements in the order of subscripts, &iCsat<2i*

894 COLLEGE ALGEBRA

There are two inversions, mz.^ b before a, and c before a, and the sign
of the term is +•

That these special rules give the same sign as the general rule of § 481
may be seen as follows :

Consider the term C8a8(24&i. Its sign is determined by the total

/» /» /7 7k

number of inversions in the two series ^ ^ a -,* Bring os to the first

2 o 4 1

position ; this interchanges in the two series c and a, 2 and 3. In each
series the number of inversions is increased or diminished by one (§ 480),
and the total is, therefore, increased or diminished by an even number.

Interchange &i and di, then interchange &i and cs ; this brings &i to
the second place, and the letters into the natural order. As before, the
total number of inversions is changed by an even number.

The term is now written a8&iC2(24, and the number of inversions differs
by an even number from that found by the general rule of § 481. Hence,
the sign given by I agrees with the sign given by the general rule.

483. The Expansion. The expansion of any determinant
may be found by forming all the possible products by taking
one element, and only one, from each row, and one element^
and only one, from each column, and prefixing to each product
the proper sign.

The number of terms in the expansion is

1 • 2 • 3 • • • (/I — 1) • w ^ \n.-> * Af \

Eor, to form the expansion of a determinant of the nth order
we make all the possible arrangements of n elements, taking
all of them in each arrangement (§ 339).

484. If all the elements in any row or column are zero, ihe
\\ determinant is zero. For every term contains one oi t&e zeros

from this row or column (§ 483), and therefore every term of

the determinant is zero.

• * A determinant is unchanged if the rows are changed to

"2- columns and the columns to rows. For the rules (§§ 478-483)

are unchanged if row is changed to column aiid oolmnn to

row.

tti 6i Ci

w

Thus,

ai

02 as

&i

62 63

Cl

C2 Ca

02 ^ C2
Og &8 Cs

DETERMINANTS

S96

485. A determinant of the third order may be conyeniently
expanded as follows : /TTN

Three elements connected by a full .line fonn a positiye
term ; three elements connected by a dotted line form a nega-
tive term. The expansion obtained from the diagram is

which agrees with § 477.

There is no simple rule for expanding determinants of
orders higher than the third.

Ezeroise 69

^

Prove the following relations by expanding :

1.

«1

a,

a,

br

<h

%

8,

S.

*,

h

a.

h

h

*i

a.

«!

2.

a,

Uo Cbi

hi h^ ftj

h

8

a.

«!

c.

Cl

= —

ft.

^

— ^8

^1

^8

a<

Find the value of :

3.

1
2
3

> r

2 3

4 4

4 6

■ 1 ^

4.

3

7
5

2
6
3

4
1
8

6,

4
-1
6

6
2

4

2

3
5

-3>l~4o-\b^

896

COLLEGE ALGEBRA

Find the value of :

6.

5 8 7

13 5

9 6 3

6 4 4

7.

2 4 6

8.

5 7 7

5 7 9

3 6 7

8 4 2

9. Count the inversions in the arrangement :

5 4,1. 3 p. 7 5 14 3 6 2. dace b.

a 5 2 3.S 6 5 4 2 13 7. c e b d a.

10. In the determinant \ai ^2 c^ d^ e^l find the signs of the
following terms :

a^^c^^^ ajb^c^^e^ e^c^aj)^

ajb^^^e^. b^c^a^e^d^. c^ajb^e^d^

11. Write, with their proper signs, all the terms of the

determinant [aj ftj Cg c?4|.

12. Write, with their proper signs, all the terms of the
determinant \a^ b^ c^ d^ e^\ which contain both cb^ and b^\ all
the terms which contain both ^3 and e^.

Expand :

13.

4.V"

0

0

0

a

0

0

b

0

a

a

b

b

b

b

a

a

16.

^•.b

C-^

a

h e 0

c

a h ^

h

e a 0

•

a

b e 1

486. Number of Terms. Consider a determinant of ihe ntibi'
order.

In forming a term we can take from the first row any one
of n elements ; from the second row any one of n — 1 " ele-
ments ; and so on. From the last row we can take only the
one remaining element.

Hence, the full number of terms is w (n — 1) • • • 1, or |ii.

DETERMINANTS

397

487. Interchange of Columns or Rows. If tm adjacent
rnlurt^njt oJ^^t^ff^ nti^jnj^.P.'n.t fgti^a nf a determinant A are inter-
changedy the determinant thus obtained is — ^.

For example, consider the determinants

A =

«!

as

as

^4

ai

a^

^2

a^

h2

^8

C4

, A' =

di

C^2

d.

d.

di

d»

d^

d.

The individual elements in any row or column of A' are the
same as those of some row or column of A, the only difference
being in the arrangement of the elements. Since every term
of each determinant contains one, and only one, element from
each row and column, every term of A' must, disregarding the
sign, be a term of A.

Now the sign of any particular term of A' is found from
an arrangement (§ 482, I) in which 3 2 is the natural order.
The sign of the term of A which contains the same elements
is found from an arrangement in which 3 2 is regarded as an
inversion. Consequently, every term which in A' has a + sign
has in A a — sign, and vice versa (§ 480).

Therefore, A' = — A.

Similarly if any two adjacent columns or two adjacent rows
of any determinant are interchanged.

488. In any determinant ^, if a particular column_is carried
over m columns, the determinant obtained is (— 1)'"A.

For, successively interchange the column in question with
the adjacent column until it occupies the desired position.
There are m interchanges made ; hence, there are m changes
of sign (§ 487). We may make each change of sign by multi-
plying A by — 1. Therefore, the new determinant is (— 1)"*A,

Similarly for a particular row.

r- o-}!i

898

COLLEGE ALGEBRA

ai 02 Os

(Kg 02 Oi

Og 02 Oi

&i hi &8

= —

6g 62 &1

—

Cg Ca ci

Ci Ca Cg

Cg C2 Ci

&g 62 61

'^

are tn^«r-

489. In any determinant A, if any two eoh
changed, the determinant thius obtained is — £i.

Let there be m columns between the columns in question.

Bring the second column before the first. The second
column is carried over m + 1 columns, and the determinant
obtained is (- l)"'+iA (§ 488).

Bring the first column to the original position of the second.
The first column is carried over m columns, and the deter-
minant obtained is (- !)"•(- 1)"'+^A, or (- 1^^+^^,

Since 2 m + 1 is always an odd number, (— !)*"•+ ^ A = — A.

\RYpii1a.r1j^^<;)y I^WO roWS.

Thus,

490. Useful Properties. If two columns of a determinant

are identical, the determinant vanishes.

For, let A represent the determinant.

Interchanging the two identical columns ought to change
A into — A (§ 489). But since the two columns are identical,
the determinant is unchanged.

.•.A = -A, 2A = 0, A = 0.

Sl Similarly if two rows are identical.

491. If all the elements in any column_ are muUipUed by
any number m, the determinant is multiplied by m.

For, every term contains one, and only one, element from
the column in question. Hence every term, and consequently
the whole determinant, is multiplied by m.

Similarly for a row.

Thus,

max Oa Og

Oi O2 Og

moi

mbi

met

mbi &2 &8

= m

61 62 bs

:^

Oa

&s

c%

mci Ca Cg

Ci Ci Cg

ag

h

c%

DETERMINANTS

899

Again,

he a a^
ca b b^
ab c c^

abc

abc a^ a'

1 a2 a«

6ca 62 &8

:=

1 62 6«

cab c^ c*

1 c2 c8

492. If each of the elements in a column or row is the
sum of two numbers, the determinant may be expressed as
the sum of two determinants.

Thus,

ai + a
6i-f /3
ci+ 7

62

as

68

dl

Oa as

a Oa as

61

62 6j

+

/3 6a 68

Cl

Ca Cs

7 Ca cg

For, consider any term, as (ai-\- a)b2Cs. This may be
written ai&a^s + ^^r^a^g. Hence, every term of the first deter-
minant is the sum of a term of the second determinant and a
term of the third determinant. Consequently, the first deter-
minant is the sum of the other two determinants.

Similarly for any other case.

493. If the elements in^any column or row are multiplied
by any number m, and added to, or subtracted from, the
corresponding elements in any other column or row, the deter-
minant is unchanged.

Thus,

ai db fTiOa aa as

ai aa as

Tiiaa aa ag

61 ± m6a 6a 63

^

61 6a 6s

:*:

fVh 6a 63

Cl ± wiCa Ca Cs

Cl Ca Cg

lAca Ca Cg

The last determinant may be written

, and therefore vanishes (§ 490).

± m

Oa Oa as
6a 63 63
Ca Ca Cs

Hence, we have only the first determinant on the right-hand side.

Similarly for any other case.

This process may be applied simultaneously to two or more
columns or rows ; bujb in this case care must be taken not to
make two columns or rows identical (§ 490).

This last property is of great use in reducing determinants
to simpler forms.

\

400

COLLEGE ALGEBRA

494. Examples. (1)

b +c a 1
c -\-a b 1
a-\-b c 1

h -\-e +a

a 1

=

e +a+b b 1
a + b +0 e 1

1 a i

= {a + h + e)

16 1

1 e

1

= 0.

Begin by adding the second column to the first.

(2)

14 16 11
21 22 16
23 29 17

=r

3
5
6

=

2

4 11

6 16

12 17

3 2 2

5 3 1

6 6-1

= 2

3 2 11

5 3 16

6 6 17

= 2(19) = 38..

Begin by subtracting the third column from the first and second col-
umns. Then take out the factor 2 (§401), subtract 3 times the first
column from the third, and expand the result by § 486.

495. Factoring of Determinants. If a determinant vanishes
when for any element a we put another element b, then a — b
is a factor of the determinant.

For, the expansion contains only positive integral powers of
the several elements, and we can write

A = Aq -\- A^a -\- A^a^ -\-A^(^ +

[1]

where A^^ Ai, A^^ >lg, . • • are expressions that do not involve a,
and consequently remain unchanged when we put b for a.
Put b for a. Since A becomes 0 by hypothesis,

Subtract [2] from [1],

A = ^i(a - b)-\-A^{a^ - b^)-{-As(a^ -&») + •••

Since a — 5 is a factor of each term of the expansion, d — &
is a factor of the determinant.

DETERMINANTS

401

The theorem also holds true when a and h are not elements,
provided a and h enter into the expansion in positive integral
powers only.

By the principle just proved, and the principle of § 490, we
can resolve into factors many determinants without expanding
them.

a^ a 1

(1) Resolve into factors

c^ c

The determinant vanishes when a = 6, when a — c^ and when & = c.
Hence, a — h^h — c^ and c — a are factors. A is of the third degree in
a, 6, c, and these are easily seen to be all the factors. It remains to
determine the sign before the product.

In A, as given, a^ft is + ; in the product (a — 6) (6 — c) (c — a) the term
a2& is — . Hence,

A = — (a — 6) (6 — c)(c — a).

(2) Resolve into factors

As in the last example a — 6, 6 — c, c — a are found to be fiactorB.
There is one other factor of the first degree.

To the third column add the second ; the result may be written

a^

a h -\-c

''.

ft2

b c -\- a

•

(?

c a -\-b

(a + b + c)

or, by Example (1), — (a + & + c) (a — b) (b ^c){c — a).

aa

a

1

6*

b

1

c^

c

1

(f^

Ez«roise 70

Show that :

1.

0 a b
a 0 e
b c 0

.\>^

= 2abe.

2.

6 H- c a a
b c + a b
e e a + b

= 4kabc.

402

COLLEGE ALGEBRA

8.

4.

1
1
1
1

0
1
1
1

a*

a'
b^

d^

a*

b*

c*

=

d*

a a*

b b^

c c^

d d^

1

1

1

0

c«

6«

e'

0

a»

' '

6«

a'

0

b

e
0
a

Find the value of :

6.

20

15

25

17

12

22

6.

19

20

16

Resolve into simplest factors :

1 aa^
S, 1 b b^ . 9.

a
b
c

11.

a

b

c

b

c

a

12.

c

a

b

1
1
1
1

a
b
c
d

a*
d^

c
b
a
0

13.

d^

3

23

13

7

53

30

7.

9

70

39

22 29 27
26 23 30
28 26 24

a^

ftc

ft«

cei

10.

c«

ab

a» ^ 1
(j» ah 1

d e h a

a

14. If all the elements on one side of a diagonal term are
zeros, show that the expansion reduces to this term.

Show that :

16.

a^ — be a 1
b^-ca b 1
c^ — ab c 1

= 0.

16.

a-\-2b a-\-4ib a-\-6b
a-\-Sb a-{-5b a -\-7b
a -\- ^b aH- 66 a + 8b

= 0.

DETERMINANTS

403

17.

18.

ja -f c* ba ca

ab c^ -{- a^ cb
ac be a^ -\-b^

(a + by c» c"

a» (b + cy a"
b^

= 4 a^ft V.

19.

b^ (c + a)'

l+« 2 3 4

1 2+« 3 4

1 2 3+« 4

12 3 4:-\-x

= 2abc(a'\-b + ey

= x* 4- 10 «».

20.

a* + 1 ^<3^ <?<3t <^<3t

ab b^-\-l cb db

ac be c* + 1 dc

ad bd cd c?^ + 1

= a» + ^^ + c« + ^^ + 1.

496. Minors. If one row and one column of a determinant
are erased, a new determinant of order one lower than the
given determinant is obtained. This determinant is called a
first minor of the given determinant.

Similarly, by erasing two rows and two columns, we obtain
a second minor ; and so on.

Thus, in the determinant

CL\ a^ cL

erasing the second row and

Ci Cs ^

the third column, we obtain the first minor ^

to correspond to the element &8» &nd is generally represented by A^, ; so

ai 03

This minor is said

that, in this case, A^, =

Cl Ct

In general, to every element corresponds a first minor
obtained by erasing the row and column in which the given
element stands.

404

COLLEGE ALGEBRA

497. If all the elements of the first row after the first ele^
ment are zeros, the determinant reduces to OrJ^^,

0 0 0

Consider the determinant A =

^1
d.

h h h

Cj C3 C4

£^3 d^ d^

Every term of A contains one, and only one, element from
the first row ; and each term that does not contain Oi contains
one of the zeros, and therefore vanishes. Each term that
contains a^ contains no other element from the first row or
column, and, consequently, contains one, and only one, ele-
ment from each row and column of the determinant

l>2 h K

^2 ^8 ^4

d^ d^ d^

, or A,

«i'

Hence, disregarding the sign, each term of A consists of a^
multiplied into a term of A„^.

Take any particular term of A, as a^^c^^\ the sign is
fixed (§ 482, I) by the number of inversions in the series
14 3 2; the sign of the term h^c^d^ of A^ is fixed by the
number of inversions in the series 4 3 2. Adding % makes
no new inversions among either the letters or the subscripts.
Consequently, the sign of the term in A is the same as the
sign of the term in ^lA^^.

Since this is true of every term of A, we have

A = OiA^.

Similarly for any determinant of like form.

498. Terms containing an Element By § 486 the sum of
the terms that contain a^ may be written ajA^. For, no one
of the terms that contain a^ can contain any one of the ele-
ments ag, ^3, ^4, • • •, and these terms are therefore unchanged

if for ag, ttg, ^4,

• in the given determinant we put zeros.

DETERMINANTS 406

If we carry the second column over the first, the determi-
nant is changed to — A. By § 496 the sum of the terms of

— A that contain ag is Og^o,? ^^^ ^® sum of the correspond-
ing terms of A is, therefore, — aa^o,*

In general, for the element of the pih row and gth column
we carry the pth row over p — 1 rows, and the qth. column
over q — 1 columns, in order to bring the element in question
to the first row and first column. The new determinant is A
if ^ -f g' — 2 is even, and is — A if p + gr — . 2 is odd (§ 488).
Consequently, the sum of the terms of A that contain the
element of the pth row and g-th column is the product of that
element by its minor, the sign being -f if ^ + g' is even, and

— it p -\- q is odd.

Thus, in

<ii di (Is (I4

61 62 &8 &4
Ci C2 Cs C4

di d^ ds d\
Here, p = 3, ^ = 3, and p -|- 5 is e»en.

we find that the sum of the terms which
contain Cg is CgAc^.

499. Co-Factors. Since every term contains one element
from each row and column, if we add the sum of terms con-
taining «!, the sum of the terms containing a^, and so on, we
shall obtain the whole expansion of the given determinant.

Thus, in the determinant \a\ &a Cg dij,

A = aiAoi — OjAa, -f asAoj — a^^a^.

The expressions A^^, — A^, A^^, — A^ are called the co-factors
of the several elements a^, ag? <^3> ^4 3^^ ^i'® generally repre-
sented by Aiy A 2, A^y A^,

Hence, in the case of \ai b^ c^d^\, we may write

A = aiA^ + ^2^2 + ^s^s + ct^A^
= b^B^ -f- b^B^ -f Ms + M«
= ai^i + b^Bi -f CiCi + c?iA ;
and so on. Similarly for any other determinant.

406

COLLEGE ALGEBRA

500. If the eleTnents in any row are mtUtiplied by the
co-factors of the corresponding elements in another row, the
sum of the prodiicts vanishes.

Thus, in the determinant \ai b^ c^ d4\f
biBi + Ma + bsBj, + b^B^ =

Oi

a.

a«

Od

h

b.

».

64

Cl

e»

«»

04

di

dt

d.

d.

No one of the co-factors B^, B2, B^, B^ contains any of the
elements &i, h^y b^, ^4. Hence, these co-factors are unaffected
if in the above identity we change bi, b^, b^ b^ to a^, a,, o^, a^.
This gives

a^B^ + a^2 -{- a^Bj, -^ a^B^= "^ "J ^ "^^ sQ.

Similarly for any other case.

501. Evaluation of Determinants. By the use of § 491, § 493,

and § 499 we can readily obtain the value of any numerical
determinant.

«!

<h

^8

a^

«!

<h

as

a^

Cl

Cj

«8

«4

d.

d.

ds

d.

Evaluate

3
1
2
4

1
3
1
3

4
2
3
2

1
1
3
3

From the first row subtract 3 times the second, from the third twice
the second, from the fourth 4 times the second. The result la

0

-8

-2

-2

\

3 2 1

0

-5 -1 1

»

0

-9 -6 -1

which, by § 486, reduces to

-8 -2

-2

8 2

2

—

-6 -1

1

^

6 1

-1

-9 -

6

-1

9 6

1

= 70.

(I486)

DETERMINANTS

407

502. Simultaneous Equations. Consider the simultaneous
equations

Write the determinant

<h *i

O2 ^2

'8

8

, and let -4i, A^, Bi, B^,

etc., be the co-factors in this determinant.

Multiply the first equation by A^ the second by A^, the third
by A^ and add.

Then (a^A^ + a^A^ + d^A^x = A^i^j -|- ^^2^13 + Aj^^s,
since (§ 500) h^A^ + M2 + hA^ = 0,

and Cjili + ^2^42 + ^s^s = ^«

Hence (§ 499),

a,

Og ^2

8 ^8

a5 =

A?! bi Ci

1^2 ^2 ^2
^8 ^8 ^8

or a5 =

rCi ©2 (/;

3

1^1 ftj ^8

In a similar manner,

__ [fl^ A;2 gsl .

z =

a^ h^ h

8

Similarly for any set of simultaneous equations of the first

degree.

503. Elimination. To eliminate Xj y, and z from the four

equations

a^x -h % + c^z 4- rfj = 0,

Oj* + % + c,« + ^2 = 0>

«rK + ^«y + ^8« + ^8 = ^J
a^ -f h^ + C4« 4- 6^4 = 0,

408

COLLEGE ALGEBRA

we substitute in the fourth equation thevalues of x, y, z found
from the first three ; viz. (§ 602),

X = —

1^1 K Csf

_ |«1 ^2 ^8

Z = —

^1 ftj Cj I

Then — a^

di h

ds

h

3 ^3

-h

Oi di
do

— c.

^2
^3

^3 ^8

*1

d,

<h

<^2

+ C^4

«a

<^3

«8

*1

= 0,

This equation, by § 499, may be written

a.

ac

a

3

a.

d,

d^

dz
d.

= 0.

Observe that this determinant is the determinant formed by
the sixteen coef&cients.

Similarly for any other set of simultaneous equations.

This determinant is called the eliminant of the system of
the given equations.

The eliminant of a system of n equations with » — 1
unknowns is the determinant formed by eliminating all tho
unknowns from the system.

(1) Eliminate y and z from the equations

2aj8 + 3y+ » = 0, .
3aj + H- y-h2» = 0,
4a;«-3y-f 4« = 0.

DETERMINANTS

409

The result is

which reduces to

205*

3

1

3aj + l

1

2

4aj2

-3

4

8a;«-

-9x

-3

= 0,

= 0.

(2) Eliminate x from the two equations

2y^-\-3x +4 = 0.
[1] is 4aj«+ Syx + 5 =0

Transpose [2], 3 x -\- {2 y^ -\- 4) = 0

Represent x^ by u, and eliminate u and x.

4 Sy 5

Then, 3 2y2^4 o = 0.

0 3 2y2^4

(3) Eliminate x from the two equations

ax^ -\-bx + c =0,
a'x + c' = 0.

[l]is

Multiply [2] by x,

[2] is

Eliminate x^ and x,
which reduces to

ax2+6x+c=0^
a'x^ + c^x =0
a'x + c' = 0

= 0,

— + -- — -0

a

6

c

a'

c'

0

0

a'

c'

[1]

[2]

This must be the condition that there exists a value of x which satis-
fies both equations, since it is assumed that such is the case when we
apply the process of elimination. ..i

We have obtained, therefore, the condition that the two given equa- )V.
tions have a common root.

t,r

In general, the eliminant of a system of n equations with
n — 1 unknowns is that fimction of the coefficients
equations which becomes zero when the equations have
mon roots, and only then.

s with I I
of the I j
e com- 1 1

410

COLLEGE ALGEBRA

Ezerciae 71

1. In the determinant [aj b^ Cg d^\ write the co-factors of a^

^2f ^4) ^1> ^4) ^29 ^8*

2. Express as a single determinant

e f 9

b 6 g

h g f

h f e

f h k

4-

c f k

-h

c k h

+

c h f

9 ^ I

d g I

d I k

d k g

3. Write all the terms of the following determinant which

contain a :

a
a
0
0
b

Expand :

4.

a

b

b

a

a

a

0

a

Find the value of :

7.

3 2 2 2

2 3 2 2

2 2 3 2

2 2 2 3

(K

'y

8.

Solve the equations :

3aj — 42^ + 2« =

10. 2x + Sy — Sz=^

5x — 5y -\- 4:Z =

0 b

b c

c b

0 0

0 0

e b

b 0

c 0

b c

c b

b

a

a

b

5.

b

b

b

b

0
a
b
c

d
0
b
c

d

d

a
0

a
b

6.

c

0

3 2 14

16 29 2 14

9.

16 19 3 17

33 39 8 38

1

a

a

a

1

b.

a

a

1

a

b

a

1

a

a

b

2

1

3

4

7

4

5

9

3

3

6

2

1

7

7

6

1 ^ ■ -
-1

\

^ 4aj — 7y + «

^^ 11. Sx
3.->*S 5 X

!e-7y + « = 16t X
-6y-3« = 10j j

DETERMINANTS

411

12.

13.

4:X-\-7y-\-3z-3w= 6l

3x + 2y-7z-4:w= 2
5x — 3y-\-z-i-5w=^13

5x -\- y ^ z -\-2w = 9
2x-i-3y — 7z-i-3w = 14:
4:X — 4:y -}- 3z — 5w = 4

>•

14. Eliminate y from the equations

x^-\-2xy-\-3x-\-4:y + l
4aj + 3y + l

::}

16. Eliminate m from the equations

mh: — 2??W5
m -i- x^

^2^1 = 01
-3waj = 0j

Find the eliminant of :
ax^ + bx -{• c =

16.

17.

18.

ctx^ -\- bx -i- c
a'x^ -h ^>'a; -|- c'

ax^ -\- bx -i- c = 0
x* -\- qx -{• r = 0

::}

Are the following equations consistent ?

3x^ + 4:xy + 4:X-i-l = 0^

^ + ^ = "L 20.
2x^-\-

4x^ + 3x + 2 = 0l ,^ .-3y-7 = 0

21. If (I) is one of the complex cube roots of 1, show that :

1 0) 0)2 1

= -4;

1 -

CD

— (I)

(I)

2

(I)

<*« 1

1 — 0)

2

(I) (I)
0)" 1

1 1

(I)

1
1

1

CD

CD

0)

S

= 3 V"ir3.

412

COLLEGE ALGEBRA

22. Show that in any determinant there are two terms
which have all but two elements alike; and that these two
terms hare different signs.

23. Show that the sign of a determinant of order 4 i7t -f 2
or 4 m -f 3 is unchanged if the order of both colnmns and
rows is reversed,

504. Product of Two Determinants. Consider the determinant

By § 491 this determinant may be expressed as the sum of
twenty-seven determinants, of which the following are lyj^es :

Or^a^ a^a^ a^,

aiOTi ^2^1 6g/5i

Oiffi iJj^i <^yi

a^a^ a^j o^j

*

a^a^ a^a^ b^

f

Oi^a ^ Cgy.

«i«8 ^Vs ^s

<h^z a^, h^

«i^s *A ^yt

There are three determinants of the first type, eighteen of
the second type, and six of the third type. Those of the first
and second types are easily seen to vanish (§§ 489, 490). There
remain the six determinants of the third iypa

Consider any one of these six determinants, as

«2^2
«2^8

1

M

(K ^

This may be written
yi «i Pi

(\

- Ci(ij)z

72

a,

2
0[8

A

or

c^o^

A

A
A

ys

It is evident that the number of interchanges required to
bring the columns into the order a fi y is the same as the
number of inversions among the letters a, fi, y; and also the

DETERMINANTS

413

same as the number of inversions among the letters a, h, c.
Hence, the sign is -f if that number is even, and — if that
number is odd. The sign before c^ajb^ is, therefore, the sign of
this term in the determinant \a^ h^ c^\ (§ 482, II).

Since the preceding is true for each of the six determinants
of the third type, the given determinant is the product of
the determinant [«! h^ c^\ by the determinant \ai P^y^]^ and
is one of the forms in which the product

«!

h

Cl

^I

A

ri

<^2

h

C2

X

«2

ft

72

^8

h

Gz

«Z

A

73

may be written.

The above proof is perfectly general and may be extended
to the product of any two determinants.

b

(1) Write as a determinant

a h
c d

The result is

ac + hd c2 + d2

a
c

d

(2) Write as a determinant the product

a

b

c

X

y

z

c

a

b

X

z

X

y

b

c

a

*

y

z

X

The result is

X

Z

Y

Y
X
Z.

Z
Y
X

where X =z ax '\- Iry -^^ cz^ F=cx + ay + 6z, Z = bz '\- cy -\- aa.

505. The notation

^1 ^ ^8 ^4
bi *2 *S ^4

2
^2

= 0

is used to denote that the four determinants obtained by
omitting in turn one of the four columns all vanish.

414

COLLEGE ALGEBRA

Exercise 72

a b

0

0

a b

c 0

c

X

c

0 c

0 b

a

b

a 0

1. Show that

2. Express as a single determinant

= -4a»^c*.

0

c b

0

c b

c

0 a

X

c

0 a

b

a 0

\

b

a 0

3. Express as a single determinant

a a a a — 1 1
a ft ^ ^
a ft c c
abed

X

0
0
1

1
0
1

0
1
1
1

0

0

1

-1

and thence resolve the first determinant into its simplest
factors.

4. Express as a single determinant

a -\- bi
c + di

c + di

a — bi

X

o^-\- pi - y -h St

where i = V— 1 ; and thence prove Euler's theorem, viz.,
the product of two sums of four squares can itself be expressed
as the sum of four squares.

Ai A2 A^

a^ 0^2 <

0^

2

5. Show that

By B2 Bq
Cj C2 C3

■ ■

ftl *2 ^8
Ci Cj C3

•

ft-fc c-f-a a -\-b

a i^ 0

6. Show that

^1 + ^1 Cj + Oi Oi + ^1

= 2

«i ^1 ^

ft2 + ^2 C2 +

^2

a2 + ft2

a, ft, ei

CHAPTER XXX

GENERAL PROPERTIES OF EQUATIONS

506. Algebraic Functions. A function of a variable x has

already been defined (§ 373) as any expression that changes

in value when x changes in value. Any expression that

involves x is, in general, a function of x. If x is involved

only in a finite number of powers and roots, the expression is

an algebraic function of x,

J 1

Thus, x*, vaj2 + X, ^ are algebraic functions of x ; but a*, logx,

are not algebraic functions of x.

507. Rational Integral Functions. An algebraic function of
x is rational as regards a;, if x is involved only in powers ; that
is, not in roots. An algebraic function of x is rational and
integral as regards x, if x is involved only in positive integral
powers ; that is, in numerators and not in denominators.

1 1 X 3 X* + 4

Thus, -r» x-«, - — — r» -T— — 5» - g . Q — -TT are rational, but not
x* '4x-f3x2-fa2 6x8 + 3x-f2

integral, algebraic functions of x ; while 4x2 + 3x -f 7, ox* + te* + ex + d
are rational integral algebraic functions of x.

508. Quantics. An algebraic function that id rational and
integral with regard to all the variables in it is called a quantic.

We shall consider in this chapter only functions of one
variable, and by quantic will be meant a rational integral
algebraic function of one variable, unless it is expressly stated
that several variables are involved.

Note. The term quantic is generally applied only to homogeneous
expressions like ox* -f hxy + cy\ This expression is obtained from

X

0x2 ^ 6g; + c by putting - for x, and multiplying through by y*. The

416

416 COLLEGE ALGEBRA

theory of the two expreniong is precisely the same, and we shall there-
fore extend the term quantic to include ezpressioxis like oa^ + 6gb + e,

ox* + te* + ex + d, etc.

The degree of a quantic that inyolyea only one yariable x is
indicated by the exponent of the highest power of x inyolved
in the qnantic ({ 122).

A quantic of the first degree is called a linear fimetion ;
qnantics of higher degrees are called quadraiies, cubics,
biqtmdratics or quartics, quintics, etc

509. General Form. Any quantic of the nth d^iee in which
X is the yariable may be written in the form

where a^ Oi, a,, . . . ^ o^n-u ^» ^^ coefficients which do not
inyolye x. Some of these coefficients may be lero, and in that
case the corresponding terms are wanting.

The coefficients may be real or complex, snrd or rational
expressions. We shall, in general, consider only qnantics
that haye real and rational coefficients. The student will
readily see what properties of such quantics are possessed by
quantics that have surd or complex coefficients.

510. Abhreyiations. For brevity a quantic that involves x
is often represented by /(x), F(x)f 4^(?^i or some similar
notation.

If any quantic is represented by /(«), it is represented by
f(a) when a is put for x.

Thus, if /(x) = 2x8 - x2 4. 3x + 4,

/(2) = 2(2)8 _2« + 3(2) + 4=:16-4 + 6 + 4 = Stt.

•

511. Equations. Every equation that contains no variables
except rational integral algebraic functions of x can, by the
transposition of all the terms to the first member, be made
to assume the form f(x) = 0, where f(x) is a quantic that
involves the one variable x. The theory of this qnantio and

GENERAL PROPERTIES OF EQUATIONS 417

that of the corresponding equation are closely related, and we
shall develop the two together.

The roots of the equation f(x) = 0 are those values of x that
satisfy the equation. These roots are also called the roots of
the quantic.

The degree of the (equation /(«) = 0 is the same as that of
the quantic f(x),

512. Divisibility of Quantics. Theorem I. If h is a root of
the equation f (x) = 0, the quantic f (x) is divisible % x — h.

For example, consider the quantic

f{x) = a^x- + ai^r"-^ + c^""' + • • • + a^^^x + a,. [1]
Now, since A is a root of the equation f(x) = 0, we have
0 = ttoA" + aiA— > + a2^--2 ^ ^ ^^_^^ _,. ^^ |-2-|

Subtract [2] from [1],

+ a»-2(«* - ^^ + a^-i (aJ - A).

Each of the expressions af — A**, a;""^ — A"""*, a;"""^ — A""*,
•••, X — h, is divisible by a; — A (§ 86). Therefore, /(a:) is
divisible by aj — A. Similarly for any other quantic. Com-
pare §§ 87, 495.

513. Theorem IL Conversely, if a quantic f (x) is divisible
fty X — h, then h is a root of the equation f (x) = 0.

For, if </> (x) is the quotient obtained by dividing f(x) by

X — hy we have

f(x) = (x-h)<l>(x).

Hence, equation /(a;) = 0 may be written

(aj - A) </> (aj) =0,

of which A is evidently a root (§ 124).

514. Synthetic Division. Divide the quantic

3a:*-4a:*-haj»-12aj« + 3aj-|-6 by a; - 2.

418 COLLEGE ALGEBRA

3aj6_4x*+ z8-12z2 + 3a; + 6|g-2

H-2ic*+ X*

+ 2x*-4x»

+ 6x8-12a^

- 2x3 + 3«

- 2x2 + 4x

- x + 6

- x + 2

+ 4

The work may be abridged as follows, by omitting the powers of x and
writing only the coefficients (§ 70) :

3-4 + 1-12 + 3 + 611-2

3-6 3+2+5-2-1

+ 2 + 1
+ 2-4

+ 6-12
+ 5-10

- 2 + 3

- 2+4

-1 + 6

-1+2

+ 4

The operation may be still further abridged. As the first term of the
divisor is unity, the first term of each remainder is the next term of the
quotient, and we need not write the quotient. Further, we peed not
bring down the several terms of the dividend or write the first terms of
the partial products. Thus,

3-4 + 1-12 + 3 + 611-2
-6
+ 2
-4
+ 5

-10
- 2

+i

-1
+2

+ 4

GENERAL PROPERTIES OF EQUATIONS . 419

If we omit the first term of the divisor, which is now useless, change
— 2 to +2, and add, we may shorten the work to

3-4 + l-12 + 3 + 6[2

+ 6 + 4 + 10-4-2
3+2+6- 2-1+4

The last term below the line is the remainder, the preceding terms the
coefficients of the quotient. In this particular problem the quotient is
3aj* + 2x' + 5a;2 — 2x — 1, and the remainder is 4.

This method is called synthetic division. For dividing
any quantic in cc by cc — A we have the following rule :

Arrange the quantic according to descending powers of x,
supplying any missing powers of x by these powers with zero
coefficients.

Write the coefficients a, b, c, etc., in a horizontal line.

Bring down the first coefficient a.

Multiply a by h, and add the product to b.

Multiply this sum by h, and add the product to c.

Continue this process ; the last sum is the remainder, and the
preceding sums the coeffi4)ients of the quotient.

Divide 2a;* - 6aj» +- 5* - 2 by « - 3.

2 + 0- 6+ 5- 2|3

+ 6 + 18 + 36 + 123
2 + 6 + 12 + 41 + 121

The quotient is 2 x* + 6 x^ 4. 12 x + 41, and the remainder 121.

515. Value of a Quantic. If we put h for x in the quantic,

f(x) = ttox* +- aiX*-^ H h a^_iX +- a^

f(h) = tto^* +- ttiA**"^ H h h^_^iX + a^.

.'. f(x) -f(h) = ao(«» - A-) +- oi (aj—i - A—')

H ha»-i(a5 — A). [1]

Divide the right member of [1] by a; — A and represent the
quotient by 4t(x).

Then, f(x) - f{h) = (x - A) </> («).

.\f{x) = (a? - A) t^ (x) +f(h).

420 COLLEGE ALGEBRA

HencBj the value which a quantic f (x) ctssumes when we put
h. for X is equal to the last remainder obtained in the operation
of dividing f (x) by x — h.

This remainder, and, consequently, the value of the quantic,
may be easily calculated by synthetic division.

The truth of the above theorem may also be shown by
another method, which has the advantage of showing the
form of the quotient and remainder.

For example, divide the quantic ax^ -}- bx^ '\- cx^ -{- dx -^ e by

a b c d e \h

ah Bh Ch Dh
a B C D R

where B = ah -\- b,

C =Bh +c =ah^-i-bh + c,

D = Ch -{- d = ah^ + bh^ -}- ch -\- d,

R = Dh+ e = ah'^-\-bh^ + ch^ -\- dh + e.

The remainder R is evidently the value which the quantio
assumes when we put h for x.
The quotient is

ax^ -f- {ah -\-b)x^-\- (ah^ + bh-\-c)x-\- (ah^ + bh^ + cA -f- d).

Similarly for any other quantic.

Exercise 73

Find the quotient and the remainder obtained by dividing
each of the following quantics by the divisor opposite it :

1. «*-3a;'^-«^ + 2a;-l. a -2.

S. «j«-3a;2 4-2a;-7. a? - 3.

3. 2a;* + 3a;8-8a;2-7a;-10. « - 2.

4. 3aj* + 2aj2-6a;-f-50. x + 3.

5. ax^ + 3bx^-\-3cx + d, x + h.

GENERAL PROPERTIES OF EQUATIONS 421

Determine whether each of the followmg numbers is ^ root
of the quantic opposite it (§ 513) :

6. (3). a;* + «*-6a; + 2 = 0.

7. (-7). x^-\-7x^-\-21x + U7 = 0.

8. (0.3). «* - 2.3 x^ + 3.6 a;2 + 4.9 a; -h 1.2 = 0.

Find the value of each of the following (juantics when for
X we put the number opposite :

9. 3x^-\-2x^-ex-\-l. (-3).

10. 2a;* + 6aj2-9a;-6. (6).

11. «» + 7a;*-2a;2-49. (-4).

12. x^ + ex^-Tx^-Sx-i-l. (-0.2).

516. Number of Roots. We shall assume that every rational
integral equation has at least one root. The proof of this
truth is beyond the scope of the present chapter.*

Let f(x) be a rational integral quantic of the nth degree,
and let f(x) = 0. This equation has, by assumption, at least
one root. Let ai be a root.

Then, by § 512, f(x) = (x - a,)f, («),

where /i (a:) is a quantic of degree n — 1.

The equation /i (x) = 0 must, by assumption, have a root.
Let a2 be a root.

Then, by § 512, f, (x) = (x ^ a,)f, (x),

where /2 (x) is a quantic of degree n — 2.

Continuing this process, we see that at each step the degree
of the quotient is diminished by one. Hence, we can find
n factors a; — ai, aj — ■ a2, • • • , x — a^. The last quotient will
not involve x, and is readily seen to be a©, the coefficient of
x" in f(x),

* See Bumside and Panton^s Theory of EquaJtwM. H. Weber's TraiU

d'Algkbre Sup&rieure.

422 COLLEGE ALGEBRA

Now, f(x) = (x- a;)f^ {x)

= (x- ai) (x - aa)/a(aj)

= ao(x — ai) (« — aj) • • • (oj — a^).

Therefore, the equation f(x) = 0 may be written

ao(x — ai) (a; — oTj) • • . (a; — a^) = 0,

which evidently holds true if x has any one of the n values
ai, as, ..., a^.

It follows, then, that if every rational integral equation has
at least one root, an equation of the nth degree has n roots,

517. Linear Factors. The factors x — ai,x — a:^ •••,« — a^
are linear functions of x (§ 508).

When f(x) is written in the form

ao(x — aj) (« — ttj) . . . (a; — a.),

it is said to be resolved into its linear factors.

From § 516 it follows that a quantic can be resolved into
linear factors in only one way.

To resolve a quantic f(x) into its linear factors is evidently
equivalent to solving the equation f(x) = 0.

518. Multiple Roots. The n roots of an equation of the nth
degree are not necessarily all different.

The equation a;' — 7 a' + 15 a; — 9 = 0 may be written

(a;-l)(a;-3)(a;-3) = 0,

and the roots are seen to be 1, 3, 3.

The root 3 and the corresponding factor a; — 3 occur twice ;
hence, 3 is said to be a double root When a root occurs three
times it is called a triple root; four times, a quadruple root;
and so on.

Any root that occurs more than once is a multiple root

519. Roots given. When all the roots of an equation are
given the equation can at once be written.

GENERAL PROPERTIES OF EQUATIONS 423

Write the equation of which the roots are 1, 2, 4, — 6.

The equation is (x — 1) (x — 2) (x — 4) (x + 6) = 0,
or X* - 2x8 - 21 x2 + 62x - 40 = 0.

520. Solutions by Trial. When all the roots but two of an
equation can be found by trial the equation can be readily
solved by the process of § 516. The work can be much
abbreviated by employing the method of synthetic division
(§ 514). (Compare § 180.)

Solve the equation a* — 3 oj' — 6 a:^ + 14 cc -f- 12 = 0.

Try + 1 and — 1. Substituting these values for x, we obtain

1 _ 3 _ 6 + 14 + 12 = 0,
1 4- 3 _ 6 - 14 + 12 = 0,

which are both false, so that neither + 1 nor — 1 is a root.

Try + 2. Dividing by x - 2,

l_3-6 + 14 + 12[2
+ 2-2-16-4

1_1_8- 2+ 8
we see that + 2 is not a root.

Try - 2. Dividing by x -f 2,

1_3_ 6 + 14 + 12 1 --2

-2 + 10- 8-12
1 _6+ 4+6 0

we see that — 2 is a root. The quotient is x'* — 6 x* + 4 x + 6.

In this quotient try — 2 again. Dividing by x + 2,

1-6+ 4+ 6|-2

-2 + 14-36
1-7 +18 - 30

we see that — 2 is not again a root.

Try + 3. Dividing by x - 3,

1 _5 + 4 + 6[3
+3-6-6

l_2-2 0
we see that + 3 is a root. The quotient is x* — 2x — 2.

424 COLLEGE ALGEBRA

Hence, the giyen equation may be written

(X + 2) (X - 3) (x2 - 2x - 2) = 0.

Therefore, one of the three factors must vanish.

Ifx+2=0, x = -2; ifx-3 = 0, x = 3; ifxa-2x — 2=0, by
solving this quadratic, we find x = 1 4- Vs or x = 1 — Vs.
Hence, the four roots of the given equation are

-2, 3, I+V3, I-V3.

Ezerciae 74

Solve :

1. a;»-7aj» + 16a; -12 = 0. 4. a;» + 9a;* + 2a5 — 48 = 0.

2. x*-5x^-2x + 24t = 0, 6. aj» - 4a« - 8a; + 8 = 0.

3. x*-6x^ + 6x-\-99 = 0. 6. «» -f 2a;« -f 4aj + 3 = 0.

7. 6x8 -29x2 + 14a; + 24 = 0.

8. 2a;» + 3a;2- 13a; -12 = 0.

9. a;*-15a;^-10a; + 24 = 0.

10. a;* + 5a;»-5a;2-45a;-36 = 0.

11. a;* + 4a;«-29a;2- 156a; + 180 = 0.

12. a;*-5a;«-2a;2 + 12a; + 8 = 0.

13. 6a;*-5a;«-30a;« + 20a; + 24 = 0.

14. 4a;* + 8a;«-23a;2-7a; + 78 = 0.

Form the equation which has for its roots :

15. 2, 6, - 7. 20. 5, 3 + V^, 3 - V^.

16. 2, 4, - 3.' 21. 2, i, 2, - f

17. 2, 0, - 2. 22. 2, 3, - 2, - 3, — &

18. 2,1,-2,-1. 23. hh-h-i-

19. 0.2, 0.125, - 0.4. 24. 0.3, - 0.2, - J^^, - f ,

25. 3+V2, 3-V2, 2+V3, 2-V3.

26. 2+V^,2-V^,l + 2V^,l-2V^.

GENERAL PROPERTIES OF EQUATIONS 426

521. Relations between the Roots and the Coefficients. The

quadratic equation of which the roots are a and ^ is (§ 193)

(a.-a)(«-i8) = 0,
or X* — (a + p)x + ap = 0.

The cubic equation of which the roots are a, /5, y is
(x-a)(x^p)(x-y)=0,
or cc» - (a H- ^ + r)a^ 4- (ctfi + «y + fiy)x - apy = 0.
The biquadratic equation of which the roots are a, fi, y, 8 is
(x ^a)(x- P)(x - y)(x - 8)= 0,
or aj* _ (a + j8 + y + S) a;» + (a)8 + «y + «S + /8y 4- iSS -f y*) «*

- (afiy + afiS + ayS + )3y8) X + afiyi = 0.

Similarly for equations of higher degree.

Take any equation in which the highest power of x has the
coefficient unity. From the above we have the following rela-
tions between the roots and the coefficients :

The coefficient of the second term, with its sign changed, is
the smn of the roots.

The coefficient of the third term is the sum of all the prod-
ucts that can be formed by taking the roots two at a time.

The coefficient of the fourth term, with its sign changed, is
the sum of all the products that can be formed by taking the
roots three at a time.

The coefficient of the fifth term is the sum of all the
products that can be formed by taking the roots four at a
time ; and so on.

If the number of roots is eveuy the last t&rm is the product
of all the roots. If the number of roots is odd, the last term,
with its sign changed, is the product of all the roots.

Observe that the sign of the coefficient is changed when an
odd number of roots is taken to form a product ; that the sign
is unchanged when an even number of roots is taken to form
a product.

426 COLLEGE ALGEBRA

522. Reduction to the p Form. By dividing the equation by

the coefficient of the highest power of x, any rational integral
algebraic equation whatever can be reduced to a form in which
the coefficient of the highest power of x is unity.

We shall write an equation reduced to this form, called the
pform, as follows :

Let a, py yy '-he the roots of this equation. Represent
by So: the sum of the roots, by Sor^S the sum of all the products
that can be formed by taking the roots two at a time; and
so on (§ 102).

From § 521 we now have

aPyB ...=(- l)"p,. p, = (- lyafiyi

• • •

Thus, let a, j9, y he the roots of the equation

x8- 7x2 -9a + 4 = 0.
Then, 2a = a4-i8 + 7 = 7,

2a/3 = /37 + 7a 4- a/3 = - 9,

aj87 = — 4.

The relations between the roots and the coefficients of an
equation do not assist us to solve the equation. In every case
we are brought at last to the original equation.

Thus, in the equation

x»- 7x2 -9x4-4 = 0,

we have a + /S + 7 = 7,

/97 + 7a + a/S = - 9,
a/87 = — 4.

If we eliminate any two of the three unknowns as /3 and 7, we have to

solve the equation

a«-7a«-9a + 4 = 0.

That is, we have to solve the given equation.

GENERAL PROPERTIES OF EQUATIONS 427

523. Symmetric Functions of the Roots. The expressions ia,
^a/S, %ocPyy ' • • are examples of symmetric functions of the
roots (§ 192). Any expression that involves all the roots,
and all the roots have the same exponents and the same coeifi-
cients, is a symmetric function of the roots.

From the relations

the value of any symmetric function of the roots of a given
equation may be found in terms of the coefficients.
If a, )3, y are the roots of the equation

a;8 _ 4a;2 4- 6aj- 5 = 0,

we may calculate the values of symmetric functions of the
roots as follows :

We have a + /8 + 7 = 4, [1]

^y + ya + a^ = 6, [2]

a^y = 6. [3]

(1) Sa2 = a2 + ^ + ^a.

Square [1], a^ + /S^ + 72 + 2/87 + 27a + 2 a/S = 16

Multiply [2] by 2, 2/37 + 270: + 2a/8 = 12

.•.a2 + /32 + 72 =4 [4]

(2) Sa2/3 = ar2/8 + a^ + /3^ + /S^a + 7^a + y^p.

Multiply [1] by [2], Za^^ + 3 ai87 = 24

Multiply [3] by 3, 3 a/37 = 16

.-. 2a2^ = 9 [5]

(3) So* = a* + /3' + 7*.

Multiply [1] by [4], cfi + ^ + y»^ J^cfip = 16

[6] is 2Qr2/3= 9

.-. a" + /88 + 7« =7

Similarly for any symmetric fimction of the roots. (Compare § 192.)

524. By the aid of the preceding sections we can find the
condition that a given relation should exist among the roots
of an equation.

428 COLLEGE ALGEBRA

Find the condition that the roots of the equation

shall be in geometrical progression.

Let p he the mean root. Then,

a + i3 + 7=-p, • [1]

/87 + 7^ + a^'- g, [2]

ai37=:-f, [3]

and ^ = ya. [4]

From [2] and [4], /87 + a/3 + jS^ = g,

or, p{y^a + p)=q.

By [1], -Pp = q>

P
Substitute in [3], fi^ for ya and — - for /3,

(-i)'--

.-. g8 = pSf>^ the required condition.

525. Complex Roots. If a complex number is a root of an
equation with real coefficients, the conjugate complex number
(§ 216) is also a root.

Let a + pi, where i = V— 1, be a root of the equation

aojc** + OiOJ**""^ -f- aao;**"^ + f- «„_!« + a^ = 0,

the coefficients of which are real.

Put a -\- fii for x in the left member of this equation^ and
expand the powers of a-\- pi by the binomial theorem. All
the terms which do not contain i, and all the terms which
contain even powers of t, are real ; all the terms which con-
tain odd powers of i are orthotomic. Eepresenting the real
part of the result by P, and the orthotomic part of the result
by Qiy we have (§ 511), since a -f- jSi is a root,

P 4- Qi :« 0.

Therefore, P = 0 and Q = 0. (§ 219)

GENERAL PROPERTIES OF EQUATIONS 429

Now put a — pi for x in the given equation. The result
may be obtained from the former result by changing i to — i.
The even powers of i are unchanged, while the odd powers have
their signs changed. The real part, therefore, is unchanged,
and the orthotomic part is changed only in sign. The result is

P- Qi = 0,

which vanishes, since by the preceding P = 0 and Q = 0.
Therefore, a — ^i is a root of the given equation (§ 611).
This theorem is generally stated as follows :

Complex roots always enter an equation in pairs.

Corresponding to a pair of complex roots, we shall have

the factors

X — a — piy X — a -{- fii.

The product of these,

is positive, provided x is real. Hence, corresponding to a
pair of complex roots, we have a factor of the second degree,
which for real values of x does not change sign (§ 220).

EzerciBO 75

1. Form the equations of which the roots are

2, 4, - 3 ; 3,-2,-4.

If a, p, y are the roots of «• — 6 x^ + 4 « — 3 = 0, find the
value of :

2. Sttl

6. Sa«)3y.

8. Sa*.

3. 2a*)3.

6. ^a^^.

9. Sa»i3y.

4. So*.

7. Sa»)3.

10. Sa^^V-

430

COLLEGE ALGEBRA

If a, p, y are the roots of cc' -hpx^ -\- qx + r = 0, find in
terms of the coefficients the value of :

16. (^ + r)(r + «)(« + i3).

a fi y

18. — ^i h 1 3 — '

py ya ap

19.

11. s«*.

12. Sa*i8.

13. Sa*.

14. Sa^)^^.

16. Sa*.
In the equation cc' + px'' -\- qx -{■ r = 0, find the condition :

20. That one root is the negative of another root

21. That one root is double another.

22. That the three roots are in arithmetical progression.

23. That the three roots are in harmonical progression.

iS + r "^ y + a "^ a-ir P

GRAPHICAL REPRESENTATION OF FUNCTIONS

The investigation of the changes in the value of f(x) cor-
responding to changes in the value of x is much facilitated
by using the system of graphical representation explained in
the following sections.

526. Coordinates. Let XX^ be a horizontal line and let
ry be a line perpendicular to XX^ at the point O,

Y Any point in the plane of the

lines XX' and YY^ is determined
by its distance and direction j&om
each of the perpendiculars XX' and
yy. Its distance from YT, meas-
ured on XX', is called the abadsaa
of the point; its distance from
XX', measured on YT, is called
the ordinate of the point.

P,

A

B«

B,

+-4

O

A

P.

■4— ♦-

Pi

Ba

Bi

'X

^3

y*

GRAPHICAL REPRESENTATION OF FUNCTIONS 431

Thus, the abscissa of Pi is OBu the ordinate of Pi is OAi ;
the abscissa of Pa is OB2, the ordinate of Pa is OA2 ;
the abscissa of Ps is OBs, the ordinate of Pg is OAs ;
the abscissa of P4 is OB^, the ordinate of P4 is 0-4.4.

The abscissa and the ordinate of a point are called the
coordinates of the point; and the lines XX' and YY' are
called the axes of coordinates. XX' is called the axis of
abscissas or the axis of x; YY' is called the axis of ordinates
or the axis of y; and the point 0 is called the origin.

An abscissa is generally represented by x, an ordinate by y.
The coordinates of a point are written thus : (x, y).

Thus, (7, 4) is the point of which the abscissa is 7 and the ordinate 4.

Abscissas measured to the right of YY' are positive^ to the
left of YY' are negative. Ordinates measured above XX' are
positive, below XX' are negative.

Thus, the points Pi, Pj, Ps, P4 are respectively (3, 4), (-4, 3),
(-3, -4), (4, -3).

527. It is evident that if a point is given, its coordinates
referred to given axes may be easily found.

Conversely, if the coordinates of a point are given, the
point may be readily constructed.

Thus, to construct the point (4, — 3), a convenient length is taken as a
uuit of length. A distance of 4 units is laid off on OX to the rights from
O to £4 ; and a distance of 3 units on OY^ downwards^ from O to A^,
The intersection of the perpendiculars erected at B^ and A^ determines
the required point P4.

Construct the points (3, 2) ; (5, 4) ; (6, - 3) ; (-4, - 3) ;
(-4,2); (-3,-5); (4,-3).

528. Graph of a Function. Let f(x) be any function of «,
where a; is a variable. Put y =/(ic); then y is a new vari-
able connected with x by the relation y =/(«). Iif(x) is a
rational integral function of x, it is evident that to every
value of X there corresponds one, and only one, value of y.

4S2

COLLBGE ALGEBRA

If different valnes of x axs laid off ae absolBBas, and the
eorreaponding values of f(x) as ordinates, the points thus
obtained all lie on a line. This line in general is a curved
line, 01 a curve, and is called the graph of tlie function f(x);
it is also called the locus of the equation y = f (z).

(1) Construct the graph of 3 — 2a:.

Put y = Z~2x. The foHowing table is teadilj compated;

= -1,

* = + 3,
» = + 4.

OSH-S, » = +!«.

Constructing the above points,
It appears HiaX the grapb of the
fimctioD 3 — 3 a: is the straight
Une MN.

^ In general, if the quantio /(z)

; the flrst powers of x and y, the gr^h is a straight Une.

Construct the graphs of the following functions :

1. 333 + 2. 6. }(J-2x).

2. x-5. 7. 3(9-3a!).
8. a; + 6. 8. i(i + Bx).

4. i(x-5). 9. C«-2)(x-3).

5. iix + 6). 10. Ba!»-17a!-ia.

GRAPHICAL REPRESENTATION OF FUNCTIONS 483.

(2) Plot the graph of ^x* - 4.

Pat y =: ^a^ ^ 4. We readily compute the following taUe :
Ifaj = + 0, y = -4; T

« = ±1, y = -3.6;

x = ±2, y = -2;

x = ±3, y = + 0.6;

a=it4, y = + 4;

« = ± 5, y = + 8.6 ;

aj = ±6, y = +14.

Plotting these points, we obtain
the curve here given.

(3) Plot the graph of
05* — a* 4- « — 5.

,^t y=:a5»-a!« + x-6. We
compute the following table :

If x=+0.6
05 = + 1,0
05 = + 1.6
05 = 4-2.0
05 = H- 2.6
X = + 0.0
X = - 0.6
X = — 1.6

y = - 4.626

y = - 4.000

y = - 2.376

y = + l.OOO

y=H- 6.876

y=:- 6.000

y=- 6.876
y = - 12.126.

X'

Interpolation (§ 433) shows that
if y = 0, X = 1.88+. Does the
result agree with the figure?

r

1

1

1

f

f

1

1

t

f

1

h

irf

529. Consider any rational integral algebraic function of x,
for example oj* + a; — ^.

Put y = X« + X - V-

Assume values of x, compute the corresponding values of y^
and construct the graph. Now, any value of x which makes

434

COLLEGE ALGEBRA

y = 0 satisfies the equation x^ -{- x — ^- = 0, and is a root
of that equation. Hence, any abscissa whose corresponding
ordinate is zero represents a root of this equation. The roots
may be found, approximately, by measuring the abscissas of
the points in which the graph meets XX'^ for at these

points y = 0.

I From the given equation the following

I table may be formed :

X/ ' " » ' I '

■ '.'I I 'X

If

» = + 0,

» = +l,
x = + 2,
x = + 3,
x = + 4,

If

X
X
X
X
X

1,

2,
3,
4,
5,

y=- 15.76
y=- 13.76
y = - 9.76
y = - 3.75
y = + 4.25.

y = - 15.76 ;
y = - 13.75 ;
y = - 9.75;
y=- 3.75;
y=4- 4.25.

The table shows that one root of /(x) = 0 lies between 3 and 4
(since y changes from — to +, and therefore passes through zero) ;
and, for a like reason, the other root lies between -> 4 and — 6.

530. An equation of any degree may
be thus plotted, and the graph will be
foimd to cross the axis XX' as many
times as there are real roots in the
equation.

When an equation has no real roots
the graph does not meet XX',

In the equation x^ — 6x 4- 13 = 0, both of ^/.^
whose roots are imaginary, the graph, at its
nearest approach, is 4 units distance from XX\

GRAPHICAL REPRESENXATION OF FUNCTIONS 486

If an equation has a double root^
its graph touches ZZ', but does not
intersect it at the point of contact.

The equation

x2 + 4« + 4 =0

has the roots — 2 and — 2, and the graph
is as shown in the figore.

XL
Ezerciae 77

-• I

. I
./

/■■

I I I i^-K I

4.x

Y'

Construct the graphs of the following functions :

1. a;2 + 3a;-10. 4. a;2-4»4-10.

2. x^-2x^-\-l. 6. a* -5a;* + 4,

3. a;*-20x« + 64. 6. x^-^x^ + x

^1.

531. Change of Origin. Consider the function

y = a* -f 4 a; - 1. [1]

Construct the graph of the given function, for convenience
using on the coordinate paper 3 spaces for 1 horizontal unit,
and 1 space for 4 vertical units.

If a;=+0
aj=+ 1
x=-\-2
a;=4-3
a; = + 4
X =4- 5

2^ = -i;

y= + 4;

2^ = 4-31
y = 4-44.

Ifa;=— 1, ya-s— 4

x=—2, y-=— 5

x = —Sf y^ — 4:

« = — 4, y=— 1

« = — 5, y = 4-4

aj=~6, y = +ll.

Now change the origin from its present position O to any
point R on the axis of abscissas, keeping the axis of ordinates
RY' parallel to its original position OY. This change does
not alter the values of the ordinates of points, but does alter
the values of the abscissas.

COLLEGE ALGEBRA

PH L i k'

The value of the given funetion ofxia altered.

For, let OR = A. The old coordinates of any point P of

the graph are OM = x, MP = y. Let x' denote the new
abscissa RM of the point P.

Then, x = x' + k.

Substitute x' + h for x in [1].

Then, y=(x' + A)« + 4(a;' + A)-l

= i'* + (4 + 2 A)a!' + A" + 4 A - 1.

Write a: for x', and we have the transformed functioD
y = a;" + (4 + 2 A) X + A* + 4 A - 1.

Hence, irhen the origin is moved along the axis of as a dis-
tance of A units, the ne^ function of x is obtained by substi-
tuting X 4- A for X in the old function of x.

If the origin is moved a distance of A units to the l^ of 0,
the value of A must of course be regarded aa negative.

EtxnolM 78
1. Transform the function j/»a;* — 6a! + 5 feo a iieir
origin, the point (5, 0).

7 .

\3-c.

4 A'

» 1 N

'- V .0.

» \

DERIVATIVES

487

2. Transform the function y = 4aj*4-3a5 — 10 to a new
origin, the point (— 2, 0).

3. Transform the function y = 3aj' — lOoj'-fOaj — 2 to
a new origin, the point (2, 0).

4. Transform the function y = a* — 1 to a new origin, the
\ point (- 1, 0). f^

-- ~ - _ _ -DERIVATIVES

H H*

532. Definitions. Let MN be a part of the graph of a func-
tion of X, as

/(aj) = 2 + V12 X - aj2

Let y = 2 + Vl2 aj - a;^ [l]

Let P be any point on the graph. Draw the coordinates OH
and HP of that point.

Let X = OH, and y = HP.

It is obvious from [1] that y =f(x). [2]

Add to X any arbitrary amount ////'.

Draw H'P' ± to XX', and draw PA' || to XX'.

Let a;' =e OH*, and y' = P'H*,

438

COLLEGE ALGEBRA

/:

It is obvious from [1] that

y' =/(*')•• [3]

It is evident that when HW is added to a, y changes to y\
and that the amount of change in y is KP\

The arbitrary amount HH^ added to x is called the incre-
ment of jr. This is written Aic
and read delta x.

Similarly, the amount KP*
added to y is called the incre-
ment of/.

Let bkX = the increment of «,

and Ay = the increment of y.

Then, £^y = KP^, tC

X and since it is added to y, the
increment is positive.

When the increment of y is
taken from y, the increment is negative.

Hence, an increment may be either positive or negative.

The increment of a variable is any arbitrary amount added to
the variable.

The increment of a function is the amount of the change pro-
duced in the function when an increment is given to the
variable of the function.

Now, x' = x -\- Ax,

Hence, by [3], y' =f{x + Ax). [4]

Agam, Ay==y' — y. ' ^ - ^.

Hence, by [2] and [4], Ay =f(x + Ax) -/(«). [6]

Therefore, to find the increment of a function when the variable
takes an increment,

Subtract the original lvalue of the function from the indue of
the function after the variable ha^ taken an increment.

DERIVATIVES 439

Divide [5] by Aa.

Then, Ay^/(x + /^)-/(«),

Now let P remain fixed and let P' move towards P along
the curve in such a way that we can make it approach P as
nearly as we please.

Then, Ax is an infinitesimal, and the fraction — ^ is, in gen-

eral, a variable, and this triable, in general, approaches a
definite limit.

When the variable does approach a definite limit this limit
is called the derivative of/ or the derivative of f(x).

The derivative of /(x) is ^^ ^ •

By [6] it is seen that the derivative oif(x) is

limit /(x + Aa;)~/(x)

The first form of this definition is the more significant when
we wish to show the relation of the increments to the deriva-
tive ; the second is the more significant when we wish to show
the relation of the function to the derivative.

The derivative with respect to x of f(x) is represented by
D^f(x) ; that of f{y) with respect to y by Dyf(y) ; that of v
with respect to w by D^v ; and so on.

The derivative of f(x) with respect to x is also represented

by fix).

Thus, DJ'(x) =f(x) ; Dyf(y) =f (y) ; and so on.

533. From [7] may be deduced the following rule for finding
the derivative of a function :

Divide the increment of the function by the increment given to
the variable.

Find the limit of this quotient when the increment of the vari-
able is an infinitesimal.

This limit is the derivative of the function.

440 COLLEGE ALGEBRA

Denote the derivative oif(x) by /'(«).

Then, /'(.). ^^t„/^£±^^:/^. X

(1) Given/(a;)=5a;2j find /'(«).

f{x + Ax) = 6(x + Ax)2 = 6x« + lOxAx + 6(Ax)«

/(x) = 6x2.

/(x + Ax) ~/(x) = 10x*Ax + 6(Ax)«.

/(x + Ax)-/(x)^^^^^^^
Ax

limit /(g + Ax)-/(x)^^^^
Ax = 0 Ax

.•./'(x) = 10x.

(2) Find i), (a;8 + 4 a; 4- 1).

The function is x»4-4x + l.
Change x to x + ^, (x + ^)8 + 4 (x + ^) + 1,
or z^ + Skz^ + 3A2x H- A» + 4x + 4 A + 1.

From the new value subtract the old,

3^x2 + 3A2x 4-^8 + 4^
Divide by h, Sx^ + 3Ax + A^ + 4.

Take the limit as h approaches 0 as a limit ;

2)a:(x8 4-4x4-1) = Sx^ 4- 4.

534. Derivative of x*. The function is aj*. Changing x to
X -{- h, we obtain (x + h)\ Now (« + A)* can be expanded
by the binomial theorem, and we obtain

(x 4- hY = a;" + nx^'-^h + ^ ? "^ '^ a?*-»A^ -h •••

From this new value of the fimction subtract a^, the old
value, and divide by h.

DERIVATIVES 441

We now have

i>.(*")=S['^-'+^^ri^ *"-'*+■••]=

nx^^K

The sum of the terms after the first approaches 0 as a limit
by § 405. Hence,

To find the derivatiye with respect to x of any power of x.

Multiply by the exponent, and diminish the exponent of x
hy one.

Thus, Dx(x*) = 4x8 ; Dx(JC-») = - 3x-*;

Note. It can be proved that this rule holds true whether n l» inte-
gral or fractional, positive or negative.

The derivative of a constant is zero, since the increment of
a constant is zero.

Exerciae 79

Find the derivative with respect to a; of :

1. x\ 4. x-\ 7. x-\ 10. {x -t- a)\

2. x\ 6. x\ ^ X

1 1 «• *' + ^- "• ^-:rz-

x' ' «»' 9. x^-\-2x\ 12. (x-\-l)-\

535. Derivative of a Sum of Two or More Functions. Let f(x)
and <t>ix) be two functions of x\ their smn/(a;)+ <A(^) is also
a function of x. Now,

_ limit r/(a! + A) -/(a;)"] Umit r«^(a! + A)- .^(a!)~|
-A = 0|_ A J * = »L A.J

Similaxly for the sum of any number of functions.

442 COLLEGE ALGEBRA

Hence, the derivative with respect to 'x. of the sum of two or
more functions ofxis the sum of the derivatives vnth respect to
X of the several functions.

The above may be formulated,

Here /is an abbreviation for/(aj), ^ for ^(aj), etc.

By means of the above and §§ 533, 534 the derivativie with
respect to x of any rational integral algebraic function of x
may be found.

Find 2)^(2 «» + 4a;2 - 8a: -f 3).

D,(2aj8 + 4x2 - 8x + 3) = 2)«(2x«) + i)*(4x2) - Da,(8a}) + J)x(3)

= 2Dxa:8 4. ^j)^^ - 8Dx« + D«8
= 2(3x2) + 4(2x)-8(l) + 0
= 6x2 + 8x-8.

536. Derivative of a Product of Two or More Functions. Let

f(x) and ^(aj) be two functions of x; their product /(«) ^ (a;)
is a new function of x.
Now,

limit

■/(« + A) <^ (« + A) - /(a; + A) <^ («)
+ f(x + h)<l> (x) - fix) if, (x)

=/(«) D,il> (x) + .^ (a;) D,/(x),
since a"^U/(* + *)]=/('«)•

DERIVATIVES 44S

The above may be formulated

Similarly for three or more functions. Thus,

J^.{f4>F) =^f<l>n,F ^fFD,4> + *i^Dj:

Hence, the derivative with respect to x of the prodtict of two
or more functions ofxis the sum of the several products found
by multiplying the derivative with respect to x of each function
by each one of the other functions,

537. Derivative of (x — a)» (See note, page 441.)

_ limit V{^ — »)** + n{x — a)"-^A -\ (x — a)**"]

"■^ = 0L h J

Thus, Dx (X - 3)* = 4 (X - 3)«.

EzerciM 80

Write the derivative with respect to a; of :

1. a;2_^4. 4. «» - 3 «* + «*.

2. a;» + 3a;"-l. 6. 4 a;* + 6 «» + 2.

3. «* + «" + 2. 6. 6a;«-7a;^ + 7aj.

7. 30^ + 40;* + «*-«^-6a;-f 6.

8. 4ic'-2a;*-a;»4-6a;»-7.

9. (a; - 2) (a; 4- 3). 12. (a; - 4)« (a; - 2) (aj + 1).

10. (a; -1) (a; -2) (a; -3). 13. {x-ccy(x-py.

11. (a;-3)»(a;-f 4). 14. (x - a) (x -- p) (x ^ y).

16. (a;-2)(aj-3)(a; + 5)(a; + 4).
16. («« 4- 2) (a;* - 4 aj 4- 8).

444 COhUSiGE AX^ODBRA

538. Successive Derivatives. TbQ derivative o£ ft functim of
X is, in general, a f unotion Qf w and ha^, in general, a derivar
tive with respect to a?.

The derivative of the derivative is called the second
derivative; the derivative of the second derivative, the third
derivative; and so on.

By derivative is meant the first derivative, unl«u tha oon^
trary is expressly stated,

The second derivative with respect to « of f(x) is wpi^
sented by pj^f(x), or ^jf\x)\ the third derivatiY§ by
D^f{x\ or by f'\x)\ and so on.

Evidently, /' («) - D,^f(x) - DJ)J{^) j

/" (x) = D,^f(x) = DJ),^f(x) = DjyjDJ^ix) ;
and so on.

539. Values of the Derivatives. The value which f(x)
assumes when we put (i for x is represented by/(a)-

Similarly, the value which f (x) assumes when we. put a
for x is represented by f (p)\ the value of f*(x) by /"(a);
and so on.

Thus, if /(x) = «« - 2x2 + X + 4,

then /'(;u) = 3ai»'-4» + l,

r(3j) = 6x-4,

and /*^ (x), /'^ (x), • • • all vanish.
If we pat 2 for s, we obtain

/(^) = 6}r(2) = &;/-{2) = 3jr'(2):^e.

Similarly for any other function,

540. Si|;Q of the Derivative. In the f unctiQil: /(«) 1^ x
increase by the ^ucces^ive addition of very small inorpm^iits.
As X increases, the value of f(x) will chaiige, sometimes
increasing, sometimes decreasing.

Suppose that X has reaohed a filed volue « ; the oone-
sponding values of f(x) ai^d /'(«) aye /(a) and /*(«)•

DERIVATIVES 445

By §532. /(a)=a[-^^"-'i^~^^'^]-

If f{x) ia mdreasinff ks. » passes through the taluc a,
f(€L-\-h)>f(a) and /'(a) is positive.

If f{x) is decreasing as a? passes through the valu^e a,
/(a -|- A) < /(a) and /' (a) is negative.

Conversely, if /(a) is positive, /(ft 4- h) —f(tt) is positive,
and f(x) is increasing as x passes through the value a.

If f'(a) is negative, f(a -f h)—f{a) is negative, and /(x)
is decreasing as x passes through the value a.

Hence, for a particular value of x, if f ' (x) is positive, f (x)
is incretliing ; and if f'(3t) is negative, f(x) is decreaBirhg ;
and conversely.

Observe that we are speaking of increasing and decreasing
algebraically.

Thus, if /(x) = x8 - 3x2 - 6x + 10,

/'(x) = 3x2-6x-6.
We find /(I) = 2, /'(I) = - 9.

.-. f(x) is decreasing as x passes through the value 1 ; for example,

/(I) = 2, /(1.1) = 1.101, ftnd 1.10K8.
Again, /(3) = -8,/'(3) = + 8.

.-. f(x) is InGreatting as x pa^efl through the value 8 j for ^joimpld,
/(3) = - 8, /(3.1) = - ir.639, feud - 7.689 > - 8.

fikercitie 81

Write the successive derivatives with respect to a) of r

1. a;*-4a;» + 2. 6. aiC« + 3 ftaj^ 4- 3 caJ + rf.

2. aj*4-4:t*-5aj. 7. aa;* + 4fta;* + 6daJ*4-4rfa8 + «

3. 2ic* + 2a;2-4a; + l. 8. (a; - a)*(a; - j3).

4. 3a;* + 3a;»-x2 + ic. 9. (a: - a) (a - )3) (« - y).
^, 4a;«-7x» + 5x-2. 10. (x-ay(x-P)\

446 COLLEGE ALGEBRA

Find whether each of the following functions is increasing
or decreasing as x increases through the value set opposite:

11. a;»-a;* + l. (2) 13. 2 a* + 3 a* -6 a. (1)

12. a;*-aj» + 6a;-l. (4) 14. 4a;*-3a;* + 4aj- 6. (—3)

541. Derivative in Terms of the Roots. Take the cubic

f(x) = a(x-a)(x-P)(x- y).

Since 2),(a:- a) = 1, D,(x-P) = l, D. (« - y) = 1 (§ 537),
we have, by § 536,

/(«) = a (a; — ^)(a; — y) +«(« — «)(» — y)+a(» — a)(«—/8)
X — a X — P x — y

Similarly, for any quantic,

^^ X — ai X — a^ X — a^ ^^x — a

542. Maxima and Minima. If, as x increases, /(^) increases
until X reaches a certain value a, but f{x) begins to decrease
as soon as x passes through the value a, the value /(a) of /(a),
when aj = a, is called a maximum value of /(a).

If, as X increases, f{x) decreases until x reaches a certain
value a, but/(a;) begins to increase as soon as x passes through
the value a, the value /(a) of /(«), when a; = a, is called a
minimum value of fix)-

From these definitions and from § 541 it follows that for
all continuous functions of x (see § 557), when/(aj) is a maxi-
mum or a minimum, /' (a;) = 0 ; and conversely, in general, if
f{x) = 0, f(x) is either a maximiun or a minimum. In other
words, the general condition for a maximum or a minimoxn
value of f(x) is /' (x) = 0.

DERIVATIVES 447

Hence, the maxima and minima values otf(x) are found by
deriving /' (x) from f(x), and then solving the equation

f(x)=0.

For, let a denote a value of x which satisfies the equation
f'(x)= 0. Then f(a) is, in general, either a maximum or a
minimum, and it may be determined by the algebraic sign of
/"(aj) whether /(a) is a maximum or a minimum.

Suppose that f(a) is a maximum. Then f(x) must be
increasing just before x = a, and decreasiny just after x = a.

Therefore, /' (a;) must he positive just before a? = a, and neg-
ative just after a; = a (§ 541). Hence, /' (x) must be decreeing
as it passes through the value 0 at the point for which x = a.
Therefore, by § 541, /" (x) must be negative when x = a; for
f"{x) has the same relation tof'(x) that /'(a;) has to/(aj).

By similar reasoning it may be proved that if f(a) is a
minimum, /" (x) must be positive when x = a.

Hence, f(a) is a ma^ximum when /" (a) is negative, and /(a)
is a minimum when /" (a) is positive.

The most important points to be determined in constructing
a graph are the points which correspond to the maxima and
minima values of the function in question.

EzerciBe 82

Find the maxima and the minima values of the following
functions of x, and plot the graphs :

1. y = x^ — 6x-\-7,

2. yT=zx*-\-6x^-x-30.

3. y = «• — 12 x.

4. y = 4a^ — 12a;-t-l.

6. y = a;* + 4a;»-20a;« + 4.

44g

COLLEGE ALGEBRA

543. Maltipk RooU. In the quantio /(ai) let a be a triple
root Then, we oan write (| 518)

f(x) = (x-ay<H^),

where the degree of <^ (x) is less by 3 than that of /(«)-

By § 53G, f(x) = (x- a)»<^'(aj) -f 3(« - «)'*(af)

= (x- ayi(x - a)<^'(aj)+ 3^(aj)],

Hence, \if(x) has a triple root a, the factor (jb t^ a)* qcciirg
in the II.O.F. of /(a) and /(a;).

Similarly for a multiple root o{ any ord^r,

To find the multiple roots of /(jr),

Find the H.CF, ofi(x) and f'(x), and resolve it into factors.
Fach root occurs once more in f (x) than the corresponding factor
occurs in the II.CF,

Find the multiple roots of

«*^ - ;c* - 5iP« -f »;' + 8»; 4- 4 = 0.

Here, /(ac) = x5 - x* - 5 jR» + as« -|- 8(b -|- 4,

.•./(x) = 6x* - 4{bI - 16x? + 2«J + 8,

Find the H.C.F. Qif(x) and/(jc) as follows j

6-4-16+ 2 + 8
5 + 0-15-10
-4+ 0+12 + 8
-4+ 0 + 12 + 8

5_6_26+ 6+ 40+ 90
5-4-16+8+ 8

-1-

-6-

10+3+32+20

60 + 16 + laO + 100

4 + 15 ^ 2 - 8

-64)-

54 +

0 + 162 + 108

-1

X- 0- 8- i -6+4

Hence, x^ - 3x - 2 is the H.C.F.

We find, by substitution, that — 1 is a root of the equation

x»-8x-2 = 0.

The other roots are found to be — 1 and 2 (i 520),

Hence, x8-3x-2 = (x + l)2(x - 2),

Therefore, — 1 is a triple root, and 2 is a double root, of the giren
equation. As the given equation is of the fifth degvee, these are all the
roots, and the equation may be written

(X + l)8(x - 2)2 = 0.

1.

a;«

2.

aj«

3.

X*

4.

a;*

6.

aj*

DERIVATIVES 449

Having found the multiple roots of an equation^ we may
divide by the corresponding factors and find the remaining
roots, if any, from the reduced equation.

Ezeroise 83

The following equations have multiple roots. Find all the
roots of each equation :

;«_ 8^2 4- 13a;- 6 = 0.
j«_-7a;2 + 16a; -12 = 0.
;*-6a;2~8a;-3 = 0.
;* - 7 a;" + 9a;2 + 27 a; - 54 = 0.
^^-^-ex^ + x^-- 24a; + 16 = 0.

6. x' - llfl)* + 19a;« + 115a;2 - 200a; - 500 = 0.

7. Resolve into linear factors

a;* — 5a;* -h 6x^ ^ 9aj» ^ 14«;* — 4a 4- 8.

8. Show that an equation of the form a* =^ a* Can have no
multiple root.

9. Show that the condition that the equation

x^ -^3qX + r* = 0
shall have a double root is 4 j* + r* = 0.

10. Show that the condition that the equation

«;* 4- Spx^ + r = 0
shall have a double root is r (4^* + r) = 0.

544. Expansion of i(x + h). Consider a qUftnUo of the

fourth degree

f(x) ^ da;* 4* 6cfc« 4-.(Ja;* 4- ef aJ 4- e.
Put «; 4- A in pl^e of a). Then,
f(x-{-h)=a{x^hy-^b{x'i-hy + c(x + hy^d(x-^h) + e.

450

COLLEGE ALGEBRA

Expand the powers of x -{- h, and arrange the terms by
descending powers of x.

f(x +h) = a

05* + 4 aA

«• + 6 ah^

a;* 4- 4 aA»

X 4- aA*

+ 3 5A

-k-Shh^

+ *A»

+ c

^2ch

-hch*

+ d

+ dh

•

+ e

But /(^)= a^*+ *^*+ ch^-{-dh + e,

f\h)= 12 ah^ + ebh +2c,
/'"(A) = 24aA 4-65,
/i-W=24a,
r W = 0.

/. /(X4- A) = /(A)4-^/W + a^^^ 4-a:»^C|^ 4-«*

If we arrange the expansion of /(» 4- h) by ascending
powers of A, we find

f(x + h)=f(x) + hf\x) + A« 4^ + A» =0^ + A« «^ .

Similarly for any other quantic.

545. Calculation of the Coefficients. The coefficients in the
expansion of /(aj 4- h) may be conveniently calculated as
follows :

Take f{x) = ax^ 4- hx^ -\- cx^ -^ dx -{- e.

Put f(x + h) = Ax^ 4- Bx^ 4- Ca;" 4- i>« 4- -ET,

where A, B, Cy D, E are to be found.
In the last identity put x — h for x.
Then, since f(x — h -\- h) =/(a;), we obtain

f(x) = A(x- hy 4- B{x - A)» 4- C{x - A)"

4-i)(a; — A)4--Er.

DERIVATIVES 451

From the last identity we derive the following rule for find-
ing the coefficients of the powers of a; in the expansion of
f{x + A).

Divide f{x) by x — h; the remainder is E, that is, f(h)f
and the quotient is

A(x- hy + B(x -hy + C(x-'h)+ D.

Divide this quotient by (« — A) ; the remainder is D, that
is, /' (A) ; and the quotient is

A(x- hf + Bix - A) + C.

Continue the division. The last quotient is A or a.
The above division is best arranged as follows (§ 515) :

a h c d e [A

ah Vh c'h d}h

a

V

c'

d'

E

ah

6"A

c^'h

a

ah

D

a

^w

C

ah

\

a B

Hence, f{x + A) = ax^ + Bx^ -\- Cx^ -{- Dx ^ E.
This method is easily extended to equations of any degree.

Exercise 84

In the following quantics put for x the expression opposite,
and reduce :

1. aj«-3aj2-f 4a;-6. (« + 2)

2. a;*-2aj2 + 6aj-3. {x -\- 4)

3. 3a;*-2a;« + 2a;2-aj-4. (« + 3)

4. 2a;*-3aj» + 6a;^-7a;-8. (x-2)
6. 2a;*-2a:» + 4a;»-5a;-4. (a; - 3)

462 COLLEGE ALGtJBRA

TRANSFORMATION oF EQUATIONS

546. The solution of an equation and the investigtbtioti of
itij properties are often f&dilitatdd bjr a chftngfe in thd form of
the equation. Such a change of form is oftlldd A OttilitOraui*
tion of the equation^

647. RoOtd with BigAd ChAnged. ^^0 foOM of the ^ueMon
f (— x) = 0 are tlwse of the equation f (X) *« 0^ «dlcA tt^t^ tii HgH
changed.

For, let tt be any rodt of equation /(«) mz 0,

Then We must have f(a) = 0.

In the quantic /(— x) put — a for 05 ; that is, a for — a;.

The result is /(«).

But we have just seen that /(a) vanishes, sinCe a is a root
of the equation f(x) = 0. Hence, /(— x) vanishes when we
put — a for X, and (§ 611) — a is, therefore,' £i foot of the
equation /(— cc) = 0.

Similarly, the negative of each of the roots of /(*) = 0 is a
root of /(— aj) = 0 ; and, since the two equations are evidentlj
of the same degree, these are all the roots of the equation

To obtain /(— x) we change the sign of all l^e odd powers
of aj in the quantic f(x).

Thus, the roots of the equation x* — 2 x^ — 13 x^ + 14 x + 24 = 0 are 2,
4, - 1, - 3 ; and those of the equation x* + 2x» - 13x« - 14» + 24 = 0
are - 2, - 4, + 1, + 3.

548. Roots multiplied by a Given Number. Goasidmr ibm

equation

ax*-\-bx^-{'CX^ + dx + e = 0, [1]

y

Put y = mXf then cc = — • Then the equation becomes

«(i)'+<l)"-(i)'+<5)^«-'^ m

TRANSFORMATION OF EQUATIONS 4$$

The left member of [2] differs from tbe left m§»be? of [1]

y

only in that — is put in place of x.

Let a be any root of [1] ; the left member of [1] vanishes
when we put a for x.

That is, aa* + ha^ + ca* + ^« + e = 0.

In the left member of [2] put moc for y.

Then, aa* -f ha^ -\- ca^ ■\- da -\- e^

which, as we haye just seen, vanishes. Hence, if a is a rgo^
of [1], ma is a root of [2]. Since the above is true for each
of the roots of [1], and the two equations are evidently of the
same degree, the roots thus obtained are all the roots of [2].

Similarly for an equation of any degree.

Equation [2] may be written in the form

Hence, to write an equation the roots of which ar© the roots
of a given equation multiplied by m,

Multiply the M^cond term of the given eqication by hi j the
third term by m^; and so on.

Zero coefl&cients are to be supplied for missing powers of x.
Write the equatioi^ of which the rogtg are double the roots
of the equation

3aj*-2ir«4-4aj*-6ic-5 = 0.

Here m = 2, and the result is

8x* - 2(2)x8 + 4(2)ex« - e(2)«a; - 6(2)* = 0,
or 8x* -4aj« + 16a;«^48aj-80 = 0.

549. Removal of Fractional Co^^^ta. If any of the ooeffi-
cients of an equation itx the form

are fractions, we can remove fractions as follows :

464 COLLEGE ALGEBRA

Multiply the roots by m ; then take m so that all of the coeffi-
cients will be integers.

Reduce to an equation, in the p form, with integral coeffi-
cients 2x« - Ja* -f |i» + I = 0.

Divide by 2, x^ - ix^ + ^^z + I = 0.

Multiply the roots by m (§ 648),

, m „ 5m2 m^ .

x» x2 + X + — - = 0.

6 12 8

The least value of m that will render the coefficients all integral is seen
to be 6. Put 6 for m, x« - x^ + 16x + 27 = 0, the equation required.

550. Reciprocal Roots. Consider the equation

ax^ + bx^-\'ex^-\'dx-\'e = 0. [1]

Put y = -; then x = -; and the equation becomes
X y

■■G)'-^'6)'-^<0"-^''(|)-^'-v P3

Let a be any root of [1].

Then, aa* + ba^ + ca^ -{- da + e = 0.

1 1

In the left member of [2] put a for - ; that is, - for y,

aa^ -f bo^ 4- ca^ + da + e,
which, as we have just seen, vanishes.

Hence, - is a root of [2]. Since the above is true for each

root of [1], and the two equations are of the same degiee, the
reciprocals of the roots of [1] are all the roots of [2].

Similarly for an equation of any degree.

Equation [2] may be written

a "i- by ■}- cy^ ■}- dy* -\- ey^ =: 0,
or, writing x in place of y,

ex^ -h dx^ -\- cx^ -\- bx -\- a = 0;

TRANSFORMATION OF EQUATIONS

456

so that the coefficients are those of the given equation in
reversed order.

Write the equation of which the roots are the reciprocals of
the roots of 2a;* - 3a;* + 4a;* - 5a; - 7 = 0.

The result is 2 - 3x + 4x2 _ 5x« - 7x* = 0,

or 7x* + 6x8 - 4x2 + 3x - 2 = 0.

551. Reciprocal Equations. The coefficients of an equation
may be such that reversing their order does not change the
equation. In this case the reciprocal of a root is another
root of the equation. That is, half the roots are reciprocals
of the other half. Such an equation is a reciprocal equation.

Thus, the roots of the equation

6x6 _ 29x* + 27x8 + 27x2 - 29x + 6=0
are — 1, 2, 3, i, ^. Here — 1 is the reciprocal of itself ; i of 2 ; i of 3.

552. Roots diminished by a Given Number. Consider the

equation

f{x) = a;* -f ««« '\-hx^-\'Cx-\-d=^0, [1]

and the corresponding graph with the point 0 as origin.

To diminish the roots of this equation by any number h is
equivalent to changing the origin from the point 0 to a point
R on the axis of x such that OR = h. The change is made
(§ 631) by writing a; -f A for a; in [1]. The result is

f(x-\-h) = (x+hy-^a (x+hy+b (x+hy+o (36+h) +d = 0. [2]

466 COLLEGE ALGfiBEA

Detioto the HoW coefficltdnts df the eqtifttion by ai, 61,^11 di.

Then, f(x -\-h) = x^-{- a^x^ + biX^ + CiX -f e^i = 0. [3]

To find the values of a^ b\, Ci, and dif transpqad the origin
back to O by writing in [3] x — h for x.
Then,

f{x) = (x-hy-{-a, (x-hy-^bi (x-hy-\-Ci (aj~A)-f c?i=0. [4]

Take out the factor x — A, and denote the quotient by Q.

Then, /(aj) = (aj-A)a + rfi. [5]

Hence, di is the remainder obtained by dividing f(x) by
aj — A. Similariy, e^ is the temainder obtained by dividing Q
by a; — A ; and so on.

Therefore, the liew coefficients are eflfcily found by the
repeated application of synthetic division to thd ooefficients
of the given equation.

Evidently the same method may be applied to an equation
qfany degree.

To increase the roots by a given number h we have only to
diminish the i*oots by the number — L

Obtain the equation the roots of which are each less by 2
than the roots of the equation

2 X* - 3 ic« - 4 x2 4- 2 aj -f 9 = 0.

The work (§ 645)
2

is as follows :

- 3 - 4

+ 4 +2

+ 2
- 4

+ 9|2
-4

2

+ 1
+ 4

- 2
+ 10

- 2
+ 16

+ 6

2

+ 5
+ 4

+ 8
+ 18

+ 14

a

+ 9
+ 4

+ 26

2 +13

The required equation is

2a5« + 13ftj8 + 8e*« + 14« + 6 = 0.

TRANSFOSMATION OF EQUATIONS 46T

553. Transformation in General. In the general problem of
transform^tioii we have given slvl ^quatiop in a?, as f(x) = 0,
and we have to form a new equation in y where y is a given
{unotion of », auob as ^(sb).

When from the equation y = fft(x) we can find a valua of
X, the transformation is giadi by ^ub^tit^ting tbi^ vplue of x
in the given equation, and reducing the result.

(1) Given the equation x' — 3 cc + 1 = 0 ; to find the equa-
tion in y where y = 9 oj — 9.

We find X = . Substitute in the given equation,

and we have (?!±^)*_ 3 (?^) + 1 = 0,

which reduces to y* + 6y^ — J6y«-19;;?0.

(2) Given the equaticm a' — 2«' -f 3 a; — 5 = 0, of which
a, p, y are the roots ; fipci the equation of whiqh tb© roots
are/3 4-y — a, y + « — A « + /3 — y.

We have y=:p + y-a = a-{-p-\-y-2a = 2^2a. (§ 621)

2-y

But, sinQf) a is a roQt Qi %he givQp ecjuation,

2 —V

Put for a, and reduce.

2 '

Then, y« — 2 y^ + 8 y + 24 = 0, the equation required.

BxerclM 85

Write the equations whose roots are the roots of the fol-

iQwmg equatipua multiplied by the mw^b^r opposite ;

1. a5«^3aBf ^-9aj^4=5 0. (« I)

8. 2at*«^3ar* + »*-flaj^4:;=:0. (^3)

458 COLLEGE ALGEBRA

4. 2a^-3«* + 6«-8 = 0. (-2)

6. 3a^-4a;»-2a;-f 7 = 0. (-2)

Transform to equations with integral ooefficients in the p
form:

6. 12aj»-4aj* + 6aj + l = 0.

7. 6«» + 10a5*-7aj + 16 = 0.

8. 10aj* + 6aj«-4aj* + 26aj-30 = 0.

9. 6«* + 3«* + 4aj»-2aj» + 6aj-18 = 0.

Write the equations which have for their roots the recipro-
cals of the roots of :

10. 3«*--2aj» + 5«*-6a; + 7 = 0.

11. 2«*-4a;»-5a;*-7«--8 = 0.

12. ««-«* + 2a;* + 4«-- 1 = 0.

Write the equations whose roots are the roots of the fol-
lowing equations diminished by the number opposite :

13. a;«-lla;« + 31a; -12 = 0. (1)

14. a;*-6a;« + 4aj2 + 18a;-6 = 0. (2)
16. a;« + 10a;2-f 13a; -24 = 0. (-2)

16. a;* + «'-16a;*-4a; + 48 = 0. (4)

17. a;* + a;«-3a; + 4 = 0. (0.3)

18. a;*-3a;«-a;« + 4a;-6 = 0. (-0.4)

19. a;«-9a;2 + 22a; -12 = 0. (3)

20. Form the equation which has for its roots the squaxes
of the roots of the equation a;* — 2 a;* + 3 a5 — 6 = 0.

21. Form the equation which has for its roots the squares
of the differences of the roots of as* — 4 as* + 2 « — 3 = 0.

SITUATION OF THE ROOTS 469

SITUATION OF THE ROOTS

554. Finite Value of a Quantic. Any positive integral power
of a; is finite as long as x is finite.

The product of a positive integral power of oj by a finite
number will be finite when x is finite.

A quantic consists of the sum of a definite number of such
products, and has, consequently, a finite value as long as x
is finite.

The derivatives of a quantic are new quantics and have,
consequently, finite values as long as x is finite.

555. Sign of a Quantic. When x is taken numericaUy large
enough the sign of a qtuintic is the same as the sign of its
first term.

Write the quantic

aoflJ" -h a^x^^^ + OjO^"* + • • • + a^

in the form ao^^fl + — + ^ + .•• -h-^\

\ a^ a^^ d^J

By taking x large enough, each of the terms in parenthesis
after the first can be made as small as we please.

If aj, is numerically the greatest of the coefficients a^a^^-^y
a„, the sum of the terms in parenthesis after the first is
numerically less than

a^ \x x^ x^J

that is (§ 280), less than -*

2*1 £

tto \ x — 1

The value of this expression can be made less than 1,
or, indeed, less than any assigned value, by taking x large
enough.

460 COLLEGE ALGEBRA

Hence^ even in the most unfavorable case, that in which all
the terms in parenthesis after the first are negative, the sum of
these terms can still be made less than 1 ; the sum of all the
terms in parenthesis is then positive. The sign of the quantic
is the same as the sign of a^^ its first term,

556. When x is taken numerically small enough the sign of
a quantic is the same as the sign of its last term.

Write the quantic in the form

\ »« »« a* J

The proof follows the proof of the last section.

557. Continttity of a Quantic A function of x^ f(x)y is ooo-
tinuous when an infinitesimal (§ 376) change in x always
produces an infinitesimal change in f(x), whatever the value
of a.

We proceed to show that if f(x) is a quantic in a;, it is a
continuous function of x.

Give to X any particular finite value a ; the corresponding
value of f{x) is f{a).

Increase x to a ■\- h-, the corresponding value of /(x) is
f(a + h)f and the increment in the value of /(as) is

/(« + A)-/(a),
or h (f'(a) + |/"(a) + • • • + ^f (a)) . (§ 644)

The derivatives /'(a), /"(a), • •>/*(<*) all have finite values
(§ 554) ; and it is easily seen from § 556 that when h is very
small the expression in parenthesis is numerically less than
2f\a), Since 2 hf{a) approaches 0 as a limit (§ 379, 1) when
h approaches 0 as a limit, the increment oif(x), whioh is less
than 2 hf'Ca), approaches 0 as a limit when h appioikahes G as a
limit.

SITUATION OF THE ROOTS 461

Since the above is true for any particular finite value of a,
we see that an infinitesimal change in x always produces an
infinitesimal change in f(x).

It follows that as f{x) gradually
changes from f(a) to f(h)j it must pass
through all intermediate values.

The derivatives of a quantic c in 05

Mill

I

are themselves quantics in x and are,
therefore, continuous.

The changes in the value of a quantic /(x)
are well illustrated by the graph of the func-
tion. Since f(x) is continuous, we can never
have a graph in which there are breaks in the curve, as in the curve here
given. In this curve there are breaks, or diaconlinuvtieSy at x = — 2
and X = -f 2.

558. Theorem on Change of Sign. Let two real numbers a and
b be put for x in f (x). If the resulting valuss of f (x) have
contrary signs, an odd number of roots of the equation f (x) = 0
lie between a and b.

As X changes from a to 6, passing through all intermediate
values, f(x) will change from f{a) to f(b), passing through all
intermediate values. Now in changing tTomf(a) to /(ft), f(x)
changes sign.

Hence, f(x) must pass through the value zero. That is,
there is some value of x between a and b which causes f(x) to
vanish ; that is, some root of the equation /(x) = 0 lies between
a and b.

But f(x) may pass through zero more than once. To
change sign, f(x) must pass through zero an odd number of
times ; and an odd number of roots must lie between a and b.

Applied to the graph of the equation, since to a root corre-
sponds a point in which the graph meets the axis of x (§ 529),
the above simply means that to pass from a point below the

462 COLLEGE ALGEBRA

axis of a; to a point above that axis, we must cross the axis
an odd number of times.

Thus, in x8 - 2a:* + 3x - 7 = 0, if we put 2 for x, the valae of the
left member is — 1 ; if we pat 3 for x, the yalue is + 11. Hence, cer-
tainly one root lies between 2 and 3, and possibly all three roots of the
equation lie between 2 and 3.

559. An eqttatian of odd degree has at lea^ one real root the
sign of which is opposite to that of the constant term.

For, if the first coeflBicient is not positive, change signs so as
to make it positive. If the last term is negative, make x posi-
tive and very large ; the sign of the left member is -h (§ 555).
Put 35 = 0 ; the sign of the left member is — . Hence, there
is at least one real positive root.

Similarly, if the last term is positive, there is at least one
real negative root.

560. Descartes' Rule of Signs. An equation in which all the
powers of x from x^ to a;" are present is said to be complete ; if
any powers of x are missing, the equation is said to be incom-
plete. An incomplete equation can be made complete by writ-
ing the missing powers of x with zero coefficients.

A permanence of sign occurs when -f follows -f , or — fol-
lows — ; a variation of sign when — follows +, or -f follows — .

Thus, in the complete equation

x« - 3aj6 + 2x* + x8 - 2x2 - X - 3 = 0,
writing only the signs

+ - + + ---,
we see that there are three variations of sign and three permanences.

For positive roots, Descartes' rule is as follows :

The number of positive roots of the equation f (x) = 0 cannot
exceed the number of variations of sign in the quantie f (x).

To prove this it is only necessary to prove that for every
positive root introduced into an equation there is one variation
of sign added.

SITUATION OF THE ROOTS 463

SuppoBe the signs of a quantic to be

+ - + + + --+,

and introduce a new positive root. We multiply by a; — A, or,
writing only the signs, by H — . The result is

+ - + + + -- +

+ -

+ - + + + -- +

- + --- + + -
+ - + ±±-T + -

The ambiguous signs ±, ± indicate that there is doubt
whether the term is positive or negative. Examining the
product, we see that to permanences in the multiplicand
correspond ambiguities in the product. Hence, we cannot
have a greater number of permanences in the product than
in the multiplicand, and may have a less number. But there
is one more term in the product than in the multiplicand, and
this term always adds a new variation. Hence, we have at least
one more variation in the product than in the multiplicand.

For each positive root introduced we have at least one more
variation' of sign. Hence, the number of positive roots cannot
exceed the number of variations of sign.

Negative Roots. Change a; to — a;. The negative roots of
the given equation are positive roots of this latter equation.

561. Hence, from Descartes' rule we obtain the following :

If the signs of the terms of an equation are all positive, the
equation has no positive root.

If the signs of the terms of a complete eqriation are alternately
positive and negative, the equation has no negative root.

If the roots of a t^fymplete equation are all real, the number of
positive roots is the same as the number of variations of sign,
and the nwmber of negative roots is the same as the number of
permaneneea of sif^

464 COLLEGE ALGEBRA

562. Existence of Complex Roots. In an inoomplete equation

Descartes' rule sometimes enables us to detect the presence of
complex roots.

Thus, the equation x^ + 6x-\-T = 0

may be written x* ±0x^ + 6x -^7 = 0.

We are at liberty to assume that the second term is positiye, or that it

is negative.

Taking it positive, v^e have the signs

+ + + +;
there is no variation, and the equation has no positive root.
Taking it negative, we have the signs

+ -. + +;

there is but one permanence and, therefore, not more than one negative
root.

As there are three roots, and as complex roots enter in pairs, the given
equation has one real negative root and two complex roots.

Exercise 86

All the .roots of the equations given below are real; deter-
mine their signs.

1. ic*+4aj'»- 43x^-68x4- 240 = 0.

2. aj8-22a:2 + 155x- 350 = 0.

3. aj*-f4a;8-35aj^-78aj 4-360 = 0.

4. x»- 12x2 -43a; -30 = 0.

5. X* - 3a;* -5x« 4-15x^4- 4a; -12 = 0.

6. a;8- 12x2 4- 47 a; -60 = 0.

7. aj*-2a;«-13a;8 4-38a;-24 = 0.

8. x« _ x* - 187 a;8 - 359 x^ 4. 186a; 4- 360 = 0.

9. a;« - lOa;^ 4- 19a;* 4- 110 a;' - 536 a;« 4- 800a? - 384 x= 0.

10. If an equation involves only even powers of Xf and the
signs are all positive^ the equation has no real root, except 0.

SITUATION OF THE ROOTS 466

11. If an equation involves only odd powers of aj, and the
signs are all positive^ the equation has the root 0^ and no
other real root.

12. Show that the equation aj' — 3aj* — aj-fl = 0 has at
least two complex roots.

13. Show that the equation x* -f 16 a* -h 7 a; — 11 = 0 has
two complex roots, and determine the signs of the real roots.

14. Show that the equation x^ -\- qx -\- r =i 0 has one nega-
tive root and two complex roots when q and r are both
positive ; and determine the character of the roots when q is
negative and r positive.

15. Show that the equation aj" — 1 = 0 has but two real
roots, + 1 and — 1, when n is even j and but one real root,
+ 1, when n is odd.

16. Show that the equation a;" + 1 = 0 has no real root
when n is even ; and but ona real root, — 1, when n is odd.

563. Limits of the Roots. In solving numerical equations it
is often desirable to. obtain numbers between which the roots
lie. ' Such numbers are called limits of the roots.

A superior limit to the positive roots of an equation is a
number greater than any positive root. An inferior limit to
the positive roots of an equation is a positive number less
than any positive root.

General methods for finding limits to the roots are given in
most text-books ; but in practice close limits are more easily
found as follows :

(1) aj* - 6a;» -f ^Ox^ - 8 aj + 23 = 0.

Writing this x«(x - 6) + 8x(6a5 - 1) + 28 = 0,

we see tha4; the left member is positive for all values of x as great as 6 ;
consequently, it cannot become 0 for any value as great as 5, and there is
no root as great as 6.

466 COLLEGE ALGEBRA

(2) aj« + 3a^-|-a;«-8ar«-51x-hl8 = 0.

Writing thiB sB»(a^ - 8) + 8x(a* - 17) + x» + 18 = 0,

we see that the left member is positlye for all vaLueB of x as great as 3 ;
oonsequently, there is no positiye root as great aa 3.

Sometimes we can find close limits by distzibatiiig the
highest positive powers of x among the negative temiB.

(3) «* + ar» - 2x« - 4a; - 24 = 0.

Multiply by 2, 2x* + 2x»- 4z2 - 8x - 48 = 0.

Writing this x*(x« - 4) + 2x(z«- 4) + x* -48 = 0,
we see that there is no positive root as great as 3.

An inferior limit to the positive roots is found \sj patting

a; =: - (§ 550)^ and then finding a superior limit to the positiTe

roots of the transformed equation.

Limits to the negative roots of the equation f{x) = 0 are
found by finding limits to the positive roots of the equation
/(- X) = 0 (§ 547).

Ezerciae 87

Find superior limits to the positive roots of :

1. a;«-2a;«-f 4a;H-3 = 0.

2. 2a:*-a:«-a;-f 1 = 0.

3. 3x*4-5x»-12x*-f 10a;-18 = 0.

4. 4a;*-3a;«-a?*-f 7a;-f 5 = 0.
6. a;*-a;«-2a?*-4a;-24 = 0.

6. 4aH^-8a:*H-22x»-f 90aj*-60aj + l = 0-

7. 5ar*-f 14a:*-7ar»4-12a:*-24aj + 2 = 0.

8. 2x'4-7a;* + 5a;»-8a;*-4a;-h3 = 0.

CHAPTEK XXXI

NUMERICAL EQUATIONS

564. A real root of a numerical equation is either commen>
surable or incommensurable.

Commensurable roots are either integers or fractions.
Kecurring decimals can be expressed as fractions (§ 280),
and roots in that form are consequently commensurable.

Incommensurable roots cannot be found exactly, but may
be calculated to any desired degree of accuracy by the method
of approximation explained in this chapter.

COMMENSURABLE ROOTS

565. Integral Roots. The process of finding integral roots
given in § 620 is long and tedious when there are many num-
bers to be tried. The number of divisors to be tried may be
diminished by the following theorem ;

Every integral root of an equation with integral coefficients
is a divisor of the last term.

Let A be an integral root of the equation

aTf" -f- ftcc--^ -f- cx*-^ H \-dx^-\-ex +/= 0,

where the coefficients a, b, c, "•, d, e are all integers.
Since A is a root,

aA» -f- ftA—^ 4- cA— 2 H [- dh^ -\- eh +/= 0, (§ 611)

or f = —eh — dh^ c7i— ^ — ftA»-i — ah\

Divide by A, ^ = -e — dh cA"-« - bh""-^ - ah'^'K

467

468 COLLEGE ALGEBRA

Since the right member is an integer, the left member must
be an integer. That is, / is divisible by h.

Hence, in applying the method of § 520, we need try only
divisors of the last term. The necessary labor may be still
further reduced by the method of the following section.

566. Newton's Method of Divisors. In the above equation

f f

Y is an integer. Put ^ = Ey transpose — e, and divide by h.

Then, ^^^ = -d cA"*"* - *A»-« - ah'^'K

fi

Since the right member is an integer, E-\-e must be divisible

by /^- „ .

E A- p

Put — 7 — = D, transpose — dj and divide by h.

Then, 7" = cA»-« — 5A»-* — aA»-».

n.

As before, D -{- d must be divisible by h.

By continuing the process we find that C -f- c, and B -\-b are

divisible by h, and for the last equation — - — = — a.

B A- h
Transpose — a, — f- a = 0, provided A is a root.

The preceding gives the following rule :

Divide the last term hjh.] if the quotient is an integ'ery to it
add the preceding coefficient, and again divide hyhy ^ tkis quo-
tient is an integer, add the preceding coefficient to it ; and bo on.

If A is a root, the quotients are all integral, and the last
sum is zero. A failure in either respect implies that h is not
a root.

From the above we also obtain

E=- (ah""-^ 4- ftA»-« + cA»-» H + dh + e),

• . . •.«•••
C =~-(ah^-^bh-{-c),
B = — (ah + b),

COMMENSURABLE ROOTS 469

so that the successive quotients, with their signs changed,
are (§ 616), in reversed order, the coefiB.cients of the quotient
obtained by dividing the left member by a — A.

Find the integral roots of

3aj* - 23cc« -f- 42x2 -f- 32a; - 96 = 0.

By substitution we find that neither + 1 nor — 1 is a root.
The other divisors of - 96 are i 2, ±3, ±4, ±6, etc.

Try +2. - 96 + 32 + 42 - 23 + 3|2

-48- 8 + 17-3
-16 + 34-6 0

H«nce, + 2 is a root. \The coefficients of the depressed equation in
reversed order are —48 — 8 + 17 — 3.

Try +2 again. -48- 8 + 17-3|2

- 24 - 16
-32+1

Since 2 is not a divisor of + 1, + 2 is not again a root.

Try -2. -48- 8 + 17-3|-2

+ 24-8
+ 16+9

Since — 2 is not a divisor of + 9, — 2 is not a root.

Try +3. _48- 8 + 17-3[3

- 16 - 8 + 3
-24+9 0

Hence, + 3 is a root. The depressed equation Is

3x2 -8x- 16 = 0,

of which the roots are 4 and — f .

Therefore, the roots of the given equation are 2, 3, 4, — |.

The advantage of this method over that of § 520 is that if the number
tried is not a root, this fact is detected as soon as we come to a fractional
quotient ; whereas, in § 520, we have to complete the division before we
can decide whether or not the number tried is a root.

470 COLLEGE ALGEBRA

567. FncdMud Roota. A ratwtuU /iyutiom fWimai he m root
of an eqwatwn with i$Ue^rraX eoeJlciemiM im, the i^fmrmL,

If possible let -9 where A and k axe integecs and ~ is in

its Icrwest ^kiin&. be a root.

Multiply by /r*~^ and transpose,
A*

-jT = - M'"* - Z^*'"** pJfc*"*-

Xow the right member is an integer ; the left member is a
fraction in its lowest terms, sinoe A* and k hare no oommon
divisor as h and k have no common divisor (§ 470, T). But
a frac-tion in its lowest terms cannot be equal to an integer.

Hence, -zt or any other rational ibuction, cannot be a root.

The real roots of an equation with integral ooeflbdentB in
the /> form are, therefore, integral or incommensaiable.

If an equation has fractional roots, we can find theae roots

as follows :

Trangform the equation into an equatum wpkk mtegral eoeffi"
events by multiplying the roots by some number m (§ 548).
Find the integral roots of the transformed equatum amd ilMdt

ea/:?i by m.

Solve the equation 36ar* - 55ar* - 35a; — 6 = 0.

Write this a:*-f Ox» - ||x« - H* " 1 = ®-

Mnltiply the roots by 6,

X* _ 55 ai* - 210 2 - 216 = 0,

of which the roots are foand to be — 2, — 3, — 4, + ••
HeDce, the roots of the given equation are

-i. -h -I1 +1; OT- -f -i. -f- +t

INCOMMENSURABLE ROOTS 471

Ezercise 88

Find the commensurable roots, and if possible all the roots,
of each of the following equations :

1. a*-4a;»-8cc + 32 = 0.

2. x^-6x^+10x-S = 0,

3. x* + 2cc»- 7x2-8x4- 12 = 0.

4. a» + 3aj*-30aj + 36 = 0.

5. a;*-12a;«4-32cc2 4-27aj-18 = 0.

6. X* - 9cc« + 17x2 + 27cc- 60 = 0.

7. x« - 5x* 4- 3x« + 17x2 -28x4- 12 = 0.

8. x*-10x» 4- 35x2- 50x4-24 = 0.

9. x«-8x*4-llaj'4-29x2-36x-46 = 0.

10. x*^ — X* — 6x«4-9x2 4-x — 4 = 0.

11. 2x*-3x»- 20x2 4- 27x4- 18 = 0.

12. 2x* - 9x8 -27x2 4- 134x- 120 = 0.

13. x« 4- 3x« - 2x* - 15x8 - 15x2 4- 8x 4- 20 = 0.

14. 18x»4-3x2-7x-2 = 0.
16. 24x«- 34x2 -5x4-3 = 0.
16. 27x»- 18x2- 3x4-2 = 0.

INCOMMENSURABLE ROOTS

568. Location of the Roots. In order to calculate the value
of an incommensurable root we must first find a rough approxi-
mation to the value of the root ; for example, two integers
between which it lies. This can generally be accomplished
by successive applications of the principle of § 558. In some
equations the methods of §§ 560-563 may be useful.

472 COLLEGE ALGEBRA

(1) Consider the equation a* — 6 a* -f- 3 05 + 6 = 0.

We find (§ 610) /(O) = + 6 ; /(4) = - 15 ;

/(I) =+3; /(5) = -6;

/(2) = -5; /(6)=+28;

/(3)=-13; /(-I) = -5.

All numbers above 6 give + ; all below — 1 give — .
Hence (§ 558), the three roots are all real ; one between 1 and 2 ; one
between 6 and 6 ; one between 0 and — 1.

(2) Find the first significant figure of each root of

a;4 - 2a:» - 11 a;« -f- 6a; + 2 = 0.

The equation has, by Descartes* rule (§ 660), not more than two posi-
tive roots and not more than two negative roots.

By (§510), /(0)=+2; /(3) = -52; /(-I) = -12;

/(l)=-4; /(4)=~22; /{-2) = -22;

/(2) = -30; /(5)=+132; /(-8) = + 20.

Hence, there are two positive roots, one between 0 and 1, and one
between 4 and 5 ; and two negative roots, one between 0 and — 1, and
one between — 2 and — 3. Plot the graph and get appTozhnate valaes
of the roots by measuring on the axis of x.

To find more closely a value for the root between 0 and 1, we find
/(0.5) = + 2.06+ . Since /(I) = - 4, the root lies between 0.6 and 1.
We find/(0.8) = - 0.9+. Hence, the root lies between 0.6 and 0.8. We
find /(0.7) = + 0.4— . Hence, the root lies between 0.7 and 0.8.

In a similar manner, we find the root between 0 and — 1 to Ue between
- 0.2 and - 0.3.

Hence, the first significant figures of the roots are 0.7, 4, — 0.2, — 2.

Exerciae 89

Determine the first significant figure of each real root of the

following equations :

1. x*-x^-2x-\-l = 0, 6. x*-6x*-3x + B = 0,

2. *»-5x-3 = 0. 6. x»-h9a:* + 24x + 17 = 0.
9 'ix"+7 = 0. 7. a;»-15x« + 63aj-60 = 0.

c^

INCOMMENSURABLE ROOTS

478

569. Horner's Method of Approximation. By this method an
incommensurable root may be found to any desired degree of
approximation. We proceed to explain the method by apply-
ing it to one of the roots of the equation

a;»-6a;* + 3a; + 6 = 0.

[1]

From Descartes' rule (§ 561) the equation has not more than
two positive roots and not more than one negative root.

Before giving Homer's process we shall construct the graph
of the function of x. In this way we not only locate the roots,
but obtain a graphical representation which enables us to
follow with ease the successive steps of the approximation,
and to see exactly how they are made.

We will first compute a number of values of f(x), writing
these values in bold type.

Value op x

VlxuB OP /(a:)

VAIiUE OP X

Value of fix)

0

+ 5

+ 6

1- 6+ 3+ 5

+ 6+0+18

0+3+28

+ 1

l_6+3+ 5

-\

1- 6+ 3+ 5

+1-5- 2
-5-2+ 8

-1+7-10

- 7 + 10- 6

+ 2

1-6+3+ 6

-2

1-6+3+ 5

+ 2-8-10
-4-5- 6

- 2+16- 38

- 8 + 19- 88

+ 3

1-6+8+ 5

-3

1-6+3+ 5

+ 3 _ 9 _ 18
_ 3 - 6 - 18

- 8 + 27- 90

- 9 + 30- 85

+ 4

l_6+3+ 5

-4

1- 6+ 3+ 5

4. 4 _ 8 - 20

_ 4 + 40-172

_ 2 - 5 - 16

- 10 + 48 - 167

+ 6

l_6+8+ 5

-5

1- 6+ 8+ 5

+ 6 _ 5 _ 10

-1-2- 5

- 5 + 55-290

-11 + 58-286

474

COLLEGE ALGEBRA

The points of maxima and minima are found by § 542.

f{x) = ic* - 6aj« + 3a: -f- 5,
f{x) = 3x^ - 12 aj + 3 = 0;
whence, a = + 3.73 or -f- 0.27,

/"(x)=6a;-12,
and is positive if aj = 4- 3.73, negative if a; = -f- 0.27.
For X = 3.73, f{x) = — 16.39, a minimum.

For X = 0.27, f(x)=^-\- 5.39, a maximum.

The graph is plotted in the figure below, for conyenience
5 spaces of coordinate paper being used for 1 horizontal unit,
and 1 space for 5 vertical units. '

X^

-3 -2 -1

"N

We will now proceed *■

-A positive root between

1 and 2 by Horner's M^ -

T)h

shows that this root 1

H»d 1-6 +3 +6 y.

1.6.

+ 1 _6 -2

Diminish the roots (

f -6-2

+ »

1 ; in 0**^' »ords, chaL^

^ +1-4

'•

its p^ *ion to ti

\ -4

-•

• ^

(§ f -«rical r

-\+'

thf

\-»

INCOMMENSURABLE ROOTS 476

The transformed equation is

x^-3x^-6x-{-3 = 0, [2]

and its roots are less by 1 than those of the original equation.

This fact is clearly shown by the new position of the origin.

As equation [1] has a root between 1 and 2, equation [2]

must have a root between 0 and 1, the new zero of course being

at the point marked 1 in the figure. The graph indicates that

this root probably lies between 0 and 0.5 and nearer 0.5 than

0. The quickest way, therefore, to find the first figure of this

root is to compute the value of f(x) in [2] for different values

of X, beginning with 0.5 and going backward 0.1 at a time till

a change of sign occurs (§ 55S), The numerical work is as

follows :

x = 0.5\l-S -6 +3

+ 0.5 - 1.25 - 3.625

- 2.5 - 7.26 - 0.626
a; = 0.4| 1-3-6 -f- 3

4- 0.4 - 1.04 - 2.816

- 2.6 - 7.04 -f- 0.184

Therefore, the second figure of the root we are seeking is 0.4.

We now diminish the roots of [2] by 0.4 ; that is, change
the origin by an amount equal to 0.4 still farther towards the
right. The new axis of y passes through the point marked 1.4.

The numerical work is as follows ;

1_3 _6 4-3 |0'4
-f- 0.4 - 1.04 - 2.816

-2.6 -7.04
+ 0.4 - 0.88

-2.2

+ 0.4

-h 0.184

-7.92

-1.8

The second transformed equation is

x^ - 1.8 x^ - 7.92 x -f- 0.184 = 0. [3]

E^

476 COLLEGE ALGEBRA

The roots of [3] are less by 0.4 than those of [2]. Snee
[2] has a root between 0.4 and 0.5, [3] most hare a root
between 0 and O.L As this root is mnch less than 1, the
yalnes of the terms in [3] containing powers of x higher
than the first power must be very small; so that we shall
probably obtain the first figure of the root, if we neglect the
terms in [3] containing 7^ and x*, and put

- 7.92 X + 0.184 = 0 ; whence, x = 0.02 +.

Hence, the root of [1], which we are seeking^ correct to two
decimal places, is 1.4 + 0.02 or 1.42.
Diminish the roots of [3] by 0.02.

1-1.8 -7.02 +0.184 [0^
+ 0.02 - 0.0356 - 0.159112

- 1.78 - 7^556
+ 0.02 - .00352

-1.76

+ 0.02

+ 0.024888

— 7.9908

-1.74

The third transformed equation is

Q^ - 1.74 ar« - 7.9908 x + 0.024888 = 0. [4]

The roots of [4] are less by 0.02 than those of [3]. Since
[3] has a root between 0.02 and 0.03, [4] must have a root
between 0 and 0.01. This root is so much less than 1 that
the first two, and even the first three, significant fig^nres of it
may be found by neglecting the powers of x higher than the
first power and simply dividing the constant term by the
coefficient of the first power of x,

- 7.9908 x + 0.024888 = 0.

0.024888 ^^oii

^ = -7:99or = ®-^^^-

Therefore, the root of equation [1], correct to six significant
figures, is ^ ^ ^^^^3j^_

INCOMMENSURABLE ROOTS 477

This process may evidently be' continued until the root is
calculated to any desired degree of accuracy.

570. Remarks on Homer's Method.

First: We diminish the roots by a number less than the
required root, and as we do not pass through the root, the
sign of the last term remains unchanged throughout the work.
The last coefficient but one always has a sign opposite to that
of the last term.

If, in [3], the signs of the last two terms were alike, the value of x
would be — 0.02+ . This would show that the value assumed for x was
too great, and we should diminish the value of x and make the last trans-
formation again.

The first transformation may, however, change the sign of the last
term. Thus, if there had been a root between 0 and 1 in equation [1],
diminishing the roots by 1 would have changed the sign of the last term.

Second : In finding the second figure of the root we make
use of the theorem or change of sign (§ 558).

Any figure after the second is generally found correctly
from the last two terms ; for, in this case, the root is so small
that powers of the root higher than the first are so much
smaller than the root itself that the^ terms in which they
appear have but slight influence upon the result.

571. It is not necessary to write the successive transformed
equations. When the coefficients of any transformed equation
have been coqiputed, the next figure of the root may be found
by dividing the last coefficient by the preceding coefficient,
and changing the sign of the quotient.

Thus, in equation [4], the next figure of the root is obtained by dividing
0.024888 by 7.9908.

For this reason, the last coefficient but one of each trans-
formed equation is called a trial divisor.

Sometimes the last coefficient but one in one of the transformed equa-
tions is zero. To find the next figure of the root in this case follow the
method given for finding the second figure of the root

478

COLLEGE ALGEBRA

The work may now be collected and arranged as follows :

-6

+ 1

+ 3
-5

+ 6
-2

11.423

-6

+ 1

-2
-4

^

-4

+ 1
-8

+ 0.^

i

-6

-6 +8
- 1.04 - 2.816

|0.4

-2.6
+ 0.4

-7.04
-0.88

+ 0.184

-2.2
+ 0.4

-7.92

-1.8

+ 0.02

-7.92 +0.184

-0.0356 -0.159112

|0.02

-1.78
+ 0.02

- 7.9556

- 0.0352

+ 0.024888

-1.76
+ 0.02

-7.9908

-i.-;

r4

- 7.9908 + 0.024888

10.003

The numbers in heavy type are the coefficients of the successlYe trans-
formed equations, the first coefficient of each equation being the same as
the first coefficient of the given equation. In this example the first coeffi-
cient is 1.

When we have obtained the root to three places of decimals we can
generally obtain two or three more figures of the root by simple division.'

572. In practice it is convenient to avoid the use of the
decimal points. We can do this as follows f multiply the
roots of the first transformed equation by 10, the roots of
the second transformed equation by 100, and so on. In the
last example the first transformed equation now is

a;8 - 30 a;2 - 600x -f- 3000 = 0,

and this equation has a root between 4 and 5. The second
transformed equation now is

a;« - 180 aj2 _ 79,200 x -f- 184,000 = 0,
and this equation has a root between 2 and 3. And so on*

INCOMMENSURABLE ROOTS

479

The complete work of the last example, for six figures of
the root, is as follows :

11.42311 +

1

-6
+ 1

+ 3
-6

+ 5
-2

-6

+ 1

-2
-4

+ 8

-4

+ 1

-6

- 80

+ 4

-600

-104

+ 8000

-2816

-26

+ 4

-704
- 88

+ 184

-22
+ 4

-792

- 178070

li

- 180

+ 2

- 79200

- 366

+ 184000
- 159112

-178
+ 2

- 79566

- 362

+ 24888

- 176
+ 2

-79908

\1

- 1740

+ 3

- 7990800
6211

+ 24888000
- 23988033

-1737
+ 3

- 7996011
6202

+ 899967

-1734
+ 3

- 8001218

•

13

- 17810

+ 1

- 800121300
17309

+ 899967000
- 800138609

-17309
+ 1

- 800138609
17308

+ 99828891

- 17308
+ 1

- 800155917

11

- 80015591700 + 99828891000 [1

We have here performed the work in full for six figures of
the root. We can find five more figures of the root by simple
division. If we divide 99,828,391 by 800,166,917, we obtain
0.124761, so that the required root to ten places of decimals
is 1.4231124761.

480 COLLEGE ALGEBRA

The reason why simple division gives five more figures of the root is
seen by examining the last transformed equation. Write this

8.00155917 X = 0.000099828391 - 1.7307x2 + x«.

As X is about 0.00001, iB» is about 0.0000000001, and x« is much
smaller. Hence, the error in neglecting the x^ and x' terms is in 8 x
about 0.00000000017, and in x about 0.00000000002. The result obtained
by division is therefore correct to ten places of decimals.

Comparing the work on page 479 with that on page 478, we see that
we have avoided the use of the decimal point by adopting the following
rule:

When the coefficients of a transformed equation hxive been obtainedj add
one cipher to the second coefficient, two ciphers to the third coefficient, and
so on. The coefficients and the next figure of the root are then integers.

If the root of the given equation lay between 0 and 1, we should begin
by multiplying the roots of the equation by 10.

573. Negative Roots. To avoid the inconvenience of work-
ing with negative numbers, when we wish to calculate a
negative root we change the signs of the roots (§ 547) and
calculate the corresponding positive roots of the transformed
equation.

Thus, one root of the equation

x8 -6x2 + 3x4-5 = 0

lies between 0 and — 1 (§ 568). By Homer's Method we find the corre-
sponding root of

x8 + 6x2 + 3x-5 = 0

to be 0.6696 + . Hence, the required root of the given equation is
-0.6696+.

Ezercise 90

Compute to three decimal places for each of the following
equations the root of which the first figure is the number in
parenthesis opposite the equation :

1. x» + 3x-5.= 0. (1)

2. ic«-6x -12 = 0. (3)

INCOMMENSURABLE ROOTS

481 0

3. aj» + «* + a- 100 = 0.

4. aj» + 10ic« + 6a; -120 = 0.
6. ic»4-9ic* + 24x-|-17 = 0.
6. aj*-12a;« + 12aj-3 = 0.

7. X^-SX*+14:X^+4:X-S

= 0.

(4)

(2)
(-4)

(-1)
(-0)

574. Contraction of Homer's Method. In § 572 the reader
will see that if we seek only the first six figures of the root,
the lajst six figures of the fourth coefficient of the last trans-
formed equation may be rejected without affecting the result.
Those figures of the second and third coefficients which enter
into the fourth coefficient only in the rejected figures may
also be rejected. Moreover, we may reject all the figures
which stand in vertical lines over the figures already rejected.

The work may now be arranged as follows :

-6

±1
-6

+ 1

+ 3
-5
-2
-4

+ 6 I 1.42311 +
-2

+ 3000

-2816

-4

-600

+ 184000

+ 1

-104

- 169112

-30

-704

+ 84888

+ 4

- 88

- 23991

-26

-79200

+ 897

+ 4

- 356

-800

-22

-79666

+ 97

+ 4

- 352

- 80

-180

-79908

+ 2

-7991

-178

- 6

+ 2

-7997

-176

- 6

+ 2

-8008

-174
-2

800
80

482 COLLEGE ALGEBRA

The double lines in the first column indicate that beyond
this stage of the work the first column disappears altogether.

In the present example we first find three figures of the
root. We then contract the work as follows:

Instead of adding ciphers to the coefficients of the trans-
formed equation, we leave the last term as it is; from the
last coefficient but one we strike off the last figure ; from the
last coefficient but two we strike off the last two figures ; and
so on. In each case we take for the remainder the nearest
integer.

Thus, in the first column of the preceding example we strike off from
174 the last two figures, and take for the remainder 2 instead of 1.

The contracted process soon reduces to simple division.

Thus, in the last example, the last two figures of the root were found
by simply dividing 897 by 800.

To insure accuracy in the last figure, the last divisor must
consist of at least two figures. Consider the trial divisor at
any stage of the work. If we begin to contract, we strike off
one figure from the trial divisor before finding the next fig^ure
of the root. Since the last divisor is to consist of two figures,
the contracted process will give us two less figures than there
are figures in the trial divisor.

Thus, in § 572, if we begin to contract at the third trial diiiaory
— 79,908, we can obtain three more figures of the root ; if we begin to
contract at the fourth trial divisor, — 8,001,213, we can obtain five more

figures of the root ; and so on.

575. When the root sought is a large number we oannot
find the successive figures of its integral portion by diyiding
the absolute term by the preceding coefficient^ because the
neglect of the higher powers, which are in this case large
numbers, leads to serious error.

INCOMMENSURABLE ROOTS 483

Find one root of «* - 3 «« H- 11 « - 4,842,624,131 = 0.

aj* - 3x2 + 11 X - 4,842,624,131 = 0. [1]

By trial, we find that a root lies between 200 and 300.
Diminish the roots of [1] by 200,

jc* + 800 x8 + 239,997 x^ + 31,998,811 x - 3,242,741,931 = 0. [2]

If X = 60, f(x) = - 273,064,071.
K X = 70, f(x) = + 471,570,139.

The signs of f(x) show that a root lies between 60 and 70.
Diminish the roots of [2] by 60,

X* + 1040 x8 + 406,697 x^ + 70,302,461 x - 273,064,071 = 0. [3]

The root of this equation is found by trial to lie between 3 and 4.
Diminishing the roots by 3, we may find the remaining figures of the
root by the usual process.

576. Any root of a positive number can be extracted by
Horner^s Method.

(1) Find the fourth root of 1296.

Here, x* = 1296,

or x* + 0x8 + 0x2 + 0x-- 1296 = 0.

Calculate the root, x = 6.

If the number is a perfect power, the root is obtained exactly.

(2) Find the fourth root of 473.

Here, x* = 473,

or X* + 0x8 + 0x2 + Ox - 473 = o.

Calculate the root, x = 4.66353 + .

577. Roots nearly Equal In the preceding examples the
chajiges of sign in the value of f(x) enable us to determine
the situation of the roots. In rare cases two roots may be so
nearly equal that they both lie between the same two consecu-
tive integers. In this case the existence of the roots will not
be indicated by a change of sign in /(«), and we must resort
to other means to detect their presence.

484 COLLEGE ALGEBRA

Find the roots of the equation x*— 515 x*+ 1155a; —649=0.

x» - 616»« + 1165X - 649 = 0. [1]

By DeBcartes* role Hub equatiOQ has no nega-
tiye root It haa, therefore, oertainly one, and
pertiapa three, poaitiYe loota.

We find /(-I) = -2320;
/(0)=-649;

/(2)=-891;
/(8) = - 17«.

The approach of f(x) towards 0 indicates either that there are two
roots near 1 or that tlie function approaches 0 without reaching it, the
graph in the latter case heing as here shown.

Let us proceed on the supposition that two roots near 1 do eziaL
Diminish Uie roots by 1. The traDsformed equation

«• - 612aB» + 128x -8 = 0, [2]

by Descartes' rule, still has either one or three positive roota, ao tliat we
have not passed the roots.

If we diminish the roots by 2, we ol>tain

x» - 509x2 - 893x - 391 = 0,
which has but one positive root ; so that we liave passed both roots.

To find the second figure of the root, neglect the first term of equa-
tion [2]. Since the roots are nearly equal, the expression

512x2-128x + 8

must be nearly a perfect square. Comparing this with a(x — ctiK or

128 2x8

0x2 _ 2 aax + a«^» we see that and — — are «mraziiiiate

2 X 612 128

values for the roots; these both give i, or 0.12.

Diminish the roots by 0.1 ; the work is as before. Contintie until
the two quotients obtained as above give different numhen for the next
figure of the root. In the present example this occurs wlien we oome to
the third decimal figure ; the transformed equation ia

x» - 61,164 x« + 61,832 x - 11,072 = 0, [«]

and the two quotients are 0.6+ and 0.3 +. To separate the roots, try
0.4; the left member of the last equation is found to he +. Since 0
gives — and 1 gives — , there is one root between 0 and 0.4, and one
between 0.4 and 1.

To calculate the first root, we try 0.3; as this gives a — flign, ire
diminish the roots by 0.3 and proceed as in { 674.

INCOMMENSURABLE ROOTS.

485

1

- 616 + 1166
1 - 614

- 649 1 1.1230914
+ 641

-614

1

+ 641
- 613

-8000

+ 7681

-618

1

+ 12800

- 6119

- 819000

+ 307928

-5120
1

+ 7681
-6118

- 11072000

+ 10884867

-6119

1

-6118

1

+ 256800

-102336
+ 163964
- 102332

- 187188

+ 184276

-2868
+ 2002

- 51170

2

- 61168

2

+ 5168200

- 1634911
+ 3628289

- 1634902

-866
+ 800

-61166
2

+ 2098887
+ 209339

- 511640

3

- 611637

8

- 611634

3

+ 20934

- 469
+ 20476

- 469
+ 20016
+ 2002

- 511681

-6116
-61

+ 200

To calculate the second root, we return to equation [3],

x« - 61,164x2 + 61,632 X - 11,072 = 0.

We have/(0.4) = +, /(I) = - ; we find /(0.6) = +, /(0.7) = + 0.383.
Since/ (0.7) is so small,/ (0.8) is undoubtedly negative.
' Diminish the roots by 7 and proceed as follows :

1

-511640

7

+ 5168200

- 3681431
+ 1681769

- 3681382

-11072000 11.1270002
+ 11072883

- 611688

7

+ 888

- 611626

7

-1999618

-200

- 511619

486 COLLEGE Al.G£BliA

Siuoe the sum of the roots (| b2l) is 515, we can find the tihird root by
0uhLra(niug from 515 the sum of tiie two roots already fonnd.

Heuoe, the 3d root = 515 -> (1.1280P14 + 1.1270002) = 512.7409064.

578. From tLe preceding sections we obtain the following
general directions for solving a numerical eqnatian :

1. Find and remove commensurable roots by §§ 565-^567, if
there are any such roots in the equation.

2. l>etermine the situation and thence the £iBt figure of
each of the incommensurable roots as in § 568.

3. Calculate the incommensurable roots bj Honier's Metliod.

Calculate to six places of decimals the positiTe zoots of the
following equations :

1. z^- 3x^1 = 0.

2. ar»-h2a-*-4jc-43 = 0.
8. 3a:»-h3a-*-h8ar-32 = 0.

4. 2a^-2Car* + 131x- 202 = 0.

6. x*-12x-\-7 = 0.

6. x*-5a:«-h2x»-13aj 4-55 = 0.

Calculate, to six places of decimals where incommensurafale^
the real roots of the following equations :

7. ar» = 35,499. 10. ar« = 147,008,443.

8. a:« = 242,970,624. 11. ^• + 2a: -h 20 = 0.

9. ir* = 707,281. 12. a:*- 10x«-h8«+120a=0.

Each of the following equations has two roots nearly eqnaL
Calculate the roots to six places of decimaJs :

13. x«-33r2_ 4x4-13 = 0.

14. 2a:* 4- 8a:»-35ar»-36aj + 117 = 0.
16. aj« 4- 11 X* - 102 a; 4- 181 = 0.

STURM'S THEOREM

487

STURM'S THEOREM

579. The problem of finding the number and the situation
of the real roots of an equation is completely solved by Sturm's
Theorem. In theory Sturm's method is perfect ; in practice
its application is long and tedious. For this reason, the
situation of the roots is in general more easily determined by
the methods already given.

Before passing on to Sturm's Theorem itself, we shall prove
two preliminary theorems.

580. Situation of the Roots of f '(x) = 0. Between any two
distinct real roots of the equation f (x) = 0 there lies at least
one real root of the equation f '(x) = 0.

Let a and p be two real roots of f(x) = 0, /? being greater
than a. Then f(a) = 0 and f(P) = 0. As aj increases con-
tinuously from a to p,' f{x) changes from 0 to 0 again ; and
must first increase and then decrease, or first decrease and
then increase. Hence, there must be some point at which
/'(x) changes from -|- to — ,
or vice versa. Therefore, for
some value of x between a
and p, /'(^) must be zero.
Hence, at least one root of
f\x) = 0 must lie between
a and p.

In the graph the curve will be
horizontal where f'(x) = 0. In

the figure here given, A, B, C, D correspond to roots of f{z) = 0.
Between A and B there is one root of /' (x) = 0 ; between B and C,
three roots ; and between C and D, one root.

It is evident that if more than one root of f'(x) lies between
a and p, the number of roots piust be an odd number.

488 COLLEGE ALGEBRA

581. Signs of f (x) and f '(^)' -^^ ^ ^^ ^^t/ ^^ ^''^^ ^f ^^
equation^ f (x) = 0, which has no equal roots.

Let X change continuousli/ from a — h, a value a little less

than a, to a -\-hy a value a little greater than a. Then f (x)

and f (x) have unlike signs immediately before x passes through

the root, and like signs immediately after x pa^es through the

root.

For, f(a -h) = - hf'(a) + ^f"(a) - • • .,

and f'(a - A) = /'(a) - hf^a) + • • • ; (§544)

since f(a) = 0, as a is a root of f(x) = 0.

When h is very small the sign of each series on the right
is the sign of its first term (§ 556); and /(a — h) and /'(a — A)
evidently have opposite signs.

Similarly, f(a -f h) and f'(a -\- h) have like signs.

Note. The above is also evident from the graph of /(x).

582. Sturm's Functions. The procei^ of finding the H.G.F.
of f(x) and f'(x) has been employed (§ 543) in obtaining
the multiple roots of the equation f(x) = 0. We use the
same process in Sturm's Method.

Let f(x) = 0 be an equation which has no multiple roots ;
let the operation of finding the H.C.F. of f(x) and /'(«) be
carried on until the remainder does not involve a;, the sign of
each remainder obtained being changed before it is used €u a
divisor.

Note. If there is a H.C.F., the equation has multiple roots. Bemove

them and proceed with the reduced equation.

Eepresent by /a (a;), fz{x)y • • yfni^) the several remauoders
with their signs changed. These expressions with f(pe) are
called Sturm's Functions.

Now, if D represents the dividend, d the divisor, q the quo-
tient, and Ji the remainder,

D = qd + R.

STURM*S THEOREM 489

Consequently, f(x) = qif (x) — /, (»),

where qi, q^, •••, q^-i represent the several quotients, or the
quotients multiplied by positive integers.

From the above identities we have the following :

1. Two consecutive functions cannot vanish for the same
value of X,

For example, suppose that /^(x) and fz(x) vanish for a particular
value of X, Qive to x this value in all the identities. By the third
identity, f^(x) will vanish; by the fourth, /s(x) will vanish; finally,
fn (x) will vanish, which is contrary to the hypothesis that /(x) = 0
has no multiple roots.

2. When we give to a; a value which causes any one func-
tion to vanish, the adjacent functions have opposite signs.

Thus, if fz (x) = 0, from the third identity ft (x) = - /4 (x).

583. Sturm's Theorem. We are now in a position to enun-
ciate Sturm's Theorem ;

If in the series of functions

f(x), t'(x), f,(x), ..., f.(x)

we give to x any particular value a, and determine the number
of variations of sign ; then give to x any greater vaJ^ue b,
and determine the number of variations of sign ; the nuniber
of variations lost is the number of real roots of the equation
f (x) = 0 between a and b.

For, let X increase continuously from a to 6.
1. Take the case in which x passes through a root of any
of the functions /'(^)> f%(?^)i • *> /i»-i(^)> for example f^(x).

490 COLLEGE ALGEBRA

The adjacent functions have opposite signs. f^(x) itself
changes sign, but this has no effect on the number of varia-
tions; for if just before x passes through the root the signs |
are + H — , just after x passes through the root they are

H , and the number of variations is in each case one.

Hence, there is no change in the number of variations of
sign when x passes through a root of any of the functions

2. Take the case in which x passes through a root of
f(x) = 0. Since f(x) and /' (x) have unlike signs just before
X passes through the root, and like signs just after (§ 581),
there is one variation lost for each root of f(x) = 0.

Hence, the number of real roots between a and h is the
number of variations of sign lost as x passes from a to h.

To determine the number of real roots, we take x first very
large and negative, and then very large and positive. The
sign of each function is then the sign of its first term (§ 555).

The reader may not understand how it is that /(x) and /"(x) always
have unlike signs just before x passes through a root.

Let a and /3 be two consecutive roots of /(x) = 0 ; let A be very small.
Suppose that at a f(x) changes from + to — ; then f^{cc) is — (§ 640).

When x = a — h, f(x) is +, /' (x) is — ;

when X = a:, f(x) isO, f(x) is — .

As X changes from a to /3, f (x) passes through an odd number of
roots (§ 580), and consequently changes sign. Hence, when x = /3 — A,
f(x) is — , /'(x) is + ; and f(x) and /(x) again have unlike signs.

584. Examples. (1) Determine the number and the signs of
the real roots of the equation

aj* _ 4a;8 -I- 6ir« - 12aj + 1 = 0.
Here, /'(x) = 4x«- 12x2+12x-12.

Let us take for f (x), however, the simpler expression

x8-3x2 + 3x-3.

STURM'S THEOREM

491

We proceed as if to find the H.C.F., changing the sign of each
remainder before using it as a divisor.

1- 3+ 3_ 3

3_ 9+ 9- 9
3+ 1

-10+9

-30 + 27

. - 80 - 10

37-

9

111-

27

111 + 37

1_4-|.6-12 + 1

l-3-l-3„ 3
-l+S- 9 + 1

-1 + 3- 3 + 3

- 6-2

8 + 1

1-10 + 87

-64

+ 64

The coefBcients of the several functions are in heavy type. In the
ordinary process of finding the H.C.F. we can change signs at pleasure.
In finding Sturm's functions we cannot do this, as the sign is all important
We can, however, take out any positive factor.

i» We now have /(x) = aj* -4«« + 6a^ - 12« + 1,

/'(x) = x«-3«« + 3x-3,
• /j(x) = 3x + l,

/,(x) = + 64.

/(«) /'(aj) Mx) A{x)

When X = - 1000 + - - +2 variation&

x = 0 + — + +2 variation&

X = + 1000 + + + +0 variations.

Hence, the equation has two real positive roots ; it must therefore have
two complex roots.

The real roots are found, by § 668, to lie one between 0 and 1, and
one between 3 and 4.

(2) Investigate the character of the roots of the equation

x^-^3Hx-\-G = 0.

We find /(x) = x« + 3flx + G,

/'(x) = 3(x» + fl'),
/i(x)=-2JJx-0,
/8(x) = -((af»+4H«\.

492 COLL£G£ ALGEBRA

Ji €P + 4 H*iB positive, we have

f(^) /'(«) Mx) Mx)
x = — 00 — + ± - 2 variations.
x = + oo + + ^ — 1 variation.

Since H may be either + or — , the sign of f^ (x) is ambiguous.
Hence, when G^ + 4 JJ' is positive there is but one real root.
If G^ + 4 H^ is negative, H must be negative, and we have

fix) r{x) Mx) Mx)

x = — oo — -H — +3 variations.
x = + oo + -H + +0 variation.

Hence, when G^ + 4 IT^ is negative there are three real roots.

Exercise 92

Find by Sturm's Theorem the number and the situation
of the real roots of the following equations :

1. aj«-4aj^-lla;-f43 = 0. •

2. a;'-6aj2-|-7a;-3 = 0.

3. aj*-4ir« + ir« + 6a; + 2 = 0.

4. aj*-5a;«-f 10ir2-6aj-21 = 0.

5. a;*-.ic«-a;2 + 6 = 0.

6. a;*-2a;«-3a;«+10a;-4 = 0.

7. a;« + 2a;* + 3ir» + 3x2-l = 0.

8. ajS + a;*-2aj2 + 3a;-2 = 0.

9. a;*-12aj»H-47a;^-66a; + 27«a

10. 9aj*-54aj« + 60aj*- 72a; + 16 = 0.

11. 2a;* - 5aj« - 17 «« + 53a; -28 = 0.

12. aj* + 2aj»-37a;^-38a;H-l = 0.

13. 121a;* + 198a;»-100a;«-36« + 4=ia

CHAPTER XXXII
GENERAL SOLUTION OF EQUATIONS

585. Numerical and Algebraic Solutions. By the methods of
the preceding chapter we can find to any desired degree of
accuracy the real roots of a numerical equation of any degree.
The methods are theoretically complete, and the solution of a
numerical equation becomes simply a question of the labor
required for the necessary computations.

In the case of a literal equation we have an entirely differ-
ent problem to solve. To solve a literal equation, we have to
find in terms of the coeflB.cients expressions which will, when
substituted for the unknown in the given equation, reduce that
equation to an identity. Thus, the roots of the general quad-
ratic ax^ -f ftx -f c = 0 have been found to be (§ 191)

2a

In the case of a particular quadratic with numerical coeffi-
cients the roots can be found by putting for a, 6, c in the above
expression their particular values, and performing the indi-
cated operations.

Similar solutions have been obtained for the general equa-
tions of the third and fourth degrees, and for certain special
forms of equations of higher degrees.

The solution of the general equation of the fifth degree-
involves expressions called elliptic functions, and is conse-
quently beyond the scope of the present treatise.

In many cases, however, the numerical values of the roots
of a particular equation are not easily obtained from the

493

494 COLLEGE ALGEBRA

general solution, and for numerical equations the general
solutions are in such cases of little value.

A general solution differs from the solutions obtained in
the last chapter in that a general solution represents not one
particular root but all the roots indiscriminately.

586. Reciprocal Equations. Reciprocal equations (§ 551),
called also* recurring equations, are of four forms :

1. Degree even ; corresponding coefB.cients equal with like
signs.

2. Degree even; corresponding coefficients numerically
equal but with unlike signs.

3. Degree odd; corresponding coefficients equal with like
signs.

4. Degree odd ; corresponding coefficients numerically equal
but with unlike signs.

The following are examples of the four forms :

1. 2x* -3x» + 4x2--3x + 2 = 0;

2. 3x« -x6-|-2x*-2x2 + x-3 = 0;

3. x6 + 3x* -2x»-2x2-|-3x + l=0;

4. 2x6-|-6x* + x8-x2-6x-2 = 0.

Every equation of the second form evidently lacks the middle term.

Every reciprocal equation of the second, third, or fourth
form can be depressed to an equation of the first form.

Second Form. Consider the equation

or a(x^ - 1) + ^«(«* - l)-\-cx^(x^ - 1)= 0.

Hence, the equation is divisible by aj* — 1 ; consequently, 1
and — 1 are both roots. The depressed equation formed by
dividing the given equation by x^ — 1 is

ax^ -f ^x* -f (a -f c) aj^ -f to + a = 0,

which is evidently of the first form.

Similarly for any equation of the second form.

GENERAL SOLUTION OF EQUATIONS 495

Third Form, Consider the equation

ax^ -f hx^ -f cx^ -\- cx^ -{- bx ■\- a =^ Of

or a{x^ -\-l)-{-hx («' + 1) + cx^(x + 1) = 0. [1]

Hence, the equation is divisible by « + 1 ; consequently, — 1
is a root. The depressed equation formed by dividing [1] by
aj + 1 is

ax* -- {a — h)x^ -{- (a — h -{- c)x^ -- (a — h)x -{- a = 0,

which is evidently of the first form.

Similarly for any equation of the third form.

Fourth Form. Consider the equation

ax^ -\- bx^ -f cx^ — cx^ — bx — a = Of
or a (x^ -l)-{-bx (x« - 1) + cx\x - 1) = 0. [1]

Hence, the equation is divisible by aj — 1 ; consequently, + 1
is a root. The depressed equation formed by dividing [1] by
a; — 1 is

ax* -{-(a + b)x^ -{-(a -{- b -{- c)x^-^(a -f ft)aj + a = 0,

which is evidently of the first form.

Similarly for any equation of the fourth form.

By the preceding, to solve any reciprocal equation, it is only
necessary to solve one of the first form.

587. Any reciprocal equation of the first form can be
depressed to an equation of half the degree.

We proceed to illustrate by examples :

(1) Solve the equation aj* - 12a;« -f 29aj« - 12 a; + 1 = 0.

Divide by x^ x2 + 1 - 12 fx + -") + 29 = 0.

x2 V x/

Solve this equation f or x ^ —

X

Then, x + - = 9 or 3.

X

496 COLLEGE ALGEBRA

Solve these eqoatioiui for x.

^. 9±V77 , 3±V6

Then, x = » and x = •

2 2

The first two roots are reciprocals each of the other ; also the second
two roots are reciprocals each of the other.

(2) Solve the equation

This is of the fourth form ; dividing by x — 1, we find the depressed

equation to he

X* - 2x« -H 3x2 - 2x + 1 = 0.

This may he written

,. + 2 + l-2(. + i) + l=0.

Extract the root, x H 1=0.

X

Solve, X = ,

2

these expressions being double roots.

Ezerciae 93

Solve the equations :

1. aj*-f 7aj*-7a;~l = 0.

2. x^-\-2x^-{-x^-{-2x-{-l=z0.

3. x^-3x^ -{-Bx^-Bx^-^-Sx-l — O.

4. aj*-5a;« + 6x2- 5ir + l = 0.

5. 2x^-5x^-\-6x''-5x-{-2=:0.

6. aj^ - 4 a:* + x' + a;2 — 4aj H- 1 = 0.

7. a:*-10ir« + 26ir«-10a; + l=:0.

8. x^ -f mx^ -\- mx -f 1 = 0.

9. x^ -\-x^ — x^ — x^-\-x-\-l = 0.

10. Sx^ - 2x^ -{- 5x* - 5x^ -\-2x -S^^^O.

GENERAL SOLUTION OF EQUATIONS 497

588. Binomial Equations. An equation of the f onn

is called a binomial equation.

We shall first consider the two equations

sr" — 1 = 0, aj» 4- 1 = 0.

If n is even, the equation a" -f 1 = 0, by Descartes' rule
(§ 560), has no real roots ; the equation a;" — 1 = 0 has two
real roots, -f 1 and — 1, the remaining n -^ 2 roots being
complex.

If 71 is odd, the equation a" -f 1 = 0 has one real root, — 1 ;
the equation a;" — 1 = 0 has one real root, 4- 1, the remaining
n — 1 roots being in each case complex.

589. Now consider the equation a;** ± a = 0, where a is

positive. Represent by Va the positive scalar Tith root of a.

Then, if a is any root of a;*» ± 1 = 0, a Va will be a root of
aj» ± a = 0.

For, (a Va)" = o^a = ^:lxa = ^a.

Since a is any root of oj" ± 1 = 0, the n roots of a;" ± a = 0 ^
( are found by multiplying each of the n roots of aj" ± 1 = 0 ^

•. by Va.

The roots of a binomial equation are all different. For
x^ ±a and its derivative na?""^ can have no common factor
involving x (§ 543).

590. If a is a root of the equation x** — 1 = 0, then a\
where k is an integer, is also a root.

For, if a is a root, a* = 1.

But (c^y = (a»)* = (1)* = 1.

Therefore, a* is a root of a;** = 1, or of a;" — 1 = 0.
Similarly for a root of a" 4- 1 = 0, provided A; is an odd
integer.

498 COLLEGE ALGEBRA

59L The Cube Roota of Unity. The equation j^ ^ 1, or
«• — 1 = 0, may be written

(x-l)(«»4-« + l)=0,

of which the three roots are

If either of the complex roots is represented by o*, the other
is found by actual multiplication to be oi^. This agrees with
the last section.

Also^ w' 4- a> + 1 = 0.

In a similar maimer, we find the roots of x* = — 1 to be

or — 1, — a>, — a>*,

592. Examples. (1) Find the six sixth roots of L

We have to solve x^ — 1 = 0,

or (x«-l)(x8+l) = 0.

Hence, the roota are ± 1» ± «i ± ft>*.

(2) Find the five fifth roots of 1.

We have to solve z* — 1 = 0,

or (X - 1) (x* + x8 + x2 + X + 1) = 0.

.-. X — 1 = 0, or X = 1 ;

or x* + x8 + x2 + x + l=0,

1 1 _ 1 4- Vs

Solve for X + -» x + - =

X X 2

Solve these equations for x. and we obtain for the remainhig foor
roots,

_ 1 4. V5 -^ Vio 4- 2 \ 5 \^^ -1 _ VsiVio-aVgVUT

GENERAL SOLUTION OF EQUATIONS 499

Ezeroise 94

Solve the binomial equations :

1. a;* 4- 1 = 0. 3. a;'^-l = 0.

2. a;« - 1 = 0. 4. aj» - 243 = 0.

5. Find the quintic on which depends the solution of the
equation x^^ = 1.

6. Show that x^±y^ = (x± y) (x ± my) (x ± w^y).

7. Show that

x^ + y^ -{- z^ — yz — zx — xy = (x -^ toy -^ ii>^z)(x -f co^y + mz).

8. If a is a complex root of a;* — 1 = 0, show that

(1 - a)(l - a')(l - a»)(l - a'')=z5.

593. The General Cubic. We shall write the general equa-
tion of the third degree in the form

ax^ + Sbx^-^Scx-\-d = 0. [1]

Before attempting to solve this equation we shall transform

it into an equation in which the second term is wanting.

z — b

Put z = ax -\- b. Then, x =

a

Substitute this expression for x and reduce,

«» + 3(ac - ft«)« + (aH -Sabc -f 2^»)= 0,

or, putting H = ac — b^, and G = a^d — 3 abc + 2 b*,

z^-\-3Hz-\- G = 0. [2]

In the transformed equation put z = u^ + v\

Then, (w* + v^y + 3 H(u^ -\-v^)+G = 0,

which reduces to

u + v-\- 3(w*y* 4- H) (u^ + v*) -f G = 0. [3]

500 COLLEGE ALGEBRA

Biuce we have assumed but one relation between u and r,
we are at liberty to assume one more relation.

Let us assume w*r* = — JET. [4]

[3] now reduces to u + r = — (7. [5]

[4] may be written wr = — H*. [6]

Eliminate r from [5] and [6], w* + Gtt = £r». [7]

Equation [7] is c-alled the rednciiii: qnadzvtk of the cubic.

Solving this quadratic, we find

-o±

Vg

2-h4if»

2

-//•

OzfV^

+ 4iy*

u =

"= u - 2

f8]

Since ax -\- b == z = u^ -^ v^, the three values of z are

where u^ is anv one of the three cube roots of tc

SiiK^e there is the sign ± before the radical, we have appar-
ently six values of z. From [4] it is seen, howerer, that there
are really but three different values of z.
The above solution is known as CardanU.

Solve, by Cardan's Method,

Here, a = 2, 6 = - 2.

Put 2 = 2x-2; then 2» -f 12z - 12 = 0.

.'. IT = 4, G = — 12, and the reducing quadratic Ja

u«-12u = 64.
Solve, u = 6 i 10 = 16 or - 4.

.\v = = - 4 or + 16.

u

GENERAL SOLUTION OF EQUATIONS 501

Hence, the values of z are

2V2-V4; 2wV^-«2V4; 2«2V^_„V4j
and the values of x are

V^ V^ V2

594. Discussion of the Solution. The above solution, while
complete as an algebraic solution, is of little value in solving
numerical equations.

In the case of a cubic there are three cases to consider.

I. All three roots real and unequal. In this case G^ + 4:H^
is negative (§ 584, Example 2), and its square root is ortho-
tomic. If we put K^ = — (G^ 4- 4 H^, we have

ax + h^[^ j+(^ y

Since there is no general algebraic rule for extracting the
cube root of a complex expression, the case of three real and
unequal roots is known as the irreducible case.

II. Two of the roots equal. In this case G^* + 4 i^' = 0
(§ 584, Example 2), and we have

'-^"-{'fi^i^fi-

III. Two roots complex. In this case G' -f 4 ^' is positive
(§ 584, Example 2), its square root is real, and we have

G 4- V(?2 4- 4 H^\} f^G- V(?^ + 4g»^*,

ax 4- h

= [ 2 ) + [ 2 -}

The value of the expression (?' 4- 4 ZT' determines the
nature of the roots. For this reason, C?' 4- 4 ZT* is called the
discriminant of the cubic.

502 COLLEGE ALGEBRA

Hence, we conclude that the general solution gives the
roots of a numerical cubic in a form in which their values
can be readily computed only in the second and third cases.

In either of these cases, however, the real roots are more
easily found by Horner's Method.

In the first case the roots may be calculated by a method
involvmg Trigonometry. (See § 616, Chapter XXXIII.)

Ezeroise 95

Find the three roots of :

1. a;' + 6aj»= 36.

2. 3a;»-6a;*-2 = 0.

3. x^-3x^-6x-4: = 0,

4. 9aj8-54aj»-f 90x-50 = 0.
6. «» 4- 3 m«' = m^ (m + 1)*.

6. In the case of the cubic, putting
show that Z» 4- M» = 2 Sa» - 3 Sa'jS + 12 afiy

\a a' a* J

2,1 G

^^^ •

a'
LM = Sa^ - l,ap

9H

^^ __ __^^^ .

and Z» - M» = - 3 V^ (P - y) (y - a) (a J P).

7. From Example 6, and the relation

(Z» - M^y = ip 4- M^y - 4 Z»M«,
show that a« ()8 - y)^ (y - «)» (a - py = - 27 ((?« + 4 IT*),
and thence deduce the conditions of § 594.

GENERAL SOLUTION OF EQUATIONS 603

595. The General Biquadratic. We shall write the general
equation of the fourth degree in the form

ojx'^ -f ihx^ -f 6caj' -f 4cfoj + e = 0. [1]

Put z=^ax-\-h.

Then, x =

a

Substitute in [1] this expression for x and reduce,

«* + 6 {ac - IP^z" + ^{aH - 3 o^c + 2 «>«)«

4- (a^e - ^a^hd + ^dU'c - 3^*) = 0. [2]
The fourth term may be written

a2(ae - 4 «>(^ + 3 c«) - 3 (ac - by.
Put H=:ac- b^,

G = a^d-3abc-\-2b*, .,
and / = ae — 4 dc? + 3 <J^.

Then [2] is written in the form

«* 4- 6 /f«2 + 4 (?« 4- a^I -SH^ = 0, [3]

in which the z* term is wanting.
To solve this equation put

Square, z^=:u-\-v-\-w-\- 2('Vuv + Vwi^ + VtJwJ).
Transpose, and square again,

«* — 2(w + v + w;)«' 4- (w 4- V + w)^

= 4 (wv 4- WW 4- vw) 4- 8 « Vw "vv Vt£?.

If this equation is identical with [3],

u-\-v + w='-3H,

uv + uw 4- vw = 3 H^ J- y

4

G

K

504 COLLEGE ALGEBRA

Hence (§ 521;, u, v, and w are the roots of the cubiG

^-^SHt^-^fsH^- ^\ - ^ = 0. [4]

This is known as Eulei^s cubie.
This equation may be written

^t^Hf- — {t + H)^ 4 = 0,

or, putting t -\- H =^ a^$, and clearing of fractional

4a«d»-/ad + /=0, [5]

where J = —^ (a^HI - r;^ - 4tH^)^ace + 2icrf- orf* — e^ — fJ^,

Equation [5] is called the redodng cnMc of the biqnadiatic.
If ^1, ^2, dg are the roots of this cubic, since t = aV — jy,
the four roots of equation [1] are given by

ax-\-h^ Va^^i - H + Va'da — JT + Va^ — £r. [6]

Since each radical may be either + or — , there are appar-
ently eight values of x obtained from the following oombiiiar
tions of signs :

+ + + + + - 4--+ - + +
4- - + - +

But Vm v^ vti! = "• "^ • Consequently, the number of

admissible combinations is reduced to four.

HencCi if x^^ x^, x^, and x^ are the roots of the equation

ax^ + 4bx* -h 6cx* -^ 4dx + e = 0,

then «i = -^ -b-^-y/at- H -Jt-\'-at-2H ■=E==\

GENERAL SOLUTION OF EQUATIONS

606

««

X,

where t

and

I = ae'-4:bd + Sc%

J = ace + 2 bed ^ ad^ — e^* — c',

The above solution is known as Euler^s.
In determinant form

H =

a

b

c

a

b

J =

b

c

d

b

c

c

d

e

596. Discussion of the Solution. Represent by a, p, y, 8 the

roots of the given biquadratic.
Then, by equation [6], we have

aa + ^ = 4" vw — "wv — "ww
afi -\- b = — Vw + Vv — "Vw
ay -^ b = — Vw — Vv + vt^;
aS + 5 = 4- Vu + Vv + -y/w

[7]

From [7], if 0i, 0„ 0, are the roots of the reducing cubic,

v=a%-H = ^(y + a-p-iy

» •

[8]

606 COLLEGE ALGEBRA

There axe six cases to be considered.

I. The four roots of the hiqvxxdratic all real and unequal.

In this case by equations [8] u, v, w are all real. Conse-
quently, $1, $29 6s are all real, and the cubics [4] and [5] fall
under the irreducible case (§ 694, 1).

II. Roots all complex and unequal.
By § 626 the roots must be of the forms

h + Tcij h — ki, I 4- mij I — mi,
and from equations [8]

w = - ~ (A; - my,

a^
v = — — (A; -h my,

a«

So that the roots of Euler's cubic are all real, two being
negative and one positive, and the cubics [4] and [6] again
fall under the irreducible case (§ 694, I).

III. Two roots real and two complex.
In each cubic two roots are complex and one is real.

IV. Two roots equal, the other two unequal.
Each of the cubics has a pair of equal roots.

V. Two pairs of equal roots.

Two roots of Euler's cubic vanish, the third being — 3 ^.

MM 2 jy

The roots of the reducing cubic are -^' -j' ^'

a a a

"VT. Three roots equal.

The roots of Euler's cubic are — H, — -fiT, — H; those of
the reducing cubic all vanish.

VII. All four roots equal.

All the roots of both cubics vanish and H =0,

GENERAL SOLUTION OF EQUATIONS 507

597. Discriminant. Comparing the reducing cubic with the

cubic

z* + 3Hz+G=:0,

we find the discriminant of the reducing cubic to be

The expression /* — 27 /^ is called the discriminant of the
biquadratic.

From the last section we obtain the following :

I. Discriminant of the reducing cubic negative ; that is^
Z? — 27 J^ positive.

The roots of the biquadratic are either all real or all
complex.

II. Discriminant of the reducing cubic vanishes ; that is^
/»- 27/^ = 0.

The roots of the biquadratic fall imder one of the following
cases:

(1) Two roots equal, the other two unequal.

(2) Two pairs of equal roots. In this case (? = 0, and

a^ a*

(3) Three roots equal. In this case 7=0 and / = 0.

(4) Four roots equal. In this case / = 0, J = 0, TT = 0.

III. Discriminant of the reducing cubic positive ; that is,
/8 _ 27 J^ negative.

Two of the roots of the biquadratic are real and two are
complex.

598. When the left member of a biquadratic is the product
of two quadratic factors with rational coefficientSi the equa-
tion can be readily solved as follows :

508 COLLEGE ALGEBRA

Solve the equation

a;* - 12a;» + 12 x^ + 176 « - 96 = 0.

Here, a = 1, 6 = — 3 ; put z = » — 3.

Then, 7^ - 42z« + 32z + 297 = 0.

Compare this with

(«a +P2 + g) («* -1)2 + rt = 0,
and we find ^ + ff— 1>^ = -42,

gg' = 297.

Eliminating g and g^, p is given by

j)8 - 84p* + 676 p« - 1024 = 0,

of which two roots are found to be ± 2.

Take p = 2, then g' = — 11, g = — 27, and the equation in ip Is

(a;a + 2« - 27)(«a - 2« - 11) = 0.

From this « = - 1 ± 2 V7, or 1 ± 2 VS.

Since x = jk + 3, we find the four yaliiee of x to be

2 + 2V7; 2-2V7; 4 + 2V3; 4-2V8.

In a similar manner, we can solve any biqiiadratic when the eobic in
p3 has; a commensurable root.

Exeroise 96 -

Find the four roots of : =..

.\

1. aj*-12a^ + 50a;«-84a; + 49 = a ,, : . vj

' 2. x* - 17 a;» - 20x - 6 = 0.

3. a;*-8aj» + 20aj2-16a;-21.«5a ;c:

4. ar<-lla;« + 46aj«-117a-h46i=0.

6, a:*-7x»-60x*-f221a-169«?:0, -

6. Show that the biquadratie can be soIt^ iQr. qaftdiiD^ttofi

if G^ = 0. * : t-

GENERAL SOLUTION OF EQUATIONS 509

7. Show that the two biquadratic equations

ax^ 4- 6 cx^ ± 4 c^aj + « = 0
have the same reducing cubic.

8. Solve the biquadratic for the two particular cases in
which 7 = 0 and J = 0.

9. Show that if H is positive the biquadratic has either
two or four complex roots.

10. Find the reducing cubic of

x* - 6 ax^ -f 8a Va» + ** + c» - 3 aJc +(12 Jc - 3a^= 0.

11. Show that / vanishes for the biquadratic

3 a(x - 2 a)* = 2 a(x - 3 a)*.

12. If the roots of a biquadratic are all real, and are in
harmonical progression, show that the roots of Euler^s cubic
are in arithmetical progression.

13. Form the equation whose roots are the squares of the
roots of ax* -f 3 hx^ -|- 3 ca; -f c^ = 0.

14. Form the equation whose roots are the cubes of the
roots of ox* 4- 3 hx^ + 3 ca; + c^ = 0.

15. Form the equation whose roots are the squares of the
roots of ax* -f 4 bx* + 6 ca:^ -f 4 c?x -f e = 0.

16. Form the equation whose roots are the cubes of the
roots of ax^ + 4 ftaj* + 6 ca;* 4- 4 c?x + e = 0.

17. Show that, if a^I = 12 H^ and aV = 8 H\ the biquad-
ratic has two distinct pairs of equal roots.

'^^ -^^

CHAPTER XXXin
COKPLSZ WBBkS

M9. Repmentation of ScaUt Ifomben. Let T.T be a atraight
line of unlimitftd length. Let 0 be a fixed point on tdiat line.

With any convenient unit of length measuie off along' tbe
line from 0 to the right and to the left a seziea of eqtial
distances.

o

Each of the points of division thus obtained reprcggata vi
integer (% 22). If the points to the right represeat posxtire
integers, those to the left represent negattve integexa.

The point 0 represents 0.

To represent a rational fraction -> where a and h

b

gers, b being positive and a either poaitiye or
divide the unit into b erpial parts, and then measore off a of
these piarts. The point obtained lies between two cxf tlie points
that represent integers.

We cannot find exact If/ the point that represents a ghren
incr>Tn mensurable number. We can, howevery ahnijs find
two fractions between which the given incommensiuaible
nurnljer lies; and the point that represents the iiiooni]ne&-
surable number lies between the points that lepiesent the
two fra/rtions.

Since the difference between the fractions can be made as
small as we please, the distance between the two points that
represent the fractions can be made as small as we please^ an^

510

COMPLEX NUMBERS 611

the position of the point that represents the given incommen-
surable number can therefore be determined to any desired
degree of accuracy.

600. The following example will illustrate the preceding
argument.

The odd-numbered convergents to the periodic continued
fraction

1 + 1 + 2'
numbering from 1 -f as the first convergent, are (§ 451)

I. ih it. Hh • ■ ; [1]

and the even-nimibered convergents are

Let K denote the complete value of the continued fraction,
and k^ denote the convergent numbered t, then (§ 449, Cor.),

and k^t-i > k2t> K

for all positive integral values of t

and k2t — ht^i > k^^ — K>0.

Now, Ajg, - k^t-i < (h - ^lYy if ^ > 1,

and k2-ki=l-^ = ^.

1

} [3]

~ ^ fC2t ^2t—i7 ^^ r ^ 1.

•12

•*• -JO* ^ -^ — ^2«— 1 ^ 0,

and ^t>^2t-K> 0.

612 COLLEGE ALGEBRA

Let M be any explicitly assigned constant number less than
K, so that jfiT — Jf > 0 ; then, since K^ M is constant and
not zero, an integer m can be found such that

Therefore, since k^m^x is the convergent numbered 2 m — 1,

and, therefore, K > kin-i > M,

Hence, if K> M> ^, there can be found in series [1] a
convergent which shall be greater than M but less than Ky
thus separating K from 3f, no matter how small K — M
may be.

Similarly, if N is an explicitly assigned number greater
than K, so that N — K>0, then an integer n can be found
such that

,\N>k^^>K.

Hence, if ^> N> K, there can be found in series [2] a
convergent which shall be less than N but greater than K^ thus
separating K from iV, no matter how small N — K may be.

Hence, there exists one number, and only one number, which
is greater than each and every convergent in the infinite series
[1] and is also less than each and every convergent in the
infinite series [2], namely, the number which is the complete
value of the periodic continued fraction.

Returning to the representation of numbers by points, the
points that represent the convergents J, \\y J|, f f f , • • • form
an endless sequence advancing from ^, and those that represent
the convergents |, f f , f j^, |^}, • • • form an endless sequence
retrograding from |. Ko poirt that lies in the first sequence

COMPLEX NUMBERS 518

coincides with a point in the second sequence or lies between
two points in it ; that is, the two seqicenoes lie wholly without
each other, as shown by [3]. Between the first sequence and
the second, but belonging to neither of them, there lies one
point, and only one, namely, the point that represents the com-
plete value of the periodic continued fraction. Every other
point between J and J either belongs to one or other of the
sequences, or lies between two points of one of them. There-
fore, the point that represents K, the complete value of the
periodic continued fraction, is completely determined by the
sequences as their sole point of section.

It is now easy to determine the number K. Since the
point K lies between the sequence

(6 19 71 266 a, \

[3 11 41 153 br J

and the sequence

7 26 97 362

^r 1

4 15 56 209

the point K^ lies between the sequence

and the sequence

{(0'(i)'(S)'(i)'-(l)'-}-

The first sequence may be written

{(-!)■ (-il=)-(-^)'-(^-^.)--}-M

The second sequence may be written

{{'^t)' (^+^«)' (^+^0' •••'(^+i)'-|f«]

(§§ 455 and 459)

614 COLLEGE ALGEBRA

Now, K + l = ^K^ ^r-1 > 3 6^

•'• ^r+i>3'"^i, and bi = 3.
2 2 1

-i<

^r+i' Q"^' 4x9*^
AlaOf rf^^ 1 = 4 (£^ — d^^i > 3 ef^

•'• ^r+i^3*'efi, and (£x = 4.
.-. c^^+l>4x3^

Let 3f be an explicitly assigned constant number less than
3 ; then, however small 3 — 3f may be, since it is greater than
0 and is constant, an integer m can be found such that

1

4x9"*

<3-3f,

and hence a point in the sequence [4] can be found that lies
between the points that represent M and 3.

Similarly, if N is an explicitly assigned constant number
greater than 3, then, however small iV^ — 3 may be, there can
be found in the sequence [5] a point that lies between the
points that represent 3 and N,

Hence, the point that represents 3 lies between the sequence
[4] and the sequence [5], and no other point lies between them;
that is, the point that represents 3 is their sole point of section.
But the point that represents K^ lies between the sequence
[4] and the sequence [5]. Hence, the point that represents
K* must be the point that represents 3, and therefore K* = 3.

It appears, then, that all scalar numbers may be represented
by points in the line XX'.

COMPLEX NUMBERS

516

Conversely, every point in the line XX' represents same
scalar number which may he integral or fractionalj eammenr
surable or incommensurable, positive or negative.

601. • The preceding method of representing numbers assumes
that the ordinal numbers, not the cardinal, are fundamental,
so that the phrase the point that represents 3 is short for the
phrase the point which is 3d in an endless sequence of points
numbered 1st, 2d, 3d, • • • ; and the phrase the point that repre-
sents I is short for the phrase the point which is 6th in a
finite sequence of points numbered |lst, 2d, 3d, 4th, 6th, 6th |,
saj the sequence Si^\ which is itself the first sequence element
in the endless sequence of sequences J5i^*, 5a% S^^\ '"\.

602. Representation of Orthotomic and Complex Numbers. An
orthotomic number (§ 206) cannot be represented by a point
on the line XX' (§ 599), since all points on that line represent
scalar numbers. We must therefore seek elsewhere for its
representative point.

Let the straight lines XX' and YV intersect at right angles
at O, and mark off OP,
OP', OP", and OP'", all
of the same length as in
the accompanying dia-
gram. A rotation coun-
ter-clockwise through a

right angle would convert

OP into OP', OP' into ^'
OP", OP" into OP'", and
OP'" into OP, so that
we may say that, taking
account of direction as

well as length,

OP' OP"

"P'

0

•P

If9

OP

III

OP

OP OP' OP"
Let i denote this common ratio.

OP

in

616

COLLEGE ALGEBRA

Then,

OP'

and

But

Also,

and

Finally,

OP^ '

OP" OP"
OP OP'

OP' .,
OP

OP"

OP ■" '^•

.-. t2=-l.

OP'" OP'"
OP OP"

OP" OP' .,

. — — . ^ t .

OP' OP '

OP'" OP"
OP OP

OP'"

op"~ *•

. • *8 •

•

OP OP
OP " OP'"

OP'" OP" OP' _ .^
' OP" OP' OP'^^'

.•.t* = H-l.

X'

■p

.ke account of direction as well as length|

we have

^ OP' = i'OP

= (V=T)OP,

OP" = ♦* . OP

= (-l)OP,

P and OP"' = i*'OP

' X =(-V^)OP.

Hence, if the point P
represents a scalar num-
ber a, the point P' zepre>
sents the orthotomio nmu-
ber a V— 1, and the point

P"' represents the negative orthotomic number — a V— 1.

Thus, exactly as all scalar numbers may be represented by

O

•P

m

COMPLEX NUMBERS 61T

points on the axis XX', so all orthotomio numbers may be
represented by points on the axis FF', which cuts the axis
XX' at right angles, or orthotamically.

Therefore, XX' is called the axis of scalars, and YT is called
the axis of orthotomics. The point 0 is called the origin.

The only point on both axes is 0, This agrees with the fact
that zero is the only number that may be considered either
scalar or orthotomic.

Again, a and ai are measured on different lines. This agrees
with the fact that a and ai are different in kind.

To determine a point that represents the complex number
a -^b V— 1, determine on the scalar axis a point A that
represents a, and on the orthotomic axis determine the point
B that represents h V— 1. Through the points A and B
draw straight lines parallel to the axes. These lines inter-
sect in a point P which represents the number a -|- 5 V— 1 in
the scale in which A represents a.

603. Vectors. When a straight line is given a definite
direction and a definite length it is called a vector. Thus,
the lines used to represent scalar numbers and those used to
represent orthotomic numbers are all vectors.

Vectors need not, however, be parallel to either of the axes;
they may have any direction.

The line AB, considered as a vector beginning at A and end-
ing at B, is in general written AB.

Two parallel vectors which have the same length and extend
in the same direction are said to be equal vectors.

604. Vector Addition. To add a I>

vector CD to a vector AB, we place / ^ ^

C on B, keeping CD parallel to its / f ^y^

original position, and draw AD. c I ^y^ /

AD is called the sum of the two / ^^ /

vectors. Al::^ /b

Then, AD — AB -h BD = AB + CD.

518

COLLEGE ALGEBRA

The addition here meant by the sign + is not addition of

numbers, but addition of vectors^
generally called geometric addition.
It is evidently identical with the
composition of forces.

From the dotted lines in the figure
and the known properties of a paral-
lelogram it is easily seen that

AD=CD + AB.

Consequently, vector addition is commutative (§ 36). It is
easily seen that it is also associative (§ 36).

605. Complex Numbers. A complex number in general
consists of a scalar part and an orthotomic part, and may be
written (§ 212) in the typical form x -|- yi, where x and y are
both scalar.

If we understand the sign -|- to indicate geometric addition,
we shall obtain the vector that represents x -\- yizs follows :

Lay off X on the axis of scalars from 0 \o M, From M draw
the vector MP to represent yL Then, the vector OP is the
geometric sum of the vectors OM and ilfP, and represents the
complex number x -f yL

Instead of the vector OP we
sometimes use the point P to
represent the complex num-
ber.

Thus, in the figure the vectors X'-\
OP, OQ, OB, OS or the pomts P,
Q, R, 8 respectively represent the
complex numbers 6 + 4 i, -6 + 5 i,
_5-3i, 3-6i.

In the complex number
x -f yif X and yi are represented by vectors. Now, vector
addition is commutative. Therefore, x -{- yi = yi -f as.

COMPLEX NUMBERS

619

This is also evident from the figure.

The expression x -\-t/iia the general expression for all num-
bers. This expression includes zero when a = 0 and y = 0;
includes all scalar numbers when y = 0 ; all orthotomic num-
bers when aj = 0 ; all complex numbers when x and y both
differ from 0.

606. Addition of Complex Numbers. Let x -f yi and x' -f t/'i
be two complex numbers. Their sum,

X -{- yi + x' -\- y'i,

may by the commutative law be written

x-\-x' -\-(y-\-y')i.

Let OA and OB be the representative vectors of » -f yt
and x' + y'i. Take A C equal to OB.

Then, OC = oJ + OB.

Draw the other lines in the figure.

Then, OH = OF -\- FH

= OF'\-OE

= X -{- x',
and HC = FA -{- KC

= FA -^EB

= yi-\- y'i.

.-. 0C = aj + aj'4-(y + y')*
= (x-\-yi)-{-(x'-{-y'i).
But OC = '0A -\-OB. o

Hence, the sum of the vectors of two coinplex numbers is the
vector of their S7im.

Since vector addition is commutative, it follows that the
addition of complex numbers is commutative.

The sum of two complex numbers is the geometric sum of
t^e sum of the scalar and the sum of the orthotomic parts of
the two numbers.

520

COLL£G£ ALGEBRA

Find the sum of 2 -f 3 1 and — 4 + •.

2 + 3 1 = OM, and - 4 + « = OJT.

If now we proceed from M^ the
ertkemi^ of OJf, in tlie dtrection
of OM' as far fts the abaolnto value

of 0M\ we reach the poinl JT^

Hence, OJT' = - 2 + 4 i, the Bum
of the two given complex numbers.
The same resolt !s reached if we
fiisifindthevalaeof2+(-4)=-2.
That is, if we count from O two
scalar units to A'\ and add to this siun 3 i + t = 4 1 ; that is, count four
orthotomic units from A" on the peipendkndar A^W,

607. Modnliis and Amplitude. Any complex number x-^yi

can be written in the form

X' A'

v^M^r

X

+

y

The expressions

')

and

- may be taken as

the sine and the cosine of some angle ^, since they satisfy the

equation

cos^<^ + sm^^ = 1.

If we put r = Vo;^ + y% the complex number may be written

r (cos ^ + t sin <^).

Since r = Vic^ + ?/*, the sign of r is indeterminate. We
shall, however, in this chapter take r always positive.

The positive number r is called the modulus, the angle ^
the amplitude, of the complex number x -f yi.

Let OP be the representative vector of aj -f yi. Since r is
the positive value of Vx^^JTp, it is evident that r is the
length of OP.

On the axis OX take OR equal in length to OP and an the
axis OY take O/J' also equal in length to OP, then ORmr and
Oi2' = W.

COMPLEX NUMBERS

521

Also, cosiZOP = -— — = -->

OH r

and

fAnROP

MP
OR'

ri

.', r(cos ROP H- I sin ROP)
= x -\' yi = r (cos <^ H- t sin ^).

Hence, the nmnerical measure
of the angle ROP «= ^ ± 2 wtt.

The above is easily seen to
hold true when x and t/ are one or both negative.

The modulus of a scalar number is its absolute value. The
amplitude of a scalar number is 0 if the number is positive,
180° if the number is negative.

The modulus of an orthotomic number ai is a. The
amplitude of this number is 90° if a is positive, 270° if a is
negative.

608. Since the sum of the lengths of two sides of a triangle
is greater than the length of the third side, it follows, from
§§ 604, 606, that, in general.

The modulus of the sum of two com^plex n/umber$ is less
than the sum of the moduli.

In one case, however, that in which the representative
vectors are collinear, the modulus of the sum is equal to the
sum of the moduli.

^s>

X.

609. MuHiplicatkm of a Complex Ifumber by ft Scalar Nttmber.

Let a; 4- y» be any complex number. If the representative
vector is multiplied by any scalar number a, it is easily seen
from a figure that the product is aa; + ayi.

Therefore,

a(x -\- yi) =: ax -\- ayi.

Hence, the multiplication of a complex iramber by a scsalar
number is distributive.

622

COLLEGE ALGEBRA

Multiply - 2 + 1 by 3.

Take 0A=z-2 on 0X\ and erect at A the
perpendicular AM = 1. Then, 0M= — 2+i,
Take OM three times, and the result is
0W= -6+3 1, the product of (-2+i) by 3.

610. Multiplication of a Complex ITom-
ber by an Orthotomic Number. We have
seen (§ 602) that multiplying a scalar number or an ortho-
tomic number by i turns that number through 90^ Let us
consider the effect of multiplying a complex number by i.
By the commutative^ associative, and distributive laws^

i X r (cos ^ + * sin <^) = r(i cos ^ — sin ^)

= r (— sin ^ -f t cos ^).

In Trigonometry it is shown that

— sin ^ = cos (90® + ^),
and cos ^ = sin (90® -f ^).

/. t X r(cos ^ + 1 gin ^) = r [cos (90® + ^) + » sin (90* +■ ^)].

Here, also, the effect of multiplying by i is to increase ^ to

w

^ + 90® ; that is, to turn the representative vector in the *
positive direction through an angle of 90°.

The effect of multiplying a complex number by an ortho-
tomic number ai is to turn the complex number through a
positive angle of 90®, and also to multiply the modulus by a.

611. Multiplication of a Complex Number by a Complex Number.
We come now to the general problem of the multiplication of
one complex number by another. This case includes all other
cases as particular cases.

Let 7*(cos ^ + i sin ^) and r'(cos ^' + 1 sin ^') be two com-
plex numbers.

By actual multiplication their product is

rr'[cos ^ cos ^' — sin ^ sin <^' + t(sin ^ cos ^' + oos ^ sin ^')3.

COMPLEX NUMBERS 523

In Trigonometry it is shown that

cos <^ cos ^' — sin <^ sin tj}' = cos (<^ -f <f>'),

and sin ^ cos ^' -f cos ^ sin <^' = sin (^ -|- tj}').

.*. r (cos ^ -f i sin <^) x r'(cos <^' 4- i sin <^')

= rr' [cos (<l> + <^') 4- ^ sin (<^ + <^')].

Therefore, the modulus of the product of two complex num-
bers is the product of their moduli, and the amplitude of the
product is the sum of the amplitudes.

Hence, the effect of multiplying one complex number by
another is to multiply the modulus of the first by the modulus
of the second, and to turn the representative vector of the first
through the amplitude of the second,

612. Division of a Complex Number by a Complex Number.

r (cos <^ 4- i sin ^)

The quotient -f. 1, , . . L

^ r (cos <^' -f t sm ^')

becomes, when both terms are multiplied by cos ^' — t sin <^',

r [(cos <^ cos <^^ + sin <^ sin <^^) + i (sin <^ cos <^^ — cos <^ sin i^')]

r'(cos>'-f sin*<^')

In Trigonometry it is shown that

cos <^ cos ^' -f sin ff> sin ^' = cos (<^ — ^'),

sin ^ cos ^' — cos ^ sin ^' = sin (^ — ^'),

and cos^^' -h sin*^' = 1.

. r (cos <^ 4- 1 sin <^) r. .^ .rx . • • /^ .int

• • "77 Tf . ' • Tk = -[cos (<^ - <^') 4- t sm(<^ - ^')].

r (cos ^'4-* sin <^') r"- ^ ^^ ^^ ^^■'

Hence, the modulus of the quotient of two complex num-
bers is obtained by dividing the modulus of the dividend by
that of the divisor; and the amplitude of the quotient, by
subtracting the amplitude of the diyisor from that of the
dividend.

524 COLLEGE ALGEBRA

613. Powers of a Complex Number. Fl^om § 611 we obtain

for the case in which n is a positive integi&t

[r (cos ^ -f- * sin <^)]" = r"[cos (^ 4- ^ H to n terms)

-f t sin (<^ -f ^ + ' • • to n terms)]
= r^(cos mj} + i sin n^y

614. Roots of a Complex Number. From § f^lS, putting ^

for mj}, and r for r", we obtain

I Vrf cos — 4- tsin— J = r(cos ^ + tsin\|^);

or

[r (cos <^ H- i sin ^)]" = Vr f cos — + t sin J J

where by vr is meant the scalar positive value of the i^t.

The last expression gives apparently but one value fo^ the
nth root of a complex number. But we must remember ttiat
there are an unlimited number of angles which have a giv^
sine and cosine. Thus, as shown by Trigonometry, the angldli

<!>,<!> + 360°, <^ 4- 720**, . . ., ^4- A; (360**),

all have the same sine and the same cosine. We have, there-
fore, the following nth roots of r (cos <l> + i sin ^) :

^(coB^ + ism^); [1]

y n nj *• ^

nr-( ^4-360° , . . ^4-360A. ^w.

V r ( cos h I sin I ; [2]

v-/ ^+(n-l)360^^ . . ^4-(n-l)860^\, . .

V r ( cos - — ^ ' h i sin - — ^^ ^ 1 5 \n\

A ^ n J ^ -^

COMPLEX J^UMBERS 626

In this series the [n H- l]th expression is the same as the
first; the [n -f 2]th the same as the second; and so on.

Therefore, there are but n different nth roots, those num-
bered [1] to [n].

From this section and the precedijig section we can obtain
an expression for

[r (cos <^ 4- * sin <jk)]'*j

7ft

where — is a rational fraction.
n

Find the twelve twelfth roots of 1.
The twelve twelfth roots of 1 are :

cos0°4-ism0° = l; [1]

cos 30° + i sin 30° = -A±l . [2]

2

cos 60° + i sin 60° = i^tl_? ; [3]

cos90° + isin90° = t; [4]

cos 330° + t sin 330° = —?^ -. [12]

615. Complex Exponents. The meaning of a complex expo-
nent is determined by subjecting it to the same operations as
a scalar exponent.

It follows that such an expression as a*"*""^, where a is a
scalar number and x-\-yi o, complex exponent, may be sim-
plified by resolving it into two factors, one of which is a scalar
number, and the other an orthotomic power of e (§ 434).

From the ordinary rules for exponents.

Put a^ = e^

Then, u = log^a*' = y log^a.

626 COLLEGE ALGEBRA

X^ . X^ . X*

Now, ^=^"*"*^"^ + [^ + fi + --- (§434)
Hence, ^^^'^^*"^Ty"'""f^"^li^"'""TF""'

By the Differential Calculus it is proved that when u is the
circular measure of an angle,

u^ . w* u^

cos« = l-J2+|^-j^ +
sin w = w— r7rH-nr — !-=■ +

[3-[6-[7

each series being an infinite series.

Therefore, e*** = cos u-\-i sin w,

and e*+ «» = e* (cos u '\-i sin «).

Also, a*+*'* = a* (cos m H- t sin u)

= g^[cos (y log^g) -f t sin (y logea)].

616. Trigonometric Solution of Cubic Equations. In the irre-
ducible case (§ 594, I) the numerical values of the roots of a
cubic equation may be found by the trigonometric tables. We
hav^ (§ 594, III)

«^ + ^ = (, 2 j-^V 2 j'

In the case to be considered G^ + 4/r' is negative (§ 694,1).

^ ,. . :^ Vg« 4- 4 iJ" .„ . ,
Put — — = -R cos ^, and ;r = %R sin ^

.'. co8^=-^, and sm <^ = ^-^^

Now, by Trigonometry, cos* <f> -\- sin* ^ = 1.

COMPLEX NUMBERS 627

. G^ G?^ 4- 4 g'

Then, R^^^-Hy,

and J? = (- fi^)^

By § 614,
ax -\-b =(— H)^ [(cos <l> -\-i sin <^)* -f- (cos ^ — t sin <^)*].

The cube roots in the right member must be so taken that
their product is 1, since in § 593 wM = — H,
The three values oi ax -\-b are :

2(-fi^)*cosf ;

3
2(-^)*cos^| + 120°Y

2(-^)*cosr|4-24oA
<^ is given by the relation •

, . sin<^ V- (G^ -f 4 H^)

tan<^ = 7 = — ^-- ^•

cos</» G

Solve the equation «• — 6 « -f- 2 = 0.

Here, C? = 2, H = - 2, and (?2 + 4fl^« =- 28.

2 ^

log 7 = 0. 84510 n, f + 120° = 156° 64' 6". *

log tan 0 = 0.42255 n, ^

0 = 110° 42' 18". 1 + 2400 = 2760 54-6".

Then the three values of z are found by logarithms to be
« = 2V2cos 36° 54' 6"= 2.2618;
2 = 2 V2 cos 156° 54' 6" = - 2.6016 ;
« = 2 V2 cos 276° 64' 6" = 0.8399.

Check : - (2.2618 - 2.6016 + 0.3399) = 0 ;

- [2.2618 X ( - 2.6016) x 0.3399] =2. (§ 521)

628 COLLEGE ALGEBRA

Homer's Method is, however, to be preferred to the method
of this section.

617. We have in this chapter extended the term number to
include complex expressions of the form a -\-h V— 1. These
expressions are often called imaginary quantities, although
when we are considering them without attempting to give
them any arithmetical interpretation, there is nothing imagi-
nary about these so-called imaginaries. The collection of
symbols 3 -f- 4 V— 1 is, as far as symbols go, as real as the
collection 3 -f- 4 V2. It is only when we seek to obtain a
result arithmetically interpretable and arrive at a complex
expression that cannot be interpreted, that such expression
can be called in a strict sense imaginary ; but under similar
circumstances a fractional number or a negative number may
become imaginary, while on the other hand a complex number
may represent as real a solution as a positive integer repre-
sents. The following problems illustrate these statements.

(1) Two clocks begin striking at the same moment ; one of
the clocks strikes 6 strokes more than the other, and the num-
ber of strokes struck by one of them is double the square of
the number of strokes struck by the other. Find the number
of strokes struck by each clock.

(2) The temperatures indicated by two thermometers differ
by 6®, and the number of degrees in the temperature indicated
by one of the thermometers is double the square of the num-
ber of degrees in the temperature indicated by the other.
Find the temperature indicated by each.

(3) Two men start to walk from the same place at the
same moment ; at the end of an hour they are 6 miles apart,
and the number of miles one of them has walked is double the
square of the number of miles the other has walked. Find
the number of miles each has walked. ^

Each of these three problems yields the equations

y = 2 aj^ and y — x = ± 6.