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HIGHER ALGEBRA
BY
HERBERT E. HAWKES, Ph.D.
PROFESSOR or MATHEMATICS IN COLUMBIA UNIVERSITY
GINN AND COMPANY
UOSTON • NEW YORK • CHICAGO • LONDON
Al'LANTA • DALLAS • COLUMBUS • SAN FRANCISCO
COPYRIGHT, 1913, BY
HERBERT E. HAWKES
ALL RIGHTS RESERVED
51G.4
gfte gtliengum jgrega
GINN AND COMPANY- PRO-
PRIETORS • BOSTON • U.S.A.
IK;
PREFACE
This text is prepared to meet the needs of the student who will
continue his mathematics as far as the calculus, and is written in
the spirit of applied mathematics. This does not imply that algebra
for the engineer is a different subject from algebra for the college
man or for the secondary student who is prepared to take such a
course. In fact, the topics Avhich the engineer must emphasize, such
as numerical com})utations, checks, graphical methods, use of tables,
and the solution of specific problems, are among the most vital fea-
tures of the subject for any student. But important as these topics
are, they do not comprise the substance of algebra, which enables
it to serve as part of the foundation for future work. Rather they
furnish an atmosphere in which that foundation may be well and
intelligently laid.
The concise review contained in the first chapter covers the topics
which have direct bearing on the work which follows. 'No attempt
is made to repeat all of the definitions of elementary algebra. It is
assumed that the student retains a certain residue from his earlier
study of the subject.
The quadratic equation is treated with unusual care and thorough-
ness. This is done not only for the purpose of review, but because
a mastery of the theory of this equation is absolutely necessary for
effective work in analytical geometry and calculus. Furthermore,
a student who is well grounded in this particular is in a position
to appreciate the methods and results of the theory of the general
equation with a minimum of eii'ort.
The theory of equations forms the keystone of most courses in
higher algebra. The chapter on this subject is developed gradually,
and yet with pointed directness, in the hope that the processes
which students often perform in a perfunctory manner will take on
additional life and interest.
Throughout the text the attempt is made to anticipate the diffi-
culties of the student, and by the use of illustrative material to make
the book readable, incidentally reducing the labor of exposition ou
iii
iv HIGHER ALGEBRA
the part of the instructor. In this connection §§ 18, 19, 69, and 89
may be consulted as furnishing instances of the method of procedure.
The exercises are for the most part new, and serve not only to
illustrate the text but to test and develop the power of the student
at every turn.
The author is under obligation to Professors G. B. Pegram and
C. H. Burnside for exercises from their special fields of science.
Especial acknowledgment should be made to my colleague, Dr. H. W.
Reddick, who has prepared a large part of the collection of exercises,
and whose criticisms, both destructive and constructive, have been
invaluable throughout the preparation of the book.
Columbia University HERBERT E. HAWKES
New York City
I
CONTENTS
CHAPTKR PAGE
I. Introductory Revikw 1
II. Functions and their Graphs 23
III. Quadratic Equation.s .'.... 37
IV^. Inkqualitiks 04
V. Complex Numbers 69
VI. Theory of E(juations 87
^'II. I'kRMUTATIONS, COMIUXATIONS, AM) I'lCOlSA lil Ml V . . . 1-il
VIII. Dktkkmixaxts 151
IX. Partial Fractions 170
X. Logarithms 17.'5
XI. Infinite Series ' 185
TABLES 211
INDEX 219
HIGHER ALGEBRA
CHAPTER I
INTRODUCTORY REVIEW
1. Factoring. The process of factoring consists in iinding two or
more expressions whose product is ecpial to a given expression.
In the type forms which are considered in this section it is
assumed that all of the coefficients are integers. Unless the con-
trary is stated, only factors having integral coefficients are required.
A prime expression has no factors with integral coefficients, except
itself and 1.
Later in this text it will be necessary to find factors whose coefficients are
not integral, but irrational, or even complex numbers. In every case of this
Kind the nature of the jiroblem in hand will indicate the type of factors
desired.
The following suggestions will prove helpful in factoring :
I. Firat look for a monomial factor common to every tmii of the
(jiven expression. If one exists, separate the expression into its yreatest
monomial factor and the corresponding iJolynomial factor.
II. Tlien determine., from the form of the polynomial factor, tcith
which of the folloiviny types it should he classed, and use the method
of factoring applicable to that type.
III. Proceed again as in II ivith each polynomial factor obtained,
until the original expression has been separated into its prime factors.
TYPE FORMS
1. ax -\- ay -\- bx -\- by =: {a -\- b){x -{. y).
2. d^-\-ZabJrb^ = {a+ b){a -\- b).
3. x'Jrbx+c = {x + p){x+q),
where y> and q are two numbers whose sum is b and whose product is c.
2 HIGHER ALGEBRA
4 . axF' -\-bx^c.
To factor expressions of this type, find two numbers whose alge-
braic sum is b and whose product is a-c. Replace hx by two terms
in X whose respective coefficients are the numbers just found, and
factor by grouping terms.
Thus 6x2-13ic- 5 = 6x2-15x4- 2x- 5
= 3x(2x- 5) + (2x-5) = (3x + l)(2x-5).
5. d'-h'={a+h){a- h).
6. a^-\-ka''b'-\.b''.
This type can sometimes be reduced to type 5 by adding and
subtracting a multiple of a%^.
' la" - &" = (a- feXa"-! + a"-2&+ a"-3&2 _^ . . . + &"-i),
when n is odd. If n is even, a" — Z»" is the difference of two squares
(type 5). In all other cases, if w is a multiple of 3, apply one of the
special types a^^h^ = {a+ b) (a^ _ aft + b%
a^-b^ = {a- b) (a^ + ab-\- b^).
8. a2 + &2 _,_ ^ _,_ 2 o& + 2 ac + 2 6c = (a + & + c)2.
EXERCISES
Factor :
1. x''-l-x + 2lx-2 kl. 9. a^ - h\
2. ^2 _ 2 _^ 2 y — 1 Solution : a" - IP - (a^)^ - {b^
= (a3 - JP) (a6 + a%^ + If)
3. r^ — lOr — 24. = (a_?,)(a2 + cr6 + 62)(a6 + a363 + 66)
4. 338 2:2-52a-,- + 2«2_ ^j^ <>^- -\- 1^-\
5. (« - 1) (2 cc - 3) - 6. 11. .ri« - yi^
6. X - a;^ 12. rrb^ - icV' + "Y - ^'«'-
7. 32-2 z\ 13. //^ - .T^ + 4 « (a - Z-).
8. «" + //. 14. x'-ia- + 9f-b'-6xij-4:aL
15. 2 (a'^ - 1) + 7 (a^ - 1).
16. abx' + «%' - (f^' + b'^)x>/.
17. «- + 9 Z-^ + 25 c^ - G ob - 10 fre + 30 be.
INTRODUCTORY REVIEW 3
18. (,s.2 _ 4)'^ _ (.s- + 2)1 22. 51a-- + 113xy-14//.
19. x^-lx^ + Ux-i^. 23. 8 x-y - 6o xhf:: + 8 ifz".
20. a-« - 7 u'V + if. 24. 2 ./•'- - 128.
21. a;8 - 4 .//^ + 8. 25. (x - l)(x - 2){x - 3)(x - 4) - 24.
2. Simplification of fractions. To simplify an expression contain-
ing fractions, it must be reduced to a simple fraction in which the
numerator and the denominator have no common factor. In reducing
fi-actions to their simplest forms, one merely performs the indicated
operations as directly as possible.
As a general rule, fractions with related denominators should be
combined and the result reduced to its lowest terms before the
whole expression is written as a single fraction. It is very desirable
to be on the alert for opportunities to cancel factors from numerator
and denominator of a given fraction. For this purpose the processes
of factoring should be at ready command.
Simplify :
EXERCISES
i.l^i±l. 5.1 + — L_
1 + i + I'S J ^ 1
J. jV^iilVs.. 1 + 1^
3
3(3.4-1.0) . (1..3y^ - (1.2)'' g ^ 1
^' .027 • .1 ■
1+
o
1 1
4.1-33^- 1 + 1
'■ a^ _ ^2 ^ 2 (a + b) ^ b -a '
a — b ((- 4- Ir a .2b
8. TT-. -TT - 7-0 -„ 7 +
2{a + b) V-a? a-h a-\-h
9. 3 ? 10.
4 x' - 1
5 H : a-- —
(5 1
7 + - .'• +
X X -\-\
HIGHER ALGEBRA
X — 1 X X x"
2 ,11 11 a- -4
12. ^^; — 77^ r^ +
5 o;-^ - 10 a; + 5 25 a; - 25 25 cc' + 100
2 ra (??i. — n) aVj — r//>^
7??. — n — ■ ^ a — b ; -— -
13. " + '' . 14. ^" + '^>
171^ -\- iv' a -\- h €? -\- IP'
mn -{-TV' a — b a} — IP
3. Roots and radicals. In applied mathematics the square roots
or the cube roots of numbers are usually found from tables of square
roots or cube roots, like those on page 215, or by use of the slide
rule. In problems where great accuracy is desired, more extensive
tables may be used, the roots may be found by the use of logaritlims,
or, as a last resort, the root may actually be extracted by the rule
found in the more elementary books on algel^ra. This rule finds its
chief usefulness, however, in the extraction of roots of algebraic
expressions.
The relation between the radical and the exponential notation is
exjDressed by the formula
where a and b are assumed to be integers, and h is not zero.
The only exception to this assumption found in the present text appears in
the chapter on logarithms, where irrational exponents are used.
In a fractional exponent the numerator indicates the power to
which the number is to be raised, while the denominator gives the
index of the root which is to be extracted.
4. Fundamental laws of exponents. The laAvs of exponents may
be stated as follows :
I. Law of Multiplication,
II. Law of Division, x° ^ x^^= x^-^.
III. Law of Involution, or raising to a power,
{x^y = (^x^y = x"^.
In these three laws the lettei's a and h may have any real value. But the
only occasion which we shall have in this book to consider any exponents more
complicated than rational fractions will arise in the chapter on logarithms.
INTRODUCTORY REVIEW 5
An important special case arises under Law IT when a = h.
We then liave 1 = 0;"-=- s-" = a'"-" = x^
which defines the iiicaiiiiii,^ of a zero exponent. Expressed in words,
this means that any iiuinhcr raised to the zero power equals unity.
When a = 0, Law II detines the meaning of a negative exponent ;
namely, -^ = a-" * and gives rise to the following rule for getting rid
of negative exponents in any algebraic expression.
Ani/ factor of tlie numerator of a fraction may he taken frovi the
numerator and written as a factor of the denominator, and vice versa,
if the sign of the exponent of the factor be changed.
„ , 1 1 2x-2 2x-2 2-10* 2-8 IG
Thus 2-8 = — =-; — = = = = --5-
23 8' </l6-8 jg-f X2 X2 x^
It is necessary to observe carefully whether the expression affected
by a negative exponent is a factor of the numerator or denominator
before transferring it.
^, a-'^+b a I + ah , ., a-'^b b
Thus — = = — , while = —
c c ac c ae
In this text the symbol V« will be used with the single meaning
+ V«, not — V^. If both values of the square root of a number are
intended, both the jjIus and the minus sign will be written.
5. Rationalization. If tlic ])vo(lu('t of two irrational expressions
is rational, each is called a rationalizing factor of the other.
Thus \^2niultiplio(l by V2i,nvc.stlie ration;il pn)(hict2._IIence they are ration-
alizinj;- factors of each other. The numbers — \ 2 and V8 are also rationalizing
factors of V2. Similarly, a + Vb is a rationalizing factor of a — V6.
The })ro('ess of rationalizing the denominator of a fraction con-
sists in multiplying both terms of the fraction by a rationalizing
factor of the denominator, so that in the simplified result the new
denominator will be rational.
For purposes of computation it is convenient to rationalize the
denominator of a fraction, since we are then able to compute the
approximate value of the fraction much more rapidly.
The numerators of fractions are also often rationalized. In this
process we multiply both tei-ms of the fraction by a rationalizing
factor of the numerator.
6 HIGHER ALGEBRA
EXERCISES
1. Is V2+V3=V5? IsV2V3=V6?
2. Is Va- + 1? = a + ^» ? Is Vic + ?/ = Vx -|- Vy?
3. Is V^2 + « = « Vl + (t ? Is V*"^ + «a;- = X Vl + a ?
4. Is ^-'^ + 7/-^ = ^^,? Isa-2.^-^ = -^?
a;^ + 3/^^ ic'^y
Simplify the following :
5. V50-V32+V98. ^ -2+vT2 , 3-Vl62
8. ;^ 1 ^
6. 4/l6+V8-V^. "^ /;
Q « — -i" + V (« — cc)'' + 4 ax
7. 4V-y--V5+V20. 2
> c > a > ac N ac
11. V3-v'2+^^-VT.-V6.
2 a 2 a
13. Show that ^ (5 ± Vl09) satisfies the equation 3 x-"^ — 5 a- — 7 = 0,
if it be substituted for x.
14. Compute the value of a?^ — Sa-^ + .r + 1, if a; = 1 — V2.
V2
15. Show that a; = — 1 ± -— satisfies the equation
2x^-lx-2 = 0.
16. Show that each of the four numbers ±"^3 ±V2, when sub-
stituted for X, will satisfy the equation
a;*-10a:2 + l = 0.
17. Compute the value of
9a;(3a;-2)4-2,
18. Show that x^ + 13 x- - 112 a; -f- 98 = 0,
if a; = -7(l+V3).
INTRODUCTORY REVIEW
Simplify tlu; lollowing and express the result with positive
exj)oneiits :
-\4
19. (a;-2\/a;«V4)
20. {^r^2^^)-\
-i
21. {x'-'yJx^^') .
22. («/>-V)-(a«^;V2)*.
23.
24.
a--U^ a-%~'
(a a-\^ .
Change into an equivalent fraction with rational denominator:
25.
26
2 a
— h — -vU^ — 4:ac
27.
28.
- ^n' - 1
'a + Va"^ - 1
a;
+ Va;-^ — if , ' 'y/x + Vy + Va; + y
Change into an equivalent fraction with rational numerator :
V .r + ?/ + V a* — y
29.
_/,_(_ VZ»"^ - 4
«c
30.
2a
Va; -\- y — Va; —
y
31.
a, 4- 2 - V.r'-^ - 4
a; + 2 + Vaj-^ — 4
Find to three decimal places the value of :
32.
33.
34.
35.
36.
37.
Vl- V5
V7 + Vs
15 + 7 V3
9 + 5V3
3 3
3 _ 2 V3 3 + 2 V3
38.
|iV3-
1
V2
iV3 +
1
V2
16
39. ^4«"S + /,° V^wFi,
when rt = — 32, /> = — 8.
Aj2-' + "
I3V15-7V2I
I3V12 -7V2J
^K^+4
40. ^/%p^-^rs-^t,
when 2> = 8, '' = 3,
s = - 1, /■ = 9.
41.
V7 4.V5_V2
V7+V5+V2
8 HIGHER ALGEBEA
6. The Binomial Theorem. The formula for the expansion of any
positive integral power of the binomial expression a + ^ is as follows ;
(« + hy = o" + ^ a»-Vy + '^ ^^\ ~ "''^ a" -2^/2
J- X • Ji
From this expansion the following rule for writing down the suc-
cessive terms of a particular expansion may be deduced.
The first term is a" and the last is b".
The second term is na"~^b.
The exponents of a decrease h[/ 1 in each term after the first.
The exponents of b Increase by 1 in each term after the second.
The product of the coefftclent of any term and the exponent of a
in that term., divided by the exponent of b increased by 1, glees the
coefficient of the next term.
The sign of each term of the exjjansion is -\- if a and b are positive ;
the signs of the even-numbered terms are — if b only is negative.
Tlie number of tervis in the expansloyi is n -\~ 1.
This formula can be established in elementary algebra only when n
is a positive integer. But the same form of expansion, except for the
last term, is valid when the exponent is fractional or negative, pro-
vided b is numerically less than a. This condition is necessar^^ in
order that the resulting series, which has an infinite number of terms,
may have a meaning. With the restriction noted, we shall assume
the validity of expansion (1) for any rational value of n without
attempting to give a proof. A rigorous demonstration demands a
knowledge of the calculus.
EXERCISES
Expand Exercises 1-3 completely ; 4-8 to five terms :
1. {2x + y)\ 4. (H-.r)-2, |j.|<l.
6.
Note. By |x| is meant the nu- a; ^ -|- 3
(l X "v^)^' inerical value of x regarding the
sign as positive. Thus| — 2|= + 2. y (!li^)
Ix I is called the absolute value of x. \x I
(m;
2
„-S+-). 8.(1--
INTKODUCTORY REVIEW 9
9. Obtain the expansion (1 — §)- = -^\ — 5'^ — -jiji^g — •• •.
10. Show also that (1 - if = i|.5 Vl5.
11. Show from each of the above results that an approximate
value of (1 - i)^ is .1^79.
7. Ratio and proportion. Tlie ratio of one number to a second
numl)er is the fraction found by dividing the first number by the
secHud.
The ratio of d to h is denoted l)y a : b or by -.
b
The dividend in this im])lio<l division is called the antecedent ; the
divisor is called the consequent.
Four numbers, n, h, c, d, arc; in proportion when the ratio of the
first pair equals the ratio of the second pair.
Cl c
This is denoted by a : 6 = c : d or by — = - .
6 d
The letters a and d are called the extremes, h and c the means, of the
proportion.
If (I, h, c, d are in proportion, that is, if
a:h = c: d, (I)
then ad = he, (II)
h:a = d: r, (III)
a:c = h:d, ' (IV)
a-\-h:h = r + </ : d, (V)
a-/>:/, = r-d: d, (VI)
a +I>: a - b = r + d : c - d. (VII)
Equation (HI) is said to be derived irijm [}) \)\ inversion.
Equation (IV) is said to be derived from (I) by alternation.
Equation (V) is said to be derived from (I) by addition.
Equation (VI) is said to be derived from (I) by subtraction.
Equation (VII) is said to be derived from (I) by addition and
subtraction.
8. Variation. The numl)er ./• is said to vary directly as the num-
ber // when the ratio of x to y is constant. This we symbolize by
xaiy, or -=/t,
where /.• is a constant.
10 HIGHER ALGEBRA
The number x is said to vary inversely as the number y when x
varies directly as the reciprocal of y. Thus x varies inversely as y
when 1 X
X cc-i or - = xy = k,
y 1
y
where k as a constant.
The intensity of a light varies inversely as the square of the
distance of the light from the point of observation. If I represents
the intensity of the light and d its distance from the point of obser-
vation, we have -. ,
Icc—z, or — = W^ = k,
cr 1
where A; is a constant.
The number x is said to vary jointly as y and z when it varies
directly as the product of y and z. Thus x, varies jointly as y and z
when X
X Qc yz. or — = k,
' yz '
where A; is a constant.
EXERCISES
1. Prove that ii a : b — c : d = e :f, then
ka + Ic — 7716 a
kh + Id — mf 1) '
2. Prove that if « : ^ = c : r/, then
(l±jr^^c^^r_d^
ah cd
3. The surfaces of similar solids have the same ratio as the squares
of their corresponding dimensions, and their volumes have the same
ratio as the cubes of their corresponding dimensions. What is the
ratio of the surfaces of two cubes if the volume of one is twice that
of the other ? What is the ratio of the volumes of two spheres if the
surface of one is twice that of the other ?
4. \i X ccy, and a; = 6 when y = 10, find // Avhen x — 15.
5. If a; cc -> and cc = 4 when y = 100, find x when y = 10.
6. If a; oc yz, and x = S when y = i and z = 5, find x when y = 20
and z = 2.
INTRODUCTORY REVIEW 11
7. A iiuin () IVct tall is walking diri-ctly away from a lamp-post
10 feet high. Find the ratio of the k^ngth of his shadow on the
ground, to the distance of the further end of his sliadow from the
lamp-post. How long is the man's shadow when he is 20 feet from
the lamp-post '!
8. The safe load of a horizontal beam supported at both ends va^
ries jointly as the breadth and the square of the depth, and inversely
as the length between supports. If a 3 by 9 ineli beam 15 feet long,
standing on edge, safely supports a weight of 1800 ])Ounds, find the
safe load of a 2^ by 6 inch beam of the same material 8 feet long.
9. The weight of a liody above the surface of the earth varies
inversely as the S(]^uare of its distance from the center of the earth,
and its weight below the surface varies directly as its distance from
the center. A body weighs 100 pounds at the surface of the earth.
What would it weigh 1000 miles above the surface ? 1000 miles
below the surface ? (Radius of the earth = 4000 miles.)
10. A disk 1 foot in diameter held 1.2 feet from the eye just
obscures a ball whose center is 13 feet from the eye. If the ball is
moved away so that the distance of its center from the eye is 25 feet,
how far from the eye must the disk be held so that the ball is
just obscui-ed ?
11. If a-"^ : 2 = 1 : .z'^, what is the value of x ?
12. A and J> are G and 16 years old respectively. In how many
years will the ratio of their ages be 2:3?
13. The time required by a pendulum to make one vibration varies
as the square root of its length. If a pendulmu 100 centimeters
long vibrates once in 1 second, find the time of one vibration of a
pendulum 81 centimeters long. What is the length of a pendulum
which vibrates once in 2 seconds ?
14. The volume of a cylinder varies jointly as its altitude and
the square of its diameter. The diameter of two cylinders are in the
ratio 3 : 2, and the volume of the second is two fifths that of the
first. Find the ratio of their altitudes.
15. Kepler's third law of planetary motion states that the square,
of a planet's time of revolution varies as the cube of its mean distance
from the sun. The mean distances of the earth and ISIercury from
fhe sun are 93 and 36 millions of miles respectively. Find the time,
of Mercury's revolution.
12 HIGHER ALGEBRA
16. The electric resistance of a wire varies directly as its length
and inversely as the square of its diameter. Its weight varies jointly
as its length and the square of its diameter. What must be the length
and diameter of a wire which is to have double the resistance but
only two fifths the weight of a wire of the same material 100 feet
long and .02 inch in diameter ?
9. Arithmetical progression. An arithmetical progression is a
succession of terms in which each term after the first, minus the
preceding one, gives the same number.
This number is called the common difference and may be positive
or negative.
The formulas for the nth. term, t„, and for the sum of n terms, S„,
respectively, are as follows :
t„=a + in-l)d,
where a is the first term, n the number of terms, and d the common
difference.
10. Geometrical progression. A geometrical progression is a succes-
sion of terms in which each term after the first, divided by the pre-
ceding one, always gives the same number. This constant quotient
is called the ratio.
The formulas for the nth. term, #„, and for the sum of the first n
terms, S„, respectively, are as follows :
t„ = ar''-\
a— ar"
'n
S„= ,
1-r
where a is the first term, n the number of terms, and r the ratio.
When r is numerically less than 1, the successive terms of a geo-
metrical series become numerically less and less, and the sum of n
terms approaches a fixed number as a limit as n increases indefinitely.
This limit is called the sum of the infinite geometrical series, and is
given by the formula q
S» = •
1-r
This formula must never be used when r is greater than unity, for
in that case the corresponding series does not approach a limit.
INTRODUCTORY REVIEW 13
EXERCISES
1. Find the 10th term and the sum of the first 10 terms of the
progression 1, ^, 2, §,•••.
2. Find the (n — 2)d term and the sum of the first n — 1 terms
of the progression a, a -{- d, a + 2d, ■■•.
3. Find the 8th term and the sum of the first 8 terms of the
progression 2, 3, |, • • • .
4. Find the sum of the infinite series 3 — 1 -}- J — • • •.
5. (a) Find the sum of the infinite series
(b) Find the sum of the series 1 + i + | + | + • • • .
6. A body starting from rest falls 16 feet the first second, 48 the
next, 80 the next, and so on. How far does it fall during the 10th
second ? How far has it fallen at the end of the 10th second ?
7. Using the information given in the preceding exercise, deduce
a general formula for the distance that a body will fall in t seconds.
8. A man standing on a cliff wishes to determine his height above
its foot. He droi)s a stone and notices that it strikes the ground in
4 seconds. How high is the cliff ?
9. The first term of a geometrical progression is 225 and the
fourth term is 14§. Find the series and sum it to infinity.
10. Twelve potatoes are placed in line at distances 6, 12, 18, •••
feet from a basket. A boy, starting from the basket, picks up the
potatoes and carries them back one at a time to the basket. How
far must he run to complete the potato race ?
11. How far must a boy run in a potato race if there are n potatoes
at a distance d feet apart, the first being at a distance a feet from
the basket ?
12 . A chain letter is written, each person receiving the letter rewrit-
ing it and sending it to two others. If the first person sends out
two letters, how many letters will have been written after all the
tenth letters of the chain have been sent ?
13. A chain letter is written, each person in the chain sending out
a letters. How many letters will have been written after all the
nth letters of the chain have been sent ?
14 HIGHER ALGEBRA
11. Linear equations in one variable. The following definitions
may be found useful for reference.
An equation is a statement of equality between two equal numbers
or number symbols.
Equations are of two kinds — identities and equations of condition.
An arithmetical or an algebraic identity is an equation in which,
if the indicated operations are performed, the two members become
precisely alike, term for term.
For example,
(a _ 6)2 = a2 - 2 a6 + &2, a • - - 6 = 0, and 2'^ - 3 • 2^ - 4 • 2 + 12 = 0,
are identities.
A literal identity is true for any value of the letters involved.
An equation which is true only for certain values of the letter or
letters involved is an equation of condition, or simply an equation.
For example, x — 2 = 0, (x — 1) (x + 3) = 0, and x"—l=y- are equations oi
condition.
A number or number symbol which being substituted for the
unknown letter in an equation changes the equation to an identity
is said to satisfy the equation.
After the substitution is made it is usually necessary to simplify the result
before the identity becomes apparent.
A root of an equation is any number or number symbol which
satisfies the equation.
We assume the following
Axiom. If equals he added to, siihtracted from, multiplied hy,
or divided hy equals, the i^esidts are equal.
As always, we exclude division by zero. In dividing each member of an
equation by an algebraic expression one must note for what values of the
letters the divisor vanishes and exclude these values from the discussion.
An equation is solved when its roots have been found. The
process of solution is the application of the foregoing axiom to the
equation in such a manner as to obtain the unknown alone in one
member of the equation.
Suppose the equation 4a:; — 7 = 17 is given, and the numerical
value of X is wanted. The validity of the equation is not affected
if 7 is added to each member ; that is, if we write 4 cc = 24, x
remains the same number as in the original equation. The same
INTIIODUCTOKY KEVIEW 15
may be said if we divide each side of the new equation by 4 and
obtain x — G. It is of tlie greatest importance to understand that
when we perform the operations mentioned in the axiom, we are
not getting anything new, but are expressing in a more available
form some symbol whose precise value did not appear clearly from
the original ('(nuitinii. In the illustration just used x is no more
truly equal to (5 after our process of solution than it was before; it
must be ecjual to 6 if this particular statement is an equation. We
have merely rewritten the equation 4x — 7 = 17 so that x appears
alone in one member. From this point of view the fact that the
number which one obtains as a result of solving an equation satisfies
the equation should not be surprising. If the work has been cor-
rectly done, it is impossible that it should be otherwise, for the
unknown is the same number at the end of the solution that it was
at the beginning.
We may say, then, that the root of an equation is obtained by
modifying the form of tlie original equation so as to display the
value of the letter, which for the time being is unknown, in terms
of what is numerically known. It cannot be too strongly insisted
that the solution of an equation consists in finding its roots, and
that the only property of a root of an equation which distinguishes
it from other numbers is that it satisfies its equation. If in the
hypothesis of a theorem a number is given as the root of a cer-
tain equation, we know that if the number be substituted for the
unknown, the resulting equation is an identity.
When we change an equation from one form to anotlicr in the
process of solving, it is assumed that x can have no value which
would reduce to zero the denominator of any fraction appearing in
the process.
EXAMPLES
1. Solve the equation
Solution. Here we assume that x cannot ocjual 0, fur tliis would cause the
denominators of two fractions to become 0.
Multiplying, we have
2x IOj 3j- 12j
16 HIGHEK ALGEBRA
Since x is not 0, we can divide it out of both the numerator and denominator oi
the second fraction and the fourth fraction. This gives
2x 10 3x
- + -+- + 3=0.
8x +
40 + 9x + 36 = 0.
17x=-76.
^ — 7 6
X -— jy.
^ +
^y-^{
v
w •
1 9 \ 2 2 8.8
Check. ^tLMl / 1 + _±^ \ + liAZLTlZ / 1 +
G _ G _ 0
TT 1 7 - "•
-w
2. Solve the equation
1 a + 1) 1 a — b
-\ = 7 +
a -\- b X a — h x
Solution. We assume that a + 6?i0, a — hjtO, x jtQ.
Clearing of fractions,
ox — 5X + a* — aW' + arh — W = ax + 6x + a^ — ab^ — a% + b^.
Transposing, — 2 6x = — 2 a^t + 2 6^.
X = a2 - 62.
In taking this step vpe have assumed that b ^Q.
Check. Substituting a? — b'^ for x in the original equation, it becomes
1 a+b 1 a—b
a + 6 ' a2-62 a-b ' a^-b~
which, under the assumption that a+6?^0, a — 67^0, reduces to
1111
+ : = r +
a+6 a—b a—b a+b
an identity. Hence a'^ — 62 is a root ii b ^ 0.
In the course of the foregoing solution, the value 6 = 0 was excluded. The
equation must now be considered for this special case. If 6 = 0, the original
equation becomes ^ 1 ^ n
- + - = - + -.
a X a X
which is an identity. Hence x is indeterminate.
Shorter solution of example 2 :
a+6 a—b 1 1
Transposing,
X
a-6
26
26
X
a2 - 62
X =
= a2 - 62,
a + 6
with the same restrictions on the letters as before.
INTRODUCTORY REVIEW 17
EXERCISES
Solve the following equations and in each ease verify the fact
that the result satisfies the equation :
1.
Ix
3
1
6^
9a:
4
2.
5a;-
6
_7
-l{-
--) =
10 /
15a; -22
6
3.
.5704
—
.20 a; =
= 19.651
.016 a;.
*• 2d "" ~ c
r"
3f//y<:- a2^>2 (2 a + iWa; „ 5a;
a + h {a + by a (a + by a
7. 1.2 a; -•^^^"•^^ = .4 a; + 8.9.
«-i(^-)-m=i(^-i)-
10. 2a; - 3 = 2.25a; - 5 - .4a; + 2.6.
11. .5555 = 5.55 a; + 333.33 - 44.4 a; - 30.91.
12. 0 = 2 a- - 3(5 + 3 a;) + 1(4 - a-) - ;J (3 a; - 16).
,o -.-. /3a;-l , 2a; + l\ ,^ /2x-5 , 7a'-l^
14- f{fV[Ul^ + 5)-10]+3}-8 = 0.
15. 7i a; -2i- [4^-^(3^-5 a-)] =18^.
7 13a- -24 10 37 13
"3 3 a; x 20 5 x
17 i(5a; + l)_3
•§(4x-l) 2'
18 ^--3-1.
1.3-3.r _ 1.8 - 8. r 5 r - .4
^ 2 ~ 1.2 ~ .3 ■
■18 HIGHER ALGEBRA
3 4^_5 5 Ta--3
20.
4 3 .r — 7 7 5 a'
21. 6(0.-6) = -^— ^(2 .•-11).
22. al> — (x — c)d = c (d + a")-
23. 7)1 {a -\-h — x) = n{a -\- h — x).
24. {a-\-c - x) {a -\- h) -\- {a - c + x) {a - h) = 2 «l
25. (ft - ^») (a — c) {a -\- x) + {a + ?y) {a + c) (a - x) = 0.
ax , ^^ I 2 ah __ (^/ 4- /;)-./•
Zb. ~7 I i ; 7 ^ ■
o a a -\- o ub
a — X b — X c — X ^
27. h ^ 1 = 3. •
a b c
28. UKiaa^ + 2)+2)4-2) + 2]=l.
29- ^[ia(K4^-i)-i)-i)-i]=i-
2 ;c - a 2 2 t .r + 2
30. - -
31.
32.
33.
f-^
3 3
f-x
1
a — -
X
1
re — -
1 a
1
1
a+ -
X 1
a
a
9
9
2
2
a. _ 51 ./■ - 15 X- 81 .r + 81
3.x -7 3(.x + l)^ ll.r + 3
2a; -9 2(.T + 3) 2a;--3ic-27'
a:: — III X — n
a'b — X , &'^c — X , c^a. — .r „
35. \ 1 = 0.
a b c
36. — — : = (a + x) : (h — x).
X X
37. {x — a -sfh) :{x — b Va) = Vft : Va.
ac (ni + nYx nx _ c ?>nx
38.
h (a — b) m brii b (a — b) in
39. — : (x — a) -\ (x — b) = 2 a(2 a i- b — x).
b a
INTRODUCTORY REVIEW 19
12. Linear equations in two variables. In the hist section it
was seen that any linear equation in one unknown may be solved ;
that is, its root may be found. The equation <ix + /> = 0, which is
the most general form for the linear equation in one variable, has
one and only one root, namely r = ? for this is the only number
which, when substituted for the unknown in such an ecj^uution, will
satisfy it.
If we consider a linear equation in two unknowns, as, for example,
2cc — 3y=(), it appears that tor any particular value of y the
equation becomes a linear equation in the single unknown x, and
therefore has one root. Hence this equation has not only a single
pair of values which satisfies it, but countless pairs.
For instance, if we let y = 1, the equation becomes 2 x — 3 = 0, which has
the root %. \i y = \, x has tlie value 9.
If we have two linear equations, each in two unknowns, each is
satisfied by countless pairs of nmnbers. The process of solving the
system of two eqiiations determines whether there is any pair of
values of the unknowns which satisfies both equations simultane-
ously. For this reason the two equations may be called a simultane-
ous system of equations. Usually such a system has one and only
one pair of common roots. Sometimes rio such values exist. In the
latter case the equations are called incompatible. If two equations
become identical when each member of one of them is multiplied
by some constant, they are called dependent equations. As a general
thing pairs of equations are independent and compatible.
EXAMPLES
1. Solve ^■^"■^^ = ^' ^^)
r3a; +
\2x-
5y = 9. (2)
Solution. Multiplying (1) by 5 and adding (2),
15x + 5j/ = 25
2 X - 5 .V = 9
17x =34
x= 2.
Substituting in (1), 0 + // = 5,
y=-l.
Check. Substitute in (1) and (2), 6-1 = 5, 4 + 5
20
HIGHER ALGEBRA
2. Solve
'ax -\- hij = («" — V^') c,
J)X — ay = 2 ahc.
Solution. Multiplying (1) by a, (2) by 6, and adding,
a^x + aby = (a^ — h^) ac
fe^x — aby = 2 ah'^c
(1)
(2)
(a2 + 62) a; = {a2 + ^2) wo
X = ac.
Substituting in (2), ajbc — ay = 2 af)^
— ay = abc,
y =—bc.
Check. Substituting the values found for x and y in (1) and (2), we obtain
a^c - b'^c = (a2 - 62) c,
abc + abc — 2 abc.
EXERCISES
Solve the following systems of equations and check the results
2.
3.
4.x-\-2y = 0,
3x — y = 15.
Sx-3y + 16 = 0,
5 y + 6 X = 17.
11.3;r +.125?/ = 1255,
10.3 x-y = 30.
7ic-10y=.l,
lla;-16y = .l.
23 a; + 15 y = 4i,
48 a; + 45^ = 18.
11. (^^ + 1)
12. {x - 2)
13. {x - 5)
7 *
2i.^ = 3^y + 4,
2iz/ = 3i 0^-47.
-Ky + i) = i,
8.
10.
K-^ + i) + f(y-i)=9-
3.5a; + 2^y = 13 + 4i £c -3.5?/,
21 a; + .8 2/ = 22^ + .7 a- - 3^ ?/.
ic : ?/ = 3 : 4,
(a.-l):(y+2) = l:2.
(a, + 4):0/ + l)=2:l,
14.
15.
(.r + 2):(y-l) = 3:l.
(y + l):(a. + 7/)=3:4:5.
(.7. + l):(.T+y-3)=3:4:5.
(y + 9):(x + 2/ + 4) = l:2:3.
(,r + 7/-4):(2a; + y + l)=l:2,
(2.,.+y_9):(a; + 2y + 7)=3:4.
;r + 1 _ ?/+ 2 _ 2 (.r — //)
~3 4~" 5 ''
3(x-3)-4(y-3) = 12(2y-a.).
INTRODUCTORY REVIEW 21
3.7- + 2// 5x--h3//
16 ^^"^^3— = ^ + ^'
' 2a- + 3// , 4./-4-3// .
15 4 . ._ .3 „
= 4, 17 X = S,
•^ V .^ '/
^^- 3 8 ^^' 4
- + - = 3. inx--=^-.
X y y
Hint. Retain fractions. l.G 2.7
H + 1=0,
18.
X 10 ^„ a; 2/
6"^ V
^ + - = 2|.
^ , ax + hy = c.
21. -^
Tinx = ny.
ax + by = (ir -b-)e,
^"^ • (a ■i-b)x+ (a -h)y = (a' + b^ c.
(a^ _ J^ (X + y) = cr + ^-'^
''•'• (^2 _ V') (2 a; + 3 y) = 2 «- + .'A + U".
(x- + r):(y + /.) = (.,+/.):(o + r).
a-:y = (a«-i«):(a»-+^.»).
X y _
2
OA
a + i a — b
a+b'
a4>
^ 1 y _
2
a + Z» a — i
a — b
25.
^-^ + 1 = 0,
a; //
X y
a 1 b
a
26.
a — X b — 1/
■ b'
b
^ 2V./- + 5-3 V//-2 = 3,
29 — _
' 3Vx + 5-4 V2/-2 = 5.
V.r-3 V« + 3
a — X b — y a V x — 3 V // + 3
31. Show that the following equations are dependent :
4^a;-375y = 2.25,
12 a; -10//- 6 = 0.
How many pairs of values of x and ?/ satisfy both equations ?
Find two pairs of values of x and y that satisfy both equations.
22 HIGHER ALGEBRA
32. Show that the following equations are dependent and lind
two pairs of values of x and y that satisfy them ;
2.125 x + 81 =.25?/,
a; + y + 4 = 1^2^ 2/.
33. Show that the following equations are incompatible :
13ic = 8 -39?/,
6y/ = 19-2x.
How many pairs of values of x and y satisfy both equations ?
34. Show that the following equations are incompatible :
2.2x + ^y=2,
5(3:c + 4y-3) = 4./- +3y.
CHAPTER II
FUNCTIONS AND THEIR GRAPHS
13. Uniform motion. Suppose a man who is taking a long walk
finds that at the end of one hour lir has covered three miles, and
that at the end of each successive hour he has advanced three
additional miles on his way. We might represent the relation be-
tween the distance which he goes and the time it takes him to do it,
by means of a graph as follows : Mark oif on a horizontal line equal
segments, each one of which represents an hour.'
On a line at right angles to this first line mark
off equal segments, each one of which represents
a mile, the point which represents zero hours and
zero miles being at the intersection of tlie two
lines. These lines we call the time axis and the
distance axis respectively. To represent the fact
that the man has walked three miles during the
first hour, we make a dot just over the one-hour
point, and three distance units above it. If he
had gone only two miles in this first hour, we
would have made the dot only two units above
the time axis. Similarly, at the two-hour point
we make a dot six distance units above the time
axis, and so on for the succeeding hours. If the rate of walking was
the same during the entire time, at the end of the first half hour he
would have covered one and one-half miles, which coidd be repre-
sented by a dot over the mid-point of the first hour segment, one
and one-half distance units from the time axis. We could insert in
a similar manner as many other i)oints as we might desire. It is
evident that all of these dots lie on a straight line.
Now the relation between the distances of any point on this line
from tlie two axes represents the iclalion between the distance which
the man walks and the time it re<[uires. If we wish to determine
how far lie luid gone at the end, say, of one and a half hours, we
23
f
0
.23
/
f
6
o
/
4
/
3
Q j
~
1
1
1 T\
ue J
Vxis
0
1
2
3
24 HIGHER ALGEBRA
have only to observe how many units of distance above the time axis
the line is at the point midway between the one- and the two-hour
points. In fact, the figure represents graphically the law which tells us
how far the man will have walked at the end of any number of hours.
The graph is not the only means we have of representing this law.
We may use an equation for the same purpose. If we represent by s
the number of miles he walks in t hours, the relation which we
expressed by a graph above we may represent by the equation s = St.
By means of this equation we can find out how far the man has
walked in any number of hours, say, two and a half, by replacing the
letter t in the equation by this number, and computing the value of s.
At first sight it might appear that the graphical method of repre-
senting the foregoing problem is less satisfactory than the other
method, on account of the unavoidable inaccuracy in drawing the
lines of the figure. For instance, it would be impossible to tell from
the figure the distance covered in a certain time, correct to a foot, or
even to a rod. It is to be observed, however, that it is equally im-
possible to measure the rate of walking with perfect accuracy, and
although we say that the rate is three miles an hour, this is only
approximate. It is a principle of great importance in applied mathe-
matics that one cannot obtain by the use of formulas results which
are more accurate than the data from which the formulas are derived.
Consequently, in dealing with problems like the one just considered,
if the drawing is carefully done, results as accurate as the original
measurement of the man's rate of walking can be obtained from the
graph.
It is often convenient to use a different scale of measurement
on the two axes, but this affects at most the shape of the graph
obtained, and not the nature of the numerical relation which is
represented.
14. The notion of function. The word quantity denotes anything
which may be measured. Distance, weight, time, volume, surface,
pressure, force, are all quantities, since each is measurable in terms
of a suitable unit. When two quantities are so related to each other
that when the first is given the second is determined, the second is
said to be a function of the first.
It is unnecessary that there should be any causal relation between the
(}uantities ; the mere correspondence of values is sufficient to establish the
functional relation. For example, the temperature on a given day depends
FUNCTIONS AND THEIR GRAPHS 25
physically on the atmospheric conditions, the angle at which the rays of the
sun strike the eaitli, and various other conditions. It does not depend causally
on the time of day. Hut ncvertheles-s, since to each time of day there cor-
responds a certain temperature, we may properly say that the temperature is
a function of the time. In § 13 the distance which the man walks is a function
of the time it takes him to do it. In applied mathematics, wherever there is
motion or change or growth a functional relation exists.
It is a matter of importance to devise simple means of representing
these functions, so that tliey may be studied and further relations
discovered. The function mentioned in the last section was repre-
sented by two means — by a graph and by an equation. Each method
was effective in displaying the fact and nature of the relation be-
tween the distance and the time, but they did it in quite different
ways. The present text will concern itself with the study of these
two methods of representing functions.
In what follows, the equation or the graph will often be studied entirely
apart from any physical meaniui,^ which the letters or the lines may have. But
it slKJuld never be forgotten that x and y in any eijuation may be the measures
of physical quantities which it is desirable to determine.
An algebraic expression involving the letter a^ is a function of x
because, corresponding to the various values of x, one or more values
of the expression can be determined.
/ 2x + 3 x^
Thus 2x2-1-1, Vx — 2x*, and — are each functions of x.
X -f- 1
Expressions involving two or more letters are called functions
of those letters. By means of equations involving such algebraic
expressions, numerical relations between the letters are defined.
Thus the equation x = 2 if tells us that x and y are so related
that x always equals twice the square of y.
EXAMPLES
1. Two towns, .1 and B, are 12 miles apart. A man walks from
^ to i? at the rate of 2 miles an hour. Express the distance s which
he travels, as a function of the time t, and represent the function
graphically. Determine from the graph what time will be required
for him to reach B. If he travels at the rate of 4 miles an hour,
what time will be required for him to reach 7?? What is the rela-
tion between the angles which the lines representing the functions
in the two cases make with the time axis ?
26
HIGHER ALGEBRA
Solution. Draw a pair of axes, the time axis and the distance axis, at right
angles, and let 0, their point of intersection, represent the first point where
^ = 0 and s = 0. Since the man's first rate is 2 miles an hour, we make a dot,
P, at a distance of 2 units above the point on the time axis where t equals one
unit. Through P and O draw a straight line. This line is the graph of the
function s = 2 i.
Now we wish to find what value t has
when s = 12, that is, how long will be re-
quired for the man to walk 12 miles.
Through the point J5, 12 units up on the
distance axis, we draw a parallel to the
time axis. This line contains all points for
which s = 12. Let D be the point where
this line intersects the graph of the function
s = 2 i. Dropping a perpendicular from J)
to the time axis, we see that for the point
D, ^ = 6, that is, when s = 12, ^ = 6 ; that
is, 6 hours are required to walk a distance
of 12 miles.
Similarly, if the man's rate is 4 miles an
hour, we make a dot, Q, 4 units above the
point on the time axis where ^ = 1, and draw
a line through Q and 0. Let this line inter-
sect B and D at C Corresponding to the
point C, f = 3 ; that is, 3 hours are required
to go 12 miles at the second rate. The func-
tion in this case is s = 4 £.
We notice that if the rate of the man is increased, the graph of the function
becomes steeper. Doubling the rate did not double the angle FOR, since Z QOR
B
F
/""
12
/
/
11
/
/
10
/
/
9
.2
/
/
8
7
9^
1
/
1
/
/
6
S
/
/
5
Q,
/
/
i
/
7
3
A
/
O
//
p
1
^
Tin
10 Axis
u
IR 2
3 i
5
6
The ratio
is not equal to 2ZP0R, but the ratio — - equals twice the ratio
OR OR
is called the slope of the line OB. and — - is the slope of the line OC. The
OR OR
slope of the line is the rate at which the man is traveling. Doubling the rate,
then, doubles the slope of the line representing the function.
2. A man starts out to ride on a bicycle at the rate of 8 miles an
hour. After riding 2\ hours he stops for 1^ hours, then continues at
his former rate. Four hours after the first man starts, a second- man
leaves the same place on a motor cycle at the rate of 16 miles an hour.
How far must the second man ride to overtake the first ?
Graphical solution. Let 0 be the intersection of the time axis and the distance
axis ; that is, the point where ^ = 0 and s = 0. Draw a line through 0 whose slope
is 8. This line is the graph which represents the relation between s and t for
the first 2^ hours. Now for the next 1 1 hours s does not increase but remains the
FUNCTIONS AND TllKlil GRAPHS
27
same. This is denoted by a line UC panilk-l to the t axis and 1 .V units long.
The man now continues at his former rate. This is denoted by a line CF having
the same slope as the line OB.
The second man starts 4 hours
later ; that is, when i = 4, but since
he starts from the same place as the
first man, s = 0. This is denoted by
the point D. Throui;li T) draw a line
whose slope is 1(5. This line is the
graph of the function which repre-
sents the motion of the second man.
The two graphs intersect at a point E
for which s — 40. Hence the second
man overtakes the first after riding
40 miles.
It sl'.ould be noticed that the lines
OBCE and UE are not the paths of
the two men but the graphs which
represent the relation between the
distance traveled and the time for
the first and second men respectively. The point E is not the intersection of the
paths of the men, but the point on the two graphs where s has the same value.
Algebraic solution. Let x be the distance in miles which the second man must
ride in order to overtake the first. In 2^ hours, at 8 miles an liour, the first
man rides 20 miles. Then x — 20 = the distance the first man rides after the
X — 20
delay, and the time required to ride this distance was — - — hours. The total
X- 20
time of the first man is 2| + 1] + hours. The time of the second man is
-/'
40
>
/
35
//
30
.2
/
7
25
4)
B
c
/
/
20
'/I
(5
i
/
1
/
15
/
/
10
/
/
/
5
^
D
' Time Axis
0
I
2
3
4 5 U
1
X
8
Hence the equation
4 H hours. These times are equal.
, , x-20 . , X
X — 20 _ x^
8 ~ 16'
2 X - 40 = X,
x = 40.
EXERCISES
Solve the following exercises both graphically and algebraically :
1. Two men start at the same time to walk a distance of 15 miles,
the lirst at 3 miles an hour, 'the second at 2^ miles an hour. How
much sooner will the first arrive than the second?.
2. A man walking 'J miles an liour loaves a town .1. He is fol-
lowed by a second man wlio Iravi's .1 I hours later, walking 4 miles
an hour. How long nmst the second man walk to overtake the first?
28 HIGHEE ALGEBRA
3. A man starts out to walk at a uniform rate and finds that at
the end of 2 hours he has walked 7 miles. If he continues at the
same rate for 3 hours longer, how far will he have walked ?
4. A stone is dropped into a pond and sends out a series of
ripples. If the radius of the outer ripple increases at the rate of
5 feet a second, what is the length of the circumference of the outer
ripple at the end of 3 seconds ?
5. Two towns, A and B, are 44 miles apart. A man leaves A for
£ at 8 A.M., riding a bicycle at a uniform rate. At 9.30 an accident
detains him for 30 minutes at a point 12 miles from A, after which
he doubles his rate. At what time will he reach B ?
6. A trip of 90 miles was made in an automobile in 5 hours. The
first part of the trij) was made at a uniform rate of 15 miles an hour
and the last part at 20 miles an hour. How much of the distance
was run at the latter rate ?
7. A man walks 5 miles in 2^ hours, then 12 miles further in
3 hours. What uniform rate would he have taken to cover the same
distance in the same time ?
8. A man starts out to walk at the rate of 3 miles an hour, and
after walking for 1^ hours he rests half an hour and then con-
tinues walking at the same rate. Another man leaving the same
place 4 hours later on a bicycle, rides at the rate of 12 miles an
hour. How far must he ride to overtake the first man ?
9. Two piers, A and B, are on opj^osite sides of a lake 12 miles
wide. A boat leaves A, crossing the lake at the rate of 12 miles an
hour. Thirty minutes later another boat starts from B to A, making
18 miles an hour. How far from pier A will the boats pass each other ?
10. Two automobiles are running in the same direction around a
circular track. They make the circuit in 1 minute 30 seconds and
2 minutes 15 seconds respectively. If they start together, after how
many minutes will they be together again ?
11. A tank has two outlet pipes. By one it can be emptied in 12
minutes and by the other it can be emptied in 4 minutes. If both pipes
are opened, find the number of minutes required to empty the tank.
12. Two automobiles leave a certain place at the same time
running in opposite directions, the first at 16 miles per hour and
FUNCTIONS AND THEIR GRAPHS 29
the second at 28 miles per hour. After going a certain distance the
second turns around, continues at the same rate, and overtakes the
first an hour and a half after the start. How far does the second
car go before turning around ?
15. Dependent and independent variables. Consider the equation
which ex])resses the relation between the area of a circle and its
radius, A = tt/-^. Of the two variables A and /•, one, the radius, can
usually be measured and the corresponding value of A determined.
In this process, the variable r is antecedent to A. Its value is found
before that of A is known, and from it the area is computed. In this
case r is called the independent variable and .1 is called the dependent
variable.
The formula for the distance .s- which a body falls from rest in a
time ^ is s = 16 1^. If the time it takes a ball to fall from rest to the
bottom of a cliff is known, the distance which it falls can be found.
Here t is the independent variable and s is the dependent variable.
If, however, we wish to find the radius of a circle whose area is
known, then it is the variable ^1 which is independent, and from it
the corresponding dependent r is found. Similarly, by the use of the
formula s = 16 t", the time which it takes a body to fall any given
distance may be computed, since t = -—- ■ In this case s is the
independent variable and t is dependent.
In general, when an expression is given involving two variables,
say, X and y, one of these is more naturally looked upon as the one
to which values are first assigned, and from which the values of the
other are determined. That one is the independent variable.
In the equations x = 2y^ — Gy + S and x = 9\^ + 4y,y is the independent
variable. In y = 2x* — 6x- + 2, x is independent.
It often occurs, however, that when the equation is not solved for
either variable in terms of the other, there is no reason for consider-
ing one as dependent rather than the other. In that case we decide
arbitrarily which one we will consider as independent. When we
have decided which we shall so consider, it is often desirable to
solve the equation for the dependent variable.
In the expression x- + y- = 4 we may equally well consider either x or y as
independent. If x is selected as the independent variable, we solve for y,
obtaining y = ± V4 — x-. This enables us to find the values of y from those of
X more readily than we could from the first form.
30 HIGHER ALGEBRA
EXERCISES
1. Express the volume of a sphere as a function of its radius. In
this functional relation, which variable is regarded as independent ?
2. Express the radius of a sphere as a function of its volume. In
this functional relation, which variable is regarded as independent ?
3. Using the relations of exercises 1 and 2, find (a) the volume
of a sphere whose radius is 3 feet, (b) the radius of a sphere whose
volume is 288 tt cubic feet.
4. A rectangle has one side 2 feet longer than the other. Express
the functional relation between the area of the rectangle and the
length of its shorter side. What is the area of the rectangle if its
shorter side is 7 feet ? In this relation which variable is regarded
as independent ?
5. Express the length of the shorter side of the rectangle in
exercise 4 in terms of the area. Which variable is regarded as
independent? Find the shorter side if the area of the rectangle
is 4 square feet.
In finding pairs of values of ./• and ij which satisfy the following
equations, which variable in each case is naturally regarded as
independent ? Eind three pairs of values of x and y satisfying
each equation.
/•
6. « - 6 7/ + 4 y2 + ^3 = 0. 10. V.r// + 1
7. x' - 8ic3 + 2 y = 0. 3-2 _ ^ji
8. ./ + 2 y - ..^ + 1 = 0. ''• —^ - "^ - '-'^'^ = ^■
9. xy + xy- + a-/ = 12. 12. .r^" + 4 = / - 4 x\
16. Accelerated motion. The distance s of a body from the ground
t seconds after it has been thrown vertically upward with a velocity v^
is a function oit: s = vt — \^t'^*
To fix our ideas, suppose the velocity with which the body is
thrown to be 64 feet per second. Then the equation becomes
i' = 64 i^ - 16 1''. (1)
If we wish to find how far above the ground the object is 3 sec-
onds after the projection, we substitute 3 for f and compute the
* In tliis equation the resistance of the air is neglected. The number 16 is the
approximate value of a constant depending on the force of gravity.
FUNCTIONS AND THEIII GRAPHS
31
corresponding value of .s-. Tlie relation between .s and / may Vje
shown by means of a graph, if the time axis and the distance
axis are taken as in § I'A. If we assign the values 0, 1, 2, 8, 4, 5 to
t, and compute the value of s correspond-
ing to each, we obtain tlie following table :
t
0
1
2
3
4
5
.s
0
48
04
48
0
-80
rfl
70
60
50
•<
x
s.
^
\
5y
-1
40
30
20
10
Time A
xisl
0
1
2
:i
•I
5
\
\
\
'
We observe that when t = T) the value of s
is negative. This means that the object would
be below the starting ])oint at the end of
5 seconds if it were not stopped. It should
be kept in mind that this graph does not
picture the path of the body; it indicates
that for the first 2 seconds its distance from
the ground is increasing, and for the next
2 seconds it is decreasing, but the actual
motion is in the same vertical line.
If we wish to know when the object is
60 feet from the ground, we may find from
the graph the value of t when s equals 60.
The result which we obtain by this means
is only ajiproximate on account of the inaccuracies of the drawing.
If an exact value for t is desired, we can replace s by 60 in (1) and
solve the resulting quadratic.
In graphing functions where the physical law represented is not in the fore-
ground it is customary to use the variables x and y and to call the axes the
X axis and the Y axis respectively.
EXERCISES
1. A stone is thrown vertically upwai'd with an initial velocity
of 48 feet per second. Plot its height as a function of the time.
What is the height of the stone at the end of one second ? two
seconds ? Determine from the graph how high it will rise, and the
time required to reach the highest point. After how many seconds
will it strike the ground ?
2. Using one set of axes, plot the functional relation between the
perimeter and the radius of a circle and also the relation between the
area and the radius. Take the same unit distance along the vertical
32 HIGHER ALGEBRA
axis to represent in the first a unit of length and in the second a
square unit of area. What is indicated by the point of intersection
of these graphs ?
What do the graphs indicate as to the relative increase in the
nmnber of units of length and the number of units of area as the
radius of the circle increases ?
3. Answer the questions in exercise 2, using the surface and the
volume of a sphere instead of the perimeter and the area of a circle.
4. The velocity of a body dropped from rest is given by the
formula v" = 64 s, where s is the distance fallen. Represent this
law graphically. A ball is dropped from a height of 256 feet. How
far must it fall to attain a velocity of 32 feet per second ? With
what velocity will it strike the ground ?
5. If a body is projected downward with a velocity v^, the dis-
tance s fallen at the end of any time t is given by the formula
s = v^ + 16 t'^. Express this law graphically when the velocity v^ is
16 feet per second. How far will the body have fallen at the end
of 2 seconds ? How many seconds will it take to fall 60 feet ? In
how many seconds will the body reach the ground if projected from
a height of 117 feet ?
6. W'ork the preceding problem under the assumption that v^ = 0,
that is, that the body falls from rest.
17. Graphs of equations. The graph of an equation is the graph-
ical representation of the functional relation which is expressed by
the equation. For the purpose of graphing equations we agree :
I. To have at right angles to each other two lines : A''OA, called
the Xaxis; and Y^0\\ called the Faxis.
II. To call the point of intersection of the axes the origin.
III. To have a line of definite length for a unit of distance.
Then the number 2 will correspond to a distance of twice the unit, the num-
ber 4^ to a distance 4^ times the unit, etc.
IV. That the distance (measured parallel to the X axis) from the
Y axis to any point in the surface of the paper be called the x dis-
tance, or abscissa, of the point, and that the distance (measured
parallel to the Y axis) from the X axis to the point bg called the
y distance, or ordinate, of the point.
The vahies of the x distance and the y distance of a point are often called
the coordinates of the point.
FUNCTIONS AND THEIR GRAPHS 33
V. That the x distance of a point to the right of tlie Y axis be
represented by a positive number, and the x distance of a point to
the left by a negative number ; also that the y distance of a point
above the A' axis be represented by a positive number, and the y dis-
tance of a point below the X axis by a negative number. Briefly,
distances measured from the axes to the right or upward are positive,
to the left or downward, negative.
VI. That every point in the surface of the paper correspond to
a pair of numbers, one or both of which may be positive, negative,
integral, or fractional.
VII. That of a given pair of numbers locating a point the first be
the measure of the x distance and the second be the measure of the
y distance.
Thus the point (2, 3) is the point whose x distance is 2 and whose y dis-
tance is 3.
Plotting or graphing an equation in x and y consists in finding
the line or curve the coordinates of whose points satisfy the equation.
The procedure is expressed in the following
Rule. When y is alone on one side of the equation, set x equal
to convenient integers atid compute the corresponding values of y.
Arrange the results in tabidar form.
Take corresponding values of x and y as coordinates and plot the
various points.
Join the points in the order corresponding to increasing values of
X, making the entire plot a smooth curve.
When X is alone on one side of the equation, integral values of y may be
assumed and the corresponding values of x coniputod.
When the equation is not already solved for either x or y, either may be arbi-
trarily selected as the independent variable and the equation solved for the other.
The resulting equation is plotted as already explained (see example 3 below).
It should be noted that we obtain the same graph whichever variable we select
as independent. The choice should be made so tiiat the labor of solving for
the dependent variable is as light as possible. For example, in case of the equa-
tion x^ — 3y^ — 4y + 3 = 0, we should take y for the independent variable.
In 4 x2 _ 7 a; _ 2 ?/ = 0 we should select x.
Care should be taken to join the points in the proper order so that the resulting
curve pictures the variation of y when x increases through the values assumed for it.
Any convenient scale of units along the axes may be adopted. The scales
should be so chosen that the portion of the curve which shows considerable
curvature may be displayed in its relation to the axes and the origin.
34
HIGHER ALGEBRA
When there is any question regarding the position of the curve between two
integral values of x, an intermediate fractional value of x may be substituted,
the corresponding value of y found, and thus an additional point obtained to fix
the position of the curve in the vicinity in question.
We shall assume witiiout proof that the graph of a linear equation
in two variables is a straight line. Hence in constructing the graph
of such an equation we only need to locate two points whose coordi-
nates satisfy the equation and then to draw a straight line through
them. It is usually convenient to locate the two points where the line
cuts the axes. If, however, these two points are close together, the
direction of the line will not be accurately determined. Error can be
avoided by selecting two points at a considerable distance apart.
The graphical solution of a system of two equations in two varia-
bles consists in plotting the equations to the same scale and on the
same axes, and obtaining from the graph the values of x and ij at
each point of intersection.
Two straight lines can intersect in but one point. Hence but one
pair of values of x and y satisfies a system of two independent linear
equations in two variables. When the two linear equations are
incompatible their graphs are parallel lines.
1. Plot 3 a- -h 4 7/ = 12.
Y
,
\
\
\
\
/
\
\
\
\
/
\
/
\
/
1
p
V
1
\
1
(>
\\y
X
1
EXAMPLES
Solution. If we set
jr = 0, we obtain 2/ = 3.
If we set y = 0, we
obtain x = 4. That is,
the points (0, 3) and
(4, 0) lie on the line
and serve to deter-
mine it completely.
2. Plot .r2-4.r + 3 = ^.
Solution. In this equation if we set x = 0, 1,
2, 3, etc., we obtain 3, 0, —1, 0, etc. as cor-
responding values of y. Thus the points (0, 3),
(1, 0), (2, — 1), (3, 0), etc. are on the curve. These
points are joined in order by a smooth curve.
N
Y
3
\
s.
O
1
\
N,
s
X
0
1
2
3
4\
X
- 2
- 1
0
1
2
3
4
5
6
y
15
8
3
0
-1
0
3
8
15
FUNCTIONS AND THKIK GRAPHS
35
3. riot L'x- + 3/ = 9.
Solution. 3 2/2 = 0-2 x-.
2/2 = 3- §x2.
y = ± \V21 -dx-.
Assiiminpj various integral valiu's for x, we obtain the following table and
jilnt ; the values nf // arc Ltivcii to the nearest tenth.
J-
.'/
-3
-2
-1 .
0
1
2
3
imaginary
±i-v/3=±.G
±iV21=±1.5
±iv'27=±1.7
±iV-2i=±ii;
±i>/3=±.C
imaginary
In tills exaniple, when x is greater than 3 or less than — 3, i/ Is imaginary.
Thus none of the curve is found outside a strip bounded by the lines x = + 3
and X = — 3.
To find exactly where the curve
crosses the X axis, the equation may
be solved for x, and the value of x
corresponding to y = 0 found. 'J'lius
Uy = 0,x=±^ = ±2A. Tlie.se
points are included in the graph.
4. Solve graphically the system
Solution. Substituting 0 for x
and then 0 ft)r // in each equa-
tion, we obtain, for 2x — y + 6 = 0,
^
•>
Y
\
1
^
1
N
\
•2. -1
0
1
)
'
\
— •^
—4^
px-y-f 6 = 0,
la- + 2 // + 8 = 0.
X
0
-3
V
G
0
and for X + 2?/ + 8 = 0,
X
0
-8
V
-4
0
Y
t
/
1
/
6
^^i
1
^/
4
V
?
• >
>
^
- 1 / -■.' 1)
X
^
\/.i. 1 !
(-
i-
'/
\
<^
%
• >
<,
^
-4
/
^^"Iv.^
/
1
1 i^"
Then constructing the graph of
each equation as indicated in the
adjacent ligure, we olitain, for the
coordinates of the point of intersection of the two lines, x =— 4 and ?/ =— 2.
36 HIGHER ALGEBRA
EXERCISES
Plot the following equations :
1.2x-37/ = 6. S.7j=(x-2y. 15. 2.r2 + 2?/2 = l3.
2. Sx + 7 7/ +14. = 0. 9.ij = 5x — x^ 16. .ry = 4.
3- ^ 2^ - i a; = 2. 10. 1/ = 5x + X-. 17. xt/ = - 4.
4. 5.5 a; + 6 y = 66. 11. y = _ 5 x - a;^. 18. x^ + / = 8.
5. 4 a; - 3 y = 0. 12. ?/ = - 5x -\- x^ 19. x^- i/ = 8.
6. 3a: + 2y = 0. 13. 9 .x^ + 4 / = 36. 20. i/^ - x' = S.
7. y = a;--3a--4. 14. 9 a-^ - 4 ^/^ = 36. 21. xY = A.
Plot the following systems and solve them graphically :
C3x-j/ = i, r3.x + y = 0, ra-2 + y^ = 25,
l2a; + 4?/ = 19. 127/ -a- =7. l3y-4a: = 0.
25 rr-4.x = 17, ra,/ = 3,
CHAPTER III
QUADRATIC EQUATIONS
18. Solution by factoring. In order to solve most efficiently all
kinds of quadratic equations, it is necessary to have two methods
at command. The lirst method, that of factoring, is simpler to
apply, and may be employed for tlie solution of many equations of
higher degree. One should always observe whether an equation
may be solved in this way before attempting the method of the
next section.
The solution by fat'toring depends on the following
Principle. The product of ttvo or more factors is zero ivhen
and only ivhen one or more of the factors are zero.
This principle is merely the formulation of the familiar rules for
multiplication by zero. We know that if we multiply any number
whatever by zero, the product is zero. If one factor of a product is
zero, it makes no difference what nunrbers the other factors are ; the
product is zero. On the other hand, unless at least one of the factors
of a product is zero, the product does not vanish.
It must be remembered that infinity i.s not a number and is never properly
considered as such.
To illustrate the method of solution by factoring, consider the
equation ^ (^ _ 3) (^ _ 4) (x - 1) = 0. (1)
"We ask what must be the value of x in order that this equation
may be satisfied ; that is, what are the roots of this equation ? Is 5
a root of the equation ? It is not unless it satisfies the equation, and
the equation is not satisfied unless at least one of its factors equals
zero. But if we replace x by 5, the first factor becomes 5 ; the second,
2 ; the third, 1 ; and the fourth, 4 ; none of which is zero. Hence 5 is
not a root of (1).
87
38 HIGHER ALGEBRA
In seeking the roots of this equation we need only to consider the
numbers which make one of the factors equal to zero. Hence x must
be a number which will satisfy one of the four equations :
X = 0, cc — 4 = 0,
cc — 3 = 0, X — 1 = 0.
These are all linear equations whose roots are 0, 3, 4, 1 respectively.
In accordance with the principle given above, these are the only roots
of equation (1).
It is observed that by this method we have reduced the solution
of equation (1), which is of the foui-th degree, to the solution of a
number of linear equations. To reduce the solution of a given
equation to that of equations of lower degree is the essence of the
method of solution by factoring. We may state the rule for solving
an equation by factoring as follows :
Rule. Transpose all the terms to the left meynher of the equation.
Factor that memher into linear factors.
Set each factor which involves the unknown equal to zero and
solve the resulting equations.
19. Solution by formula. In order to obtain the formula which
we shall use, it is necessary to solve the general quadratic equation
ax^ + hx + c = 0, (Q)
where a, h, and c are real numbers, and where a =^ 0. This we do
as follows :
Transposing c, ax^ + bx — — c. (1)
he
Dividing by a, x^ -\ — x = — —•
^ -^ ' a a
Adding \-^) to both members to make the left member a perfect
square.
, h h^ c V - 4 «c 4- ^'
X^ + - X + — -; = h —— , =
ft "" ' 4 d^ a 4:a^ 4 a^
/ , bV b''-A ac
Jiixpressmg as a square, yx -\- ^r— ) = — ^— ^ — (J)
Extracting the square root.
QUADRATIC EQUATIONS 39
_ ft ± V&2 _ 4 ac . ^
Transposing, Jf = — (F)
The roots arc
^ /, _l_ V/r - 4 ru- _ - /, — V//- — 4 ufi
In equation ((^) tlic nnniber x appears in a form \vhi(-h gives us
no idea of its value in terms of a, h, and c. It is indeed unknown.
But each step of the solution brings us nearer to an equation in
which a; stands alone in one member, wliich is the object of the proc-
ess. The critical point in the pidcrdure is passing from (2) to (3).
Since the square root of any number or expression (not zero) has
not one but two values, the necessity of extracting the square root
in order to find the value of x carries with it the existence of two
roots of the quadratic equation. As in the ease of the linear equa-
tion, the process of solution does not change the value of a: ; it dis-
covers it. The value of x in (Q) is knotted up in the equation,
and we merely untangle the knots to display it in terms of known
constants.
The solution of quadratic equations ^\■lli(•ll involve fractions or
radicals often necessitates the operation of multiplication by an ex-
pression involving x, or that of raising both sides of the equation to
a power. Either operation may introduce into the equation roots
which it did not originally possess, and lead to values of x which do
not satisfy the original equation. Such results are called extraneous
and should never be retained as roots. A certain method of detect-
ing extraneous roots is to substitute in the original equation all the
values of x which have been obtained, and retain only those which
satisfy it.
To solve a quadratic equation in x by formula we proceed as
follows :
Rule. Write the equation in standard form (ff).
Suhstitute the coefficient of i\ the coefficient of x, and the constant
term for a, b, and c, respectiveh/, in (-^)-
If, in getting the equation into standard form, each member has
been multiplied by an expression involving the unknown, or has been
raised to a power, substitute in the original equation all the values
which have been obtained, and reject the extraneous roots.
40 HIGHER ALGEBRA
Even if neither of these operations has been employed, the substi-
tution in the original equation of the values found should be per-
formed in order to afford a check on the accuracy of the solution.
A check which is more convenient for many cases will be derived
in § 24.
In the following exercises the quadratic equations should always
be solved by factoring when possible.
EXAMPLES
Solve and check :
1. 5x^ + 4:x = 12.
Solution. Transposing, 5x2 + 4x — 12 = 0.
Factoring, (5x - 6) (x + 2) = 0.
Hence x must satisfy one of the equations
5x-6 = 0, x+2 = 0.
X = |, or X =— 2.
Check. 5(1)2 -H 4(1) = 12, 5(- 2)- + 4(- 2) = 12,
¥- + -¥- =12, 20 - 8 = 12.
-%^- = 12.
2. a-' + 2 ab (a^ + h^) = (« + hf x.
Solution. Transposing, x^ — (a2 + 2 a* + 5-)x + 2 ab{a^ + W-) - 0.
Factoring, [x - (a^ + b-)]{x — 2 ab) = 0.
X = a^ + 62^ or x = 2 ab.
Check. (a2 + 62)2 ^2ab (a'^ + b^) = {a + b)- (a2 + 62).
Dividing by a"^ + 62, a^ + 2 a6 + 62 = (a + 6)2.
Also (2 a6)2 + 2 a6 {a^ + 62) = {a + 6)2 . 2 ab.
Dividing by 2 ab, a^ + 2ab + 62 = (a + 6)2.
3. Vl3+3;+Vl3-a; = 6.
Solution. Transposing, Vl3 + x = 6 — Vl3 — x.
Squaring, 13 + x = 36 - 12 Vl3 - x + 13
Transposing and dividing by 2, x — 18 = — 6 Vl3 — z.
QUADRATIC EQUATIONS 41
Squaring, x^-S6x+ 324 = 30 (13 - x) = 408 - 30 x.
x2 = 144.
x = ±12.
Vl3 + 12 + Vl3- 12 = 0, V13-12 + V13 + 12 = 6,
5+1 = 6, 1 + 5 = 6.
Therefore 12 and — 12 are roots.
Va -\-Vx_ 2 Vx (x + af
* V« - Vo; Va-\-Va: o(x — a)
Solution. Rationalizing the denominators,
a + 2 Vox + X _ 2 Vaj — 2 x (x + a)2
a — X a — X a(x — a)
Now a — X cannot equal 0, for this would give zeros in the denominators.
Hence we can divide a — x out of each denominator ; then, multiplying through
by a, we have
a^ + 2 a Vox +ax = 2a Vox — 2 ax + x^ + 2 ax + a^,
X' — ax — 0,
X (x — a) = 0,
from which x = 0, or x = a.
But x = a has been excluded. Hence x = 0 is the only root.
Check. Substituting x = 0, we have 1 = 0+ 1.
5. 3x''-5x = l.
Solution. Writing in standard form,
3x2 _ 5x-l = 0.
Here 3 corresponds to a, — 5 to ft, and — 1 to c in the general quadratic
ax- + 6x + c = 0. Substituting these values in (F),
_ (_ 5) ± V25 _ 4 . 3 (- 1)
gives X = — ^ ^
^ 2.3
Check.
_ 5 + \/25 + 12 _ 6 + V37
~ 0 ~ 6
25 ± IOV37 + 37 , 5 ± V37 ,
5 = 1,
36 0
31 ± 5 V37- 25 T 5 V37 = 0,
31 - 25 = 0.
42 HIGHER ALGEBKA
6. 2h~x- = ]cx + 2.
Solution. Writing iu standard form,
2k^x^ - kx - 2 = Q.
Then a = 2 A;2, h=- k, and c = - 2.
Substituting these vahies in the formula (F),
^ ^ - (- A:) ± V(- ky - 4 ■ 2 A:^ (- 2)
2 . 2 A:2
k±\^+16k'^ k±kVr: 1 ± a/i7
4A:2 4A;2 4A;
18±2Vr7 l±Vl7 +
2.M^4^T=^^4^+2,
8 4
9 ± \^ = 9 ± Vl7.
7. V^: + l + V3.'+l-2 =
= 0.
x + 1.
(1)
Solution. Transposing,
V3x + 1 = 2- Vx + 1.
(2)
Squaring both members of (2),
3x + l = 4-4 v'x + 1 +
(3)
Transposing and collecting.
2x-4=-4a^x + 1.
(4)
Dividing (4) by 2,
X — 2 = — 2 \ X + 1 .
(5)
Squaring both members of (5), x^
— 4x + 4 = 4x + 4.
Transposing,
x2-8x = 0.
Factoring,
X (x - 8) = 0.
Therefore
X = 0 or X = 8.
Substituting 0 for x in (1),
1 + 1-2 = 0.
Therefore 0 is a root of (1).
Substituting 8 for x in (1), Vs + 1 + V24 + 1 - 2 = 0,
3+5-2 = 0,
or G = 0.
Therefore 8 is extraneous and 0 is the only root of (1).
Note. If both members of an equation are multiplied by an expres.sion con-
taining the variable, or are raised to a power, extraneous roots may be intro-
duced. When either of tliese operations enters into a solution, the substitution
of the results in the original equation is properly a part of the solution, as in
examples 3 and 7 above. When these operations are not used, the substitution
is merely a check on the accuracy of the work.
QUADRATIC EQUATIONS 43
EXERCISES
Solve the following equations :
1. a;2_4a;-21 = 0. 9. 12a-2 - 71 r - 6 = 0.
2. y2-10y + 24 = 0. 10. H j^ + 5a; = 56.
3. !!' + 10^^-24 = 0. 11. *)/ = 6y + 26.
4. if - 10 y - 24 = 0. 12. ./•- - .03 ./• + .018 = 0.
5. «2_j_io~4-24 = 0. 13. ;iw-+ 7 = 8a-.
6. a;^ - 2 x + 2 = 0. 14. .03 ./ - - 2.23 x + 1.1075 = 0.
7. 6 a-- - 7 a- + 2 = 0. 15. hcj-' - hex + adx = hd.
8. s"" - 10 .s + 18 = 0. 16. 4 ,i + <u'- = 2 x- + 2 a'x.
17. y^ -2ax + a- + lr = 0.
18. 14 a-- + 45.5 a- = - 36.26.
19. {x - a -{.h){x-a + c) = (a- If - x\
20. a\h-xf = b\a-xf.
21. (a; -6)- -(2 a: -5)2 = 6.
22. (2 a; - 17) (.r - 5) - (3 a; + 1) {x - 7)= 84.
23. vi^x- — //I (a — b)x — ab = 0.
24. a; + a = (./-^ - a- + 1) (a- + ^0-
25. (2a;-(/)- = ^-(2a;-o)+2i-.
26. (3a; - 2 a + bf + 2b{3x-2 a + b)= a- - b-.
). k\lz^ _ 1) = ^A ,t2 + .^ - ^
27. X + ~ = a -\ 30.
a- a
. 31.4 mnx + (nr - ir) (1 - a--) = 0.
28. a + x = - + -- 32^ (^ _ ly^ = „ (^2 _ ^^
29. c(/(l + a-) = (<•- + (/•-) a-. 33. (7 - 4 V3)ar + (2 - V3)a- = 2
b 2x-b
1 1 ''a-
35. = + = =— •
2 + Vl - .,• 2 - V4-a- y
Suggestion. Rationalize tlie denominators.
44 HIGHER ALGEBRA
1 + Vl - ic 1 - Vl - X 2
37. Vll - X + Va--2 = 3.
38. x-\/x - 2 + 2 Va; + 2 = Vcc« -f 8.
39. Va(a; — i) + V^'(a; — u) = x.
40. V(x -l){x- 2) + V(a" - 3) (ic - 4) = V2.
A— 1 ic — 4 X — I X — 6
42.
2 3 4 5
4— a; 5 — X 6 — x 1 — x
wa — X \a — X
JO V (; — X \ a — X I—
43. = Va;.
X a
44. 2 Va-- - 9 a- + 18 - Va;'^ -4a;-12 = a--6.
Suggestion. Factor the expressions under the radicals.
45. J-Z,' + "Lz} = 4 v^:^
Suggestion. "Write the numerators 2 — 3 — 4 and x — 4 — 1.
. 46. .fi!±i+.EEi = 2x(^.
\h -^X^ \h -X \h
20. Quadratic form. An equation is in quadratic form if it may-
be considered as a trinomial consisting of a constant term and two
terms involving the unknown (or an expression which may be con-
sidered as the unknown), the exponent of the unknown in one term
being twice that in the other.
Thus j;-8V^+13 = 0, x~t + x~t-3 = 0, a^x-S" - (a + 6)x-« + 62 = 0,
x2 — 2x — 3 — Vx2 — 2 X — 3 + 17 = 0 are all in quadratic form. In the last the
expression V x- — 2 x — 3 is taken as the unknown.
It is often convenient to replace by a single letter the lower power
of the variable or expression with respect to which the equation is
in quadratic form, and proceed as in the case of the ordinary quadratic
equation.
QUADRATIC EQUATIONS 45
EXAMPLES
1. Solve cr-8V^ + ir) = 0.
Solution. This is a quadratic in Vx and may be written
(Viy-SV^+ 15 = 0.
Factoring, ( Vx — 5) ( Vx — 3) = 0.
Vx = 5, Vx = 3.
The roots are x = 25, x = 9.
Check. 25 - 8 • 5 + 15 = 0 ; 9-8-3+15 = 0.
2. Solve (x''-xf + x''-x-Q = 0. (1)
Solution. This is a (juadratic in x^ — x.
Factoring, [{x' - x) - 2] [{x"- - x) + 3] = 0,
or x2 - X - 2 = 0, (2)
x2 - X + 3 = 0. (3)
Factoring (2), (x + 1) (x - 2) = 0.
Hence x=— 1, x = 2.
Applying the formula to (3),
1 ± Vl - 4 . 3 1 ± V-11
X = = •
2 2
The roots of (1) are x =— 1, 2,
Check. Substituting — 1 in (1),
(1 + 1)2 + (1 + 1) - 6 = 4 + 2 - 6 = 0.
Substituting 2 in (1),
(4 _ 2)2 + (4 - 2) - 6 = 4 + 2 - 6 = 0.
Substituting in (1),
[C-^p-^H-^y-
±V-il
6=9-3-6 = 0.*
* The product V^ ■ V^= Va (- 1) • Va (-1) = a(V^)^= - a, where a is any
positive number. The operations on complex munbers will be explained fully in
Chapter V.
46 HIGHER ALGEBKA
EXERCISES
Solve the following equations :
1. 2 .r + 7 V^ - 4 = 0. 14. 13 ;7'^ = .rl + 36.
2. xi + 2xi = l. 15. cc^ + 3.r-l-V2x' + 6x+l=:0.
16. x^ + 5x-10 = Va^~+5xT~2.
18. x'^ + -^ + x + - = i.
x^ x
3.
x-^ + x-^- = a.
4.
a:''_12a;2 + 27 = 0.
5.
.ri - x-i = 11.
6.
.T- + .-):•" ^ = a'' + «~-.
7.
.r + 3 VS^ = 50.
8.
7 .r^ - 3 *^ - 2 = 0.
9.
a; - 1 - Vif + 5 = 0,
o 5
19. 5,r + \\5x
^ 20. (2./---3.r+l)2 = 22a;2-33a;+l.
10. X + V.t- + 3 = 4 X - 1. V- - ^/
n. 3x-aV^^:=^ = 2(x + 2). 21- -V^^(;7f^"
12. x + 1215 = 49 V615 + x. V^-^2a-h 3 a - 6
22 ■ = -
13. 19xi = x^-21Cx ' a Vic
23. V.r+ ^= =— =+ ^
VZ* V.c Va
24. 8 (8 ,r - 5f + 5 (5 - 8 .r)« = 85.
25. {x^ + 2)'^ + ^ = 4 ;r- + 8 .
V .//"^ + 2
21. Number of roots. In § 19 the general quadratic equation
ax"^ -\- l)x -\- c = 0 was solved, and it was found that it has two roots.
Reference to this solution shows that the roots found are the only
ones possible, for none of the operations which we j)erformed in
the course of this solution affected the character of the result.
For instance, if x satisfies (Q), it must also satisfy the equation
ax^ -\- bx = — c, for this is obtained from (Q) by adding — c to each
member. We may follow through all of the equations which we
obtain in the course of the solution and see that any value of x
which satisfies (Q) must satisfy each of them. But we finally obtain
the equation , z-^; —
— b ± -w¥ — 4 ac
x = .
2 a
which tells us that x must have one of the two values which we
obtain by taking separately the signs in the numerator before the
QUADRATIC EQUATIONS 47
radical. Since this is the same equation whieli we had at first,
except for its form, and since in this etjuation x can have only
two values, the same is true for (Ci), and the general quadratic has
only two roots.
22. The factor theorem for the quadratic. Altliough this theorem
will be proved later for equations of higher degree, a demonstration
for the special case of the quadratic is included here.
Theorem. Jf x^ is a root of the equation
aa:' + /'.r + ^ = 0, ((?)
then x — x^ is a factor of its left member.
The hyiiothesis in tliis tiieoreni is that x^ is a root of ((^); that
is, the equation must be satisfied when x is replaced by x^. Hence
the hypothesis is equivalent to the statement that ax^ -\- bx^ -(- c = 0
(see § 11).
Consequently,
ax^ -\-hx -\- e = ax" + hx -\- c — («rf + hx^^ + c),
since ax^ + f'j-^ + e = 0.
But ax^ + bx -\- e — (ax^ + bx^ + c) = a (y:- — .'-i") + b (x — x^).
Taking out the common factor, we obtain
(x — x^) la(x + cCj) + b'].
Hence cc — a-j is a factor of the left member of (Q).
By means of this theorem we are able to write down a quadratic
equation if its roots are known.
EXAMPLE
Form the equation whose roots are 4 and — 2.
Solution. By the precedinnj theorem the factors of the left member of the
equation must be x — 4 and x + 2.
Hence the equation i.s {x — i) {x + 2) = 0, ov x- — 2 x — 8 — 0.
EXERCISES
Form tlio equation whose roots are the following :
1. - 3, 2. 5. _ 1, _ 4. 9. 1 + V^, 1 - V^.
2. VS, - Vs. 6. 0, 2. 10. V^, - V^.
3. 2V2, -V8. 7.1,1. 11.^/4,-^8.
4.1,2. 8. 2 + V3, 2-V3. 12. "^ ±^--^1.
48 HIGHER ALGEBRA
23. Reduced form of quadratic. If we multiply each member of
an equation by a factor which involves the unknown, we obtain a
new equation which has more roots than the original. The roots
gained by this process are the values which reduce to zero the
expression by which we multiply.
Multiplying x^ — 3 x + 2 = 0 by x — 4, we have (x — 4) (x^ — 3 x + 2) = 0,
which is satisfied not only by the roots of the original equation but by the
number 4 in addition.
Similarly, if we divide each member of an equation by a factor
which involves the unknown, we obtain an equation with a less
number of roots than the original. Here the roots which are lost are
the values which reduce to zero the expression by which we divide.
For example, if x^ — Sx^ + 6x = 0 be divided by x, the equation loses the
root x = 0.
We may, however, multipl}^ or divide an equation by a constant,
not zero, without affecting the number of its roots or the value of
the unknown.
If we multiply both members of the equation x^ — 3 x + 2 = 0 by 4, we have
4(x2 — 3x 4- 2) = 0. This equation cannot be satisfied unless one of the factors
of its left member becomes zero. The same values of x will make the expression
inside the parenthesis vanish, regardless of any other factor which may be present,
while the constant factor 4 is never zero. Hence the roots of the equation after
multiplication by 4, or by any other constant, are the same as they were before.
We may accordingly divide the equation (Q) by the constant a
without affecting the values of the roots of the equation.
h c
The equation cc^H — x -\ — = 0
a a
may be written in the form x^ -\- jJ^ -\- q = 0, (IV)
where » = - > and n = -'
a a
We shall call {R) the reduced form of (Q).
24. Relation between the roots and the coefficients. The equa-
tions (Q) and (R) have the same roots,
-Jjj^ ^y" - 4 ac , -b- -Vlr - 4 ac
X = and X = •
^ 2 a ^ J, a
Adding these roots, we obtain
- ^ '4- V^,2 _ 4 ac -b- ^li' - 4:ac _-2h _ h _
^i + ^'^~ 2a ^ 2a ~ 2a ~~ a~~^'
QUADRATIC EQUATIONS 49
Multiplying the roots we get
/, ^ V//--4^^A/- h - V//^ - 4 nA IP- -l?^-\ ac
^i''*'2"\ 2« /\ 2a / 4«^
These results we may state in the form of a
Theokp:m. The sum of the roots of a reduced quadratic equation
equals the coefficient of the term in x, with its sign changed.
The product of the roots of a reduced quadratic equation equals
the constant term.
When roots of a quadratic equation are given in somewhat com-
plicated form, it is simpler to form the equation by use of this
theorem than by means of the theorem of § 22.
The above theorem also serves as a convenient check for the
solution of a quadratic, especially when the roots are complicated.
EXAMPLES
1. Porm the equation whose roots are 3 -\- V5 and 3 — v5.
Solution. Let Xj = 3 + VB, Xj = 3 — VB.
— p = Xj + x.j = G, or j3 = — 6.
7 = Xj • x, = 4.
The equation is x^ — 6 x + 4 = 0.
2. In the equation x- — 2x + Ic = 0,
what must be the value of k in order
(a) that one root shall be double the other ?
(b) that the difference of the roots shall be half their sum ?
Solution, (a) Let r and 2 r represent the roots.
Then r + 2r = 3r = -p = 2, (1)
and r-2r = 2r^ = q = k. (2)
From (1), '"'=%'
Substituting in (2), * = g-
Check. Putting fc = ^ in the original equation, we have
9x2- 18x + 8 = 0,
(3x-4)(3x-2) = 0,
— 3» or X — ij,
and one root is double the other.
50 .HIGHER ALGEBRA
(b) Applying the formula, we have
2
The roots are Xj = 1 + Vl — A;, Xg = 1 — Vl — k.
Since the difference of the roots is half their sum, and the sum of the roots
is 2, we have ^ a:^ + J-q _ 2
or
1 4- Vl-k - (l - Vl-fc) = 1.
2Vl-A: = l,
1-A;=l, A:=|.
Check. Putting & = ^ in the original equation, we have
4x2- 8x + 3 = 0,
(2x-3)(2x-l) = 0,
X = ^^ or 1, and } - J = J (^ + i).
EXERCISES
1. Form the quadratic equation wliose roots are — 2+V6 and
- 2 - Ve.
2. Form the quadratic equation whose roots are 5 + 2 V— 1 and
5-2 V^.
1 + Vt
3. Form the quadratic equation whose roots are
4. Form the quadratic equation whose roots are
o
3 ± Vl3l
10
5. Find the value of the literal coefficients in the following
equations :
(a) a;^ -f ^a: — 9 = 0. One root is 3.
Hint. Since 3 is a root, it must satisfy the equation.
(b) a-^ + 4 a- + c = 0. One root is 2.
(c) ax^ + 3 a- — 4 = 0. One root is 2.
(d) x^ — bx — 6 = 0. One root is — 3.
(e) 2a'^ — 6a; — c = 0. One root is — 4.
(f ) x^ — 6 a; + 0 = 0. One root is double the other.
(g) a-^ + e = 0. The difference between the roots is 8.
(h) x^ — 5 x -{- c = 0. One root exceeds the other by 3.
(i) a;^ — 7 a; + 0 = 0. The difference between the roots is 6.
(j) x^ — 6 X + c — 0. The difference between the roots is 4.
QUADRATIC EQUATIONS 51
6. In the equation r/.r- — 3.r -f Jc = 0, what must be the value of
k in order Miat the product of the roots sliall be twice their sum ?
7. Form the equation wliose roots are the reciprocals of the roots
of the equation 2x^ — 5x -\-S = 0.
8. If ./'j and x^ are the roots of the equation ax'- + /u- -f- c = 0,
show that 11 J,
a-, x~ c
9. Find the condition that one of the roots of the equation
x^ + ]px + y = 0 is double the other.
10. I'ur what values of k is one of the roots of the equation
(A- - 4) x^ - (^1 k - 1) .r = 7 - 5 /• double the other ?
11. Find the condition that one of the roots of the equation
ax- + hx + r = 0 is the reciprocal of the other.
Hint. Let a;, be one root and — be the other.
a-i
12. For what values of k and / will one of the roots of the equation
kx'- -f- Ix + /.• = 0 be the reciprocal of the other ?
13. For what values of k will the difference of the roots of the
equation 5 a*" + 4 a- + k = 0 equal the sum of the squares of the roots ?
14. Find the equation whose roots are the reciprocals of the roots
of the equation ax^ -\- bx -{- r = 0.
15. Find the equation whose roots are double the roots of the
equation x- + j)x + y = 0.
16. Find the equation whose roots are n times the roots of the
equation x- + px -f y = 0.
17. Find the equation Avhose roots are the negatives of the roots
of the equation x- + ^^a* + y = 0.
18. Given the equation x" — 3.v + 5 = 0. What is the equation
whose roots are (a) the negatives of the roots of the given equation ?
(b) three times the roots of the given equation '.' (c) the reciprocals
of the roots of the given equation ?
19. Show that the condition that one root of ax" -\- hx + r = 0
shall be n times the other root is
, (n + If
Ir = ■ ar.
n
62 HIGHER ALGEBEA
20. From the result of the preceding exercise find the condition
(a) that the two roots of the equation ax'' + bx + c= Q shall be
equal ; (b) that one shall be 'twice the other ; (c) that one shall be
three times the other. Write an equation illustrating each case.
21. Find the equation whose roots are each less by 2 than the
roots of the equation a;^ — 5 a; + 4 = 0. Set the left member of each
equation equal to y and plot.
22. Find the equation whose roots are each less by k than the
roots of the equation cc" + j9a; + 2' = 0.
Hint. Let the roots of the given equation be Xj^ and x^. Then the roots of
the required equation will be x^ — fc and x^ — k.
23. Find the equation whose roots are each greater by 1 than the
roots of the equation cc^ — 2 .t — 3 = 0. Set the left member of each
equation equal to y and plot.
24. Find the equation one of whose roots is less by k than the
smaller root of the equation sr- + 2^^- + (7 = ^5 ^iid the other of whose
roots is greater by k than the larger root of the given equation.
25. Find the equation which has for one root a number 2 less
than the smaller root of the equation x'^ — x — 2 = 0, and for the
other root a number 2 greater than the larger root of the same equa-
tion. Set the left member of each equation equal to y and plot.
25. Classification of numbers. All the numbers of algebra are
in one or the other of two classes, real numbers and complex numbers.
The real number will be left undefined, since an accurate definition
involves questions too delicate to be considered here. Any number
which can express the measure of a distance is real, as, for instance,
2, I, V5, V3 - 7 V2, and tt.
Real numbers are of two kinds, rational and irrational.
A rational number is a positive or a negative integer, or a number
which may be expressed as the quotient of two such integers.
Any real number which is not rational is an irrational number.
A complex number is the indicated sum of a real number and a
pure imaginary, where by a pure imaginary number we mean the
indicated square root of a negative number.
26. Character of the roots of the quadratic. The determination
of the character of the roots of a quadratic equation consists in find-
ing to which of the foregoing classes of numbers the roots belong.
QUADRATIC EQUATIONS 63
Consider the equation (Q) and its roots,
2~a (^)
The expression Ir — 4 ac which appears under the radical sign is
called the discriminant of the equation. An inspection of its value is
sufficient to determine the character of the roots of the quadratic. No
formal proof is necessary to see that the following statements are true.
I. Wlien Ir — 4 ac is riegative, the roots are comj^lex numbers.
II. WJien Ir — 4 aa = 0, the roots are real and equal. In this
b
case X, = .r, =— — .
' - 2a
III. W7ien b" — 4: ac is positive, the roots are real and distinct.
IV. WJien b^ — 4:ac is positive and a perfect square, the roofs
are real, distinct, and rational.
The converses of these four cases are also true. For instance, (1)
can only be complex when the expression under the radical sign is
negative ; that is, when the discriminant is negative.
27. Parameters. We often meet quadratic equations whose coeffi-
cients are not numerical, but involve one or more letters. For
instance, the equation x- + 2x + k = 0 is of this type. Several of
the equations on pages 50-51 are of the same kind. The letter k
might conceivably have any value we choose to give it, but after
we have once assigned a value, it is a fixed constant. Before we
have decided what value to assign, it is indefinite. For each value
of k there is a perfectly definite quadratic equation. If, for ex-
ample, we give k the value — 3, we have the corresponding equa-
tion a" + 2 .r — 3 = 0, whose roots we can find. But the equation
x^ + 2x -\- k = 0 really represents an infinite number of numerical
equations corresponding to the infinite number of values which k
may take on. Some of these may have complex roots and others
may have real roots, and it is often necessary to select from this
infinite set of equations the one, or the few, whose roots have a
certain character. Symbols like this letter k are called parameters,
to distinguish them I'lom the letter x, which, we have called the
unknown or the variable. The whole set of equations which we
obtain by letting the parameter take on a set of values, we call a
family of equations.
54 HIGHER ALGEBRA
EXAMPLE
For what values of k are the roots of »-^ + A;(a? — 1) + 3 = 0
equal ? real ? complex ?
Solution. Writing the equation in tlie form (Q), we have
a;2 + A:x + 3 - fc = 0,
where a = 1, b = k, and c =: S — k.
The discriminant b" - i ac = k"^ - 4(3 - k) = k^ + 4:k -12 = (k + 6){k - 2).
The roots will be equal when (k + C^) {k — 2) = 0 ; that is, when A: =— 6 or 2.
The roots will be real when {k + 6) (^ — 2) > 0. In this case the factors k + 6
and k — 2 must be both positive or both negative ; that is, A; > 2 or k< — 6.
The roots will be complex when {k + 6) (fc — 2) < 0. Hence one of the factors
k + 6, k— 2, must be positive and the other negative. It appears that the
factors are both laositive when fc > 2, and both negative when k< — 6. But for
values of k between these numbers the first factor is positive, while the second
is negative. Hence the roots of the original equation are complex for values
of k such that — 6 < fc < 2.
The situation may be illustrated by representing the values of k on a line as
follows :
"3 "3
S 3
Real and (liatinct H Complex M Real and distinct
-(j 0 2
EXERCISES
1. Find, by use of the discriminant, the character of the roots of
the following equations :
(a) 2x' - 7.r + 3 = 0. (d) 2.^2 - 4a; + 3 = 0.
(b) 9 ic- - 4 ,r + -4 = 0. (e) 18 .r^ + 63 x + 40 = 0.
(c) x2 + 6 ./• -8 = 0. (f ) 3 ^- + 15 f + 19 = 0.
2. For what values of k are the roots of 9.r^ + (l + A-)a; + 4 = 0
equal ? real ? complex ?
3. For what values of n are the roots of Sn^x(x -{- 3) = n — 5
equal ? real ? complex ?
4. For what values of k are the roots of x^ -\- k (x -\- V) -{- 3 = 0
equal ? real ? complex ?
5. What can be said of the character of the roots of the equation
ax^ -\- hx -\- c = 0 if a and c have opposite signs ? Would changing
the sign of h affect the character of the roots ?
QUADRATIC EQUATIONS 55
28. The special quadratics. In tlic; general quadratic (Q) the
letters h and c were supposed to have any real values. Since zero is
included among these values, all our Avork has tacitly included the
cases where one or more of the coefficients vanish, provided zero does
not appear in the denominator of a fraction. If a — 0, the quadratic
equation degenerates into a linear equation. This ciisc will he con-
sidered in the next section.
Cask I. We first consider the case where c = 0, and the equation
reduces to the form 2 _|_ /■;,._ n
Factoring, we obtain x (ax -\- h) = 0.
Hence the roots are x = 0, and .r =
a
That is, one of the roots is zero.
Conversely, if cc = 0 is a root, x — 0 is a factor of the quadratic
(see § 22), and the equation can have no constant term.
This result we may express by
Theorem I. A quadratic equation has a root equal to zero when
and only ivhen its constant term is lacking.
Case II. In case i = 0 we have the special equation
ax^ -f r- = 0.
Writing the equation in the form
a
we obtain the roots x^= \l > .r, = — xj , which are equal
numerically but have opposite signs.
Conversely, when the roots of a quadratic equation are equal
numerically but have opposite signs, the equation has no linear term
in the unknown. For if we represent the roots by x^ and — a-,, the
corresponding factors are x — x^ and .7- + x^. The resulting eqxia-
tion is x^ — x^ = 0, which does not contain a linear term in x.
This affords
Theorem II. The roots of the general quadratic equation are
equal hut with opposite signs u'hen and only when h = 0.
Case III. If c = h = 0, we have the special case ax^ = 0, where
the roots are equal by Case II, and hence each is zero by Case I.
56 HIGHER ALGEBEA
EXERCISES
1. Prove Theorem I by considering the expressions for the roots
in terms of the coefiicients given in § 24.
2. (a) If the equation formed by setting a function of x equal to
zero has a zero root, what is the characteristic i^roperty of the graph
of the function ?
(b) If the equation formed by setting a quadratic function of x
equal to zero has roots equal but opj^osite in sign, what is the char-
acteristic property of the graph of the function ?
3. Determine k so that each of the following equations shall have
one root equal to zero :
(a) 3a;2-2x + 2A-2-2 = 0.
(b) a;2 - .r + 2 A-2 + 3 A- - 2 = 0.
(c) 5*2 - 3a- - A-2 - 12 A- - 5 = 0.
(d) {X - Tif + 3 (.r - 2 A;) = 0.
(e) (3a- + A- - 1)- - 2(3a; + A: - 1) + 1 = 0.
4. Determine k and m so that each of the following equations shall
have both roots equal to zero :
(a) 2x^ + ^kx + l mx — x -\- k -\- m. + 1 = 0.
(b) x^ + Skx + 4: mx = m — 5k + 253 x.
(c) 3 a-- + A; {x - 2) + mx + 1 + A-^ = 0.
(d) m {x^ — x + l) + kx = x — vr + 2.
(e) m^x{l + x) - (1 + 3 wr) x - m (2-3 x) + k = 0.
5. Determine k so that each of the following equations shall have
roots numerically equal but opposite in sign :
(a) X' — 2 k'-x = kx + 1.
(b) (9 a- + 5 k^ (x + A-) = 2 x.
(c) 2 k\x^ -\- X -\- 1) - 5 (kx + 3) + 2 .r = 0.
(d) x^ + k\k - l).r - 6(kx + 1)= 0.
(e) (a- + 1) (A-a; - 1) = (1 - a-) (1 + A-a-).
29. Degeneration of the quadratic equation. If in the equation
(Q) we set a = 0, while b and c are not zero, we no longer have a
quadratic, but a linear equation, which has but one root. If instead
of substituting 0 for a we let a take on smaller and smaller values,
we shall obtain a number of equations of the same family whose
QUADRATIC EQUATIONS 67
left members differ from the left members of the linear equation
1,x -f ^' = 0 by just as little as we please. In this way we can find
out what becomes of one of the roots of the quadratic when a
vanishes. The roots of (Q) are
], _f. V//- - 4 ac —h- ^/r - 4
ac
1 2 II 2 II
Since division by 0 is ruled out of algebra, we cannot replace a
by 0 in these formulas. We can only let a approach 0. liut even
then a-j approaches the form §, which is also meaningless. To avoid
this difficulty we rationalize the numerators of both cc^ and x^ by
multiplying both numerator and denominator by a rationalizing
factor of the numerator. We obtain
^1 =
(_ A + V//-^ _ 4 ac) {-h- V6- - 4 ac)
2 a i-b- -s/b^ - 4 ac)
IP- ^h--\-\ac 2 c
2a{-b--\/b'-lac) -b- -y/b- - 4
ac
^ ^{-b- V/>- - 4 ar) (- b + ^b- - 4 ac)
- 2 a {-b + -s/b- — 4 ac)
fp _ i;i _|_ 4 ac 2 c
2a{-b-i- Vi- - 4 ac) -b + V^»- - 4 ac
As a approaches 0, b''^ — 4 ac approaches b'-^, and x^ approaches
— or — 7" But since the denominator of x,^ becomes very
small as a approaches zero, x^ increases without limit, that is, becomes
infinite. Thus in the equation ax- + bx -f r = 0, when a is allowed
to approach 0, one of the roots of the quadratic approaches the root
of the linear equation bx -\- c = 0, while the other becomes infinite.
The graph must then approach a straight line as a limit as a
approaches 0. This is made clear from the following figure, which
represents the equations of the family
«cc^-?-2 = .v, (1)
2 •"
correspondiuLT to the values a = 1, I, ^\^, 3\j, 0, — jV-
58
HIGHER ALGEBRA
In the figure the curves represent the following equations
y-
5
^-2
2
X
2
2
I/-
X X ^
10-2-2 = ^-
50
X
X
2
X
y-
-t — y.
32 2
2/-
(I)
(H)
(III)
(IV)
(V)
(VI)
In a similar manner we can show that when in the equation
bx + c = 0, h approaches 0 as a limit, the root of the linear equation
becomes infinite (see
II, § 28).
30. Families of
curves. Equation (1)
of the preceding
section represents a
family of equations,
and the graph shows
six of the corre-
sponding family of
curves. If we may
judge by the four
curves (I)- (IV), which appear in the figure, all the curves of the
family are tangent to the same line. We might wish to know whether
any equations of the family have equal roots. This algebraic question
corresponds to the geometric question whether any of the curves
of the family are tangent to the A' axis. Apparently none of the
curves tangent to the line (V) on its upper side has this property, but
if we use the method of § 26, we find that if a = — ^l> equation (1)
o
X" X
II
l\
ji
H
r
1
I
"1/
\liv
\
V
\
\
/
y
\
\
\
^
\
/
/
\^
V
\
s
\
N
/
>
/
\
N
\\
\
/
/
I
>
r<
^\\
\
j
/
/
/
f
VI
^vV;;
^
^
<^
/
/
.A
/
X
<
y
/
^
/
^
/
■^
^
y\^
r^
^
^
■ — ,
1
^
^ — '
-~^
IV
Nk.
V
CK"-
32 2
- 2 = y is
has equal roots. The graph of the equation
denoted by (VI) in the figure. Since the constant term is not 0,
no member of this family has a root equal to 0 ; that is, none of the
curves passes through the origin.
QUADKATIC EQUATIONS 59
EXERCISES
y
1. Sliow tliat all the curves of the family ij = ax^ — - — 2, coii-
sidered in the preceding section, are tangent to the straiglit line
?/ = - ^ - 2 at the pohit (0, - 2).
2. Plot several curves of the family ij = <ix} -\- x -{- \. Discuss
the behavior of these curves as a approaches zero. Will any of the
curves go through the origin ? For what value of a will the equation
ax''^ -(- .r + 1 = 0 have equal roots ? What is the common tangent line
of the family ? What are the coordinates of the point of tangency ?
3. Draw several curves of the family y = ax" — 4 and discuss the
behavior of these curves as a varies.
4. Draw several curves of the family y = ax'- — 2./- + 2 and dis-
cuss the l)ehavior of these curves as a varies.
5. What value must A- approach so that one root of each of the
following equations may become infinite ?
(a) 2L-x--(ix--\-7x-k = 0. (c) (k + 2)(x^ + l) = 2x(x-l).
(b) (.,• + !) (/.■.'• + .'■-1) = 1. (d) Px^ = 3(2k~S)(x--x-l).
31. Graphical solution of the quadratic equation. Let
1/ = (fx- + kr + c, (1)
where, as usual, (t, h, and c represent real numbers and a is positive.
If we let X. take on various values, y will have corresponding values
and we may plot the equation as in § 17. A root of the quadratic
equation aj- + hx + r = 0 (2)
is a number which substituted for x satisfies the equation and there-
fore gives the value // = 0 in (1). Thus the points on the graph of
(1) Avhieh represent the real roots of equation (2) are the points
for which y = 0 ; that is, where the curve crosses the X axis. The
numerical values of these roots are the measures of the distances
along the A' axis from the origin to the points where the curve cuts
the axis. The existence of complex roots of (1) is determined by
the following
Theorem. If the graph of (1) has no point in common icith
the X axis, equation (2) has complex roots, and conversed i/.
Every equation of form (2) has two roots either real or complex
(§ 21). If the graph of (1) has no point in common with the A' axis,
60 HIGHER ALGEBRA
there is no real value of x for which y = 0, and consequently no real
root of (2). The roots must then be complex.
Converse/ 1/, if (2) has complex roots, there is no real value of x
which satisfies it, and which makes ?/ = 0 in (1). Thus the curve
has no point in common with the X axis.
32. Maxima and minima. Consider the equation
y = 2x' + 7x + 2. (1)
By substituting for x a very large positive or negative number,
say, X = ± 100, y is large positively. Thus for values of x far to the
right or left the curve lies far above the X axis, but for one value
of X we get only one value of y ; that is, there is only one point on
the curve above (or below) any specified point on the A' axis.
If, however, we assign to y a certain value, we can find the corre-
sponding values of x by the solution of a quadratic equation ; that
is, the curve has two points, whose abscissas are either real, coin-
cident, or complex, on the same horizontal line with any point on
the Y axis. In equation (1) let y — 2.
Then 2 = 2 x^ + 7 .r -f- 2,
or 2 x^ + 7x = 0.
The roots are x^ = — 3^, .r^ = 0.
Hence the points (— 3^, 2) and (0, 2) are on the curve (§ 17); that
is, if we go up two units on the Y axis, the curve is to be found
3^ units to the left and also again on the Faxis. If in (1) we let
?/= — 4, the corresponding values of x, namely,— 1|- and —2, are very
nearly equal to each other, which means that the curve meets a line
parallel to the A' axis and four units below it at points near together
We may now ask. Where is the lowest, or minimum, point of the
curve ? This lowest point certainly has as its value of y that number
to which correspond equal values of x. Hence we must determine _-:r
what vahie of y the equation (1), which we now write in the form
2x'' + 7x-\-{2-y)=0,
has equal roots. Comparing this equation with ax^ -{- bx -\- c = 0, we
liave 2 = a, 7 = h, 2 - y = c.
Thus the condition li^ — ^ ac = 0 becomes
49 -4-2(2- 2/)= 0,
49-16 33
QUADRATIC EQILVTIOXS
61
Substituting tliis value of ij in (1), we get — | as the correspond-
ing value of X. We may express the foregoing results in tabular
form, and draw the curve.
V-
X
0
2
-4
— 4'
^8
— .3 + or - 3.2 +
0 or - Z\
- 1 .\ or - 2
This gives a single value of // Inr wliidi the values of x are
equal ; hence the graph of (1) is a single festoon, as in the figure.
If we take the equation
ax?' + /'.'• -f- c = ?/,
where a is positive, we find in a similar manner
that the coordinates of the niiniiiium point of the
curve are ,., ,
y= — TT— ' (-)
4«
X =
2 a
(3)
1
Y
1
\
1
--
X
\
O
\
\
1
\
J
The results of this section enable us to deterniinc from the coeffi-
cients the value of ?/ for the lowest point of the curve, and hence to
show beyond question whether the equation has real or complex roots.
The ordinate of the lowest point is the least value that the function
ax^ -\- bx -\- c takes on for any value of the variable.
When a is negative, the form of the graph is the same as that
illustrated in tlie figure, but the curve is inverted, as in the case of
curve (VI), p. 58. When a is negative we find the highest or maxi-
mum point of the curve in the same way that we liave just found
the minimum point for the case where the coefficient of x^ is positive.
The coordinates of the maximum point of the curve are the same
as given in (2) and (3).
When X =— -^r-) the function ax'^ + h.r + r takes on its minimum
2a
value if a is positive, and its maximum value if a is negative. These
maximum and minimum values of the function are the values of y
given by tlio foi-niula
b- — 4 a/;
4a
62 HIGHER ALGEBRA
EXERCISES
Plot the following equations and determine the points where the
graphs cut the A' axis. Find in each case the lowest or highest point
according as the curve is concave upward or downward.
1. l/ = x''-6x-\-5. 6. x^ + 4 a- - 2 y = 0.
2. y; = .T^ -f- 4 .T + 4. 7. 9j = 2x — x\
3. y = a-2 - 6 .7- + 10. 8. i/ + x^ + 2x + 2 = 0.
^. y = 2x'-x-?,. 9. ic'=-4a; + 4 + 47/=0.
5. 7/ = 1 — X — 2 x-l IQ. X- — ^x — y =0.
11. Divide 10 into two parts such that their product shall be
a maximum.
12. Divide 10 into two parts such that the sum of their squares
shall be a minimum.
13. Divide 12 into two parts such that the product of half one
part by a third of the other part shall be a maximum.
14. Find the number of acres in the largest rectangular field that
can be inclosed by a mile of fence.
15. A window is to be made in the shape of a rectangle surmounted
by an equilateral triangle one of whose sides is the upper base of the
rectangle. The perimeter of the window is to be 22 feet. Find its width
and height in order that it may admit the maximum amount of light.
Solution. Let the base of the rectangle be 2 a; and its altitude be y. Then
the perimeter of the entire window is
2i/+ 6x = 22,
from which y = \\ — Zx.
The area of the window is 2 jy + VSx^,
or, substituting the value of y found above,
22x- 6x2 + V3x2.
The question is, For what value of x will this function take on its maximum
value ?
The coefiBcient of x^ in this quadratic function is v'3 — 6 and the coeflBcient
of X is 22. Substituting these values for a and h respectively in formula (3),
we have , /- / /- \
__ b _ 22 _ 11 V3 + 6_ ll(V3 + 6)
2 « 2 ( V3 - 6) V3 - 6 V3 + 6 ~ - 33 _
= ^(V3 + 0)= ^(1.7321 + 6) = 2.5774.
2x = 5.15 feet, the width of the window.
QUADRATIC EQUATION'S 63
The height of the window is
y + VSx = 11 - 3x + VSx
= 11 _ (;^ _ V3) z = 11 - (1.2679) (2.5774)
= 11 - 3.27 = 7.73 feet.
16. Solve the same problem for a window in the shape of a rec-
tangle and an isosceles right triangle wliosc hyjxjtiMuise is the upper
base of the rectangle, the lu'i'inicter of the window being 28 feet.
17. Solve the same i)roblem for a window in the shape of a rec-
tangle surmounted Ijy a trapezoid each of whose legs and upper base
are equal to half the upper base of the rectangle, the perimeter of
the window being 37.3 feet.
18. Solve the same problem for a window in the shape of a rec-
tangle surmounted by a semicircle, the perimeter of the window being
32 feet.
19. Find the dimensions of the rectangle of largest area that can
be inscribed in an isosceles triangle whose altitude is 20 and whose
base is 14, one side of the rectangle lying on the base of the triangle.
20. Find the dimensions of the rectangle of largest area that can
be inscribed in a right triangle whose legs are a and b, one angle of
the rectangle coinciding with the right angle of the triangle.
CHAPTER IV
INEQUALITIES
33. General theorem. We say that a is greater than h, ov a> h,
when a — ^ is positive. If a — h is negative, then a is less than h,
or a < h. As we distinguished between identities and equations of
condition in § 11, so in this discussion we observe that some state-
ments of inequality are true for any real value of the letters, while
others hold for particular values only. The former class may be
called unconditional, the latter conditional, inequalities.
Thus a^ > — 1 is true for any real value of a and is unconditional, while
X — 1 > 2 only when x is greater than 3 and is consequently conditional.
The two inequalities a > h, c > d, are said to have the same sense.
Similarly, a < b, c < d, have the same sense. The inequalities a > b,
c < d, have different senses.
Theorem. An^ real number may he added to or subtracted
from each member of an inequality without affecting its sense. Both
members of an inequality yyiay be multiplied or divided by any posi-
tive number without affecting the sense of the inequality.
Let a > b, that is, let a — b = k, where A; is a positive number.
If m is any real number, evidently
a ± rn — (b ± ?») = k,
or, a ± in > b ± vi.
Similarly, ma — vib = ink,
or, if m is positive, 7na > mb.
When m is less than 1 this amounts to dividing both members by a positive
constant.
If each member of an inequality is multiplied by a negative number, that
is, if m is negative, the sense of the inequality is changed.
Corollary I. Terms may be transposed from one member of
an inequality to the other, as in the case of equations.
64
INEQUALITIES 65
Corollary 11. If both members of an ineqvalifi/ are positive,
each member may be rained to any power tvithout chanyiny the sense
of the inequality ; if both members are neyative and each is raised
to the same even potver, the sense of the inequality is chatiyed ; if
both members are neyative and each is raised to the same oddpoiver,
the sense of the inequality is not chanyed.
Thus, when both a and b are positive, if a < b, then a" < 6", but since — a > — 6,
(_ rt)3 > (_ i,)» and {- «)- < (- 6J2.
Siniilariy, Va < Vb, but — Va > — Vb. That is, if the negative signs are
taken in extracting the square root, the sense of the inequality is changed.
EXAMPLES
1. Show that — • > X + ij, unless x = y, where x and y
represent positive real numbers.
Solution. Multiply both sides of the iiietiuality by 2y. This will not change
the sense of the inequality since 2y is positive.
x2 + 32/2 >2xy + 22/2.
Subtracting 2xy + 2y^ from both sides,
x2-2x2/ + 2/2>0,
or {x - ?/)■- > 0.
This last inequality is true unless x = y, since the square of any real number
except 0 is positive.
If now the steps are performed in the reverse order, the original inequality
is established, and therefore holds for all positive real values of x and y
unless X = y*
2. Show that a^ + h^ > trJi + ah-j unless a = h, where a and h rep-
resent positive n-al numbers.
First solution. Divide each side of the inequality by a + 6. Since a + 6 is
positive, the sense of the inequality is not changed.
a"- - ab + 62 > ab.
Subtracting ab from each side, (a — b)' > 0,
which is true unless a = b. llonce, reversing the order of the operations, it
appears that the lirst inequality holds.
*In this ni«'th(i<l of proof we first assuino that the inequality in question is true
and then pass from it, by lejiitiinatc optTutions, to a sclf-eviiU-nt ini'ijuality. But this
proi-ess does not establish the validity of tlic oriiiinal cxpri-ssion. Tlic proof is not
complete, until, starting with tlie evident inequality, we perform the operations which
will lead back to the original. It is usually sulheient to observe that it is possible to
go through this retrograde process without actually doing it.
Q6 HIGHER ALGEBRA
Second solution. Subtract a% + a62 from each side.
a3 _ a^b - aff~ + b^>0.
Factoring, a^ {a — b) — Ifi {a — h)> 0,
(a2 - ^2) (a - 6) > 0.
If a > b, both factors are positive. If a < 6, both factors are negative. In
either case their product is positive. Hence the inequality holds unless a = b.
3. Show that Vs + Vlo <
Vio + Ve.
(1)
Solution. Squaring,
18 + 2 Vis < 16 + 2 V60.
(2)
Transposing and dividing by 2,
1 + V45 < Veo.
(3)
Squaring,
•_i. =_ - 1 i_i:„„
46 + 2 \''45 < 60,
2 \/45 < 14,
V45<7,
(4)
which is a known relation.
Now performing these operations in the reverse order, and taking the positive
square root in passing from (4) to (3), and from (2) to (1), we find that the first
inequality holds.
EXERCISES
Show that the following inequalities hold where the letters rep-
resent positive real numbers :
2sr>/ .r 4- '/ 1
1. ■ '— < — 7r-^> unless X = ?/.
x + i/ 2
2. (-4-l)>4-> unless a = J.
3. 1 — X — X- -i- x^ > — 4 X — 4: x^.
X- ir ,
4. a; + V < h — ' unless x = y.
5. a? -\- Ir -\- c'^ > ah +- he +- ca, unless a = h = c.
6. (a -\- h) (h +- (') (c -\- a) > 8 ahr, unless a = h = c.
7. ci^ + h^ > a^h +- ah'^, unless a = h.
8. x^ + 1 > X- +- X, unless x = 1.
9. a--" +- 1 > .t2«-i + X, unless x = l.
10. Vr + Vn > V5 -f VTs.
11. Vt + 2 Vs < Ve +- Vi4.
12. V3 + V2I < 2VT0.
13. V2 + V3 + VS < V30.
14. Vt 4- ViT < Vs + Vli.
15. If (r -\- h"^ z=l and x- +- y- = 1, prove that ax + h)/ < 1.
IXEQUALITIE.S 67
16. Show that the sum of any positive number and its recii)rocal
is never less than 2.
17. Show that a; + 1< 2 a-'' if r > 1, and that x + l>^2 ./-'' if :r < 1.
34. Conditional linear inequalities. W'c li;ive solved the equa-
tion ax -\- b = 0, and found that x = But if we consider the left
a
member as a function of x, we see that for various values of x the
expression ax + 0 takes on different values, some of which may be
greater and otliers less than 0. We now seek to determine the class
of numbers which make (,x + /y < 0. (1)
That is, we wish to solve this inequality.
First, let a be positive.
By Theorem, § 33, we have ax < — b,
hence x<
a
Now let a be negative and crpial to — .1, where A is positive.
Then (1) may be written
— Ax + h<0, or A X — /> > 0.
Solving as before, we olitain
f> _ h _ h
A —a a
We may solve in a similar manner the inequality
ax + /> > 0. (2)
35. Graphical interpretation of the linear inequality. If we set
ax -\- b = y, we see that if y is 0 the corresponding value of x must
be the root of the equation ax + b = 0. But all the values of x
which give y a negative value satisfy the inequality (1) ; that is,
the values of x for which the line ax -\- b = y is below the A' axis
satisfy (1), while the values of x for which the line is above the
X axis satisfy the inequality ax -\- b>0. Hence to solve an in-
equality of type (1) or (2) graphically we may plot the function
represented by the left member, and determine for what values of
X the graph is respectively below or above the A' axis.
36. Conditional quadratic inequalities. Consider the expression
y- + 4 .«• — 5 = //.
Construct the gra])h of tliis ('(piation. From the figure it appears
tliat 1/ is nt>gative for values of ,/■ lu'twci'n the roots of ./•- -|- 4 .r — 5 = 0,
68
HIGHER ALGEBRA
and. positive for other values of x except the roots themselves.
Since the roots are — 5 and + 1, we can say that the inequality
a;^+4a; — 5<0is satisfied when — 5 < a? <1.
This example shows that if the equation
av? + ia; + c = 0 has real roots, so that the
corresponding graph cuts the X axis, and
if a is positive, so that the curve is con-
cave upwards, the inequality
ax^ ■\-hx + c<^ (1)
is satisfied for values of x which lie be-
tween the values of the roots.
When the roots of ax?" -|- ^.r -|- c = 0 are
complex, so that the graph lies entirely
above the X axis, there is no real value
of X which satisfies (1).
When the roots are real and the sign of
a is negative, so that the graph is concave downwards, the curve
will be above the X axis for values of x between the roots, and hence
(1) will be satisfied only by values of x exceeding the greater or less
than the smaller root.
EXERCISES
Solve the following inequalities. Illustrate graphically :
\
\y
1 /
\
/
\
1
\
\
/
>
0
/
X
/
\
/
\
/
'
,
'
\
1
\
/
\ 1
/
/
i I
2x
5>0.
3 ^ < 5 j; -f 2.
3(x -f 1) > X -f 3.
3 — 4a->2 — a-.
(./• + 1) (x -f 6) < (x + 2)(x + 4).
x^< 9.
1 + 4 ,/■ + 4 .7-2 > 0.
1 -f X -{-2x''> 0.
For what values of x do the following pairs of inequalities hold ?
Illustrate graphically :
+ 7<9 + a;, Cx^ + 5<6x.
>9-Ax. * t3a'-6>4x-9.
■2(l-x)>3x-^T, Cx' + 2x<8,
+ 2 < 5 .r + 8. ■ 12 a- + 8 > ic^
1.
2.
3.
4.
5.
6.
7.
8.
9.
a; + - < 2.
X
10.
(a- - 1) (a- - 4) < 0
11.
a;^ — a* < 6.
12.
x'' + 4tx + 3<0.
13.
(1 - x) (x - 4) > 0.
14.
6-\-x>x\
15.
x^-Sx + 22>6.
16.
x- + 3>3x.
17
18
f3.
1 5 a;
CHAPTER V
COMPLEX NUMBERS
37. The imaginary unit. When we approached the solution of
quadratic equations we saw tliat many equations, as, for example,
the ecjuation x'- = 2, are not solvable if we are at liberty to use only
rational numbers. It is necessary to introduce irrational numbers in
order to solve them. The excuse for introducing such numbers is not
that we need them as a means for more accurate measurement, — the
rational numbers are entirely adequate for all mechanical purposes,
— but that they are a mathematical necessity if we propose to solve
equations of this type.
A similar situation demands the introduction of still other num-
bers. In attempting to solve the equation
a--' = - 1 (1)
we may write the result in the form
X =± V — 1.
But we realize that there is no real number whose square is — 1.
We may write the si/mhol V— 1, but its meaning must be somewhat
remote from that of V2, for in the latter case we have a process by
which we can extract the square root and get a number whose square
is as nearly equal to 2 as we desire. This process is not applicable
in the case of V— 1. In fact, this symbol differs from 1 or any real
number not merely in degree but in kind. One cannot say that V — 1
is greater or less than a real number, any more than one can com-
pare the magnitudes of a quart and an inch.
V— 1 is symbolized by i and is called the imaginary unit. The
term " imaginary " is jierhaps too firmly established in mathematical
literature to warrant its discontinuance. It should be kept in mind,
however, that it is really no more and no less imaginary than the
negative or the irrational numbers. So far as we have yet proceeded
it is merely something which satisfies equation (1). lint when we
have defined the various operations on it and ascribed to it the
m
70 HIGHER ALGEBRA
characteristic properties of numbers, we shall be justified in calling
it a number.
Just as we built up from the unit 1 a system of real numbers, so
we shall construct from V— 1 = i a system of imaginary numbers.
The fact that we cannot measure V— 1 on a rule will cause no more
confusion than our inability to measure v 2 exactly. As we are able
to deal with irrational numbers as readily as with integers when we
define what we mean by the four rational operations on them, so
will the imaginaries become indeed numbers with which we can
work when we have defined for them the corresponding operations.
38. Addition and subtraction of imaginary numbers. We write
0 = 0 i,
1 = 1 i,
i -\- I ^ 2 i,
i -\- I -\- l = 3 /,
i -\- l -\- ■ ■ ■ -\- i = ni. (I)
V- -^ /
n terms
Also we write a V — 1 = « i, where a represents any real number.
Consistently with § 3 we write
± V^7- = ± Vr/- . (- 1) = ± V^ • V^ = ± a V^T= ± al. (II)
We speak of a positive or a negative imaginary according as the
radical sign is preceded by a positive or a negative sign.
We define addition and subtraction of imaginaries as follows .
ai ±bl = (a ± h) i, (HI)
where a and b are any real numbers.
AssUMPTlOX. The commutative and associative laws of multi-
plication and addition of real numbers we assume to hold for
imagiyiary numbers.
39. Multiplication and division of imaginaries. We have already
virtually defined the multiplication of imaginaries by real numbers
by formula (I). Consistently with § 3 we define
V^ . V^ = 1.1= i- = - 1.
Thus V^i ■ V^ = V^ . V^ i . i = Vab .(-!) = - V^.
CO.MPJ.EX NUMBEKS 71
The law of signs in niultii)lication may be expressed verbally as
follows :
The product of two imaginnries with like signs before the radical
is a negative real number. The product of two iniaginaries with
unlike signs is a 2Jositive real number.
For instance, — V^ • V- 9 = - 2 • 3 • i^ = 6.
We also note that i^ =_ i, r'' = — i, i* = 1, j^ = j^ .
And, in general, ^n + k = [k^ ^ = o, 1, 2, 3, • • •
We define division of imaginaries as follows :
V« ■ i
In operating with imaginary numbers, a number of the form
V— a should always be written in the iorni Vcf i before i)erforming
the operation. This avoids temptation to the following error :
V— a- -yZ—b = V(— a) ■ (— b) = Vab.
EXAMPLES
Simi)lify the following :
1. V^"V-2.
Solution. A^^ • a/^ = VS • I • Vi • i = V2^ • i- = 4 • (- 1) = - 4.
1
2- .ft
lit
Solution. — = - = =— I.
i^ i - 1
EXERCISES
Simplify the following:
1. V^9.
2. V^)d.
3. V3 V-48.
4. V^~3 V-27.
5. V-12fl"-'^»V.
6. V2 ax - (a- + .r-).
7.
V— a--" V— X-
:» + 1
8.
(3V-4f.
9.
(-v^O-
10.
V-32
V-12
11.
i''- i'\
72
HIGHER ALGEBRA
12. i'^-i^K
13. ^,-
u.(?-i
15.
17. -7^ +
t^ ' ^27
18. (-V-Sv^y.
19. i^^ + 9V3r0.3V32.
(-0-^+(-0^
1-i
16. ; + 2r+3i-^+4il
20.
3V^
2 V^V-18
V^V^V-12.
40. Complex numbers. The solution of the quadratic equation
with negative discriminant, § 19, affords us an expression which
consists of a real number connected with an imaginary number by
a + or a — sign. Such an expression is called a complex number.
It consists of two parts which are of different kinds, the real part
and the imaginary part. Thus 6 + 4 1 means 6 I's + 4 i's. To any
pair of real numbers {x, y) corresponds a complex number x + iy,
and conversely.
41. Graphical representation of complex numbers. We can rep-
resent all real numbers on a single straight line. When we wished
to represent two variables simultaneously (§ 17), we made use of
the plane, and assumed a one-to-one
correspondence between the points
on the plane and the pairs of num-
bers (x, y). The general complex
number x + iy depends on the values
of the independent real numbers x
and y, and may then proj^erly be
represented by a point on a plane.
We represent real numbers on the
horizontal axis, imaginary numbers
on the vertical axis, and the complex number x + iy by the point
(x, y) on the plane. Thus the complex numbers 6 -}- i 3, — 4 -f i 4,
7 — 1 5, — 2 — i 4 are represented by points on the plane as indicated
in the figure.
42. Equality of complex numbers. We define the two complex
numbers a + ih and c -\- id to be equal when and only when
a = c and b = d.
-4H
vU
bl
a
6 +
is
o
.22
<
0
A
xis
ol
He
alsi
-2-
iA
7 —
is
COMPLEX NLMI'.KKS 73
Symbolically, a -\- lb = c -\- ul
when and <»nly when a = c, and h = d.
The definition seems reasonable, sinee 1 and i are different in
kind, and we should not expect any real multiple of one to cancel
any real multiple of the other.
Similarly, if we took for units not abstract expressions as 1 and i, but con-
crete objects as trees and streets, we should say that
a trees + 6 streets = c trees + d streets
when and only when a = c and b = d.
The foregoing definition may be stated in the form of the
Principle. When two immerical ezpressio/is involving imagi-
naries are equal to each other, we may equate real parts and
imaginary parts separately.
The graphical interpretation of the definition of equality of complex num-
bers is that equal complex numbers are represented by the same point on
the plane.
From the definition given we see that a -\- lb = 0 -when and only
when a = b = 0.
Assumption. We assume that complex numbers obey the com-
mutative, the associative, and the distributive latvs.
This assumption includes the usual rules for the removal of parentheses.
We are now able to define the fundamental operations on complex
numbers.
43. Addition and subtraction. By applying the assumption just
made we obtain the following symbolic expression for the operations
of addition and subtraction of any two complex numbers a + lb and
c + id: a + lb ± (c + Itl) = a ±r + I (b ± d). (1)
Rule. To add (subtract} complex numbers, add (subtract} the
real and imaginary parts separately.
44. Graphical representation of addition. We now proceed to
give the graphieul interpretation of aiUlitiun and subtraction.
Theorem. The sum of two runnbers A = a + ib and H = r + id
is represented hy the fourth vertex of the parallelogram formed on
OA and OB as sides.
74
HIGHER ALGEBRA
Let OX represent the axis of reals and OY that of imaginaries.
Let A and B represent a + Ih and c + id respectively, and let OA SB
be the completed parallelogram of which OA and OB are two of
the sides. We have to prove that the coordi-
nates of 5^ are a -\- c and h -\- d (see (1)).
Draw ES A. OX, AH _L ES, DB _L OX,
FA ± OX. A AHS = A ODB since their
sides are respectively parallel and OB = A S.
Then DB = HS = d,
and ES = EH -j- HS = h -\- d.
Also on ^ AH = FE = c,
and OE = OF ^ FE = (f -{- c.
Hence the point .S' has the coordinates a -\- c and h + d, and there-
fore represents the sum of .1 and B.
E X
EXERCISES
1. The difference A — B of two niunbers .4=1^/ + Ih and B = c-\- id
is represented by the extremity D of the line OD drawn from the
origin in the direction BA, and equal to BA.
2. Represent graphically the following expressions:
(a) 3 - /.
(b) 21 + 7.
(c) _4-2i'.
(d) i - 1.
(e) i + 5i.
(f)
(g)
3i
(h) (2 + /) + (3 - 2 /).
(i) (l-l)-il+l).
(j) 2(3-40-4(1-20-
(k) (4t-5) + (4-50.
(1) (-1 +20-(|-40.
(m) (- 3 - § 0 + (1 - I 0-
(n) 3/ + ^-
In the following exercises apply the principle of § 42. Find the
real values of x and y satisfying the equations :
3. (3 + 0 ^ + (1 - 2 0 // + 7 ; = 0.
- ,.'.. ^ 4.3?+ hnj — ?/ = 5 + 36 i.
5. 2x-?,y + i{x''- y-) = 4.
6. (3; + i)(x-y) = (l-l)x-{-2i(l~y) + 3:
COMPLEX LUMBERS 75
45. Multiplication of complex numbers. The assumption of i? 42
enables us to niulti[)ly coinijlfx iiunihers as follows:
<i -\- II,
c -\- hi
ac -\- id) -\- lad 4- i'l>'l = iic — fjfi + i(r/, -f- aJ).
46. Conjugate complex numbers. Complex numbers differing
only ill lilt' »iL;ii <>1 tlirir iiiuiginiiry parts an; called conjugate complex
numbers.
TuEoiiEM. The sum and the product of conjugate complex num-
bers are real numbers.
Thus a + Ih 4- a - lb = 2 a,
(a + Ih) (>i — Ih) = n^ + //-.
47. Division of complex numbers. The quotient of two complex
numbers may always be expressed as a single complex number.
Rule. To express the quotient — in the form x + iy, rationalize
the denominator, using as a rationalizing factor the conjugate of the
denominator.
a -\- ih a -\- ih c — id
Thus
C + id (■ + id (' — id
ac + bd — i{(id — />f)
^ c- + d'
ac -\- hd . ad — he
f^ + d-' ' c' + d'
(1)
We have now defined the loui' iuudamental operations on com-
plex numbers and shall make frequent use of them. If the question
remains, "After all, what arc these so-called numbers?" we may
reply : " They are expressions for which we have defined the funda-
mental algebraic operations. And, since they have the properties of
muubers, they must be recognized as such, just as a flower Avhich has
all the chara(^teristic 2)roi)erties of a known species is thereby deter-
mined to belong to that species." Furthermore, our oi)erations have
been so defined that if the imaginary parts of the complex numbers
vanish and the numbers become real, the equation expressing any oper-
ation on complex numbers reduces to one expressing the same opera-
tion on the real parts of the numbers. Thus in (1) above, if h = d — Oy
the ciiuaiioii reduces to a _ a
■c c
76 HIGHER ALGEBRA
EXAMPLES
Perform the indicated operations and simplify ;
1. ' (2 + V^r2)(4 + V^^).
Solution. 2 + V^^ = 2 + V2(- 1) = 2 + iV2
4 + V-Ts = 4 + V5 (- 1) = 4 + i V5
8 - VlO + i 4 V2 + i 2 VS
= 8 - VlO + (4 V2 + 2 VI) i.
2. 5--(V2- ;V3).
Solution.
5 _^ 5(V2 + iV3) _ ^5(V2 + zV3)^^;^ ^ .^/g
V2-iV3 (V2-zV3)(V2 + iV3) 2 + 3
EXERCISES
Perform the indicated operations and simplify :
1. 'tzll. 12. (-!+.• V3f + (-!-.- V3f
1 + *
4 13. fl + ^'^^
1+- v^
3 14.
2
4
V2"
1+i
■ V2 + t
4. (V3 + .-V2)(V2 + .-V3). 15. ^+^^.
V3 — i V2
5. (aV^» + icV^)(a.V^-<vV^).
1 + 2 ; + 3 r
6. {■yfT+'i + Vr^)l •^''' 1 _ 2i + 3 i''
1. {x + iy)\ /2-iY /2 + A-
17 ' ' ' '
8. (3 + iy + (3 - iy. ' V2 + 1/ \2 - I
1 1 l+,- + 2r + 3/«
^- (r31)-2-(i+i)2- • ^ + 2i^ + 3i« + 4i*
87 ^^ V^ - V^
11. j=- 20. . p=-
4 + 7V~5 V^ + V^
COMPLEX KUMBEllS
Prove the following relations :
77
21. (2+0' =
22. (l-2i7 =
7±}
11 -2t
2i-l ■
24. (Vii+V2/)'
25
l_V-24
25 (l-0' = j
-4
+ i
1 .. ^ .vo 36 i — 77
26.
(1 + if ^ V5 - i
VS + i (1 - if
Perform the indicated operations and simplify :
a + ib a — ih
27.
a — ih a-\- lb
29.
a
+ ^Vl-«'
-iVl-«2
28. ^^+;'^^ + :i^^-
e + tc/ r — id
12 -5i
31. Reduce
30.
Vl+a-4-tVl— .r Vl— .r + /Vl+.r
VT+aJ — * Vl — a; Vl — x — iVl+x
and
5
to simplest form and represent
2-3i l + 2t
their sum and their difference graphically.
5 -Hi J -(T + 9/) , . , , ^
32. Reduce 7—; r- and -777; zr- to simplest form and repre-
2(1 — i) 2(1 + i)
sent their sum and their difference graphically.
48. Polar representation. The graphical representation of com-
plex numbers given in § 41 suggests the graphical interpretation
of the operations of addition and subtraction
given in § 44. Rut the graphical meaning of the
operations of multiplication and division may be
shown more clearly in another manner. "We have
seen that Ave may represent x + ly by the point
P (x, y) on the plane. Let us represent the angle
between OP and the axis of reals by 6 (read theta). This is called
the angle of the complex number ./• + ///. ^^'e will denote the length
of the line OP by p (read rho). This is called the modulus of a- + ///.
Then from the figure
X = p cos 6, (1)
y = 9 sin 0, (2)
and x^ + y- = p". (3)
Y
A
x^iy
>
/
y
A'
0
X
X
78 HIGHER ALGEBRA
Hence the complex number x -f ly may be written in the form
X + 11/ = p (cos 6 -{- i sin 6), (4)
where the relations between x, y and p, 0 are given by (1), (2), and
(3). When the value of p is found by the use of (3), the positive
sign is always taken. A number expressed in this way is in polar
form, and may be designated by (p, Q). We observe that a complex
number lies on a circle whose center is the origin and whose radius
is the modulus of the number. The angle is the one which the line
representing the modulus makes with the axis of real numbers.
When the values of p and Q are given we can find the values of x
and y for the corresponding complex number by means of (1) and (2).
When a number is given in polar form- it should be kept in mind
that the modulus, or the value of p, is the coefiicient of the expression
cos Q -{- i sin Q.
Thus ill the number 2 (cos 30'' + i sin .30^), 2 is the modulus and 30° is the
angle.
EXAMPLE
Find the modulus and the angle of the number 1 + i VS and
write the number in polar form.
Solution. Let 1 + iv3 = x + ?y ; tlien x = 1, xj = VS.
By (3), p = ^>x'- + if- = Vl + 3 = 2.
X 1
By (1), x—p cos (9, or cos ^ = - = - .
p 2
Hence 0 must equal either 60° or 300°. But since the number 1 + i^/3 is
represented in the first quadrant, the only possible value is ^ = 60°, and we have
1 + i V3 = 2 (cos 60° + i sin 60°).
EXERCISES
Find the modulus and the angle of each of the following numbers
and write them in polar form. Plot the numbers.
1. -l + iV3. 5. -Vs+iVs.
2- !-*• 6. 1 - i VSi.
3. -3-3i. 7. _3l4i.
4. V3 + i. 8. 5 + 12i.
roMPT.KX XUMBERS 79
Change the following complex numbers from the polar form to the
form X + ///. Plot the iminbers.
9. cos 225° + i sin 225°. 12. 2 V2 (cos 135° + i sin 135°).
10. 2 (cos 300° + i ain 300°). 13. ^ (cos 180° + i sin 1S0°).
11. O(cos 120° + i sin 120°). 14. cos 270° + I sin 270°.
49. Multiplication in polar form. If \\c Ikivc two munbers
p (cos ^ + i sin ^) and p' (cos 6' + i sin ^'), we may multiply them
as follows :
p(cos^4- /sin^)p'(cos(9'+ /sin &)
= pp'(cos 6 cos 6' + i cos 6 sin 0'
+ I sin 0 cos d' + /- sin 6 sin 6')
collecting terms, = pp' [(cos 6 cos 6'— sin 6 sin 6')
. ^- ,,.,. ^, +t(sinecos^'+c()s^sin^')]
by the addition theorem
in trigonometry, . =p,T»s(« + «')+ ;sin,» + »')]. (1)
In this product pp' is the new modulus and 6 + 6' the new angle.
We may now state the following
Theoke.m. lyw product of the two numhers p (cos 6 + i sin 6) and
p'(cos d'-\- isin 6'') has as its modidus pp' and as its angle 6 + 6'.
It is observed that the product of two numbers is represented on
a circle whose radius is the product of the radii of the circles on which
the factors are represented. The angle of the })roduct is the sum of
the angles of thi^ factors.
50. Powers of numbers in polar form. When the two factors of
the preceding section, (p, 6) and {p, 6'), are equal, that is, when p = p'
and 6 = 6', equation (1) assumes the form
Ip (cos 0 + i sin 6)f = p- (cos 2 6-\-i sin 2 6). (1)
This suggests as a form for the nth power of a complex iiuinl>er
[p (cos 6 + i sin 6)y' = p" (cos n6 + i sin n 6). (2)
The theorem expressed by (2) is known as De Moivre's theorem.
Stated verbally it is as follows: The modulus of the iii\\ power of a
number is the nth power of the modulus of the number. The angle
of the nth power of a number is n times the angle of the number.
80
HIGHER ALGEBRA
51. Division in polar form. If we have, as before, two complex
numbers in polar form, (p, 6) and (p', 6'), we may obtain their quotient
as follows : p (cos 0 + i sin 6)
p'(cos6' + isinO')
p (cos e + i sin 6) (cos 0' - i sin 0')
(rationalizing) = ^'(cos ^' + isin ^') " (cos ^' - i sin ^')
p[cos(g-^')+/sin(g-^')]
~ p'(cos2^' + sin'-^^')
^"s"" <-• + ™s= .■ = 1) = ,^ [™^ («-«■) + ; sin (« - .■)].
We may now state the following
Theorem. The quotient of two complex numbers lias as its mod-
ulus the quotient of the moduli of the numbers, and as its angle the
difference of the angles of the nuinbers.
EXAMPLES
1. Find the moduli and angles of the numbers 2 — 21 and V3 + i
and of their product. Plot the three numbers.
Solution. Let 2 — 2 i = x + t?/.
Then cc = 2, ?/ = — 2,
and p = Vx2 4- y^ = V4 + 4 = 2 V2.
^ X 2 1
cos 0 — - = = —- ■
P 2V2 V2
Hence 0 - 45° or 315°.
But since 2 — 2 iis represented
in tlie fourtli quadrant,
(9=315°.
Let V3 + i = x' + iy'.
Then x' = Vs, y' = 1,
and . p' = Vx'2 + 2/'2
= V3 + 1 =: 2.
cos 0' --
2-21
x;_ V3
p'~ 2 ■
Hence 6I' = 30° or 330°.
But since V3 + i is represented in the first quadrant, 6' = 30°.
By the theorem of § 49 the moduhis of the product is
pp' =: 2 ^ • 2 = 4 a/2.
The angle of the product is (9 + (9' = 315° + 30° = 345°.
Hence P = (2 - 2 i) (Vs + i) = 4 V2 (cos 345° + i sin 345°).
COMPLEX NUMBERS
81
2. Find the moduli uiid angles of tlu^ numbers 2 — 2 Vi^ i and
1 + / and oi' their (|Uotient. Tlot the three numbers.
Solution. Let 2 — 2Vsi = x + iy.
Then x = 2 y =- 2 Vs,
and p = Va:2 + y^ = V4 + 12 = 4.
. X 2 1
cos^ = - = - = -.
p 4 2
Hence 6 = 00° or 300°.
But since 2 — 2 Vs i is represented in the
fourth quadrant, 0 - 300°.
Let 1 + i = x' + iy'.
Then x' = 1, y' = 1,
and
p' = Vx'- + J/'- = VTTl = V2.
1
P' V2"
cos 6' = — = — =
(2-i2V3)
Hence ^' = 45° or 315°.
But since 1 + i is represented in the first quadrant, 6' = 4.5°.
By the theorem, § 51, the modulus of tlie (luoticnt is - = — = 2v2.
The angle of tlie quotient is 0-0' = 300° - 45° = 255°,
^"■^ =2V2(cos255°+ z sin 255°).
Hence P
l + i
EXERCISES
Find the moduli and angles of the following numbers and of the in-
dicated products, quotients, or powers. Plot the numbers in each case.
1. (2 + 2 0(- 1 + Va . /). 10. (-2 + 2 0^
4. 6i -
2. (- V3 + 0(-l-0•
3. (i + ^V3i)(^-iO-
V2 V2
2 ~ 2
5. (3-3i)(-' + /VT2).
6. (4 + 3t)(l + V-0-
7. (1 + 0'-
8. (1_^
\2 2
11.
12.
13.
14.
15.
V3
_ V2 - V2 i
2 - 2 V3 i
— i
_3_3V3^
2 + 2i
9. (- 3 - V3 /)•'.
Si
- T + 24 t
3 + 4i
16. (- 1 + 0'°-
17. (3 + 3 /) (- 1- + .V V3 /)(- 2-2 i).
82 HIGHER ALGEBEA
/ 1 VS V 19. [2(cos60°+ ^'sin60°)]-l
18. ^ " "_ 20. [1 (cos 15° + i sin 15°)]-.
1 V3 .
- o - -^T *- 21. (cos 45° + i sin 45°)^^
22. [3 (cos 75° + i sin 75°)] [f (cos 15° + i sin 15°)].
^^ Meosl80°+^sinl80°) 24. P(eos 135°+ / sin 135°)]. .
• i (cos 100°+,: sin 100°) [i^(cos315°+isin316'')l'
25. For what values of n will (l + V— 3)" be a real number ?
52. Roots of complex numbers. We have seen that the square
of a complex number has as its modulus the square of the modulus
of the number, and as its angle twice the original angle.
Thus the number (1, 30°), or 1 • (cos 30°+ i sin 30°), has as its square
the number (1, 60°). Also the number (1, 210°) has as its square
(1, 420°). But (1, 420°) = (1, 60°) ; for if two complex numbers have
the same modulus and their angles differ by a multiple of 360°, they
are represented graphically by the same point, and are thei'efore
identical. For example, (1, 60°) = (1, 420°) = (1, 780°), etc. Hence
it appears that any complex number with an angle greater than
360° is equivalent to one with an angle less than 360°.
We have, then, found two numbers (1, 30°) and (1, 210°) each of
whose squares equals the number (1, 60°), that is, we have found the
square roots of this number.
In general, the modulus of the square root of a number is the
positive square root of the modulus of the number. The angle of
the square root of a number is one such that if we double it we
get either the original angle or one which differs from it by 360°.
Expressed in symbols, if the original angle is Q, the angles of the
square roots are a a
^ and ^ + 180°.
This may be denoted more compactly as follows :
I + A- 180°, (7. = 0,1),
by which we mean that we substitute in the expression indicated,
first, the value k = 0, and then the value k — 1, obtaining the same
9 6
values - and — + 180°, for the angles which were given above.
COiMPLEX NUMBERS 83
III ;i similar manner, if we seek all the numbers with angles less
tlian 360° wlii(;li cubed give a certain number, we must find three
angles which multiplied by 3 give the original angle or one which
differs from it by a multiple of 360°.
For example, the three numliers (1,20°), (1,140°), and (1,260°)
have as their cubes the three numbers (1, 60°), (1, 420°), and (1, 780°)
respectively. But since all of these numbers have the same modulus
and their angles differ by either 360° or 720°, they are really the
same number ; that is, the numbers first given are the three cube
roots of the number (1, 60°).
In general, the modulus of a cube root of a complex number is
the real cube root of the modulus of the ninnbci'. The angle of a
cube root is an angle such that if we multiply it by 3, we obtain
cither the original angle or one which differs from it by a multiple
of 360°. If the original angle is denoted by 6, the three angles of
the cube roots are fl
r. + k 120°, {k = 0, 1, 2).
We may treat the problem of finding tlu>, 7;th roots of a number
(p, B) similarly. The modulus of the ?ith root of (p, 6) is the real
l)ositive nth root of p, namely yp. The angles are those angles
which, multiplied by n, give either B or an angle which differs from
B by a multiple of 360°. There are n such angles less than 360°.
In the notation which we have adopted the angles of the nth
roots are q q^./^
^ + /r— , (A=0,l,2,...,n-1).
n n
Thus ^/p (cos B + i sin B) = y^ cos ( -
, 360\ . . (B , 360
+ A- + 1 sm - + A-
n I \n 11
where for a given value of n, k takes on the values 0, 1, • • •, n — \,
and where yp indicates the real positive nth. root of p.
For example, the five angles of the fifth roots of a number whose angle is 60° are
60° ^ 360
obtained by adding to — = 12°, the angles k . where A; = 0, 1, 2, 3, 4 ; that
5 5
is, the angles are 12°, 12° + 72° = 84°, 12° + 2 • 72° = 156°, 12° + 3 • 72° = 228°,
12° + 4 • 72° = 300°.
"When, in the following exercises, the radical sign, \'' , is used over a complex
nnnibor, or over a real number which is thought of as a complex number with zero
imaginary part, all n of the roots are meant. When only the aritlunetioal square
root of a real number is intended, the usage explained on page 5 is followed. The
context will make it clear in each case which meaning of the radical is to be taken.
84
HIGHER ALGEBRA
EXAMPLES
1. Perform the indicated operation and plot:
V- 2 + 2 V3 i.
Solution. "We first express the number —2 + 2 Vs i in polar form.
p = V(- 2)2 + (2 V3)'^ = V4 + 12 = 4. r
cos^ = — =--, (9 = 120° or 240°.
4 2
But since —2+2 Vs i is in the second
quadrant, 0 = 120°.
Hence - 2 + 2 Vs i = 4 (cos 120° + i sin 120°).
Now V^ = Vi = 2,
which is the modulus of the square roots.
l+iVs"
The angles are
and
e _ 120°
120°
_+ 180° = —
2 2
= 60°,
+ 180° = 240°.
-i-;V3
Hence \/- 2 +V3i = 2 (cos 60° + i sin 60°) = 1 + Vs i,
or 2(cos240° + isin240°)= -1- V3i.
It is sometimes more convenient to find the roots of a complex
number in another way. For example,
2. Find V3 + 4 i.
Solution. We see from § 52 that a root of a complex number is always a
complex number.
Hence we may write V 3 + ii = x + iy.
Squaring, 3 + 4z = x2 + 2 ixy — y^.
Then, by § 42, x'- - y'' = 3, (1)
2x2/ = 4. (2)
Squaring (1) and (2), x* - 2x^y- + y^ = 9
4 x^yg = 16
Adding,
Hence
X* + 2 x2;/2 + 2/4 = 25
x2 + 2/2 = ± 5.
But since x and y are real, the sum of their squares must be positive. Then
we must take 3.2 ^. ^^2 -_ 5
Adding (1), x2 — 2/2 = 3
we obtain 2 x2 = 8
x2 = 4,
X =± 2.
Substituting this value of x in the equation 2xy = 4, we find y =±1.
.: VS + 4 i =2 + 1 or — 2 - i.
COMPLEX NUMBERS
85
3. Solve the equaUuu x^—l = 0, and represent the roots graphically.
Solution, x^ — 1, or x = Vl.
Let 1 = 1 + 0 • I = /3 (cos 6* + i sill 0) . Tlieii p = 1 , B-^^.
x = \/l(cosO° + isinO°)
- 6/Tr /0° , 360°\ ... /O^
+ A:
300
■)]
(where k takes on the values 0, 1,2, 3, 4)
' cos 0° + i sin 0° = 1, when fc = 0,
cos 72° + i sin 72°, when i = 1,
= \ cos 144° + i sin 144°, when A; = 2,
cos 210° + i sin 210°, when A; = 3,
^ cos 288° + I sin 288°, when fc = 4.
These numbers we observe lie on a circle of unit radius at the vertices of a
regular pentagon.
EXERCISES
In the following exercises perform the indicated operations and
represent graphically the complex number and its roots :
1. Work example 2 above, using polar representation.
2. \/4 + 4V3i.
3. V7.
10. V5 + 12 I
11. V-1 +4V^.
4. V^".
5. Show that
V7 + V^- = ± V2, or ± l V2.
6. \/-l + 2V2i.
7. -v^3 + 3 I
8. \/V3 - I
9. ^4-4 i.
12. -^8(008 15°+ i sin 15°).
13. -v/16 (cos 200° + i sin 200°).
14. ^/S (cos 60° 4- i sin 60°).
15. Ml - ^V^.
16. Vit;-.
17. ^8.
18. S/- 128.
Solve the following equations and illustrate the results graphically:
19. .r'^-32 = 0. 23. r^ + 1 = 0.
20. a-»-l = 0. 24. .'•«- 1 = 0.
21. .r" + 1=0.
22. a.-* -16=0.
25.
-1 = 0.
26. .'■•' + 1 = 0.
86 HIGHER ALGEBRA
27. Show that either of the complex roots of the equation x^ = 1
is the square of the other, and that the sum of the three roots is zero.
Represent the three roots and their sum graphically.
28. Show graphically that the sum of the roots of the equation
x^ = 1 is zero.
29. Show graphically that the sum of the roots of the equation
x" = 1 is zero if n is & positive even integer.
Note. This is true when n is any positive integer, whether even or odd, as
we shall see in the next chapter, § 62.
30. Show that the product of the three cube roots of 1 is 1.
31. Prove that the product of the n nth roots of 1 is 1, if n is
odd, and — 1 if ?i. is even.
CHAPTER VI
THEORY OF EQUATIONS
53. Introduction. As a preliminary to the development of the
methods and tlicorcnis of this chapter, a few definitions are necessary.
A term is rational if it may be obtained in its simplest form from
unity and the letters concerned by means of the four operations of
addition, subtraction, multiplication, and division, without the extrac-
tion of any root. If each of the terms of an algebraic function is
rational, it is called a rational function.
_, , . a b I 4x 8a -6x2 ax^ + bx -^ c
The functions - + — ,» -^ r' ' — ^ ^ "• 7 — '
X x^ x^ 5 cx" — 1 cx—f
are eacli raliuuiil in x.
A term is integral if the letter which is taken as the variable does
not appear in the denominator. A term may be integral and still
involve a radical sign. If each of the terms of an algebraic expres-
sion is integral, it is called an integral function.
4x r
Tlie functions — , 8 x''^ — Vx, ax- + 6x + c,
are each integral in x.
An integral function is not necessarily rational, nor is every
rational function integral. Thus — + V.r -f 8 is integral but not
•'-1
rational, while "— H [- 8 is rational but not integral.
4 ./■
An algebraic function is rational and integral if each of its terms
is rational and integral. Such an expression is frequently called a
polynomial. The polynomial /(.r) = a^"" + o^.r""* H h "„, where
Og, Wj, rt^j-- ■■> "nj ^I'e all integers, n is a i)ositive integer, and a^ ^ 0,
we shall call the general polynomial of the «th degree, or the poly-
nomial in o-form.
The equation ^^3." _(_ ^,^.,."-i + \-a„ = 0 {A)
we shall call the general equation of the /ith degree, or the equation
in a-form. The subject of our study in this chapter is the rational,
integral equation of degree n in one variable.
87
88 HIGHER ALGEBEA
It is necessary to keep in mind that the symbols a^, «!, • • • , a„, stand for
numbers. Since they are all coefficients in tlie same expression, we denote them
all by the same letter, a ; but since they are the coefficients of different terms,
they must be distinguished from each other in some way. This is done by giving
each a a subscript equal to one less than the number of the term in which it is
used. In this way we know that a^ is the coefficient in the fourth term, a^ that
in the third term, and so on.
The notation f{x), read "/of x," is simply a symbol denoting that the expres-
sion in question is a function of x. Other letters are sometimes used to denote
functions of x, as, for instance, F{x), ^(x), and Q{x).
If in f(x) the variable x is replaced, for example, by the number 3,
the resulting expression is denoted by /(3). We may similarly
replace x all through f(x) by a letter, as, for instance, c. The result-
ing expression a^c" + a^c"~^ + • • • + '^'« ^^& denote by /(c).
For example, if /(x) = 2x^ — 3x- - 7 x + 5, f{2) = 2(2)3 _ 3(2)2 _ 7 . 2 +
5 =- 5, while /(O) = 5 and/(- 1) = 7.
The equation x" +2^^"'^ -^ 2\^"~^ + • • ■ + P,, = ^, (P)
where 2^^, 2^-21 ' ' '^Vni ^^6 all rational numbers, and 7i is a positive inte-
ger, we shall call the p-iaxm. of the equation of the wth degree. It is
observed that any rational integral equation with rational coeflB-
eients may be brought into the a-form, (.4), by transposition and
multiplication by the least common denominator of the coefficients.
Furthermore, any general equation may be put into the /»-form
by dividing by a^. Since a^ is assumed to be different from zero,
this can always be done.
For example, the equation §x* — | x = § x^ — x* + ' , after transposition and
multiplication by 24, becomes
40x* - 32x2 -21x- 6 = 0,
which is in the a-form. Dividing by 40, we have the equation
•*' 5"'' 40'*' 50 — ")
which is in the p-form.
EXERCISES
Reduce the following equations to the a-form and to the p-ioxnx :
xi' x} , X ?>x^
2. -^ -H 1 = rr^ -f 4.
5 15
3. .bx-.15x^ + .2h = x\
THEORY OF EQUATIONS 89
4. 1.4 X* = 2.8 :r' - .7 J- + 2.1 x\
5. If /(a:) = 2x-«-4x+6, find/(0),/(l),/(-2), /(«), and /(-a:).
6. It /(a;)=a;2 + 3x-3,
find /(3), /(^O, /(^^ + 1), /O'-l)-
7. If f(x)=2x*-x\
find /(V2), /(O), /(0, /(-^)-
8. Keducing tlie equation — + - — 1 = 0 to the a-form and to
o O
the jy-forni,
we
get
2x^ + x-6 = 0,
and
a;2 + ^ - 3 = 0.
Let
/2(:r)= 2 0^^4-0: -6,
(a) Graph the three functions J\(x), /^(j'), and Jl(x).
(b) What relation do you notice between these graphs ?
(c) What are the roots of the three equations /^(x) = 0, /^(x) = 0,
and/3(a;) = 0?
(d) Are the roots of an equation changed by reducing it to the
a-form and to the ^>form ? Why ?
9. Reduce the equation .75x- + 5.875x = 1 to the «-form and
find its roots.
10. li f(x)= a^x'^ + a^x -\- a^, determine a^, a^, and a.^, so that
/(I) =4, /(2)=12, /(-1)=G.
11. U f(x)= a^x^ + a^x' + (?.,j- + a.^, determine a^, a^, a,,, and a^,
so that /(O) = fa) = 2, /(- \j=l, /(2) = 8.
12. Given f(x) = a^x" + a^x + r/.^, r/^ ^ "i' '^'^^^ /(''o) =/(«J
Express a^ in terms of a^.
13. Find a polynomial of the second degree which has the value 0
when a; = — 1 or 2, and the value — 2 when x = 0. Graph tlie poly-
nomial. Is it always possible to find the polynomial however we
select the three values of x and the corresponding values which
the polynomial is to have ?
90 HIGHER ALGEBRA
14. Find a polynomial of the second degree whose graph passes
through the point (2, 4), and such that the equation formed by
equating it to zero has roots — 2 and 4.
15. Eind a polynomial of the third degree which vanishes when
a; = 0, 1, or 2, and which equals 2 when x = 3.
16. Eind a polynomial of the fourth degree which equals 1 Avhen
X is 0, equals 0 when a; is 1 or — 1, and equals 21 when ,r is 2 or — 2.
54. Remainder Theorem. The following theorem lies at the basis
of most of the work of the present chapter :
Theorem. //' ./(.r) i^ divided hy x — c, the remainder is f(/)-
Illustration. Let /(x) = 2x3 + Sx^ - 4x - 6, and let c = 2. The theorem
will be verified for this case if the remainder obtained by dividing /(x) by
X - 2 is /(2) = 2 • 23 + 3 • 2- - 4 • 2 - (3 = 14. If the division is actually per-
formed, the remainder is found to be 14. The result of the division may
be expressed thus :
2x3 + 3x2-4x-6 ^ ., „ ,^ 14
= 2 x2 + 7 X + 10 +
Proof. If /(.'•) is divided hj x — c, let us call the quotient Q(x).
We must prove that the remainder is /(f), that is, that
li
X — C X — c
Consider the expressions
/(^) = %^" -I- fV'^""^ -\ han~\X + ^„,
and f(c) = a^(f + o^c"-i -\ \- a,^_^<' + «„.
Subtracting, we get
= a^x- + a^x^-^ + • • ■ + «„ _i,r + r7„ - (rrr« + a^-- "^ + • • • + "n-i<' + <'n)
= a^(a-» — c") + ftj(.x»-i — c"-^) H h a„_i(x — r).
But since x — c is a factor of each term in the right member
(type form 7, p. 2), we may take x — c outside a parenthesis, and
call what remains inside Q(.r). We have then
f(x)-f(c^ = (x-r)Q(x), (1)
or, after transposing /(c) and dividing by x — c,
X — c X — c
which was to be jiroved.
THEOUY OF KQLATKJNS 91
Factor Thkoiieim. If c U a root of f(x) = 0, then x— <: is a
factor off(x).
If c is a root oif(x)= 0, i\m\f(c)= 0, and from (1) we have
f(x) = (x-c)Q(x).
That is, f(x) is expressed in factored form, with a; — c as one of the
factors.
EXERCISES
1. State and ])i'ove the converse of the Factor Theorem.
2. By use of the Remainder Theorem, find the remaindei- when
2 X* -\-x^ — G .'•■' + 1 is divided by ic — 1 ; by a; + 2.
3. By use of the Kemainder Theorem, find the remainder when
3 a-* + 2 x^ — 1 is divided by ./■ — ^ ; by x.
4. l^y use of the Remainder Theorem, find the remainder when
2 x^ — x^ — x'^ -\- Ax — 1 is divided by a- — 3; by a; + 3.
5. By use of the Remainder Tlieorem, find tlio remainder when
a;'' + 1 is divided by a' + 1 ; by x — 1.
6. By use of the Remainder Theorem, find the remainder when
^18 _|_ ^,13 ^g divided by a; + o ; by x — a.
7. Sliow tliat 2 is a root of the equation 2 x^ — 3 x" — 4 a- -|- 4 = 0.
8. Show that if /(c) ^ 0, then /(a-) is not divisible by x — c.
9. Find a polynomial /(a) of the second degree such that 1 and
2 are roots of /(a-) = 0, and /(a;) has the value 8 when x = Q.
10. Find a polynomial /(a-) of the third degree such that 0, — 1,
and 3 are roots of f{x) = 0, and /(.'•) has the value 12 when a- = 1.
55. Synthetic division. In plotting the function
/(a;)=a„x" + a,a;"-> + ■■■ + «„,
where the a's are integei*s, by the method of § IT, it is necessary to
find the values of the function for various values of x ; that is, we
must obtain the values of /(I), /(2), /(3), etc. The Remainder
Theorem tells us that these are nothing else than the values of the
remainders after dividing f{x) by a; — 1, a; — 2, x — 3, etc., which
we may find rapidly if we make us(> of the following abridgfd
method of division.
92
HIGHER ALGEBRA
The method may be illustrated by the following example :
Let /(.r) = 2 *^ - 3 x"" + x- - x - 9, c = 2,
and let us divide f(x) by ic — 2.
By long division we have
2x'
2x*
Sx^+ x^- x- 9
x-2
2x'
+ .<.:"- + 3 J3 + 5
1 x^ + x^
lx«-2a;2
3 X- - X
2,x^-Qx
bx-
9
5x —
10
+ 1
We can abbreviate this process by observing the following facts.
Since x may be regarded as merely the carrier of the coefficient, we
may omit writing it. Also we need not rewrite the first number of
the partial product, as it is only a repetition of the number directly
above it in full-faced type. Our process now assumes the form
2-3+1-1-
-4
2+1+3+5
+ 1
-2
+ 3
-6
+ 5
-10
+ 1
Since the minus sign of the 2 changes every sign in forming the
partial product, if we replace — 2 by + 2 we may add the partial
product to the number in the dividend instead of subtracting. This is
also desirable, since the number which we are substituting for x is 2,
not — 2. Thus, bringing all our figures on one line, dropping the 1 in
the divisor, replacing — 2 by 2, and omitting the coefficients of the
quotient, we have 2 - 3'+ 1 - 1 - 912
+ 4 + 2 + 6 + 10
2+1+3+5+ 1
THEOliV OF EgiATloNS 93
The last number in the lowest line is the remainder in the division.
Hence it is the value of the function f(x) when x is replaced by 2,
that is, /(2).
It is also to be observed that, from the nature of the operation, the
numbers preceding the remainder in the last line are the coefficients
of the quotient in the division. In this case the quotient is
2a;» + a--^ + 3.i- + 5.
We have illustrated the following rule. Since the process may be
looked upon as merely a convenient arrangement of the operation of
long division, no formal proof will be given.
Rule for synthetic division. Write the coefficients of the
polynomial in order, supplying 0 ivhen a term is lacking.
Multiply the number to he substituted for x hy the first coefficient,
and add (algebraically^ the product to the second coefficient.
Multiply this sum by the number to be siibstituted for x, add to
the third coefficient, and proceed until all the coefficients are used.
The last sum obtained is the remainder and aho the value of the
polynomial when the number is substituted for the variable.
The method of synthetic division is useful not only in finding the
values of the function for purposes of i)lotting, but also in deter-
mining whether the function has any factors of the form x — r. For
if by the process of synthetic division the remainder /(c) comes out
zero, then the function has a factor x — c. It should be noted that c
may be integral, rational, or, in fact, any kind of a number.
The process of synthetic division in the foregoing example may be looked
upon as a reduction of each term in the polynomial to one of the next lower
degree by replacing one of the x's by 2 and combining until the numerical
value of the function is obtained.
Thus in 2x'' — 8 j^ + J- — j — 0, if j = 2, we have the first term 2x* = 2 • 2 j''
= 4x^. Adding to the second term, we have 43"' — 3a;* = x*. Letting one x = 2,
x^ = 2x2. Adding to the third term, 2x^ + x- = Sx^. Substituting 2 for one x,
3x2 = Ox. Adding to the fourth term, 6x — x = 5x. Substituting 2 for x,
5x = 10. Hence the value of the function for x = 2 is 10 — 0 = 1, which agrees
with the result already obtained.
This process is similar to that of converting a distance expressed in yards,
feet, and inches into one expressed in inches, by reducing the yards to feet,
adding to the number of feet given, reducing this to inches, and adding to the
number of inches given.
94 HIGHER ALGEBRA
EXERCISES
1 . Find the remainder when 2 a-^ — 5 x* + 4 .x^ — 54 x^ — 32 a; — 30 is
divided by a; — 4. Do this by direct substitution, as in § 54, and also
by the method of synthetic division. Which method is preferable ?
2. Find the remainder when x* — Sx^ -\- 6 is divided by x — 2.
Do this by both methods mentioned in the preceding exercise. Which
method is preferable ?
3. Find the remainder and the quotient when 3 x^ — 2 .t' + .r + 6
is divided by a; + 3 ; by x — 3.
4. Find the remainder and the quotient when 4 x^ — 4 x" — 3 x + 2
is divided by x — i ; by a; + ^.
5. Given /(a-) = 8 x^ - 24 a;^ - 16 a- + 40, find /(I), /(i), /(- 2),
/(8),/(V2).
6. Find the value of the function 3 x" — 11 x^ — 18 x — 24, when
X = I ; when x = — J.
7. Show that x — 2 and x + 5 are factors of x* — 23 x^ + 18 x + 40.
What are the other factors ? What are the roots of the equation
a;4 _ 23 x^ + 18 X + 40 =^ 0 ?
8. Show that — 3 is a root of the equation x^ + 4 x^ — 17 x — 60 = 0.
What are the other roots ?
9. Show that x — 1 is twice a factor, that is, that (x — 1)- is a factor
of ic^ — a;^ — 2 X + 2, and hence that 1 is a double root of the equation
X* - x~ -2x + 2 = 0. What are the other roots ?
10. Find the value of k if 2 is a root of the equation 2x* — 6x^
+ 4 kx + 13 = 0.
11. Find the values of k and Mf — 1 and 2 are roots of the
equation 3 x* - 3 x^ - 10 x" -j- 2 kx - 2 1 = 0.
12. For what values of k will 1 be a root of the equation 5 x^ — 4 x^
+ 2 k-j-' -\-k = 2?
56. The graphing of functions. AYe are now in a position to find
the graph of a polynomial in the most expeditious manner. We shall
symbolize the function f(x) by ij and find the values of ij corre-
sponding to various values of x.
In plotting, if the table of values consists of numbers which are
large or are so distributed that the plot would not be well propor-
tioned if one space on the paper were taken for each unit, a scale
should be chosen so that the plot will form a graceful curve.
THKOKV OF ISOLATIONS
9o
EXAMPLE
Plot y = ./■•' + 4 ./■- - 4.
Solution. Wo liiid b}' syiitliclic division tlie values of y corresponding to
various values of x.
Tlie value of ?/ when x = 0 is found by direct sub-
stitution.
1 + 4+ 0- 4(J.
+ 1+ 5+ 5
X
- 5
-4
-3
-2
- 1
0
1
2
y
-20
-4
+ 5
+ 4
-1
-4
+ 1
+ 20
Since the luimerical values of y for x = 2 and for
X = — 5 are so umcli larger than the other values of y,
and since, as we shall see in the next section, they give
us points on the curve which we are not interested in,
we shall not include them in making the graph.
In this figure two spaces are taken to repre-
sent one unit of x. A single space is taken for a
unit of y.
By referring to the graph it is seen that the
curve crosses the X axis at about the points x = .8,
x=— 1.2, and x = — 3.7. Since these are approxi-
mately the values of x whirli make y, or /(x), equal
to zero, they are approximately the roots of the equa-
tion x3 + 4x2-4 = 0.
By performing mentally the algebraic additions in the
process of synthetic division, the work in this example
may be compressed to the following form
1+5+ 5+ 1
1 + 4+ 0- 4|_2
+ 2 + 12 + 24
1 + 0+ 12 + 20
1 + 4+ 0- 41-
1
- 1 - 3 + 3
1+3- 3_ 1
1 + 4+ 0- 41-2
-2
4+8
1+2- 4+ 4
1 + 4+ 0- 4|-3
-3- 3+ 9
1+1- 3+ 5
1 + 4+ 0- 4|-4
- 4+ 0+ 0
1 + 0 + 0- 4
1 + 4+ 0- 41-5
-5+ 5-25
1 _ 1 + 5-29
X
y
0
1 + 4 + 0
- 4
1
f) r,
1
2
0 12
20
- 1
3-3
- 1
-2
2-4
4
-3
1 -3
5
-4
0 0
- 4
-5
-1 5
-29
•-
V,
Y
1
/
\
y
1
\
/
/
1
/
/i
V
/
"'
-3
-2
\
-1
0
/
I A'
,/
V
/
i
\
/
1
\
y
f
57. Extent of the table of values. Since the object in plotting a
curve is to obtain information regarding the roots of its equation,
stretches of the cuivc bryoiid all ci'ossings of the A' axis are of no
interest for the present purpose. Hence it is desirable to know when
a table of values has been formed extensive enough to afford a i)lot
96 HIGHER ALGEBRA
which includes all the real roots. If for all values of x greater than
a certain number the curve lies wholly above the axis, there are no
real roots greater than that value of x.
By an inspection of the preceding example it appears that if for a
given value of x the signs of the partial remainders are all positive,
thus affording a positive value of y, any greater value of x will afford
only greater positive partial remainders and hence a greater positive
value of y. From this point on, the curve must rise as x increases.
Hence none of the roots can lie to the right of this point.
Thus when all the ■partial remainders are positive^ no rjreater positive
value of x need he suhstittited.
Similarly, when the partial remainders alternate in sign, beginning
with the coefficient of the lilghest p)Ower of x, no value of x, greater
negatively, need he suhstituted.
EXERCISES
Graph the following functions. Set each equal to zero, and deter-
mine between what consecutive integers the real roots of the resulting
equations lie :
6. .T* + 19a;"^ + ll.
7. .r*-3sc"'-9.r-31.
8. a-* - 2 ;r'' - x + 2.
9. .T* _ 2 x-^ 4- 3 a-2 - 20 X - 47.
10. 6 x' - 13 .r^ + 20 x~ - 37 x + 24.
11. A rectangle whose perimeter is 36 inches is rotated about a
line joining the mid-points of two opposite sides, thus generating a
cylinder whose volume is 550 cubic inches. Find the dimensions of
the rectangle.
Note. Take ir = -y^-. Use the graph to obtain an approximate result.
12. A rectangle whose perimeter is 33 inches is rotated about a line
joining the mid-points of two opposite sides, thus generating a cylin-
der whose volume is 385 cubic inches. Find the dimensions of the
rectangle.
13. Each dimension of a rectangular tank 6 x 8 x 10 feet is to be
increased by the same amount, so that the tank will have a capacity
of 1000 cubic feet. Estimate from a graph the length each edge is
increased.
1.
x^-2,x + l.
2.
a;3_ 2x^-10.
3.
a;^ - 17 a; + 100.
4.
2a-^_3.r^--7./- + 5.
5.
5 x^ - 7 ./•- + 3 ,/• + 9.
TllKOin' Ol' EQUATIONS 97
14. Till! radii of four spheres are in arithmetical progression, hav-
ing a coninion dii'tVrence of 1 inch. If the largest sphere is equal in
volume to the sum of the other three, find their radii.
58. Number of roots. It appeared from the solution of the quad-
ratic equation that every equation of the second degree has two and
only two roots. In the preceding exercises it may have been noted
that the graph of a function of degree n never crosses the A' axis
more than n times, that is, none of these equations has more than
n roots. Whether the general equation of degree n has any roots at
all is a problem which remained unsolved until about a hundred years
ago, when it was proved, by methods which we will not reproduce
here, that every rational integral equation of degree n possesses at
least one root. That this root may not be a real number is indicated
by problem 6, p. 96, in which the graph does not cross the X axis.
But in such a case the theorem would demonstrate the existence of
a complex number Avhich satisfies the equation. This Fundamental
Theorem of Algebra we shall assume. We can then prove
Theorem I. Every equation of degree n in general form has n
roots.
Given the equation a^^^" + a^x"~^ + • • -f "„ = 0,
where the a's are integers, ((^ ^ 0, and n is a positive integer. Since
the multiplication of an equation by a constant does not affect the roots
of the equation in any way (§ 23), we may multiply each member of
the equation by the constant — ? thus throwing the equation into the
^-form. It should be kept in mind that the coefficients 7?^, j^^j • ••■>Pn^
are really nothi
We have, then, ^.„ ^ ^^^ ^,, - 1 ^_ . . . ^ ^, _ _ 0,
an equation which has the same roots as the original equation.
By the Fundamental Theorem of Algebra this equation has at least
one root, which we will call a-j. By the Factor Theorem x — x^ must
be a factor of the left member. Hence if we write it in factored form,
we have ^^ _ ^^^ ^^.„_i _^ ^^^.,._, ^ _^ ^^^ _^^ _ 0
Now reasoning as before, the expression inside the second ])aren-
thesis, set equal to zero, has at least one root, which we will call .r,.
are really nothing else than — > — > • • • ,
% % %
98 HIGHER ALGEBRA
and by the Factor Theorem the expression must have x — x^ as a,
factor. We may then write the expression within the second paren-
thesis in factored form. Hence
(x - x;) (x - x^) (a'«-2 + .s-jj'"-3 + . . . + .s„_2) = 0.
We may continue this process until the last factor is of the first
degree, which, set equal to zero, will have a root which we may call
.T„. We have then the second equation with its left member ex-
pressed as the product of ?i. linear factors,
(x — x^) (x — .T^) ■ ■ ■ (x — x„) = 0.
The roots of this equation are x^, x^, ■ ■ ■, x,^, which are evidently n in
number.
Not all of these roots need be distinct. If tAvo of the roots, say x^
and x^, are equal to each other, /(«) will have (x — xj- as a factor.
We say that x^ is then a double root of f(x) = 0. If /• roots are
equal to each other, /(.r) will have r equal linear factors, and f(x) = 0
will have an r-fold root, or a multiple root of order r. Multiple roots
may be regarded as limiting cases of roots which have been approach-
ing each other and have finally become equal.
It should be particularly noted that certain of these roots x^, x^,
. ■ •, x„, may be complex numbers, so that these linear factors are not
necessarily of the simple type considered in § 1. It was stated in
that section tliat it is sometimes desirable to find factors whose co-
efficients are not integers, rational fractions, or even real numbers.
With this understanding we may state as a result of our theorem
that the general polynomial may be expressed as the product of
linear factors.
Assumption. If x— a, x—h, x — c, • ■ ■, x ~ !<; are each factors
of a jiolynomial, then their product is a factor of the jyolynomial.
Theorem IT. The general equation of the ntli degree^ a^y^ -\-
a a:" ~ ^ + •••+'/„ = 0, has no more than n roots.
For if x^, x^, •••,»"„, a-„^i, are each roots of the given equation, then,
l)y the Factor Theorem, x — x^,x — x,^, ■ ■■,x — x„,x — a-„ + 1, are each
factors of the left member. Hence, by the preceding assumption,
their product is contained in the left member, which would there-
fore have to be at least of degree n + 1, which is contrary to the
hypothesis.
T1IJ:(.)KV Oi' EQUATION'S 99
THEOr.E>r III. //" the equation a^x" + n^x"'^ -f . . . -|- a^ = 0 in
satiisfied by more than n distinct values of x, (til of its coefficients
must vanish.
The distinction between this theorem and the preceding lies in
the hypothesis. There it was assumed that a^ =r= 0 (§ 53), and we
found the nunil)er of vahies of x which could satisfy the equation.
Here our hypothesis states that the equation is satisfied by more
than n values of x, and we ])ro])ose to d(;termino what hajipens to
the coefficients.
Illustration. This theorem find.s no application when we are dealing with
e(luiUi(ins with numerical coefficients, for in tiiat case if any term has a zero
coefficient it siniplj' drops out. I5ut we niiglit liave the quadratic etjuation
(a - 2)x2 + {cfi _ 4)x + a- - 3 a + 2 = 0,
wliich we had found in some way was satisfied by tlie three numbers
X = 3, X = 2, and x = 1 .
The theorem tells us thatanuist have such a value tliat all nf the coefficients
vanish ; that is, a nuist equal 2.
Proof. Suppose that not all of the coefiicients vanish. Then the
degree of the polynomial will be n or perhaps less than n. Therefore,
by the })receding theorem, it cannot have more than n roots. But by
the hypothesis it has more than n roots. Therefore the assumption
that not all of the coefficients vanish is false.
Corollary. If tivo polynomials in x of degree n are equal to
each other for more than n values of x, the coefficients of like powers
of x are equal to each other.
We have given
a,a;» + a^x^-^ + . . . + ^,, = h^^x" + \y' - 1 + . • . + ^,„
for more than n values of x. Transposing, we have
{% - ^o) ^•" + (": - \) •'•" -' + •••+(««- ^.) = 0.
By Theorem III, a^ - A, = 0, or n^ = \,
"i - K = ^' "'■ "i = ^'v
I'n = ^\ <>!■ "n = f'.
100 HIGHER ALGEBRA
EXERCISES
1. If the equation a'^(.c'' + x + 1) + 3 a {x" + 2) - 9 (a- - 1) = 0 is
satisfied by a; = 2, a- = 3, and x = 5, find the value of c
2. If the polynomial (3a-{-b — S c) x^ + (a + h + c) x - (2 a + h-c)
vanishes when x has three different values, what must be the values
of a, h, and c ?
3. If the equation ax^ + 1/ {x^ -\- x)+ h(x^ -\- 2x) + x — 1 -\- c^ = 0
is satisfied for four different values of x, what are the values of a, b,
and c?
4. If 19 a; 4- 1 = ^ (3 a; - 1) + i3 (5 x + 2) for all values of x, find
the values of A and B.
5. If a-' - 2 = Ax (x - 2) + Bx (x - 1) + C (a; - 1) (a; - 2) for all
values of x, find the values of .4, B, and C.
59. Complex roots. In the exercises on page 96 it was noted that
the graphs of some of the functions cross the A' axis fewer times
than the degree of the corresponding equation; for instance, the
graph for the second exercise crosses but once. Since crossing the
axis indicates a real root, and since every equation must have ?i roots,
real or complex, we can tell from the graph how many complex roots
an equation has.
Theorem. If a rational integral equation with real coefficients
has the complex manber c + id for one of its roots., it must also have
the ymynher c — id for a root.
Given the equation
/(•^■)= V" + "i'^'""' +••• + "» = ^'
where the «'s are real numbers, and given that c + id is a root of
this equation, it is required to prove that c — id is a root.
To say that c + id is a root of the given equation means that if
c -\- id is substituted for x, the equation is satisfied ; that is,
f(c + id) = a^(c + idy + a^(c + uiy-' + • • • + «„ = 0.
Now if we expand each of the powers oi c -j- id by the Binomial Theo-
rem,* we obtain an expression some of whose terms contain no i,
while others contain i to various powers, from i to i". But, by § 39,
any power of i reduces to 1, — 1, i or — /, so that finally each terra
* Let the studeut write out these expansions to several terms.
THEORY OF EQUATIONS 101
in tlie expansion will citlici' contain no I at all or it will contain i to
the lirst ])o\\ci-. Now let. us i;rou]) togotliLT all terms of the expan-
sion which do not contain /. Denote this group of terms by P. Then
group together all terms of the expansion containing I and denote the
expression representing the complete coefficient of i by Q. Then we
'"■'^y ^^'^'ite /(,. + /,/) = /> + iQ = 0.
Hence, by § 42, we must liav^e 7' = 0 and (i — 0.
Now we have to show that c — id is a root of the given equation ;
that is, we must show that f(c — id) = 0. T^et us form the expression
f(c — id). This may be obtained from the expression we have found
for/(c + /'/) by changing i to — /; that is, Ave have
/(c _ id) = P - iQ,
where P and Q represent the same expressions as before.
But we proved above that these expressions P and Q must each
equal 0. Hence j-q. __ ij^^ ^ q .^
that is, c — id is a root of the given equation.
Illustration. Consider the equation
/(x) = x8 + px + (/ = 0. (1)
Let c + id be a root of (1) ; we will prove that c — id is also a root. Since
c + id is :i root, we have
f(c + id) = (c + idf + p (c + iO) + 7 = 0.
ExpandiuL: llie first term by the IJinoniial Tiu'orfni.
f(c + id) = c» + 3 cHd + 3 c {id)- + {idf + pc + pid + q
= c3 - 3 cd2 + pc + 7 + i (3 e-d- d^ + pd)
= P+iQ = 0,
where P = fS - 3 cd- + pc + q; Q = 3c^d-d« + pd.
By § 42, P = 0, Q = 0.
Now /((• — id) = (c — id)^ + p(c — id) + (j
= c3 - 3 cd- + pc + q- i (3 c-d -d^-itpd)
= P- iQ.
But we have shown that P = 0 and Q = 0. Hence /(c — id) = 0 and c — id
is a root of (1).
Corollary. Every equation of odd degree with real coefficients
has at least one real root.
The roots cannot all be complex, else the degree of the equation
would be even by the preceding theorem.
102
HIGHER ALGEBRA
60. Multiple roots. Wlieu we plot the equations
y = a;3 + 4a;^-4, (1) 7/ = x-^ + 4 ic^ - 1, (2)
y = x^-^4.x% (3) 2/ = x3 + 4x^4-1, (4)
we see that corresponding to the increase of the constant term there
is a corresponding elevation of the curve with respect to the X axis.
In every case the curve is the same, but the corresponding values
Y
^
s
t
/
N
/
\
\
/
1
/
0
i/
X
/
V
1/
/
\
/
V
/
2/ = a:3 + 4x2-4, (1)
2/ = x3 + 4x2-l, (2)
Y
f^
s
N
\
\
i
)
1
/
\
/
\
1
/
V
\
/
\
/
l'
0
X
1
?/ = x^ + 4 x^,
(3)
1
1^
>,
}'
'
^
L
/
\
1
\
/
^
V
/
\
/
\
/
V
/
/
\
/
f
V
y
1 ^
1 X
2/ = x3 + 4x2 + l, (4)
of y are different. In (1) and (2) the curve crosses the X axis three
times, in (3) it touches the X axis, and in (4) we have only one cross-
ing. Hence equations (1) and (2) each have three real roots ; (3) also
has three real roots, where one root is counted twice ; and (4) has
only one real root. As the curve is raised, an inspection of the right
elbow of the figure shows that two of the roots approach nearer to
THEOKV OF EQUATIONS
103
eafh other, and finally coincide in (3), forming the double root. As
it is further raised, this elbow fails to intersect the X axis, and the
pair of roots has ceased to be real. But since a cubic equation always
has three roots, a pair of roots must have become complex. Conse-
quently, a double root is, in a certain sense, a limiting case between
two roots which are real and distinct, and a pair of complex roots.
Thus we have the
Principle. Corresponding to every elbow of the curve that does
not intersect the X axis there is a pair of complex roots of the equation.
The converse is not always true. It is not always possible to find
as many elboAvs of the curve which do not meet the A' axis as there
are pairs of complex roots.
Consider the equations
x^-16x.
y
x" — .r.
(^^)
?/ = y^.
(")
y = CF° — iba*, (5)
The equation .r^ — 16 a; = 0 has the routs — 4, 0, and + 4. The
graph of (5) consequently cuts the A' axis at the points — 4, 0, and
1'
Y
1
■
f
/
V
/
1
/
s
\
/
/
/
>
/
1
_/
/
0
\
X
4
u
X
i
\
\
\
/
/
/
X
-^
i
f—
/
,
/
1
f
/
i
1
'
y = x^ — \C)X
y z= x" — X
+ 4, and lias two elbows. The equation .r^ — x = 0 has the roots
— 1,0, and + 1, and the graph of (6) also has two elbows. But
since the points where it crosses the A axis are closer together than
in the case of (o), the elbows are less prominent. Finally, in equa-
tion x^ — 0 the roots are equal to each other, and the corresponding
curve has no elbows at all, but crosses the axis only at the origin.
104
HIGHER ALGEBRA
Y
1
/
/
/
/
/
>
f
/
_^
'
/^
'o
X
/
/
/
/
1
'
/
/
y = x"
The forms of the graphs of the equations here considered indicate
that as three real roots of an equation become more nearly equal, two
elbows of the corresponding gra^Dh
become less and less prominent until,
when the roots are equal, the elbows
have faded away entirely. Hence
the graph of an equation crosses
the A' axis only once for a three-
fold root.
It should be observed that the fore-
going curves belong to the system whose
equation is x-^ — ax = y, and that the
values of a are 16, 1, and 0.
Similarly, Ave could show that if
an equation has an T^-fold root,
the corresponding graph crosses
the A' axis if n is odd, and touches
it if n is even.
61. Binomial quadratic surd roots. By a binomial quadratic surd
is meant a number of the form V« + Vz*, where a and b are positive
rational numbers but are not both perfect squares.
Theorem I. If a binomial quadratic surd of the form a + Vi is
equal to zero, then a = 0 and h = 0.
If (I + Va = 0 and either a = 0 or i?» = 0, clearly both must equal
zero.
Suppose, however, that neither a nor b equals zero. Then, trans-
posing, we have a = — V^, and a rational number is equal to an
irrational number, which is impossible.
Hence the only alternative is that both a and b equal zero.
Following the terminology used in the definition of conjugate
complex numbers (§ 46), a -f V^ and a — Vz* are called conjugate
binomial surds.
Theorem II. If a given binomial surd a-i-Vb is a root of an
equation with rational coefficients, then its conjugate is also a root
of the same equation.
The proof of this theorem may be made analogously to the proof
of the theorem on page 100.
TIIEOKY OF EQUATIONS 105
62. Relations between roots and coefficients. Tf the equation in
has for its roots the numbers .r^, ji\^, . . ., .r„, we have seen in § 58 that
it may be written in i'uctored f'oi'in as follows :
{X - x^) (x - x^) ...ix-x^ = 0.
If we multiply out the left member of this equation, collect like
powers of x, and compare the various coefficients obtained with the
coefficients in (/'), we sliall hiid certain relations between the roots
and the coefficients.
Illustration. Lot n = A ; that is, let the e(]uation be
X* + p,a;3 + p.^x^ + V^i + Vi = 0, (P)
or, ill factored form, (x — x^) (x — x.^ (x — x.^) {x — x^) = 0.
Multiplying out, we have
{X — X,) (X — X2) (X — X3) (x — x^) = x'' — (x^ + X2 + x^+ x^)x^
Equating coefficients of like powers of x in this equation and in (P), we have
Xi + X2 + X3 + a;4 = — Pv (1)
X^X„ + XjXg + x^x^ + x,X3 + x.,x^ + x^x^ = p.,, (2)
*1*^2 3 "*" **'1*^2 4 * '*^i*^3'^4 1 '*^2 3*^4 ~~ -2^3' \^/
X1X2X3X4 = P4. (4)
This result suggests the following general theorem for the equatiou
of degree 71 in ^j-forni :
Theorem, (a) The sum of the roots equals the coefficient of
the second term with its sign changed.
(b) The sum of the products of the roots taken two at a time
equals the coefficient of the third term.
(c) The sum of the products of the roofs taken three at a time
equals the coefficient of the fourth term tvith its sign changed.
(d) The product of the roots equals the constant terin, with its
sign changed if n is odd.
"W'e will now prove parts (a) and (d) for an equation of any de-
gree'. AVe will first show that if (a) and (d) are true for an ecjuation
of any degree k, they will also be true for one of degree A- + 1.
106 HIGHER ALGEBRA
Assuming, then, that (a) and (d) are valid for any equation of
degree h, we may write the equation in the form
x^ - (a-, + .7-2 + ■ ■ ■ + .r^)./''- - ^ + • • • ± (a-,.r.^ - . • .r,.) = 0. (5)
Let us now multiply each member of this equation hy x — a"t + i,
thus obtaining an equation of degree 7v + 1 with the roots a-^,
a;* _ {x^ + :r_^ + • • • + ■'■i>'- ' + • • ■ ± (a-^a^a " " " ■'^'.) = ^
-,.X; + 1
X
(.r^ + ./•, + ■ • • + ;r, ) ./-^ + • • ■ ± (.''ir, • • - x^) x
Xi.X
k-' k + 1
ic- - - — (a-^ + .'•, H h a'i- + x^ + 1) ./■'^- H ^ x^c,^ ■ ■ ■ x^-;, + 1 = 0. (6)
^fc + i
It is observed that the coefiicient of the second term and the con-
stant term of (6) are of the form required by (a) and (d). We have
shown, then, that if we can write an equation of degree /.:, with roots
x^, x^, ■ • •, .r^., in form (5), that is, if (a) and (d) are true for an equa^
tion of degree k, then (a) and (d) must also be true for an equation
of degree A' + 1.
But we know from (l)-(4) that (a) and (d) are true for an equa-
tion of degree /.■ = 4. Hence the demonstration shows that (a) and
(d) are true for an equation of degree I- -\-l = 5. But if they are
true for an equation of degree 5, they must also hold for one of
degree 6, and so on, for an equation of any higher degree.*
In applying the above theorem the coefficient of a missing term is taken as 0.
This method of demonstration by which a fact which is known to
be true for a certain value of k is j)roved to be true for the value
/.- + 1, and hence for all succeeding integral values, is called the
Method of Complete Induction.
Corollary I. If x^ is a root of an equation in p-forni with
integral coefficients, it is a factor of its constant term.
This follows immediately from (d).
a
Corollary II. If the rational number - is a root of the equation
in a-form with integral coefficients, f(^x^=a^x"-{- a^x" " ' H \-a,, = 0,
then h is a factor of a^, and a is a factor of «„.
*For a proof of the complete theorem, see Hawkes, Advanced Algebra, p. 177.
THEORY OF EQUATIONS 107
Illustration. Suppose that the numbers \, ^, and -^ are roots of a cubic equa-
tion. Then the equation must be expressible in factored form as follows :
Each of these factors may bo written witli a common denominator, and the
equation becomes
r-^)(^)(^)=».
or, after multiplying through by 30, in order to make the coefficients integral,
(2x- l)(3x-2)(5x-3) = 0.
Finally, multiplying out,
30x-''-53x2 + 31x-G = 0. (1)
From this form it appears that the coefficient of x^ in (1) is exactly the prod-
uct of the denominators of the fractional roots, and that the constant term 0
is the product of the numerators of the roots, except for sign. Furthermore, if
eciuation (1) had been given and wo wished to detcriniiie its fractional roots,
we would only need to consider fractions whose denominators are factors of 30
and whose numerators are factors of 6.
Proof. Let the equation be denoted by /(a-) =0. If -r is a rational
root, then x — j must be a factor oif(x). Hence
/(^) = (•'• - t) ^l (-) = 0, or /(.r) = i^^f^^ Q (^) = (l^x - a) ^ •
But since f(x) has integral coefficients, — —■ must be a polynomial,
of degree 7i — 1, with integral eoeflB.cients. Call this polynomial Q' (x).
Now since
f(x)=(bx-a)Q'(x) = (bx-a)(q^a-~'-\-. . . + q^_^) = bq^x'' + ■ ■ ■~a>j„_„
it appears that i is a factor of a^, and a is a factor of a„.
/I 1
Illustration. Let /(x) = 0 bo (1) above, and let - = - . Then dividing the
6 2
K'ft member of (I) by x — J, we find that Q(x) = 30x- — 38x + 12. Hence
(6x— a)Q(x) = 0 becomes in this case (2x — l)(15x- — 19x + G) = 0. It is
observed that in dividing the polynomial /(x) by x , each coefficient of the
b
quotient Q (x) contains 6 as a factor, which may be divided out when the ex-
pression is set equal to zero.
63. Formation of equations with known roots. If we know all
of the roots of an equation, we may form the equation in cither
of two ways : thi; first method uses the princi})lo of the Factor
Theorem ; the second employs the relations between the roots and
the coefficients derived in the preceding section.
108
HIGHER ALGEBRA
First method. If x^, x,^, ■ • -, x^, are the roots, multiply together
the factors x — x^, x — x^, • • •, x — x^^, and set the product equal to 0.
Second method. From the roots find the coefficients, using the
relatio7is of ^ 62.
If the equation and all but one of its roots are known, that root
can be found most readily by the solution of the linear equation
obtained by setting the sum of the roots equal to the coefficient of
the second term with its sign changed.
If all but two of the roots are known, the unknown roots may be
found by the solution of a pair of simultaneous equations formed
by using the coefficient of the second term and the last term.
In using the second method the equation must always be in
2?-form.
EXAMPLES
3, ±a
1. Eind the equation having for roots the numbers 4, ^, _i_ g-
Draw the graph of the function forming the left member of the
equation.
First solution. Applying the Factor Theorem, we may write the equation in
the form (x - 4) (x + 3) (x - |) (x + |) = 0,
or (X - 4) (X + 3) (2 X - 3) (2 X + 3) = 0.
By multiplication we find
4x4-4x3- 57x2 + 9x + 108 = 0. (j)
The graph of the function
1/ = 4x4 -4x3 -57x2 + 9X + 108
crosses the- X axis at the points where x = 4,
3
5''
— -J, — 3. These numbers are the roots of
equation (1).
Second solution. We may find the equation in p-iovm. by applying the results
of § 62. We have
Pi
(4-3+ J-|)=_l.
p, = 4{- 3) + 4 (J) + 4 (- I) + (- 3) (t) + (- 3) (- I) + (t) (- I)
— — 19— 9 — — 5 7.
— 1- 4 — ^ .
p,=- {4(- 3) (I) + 4(- 3) (- 1) + 4(1) (- 2-) + (- 3) (•^) (- %)}
P4 = 4(-3)(i)(--i-) = l|&.
Hence the equation is x* — x^ — -^y x- + | x + J^^s. = o.
Writing this in the a-form, we have
4x4-4x3-57x2 + 9x + 108 = 0.
THEORY OF EQUATIONS 109
2. Solve the equation
2x*-\-7x'' + Ux^ + llx-10 = 0,
given that one root is — 1 -{-2 i.
Solution. Writing the equation in p-form,
x*+ ^, j^ 4- 7 /•■= + yj 1-5 = 0.
By the theorem of § 69 the equation has a root — 1 — 2 i. Since the
equation i.s of the fourth degree, it has four roots. Let tlie unknown roots be
denoted by r and s. The sum of the two known roots, — 1 + 2 i and — 1 — 2 i,
is — 2 and their product is 5. Then, by § 62, we have
the sum of the roots, r + s — 2=— I,
the product of the roots, 5rs =— 5 ;
or r + s =— S (1)
and rs=— 1, (2)
a set of two equations, to be solved for r and s.
From (2),s = Substituting this in (1),
r
_ 1 __ 3
^ r~ V
2r2 + 3r-2 = 0,
(r + 2)(2r— 1) = 0, r=-2 or \;
and substituting these values of r in (2) we obtain
s= \ or — 2.
The other two roots are, then, — 2 and \.
Therefore the four roots of the original equation are
-2, >, -1±2/.
Check. The equation whose roots are — 1 ± 2 iis x^ + 2 j + 5 = 0.
The equation whose roots are — 2, \ is 2x- + 3x — 2 = 0.
(x2 + 2x + 5)(2x2 + 3x-2) = 2x* + 7x3 + 14x- + 11 x- 10 = 0,
which is the given equation.
EXERCISES
Find the equations having the following roots. One method of § 63
may be used to check the other. Draw roughly, without making a
table of values, the graph of the function forming the left member
of each of the equations in exercises 1-15, noting that each real
root of the equation represents a point where the graph crosses or
touches the A' axis. The graph crosses the X axis if the real root is
a single root or multiple root of odd order, and touches it if the root
110 HIGHER ALGEBKA
is a multiple root of even order (see § 60). If tlie coefficient of the
highest power of x is positive, the value of // is positive for large
positive values of x.
1. 1,-2. 13. 1 ± V2, - 2.
2- 2, 3,-4. 14. 2 ± V3, - 2 ± Vs.
3. 2, 1, 1, 0. r-
' ' 15. 0, 0, 0, - 1 ± Vs.
4. + 3, + C. ' ' '
5. 0, 0, 0, 2. 16. 1, ±W2.
6. - 1, - 1, 0, 0, 0. 17. - 1, 1 ± i.
7. 0, 0, 0, 0, - 3. 18. 2 ± /, - 2 ± i
8. 1, 2, + 1 19. 0, 0, ± i, ±2t.
^' S' B5 — 1- -- -, — 1 ± i Vs
10. 2, i,3, 1. 20. 1, 2
11. ±V2, 0, 1. _ 1±/V3 -l±iV3
12. ± V3, ± v:
o.
21.
9
22. Form an ecjuation of the second degree one of whose roots is
l4-2V^r5
3
23. Form an_equation of the third degree two of whose roots are
-l-iV23
"'*' 2
24. Form an equation of the fourth degree two of whose roots
are .; l±I^.
2
25. Form an equation of lowest possible degree with real coeffi-
cients having the two numbers ±1+1 for roots.
26. Solve the equation :r^ — x'^ — 22 x + 40 = 0, given that one
root is double another.
27. Solve the equation a.-" — S.r^ + 11 ./■- + 32.<' - 60 = 0, given
that the sum of two of the roots is 0.
28. Solve the equation x^ — 12.^- + 23 .r + 36 = 0, given that the
roots are in arithmetical progression.
29. Solve the equation Ax^ + 12x'^ — 67 x + 30 = 0, given that
the sum of two of the roots is 3.
30. Solve the equation .^^ — 14a;^ — 31ic — 16 = 0, given that
two of the roots are equal.
THEORY OF EQUATIONS 111
31. Solve the equation ix" + 9^*^ — 30 a; — 8 = 0, given that one
root is the reciprocal of another.
32. Solve the equation x* — 4 ./-^ + 5u-^ + 8 .r — 14 = 0, given tliat
one root is 2 — i Vs.
2
33. Solve the equation (./■ — 4)^ + 2 (cc — 4) = 1, given that
one root is 2 + VS.
34. Solve the equation x* + 12 r' -\- 78 x^ + 252 x + 272 = 0, given
that — 3 + 5 i is a root.
35. Solve the equation x^ — 2 x* — u-" + 2 ./•- + 10 ./• = 0, given that
2 — t is a root.
36. Solve the equation x^ — 12a;^ + 4(5./- — 85 j- -f 50 = 0, given
that two of the roots are 1, 1 + 2 i
37. Solve the equation 2x'^ + 13x-- — 2Gx' — IG = 0, given that
the roots are in geometrical progression.
38. (a) Solve the equation a'* — 6.'-^ + 7:'-"- -|- G« — 8 = 0, given
that the sum of two of the roots is equal to the sum of two others,
and that one root is the negative of another.
(b) Determine another equation of the fourth degree having, like
the above equation, the sum of its roots equal to + 6, their product
equal to — 8, the sum of two of its roots equal to the sum of two
others, and one root the negative of another. What are the roots of
this equation ?
39. What must be the value of /.• if the sum of three of the roots
of the equation x* — 3x^ = kx — 9 is 0 ?
40. Show that an equation Ax* -\- Bx + C = 0, where the coef-
hcients are real, cannot have four real roots unless B and C are
both zero, in which case all the roots are zero.
64. Detection of rational roots. In the following sections we
shall apply ourselves to the problem of finding the numerical values
of the roots of a rational integral e(i[uation with integral coefficients.
The simplest type of root, and the one which is easiest to find,
is the integer. By Corollary I, § G2, any root of an equation in
j9-form with integral eoettieients is a factor of the constant term.
Hence no integers other than such factors need be tried in any
particular case of this kind.
For example, if the constant term in an equation in j>-hn-in with
integral coefficients is 3, the only possible integral roots of the
112
HIGHER ALGEBRA
equation are ± 1 and ± 3. If it is found by synthetic division that
none of these is a root of the equation, we must conclude that the
equation has no integral roots.
The existence of any rational root may be determined by the use
of Corollary II, § 62. For example, consider the equation
6 x-^ - x-2 _ 3 X - 20 = 0.
First obtain the table of values for the function in the left member as
if to plot it.
x
0
1
2
-1
!/
-20
-18
+ 18
-24
From this table it appears that there is a real root between x = 1
and X = 2, since ?/ is negative for x = 1, and positive for a; = 2.
Hence we seek fractions of the form -? of value between 1 and 2,
such that a is a factor of 20, and h is a factor of 6. The factors of
20 are ± 1, ± 2, ± 4, ± 5, ± 10, ± 20. The factors of 6 are ± 1,
± 2, ± 3, ±6. The only fractions which satisfy the conditions of
the problem are | and ^. These values we try by synthetic division.
3_20|f 6-
6-1
+ 8 + -2 8 ^
7 6.
6 + 7 + -y- - igA
1_ 3-20^
+ 10 + 15 + 20
6+ 9 + 12+ .0
Since the fraction | gives the remainder 0, we have shown that
this is a root of the equation. The remaining roots of the equation
may now be found by solving the quadratic equation
6a-2 + 9x + 12 = 0,
formed by setting the quotient of the division equal to 0. They are
- 3 ± V- 23
When the coefficient of the highest power of x in an equation is
equal to unity, the denominator of any fractional root must have
the value 1, since this is the only integer which is a factor of that
coefficient. This is equivalent to the statement that if an equation
in ^^-form with integral coefficients has any rational roots, they
must be integral. From this point of view Corollary I, § 62, is a
particular case of Corollary II.
TllEOKY OF EQUATIONS
113
EXAMPLE
Solve the equation
2x* - x^ - r> x^ -(- 7 x - 6 = 0.
Solution. First obtain a table of values for the function.
X
- 2
- 1
0
1
2
y
0
- 15
- f)
-3
12
2-1-5+7- G[2
+ 4 + 6 + 2+18
2 + 3 + 1 + '.> + 12
2- 1 - 5 + 7 - (i|- 2
- 4 + 10 - 10 + ()
2 _ 5 + 5 - 3 + 0
The values of the function when x = 0, 1, and — 1 are obtaiued by inspection
on substituting these values of x in the function. The values of the function
■when X = 2 and — 2 are obtained by synthetic division. When we try 2 by
synthetic division we notice that the numbers in the last line are all positive.
Hence the equation caiuKit have a root larger tlian 2 (§ 57), and it is unnecessary
to try larger numbers.
When we try — 2 by synthetic division we notice that the luimbers in the
last line alternate in sign. Hence the equation cannot have a I'oot smaller than
— 2, and it is imnecessary to try smaller muubcrs. This division also shows that
— 2 is a root.
The remaining roots of the original equation must be the roots of the reduced
equation formed by setting the quotient of the division by x + 2 equal to zero ;
namely,
2x3-5x2 + 5x- 3 = 0.
This equation nmst have a root between 1 and 2, since we see from the table
that the original equation has a root between 1 and 2, as the function is neg-
ative when X = 1 and positive when x = 2. If this root between 1 and 2 is
rational, it nuist be a fraction whose inuncratdr is a factor of 3 and whose
denominator is a factor of 2. The only possibility is ■^. We try this value by
synthetic division.
2-5 + 5-3U
+ 3-3 + 3 '
2-2+2+0
Hence iy is a root. The quotient of this ilivision, set equal to zero, is a (juad-
ratic equation which gives the remaining roots
2x2 -2j + 2 = 0,
x^ — X + 1 = 0,
1 ± V^
Therefore the roots of the original cciuation are — 2, -,
3 i± V_3
6.
16x^-5x-~3 = 0.
7.
2x'-T x^-x'-\-21 .X - 15 = 0.
8.
,,.i 4- 9 ,,.3 + 5 a.2_ 23 :r + H = 0.
9.
2.r* + 2a-^-a-2 + l = 0.
10.
10x^-21 a;^-21cc-10 = 0.
114 HIGHER ALGEBRA
EXERCISES
Solve the following equations :
1. 4:X^-8x^-x-{-2 = 0.
2. 3x^ + 13:r + llx-U = 0.
3. 4x' + 3ic2- 20a; -15 = 0.
4. 27 cr'' 4- 63 a-'^ + 30a; - 8 = 0.
5. 2 a-^ - 15 ,r- + 46 a; - 42 = 0.
11. 4 a'" - 23 a-' + 15 .r + 9 = 0.
12. 9 a-* + 15 a;8 - 143 a-'- + 41 a; + 30 = 0.
13. a;5-10a;- + 15a--6 = 0.
14. 12a-^ + 4a-* - 17a;'' + 7a- - 2 = 0.
15. a-^ — a-* + a;^ — a;^ + a; — 1 = 0.
16. Show that the equation 3a-^ + x — 1 = 0 has no rational roots.
Determine how many real roots the equation has.
17. Find a rational root of the equation 2 a;* — 13 a-' + 16 .r^
— 9 a; + 20 = 0. Show that the equation has only one rational root.
18. Show that the equation 3 a;* — 2 x^ — 21 a;^ — 4 a; + 11 = 0 has
four real irrational roots.
19. A beam of span I is fixed at one end and rests on a support
at the other end. The distance of the load P from the supported
end being Id, where k is jDOsitive and p
less than 1, the position of P which '^
gives the maximum positive moment is * '''
given by the equation 2A;^ — 3A; + 1 = 0. Show that for the maximum
positive moment the load is at a distance .366 1 from the supported end.
Sdggestion. In the three following exercises take tt = ^^-.
20. A rectangle whose perimeter is 34 inches is rotated about a
line joining the mid-points of two opposite sides. If the volume of
the cylinder generated is 550 cubic inches, find the lengths of the
sides of the rectangle.
21. The altitude of a cone exceeds the radius of the base by
2 inches and its volume is 462 cubic inches. Find the altitude of
the cone and the radius of the base.
22. The sum of the radius of the base and the altitude of a right
circular cone is 10 inches and its volume is 66 cubic inches. Find
the altitude and the radius of the base.
TPIEORY OF EQUATIONS 115
23. A spherical shell an inch thick, whose outer diameter is
12 inches, is equal in volume to the sum of two spheres whose
radii differ by 1 inch. Find the radii of the spheres.
24. It is desired to double the capacity of a tank G x 8 x 10 feet
by making equal elongations of its dimensions. Find the elongations.
25. The volume of a rectangular parallelejnped is 60 cubic feet.
Its total surface is 94 square feet and the total length of its edges
is 48 feet. Form the equation whose roots are the dimensions of the
parallelepiped and find these dimensions.
65. Multiplication of the roots of an equation by a constant.
Suppose we have given the equation
/(a-) = a^x" + fl j.r" - 1 +...+.,„ = 0. (.1)
Oall its roots x^, x^, • • -, x„. It is required to find an equation whose
roots are equal to these numbers each multiplied by k ; that is, we
seek the equation whose roots are
/.-o-j, kx^, • • •, Aa-„. (1)
Consider the equation
/©=".(t)'-'.(0"'- — «. <^)
where x' is the variable. We shall show that this equation is satis-
fied by the numbers (1). Replacing x' by any one of the numbers (1),
say by kx^, the polynomial in the left member becomes
/(f) =/w.
But f{x^ = 0, since x^ is a root of the equation f(x) = 0. Hence
equation (2) is satisfied by the numbers (1).
If we remove the parentheses in (2), multiply through by k", and
drop the primes, we have
a^x" + ap-"-' + a Jr.,-" - ^' + . . . -f k"a„ = 0, (3)
which is the equation sought, having roots each /.• times the roots of
equation (A).* We may express the result in the following
Rule. To find an equation tvJiose roots are equal to the roots of
(^) each mnJtipJied by the constant k, midtipli/ the terms of (.1),
hefihinin;! with the second. In k. A"", ■ ■ ■, k" respective! i/.
* It should 1)0 kept in inin<i that a given equation has the same roots whether the
variable is called x or x\
116 HIGHER ALGEBRA
Tlie special case of this rule for the value k = — l may be expressed
as follows ;
To find an equation in general form whose roots are the negatives
of the roots of a given equation, change the signs of alternate terms.
Care must be taken in applying the above principles to supply missing terms
by zeros.
66. Descartes's rule of signs. A pair of successive like signs in
a polynomial is called a continuation of sign. A pair of successive
unlike signs is called a change of sign.
In tlie polynomial f{x) = 2 x« - 3 a"" + 2 x2 + 2 x - 3 (1)
are one continuation of sign and three changes of sign. This may be seen more
clearly by writing merely the signs, -i h H •
Let us now determine the effect on the number of changes of sign
in a polynomial if it is multiplied by a factor of the form x — a
where a is positive ; that is, where the number of positive roots of
the equation f(x) = 0 is increased by one.
Illustration. Let us multiply (1) by x — 2. "We have then
2x-'-3x3 + 2x2 + 2x -3
X -2
2 x5 - 3 x4 + 2 x3 + 2 x2 - 3 X
-4x^ + 6j»-4x2-4x + 6
2x5- 7j;4 ^ 8x3-2x2- 7x + 6
In this expression the succession of signs is H 1 F , in which tliere
are four changes of sign; that is, one more change of sign than in (1). If an
increase in the number of positive roots always brings about at least an equal
increase in the number of changes of sign, then an equation in general form
cannot have more positive roots than there are changes of sign in its left
member. This is the fact, as we now prove.
Descartes's rule of signs. An equation in general form,
f(x) = 0, has no 7nore real positive roots than f(pr^ has changes
of sign.
We shall show that if we multiply each member of an equation of
degree n by x — a, where a is positive, thus forming an equation
of degree n -\-l, the number of changes of sign in the new equation
always exceeds the number of changes of sign in the original equa-
tion by at least one ; that is, the number of changes of sign increases
THEORY OF EQUATIONS
117
at least as rapidly as the increase in tin; number of positive roots
when such a multiplication is made.
Let f(x) = 0 represent any particular equation of the nth degree.
The first sign of /(a-) may always be taken as +• The remaining
signs occur in successive groups of + or — signs which may contain
only one sign each. If any term is lacking, its sign is taken to be
the same as one of the adjacent signs. Thus the way in which the
signs of /(.f) niay occur is rc])resented iu the following table, in
which the dots represent an indefinite number of signs. The multi-
plication of ./'(■'■) hy a; — a is represented schematically, only the
signs being given.
All + signs
All-
signs
A11 +
signs
All — signs
Further
groups
All-
-signs
/(x), + ■■■■ +
_ . .
. . _
+ ••
•• +
_ _
+ ■••• +
_ . .
. . _
X — a,
+
—
xf(x), ++•■• +
. . . —
+ +
•• +
... —
+ +•■• +
.
. . —
- af{x),
- +
••• +
+ -
. . _
-+ ••• +
+
-+ •
••+ +
{x-a)f{x), +±...±
-±
... -t
+ ±
••±
-± •••±
+ ±---±
-± •
•• ± +
The ± sign indicates that either the + or the — sign may occur
according to the values of the coefficients and of a. The vertical lines
denote where changes of sign occur iu/(a-). Assuming that all the
ambiguous signs are taken so as to afford the least possible number
of changes of sign, even then in (x — a)f(x) there is a change of sign
at or before each of the vertical lines, and in addition, one to the
right of all the vertical lines. Hence as we increase the number of
positive roots by one, the number of changes of sign increases at
least by one, perhaps by more.
The only possible variation which could occur in tlie succession of
groups of signs in f(x), namely, where the last group consists of
+ signs, does not alter the validity of the theorem.
Illustration. Let /(x) = x^ — 4x8 — x + 2, and let a = 2.
f{x\
x-2,
xf{x\
- 2./-{j),
(x-2)/(x),
1+0-4-0-1+2
1-2
1+0-4-0-1+2
-2-0+8+0+2-4
1
2-4+8-1+4-4
2 changes
5 changes
118
HIGHER ALGEBRA
Since /(— .r) = 0 lias roots opposite in sign to those of f(x) = 0
(§ 65), we can state
Descartes's rule of signs for negative roots. TJie
general equation fQx) = 0 has no more negative roots than there
are changes in sign in /'(— a').
If by Descartes's rule it appears that there cannot be more than
a positive roots and b negative roots, and it a + h < n, where n is
the degree of the equation, then there must be complex roots, at
least n — (a + h) in number.
In applying Descartes's rule no signs need be supplied for the missing terms.
EXAMPLES
Obtain all the information possible concerning the roots of each
of the following equations by the use of Descartes's rule and by
inspection of the constant term.
1. a'^ + 3a'- + l = 0.
Solution, /(j;) : + + + , no change ; therefore no positive root.
/(— x) : h + , one change ; therefore not more than one negative root.
Since the equation has three roots, it has one negative and two complex
roots.
X"
ic + 1 = 0.
Solution. f(x) : H J- , two changes ; therefore not more than two positive
roots.
/(— x): 1- +, one change ; therefore not more than one negative root.
There are five roots in all and there must be an even number of complex
roots. Hence there are three possibilities which may be represented by the
following table :
+
—
comp.
2
1
0
1
0
1
2
4
4
Of course only one of these combinations actually occurs, but Descartes's
rule does not tell us which one.
Since, however, the constant term, with its sign changed, equals the product
of the roots, and since the product of conjugate complex numbers is always
positive, the second combination cannot occur ; that is, the equation surely has
a negative root.
THEORY OF EQUATIOXS 119
EXERCISES
Obtain all tlu; information possible concerning the roots of each
of the following equations by the use of Descartes's rule and by
insi)ection of the constant term :
1. x'-+'3x- + 2 = 0. 8. x^-:f:*-2x^ + 3j- + 2x + l = 0.
10. n.r'5-3a;-l = 0.
NoTi;. Ill the following exercises n is to
4 J-'' 4- Q y- 4-3^0 ^^ regarded as a positive integer.
11. .--" + 1 = 0.
12. .1-2" -1 = 0.
6. a^^ - 1 = 0. j3_ ,^.2„ + i + 1^0.
7. a;« + 1 = 0. 14. a--''+i -1 = 0.
5. Sx"- X- + 2 X- - 1 = 0.
15. Find the equation whose roots are twice the roots of the
equation a;'' — 2 a;'' — 3 a: + 1 = 0.
16. Find the equation whose roots are one third the roots of the
equation 2 a;^ — 6 x^ + 3 = 0.
17. Find the equation whose roots are equal to the roots of the
equation x^ -\- 2 x'^ — 8 a- + 8 = 0, each multiplied by — f .
18. Find the equation whose roots are four times the roots of the
equation x^ — 2 x* -\- ^^_^ x — ^^^^ = 0.
19. Form the equations whose roots are the negatives of the roots
of the equations in the four preceding exercises.
20. Transform the equation x* — ^x^-\- i x'- + 2 .r — 1 = 0 by multi-
plying the roots by the smallest number which wall make the coeffi-
cients of the transformed equation integers, and the coefficient of
the first term unity. Solve the transformed equation, and hence
obtain the roots of the original equation.
67. Diminution of the roots of an equation. Before proceeding
with the determination of the irrational roots of an equation it is
necessary to show how to form an equation whose roots differ from
the roots of a given equation by a constant.
Sui)pose the general equation
/(a-) = a^x" + a^x" -' + ■■■ + a„ = 0, (A)
120 HIGHER ALGEBRA
with the roots a-^, a-^, • ■ -, a:*„, is given, and it is required to find an
equation whose roots are less than these numbers by the constant a ;
that is, we seek an equation which is satisfied by the numbers
^'i - *' ^2 - «' • ■ •' ^'n - «• (1)
Let X =x' + a and consider the equation
f(x' + a) = a^ (x' 4- «)" + a^ (x' + a)" -' + ...+ r/, = 0, (2)
where x' is the variable. We shall show that this equation is satisfied
by the numbers (1). Replacing x' by any of the numbers (1), say,
iCj — a, the polynomial in the left member of (2) becomes
But /(^i) = 0, since x^ is a root of the equation f(x) = 0. Hence
a-j — a is a root of equation (2). Similarly, all the numbers (1) are
roots of equation (2).
To express equation (2) in general form, it is only necessary to
remove the parentheses and collect powers of x'. The result may be
written as follows :
F(x') = A^x"' + A^x'"-'^-{ h A,, = 0, (3)
where the .4's are the coefficients which we obtain by collecting like
powers of x'. Since the coefficients in this function are different from
those in (A), we denote it by a different symbol, F(x').
niustration. Consider the equation
f(x) = x^-6x'^ + nx-6 = 0, (4)
whose roots are 1, 2, and 3. Let us find the equation whose roots are less by 2
than those of (4). Let x = x' + 2, and form the equation f(x' + 2) = 0. We
obtam j.^^, + 2) = (x' + 2)3-6 (x' + 2)^ + 11 (x' + 2) - 6 = 0. (5)
Simplifying (5), we get ^i^') = ^'^ — ^' — 0-
We see that the roots of this equation, — 1, 0, and 1, are less by 2 than the
roots of equation (4), namely, 1, 2, and 3.
We will now derive a method of obtaining the coefiicients of (3)
more rapidly than they can be computed by expanding the binomials
in (2). It must be kept in mind that x^ -\- a and x are merely different
symbols for the same thing ; that is,
X — x' -i- a, or x' = X — a, (6)
THEORY OF EQUATIONS 121
and we may at any time use the notation wliidi is uiost convenient
for us. Since (2) and (3) are identical, we have
F{x')=f{x' + a)=f{x).
We wish to compute the coefficients in tin- expression
F{x') = A^x"' + A^x'" -• + ... + .-!,,. (7)
If we divide the right member of (7) by x', we obtain A^ as the
remainder. But since p/^i\—.f/^\
and x' = X — <i,
the result of dividing Fix') by x' is the same as that of dividing
f(x) by a; — a. Since f{x) is given, we can readily divide it by a; — a
by the synthetic method, and in this manner find the numerical value
of .!„.
The quotient of dividing (7) by x' is A^x'''-'^ + A^x"*-"^ -\ h ^<„_ i-
If we divide this quotient by x', we obtain the coefficient /l„_i as a
remainder. But this division is precisely equivalent to dividing the
quotient of ^ ^ by x — a. Proceeding similarly, we may obtain
in order A,^_.,, • ■ ■, A^. This method for computing the coefficients of
the equation whose roots are less than those of f{x) = 0 by the
constant a we may express by the following
Rule. The constant term of the new equation is the remainder
after dividhuj f(x) hy x — a.
The coefficient of x' in the new equation is the remainder after
dividing the quotient just obtained hy x— a.
The coefficients of the higher powers of x' are the remainders after
dividing the successive quotients obtained by x — a.
Illustration. Let us compute by this method the coefficients of the equation
whose roots are less by 2 than those of (4). We lirst divide by x — 2 syn-
thetically. l_0 + n-0[2
+ 2- 8 + 6
1_4+ 3+0
Hence 0 is the value of the constant term in the new equation. By § 55 the
coefficients of the quotient in this division are the numbers in the last line of
tlio division up to the remainder. Dividini,' tliis (piotient by j — 2 synthetically,
we obtain — 1 as the next to the last coefficient of the new equation.
l-4 + 3[2
+ 2-4
1-2-1
122 HIGHER ALGEBRA
The coefficients of the new quotient are 1 and — 2, anel, performing the next
division, we get 1 — 212
+ 2~
1 + 0
This process may be arranged more compactly as follows, where the full-
faced type shows the coefficients of the transformed equation :
1_6 +11 -6[2
+ 2 - 8+6
1-4 + 3| + 0
+ 2 - 4
1-2
+ 2
- 1
1 + 0
Hence the new equation is x'^ — x' = 0.
We may now drop the primes and write the new equation,
x^-x = 0.
68. Graphical meaning of the transformation. If an equation
has each of its roots decreased by the positive number a, then the
graph of the function in tlie new equation will cross the A' axis a
units farther to the left than the graph of the old one. In fact, the
new graph is just the same as the old one, except that its position
is a units to the left. This is expressed by the relation x' — x — a,
which indicates that the abscissas for points on the new curve are
each a units shorter than those for the corresponding points on
the old one.
By means of this transformation we may bring any crossing of a
graph within one unit of the origin. This corresponds to decreasing
the roots of the original equation by a number such that one of the
roots of the new equation falls between 0 and 1.
Decreasing the roots by a negative number is equivalent to increas-
ing them and to moving the graph to the right.
EXERCISES
1. Transform the equation x^ — 4:X^-\-x-\-6 — 0 into an equa-
tion whose roots are less by 2 than the roots of the given equation.
Plot the function forming the left member of each equation.
2. Transform the equation x* -\- x^ — 5 x'~ + 3.t = 0 into an equa-
tion whose roots are greater by 1 than the roots of the given equation.
Plot the function forming the left member of each equation.
THEOUV UF EQUATIONS
128
1.5.
o
Transform each of the lullow iii.i;- •■(luat ions into one whose roots are
less by the numhcr opposite than the roots of tlie given equation:
3. X-'* - 15 x^ +7^-1- 1-5 = 0, 5.
4. a-''-2.<'^4-l = 0, .2.
5. X* + 63" + 10a;'' + .r - 1 = 0, - 1.
6. 2x^- 5 x' + x + 2 = 0,
7. ]()a''-13./- + 9 = 0,
8. ;/••'- 1.5/- + 2.'- -2.5 = 0,
9. Transform tlic tMiuation 36 a-" — 108./-- + 107 ./■ - 35 = 0 into
an equation whose roots are less by 1 than the roots of the given
ecjuation. Solve the transformed equation and thus determine the
roots of the given equation.
10. Transform the equation 16 x* — 72 x^ — 61 a- — 15 = 0 into an
equation whose roots are greater by .5 tlian the roots of the given
equation. Solve the transformed equation and thus determine the
roots of the given equation.
11. How mnch must the roots of the equation x* — 8x^-{-9x^
_^ 38a; _ 40 = 0 be diminished in order that the sum of the roots of
the transformed iMiuation shall be 0 ? Find the transformed equation.
Hint. Tlu' sum of the roots must be dimiuislied by 8.
12. IIow much must the roots of the equation /* + 4a;^ — 3a; +7 = 0
be diminished in order that the coefficient of x in the transformed equa-
tion shall be 0 ? Find the transformed equation.
Hint. Decrease the roots by h, and determine h so that the coefficient of x'
is zero.
69. Location principle. If in plotting a function i/=f(a-) the
value x = a gives the corresponding value of // positive and equal to
c, while the value x = h gives the corre-
sponding value of // negative, say, equal to
— d, then the i)oint (a, c) on the curve is
above the X axis, and the point (b, —d) on
the curve is below the A' axis. If the curve
is unbroken, it must then cross the X axis
at least once between the values x = a and
x^=b, and hence the tMpiation _/'(^.r) = 0 must
have a root between these values of x. The
shorter we can deteiniine this interval between a and b, the more
accm-ately we can find the root of the equation. Horner's Method of
y
~f"1
1
1
c
\
I
^
\f
0
«-a->i >s^
k.^
< —
-b ->
(P,-d)
124
HIGHER ALGEBRA
approximation, which we shall explain in the next section, is noth-
ing but an ingenious process for making the interval in which we
know a root must exist as small as we wish. We have throughout
this text assumed the property of unbrokenness or continuity of the
graph of ?/ = a^x"^ + o.^.x;" ~^ -\- • • • -j- a^.
This geometric assumption may be expressed in algebraic language
in the following
Location principle. When for two real unequal values of x,
say, x = a and x = b, the values of y =/(x) have opposite signs, the
equation f Qx^ = 0 has a 7'eal root between a and b.
The interval between x — a and x = h we shall call the location
interval.
70. Horner's Method of approximating irrational roots. We are
now in a position to determine the real roots of an equation to any
desired degree of accuracy.
It is assumed that all rational roots have been found by the methods
of § 64, and that all the roots which remain are either irrational or
complex.
Consider the equation
tc^ + 3 ct- - 20 = 0.
(1)
Let us find its real roots to two decimal
places.
First form a table of values for the
function y = x^ + 3 x — 20, and plot the
function.
,/'
0
1
•>
o
- 1
//
- 20
-16
-6
+ 16
-24
•
Y
'
j
/
/
/
O
/
X
1
/
/
/
/
f
By Descartes's rule it appears that (1)
has no negative root. By the location
principle it appears that there is a root between 2 and 3. The
whole point of Horner's Method consists in decreasing the roots
of the successive equations which we meet in the course of the
process by the lesser of the two numbers which bound the location
interval.
TirEOUY OF EQUATIONS 125
Here we decrease the roots of equation (1) by 2, as follows ;
1 + 0 + 3 - 20 [2
+ 2 + 4+14
1 + 2 + 7
- 6
+ 2+8
1 + 4 +15
+ 2
1 + 6
Tlu! resulting equation is
x^+6x'
+ 15 a; - 6 = 0
(2)
We know that equation (2) lias a root between 0 and 1, since
equation (1) has a root between 2 and 3. From the graph we can
estimate in tenths the position of this root. Ilav^ing made an esti-
mate, say .3, it is necessary to verify it and to determine by synthetic
division precisely between which tenths the root lies. Thus, trying
.3, we obtain i + 6.0 + 15.00 - 6.00 [^
+ 0.3 + 1.89 + 5.07
1 + 6.3 + 16.80 - 0.93
In the computation by Horner's IMetliod it is usually unnecessary to preserve
more decimal places than are called for in the root which is sought. That is,
we avoid carrying through the process figures which have no effect on the
result. In the present example we shall write down the decimals correct to
two places. For instance, in the last multiplication by .3 above, the product
is 5.0G7, but we write only 5.07, the approximate value to two decimal places.
When the figure in the tliird decimal place is 5 or more, we add 1 to the figure
in the second decimal place ; when it is less than 5, we drop it. If three
decimal places had been required in the root, we would have preserved the
decimals in the computation correct to three places. This method of shorten-
ing the computation is suiEciently accurate except in rare cases, where the
remainder by synthetic division is so near 0 that its sign would be changed
if all the figures of the decimals were retained. In such a case we would
perform the synthetic division retaining all figures of the decimals.
Since the renuiinder in this synthetic division is negative, it
appears that for x = .3 the curve is below the X axis, and that the
root is greater than .3. r)nt we are not justitifd in assuming that
the root is between .3 and .4 until we have substituted .4 for x. This
we proceed to do. i + 6.0 + 15.00 - 6.00 [A
+ 0.4 + 2.56 + 7.02
1 + 6.4 + K.nc. + 1.02
126 HIGHER ALGEBRA
Since the remainder is positive for x = A, the location principle
shows that (2) has a root between .3 and .4 ; that is, (1) has a root
between 2.3 and 2.4.
To find the root to two decimal places, decrease the roots of (2)
by .3, the lesser of the two numbers between which tlie root of (2) is
now known to lie. The new equation has a root between 0 and .1.
This process is performed as follows :
1 + 6.0 +15.00 -6.00[^
+ 0.3 + 1.89 + 5.07
1 + 0.3 + 16.89
+ 0.3 + 1.98
-0.93
1 + 6.6
+ 0.3
+ 18.87
1 + 6.9
Thus the new equation is
x' + 6.9 x^ + 18.87 X - .93 = 0. (3)
This equation has a root between 0 and .1. We can find an
approximate value of the hundredths' place of the root by solving
the linear equation 18.87 x — .95 — 0, obtained from (3) by dropping
all but the term in x and the constant term.
93
Thus X = -^ = .04.
This suggestion must be verified by synthetic division to deter-
mine between what hundredths a root of (3) actually lies.
1 + 6.90 + 18.87 - 0.93 |.04
+ 0.04+ 0.28 + 0.77
1 + 6.94 + 19.15 - 0.16
Thus the curve is below the X axis at x = .04 and hence the root
is greater than .04. We must not assume that the root is between
.04 and .05 without determining that the curve is above the X axis
at a; = .05. ^ _^ 6 90 + 18.87 - 0.93 1.05
+ 0.05 + 0.35 + 0.96
1 + 6.95 + 19.22 + 0.03
Thus the curve is above the A' axis at a- = .05. B}- the location
principle, (3) has a root between .04 and .05 ; that is, (1) has a root
between 2.34 and 2.35. Hence the root to two decimal places is 2.34.
TIIEOKV OF KQIATIOXS
12
>T
Tlic ])rece(liii<,' (•(unpuUitioii may be anungod more comijactly
as follows :
1 + 0 + 3 - 20 [2
+ 2 + 4+14
1 + 2 + 7
+ 2+8
- 6
1 + 4
+ 2
+ 15
1 + 6.0+ ir).00- 6.00 1^
.4 + 'J.r>(\ + 7.02
1 + 6.4 +17.56 + 1.02
.93
X =
18.87
.04.
1 + 6.0 +15.00 - 6.00[^
+ .3 + 1.89 + 5.07
1 + 6.3 + 16.89
+ .3 + 1.98
- .93
1 + ij.(\
.3
+ 18.87
1 + 6.9
[ioot = 2
+ 18.87
.34+.
-.93
1 + 6.90 + 18.87 -.93 [.04
+ .04 + .28 + .77
1 + 6.94 + 19.15 -.16
1 + 6.90 + 18.87 -.93|.05
+ .05 + .35 + .96
1 + 6.95 + 19.22 +.03
The foregoing i)rocess affords the following
Rule. Obtain all possible iyifoj'mation about the roots by Descartes' 8
rule.
Plot the function. Apply t/ic location principle to determine
between what consecutive positive inteyral values a root lies.
Decrease the roots of the equation by the lesser of the two numbers
which bound a location interval.
Estimate from the plot the nearest tenth to which the desired root
of the new equation lies, and determine by synthetic division precisely
the successive tenths between ivhich the root lies.
Decrease the roots of the 7U'W equation by tlic lesser of the tivo
numbers which bound its tenths' location interval, and estimate the
root of the residting equation to the nearest hundredth by solving
the linear equation formed by dropping all hut the last two terms
of the equation.
Determine precisely by synthetic divisio)i the hundredths^ location
interval.
Proceed similarly to find the root to as many places as may be
desired.
128 HIGHER ALGEBKA
The sum of the integral, tenths, and hundredths values obtained
in the foregoing process is the approximate value of the root.
To find the negative roots of an equation f(x) ~ 0, determine the
positive roots of /(— x) = 0 and change their signs.
When all the roots of an equation in the ^-form are real, a check
to the accuracy of the comj)utation may be found by adding the
roots together. The result should be the coefficient of the second
term with its sign changed.
Sometimes an equation lias roots so nearly equal that the table of values
formed for integral values of x gives no information as to whether there
are roots between two consecutive integers or not. For instance, the table of
values for the equation x^ + ITx^ — 46x + 29 = 0 does not tell us whether the
equation has three real roots or only one. In such a case we might form a table,
using values of x differing by .1 or .01, and thus locate the root between two
successive tenths or hundredths. But since such equations occur very rarely in
practice, and since the calculus affords a very simple method of determining a
number between the roots if they are real and distinct, the complete discussion
of this case will not be given here.*
EXERCISES
1. Find to two decimal places a positive root of x^-{-3x^ —
2 cc - 1 = 0.
2. Find to two decimal places a positive root of x^—6x^-{-
10.r- 9 = 0.
3. Find to two decimal places all the real roots of 2x* — Ax^ +
3 a-2 - 1 = 0.
4. Find to two decimal places all the real roots of .7'^+4.7--— 7 = 0.
5. Find to tAvo decimal places all the real roots of x* — ix^ +
14:X--4x-U7 = 0.
6. Find to three decimal places a positive root of x^—9x^-\-
25 a; -18 = 0.
7. Find to three decimal places a positive root of x^ — 2 x' -{■
2 a; -101 = 0.
8. Find to three decimal places all the real roots of Sx^ — 5x^=Sl.
9. Find exactly a real root of 4a-^ + 23 a-- — a* — 377 = 0, and
show that the other roots are imaginary.
* See Hawkes, Advanced Algebra, p. 200.
TIIEUKY' OF EQUATIONS 129
10. Sliow that tlie equation ./•' — 7.t + 7 = 0 has two roots l>e-
tween 1 and 2 and one negative root. Find each of the roots to three
decimal places.
11. Solve exercise 11, p. 96, getting the result to two decimal places.
12. Solve exercise 13, p. 9G, getting the result to two decimal places.
13. If a wooden simple beam x inches square and of 12 feet span
carries a load of 300 pounds at the middle when it is also subject to
a longitudinal tension of 2000 pounds, the allowable tensile strength
being 1000 pounds per square inch, the safe value of x is given by
the equation x^ — 2x— 64.8 = 0. Find to two decimal places the
size of the beam.
14. A beam of span I, fixed at one end and resting on a support
at the other end, is subjected to a uniformly distributed load. The
point of maximum deflection for a safe load is given by the equation
8 x^ — 9 /a;' + /^ = 0, where x is the distance from the supported end.
Show that a- = .4215 I.
15. A hollow cylindrical shaft 17 inches in outside, and 11 inches
in inside, diameter is to be coupled by 12 bolts placed with their
centers 20 inches from the axis. The proper
diameter of the bolts is given by the e(]ua- *
tion d' + 3200 (/■-- 337.6 c/- 13,500 = 0. Find ^ '^ t,
the diameter of the bolts to one decimal place.
16. A wooden column x inches square and 12 feet long, having
fixed ends, is to carry an axial load of 50 net tons with a factor of
safety of 10. The size of the column is given by the equation
a-'' — 125 ur = 10,368. Find the size of the column to one decimal
place.
17. If the column in the preceding exercise has round ends, its
size is given by the equation x* — 125 x^ = 41,472. Find the size of
the column to one decimal place.
18. In exercise 16, if the eccentricity of the load is 2.5 inches, the
size of the column is given by the equation x* — V2ox^ — 1875 a; —
10,368 = 0. Find the size of the column to one decimal place.
19. The Gas E(juatiun of \'an (.lor Waals is (p + —A i^r — b) = 1,
where v is the volume of the gas, j) tlie ])ressure, and a and f> are
^
130 HIGHER ALGEBRA
constants depending on the gas. For carbonic acid gas a = .00874 and
i = .0023. Find the value of v to two decimal places whenp =1.
Hint. Reduce the equation to one of third degree in v with numerical coeffi-
cients and multiply the roots by 10 before solving.
20. The following equation occurs in the theory of chemical
actions*: \x^ -\- ^x^ -\- %x^ —I.IQQ. Eind the value of x to two
decimal places.
21. The cubical coefficient of thermal expansion of paraffin is
.000584 per degree centigrade. If t be the temperature on the cen-
tigrade scale, the linear coefficient of expansion of the j)araffin is to
be found from 3 ^^^ ^ ^..^ ^ 3 ^^3^2 ^ .000584.
Find to two decimal places the linear coefficient of expansion a
at 30° C.
22. An empirical formula for the volume of one gram of water
at temperature t degrees centigrade is
V = 1 - .00009417 t + .000001449 1' + .0000005985 t\
where v is the volume in cubic centimeters. Find correct to tenths
of a degree the temperature at which the volume of 1 gram of water
will l>e 1.0002 cubic centimeters.
71. Solution of the cubic. In § 19 exact expressions for the roots
of the quadratic equation ax^ -\- hx -\- c = 0 were found in terms of
its coefficients, a, 1>, and c. The only process of approximation needed
in order to find an irrational root of such an equation is that of
extracting the square root. Hence it is never necessary to use
Horner's Method to find the roots of a quadratic equation.
In this section we shall see that it is possible to find an exact
expression for the roots of the cubic equation in terms of its coeffi-
cients, and that the formulas obtained may be used in certain cases
as a substitute for Horner's Method. A similar but more laborious
solution for the equation of the fourth degree exists, but will not
be given here. It is, however, impossible to obtain any general
solution of equations of higher degree than the fourth by means
of algebraic operations.
The cubic equation in the ^^-f orm is
x^Jr]\x^+T>.4c+i\ = 0. (1)
* J. L. R. Morgan, Elements of Physical Chemistry, 4th ed., i>. 506.
THEORY OF EQUATIONS 131
If we increase each of the roots of this equation by 7-', iu order
to remove the term in x'", we obtain the equation
lacking- the term in a;^, and wliich we may write in the form
a.«+^,.r + y = 0, (3)
^ OS
where P=p.2-Y, and q = -^-J-^+p^. (4)
The cubic in I'orm (3) we can solve as follows : Let
x=// + z. (5)
This amounts to replacing the single varial)le x by the two vari-
ables y and z. We deliberately complicate the problem in this way,
because we shall obtain a relation between y and z which will enable
us to find the values of the two more easily than we could determine
the value of x alone.
Substituting (5) in (3),
(>/ + ^f+J>(!/ + -")+'/ = y' + ^' + (3 2/^ -i-p)(y -h!^)+q = 0. (6)
Having introduced an extra variable, we are at liberty to impose
a condition on 7/ and z.
Let 3 7/,-+y- = 0, or yz = ^- (7)
Then (<")) will reduce to 7/ + z^ = — q
and from (7) we have ?/-.'^ = — ^•
By reason of the relations between the roots and the coefficients of
a quadratic equation (§ 24), it appea'rs that y^ and ?} must be roots of
the (piailratic equation whose coefficients are 1, y, and — ^; that is,
of the equation >,»
''+'/'-^ = 0. (8)
Solving (8), Ave find
f
= -2 + ^4+27-^'^' ' =~2-N4+27-''-
Hence y = ^/J , w "v^, w- V!T ; z = -VT:, w ^v^, <d^ -Vb, Avhere w
represents one of the complex cube roots of 1 (exercise 27, p. 8(5).
The roots of (3) would at first siglit seem to be nine in number,
namely, the values which wo olitain l)y adding each of the three
values of y to each of those of z. But reference to (7) reminds us
132 HIGHEK ALGEBRA
that the product yz must be real ; hence all of these nine values
are ruled out excejit the following, which are the roots of (3) :
These expressions are called Cardan's Formulas for the solution
of the cubic. „ ,
When the value of — + — is positive we can extract its square
root and compute the values of A and B readily. In this case we
find but one real root, the other two being complex. If the value of
2 3
4" + Tir is negative, in which case there are three real roots, it is
necessary, in order to solve the equation by this method, to extract
the cube root of a complex number. This is more laborious than it
is to find the roots by Horner's Method. Consequently, we shall use
these formulas only when -r + 77= > 0. In this case there is only
one real root.
If an equation of the third degree in p-fonn has three real roots and they
have been found by Horner's Method, we have seen that the work may be
checked by adding them together. The result should be the coefficient of the
term in x^ with its sign changed. If tlie equation has only one real root, this
clieck is not available, and the result of Horner's Method may be checked by
the method of this section.
The foregoing method of solving the cubic is due to the Italian, Tartaglia, but
was first published by Cardan in 154.5. At this time the operations with com-
plex numbers were imperfectly understood, and every effort was made by mathe-
maticians to avoid them. It must have been not a little irritating for the early
algebraists to realize that the only case in which Cardan's Formulas solve a cubic
without extracting a root of a comiDlex number is that in which tlie equation has
a pair of complex roots. To find tlie tliree real roots of a cubic when they are irra-
tional, the case in which they were chiefly interested, the cube roots of complex
numbers are necessary.
EXERCISES
Solve the following cubics, using Cardan's Formulas :
1. ic^ -f 3 a;2 -f 3 .i' + 2 = 0. 3.4 x^ + 2 x- -1 = 0.
2. cc'^ - 11 X + 20 = 0. 4. 2 ic' - 9 x-- + 2 a- + 30 = 0.
Find to two decimal places the real root of each of the following
cubics, using the tables (pp. 215-217) to evaluate the radicals :
5. a.3 + 3a; - 20 = 0. 8. x^ -f- ^x' -f 5a; - 17 = 0.
6. ^3 _ ^ _ 33 ^ 0. 9.2 x^ + 12 x^ + 27 a; - 68 = 0.
7. a:^ _ 8a; _ 24 = 0. 10. x^ - 9a;- + 25a; - 18 = 0.
TJIKOUV OF EQUATIONS
133
72. Graphical solution of the quadratic equation. When the real
roots of an equation are desired to wo more than one decimal place, the
equation may be solved graphically. Take, for example, the equation
a;2_3.r + 2 = 0. (1)
We seek the values of x which satisfy this equation.
Consider y = x^, and y = 3.r — 2. (2)
If we solve these equations, we find two values of x such that the
corresponding values of y in equations (2) are equal to each other, since
for these values of x we would have
x^ = 3x - 2, or a-'' - 3a; + 2 = 0.
To solve this problem graphically
we plot equations (2) on the same
axes and note the values of x where
the straight line ?/ = 3 a; — 2 inter-
sects the parabola y = x^. These are
the values of x which afford equal
values for // in equations (2), the
coordinates of both curves being
identical at a point of intersection.
In the adjacent graph the abscissas
of the points of intersection are 1 and
2 respectively. Hence these numbers
are the roots of (1).
The advantage of this graphical
method lies in the fact that for all quadratic equations in p-forni
the first equation of (2) is the same, and hence the parabola may
be drawn once for all in ink. This leaves only the necessity of
drawing in pencil one straight line for each solution and noting
the points of intersection.
-\
Y
//
r
j
(2.4)
1
/
\
y \
(_)
y
A'
I
EXERCISES
Solve graphically the following quadratics :
1. a;2 + 2 a; - 8 = 0. 6. 2 x- - C. .r + 1 = 0.
2. x^ + a; - 3| = 0. 7. 3a-- + 5.r + 2 = 0.
3. X- - 7 .r + 12.25 = 0. 8. ./' + -3 .r - 2.4 = 0.
4. a- - 3 X - 2 = 0. 9. .r - 3.7 a: + 3.3 = 0.
5. a-^ + 4a- + 3 = 0.
10. X- - W X
3.5
0.
134
HIGHER ALGEBRA
73. Graphical solution of the cubic equation. It is assumed in
what follows that the term in x^ has been removed, from the cubic
in p-form by increasing each of the roots by ^ ? leaving the equation
in the form ^3 r , n r-w
X + px + '/ ^ 0. (1)
This transformation should be performed by synthetic division,
as in § 67.
We then plot the cui've y — x^ and the straight line y — — {j^-c + q)
on the same axes, and note the x distances of their points of inter-
section. These will be the real roots of (1). The plot of the curve
?/ = a-^ should be made carefully on a large scale in ink, so that the
line may be drawn in pencil, as in the preceding section. In this way
the same curve may serve for many problems. This method gives
only the real roots of the cubic, and there will be one or three real
roots according as the line y —— {px -\- q) cuts the curve ?/ = a-^ in
one or three points.
EXAMPLE
Solve graphically x^ + 6 x^ -\- 8 x — 1 = 0.
Solution. Here p^ = 6. We must then increase the roots by 2,
1 + G + 8 - 1 1-2
-2-8+0
(1)
1+4+0-1
-2-4
1 + 2-4
_ 2
1+0
The transformed equation is
•r^ — 4 a; — 1 = 0.
Plot the equations
(2)
y = a-'
and ?/ = 4 x + 1
on the same axes.
The abscissas of the points of intersection are
approximately 2.1, — .3, and — 1.8. Hence these
are approximately the roots of (2). Since the roots of (1) were
increased by 2 to form (2), we obtain the numbers .1, — 2.3, and
— 3.8 as the approximate values of the roots of (1).
Y
1/
f
J
f^^f
*>/
^/
h
^
0
A'
n
'\
THEOia OF EQUATIONS 185
EXERCISES
Find graphically the real roots of the following cubics :
1. Equation of exercise 10, \>. 1 '■'>-. 5. j-'' + .r — 20 = 0.
2. Equation of exercise 4, p. lUS. 6. ./■'' -{- 'A x- — 2 x — 1 = 0.
3. Equation of exercise 5, p. I'A'J. 7. .'•'* — ('>y- + ;">./• -{-11 =0.
4. Equation of exercise 8, ]>. I'.VJ. 8. ./•' — \) x'- — 2x -{■ 101 = 0.
74. Derived function of the cubic. Let us ronsidcr tlic culnc
function y'(^.,.^ = ^.^f _|_ ax- + Ox + r.
In this expression replace x by x -{- h, wIutc // is a real number.
We then have
f(x -{-/>) = (x + hy + a (x + 70' + A fr + //) + ..
Expanding the terms of this function by tlui lUnomial Theorem
and collecting like powers of h, we have
f(x + /') = ^'^ + (>.r- + //.,■ + r -f (3 x-' + 2 ^M' + ^'; /' + (3,>- + ") h- + h\
The coefficient of // in this expansion is called the first derivative
of f{x), and is symbolized by /'(.r). If we write f{x) and f'{x) in
separate lines, we can observe the relation which thoy lioar to each
other. y'(^) = ^fl j^ ax- + hx + <:
/'(.r) = 3.x-2 + 2a.r + i.
We see that the first term of ./"'(.'■) has as its coefficient 3, which
is the exjwnent of the first term of f{x), while its exponent, 2, is one
less than the exponent of the first term of f(x). We may obtain the
second and third terms of f'{x) from those of f{x) in a similar
manner. The last term, c, oi f(x) may be regarded as cx° ; then the
corresponding term of f'(x) is 0 • c • a;~ ^ = 0.
This procedure suggests the following rule for finding the deriva-
tive of f{x), the general validity of which we shall establish in the
next section.
Rule. If the kth term of f(x) is multiplied by its exponent^ and
its exponent is decreased by ane, the residt is the kth term of f(x).
EXAMPLE
Find the first derivative of f(x) = x^ + 3 x'- — 1 x -\- 4.
Solution. By the rule we obtain
/'(J) = 3x2 + Ox- 7.
136 HIGHER ALGEBRA
EXERCISES
Find the first derivatives of the following functions :
1. Sx'- 2x^ + X - 1. 6. i.r=^ - 1 x"" + h
2. x^ + X- + :/■ 4- 1. 7. 4:x(x'' - 2x + 3).
3. 7x'-Gx-5. 8. (.r-1)^
4. _3.r^ + 2.r--8. 9. (./- + 3) (.r - 2).
5. a^x^ + a^x- + a^x -\- a^. 10. 2(,/' + l)-^(a3 - 1).
11. If /(x) = x^ + 3x^ + 6.r + 6, show that f'(x) =f(x) - x\
12. If /(a-) = a-^ + 6 x'' + 12 a- + 8, show that ^f{x) ={x + 2)/'(x).
75. Derivative of a polynomial. Instead of confining ourselves to
the cubic or biquadratic function, let us now consider the polynomial
of order w, f(x) = a^x"" + a^x"- ^ -\ + a„.
Replace a^ by cc + k in this function, expand each term by the
Binomial Theorem, collect the terms free from h, and also those
containing h to the first power. We then obtain
fix + h)
= a^ (x + h.y + a^(x + hy-' + ■■■ + a„_,(x + h) + a„
= a^lx'' + nx^-'h + h 7i{n - l)A-"--/r -{ h /'"]
+ a^ [.<•" -^ + {n — 1) X" - -h
+ ^(n-l)(7i-2)x''-Vi2+ . + /i»-i]
H H ('n_^(x 4- /0+«n
= a^a?" + fl^.x" " ^ + ••• + «„_ P-f- + r/„
+ lo.nx--' + 0^(71 - l).r"-2 H + ^', _ j] • A
+ F(x) . h'' + Fj(.r) . ]i;' H + 7^"
= ./'(^) +/(■'•) ^^ + ^(^O • ^^" + ^iC^-)^'' + ■ • + /^"- (1)
In this expansion the coefficient of h is called the first derivative
of f{x). It is symbolized by f'{x).
Writing f{x) = a^.T" + a^x"-^ + • ■ ■ -\- fin-\^ + '^n
and fix) = ftpwa;"-' + a,(7i — l)^-''-^ -\ h fir„_i,
we see that /' {x) may be obtained from f{x) by the rule stated in
the preceding section.
The successive terms of the expansion (1) which contain h to
powers higher than the first are not written out in detail. These
coefficients are here represented by F(x), etc.
TliEUKY OF EQUATION'S 137
EXERCISE
Denoting by f"{x) the first derivative of f{x), show that Fi:r),
the coefficient of h'^ in the expansion (1), is equal to ' ^ '^ - The
expression /"(.r) is called the second derivative of /(a-).
76. Double roots. The expansion (1) in the preceding section
may be written in the following form by replacing x by x^, h by
X — Xj, and recalling the result of the jjreceding exercise :
An inspection of (2) shows us that if a-, is such a number that
f(x )= 0, and at the same time f(.r^)= 0, then the first two terms
of (2) vanish, leaving all the successive terms divisible by (x — xj-.
That is, if a-, is a root oif(x) = Oand of/' (a:) = 0, but not oif"(x) = 0,
it is a double root of f{x) = 0. This is equivalent to the statement
that if X — a-j is a common factor of f(x) and f (x) but is not a factor
of f"(x), then x^ is a double root of f{x)= 0. Hence we have the
Rule. The monhcr x^ is a double root of the equation f(^x')= 0
iffC^O = 0, / (-r,) = 0, andfXxO ^ 0.
In the above exercise it was determined that the coefficient F(x)
f"(x)
of h^ in (1) is ' — ^~- • It may also be shown that the coefficient of A*
in CI) is • ^^^ J where /'*'(a;) is the kth. derivative of f(x). We may,
then, write (2) in tlie form
/(x) = f(x^) + (x- a-,)/' (a-,) + ^^^^^f"(x,)
+ ■■■+ ^''~r'^V'^'(^i) + • • • + (a- - a-,)".
From this expansion we obtain the rule for finding an ;--fold root
of/(a-)=0.
Rule. If a monher is a root of f(x)=^ 0 and of its first r—1
derivatives, but is not a root of the rth derivative, each set equal to
zero, it is an rfold root off(x') = 0.
In computing the successive derivatives of y"(a-), it must be kei»t
in mind that the derivative of any constant term is zero.
138 HIGHER ALGEBRA
EXAMPLES
1. Find a double root of the equation
16a;3-12ic'^ + l = 0.
Find also the other root.
Solution. f{x.) = 16 x3 - 12 x2 + 1,
f'{x) =:48x^- 24x = 24x(2a;- 1),
/"(x) = 96x-24.
"We see that J is a mot of f'{x) — 0. It is also a root of /(x) = 0, since
/(I) = 2 — 3 + 1 = 0. But \ is not a root of f"{x) = 0. Hence | is a double
root of /(x) = 0.
Since the product of the roots of /(x) = 0 is — /,-. , the other root, r, must be
such that ^ . ^ • r = — y'j, . Hence ?• = — ] .
Therefore the three roots of /(x) = 0 are \, i, — |.
2. Show that 1 is a triple root of the equation
x^ _ 5 ,r3 + 9 .T^ - 7 .r + 2 = 0,
and find the other root.
Solution. /(x) = x« - 5 x3 + 9 x2 - 7 X + 2,
f'(x) = 4 x3 - 15x2 + 18 X - 7,
/"(x) = 12 x2 - 30 X + 18 = 6 (X - 1) (2 X - 3),
/"'(x) = 24x-30.
We see that 1 is a root of f"{x) = 0. Also 1 is a root of /(x) = 0 and /'(x) = 0,
since /(I) = 0 and /'(I) = 0. But 1 is not a root of f"'{x) = 0, since /"'(I) ^ 0.
Hence 1 is a triple root of /(x) = 0.
Since the product of the roots of /(x) = 0 is 2, the other root is 2.
EXERCISES
Find the double roots and other roots of the following equations :
1. x^-a;2-5a'-3 = 0. 4. .x^ + 8.r- + 20.r + 16 = 0.
2. 27x^-90: + 2 = 0. 5. 16^'^ - 60.r- + 125 = 0.
3. 2:r3-15a;2 + 24x + 16 = 0. 6. 63a;^ + 321x2 + 469^' + 147 = 0.
Find the triple roots and other roots of the following equations :
7. x^ - 2 .r^ + 2 X -1 = 0.
8. 2x* + 11 ./-^ + IS ;r- + 4 x - 8 = 0.
9. 16a;*-24x'^ + 16x-3 = 0.
10. 3 x" - 32 .r^ + 96 .r^ - 256 = 0.
TllEOKV OF EQUATION'S 139
11. Show that 1 is a fourfold n^ot of tlu; tujuatioii 1 j-} — 5a;* +
10 a,-'"' — 10 X + 3 = 0, and iiiid tli(3 other root.
12. Sliow that — 1 is a iivcfold I'oot of the equation ./•'"' + .3 x-^ —
10x-=*-15x'''- 9x - li = 0, and iiml the other root.
77. Error in computation. Suppose the values oi f{x) are to be
computed by substituting values of x which are the result of measure-
ment and hence not known exactly. By means of the derivative we
can find the approximate error in the function when the error in x is
known, provided that error is small.
Consider, for example, the expression for the volume of a cube in
terms of one of its edges, V= x^. If we could measure the edge with
perfect correctness, we could find accurately the volume of the cube ;
but wlu'u our lule seems to read, say, 2.25 inches, we know that there
may be a slight error in the reading, due to slight inaccuracies in the
rule, our vision, and our method of using the rule.
Let the measured value be x^ and let the small error be denoted
by h. Of course we do not ever know ]ust how great h is. We
may usually assume, however, that it does not exceed some definite
small number. Then letting x = x^ + h, and expanding the function
V=f{x)—x^ by formula (1) of the preceding section, we obtain
T ' = /(.r^ +h) = (x^^ hf = .r,f + 3 xf;h + 3 xjc' + h\
Now since A is small, /r and Ii'^ will be much smaller, and may be
neglected, as they would not affect the result appreciably. Erom this
expression it appears that the value of the volume differs from the
value of x^ by 'Sxj/t, if we neglect the last two terms. Hence, if
we assume that h = .02, and x^ = 2.25, the approximate error for ]'
is 3 . (2.25)"-' . .02*= .3 cubic inches.
In general, let x^ be the measured value, and /i the error of the
measurement. We may write (§ 75)
Here f(x^) is the value of the function if our measurement were
correct, while the ai)proximate error in our result, omitting all terms
containing ])owers of h higher than the first, is h -/'(x^). We may
now state the following
RlTLE. To find the approximate error in the J'ltnrtinn t\.r) due
to a snudl error, h, in the measurement x , midtiply h l>y fi^i'^-
140 HIGHER ALGEBRA
EXERCISES
1. The edge of a cube is found l^y measurement to be 3.2 inches.
Find the approximate error in the computed volume due to an error
of Jq of an inch in measuring the edge.
2. If the diameter of a sphere is found by measurement to be 10.3
inches, find the approximate error in the computed volume due to an
error of .1 of an inch in measuring the diameter. (Take tt = '^^-.)
3. The height of a cylindrical column is known to be 10 feet.
What is the approximate error in the volume computed from a
diameter measurement of 50 inches if this measurement is a half
inch in error ?
4. A surveyor measures a square field with a 50-foot chain which
is 1 inch too long and finds the area to be 62i acres. Find the area
of the field in acres correct to 2 decimal places and show that the
amount neglected does not affect the second decimal place.
5. Find the approximate ei'i'or in the function x^ — 2x^ -^ x — 'S.
due to an error of .03 in a value 1.25 taken for x.
6. Find the approximate error in the function 7-x^ — 2 }^x^ if r is
known to be .1 and the value of x, 8.1, is inaccurate by .1.
7. The diameter of a right circular cylinder whose altitude is 5
feet is measured and found to be 8.2 inches, but the measurement is
inaccurate by .1 of an inch. Find the approximate error in computing
the total surface.
8. A right circular cylinder is capped by a hemisphere. The height
of the cylinder is 50 inches. Its diameter is found by measurement
to be 10| inches. Find the approximate error in computing the total
surface of the solid from a diameter measurement which is -^^ of an
inch in error.
9. A Norman window is in the shape of a square surmounted by
a semicircle. Its width is measured to be 40.5 inches and its area is
computed. Find the approximate error in the computed area due to
an error of -i- of an inch in measuring the diameter.
CHAPTER VII
PERMUTATIONS, COMBINATIONS, AND PROBABILITY
78. Introduction. The formulas which will be used in this chapter
depend on the following
Theorem. If an act tvhich may he performed in p way a is
followed by an act which may he performed in q ivays^ the total
number of ways in which the two acts inay be performed in
sticcession is p • q.
For with each of the p ways of performing the first act one has a
choice of q methods for the second. Hence with the entire p ways of
performing the first there will he 2^ • q ways of performing both acts.
For example, if there are 6 roads from A to B, and 4 from B to C, one has
the choice of 6 • 4 = 24 routes iu going from A io C through B,
EXERCISES
1. A room has 6 doors. In how many ways can a person enter and
leave by a different door ?
2. A man has 4 suits of clothes and 7 neckties. How many wa3-s
can he dress, not wearing tlie same tie twice with the same suit ?
3. Two dice are thrown. In how many ways can they fall ?
4. In presenting 8 men to G women how many introductions are
made?
5. Tliree coins are pitched. In how many ways can they come up ?
6. A tow 11 has () hotels. Three people wish to stay at different
hotels. In how many ways can this be done ?
7. In how many ways can two letters be posted in 4 letter boxes ?
8. There are 25 stations on a branch line of a railroad. If both
one-way and return tickets are sold between all stations, how many
different kinds of tickets must be printed ?
141
142 HIGHER ALGEBRA
79. Permutations. Each different arrangement of a number of
tilings is called a permutation. The letters A, B, and C may be
arranged in the six different orders, ABC, ACB, BAC, BCA, CAB,
CBA, each one of which is a permutation of the letters, distinct
from the others. In determining how many permutations of these
letters there are, we may employ the idea of successive acts as
explained in the preceding section. Thus let the first act consist in
filling the first place. This may be done with any one of the three
letters, and hence in three different ways. With this place filled,
there are only two letters left with which to fill the second place,
which may be done in two ways. This affords 3-2 = 6 ways of fill-
ing the first two places. But with the first two places filled with two
of the letters there is no choice in the way of filling the third place,
as there is only one letter left. Hence the number of permutations
of the three letters is C.
Theorem. The 7iumher of permutations of n things taken r at a
time is j>^ = n(n - l)(n - 2) . . . {n - r + 1).
If only r of the 7h things are to be used at a time, there are only
r places to be filled. Since the first place may be filled by any one of
the n things, and the second place by any one of the ?z- — 1 remaining
things, we see that the first two places may be filled in 7i(n — 1) ways.
The third place may be filled by any one of the 7i — 2 things which
are left ; hence the first three places may be filled in ?i(n — l)(7i — 2)
ways. Proceeding in this way, it appears that when r — 1 places
have been filled, we have left n — (r — 1) letters with which to fill
the last place. Hence the rth place can be filled in w — (;■ — 1) ways,
and n — r -\-l is the last factor in the expression „ P^.
Corollary. The number of permutations of n thi)i[/s taken all
at a time is w(w - 1) (w - 2) • • • 2 • 1 = w !.
The symbol nl is read "factorial ?i." It is sometimes represented by|n. In
the foregoing theorem it is assumed that the elements are distinct, and that no
element is used more than once in a given permutation.
EXAMPLE
If one has eight flags of different colors, how many signals can be
displayed by showing them four at a time on a vertical line ?
Solution. Here n = 8, r - i. Hence gP^ = 8 • 7 • 6 • 5 = 1680.
PEK.MITATIOXS 143
EXERCISES
1. llow many arnuigijiuciits ut llic IctLurs in the word "C'uhimljia"
can be made, using in each arrangeinorit (a) 4 letters ? (b) all the
letters ?
2. J^'oiir people enter a room in which there are 7 vacant chairs.
In how many ways can they be seated ?
3. lu liuu' many orders can a liaml (if G cards be played ?
4. With 5 flags of different colors, how many signals can be dis-
played by showing them any number at a time on a vertical line ?
5. How many different numbers less than 1000 can be formed
from the digits 1, 2, 3, 4, 5 ?
6. What is the number of permutations of the letters of the alpha-
bet, taking three at a time ?
7. How many numbers of 7 figures having 0 as middle digit can
be formed from the digits 0, 1, 2, 3, 4, 5, 6?
8. In how many ways can 5 red books and 4 blue ones be arranged
on a shelf so that all the books of each color are together?
80. Permutations with repetitions. Let us determine how many
numbers of three digits can be written making use of the digits 2, 3,
4, and 5, where each digit may be used repeatedly. Here we have
three places to fill. The first may be filled in any one of 4 ways, and,
since repetition is allowed, the second and the third place may each
be filled also in 4 ways. Hence all three places may be filled in
4 • 4 • 4 = 4*^ ways. By similar reasoning we establish the following
Theorem. The numbers of permutations of n tlttn;/.^ ttd-en r nf
a time tvhen repetition is allowed i>< )i'\
This theorem assuuios that onoh thing may be used r times. If restriction
is placed on the amount of repetition of one or more of the objects, the theorem
is modified.
81. Permutations of things not all different. In the foregoing
sections it has been assumed that all of the things to be permuted
are different. If this is not the case, we have a modification of the
formulas derived. In order to find the number of permutations of
the letters in the word " algebra," taken all at a time, it is necessary
to note that the lett(?r a occurs twice. If for the moment these o's
are considered as distinct, we shall have 7 ! permutations. But if in
144 HIGHER ALGEBRA
eaeli permutation the a's are treated as not distinct, we can inter-
change them witliout affecting the permutation ; that is, the number
. 7!
of distinct permutations is — •
Theorem. //", of n things, n^ are alike, ii^ are alike hut of
another kind, n^ are alike hut of still another kind, etc., the num-
her of distinct permutations of the n tilings, taken all at a time, is
n\
n^\ nA nA • • •
For since the n^ ! permutations of the n^ equal things are exactly
alike, there will be only — : times as many distinct permutations as
there would be if these n^ things were distinct. For a similar reason
the total number, n\, oi permutations of n things is divided by n^\
because of the equality of the n.^ things, and so on.
When only a part of the n things are taken at a time, and some of them are
alilce, the situation is mucli more complicated and will not be considered here.
EXERCISES
1. How many different numbers less than 1000 can be formed
from the digits 1, 2, 3, 4, 5, where each digit may be repeated ?
2. Three dice are thrown. How many ways can they fall ?
3. Find the number of distinct permutations of the letters of the
word " mathematics," using all the letters in each permutation.
4. In how many ways can 4 coins be given to 10 boys, if each boy
may receive any number of the coins ?
5. Find the number of integers having 5 digits.
6. In how many ways can 6 letters be posted in 3 letter boxes ?
7. (a) Find the number of distinct arrangements of the letters
of the word " sophomore," using all the letters in each arrangement.
(b) In how many of these arrangements do the 3 o's come together ?
(c) In how many of these arrangements do the 3 o's come at the end ?
8. (a) Find the number of distinct arrangements of the letters
of the word " engineering," using all the letters in each arrange-
ment, (b) In how many of these arrangements will the i's not
occur together ?
CO:\rP,INATIOXS 145
82. Combinations. A group of objects which is independent of
the order of its elements isj called a combination. For example, a
committee consisting of three men, A, 13, and C, is the same com-
mittee whether we think of them as standing in the order ABC
or CBA. It is evident, then, that there are more permutations
of n things taken r at a time than there are combinations. The
combination depends merely on the selection of the objects them-
selves and not at all upon the order in which they are arranged in
the final groups. Since each combination of r things gives rise to
r\ permutations, it appears that there are ?•! times as many permu-
tations of n things taken r at a time as there are combinations. This
leads us to the
Theorem. The number of combinations of n things taken r at
« '^""'^ ^^ n{n- \.){n-2) . ■ ■ {n- r + I)
nCr= y^ • (1)
This formula is easily remembered if one notices that there is the same num-
ber of factors in tiu' numerator as in the denominator ; that is, just r.
From the definition of combinations it is seen that the number of combi-
nations of 71 things taken all at a time is 1.
Theorem. The number of combinations of n things taken r at a
time is the same as the nu)nber of combinatinns rf n things taken
n — r at a time.
Expressed symbolically, „f^ = „C
n n — r'
r ^n(n-l)(n-2)...[n-(n-r) + l^
^^' " "-'• {n-r)l
n(n-l)(n-2) •••(/' + !)
(n-ry.
multiplying numerator and denominator by r!,
^n(n-l)(n~2) ••■ (r -{- 1) >•(;•- 1) ...2.1
(71 - r) ! r !
dividing numerator and denominator by (n — /•)!,
n(n - l)(n - 2) ■ ■ • (n - r + 1)
/•!
= ..r..
The utility of this theorem will be appreciated if one compares the amount
of computation involved in finding .,^C^ and its ei]ual .iiCjg.
14G HIGHER ALGEBRA
For the solution of the exercises which follow, no specific rules
can be laid down. In general, one should first observe whether the
question involves combinations or permutations. If the latter, any
possibility of repetition or equality of elements should be noted.
So far as possible it is advisable to fall back on the principles on
which the various formulas depend rather than to form the habit
of using the formulas blindly.
EXERCISES
The first eight exercises involve only combinations.
1. Find the value of (a) ^C^; (b) ^/.\^.
2. How many committees of 9 can be selected from a group of
12 men ?
3. How many crews of 8 men can be selected from a squad of
13 men ?
4. How many straight lines are determined by (a) 7 points, no
three of which are in the same straight line ? (b) n points, no three
of which are in the same straight line ?
5. How many planes are determined by (a) 10 points, no four
of which are in the same plane ? (b) n points, no four of which are
in the same i)lane ?
6. Find n, if (a) „C, = 28 ; (b) „C^ = 84.
7. Find n, if (a) „C^ = /'., ; (b) „C„_3 = 35.
8. How many different sums can be made up from a cent, a nickel,
a dime, and a quarter ?
9. If „P,. = 110 and „r',. = 55, find ?t and r.
10. li„C^=h1-,r„^nd7i.
11. U,,P^ = Gj\,find7i.
12. In how many ways can 7 coins be given to two boys so that
one will get 3 and the other 4 ?
13. With 12 cadets, (a) in how many ways can a guard of 6 be
chosen ? (b) in how many ways can a guard of 6 be arranged
in a line ? (c) in how many ways can the 12 be divided into two
equal groups ?
COMBINATIONS 147
14. A committee of 7 is to lu- clioscn from S Eiiglislimen and 5
Anierieaiis. In liow many ways can the committee l)e chosen if it is
to contain (a) just 4 Englislimcu ? (b) at least 4 Englishmen?
15. How many signals can be made l)y hoisting 8 flags, all at a
time, on a staff, if 2 are white, 3 black, and the rest red ?
16. How many signals can be made with the flags of exercise 15,
using them all at a time, if a red flag is always at each extreme ?
17. Show that the number of orders in "\Yhich n things can be
arranged in a circle is (n — 1)!.
18. In how many orders can 7 men sit around a circular table ?
19. In how many orders can 4 men and 4 women sit around a
circular table so that a man is always between two women ?
20. Out of 8 consonants and 3 vowels how many arrangements of
letters, each containing 3 consonants and 2 vowels, can be formed ?
21. How many handshakes may be exchanged among a party of
12 people, no two shaking hands more than once ?
22. How many numbers greater than 100,000 can be formed by
arranging the digits 1, 3, 0, 3, 2, 3 ?
23. The Greek alphabet has 24 letters. How many fraternity
names can be formed, each containing three letters, repetition of
letters being permitted ?
24. In how many ways can a baseball team of 9 men be selected
from 14 men, if only two of them can pitch, and these two can play
in no other position ?
25. How many telegraphic characters could be made by using
3 dots, 2 dashes, and 1 pause?
26. In how many ways may 15 passes and 5 failures be adminis-
tered to a class of 20, taking them all at a time ? »
27. In how many ways can 7 men stand in line so that 2 particular
men will not be together ?
28. How many different sets of 4 hands can be dealt from a pack
of 52 cards ?
29. In how many ways may a football team of 11 men line up if
the center and quarter back keep theii* positions, no line man being
called back and no back being put in fhc line? It is assmued that
in each line-up there are three men on each side of the center.
148 HIGHER ALGEBRA
30. (a) How many diagonals has a decagon ? (b) How many
diagonals has a polygon of n sides ?
31. Of 12 musicians 10 play the violin, 7 of these 10 also play
the viola, and the remaining 5 play the cello. How many trios of
different kinds of instruments can be made up ?
32. How many triangles can be drawn, taking as vertices 8 points,
just 3 of which lie in a straight line ?
33. For a given value of n, what value of r affords the greatest
value of „CV ?
HiKT. Since there are the same number of factors in the numerator and
71 — 7* -f- 1
denominator of nCri the smallest value of r which makes > 1 will be
the value sought.
83. Probability. If a bag contains 3 white balls and 4 black balls,
and 1 ball is taken out at random, what is the chance that the ball
drawn will be white ?
This question we may answer as follows : There are 7 balls in the
bag and we are as likel}^ to get one as another. Thus a ball may be
drawn in 7 different ways. Of these 7 possible ways 3 will produce
a white ball. Thus the chance that the ball drawn will be white is 3
to 7, or f . The chance that a black ball will be drawn is f .
We may generalize this illustration as follows : If an event may
happen in p ways and fail in q ways, each way being equally prob-
able, the chance or probability that it will happen in one of the p
ways is jj
p + q
The chance that it will fail is
(1)
^ (2)
P + q
The sum of the chances of the event's happening and failing is 1,
as we see by adding (1) and (2).
The odds in favor of the event are the ratio of the chance of hap-
pening to the chance of failure. In this case the odds in favor are
'-■ (3)
q
The odds against the event are - •
p
PROBABILITY 149
EXAMPLES
1. li' the cliiuice of iiii event's haiipciiiiig is j'^^, what are the odds
in its favor "'
V 1
Solution. By (1), —i-— = — .
Hence lOp = p + q,
or 9p = 9,
P 1
or — = - » which by (3) are the odds in favor.
q 9 '
2. From a pack of 52 cards 3 are missing, ^^'hat is the chance
that they are all of a particular suit ?
Solution. Tlie number of combinations of 52 cards taken 3 at a time is
jjC'g = — . This represents p + q. Tlie number of combinations of the
1 • 2- 3 13. 12 • 11
13 cards of any one suit taken 3 at a time is J3C3 = This repre-
sentsp. 13- 12- 11
p 1.2.3 13.12.11 11 11
Thus —^ — = — = =
p + q 52 • 51 . 50 52 • 51 ■ 50 17-50 850
1-2-3
3. What is the chance of throwing one and only one G in a single
throw of 2 dice ?
Solution. Tliere are 36 possible ways for the two dice to full. This represents
p + q. Since a throw of two 6's is excluded, there are 5 throws in which each
die would be a 6; that is, 10 in all in which a 6 appears. This represents p.
» 10 5
Thus
p + q 36 18
EXERCISES
1. A bag contains 8 white and 12 black balls. What is the chance
that a ball drawn shall be (a) white ? (b) black ?
2. A bag contains 4 red, 8 black, and 12 white balls. What is the
chance that a ball drawn shall be (a) red? (b) white? (c) not black ?
3. In the previous problem, if 3 balls are drawn, what is the
chance that (a) all are black ? (b) 2 red and 1 white ?
4. \\'lKit is the chance of throwing neither a 3 nor a 4 in a single
throw of 1 die ?
5. \\'hat is the chance of throwing 7 in a single throw with 2 dice?
150 . HIGHER ALGEBRA
6. What is the chance of throwing three 5's in a single throw
with 3 dice ?
7. AVhat is the cliance of throwing 2 heads in a single throw
with 2 coins ? in 2 throws with 1 coin ?
8. If 3 coins are throAvn, what is the chance that just one will
be a head ?
9. Four men seated at a table match coins, agreeing that the odd
man shall pay for the dinner. A remarks that it is likely to require
several trials before one coin comes up different frOm all the others.
B replies that the chances are even that this will happen on the first
trial. Which is correct ?
10. Three cards are drawn from a suit of 13. What is the chance
that they will be ace, king, and queen ?
11. Two cards are drawn from a pack of 52. What is the chance
that they are both aces ?
12. Four cards are drawn from a pack of 52. AYhat is the chance
that they are all clubs ?
13. If 12 men stand in line, what is the chance that A and B are
next to each other ?
14. A man selects by lot 3 from a list of 10 friends to make up a
dinner party. The list contains just 2 brothers. What is the chance
that they are both invited ?
15. If 3 dice are thrown, what are the odds in favor of at least
2 turning up alike ?
16. Four men throw rackets for choosing partners in a game of
tennis doubles. The 2 '^ smooths " and 2 " roughs " are to be partners.
What are the odds against the choice being made on the first throw ?
CHAl'TKIt \'III
DETERMINANTS
letters 1\ = Y'' f'
84. Determinants of the second order. As a matter of notation
it is agivt'd l)y niathenuitical writers to give the arrangement of
the meaning a^),, — ah^, where these letters may
represent any numbers. The arrangement is called a determinant.
Since there are two rows and two columns, the determinant is said
to be of the second order. The expression a^l\^ — aj>^ is called the
development of the determinant. The value of the development of
a determinant is often spoken of as the value of the determinant.
The symbols a^, h^, o,^, h,„ are called elements, and a^, h_^, are said to
comprise the principal diagonal of the determinant.
Thus
1 3
2 4
1.4_2.3 = 4-6=-2:
X 1
y 0
= x-0— ?/-l=— y.
The historical reason for this apparently artificial notation is the
appearance of numbers in the form of the development a^>_^ — ajj^ in
the solution of a system of two linear equations in two variables.
Thus if we have given „^^ ^ /^^^^ ^ ^^ . ^^^
and a./^ + h.j/ = c,^, (2)
we obtain by the usual method of solution,
Cll>2 — '"o^l
X
"A -~ "J'l
and 1/
"l^2 — ^2^1
"A - "A
Using the determinant notation, we may write these results as
follows :
X =
'\
^
"i
^"i
'■•2
^
» y =
"2
^2
«1
K
"1
K
«2
h
«2
h
These expressions may be used as formulas for the solution of
linear systems of equations in two vai'iables.
151
152 HIGHER ALGEBRA
The analogy between the solution of this simple system and the
more complicated cases which follow will be seen more clearly if we
observe that
I. The determinants in the denominators are identical, and each
consists of the coefficients of x and y as they stand in the original
equations (1) and (2).
II. The determinant in the numerator of the value of x is formed
«.
1
from the denominator hy reijlacing the coefficients of x, namely
C 2
hij the constant terms \
III. The determinant in the numerator of the value of y is formed
from the denominator hy replacing the coefficients of y, namely ,',
c 2
Inj the constant terms ^
85. Determinants of the third order. The arrangement of letters
^3 =
(I .
1 ^ ^1
«2 h ^2
% h ^
(1)
has been given the meaning
''A''z + ''M + "J\''2 - ''h'x - "I'^z - «i¥2' (2)
where the letters may take on any numerical values. The expression
(1) is really an abbreviation or symbol for (2).
Since D^ contains three rows and three columns, it is called a
determinant of the third order, and (2) is called its development.
The letters a^, h^, and c^ constitute the principal diagonal. Similarly,
we may have a determinant with n rows and n columns. This is
called a determinant of the wth order.
Comparing this development with that of the determinant of the
second order, we observe the following principle which will serve as
a part of the rule for the development of determinants of orders
higher than the third.
Each term of the development of a determinant of order 3 consists
of a product of 3 elements, one from each row, and one from each
column of the determinant.
IJETEKMINANTS
lo3
This is verified in the case of (2) by observing that every one of
the hitters a, h, and c occurs in each term, and that every one of the
subscripts 1, 2, and 3 also occvu-s
in each of the terms of the devel-
opment.
This statement gives us the law
of t'oi inatioii for the development
of a determinant of any order.
The only feature which it does
not cover is the determination of
the signs of the terms. It so hap-
pens that for the determinant of the tliird order there is a simple
rule for the determination of these signs. In the above figui-e the
continuous lines indicate the right diagonals, while the dotted lines
indicate the left diagonals. We may then state the
Rule. To evaluate a determinant of the third order, midtipjy
the numbers in each of the three right diagonals ; mxdtiply the ninn-
hers in each of the three left diagonals changing the sign of each
product; then add the six products.
It should be kept in mind tliat this rule does not apply to determinants of
higher order than the third.
Evaluate
3
2 1
4-6 2
,
1 0 1
3 2 1
4-0 2
1 U
1
EXAMPLE
:- 18-f- 0 + 4 + <>- 0
10.
EXERCISES
Evaluate tlie followinti: determinants:
2.
4
1
G
a
b c
5
o
.S
•>
1
,
3.
h
c a
.
5.
1")
<»
L'l
1
0
1
c
a b
3
1
1
1
1
0
-(' -b
'>
-1
1
4
3
0
4.
('
0 -a
6.
4
0
-3
6
1
—
>
b
a
0
1
<>
3
154
HIGHER ALGEBRA
a X
y
0 b
c
.
0 c
h
a
h
c
— a
h
m
— a
-h
c
9.
10.
2 3
-1 4
+
4
-1
Solve the following equations
V2 — X X
X V2 -f- a;
11.
= 0.
X
1
0
1
J'
1
0
1
X
12.
15. Show that
= 0.
13.
14.
1
9
3
1
CC
X
3
9
^
1
1
1
1
r;
X'
(•
b
^^
a-
0.
0.
a
h
9
b' f
9 f
= -{hrj±ch)\ if
l>' f
f <^'
0.
86. Solution of linear equations in three variables. If we solve
the equations ^ ^ i ? . i 7
(3)
«3^ + hy + «3- - -3'
for X, y, and s, we obtain results which may be put in the form of
determinants, in the following manner :
X =
^
K
c
1
d.
K
^2
^3
h
^3
«i
h
^1
^'2
K
^2
^'3
K
^3
?/
"1
d.
'\
'h
d.
^
"3
d.
^3
'\
K
^'1
%
f'.
^2
%
^
^'3
z =
«1
h
d.
«2
h
d,
«3
h
d^
«1
K
^1
^'2
K
^2
''3
K
''3
(4)
We could show by substitution that these values satisfy equa-
tions (3). It is customary to use them as formulas for the solution
of a system of three linear equations in three variables. They may
be remembered without difficulty if the following properties are
verified from equations (4), and the analogy with the statements
on page 152 is observed.
DETERMINANTS
loo
I. The determinants in the denominators are identical^ and each
consists of the coefficients of x, t/, and z, as they stand in the orifjinal
e(/nations.
II. j!Jach determinant in the numerator is formed from the de-
nominator hy putting the column of constant terms (as they stand
in the oriyinal equations) in place of the column of the coefficients of
the variable ivhose value is sought.
EXAMPLE
Solve the following equations by determinants :
X + y + ". = 2,
x + 3y-4 = 0,
y~2z = (j.
Solution. Rearranging so that terms in the same variable are in the same
colunui, and supplying the zero coefficients, we get
X + 2/ + z = 2,
a; + 32/ + 02 = 4,
Ox+ y-2z = Q.
By (4), p. 154, X =
y =
z =
2
1
1
4
3
0
6
1
-2
1
1
3
0
1
-2
2
1 1
4
0
P
6-
-2
-3
1
2
3
4
0
1
0
-12 + 0+4-18 + 8 + 0 -18
-G+0+1^0-0+2
3
= 6,
-3
-8+0+6—0+4+0
^3
18 + 0 + 2-0-6-4
^^1
2
10
2
3'
10
"3 '
Check. Substituting the values found in the original equations, we have
0 + (- Vl + (- -'3") = 6 - J32. = G - 4 = 2,
^ + 3(- 5)-4 = 6-2-4 = 0,
2 o I 1 o\ 2 1 2 0 _ n
156
HIGHER ALGEBRA
EXERCISES
Solve the following systems of equations :
x + y + z = 2,
1.
a; + 2y + 3s = 14,
x + 3?/ + 6^-20.
2:^ + 3^ = 12,
2.
3a; + 2^ = ll,
1
3 y + 4 « = 10.
3 a; + // - .t - 8 = 0,
3.
33/-2a- + s-5 = 0,
a; - 2/ + 2 ." + 6 -: 0.
2x + ii = z,
4.
15a--3// = 2|,^,
4a; + 3y/ + 2:^ = 18.
i^-li/ = o,
5.
lx-\z = l,
hy
o
4a;-3y + 8s = -4,
2x + Sij-3z-5=-0.
Sx — 5>/-\-7z = 28,
7. 2a; + 6?/ -9s; + 23 = 0,
4a;-2?/-5;? = 9.
a; + ?/ + « = 1,
8. 1 + 1 + 4. = 1,
5x 3 y ^ _ -I
"3" + It ~ 2 ~ ^°
a; + ?/ = ;s + a,
x + z = ij -\- a,
y -\- z = x -\- a.
6 12 10 ,
- H = 4,
X y z
5 + 2 + 5 = 4,
x y z
6 4 5,
- + - + - = 4.
a; ?/ s;
10.
87. Inversion. In order to find the development of a determinant
with more than three rows and columns, the idea of an inversion is
necessary. If in a series of positive integers a greater integer pre-
cedes a less, there is said to be an inversion. Thus in the series 12 3 4
there is no inversion, but in the series 12 4 3 there is one inversion,
since 4 precedes 3. In 1 4 2 3 there are two inversions, as 4 precedes
both 2 and 3 ; while in 1 4 3 2 there are three inversions, since 4
precedes 2 and 3, and also 3 precedes 2.
88. Development of the general determinant. We may write a
determinant of the nih. order as follows :
D„ =
«1
K
^1 •
••^1
%
K
^2 •
••^
%
h
^■3 •
■■^
Ik
/.,
(1)
We can now define completely what is meant by the development of
such a determinant.
DETERMINANTS 157
Rule. The development of a determinant of the nth order con-
sists of the aJ(jebraic sum (f nil the terms ichlch can he formed
possessing the following properties :
I. Each term consists of the product of 71 elements^ one from each
row, and one from each column of the determinant.
II. The sign preceding each term is + or — according as the mim-
her of inversions of the subscripts of that term is even or odd, the
order of the letters being the same as that of the principal diagoyial.
According to this rule, the sign of the term oji^c,^ in the (leveloj>-
ment of tlie determinant of the third order should be -f, since the
numbers 312 have two inversions. Reference to (2) of § 85 verifies'
this result. The signs of the other terms in (2) may be similarly
obtained, and the use of the diagram on page 153 for the determi-
nation of the signs may be justified in this way.
An application of the first part of the preceding rule to a deter-
minant of higher order than the third shows that the development
of such a determinant contains more terms than would be obtained
by taking the diagonals as explained for the determinant of the
third order.
For example, if n — 4, the rule requires that a.^h.,c^d^ is a term of the devel-
opment, although this set of elements does not occur in any diagonal of the
determinant.
Since there are as many terms in the determinant of order n as
there are arrangements of the subscripts 1, 2, • • •, n, it appears that
the number of terms equals the number of permutations of n things
taken all at a time, which is, by § 79, equal to n\.
For example, there are 5 • 4 • 3 • 2 • 1 = 120 terms in the development of a
determinant of the fifth order.
89. Properties of determinants. The meaning of the following
theorems should be studied in the illustration with a determinant
of the third order before the general proof is read.
I. If each element of any row or coluynn is midtipUed by a con-
stant, the value of the determinant is multiplied by that constant.
Illustration.
ma„ b„ c.
'""3 ''3 C3
= m(a^b.yC^ + a.J>^c., + aJ>gC^ — n^b.yC^ — a^b^c.^ — ajb^c^) = m ■ D^.
158
HIGHER ALGEBRA
Proof. Since by the definition of the development of a determinant
of the nth order given in the preceding section, each term must con-
tain one and only one element from each row and each column, the
factor ??i will appear just once in each term. If m is written out-
side a parenthesis, it appears that the parenthesis itself contains the
development of the original D„.
II. The value of a determinant is not affected if the rows and
the columns are interchanged.
= "1^*2^3 + '^3^1^2 + "2^3^! ~" '^Z^-l''l ~ ^1^3'^2 ~ ^Jh''-Z (1)
Illustration.
«i
^
Cl
■03 =
«2
K
^2
«3
h
^■3
«1
(In
"3
h
h
^^3
Cl
c„
C3
= a^b„r^ + a^h^c^ + ftJ^^^x ~~ ^3^2'^! ~" ^i'^3'^2 ~ ^2^i^3 — -^3
Proof. If the rule for the development of the determinant i)„ (§ 88)
is applied to the determinant
«i a.^ (>,■■■('„
^1 '-2 ^ ■■■ ^»
^1 ^2 ^3---'«
formed by interchanging the rows and columns of Z)„, the two
developments are identical. The signs of the corresponding terms
are identical, since the principal diagonals in the two determinants
are the same.
III. The value of a determinant is changed in sign if two roivs,
or two columns, are interchanged.
Illustration. Compare (1) with
ftj
^1
^1
ag
^3
C3
«2
b.
c^
= ct-^b^c^ + a.p-f^ + (^zb-2pi ~ ^^'ih'^i ~ '^i^^'^s ~~ '^h^i'^i
D,
Proof. We need consider only the interchange of two rows, since,
by II, rows and columns may be interchanged without affecting the
value of the determinant.
An interchange of two adjacent rows does not affect the order of
tlie letters in the principal diagonal or in any term of the develop-
ment ; it merely interchanges two adjacent subscripts in each term,
DETERMINANTS
1.-9
that, is, it alfuids one iiion; or one less inversion in the subscripts of
each term, and hence changes tlie sign of each term. An intercljange
of adjacent rows or of adjacent colunnis is called a transposition.
An interchange of two rows separated from each other by vi
intermediate rows requires ?/i transpositions to place the lower of
the two rows next under the upper one, followed by m + 1 trans-
positions to place the u})per row in the place formerly of-cupied by
the other, that is, 2 ?;t -|- 1 transpositions in all. But since each trans-
position causes a change of sign in the determinant, and since the
whole process involves an odd number, 2 m -f 1, of such changes, we
see that the determinant is left with a sign opposite to that of the
original determinant.
IV. If a determinant has two roivs or two colum)n< lilruticaJ^ the
value of the determinant is zero.
Proof. Let /)„ be the value of the determinant, and let the two
identical rows be interchanged. Then, by III, the sign of the deter-
minant is changed. But since the rows which were interchanged were
identical, J\ is really not affected at all ; that is, D„ = — D„, or 7)„ = 0.
V. If each of the elements of any row or any column of a determi-
nant consists of the sum of two numbers, the determinant may he
expressed as the sum of two determinants.
Illustration. Let the elements Oj, «.,, and a.^ in J)^ be replaced by a[ + a[',
a^ + c4', ^"*^ ^s + ^3 respectively. Then
"i + "i
U.^ + «2
«3 + «3
^1
= {a{ + ay) h.f^ -1- («o 4- a.'.') h^c., + {cu, + a.^) h^c.y
- (c% + a'.^) (JoCi - (<i{ + a[' ) l).jC.-, — («.', + cC) b^c^
i '^1 . "i • "i ^^1 "j "i
"i
'h
^i
(h
K
(".,
+
«3
h
'•li
rt..
Proof. In the determinant y>„, if each of the elements of any
column, say the first, consists of the sum of two numbers, as a[ -\- a[',
a'i-\-a'2, flg + rts', • • •, «^ + o^,', each term of the development will contain
one of these binomials. If the parentheses containing these binomials
in each term are removed, and the terms containing the a"s and the
160
HIGHER ALGEBRA
a"'s are considered separately, it appears that we have the develop^
ments of two determinants whose first columns are the a"s and the
a"'s respectively, and whose remaining columns are the same as those
of the original determinant 2)„.
VI. If in a determinant the elements of a row or column are
replaced hy those elements plus the corresponding elements of another
roiv or column, each multijMed hy the same constant, the value of
the determinatit is unchanged.
Illustration.
«3 + '"^3 ^3 '"S
I», + ?M-0 = !»,.
The proof for the general case follows immediately from V, I,
and IV.
90. Development by minors. The determinant which remains
when the row and the column which contain a certain element are
erased is called tlie minor of that element.
', b, c
«1
\
«2
b.
«3
h
«1
\
a^
b.
a.
b.
Ci
mb^
5, c,
Co
+
mb.^
h ^2
C3
mb^
^3 C3
Cl
h
6, f,
C2
+ 711
b.
^2 <^2
^3
f^z
^3 ^'S
1
For example, in the determinant
that of ttj is
a.-,
a.
bo
5,
the minor of c„ is
6,
Since the development of a determinant contains only one element
from any particular row, it appears that a certain number of the
terms of the development contain, for example, a^, but no other ele-
ment of the first column ; other terms contain a^, still others a^, and
so on. In (2), § 85, for example, the first and last terms contain a^,
the second and the next to last contain a^.
The essence of the method which follows consists in determining
the exact nature of the coefficients of the elements of any row or
column.
Theorem. The development of a determinant of order n may be
written as the algebraic sum of n terms, each term consisting of two
factors.
The first factor of each term is an element of a certain row or
column, each element being used but once.
DETERMINANTS
161
The second factor of each term is the minor of the first factor of
that term.
The element in the first factor is ivritten with its sign changed,
or unchanged, according as the sum of the number of the row and
of the column in which it lies is odd or even.
Illustration.
- 1
3
1
0
6
- 1
. 3
- 1
3 6
1
= - 1
— 2
+ 0
3
4
1
4
1 3
4
= -(24 + 3) -2(12 + 1)=- 27 -26 =-53;
or, developing with respect to the third coluimi, we have
- 1 2
3 6
1 3
' = _ 5 _ 48 = - 53.
0
-12
-1 2
1
= 1
1 3 +^
3 6
4
(_3-2) + 4(-0-6)
If any of the elements of a determinant are 0, it is shorter to develop the
determinant with respect to a row or column having the most 0 elements.
Proof. Consider the coefficient of a^ in the development of determi-
nant (1) of § 88. The coefficient of a^ consists of terms in each of
which all of the other letters occur, and since all i)ossible inversions
of the subscripts 2, 3, • • •, «. are present, it must contain all of the
terms of the minor of a^. Since the removal of tlie subscript 1 from
the first place in the series of subscripts 1, 2, • • ■, n, leaves the same
number of inversions in the subscripts 2, 3, • • •, n, that were origi-
nally present, it appears that each term of the coefficient of a^ has
the same sign as the corresponding term in the minor of a^ Hence
the coefficient of a^ is precisely its minor.
Consider now any other element of the determinant, say the ele-
ment in the /rth row and lih. column. By successive transpositions
of rows and of columns we may bring this element into the leading
position at the upper left corner. It requires k — 1 transpositions of
rows to bring the ^-th row to the top, and then l—l transpositions of
columns to bring the element in question into the leading position ;
that is, I- -\-l — 2 in all. Hence the sign of the determinant will have
been changed /.+-/ — 2 times, and will be the same as it was origi-
nally if this number, or what amounts to the same thing, k -\- 1, is
even. Tf /.• + / is odd, the sign of the determinant Avill have l)oen
changed by this process.
162 HIGHER ALGEBRA
When the element in question is once in the leading position, its
coeificient is its minor, by the reasoning which we went through with
a^ ; but if the sign of the determinant has been changed in getting
it there, the term of the development which contains this element is
changed in sign. Hence the theorem is established.
It is important to note that a given element has the same minor
after a transformation of this kind that it had originally.
Corollary. If in the development of a determinant hy minors
with respect to any roiv, the elements of this roiv are replaced hy the
elements of another row, the resulting expression vanishes.
This follows from the fact that the expression which we obtain is
virtually a determinant with two rows identical.
For example, if in the development of the determinant
2)3 =
'1
^2 Co
63 C3
vfheve A■^^, A„, and yl 3 represent the minors of Oj, a^, and a^ respectively, we
replace the a's by the 6's, we have
b^Aj^-b.,A., + /93.-I3,
which is equal to
h, b^ q
'^2 ''2 ^2
''3 h ^3
Hence by IV, § 89, ^^ j ^ _ j,^.^^ ^ ^^^4^ ^ 0.
91. Directions for evaluating determinants. In finding the value
of a determinant with numerical elements it is frequently desirable
to transform it in such a way that as many elements as possible in
some row or column are zero, so that as many terms as possible in
the development by minors will vanish. To this end the determinant
should be scrutinized, with the following points in mind :
First, are there two rows which contain identical elements in like
columns ? If so, replace one of these rows by its own elements minus
the corresponding elements of the other row.
Second, are several elements of any row each m times the corre-
sponding elements of another row ? If so, replace this row by its own
elements minus m times the corresponding elements of the other row.
The mastery of these principles will carry with it the ability to
use others in reducing numerical determinants to manageable form.
IJETEILMI.NA.NTS
163
Evaluate the determinant
EXAMPLE
35
05
25
12
17
2G
12
5
4
5
9
— /
3
5
2
1
Solution. Subtracting 12 times the fourth row from the first, subtracting 5
tinios the fourth row from the second, and afhlinu 7 times the fourth row to the
tliird, we have
- 1
5
1
0
2
1
2
0
25
40
1(5
0
3
5
2
1
Now developing by minors with respect to tlie fourth column, the element 1,
being in the fourth row and fourth (■(ihuiui. retains its sign +, since 4 + 4 is an
even number, and the determinant reduces to one of the third order
- 1
5
1
2
1
2
25
40
16
Now subtract twice the first row from the second ami 10 times the first row
from the third ; this gives
- 1 5 1
4-00
41-40 0
4-0
41 -40
:- l(iO + 309 = 200.
It should be noted that a row whose multiple is condjined with another is left
unchanged.
EXERCISES
Evaluate the following determinants ;
1.
13
17
4
28
33
8
40
54
13
k
a h
4-^-
k
h c
+ «
k
c a
4-6
1
1
0
1
1
0
1
1
0
I
1
1
1
1
•)
I
o
4
•>
3
3
3
•)
1
3
• »
•>
164
HIGHER ALGEBRA
5.
7.
9.
10.
1
15
14
4
2
6
7
9
8
10
11
5
13
2 16
-1
1
1
1
1 ■
-1
1
1
1
1
-1
1
1
1
1
-1
0
-1
-1
1
4
5
1
1
3
9
4
1
-4
4
4
1
1 2
3
a
2 3
4
h
3 4
5
c
•
4 5
6
d
1 1
1
1
1 2
3
4
-
1 4
9
16
1 8
27
e
)4
9 13 17 4
18 28 33 8
30 40 54 13
24 37 46 11
11.
12.
13.
14.
15.
16.
47 5
2 91
54 6
3 92
28 3
3 93
0 0
0 5
11 12
8 1
10 17
21 3
15 38
19 2
6 11
8 1
X + 1
1 1
1
1 U
' + 1 1
1
1
1 z + 1
1
1
1 1
1
a X
y ^
X, h
0 0
/A 0 c 0
z. 0 0 d
1
_ 9
-3
-2
— *?
a
r/
X
0
3
2
9
9
0
2
3
9
9
1
2
2
3
9
0
h
0
ar
0
d
X
e
0
X
0
0
a*
h
i
J
0
0
0
0
17. Show that
X xij y
2x X -\- 1/ 2 1/
111
(x - yy
18. Show that if all the elements of a determinant of order n
which lie on one side of the principal diagonal are zero, the value
of the determinant is equal to the product of the elements of the
principal diagonal.
DETERMINANTS
165
Solve the following systems of equations :
a.r + />;/ + c': = a, ./• + //
l/.r + (-1/ -\- az = b, 21. X + -.'
19
ex + a// -\- bz = c.
y + z =
X -{-2 a.
^ + II 4- ~ = ''' + ^' 4- ^,
20. hx -\- cy -(- az = ah + In- -\- ca,
ex + ay + hz = ah -(- /«• + ca.
(i.r -\- hi/ — rz ^ 2 ah,
22. A// + rv — ax = 2 be,
cz -\- ax — by z= 2 ac.
92. Solution of systems of linear equations. Suppose that we
have given n linear equations in n variables. We seek a solution of
the equations in terms of determinants. For simplicity let n = A.
Given
a^x + h^y + c^: + <l^a'=f^, (1)
(2)
(3)
(4)
The coeflficients of the variables in the order in which they are
written may be taken as forming a determinant 1)^, Avhieh we call
the determinant of the system. We assume that Z)^ ^ 0. Thus
a.
^.
c.
d.
1
1
1
1
a^
b.
6'„
<^o
A =
2
2
2
2
4
a
b
C„
d.
8
3
8
8
«4
^4
''a
'^4
Symbolize by ^4^^, B^, etc., the minors of a^, b^, etc., in this deter-
minant. Let us solve for x. Multiply (1), (2), (3), (4) by A^, — .1.,.
A^, — A^ respectively. We obtain
A^a^x + A^b^y + A^e^z -f A^i\ir = A^f^,
^8«8^ + ^^8^3.V + '-^s'V- + 'K'h"' = 'KA>
- A^a^x - A J, J/ - A^<Y - A^djc=-Aj\.
If we add these equations, the coefficient of x is the determinant
Z)^, while the coefficients of y, z, w, are zero (by the corollary of § 90).
166
HIGHER ALGEBRA
The right member of the equation is the determinant D^, except
that the elements of the first column are replaced by f^, f^, f^, f^
respectively. Hence
X =
A
h
'\
'h
U
h
^2
d.
/s
h
^3
^
./;
\
'■4
'h
«i
K
^'l
<h
«2
K
'-.,
^h
%
h
^
a.
K
^4
<h
In a similar manner we can show that the value of any variable
which satisfies the equations is given b}' the following
Rule, The value of one of the variables in n linear equations in
n variables consists of a fraction whose denominator is the determi-
nant of the system and ivhose numerator is the same determinant,
excejyt that the column tvhich contains the coefficients of the given
variable is replaced by the column consisting of the constant terms.
When the determinant of the system is zero, we cannot solve the
equations without further discussion, which may be found in more
advanced treatises.
EXAMPLE
Solve for x the following system of equations :
ax -\- 2hy ^= 1,
2 by + 2, cz = 2,
3 c,i + 4 div = 3,
4 dw + 5 ax = 4.
Solution. Rearranging, we obtain
ax + 2bi/ =1,
2by + Scz = 2,
Scz + 4dw = 3,
5 ax + idw — i.
DETERM1NA>TS
hy
X =
1
2 b
0
0
2
21)
3c
0
3
0
3c
4d
4
0
0
4d
a
26
0
0
0
26
3c
0
0
0
3 c
4d
5a
0
0
4d
1
1
0
0
246c£!
2
3
1
0
1
1
0
1
4
0
0
1
1
1
0
0
24 a6c(Z
0
0
1
0
1
1
0
1
5
0
0
1
1
1
0
0
1
0
1
0
3
U
1
1
4
0
0
1
1
1
0
0
0
1
1
0
0
(J
1
1
0
-5
0
1
1
1
0
—
3
1
1
4
0
1
1
1
0
a
0
1
1
-5
0
1
1
1
0
—
2
0
1
4
0
1
1
1
0
«
0
1
1
0
5
1
2 1
4 1
1 1
5 1
-2
-4a
2a
(1)
93. Solution of homogeneous linear equations. A homogeneous
equation is one iu which all the terms are of the same degree in the
variables.
The equations considered in tlie previous section become homo-
geneous if f^=f_^= /g = f^ = 0. We have then
«4^ + f'^f/ + '^4-- + ifi"- = 0-
These equations are evidently satisfied hy x= 7/ = z = ir = 0. This
we call the zero solution. "We seek the condition which the coeiti-
cients must fulfill in order that other solutions also may exist. If
we carry out the method of the previous section, we observe that
the determinant in the numerator of every fraction which affords
the value of one of the variables equals zero. Thus if D^is not equal
to zero, the only solution of the above equations is the zero solution.
This gives us the following
Principle. A system of n linear homogeneous equations in n
variables has a solution distinct from the zero solution only when
the determinant of the system vanishes.
Whether a solution distinct from the zero solution always exists
when the determinant of the system equals zero, we shall not
determine, as a (^onij^lete discussion of the question is beyond the
scope of this chapter.
168 HIGHER ALGEBEA
In any particular case a system of linear equations whose deter-
minant has been found to equal zero may usually be solved by
dividing all of the equations by one of the variables, as w in (1), and
solving the fii'st n — 1 equations for the ratios _,_,_,.... When
tv w w
values of these ratios can be found, they will always satisfy the
remaining equation, and in this way an infinite number of sets of
roots may be obtained.
EXAMPLE
Solve the following system of homogeneous equations :
5a; + 4?/+ z = 0,
6x -\- >/ -\- 5 z = 0.
Solution. Here the determinant of the system
I) =
.5 4 1
5 3 2 1 = 0.
6 15
Dividing each of the equations by z, we have
55 + 4^ = -l,
z z
55 + 3^=-2,
z z
6?+ ^=-5.
z z
Solving the first two of these equations for - and -,
z z
?=-l, ^ = 1.
z z
These values satisfy the third equation. Hence x, ?/, and z may be any
numbers which satisfy the relation
x:?/:2 = l: — 1: — 1,
or x=— y =— z.
Thus the system is satisl5ed by the sets of numbers,
1, - 1, - 1 ; 2,-2,-2, etc.
DETERMINANTS
1G9
EXERCISES
Solve the following systems of equations :
1.
2.
— ir + r + // -(- ,- = 8,
w — X + ;/ -\- ." = 6,
w + x — i/-\-rc = i,
w + X + y — :: — 2.
V + 3 ?<; - 11 = 0,
w + 3 a- - 15 = 0,
a; + 3?/ -19 = 0,
2/ + 3«-8 = 0,
s + 3v-7 = 0.
4.
a; + 32/ — « =1,
7/ + 3 « — «• = 4,
z + 3w— cc = 11,
t^; + 3 X- — // = 2.
w + a; — « = 2,
'jc + 2y-?,,r = A,
3a--5// + 2«=-l,
2«' + y-;v = 0.
Solve the following systems of homogeneous equations
3 a; + 5 ?/ + 6 s = 0,
5. 2x-\-Ay + 5z = 0,
a, + 2y + 3." = 0,
6. 2a;4-3?/ + 4;s = 0,
3x- + 4y + 5^ = 0.
2x + 3?/ + 2;s = 0,
7. 3a- + Z/ + -i.~ = 0,
4a;-2y-s = 0.
6 a; + y - 7 -v = 0,
8. 5a: — 10// + 5.t' = 0,
4a; + 3y — 7^ = 0.
CHAPTER TX
PARTIAL FRACTIONS
94. Introduction. In the integral calculus it is often necessary to
express a fraction in which the denominator is an integral rational
function of one variable, as the sum of several fractions each of
which has a linear, or at most a quadratic, function in the denomi-
nator. It is always assumed in what follows that the denominator of
the original expression is of higher degree in x than the numerator.
Whenever this is not the case, it is necessary to reduce the frac-
tion by long division to a mixed expression in which the fractional
part is in the desired form. This step, which is preliminary to all of
the cases, we state as follows :
f(x)
If in ~r\i f(x) is of higher degree than (f) (a;), express hy means
of long division the fraction in the form = Q (a:) + ,
ivhere R (x), the remainder in the divisioji, is of lower degree
than ^ (a-).
95. Denominator with distinct linear factors. The essence of
the following method consists in assuming that the fraction can .be
expressed in the following form, and then seeking to determine the
numerators which in the assumed form are left undetermined.
EXAMPLE
Separate into partial fractions
■X + 4
(x - 1) {x -2)(x- 3)
Solution. Assume
- + 4 ^ +_^+^^. (1)
(x - 1) (x - 2) (X - 3) X - 1 X - 2 X - 3
To determine what numerical values A, B, and C must have, multiply both
sides of the equation by (x — 1) (x — 2) (x — 3).
We have x + 4 = ^(x - 2) (x - 3) + iJ(x - 1) (x - 3) + C'(x - l)(x - 2).
170
PARTIAL FRACTIONS 171
Since this expression is assumed to be an identity, it must be true for all
values of x (§ 11). If, then, we let x take on the value 1, all of the terms in the
right member vanish excepting the one containing A. We can then lind the
value of ^. 1 + 4 = yl (1 - 2) (1 - 3), or ^ = ^.
Similarly, yjo find the values of B and C by letting x take on the values 2 and 3
respectively. Tims />' = — (5, C — \.
Hence a; 4. 4 5 g 7
+
(x - 1) (x - 2) (x - 3) 2 (X - 1) X - 2 2 (X - 3)
In this text we shall not prove that the assumption analogous to
(1) always leads to the partial fractions which we seek. In any par-
ticular case the fractions obtained should be added as a clieck on the
work, and the validity of the result in this way determined.
EXERCISES
Separate into partial fractions :
1.
.t2-2
3 + 2a;
(2ic-3)(5a;-4)
2a-2-l
• (a;_i)(a;-^ + 3a: + 2)
C ^
X ix - 1) {x - 2)
0-24-4
{x-2){x + 2){x-
X
-1)
x" + 11 a; + 30
"• {x + V){x + ?,){x^o)
3.
96. Denominator with distinct linear and quadratic factors. The
method of treating this case is similar to that shown in the preceding
section. The assumption is slightly different.
The numerator of a partial fraction with a linear denominator is
always assumed to be a constant. The numerator of a partial fraction
with a prime quadratic factor (see p. 1) in the denominator is always
assumed to be linear and of the form Ax +£.
EXAMPLE
Separate into partial fractions :
5a-2+8.r + ll
{x'--^V){x-2,){x + V)
Solution. In this example we make the assumption
5x2 + 8x4-11 Ax^B C D
(x2 + 1) (X - 3) (X + 1) X- + 1 X - 3 X + 1
Multiplying by (x"- + 1) (-c - 3) (x + 1), we have 5x- + 8x + 11 = ^Ix (x - 3)
(X + 1) + /?(x - 3) (X + 1) + C(x- + 1) (X + 1) + Z)(x2 + 1) (X - 3).
172 HIGHER ALGEBRA
Letting x =— 1, we obtain D — — 1 ; letting x = 3, we obtain C = 2.
Substituting these values for C and D, and letting x = 0, we obtain
B = -2.
Substituting the values already found, and letting x = 1, we obtain
A =- 1.
„ 5x2 + 8x + n -x-2 2 1
Hence • =: h
(x2 + 1) (X - 3) (X + 1) x^ + 1 X - 3 X + 1
EXERCISES
Separate into partial fractions :
1 + .r + a"^ X
4.
x(^x^ + 4) (x-\- 3) (2x^ - a- - 4)
a;2 + 15 ^ 3.r + 4
2. .. . ., . ^ — — • 5.
(x - 1) (x" + 2 a; + 5) (x^-x) {x' + 1)
3 - a-2 x^-\-5
6.
2 + X + 2 a-'^ + .r'^ ,r^ + a;^ -1- a;"-^ + a-
97. Denominator with repeated factors. If the fraction is in
the form \„' replace x by ?/ + « in both numerator and de-
nominator, and simplify the numerator. The partial fractions are
directly obtained, and may be expressed in terms of x by replacing
y hy X ~ a.
EXAMPLE
Separate into partial fractions :
3a;2_4a-4-3
{x-2f
Solution. Letting x = y + 2,
3x2 - 4x + 3 _ 3(?/ + 2)2 _ 4(y + 2) + 3 _ 32/2 + 82/ + 7
(X - 2)3 {y + 2- 2)3 2/
3 8 7 3 8
3
y y- 2/3 x-2 (x - 2)2 (x - 2)3
When a factor of the form (x — a)^' appears together with other
factors in the denominator of a fraction which is to be broken up
into partial fractions, the assumed form is taken as in the following
example, and the coefficients may be found by replacing x by conven-
ient integers. It is frequently imjDOSsible to find integers which will
enable us to determine a coefficient at each substitution, but systems
of equations are obtained from which the coefficients may be found.
PARTIAL FRACTIONS 173
EXAMPLE
Separate into partial fractions :
5 a-'^ — 6 at — 5
(x-iy(x-^2)
Solution. Assume
5x2 -6a; -5 ^ ^ q j,
+ T ^ + ^ T-,+
(x-l)3(x + 2) x-1 (x-1)'^ (J--1)'' x + 2
Multiplying by (x - 1)3 (x + 2),
5x2_6x-5 = ^(x-l)2(x + 2) + -B(j!-l)(x+ 2) + C(x + 2) + 2)(x- 1)3.
Letting x = — 2, we find D = — 1 ; letting x = 1, we find C = — 2.
Substituting these values of C and D, and letting x = 0, we obtain
A- B=-\.
Letting x = 2, we obtain A + B = 3.
Solving, A = \, B = 2.
Hence ^j.2_c,x^5 12 2 1
(z - 1)3 (X + 2) x-1 (x - 1)2 (x - 1)8 X + 2
EXERCISES
X
2a;2-6a;-5
*• (2a:4-3)«
(X - 4)«
3-\-x
(5 - a-)2
2.T''-1
2x + o
a.'»_6a-«-f-4.r--2.r + 3
a:8(x2 + 1)
(.r-3)(x-
l)"
1.
2.
3.
98. General directions. The following statements indicate the
form of the assumed partial fractions in each case, corresponding to
the types of factors in the denominator of tlie original fraction.
I. Corresponding to the factor x — a, assujne the partial fraction
A
X — a
II. Corresponding to the pri^ne factor or" + hx + c, asstime the
partial fraction a .l. n
ax^ + hx + c
174 HIGHER ALGEBRA
III. Corresponding to the factor (x — a)^", assume the sum
A B K
+ - —; + •■• +
X— a (x — a)'^ (^ — ay
IV. Corresjjonding to the factor (cix~ + hx + c)*, assume the sum
Ax + B Cx+ D Mx + X
ax' ■\-bx + c {(ix^ + hx + c)- (j.ix' + hx + cf
EXERCISES
Separate into partial fractions :
^ x' + x + l , , 2 r-^ - 1
2.
3.
4.
5.
a-^ + 1
a-^ + l
a;^ + .T + 1
a;" — 4 a;^ + cc + 6
2 .r^ + 3 .r^ + a-
6
a-^
(a; - l)'-^ (a- + 2)
a*
7.
a'^ — a-^ + 2 r/?«,a:
(jim — a-) (r/^ -f- x^)
2x^ + o x^ - 6
^' x* + 2a,'3-a,--^-2a;
1
10.
XX.
3 0^-3 V
12.
f + .?/ + 1
(« + l + y)[a(y + lj + ?/]
13.
x^ - a'^ - 1
(X - 1)-^
14.
.T* + a-2
a;* + a;"^ + 1
15.
x^ 4- a;
(a- - 1)-^ (a-^ + 4)
16.
4 a-s _ 5 x + 2
^6 _ ^5 _ ,,.4 _|_ ^3
17.
18.
3 a-3 + a-2 + a- + 1
(5 a,-2 - a- + l)'^
2
3(2a; + l)(a'- + a- + 3)
19.
:,^ + 1
(a;'^ - 1) {x^ - 5 X + 6)
20.
2 + a-^
yy-» -, 0 /-* . Ox
a;(a;-^ + l/ *"' (2 - a-)^ (1 + x'^)
CHAPTER X
LOGARITHMS
99. Introduction. Up to the present point in this book only rational
numbers have been used as exponents. The fractional exponent was
assumed in § 2 to obey the laws of operation which govern the
integral exponents, and its meaning was defined. Biit the mean-
ing of a number with an irrational exponent, like 3"^-, has not as
yet been considered. To treat this subject and all of the delicate
questions which are connected with it is at present impracticable,
but the general meaning of an irrational exponent is not diflficult to
understand. Every irrational number is capable of approximate ex-
pression in terms of decimals, and can be computed to as many places
as necessary by the api)lication of some numerical law. For example,
V2 may be found to as many decimal places as we desire by the or-
dinary process of extracting the square root. None of the approxima-
tions 1.4, 1.41, 1.414, 1.4142, is exactly Vl^, but each one is correct
to as many places of decimals as it contains, and consequently each
number differs by less than the preceding one from the exact value
of ■v2. The value of a number with an irrational exponent may be
approximated in a similar manner. For example, 3"^ msiy be com-
puted with increasing accuracy by allowing V2 to take on a suc-
cession of approximations like that given above. The value of
31.414 _ 3^5^ _ °^3i"4 could be computed directly only with difficulty.
The simplest method which has been explained would be the solution
of the equation a;^°°° = 3"^* by Horner's ]\[ethod.
In what follows we shall assume that exponents may be irrational
numbers, and that the formal laws of oi)eration with these exponents
are the same as already given on page 4 for the case of rational
exponents.
100. Definition of logarithm. I u the preceding section the possi-
bility of finding a in the expression
a = b', (1)
175
176 HIGHER ALGEBRA
when b and x are given, was discussed. The reverse process of finding
X when a and b are given gives rise to logarithms.
Definition. The logarithm of a given number is the exponent
in the power to which a number, called the base, must be raised in
order to equal the given number.
In (1) X is the logarithm of a for the base b, or, symbolically
expressed, x = log^a. Thus
h^ = a and x = log^a (2)
are equivalent relations. The foregoing definition assumes that when
a and b are given, the real number x always exists, an assumption
which is justified when both a and b are positive and b ^ 1.
Although, theoretically, any positive number excepting unity could
be taken for the base of a system of logarithms, the only ones
which are ever used in computations are 10, and the number
e = 2.7182 • • • (the Napierian base), which we shall meet later. When
the base is not expressed, as in log 2, the base 10 is understood, since
10 is the usual base for purposes of computation.
EXERCISES
1. If 2* = 16, what is the value of a- ? log^lG = ?
2. Eind the values of log327, logjolOOO, log^i, log^49.
3. If b^ = 343, what is the value of b ? If log, 343 = 3,b = ?
4. Eind the base b in each of the following: logj32 = 5;
logi,100 = 2; log6.008 = 3.
5. If 4^ = a, what is the value of a ? If log4a = 2, a = ?
6. Eind the number a in each of the following : log^a = 2 ;
logio* = 4; log.5a = 3.
7. Eind the value of loggl; log23l. Show that log^l = 0 for any
positive base b ; that is, in any system of logarithms log 1 = 0.
8. What is the value of log^l ? logj2 ? log^a ? Why can 1 not be
used as base for a system of logarithms ?
9. What is the value of log_2 4 ? log_2 8 ? Why can a negative
number not be used as base for a system of logarithms ?
10. What is the value of log.^ (- 4) ? log_o (- 8) ? log, (- a) ?
Show that negative numbers have no real logarithms for positive
bases.
LOGARITHMS 177
11. Wliat is the value of log 100? log 10 ? log 1 ? log .1 ? log .01 ?
Which numbers have positive and which have negative logarithms ?
12. Wliat is the value of log.^2 ? log^o ? log,// '! Show that in any
system the logarithm of the base is 1.
13. IfV;^=V6.^<^findlog,a. ^^ Show that log, a = .-^.
14. If r;- =100-^10, find log a. ^"^'^ '
2 ' 17. Show that W''^b'^ = a.
15. If ae-^=(e')3e3j findlog^a.
18. Show that logi« = log, - = — log, a.
19. Show that log, a • \o^J) • log„c = 1.
101. Operations with logarithms. The fact that a logarithm is
an exponent lies at the basis of its usefulness, since it enables us to
employ the laws of exponents with telling effect. "We now prove
four theorems which enable us to apply logarithms in numerical
computations.
Theorem I. Tlie logarithm of the product of two numbers is
the sum of their logarithms.
Symbolically expressed, \ogf,(ar) = log^a + logj,c.
Let log(,a = a",
logft^ = y.
Then by (2), p. 176, //" = a,
h" = c.
Multii)lymg, Ij''+» = a • c,
or by (2) ^og^(ac) = x -\- ;/ = log,« + log^c
Theorem IL The logarithm of the nth power of a number is n
times the logarithm of the number.
Symbolically expressed, log, a" = n log,«.
Let log, a = X,
or i"" = a.
Raising both sides to the nth power,
(b'^y = b'"' = a",
or log,«" = nx = ?!log,rr.
178 HIGHEFv ALGEBKA
Theorem III. The logarithm of the quotient of two numhers is the
logarithm of the numerator minus the logarithm of the denominator.
Symbolically expressed, log^- = log^a — log^c.
Let logj,« = X,
log6C = y.
Then Z*^ = a,
b" = c.
a
Dividing, y^-y =
c
or logft - = a- - y = log(, a - log^ c.
Theorem IV. The logarithm of the real ntli root of a number
is the logarithm of the number divided hy n.
Symbolically expressed, log^ ^a = — ^^^^-~ •
Let logj,a = X,
or ' If =L a.
If -
Extracting the nth root, {Ify = ^" = Va,
nr- X logftft
or ' log.Vft = -= •
n n
EXAMPLE
Given log2 = .3010, log 3 = .4771, logo = .6990, log 7 = .845L
Find log (-^-VH).
Solution. log (^78 • Vb) = g log 7 + ^^ log 5,
3 log? =^(.8451) = .5071,
I log 5 = I (.6990) = .3495,
log (-v't^ . VI) = .8566.
Note. In using four places of decimals, when the number in the fifth place
is'less than 5 it is dropped ; when it is more than 5, 1 is added to the number
in the fourth place. When it is exactly 5, 1 is added to the number in the
fourth place in case that number is odd ; otherwise it is dropped. Thus | log 7
came out .50706, which, to four places, is ,5071.
LUGAUlTliMS 179
EXERCISES
Using the logarithms on the preceding page, find :
1. log 210. 3Vl5
9. log —
2. log 32. ^ 7
3. log 18. ,. 1 81
4. log 1225. (V7)
5. log2v^. 8^.9^"
6. log9VlO. ^''^""S^^TT^'
7. log(^.^). ^2. log FVI
8. log (^U. ^7). N^05"'
13. If the edge of a cube is a, its surface S, and its volume V, show
that (a) log 5 = 2 log a + .7781 ; (b) log F = 3 log a.
14. If the edge of a regular tetrahedron is a, its surface S, and its
volume T', show that
(a) log 5 = 2 log a + .238G ; (b) log T' = 3 log a - .8286.
Deduce the following relations :
15. log(/; -jj = log (b + a) + log (b - a) - log/>.
/ 2 b b'^V
16. log (l - — + -2 j = 6 [log (a - b) - log a].
''' '^^ [271^ + 2(TT^1 = - ^'^'^ ^^ - ^) + ^°^' (^ + ^)>
18. log V(2x-3)(a:-l)-6 = ^ [log (2 a; + 1) + log (x - 3)].
19. log ;^E^±1 = ^^ ^1,.. (., + 2) _ log (.r
!)]•
20. loff
^
7TT " a" + .^^ = " '^^^^^•' + ^ '''^^ ^-^ + ^^^•
+
21. Show that if a and i are the legs of a right triangle and c is
the hypotenuse, then
log a = ^ [log (r + /') + log {c - /')] ;
log^.= Klog('' + ") + log (^'-'Oi-
ls there a similar formula for logc?
180 HIGHER ALGEBRA
22. In a right triangle given c = 285, h = 215, find a (see
exercise 21).
23. In a right triangle given c = 34.69, a = 26.21, find b.
102. Tables of logarithms. Explanation of the method of looking
up logarithms and antilogarithms in tables will not be given here.
The student who is unfamiliar with the use of tables is referred to
books on elementary algebra or trigonometry for a detailed discussion
of this procedure. A four-place table is found on pages 212-213,
together with a rule for its use.
103. Exponential equations. An equation in which the unknown
occurs in an exponent is called an exponential equation. The use of
logarithms is usually required to solve such equations. In these
solutions it should be remembered that logarithms are nothing but
numbers, and should be treated as such.
EXAMPLES
1. Solve 2"-'* = 5.
Solution. Taking the logarithm of each member of tlie equation,
log (2^- 4) = log 5.
By Theorem II, p. 177, (x — 4) log 2 = log .5,
, log 5 , .6990
or a; = 4 + -^ = 4 +
log 2 ..3010
= 4 + 2. .322 = 6.322.
log .5
It should be kept in mind that in the fraction — - — both numerator and
log 2
denominator are numbers, and that it is the quotient of these numbers which
loc 5
is called for. Log| is a very different number from —2— and should not
be confused with it. °
2. Solve 4^'- 2^= 64.
Solution. 4x2-2x — 43
Taking the logarithm of each member of the equation,
(x2-2a;)log4 = 31og4.
Dividing by log 4, x^ — 2x = 3,
or (x - 3) (x + 1) = 0.
Hence x == 3 or — 1.
LOGARITHMS 181
EXERCISES
Solve, obtaining results to three figures :
1. 2^ = 19. I
16. (7.2y=(5.9Y.
2. (3.1)- = 90.7. \ J \ J
3.10-^ = 20. 17. a)-^ = 2o3-^.
4.3-^ = 21.45. 18. 18«-=(54V2r-.
5. 5^ + 2 = 7-3 19- 4^^-8(4^ + 12 = 0.
6. (2.2)«-i = (3.3)- + *. 2°- ^^'"^ - 16(64-)+ 64 = 0.
7. 2-'-3- = lG. 21. Va'^- + '' . -Va«-+io = a^ . V^.
8. 6--' = 36- 2a. ^'^'^'.
9. (4^')^~^ = 2* ^ \t'/ ^ \«/
10.' 2^^ = 64-.''" 23. (Uy—'=(W='-'-
<, > n ! g + 5 1+17
11. a^--' = b-^-\ 24. 32- -7 =(.25)128^^.
12. a3^-Tx + 2^i 25_ ,^/___^5^^^__^
13. Vo^ = V a- Va--3. 26. 8*^ = 6^^
14. -v^=^~Va-. 27. 3- — 5- + 2 = 3a: + 4 _5x + 8_
15. '-V243 = -v^. 28. 9.7-- • (1105.8)- = 57— ^
29. 5- + 5- + 1 + 5- + 2 = 3- + 3- + 1 + 3-+-.
30. Solve the simultaneous equations :
3- = f,
18 y- - i/- = 81.
31. What dithculties are met if one attempts to solve such equa-
tions as the following : (a) 2- + 3- = 5-; (b) x^ =2-; (c) ar= = 2.
104. Compound interest. Let P represent the number of dollars on
interest and /• the rate, expressed in hundredths. Thus if the interest
is 6%, r = .06. Then the interest at the end of one year is /• • /* dol-
lars, and the accumulation at the end of the year is
1> + rP =P(1+ r) dollars.
The interest for the second year is P (! + ?•)• /•, and the entire
accumulation at the end of two years is
P (1 + ;•) + /'(! + r) . /• = (1 + /•) (P + ,-P) = P (1 + /•)- dollars.
Similarly, at the end of ii years the accumulation is
.1=7'(1+;-)" dollars.
182 HIGHER ALGEBRA
By means of this formula A , P, r, and n are related, and if all but
one of these are given, the remaining one can be found.
For example, if A, P, and r are known, n can be expressed in terms
of these by first taking the logarithm of each side of the equation
and solving as follows :
loo"^4 — logP
logyl = logP + ?ilog(l+ r), or n= ^^ n i /a '
If the interest is compounded semiannually, at the end of the first
half year it is ( -]p, and the accumulation at that time is
p(l + ^) dollars.
Proceeding as in the case where the interest was assumed to be
compounded annually, it is found that the accumulation for n years
when the interest is compounded semiannually is
A=pIi + ]A dollars.
log.4-logP
In this case n = — - — —
21og(l + |
EXAMPLE
Find the accumulation at the end of 10 years on $1500 at 4'y^, com-
pounded semiannually. Find the limit of error of the computation.
Solution. A-P(l +
' 2
•In
P = 1500, r = .04, ^1 = 10.
/ 04\20
^=1500(1 + ^) =1500(1.02)20.
log 1500 = 3.1761
20 log 1.02= .1720
log^ = 3..3481
2229 dollars
To determine the limit of error of this computation it is necessary to observe
that the limit of error in the table of logarithms is .00005 ; that is, the true value
of any logarithm may be greater or less than the one given in the table by not
LOGARITHMS 183
more than tliis number. Hence in multiplying log 1.02 by 20, the possible error
is 20 X .00005 = .001. In the logarithm 3.1761 there is a further possible error of
.00005. Hence the total limit of error in log A is .00105 ; that is, the true value
of log A is between 3.3470 and 3.34it2. Keference U) the table shows that this
amount of error in the logarithm would correspond to an error of 0 in the fourth
significant figure of the antilogarithm. Hence the limit of error in the result
is 0 dollars. If a result correct to cents is desired, seven- or eight-place tables
would be necessary.
105. Change of base. As we shall see on page 207, the computa-
tion of logarithms is actually carried out, not for the base 10 which
we ordinarily use in our tables, but for the base e = 2.7128 • • •. In
order to pass from logarithms for one base to those for another we
need the following
Theorem. logx = ^i^. (1)
log^e
Suppose that the logarithms of all real iiuniliers have been found
for the base b.
Let a; be a number whose loejnritlim for the new base, c, is desired.
Suppose that logc« = z ; that is, r — x. (2)
Taking the logarithm of each member of this equation for the
or
z logjC = log^a;,
logft.r
Hence by (2)
1 logft-^
log^.r = .j
It will be necessary to use the foregoing theorem on page 210 in order to
obtain the logarithms for the base 10 from those for the base e.
EXAMPLE
Find log^ 5.
Solution. Here X = 5, 6 = 10, c = 2.
log 5 .6900
Applying (1), we find log„ 5 = = = 2.322.
^^•^ '^ ' - log 2 .3010
It is observed that this question is equivalent to the following : Find the
power to which 2 must be raised so that the result will be 5.
184 HIGHEE ALGEBRA
EXERCISES
Eind the accumulation on each of the following :
1. P dollars for n years at the rate r compounded quarterly.
2. ^1200 at the end of 8 years at SfJ^ compounded annually.
Find the limit of error of the computation.
3. $850 at the end of 12 years at 6% compounded semiannually.
Find the limit of error of the computation.
4. |1500 at the end of 10 years at 4% compounded quarterly.
Eind the limit of error of the computation.
5. $75 at the end of 6 years 8 months at S'y^ compounded
annually.
6. In what time will a sum double itself at 4^ compounded
annually ? at 5% ?
7. At what rate will a sum double itself in 20 years, interest
compounded annually ?
8. At what rate will a sum treble itself in 15 years, interest
compounded annually ?
9. In what time will a sum double itself at 5^ compounded
semiannually ?
10. A certain society offers a life membership for $50, which ex-
empts the member from further dues. Other members must pay
$5 annually. Counting interest at 5'y^, show that if a member lives
more than 13 years after joining the societ}-, it pays him to take
out a life membership.
11. What rate of interest payable annually is equivalent to 5%
payable semiannually ?
12. A house worth $5000 is let for $400 a year, payable at the
end of each quarter. If the tenant wishes to pay at the end of the
year, how much must the rent be raised in order that the landlord
may obtain the same rate of interest as before ?
13. Eind (a) log^S; (b) log^S; (c) log,.3(11.98).
14. Eind the value of the product
logs 4 • log4 5 . log5 6 . logs 7 • log, 8 • log, 9.
15. Seventeen is what power of 3 ?
16. To what power must 2 Vs be raised to obtain V7 ?
CHAPTER XI
INFINITE SERIES
106. Variables. A letter which, during a given discussion, may
take oil several distinct values is called a variable. A variable need
not take on all or even many numerical values. It is not uncommon to
speak of x in the equation ax^ + fta; + c = 0 as a variable, although it
can take on only two values and at the same time satisfy the equation.
It should be noted, however, that in considering the function ax^ + 6x + c,
as we do, for example, when we plot it, x is a variable which takes on all real
values.
In equations like 2 .r + 3 y = 4, both x and y take on countless
values and both are called variables. Usually the values which a
variable may take on are limited by some law which is frequently
expressed by means of an eciuation.
In the e(|Ui\tion 2x + 3?/ = 4 tlie variation of x and y is limited to those
values which satisfy the equation. For example, if x equals 8, the corresponding
value of y is determined by the equation to be — 4.
107. Infinity. If a variable takes on the succession of integral
values 1, 2, 3, • • •, we can think of no greatest value of the variable;
for when we imagine a certain integer as the last one, we can immedi-
ately think of a greater. We express this condition by saying that as
the variable takes on the positive integral numbers in order, it be-
comes infinite. To say that a variable becomes infinite is a short way
of saying that a value of the variable exists which is greater than an
arbitrarily chosen number M, however great .1/ may be. Infinity is
not a number, and must not be used in operations as if it were. It
is merely a name to indicate that a variable has become greater than
any number. It is often symbolized by oo .
108. Limits. Consider the set of numbers
1 3 7 1 .1 . . . CW
From the numbers which are written one can determine as many
more as desired following the same law. As one reads toward the
185
186 HIGHER ALGEBRA
right, the numbers increase, each one being the arithmetical mean
2" — 1
of its predecessor and unity. The nth. term is — — — However
far we continue the set of numbers we never find one which ex-
ceeds or even equals 1. But however small a number we may think
of, say .01, we can find a number in the set which differs from 1
by less than this number. The number ||| is the first in the set
which differs from 1 by less than .01. If we had thought of .001,
.0001, or any other small number instead of .01, we could have found
a number in the set further to the right which differed from 1 by
less than it. The numbers of this set may be considered as different
values which a variable x assumes. We have the following
Definition. If a variable x takes on values in order, such that
the difference between x and some fixed number A becomes and re-
mains numerically less than d, hoivever small d may be taken, then
X is said to approach A as a limit.
In the case mentioned, the value we first took for d was .01, and
we saw that i|| differed from the fixed number 1 by less than .01.
And not only this, but all numbers in the set further to the right
differ from 1 by even less. As a matter of fact, this set of numbers
actually approaches 1 as a limit. This can be proved by showing
2" — 1
that for a sufficiently large value of n the value of — -^-— differs
from 1 b}^ as little as we please.
The reason for including the words " and remains " in the definition may be
appreciated if the set of numbers (1) be replaced by a set in which the same
numbers are found, except that every alternate number is replaced by the
number 2, tfuis: 1, 2, |, 2, § .^, 2, • • •. Then the set would not have 1 as a limit,
for although a number of the set could be found which differs from 1 by less
than any assigned value, yet the very next number of the set differs from 1 by
unity. These 2's may occur in the early part of the set and not affect the ap-
proach to a limit, but they cannot appear throughout the whole extent of the set.
The approach of a variable to a limit may be illustrated geometri-
cally by considering the numbers of the set (1) as measuring distances
on a line.
0 14 % '/s ^Yi6 1
— H 1 I 1 I III ■
We have already stated that if, after taking on a certain finite
number of values, the variable always remains less than the distance
INFINITE SERIES 187
d from A, however small d may be taken, it approaches A as a limit.
This condition is illustrated geometrically by a bunching of the
points near the point representing the limit.
If, instead of considering the variable x, we deal directly witli the
set of numbers, the foregoing detinition may be given in another
form, which lends itself more conveniently to symbolic expression as
follows :
If, in the set of numbers^ Mj, Wa, Wj, • • •, m„_i, m„, • • • a subscript
m can he found such that the difference hetiveen any m„ and A (jvhen
n is greater than rn) is numerically less than d (ivhere d is an arbi-
trarily small positive riumber^ then the set of lis has A for a limit.
Expressed in symbols.
If |//,, — .1 |<rZ, for n^m, then limw„ = ^.
n=oo
Many variables cannot take on their limiting values. Most, if not all, of the
variables which one meets in elementary geometry are of this character. Such
limits are sometimes called inaccessible. Other variables do take on their limit-
ing values. For example, the distance from a falling particle to the ground is
a variable which approaches and takes on the value zero. Such limits are often
called accessible. But whether the limit is accessible or not, is of no consequence
in the definitions given above.
We are nmv in a position to see that the fraction - approaches zero as a
limit if a is a constant and n is a variable which becomes infinite. Consider,
2
for example, the fraction - • The process of determining whether this fraction
approaches zero may be explained clearly by means of the following dialogue, in
2
which Henry claims that - does not approach zero, and John contends that it
does. „
Hairy. "I see no reason why this fraction - approaches zero as a limit."
John. "You must admit that it does approach zero if, -when you name any
number as small as you like, I can find a value of n .so large that for my n and
2
all larger values the fraction - is less than your small number."
Henry. "' Yes, I admit that, for it is in accordance with the definition of the
limit of a variable."
John. "" AVcll, then, name a small number."
Henry. " I challenge you to find an n which will make the fraction less
than .0001." ^
John. " If Ji has the value 100,000, you will find that - is less than your .0001.
Henry. " I see that ; but suppose I name .000001 ? "
John, '"rhcn I would k-t n have the value 10,000,000."
188 HIGHER ALGEBRA
Henry. "There is no use in continuing this further, for I see that whatever
10 2
small number, as A:, I may name, if you take n equal to — , then the fraction -
k k n
becomes -, which is certainly less than the k which I named."
5
In a manner similar to that outlined in the foregoing dialogue it may be seen
that - becomes and remains less than any small number k for all values of n
equal to or greater than Hence the fraction - approaches zero as a limit.
k n
109. Infinite series. We are familiar with sums, like a-{-b-\-c,
which have a detinite number of terms. We have also used sums,
like x" -\- r/j.T""^ + • • • + «„, which have an indefinite number, n, of
terms ; but we have always assumed that n has a finite value, so that
the operations which are indicated in any such function can actually
be performed in a finite length of time. An Infinite series is the indi-
cated sum of a never-ending or infinite set of terms. Since we can
never write down all of the terms of an infinite series, it is essential
that from the few which we do write the law may be apparent by
which we can find as many more as we desire.
The infinite series whose terms are u^, u,^, n^, • • •, ?/„, • • • is often
denoted by "V^^, read "summation ?/„," or by ^"nj ^^^^ "summa-
n = 1
tion «„ from ?i = 1 to ?i = oo ." Thus we write
Z^ "« = "i + ^'2 + "3 "^ ^ "« ^ •
7i = l
The nth. term of this series is «„; that is, the subscript is the same as
the number of the term. The sum of the first n terms is denoted by 5„.
Thus S„ = u^ -H «2 + "3 + h w„,
and S^ = «j, S.^ = v^ + u^, S^ = u^ + »,, -f u^, etc.
Sometimes an infinite series is written in the form
CO
2
"„ =
^'0 + "1 + "2 -I + "« - 1 + ^'« +
In this series the nth. term is it„_i, the subscript being one less
than the number of the term, and ?/„ being the (n + l)st term of
this series. The sum of the first n terms is then
Sn = "0 + ^'1 + ^h -^ 1- "n-i-
Throughout this chapter ?i will represent any positive integer
or zero.
INFINITE SEIJIKS 189
EXAMPLES
oo f) -|
1. Write down the first live terms of the series X "~. ^ AVhat
is the 10th term ? the nth term ? "=^
Solution. > =r.J I I I
^. 2" 2 22 2" 2* 25
n= 1
^. ,^ . . 2-10-1 r.) ^, , . 2 71-1
The lOth term is = The nth term is -•
210 210 2'»
gp O I -1
2. Write down the hrst five terms of the series 2^- ;-r,-
What is the 8th term ? the nth term ? the (n ■+■ l)st term ?
2n+ 1,3579
-' + 2^ + 3^ ■ •
2-7 + 1 15
Solution. > Z — =1J 1 1 1
^o(n + 1)2 ^2"' 32 ^ 42 ^ 52
The 8th term is
82 82
rr, „ , - 2(ji-l) + l 2)1-1
The Tjth term is —
The {n + l)st term is
[(n - 1) + 1]"^ n2
2n + l
(H+l)2'
EXERCISES
Write down the first five terms of the following series
*o
1.
11 = i
5.
Z (2 'O^
»•= 1
2.
00 M
-^ On
,,= 0"
6.
Z(2ri + 1)!
3.
CO ^
n = l
+ ^)-
7.
00 -|
,:f/ ') 2.-1
4.
8.
»■ St-
"> Qn + l
10. Sc-i)"^^-
H = 0
00
11.2;
2 ?i - 1
^.n — S
2w
12. t(-i/"'i
il = l -<
13. ^^'rite down (a) the 7tli term in the series of exercise 3;
(b) the 8th term in the series of exercise 6 ; (c) the 9th term in
the series of exercise 10 ; (d) the 10th term in the series of
exercise 12.
14. Find (a) the sum of the first five terms of the series in exer-
cise 2; (b) the sum of the first four t.'iius of the series in exercise
7 ; (c) the sum of the first six terms of the series in exercise 5.
190 HIGHER ALGEBRA
15. How many terms of the series of exercise 2 must be taken in
order to make .S',^ differ from 2 by less tlian .001 ?
16. How many terms of the series of exercise 5 must be taken in
order to make <S„ greater than 1000 ?
Find the nth. term in each of the following series and express the
series in the 2 notation :
17. 1 + 2 + 3 + 4 + ....
i« i-i-l - ^-
18. 2'2 + 32 + 42 + 5'2 + ••••
19. 12 + 32 + 52 + 72 ^ ,
20. 1 • 2 - 2 • 3 + .3 . 4 - 4 . 5 + . . ..
11 1
^^- ^ + 172 + 17273 + 1. 2. 3. 4 + -*"
22. X + x^ + x^ -\- X' -\ .
1111
23- - + -T4 + -9 + ^+---
tAj «A. »^ «A/
3 3'^ 3^ 3*
04. _ -I . -J I 1- . . .
52 ^ 10'^ 15- 20^
^.2 ^A ^,G ^,8
25. — = + -^ + ^ + — ^ + . . ..
V2 Vl V(5 Vs
x'^ X* x^
^^' 27 + 4] ~6T"^ ■
12 3 4
27. 1 1 1 h • . ••
2.3 3.4 4-55.G
2J 3] _ £!
*°' -^ 92 |- 02 42 "■" ■ ' ■■
29. In the series of exercise 23, compute .S'^ if cc = 2.
30. In the series of exercise 26, compute S^ it x = ^.
110. Convergence and divergence. An inspection of the two
following series indicates that they are of quite distinct types:
2;2" = l + 2 + 4 + 8+.... (2)
n = 0
INFINITE SEIMES 191
111 (1) each term adds only oiu; half the difference between the
sum of tlie preceding terras and 2. Consequently, however many
terms we may add together, we can never obtain a sum which exceeds
or even equals 2.
Since there is an infinite number of terms in an infinite series, it
would be impossible to compute their sum in less than an eternity
of time. But since this is not at our disposal, it is, in strictness,
without meaning to speak of the sum of the terms of an infinite
series, for such an operation could never be performed. In the case
of series (1), 2 is not the sum of any number of terras which we
could write down ; it is greater than any such sum. But it is
approached as a limit by the sums of increasing numbers of terms.
In spite of the fact that ^ is not really the sum of the infinite num-
ber of terms of (1), but the limit of S„ as n becomes infinite, never-
theless it is called the sum of the series.
Definition. WJien the sum S„ of the first n terms of an infi-
nite series approaches a limit, as n becomes infinite^ the series is said
to he convergent. This limit is called the sum of the series.
In most cases it is a simpler matter to find that this limit exists, and hence
that the series in question is convergent, tlian it is to determine the exact value
of the limit.
In series (2) each term is greater than the preceding one, and by
adding a sufficient number of them a sum can be obtained greater
than any number which we may name.
Definition. When the sum **?„ of the first n terms of an infi-
nite series does not approach a finite limit as n becomes infinite, the
series is said to be divergent.
Consider the series 1 -f 2 -f- .'? -f- • • • + « -f- • • • .
In this case a value of n can be found so great that the value of
S„ is greater than any value which can be assigned. Hence S„ does
not approach a finite limit. In the case of all the divergent series
considered in this text, the value of 5„ becomes greater than any
assigned number, provided we take 7i large enough.
There is another kind of divergent series of ^vhich
1-1 + 1-1+--- (1)
is the type. Tliis is called an oscillating series, because the values of S„ oscillate
between certain values, but never settle down to a limiting value. In series (1)
iS„ is either zero or 1, according as n is even or odd.
192 HIGHER ALGEBRA
Theorem. If each term of an infinite series ivith positive
terms is greater than a fixed ^lumber, however small, the series
is divergent.
Eor a sufficient number of terms, each greater than this fixed small
number, would add up to a sum greater than M, however great M
might be.
This theorem assures us that none of the following tests for convergence are
necessary unless the terms of the series approach zero as n becomes infinite.
111. Comparison test for convergence. The problem of finding
whether a given series converges or not, and that of finding the exact
value to which it converges, are quite distinct. We shall give some
of the most important methods for attacking the former problem,
but shall content ourselves with computations for obtaining the
approximate value of the sum of the series.
In what follows we shall make use of the following
Assumption. // S^ is a variable which ahvays increases when n
increases, but which Jiever exceeds some finite number D, then S^
approaches a limit A, ivhich cannot be greater than D.
The only type of series for which we have hitherto derived any
test for convergence is the geometrical series
S= a-\- ar + ar^ + or^ + • • • + r//-" + . . . = —^— , (1)
1 — r ^ ^
where r is numerically less than 1 and a is any real number (see § 10).
Consider the two series
- 1 1 ^ 1 ^ 1 ^
1
•+ h +
71 1
" "'2'2.32.3.
4 '
'•-^+1+ i + §
+ •
••+2-' +
An inspection shows that each term of S after the second is less than
the term of S' which is directly below it. Furthermore, the nth. term
of S, namely — > is less than the nth term, - — r > of S', since
' ^ nl 2"~^
2 • 3 • • . n > 2 • 2 • • • 2, to 7i - 1 factors, if ti > 2.
TXFTXTTl-: SFJJTKS 193
But S' is a geometrical series with the limit 2. Hence we would
expect S to converge to a limit not greater than 2. That this is the
case follows from the following general
Theorem. Let u + "., + "g + • • • ''<-' <itt htjinlte series of positive
terms ivhich is to be tested. If a series of positive terms v + /• -j-
V -\- • • ■ can lie found, which is known to converge, and is such
that each term of the u-seHes is equal to or h'.s.s than the corresjjond-
ing term in the v-series, then the u-series must converge, and its sum
is equal to or less than the sum of the v-series.
Let the sum of the v-series be A.
Let >'„ = Wj + )i., + "3 + • • • + "„
and ,S"„ = v^ + r,, -I- ,-3 + • . • 4- r,„
where n is any positive integer. Then since the second series con-
verges to A, we have -i- e' _ 1
n = 00
Since all of the terms of the r-series are positive, we have
s:<A.
But, by hypothesis, 5„ ^ S',^.
Hence 5„ < ^-1 ;
that is, the sum of any number of terms of the ?<-series is less than
a fixed number. Hence by the assumi)tiou on page 192 the limit of
.s'„ exists and is not greater than .1 ; that is, the ?<-series converges.
It is often necessary to disregard some of the first terms of a series in order
to apply tills theorem. But since the sum of any finite number of terms must
be finite, it is sufficient to sliow that the series after a certain number of terms
converges.
To test a series of positive terms for convergence, we write down
the nth term of the given series, or the nt\\ term of the series which
remains after omitting some of the first terms from the given series.
Call this term -?/„. Kow compare this with v„, the «th term of a series
known to be convergent. If «„ ^ r„ for every value of n greater than
any particular integer, the ?/-series is convergent. This is called the
comparison test for convergence. The i?-series is called the comparison
series. If //„ does not turn out to be equal to or less than r„, this does
194 HIGHER ALGEBRA
not prove that the ?<-series is not convergent ; it merely shows that
the y-series used is not effective as a comparison series. Any series
derived from (1), p. 192, by substituting any real number for a and
any positive number less than 1 for r is known to be convergent and
can be used as a comparison series. After any series has been proved
convergent it can be used as a comparison series for proving other
series convergent.
EXAMPLE
Test the series
^■=-+l + 2-+l + l + l + --
Solution. Disregarding the first two terms, the nth term of the remaining
series is
1
?t„ =
{n + 1)"
Use as a comparison series the geometric series (1), p. 192, where a = 1, r = i,
and tlie first term is dropped.
1111
2+2^ + 2^-
Then ^" = z + _ + _^ + ^ + . .
This series is known to be convergent and its nW\ term is r,, = — We must now
2"
show that M,j ^ r„ for all values of n greater than some integer ; that is, that
1 ^1
{n + 1)»'^2"'
or (n + 1)« ^ 2". (2)
This is tnie for all values of n>0, for if n = 1, (2) becomes 2 = 2; if n>l,
evidently (?i + 1)" > 2". Hence S is convergent.
EXERCISES
Test the followins: series :
]. 1 1 , .1,1.1.1.1
22 + 33 + 44
1. l + ^ + ^ + ,,+--. 4. ^ + - + ^ + ^ + ^ +
2 "*" 2* "^ 2' "^ 2^*^ "^ ' 1 • 2 "^ 3 • 4 "*" 4 . 5 "^
.111 ,111
3-1 + 23 + 33 + 43'^ • ^- 1 +^ + 3"2 + 42-1 •
INFINITE SERIES 195
11. State an assumption siniilur to that of the preceding section
regarding the limit of a variable which continually decreases, but
which remains greater than a fixed number.
12. State and prove a theorem similar to that of the preceding
section regarding the convergence of a series, each of whose terms
is negative.
112. The Harmonic Series. One of the most important series for
the pur])()ses of testing divergence is the Harmonic Series,
1 -U .1- -1- .1 J- J_ -I- . . .
The terms of this series become smaller and smaller and approach
zero as a limit. It is difficult to believe at first sight that the sum
of terms of this character can add up to a value greater than any
number which we can assign. lUit we can prove the
Theorem. The Harmonic Series is divergent.
Consider the terms of this series grouped as follows, the successive
parentheses containing 1, 2, 4, 8, 16, • • • terms respectively :
Since in the second parenthesis \ is greater than ], their sinn is
greater than twice |, or ^. Similarly, in tlie third parenthesis, the
sum is greater than four times ^, or I ; that is, by arranging the
series in this way we see that it consists of the sum of an infinite
number of groups of terms, each of which is greater than J. Hence
the sum of the series does not exist, and the series is divergent.
113. Comparison test for divergence. We can now compare a
series with the Harmonic Series and obtain a test for divergence
similar to that for convergence in §111. For example, consider the
two series 111 1
2 + 3 + 4 +••■+ » +••■
and — p H ^ -\ p + • • • H p + • • •.
V2 V3 V-4 Vn
196 HIGHER ALGEBRA
Since the denominator of each fraction in the second series is less
than the denominator in the term directly above it, each term of the
second series is greater than the corresponding term of the first.
But the first series diverges ; hence one would expect the second to
diverge also. That this is the case appears from the following
Theorem. Let u^ + u^^ + w^ -^ ■ • • he an infinite series of positive
terms which is to he tested. If a series of positive terms, v + v +
V -\- ■ • •, can he found which is Icnown to diverge, and is such that
each ter7n of the u-series is equal to or greater than the corresponding
term of the v-series, then the u-series must also diverge.
Suppose we have ti^ + u,^ + ^'g + • • • + "„ + • • •
and ^1+^2+ "3 -I ^ '■« + • • •>
and suppose that the w-series diverges ; that is, that a value of n can
be found such that
l\ + V.^ + ^'3 + • • • + ^-n > -V,
where M is a number taken arbitrarily large.
By hypothesis % = v,. for every integral value of k. From these
hypotheses it follows immediately that
^'1 + "2 + "3 + ^ '"» ^ '^^5
that is, that the ^-series diverges.
The preceding theorem shows that in order to prove that a series
is divergent we may show that its nth term, u„, is equal to or greater
than the nth. term, v„, of a known divergent series, for all values of
71 greater than some integer. This is called the comparison test for
divergence.
Besides the Harmonic Series a useful one to employ in testing for
divergence is the geometrical series in the case where the ratio is
greater than unity. This may be written in the form
a + ar + ar + ar^ -\ h 01" + • • •, r > 1.
Here a may be given any convenient numerical value and r any value
> 1. The series is then knoAvn to be divergent.
As in the test for convergence it is often necessary to neglect the first few
terms in the application of the theorem of this section. In so doing we merely
recognize that if a series from a certain term on diverges, the entire series
must diverge.
I^'MNITE SEUIES 197
EXAMPLE
Test the series 1 + i + 5 + 4 + ?, H •
Solution. Till- ?itli fcnii <>f tins series is u„ =
n J
Comparing this willi the nth term of the Harnionic Series, »„= -, we will
show that Ti— 1 1 f-,.
n n
when n is greater than some particular integer. This is true when n>l; for
when ri = 2, (1) becomes ^ = ^, and wlien n>2, evidently n — 1 > 1. Hence the
series is divergent.
EXERCISES
Test the following series :
^•^!-^^— ■ 3.i.^i.^.....
3 r> 7 ft
^- 1 + 5—2 + ^ + 3:4 + 1:5+ ■■■•
>• 2j t~> — i"
114. Indeterminate forms. In the sections which follow it will
be iR'cessary to consider the limit approached by a fraction when
both numerator and denominator become infinite.
n -\- 2
Consider, for example, the fraction — r • If n takes on the posi-
Zn -\- X.
tive integral values in order, we can find a value of n so large that
both numerator and denominator of the fraction are greater than
any definite number, like 1000 or 1,000,000, whicli we can imagine ;
that is, both numeratoi' and denominator of the fraction become
intinite as n becomes infinite. The limit of the fraction, when written
in this form, does not seem to have any definite numerical value.
But let us divide both numerator and denominator of the fraction
by 11 before we begin to let n increase in value. We then obtain
i+?
11 -if- '2 n
= If in the fraction in the right member of this
271 + 1 .,^1
^ "T"
n
equation we let n take on larger and larger values, it appears that
198 HIGHER ALGEBRA
12 ... 1
the fractions - and - approach zero as n becomes infinite, leaving —
as tlie value which the original fraction approaches. This fact we
write as follows : lim 7: ^ = - •
Sometimes we can accomplish a similar result by other means.
2 ra + 4
Suppose we have given the fraction — — • If n becomes infinite,
both numerator and denominator of the fraction also become infinite.
But if one observes that = ~ ? and the factor n + 2 is
?i + 2 7i + 2
canceled from numerator and denominator before n begins to take
on the increasing values, it appears that the value of the fraction is
always 2, and that this value is entirely independent of what value n
may have. Hence lim -r- = 2.
^ » = «> /i + 2
The point in each of these methods consists in throwing the
variable into a position so that the limit of the fraction can be
found when the variable finally becomes infinite.
1. Find lim
EXAMPLES
li" -1 71 + 2
3 n' 4- 1
Solution. Dividing numerator and denominator of the fraction by rfi, wc
obtain
-■ 7 2
1 + —
n^ — 7 n + 2 _ n rfi
^^' + ' ^ 3 + i
7 2 1
Letting n become infinite, each of the fractions - , -- , and -^ approaches 0,
and the original fraction approaches \.
Hence „ „
,. n2 — 7n + 2 1
hm
2. Find lim
3 n^ +1 3
n' + 2/1+1
71^-3
Solution. Dividing by n^, we obtain
- + - ~
n- + 2 ?i + 1 _n n'^'^ V?
v?-Z 3
INFINITE SERIES 199
Letting n become infinite, the numerator approaches zero, while the denomi-
nator approaches 1. o r> , «
Hence lim = - = 0.
,i = x. n* — 3 1
3. riiiil lull j-^--
" = * ^''- (n + l)!
Solution. -^ — — '— = n + l.
Iim(^i + ili =
Hence lim ^^ — = oo.
n = to 71 !
EXERCISES
Evaluate the following limits :
2 71 ' ^' -^^"^
n = «
n = * (n + 1) V?J, + 1
, »
^ T ^^! _ ,. 5(4/1?
4. lim 7 — — —-• 11. hm — ^ ^
n = x(^^ + l)! ^^- r:4(7i + l)«
5- ii"^?— rTr2* 12. Inn ... ^ , — ., ., ; 4x •
K = 00 {n -f- 1) „ = x7i-(n* + ii'a- + «^)
3 ?r - 6 «(l-3-5---2?i-l)
6. hm-^ —• 13. hm '; ^ , -^
n = X
,. (>^ + l)V^rTT 2-4.6---271 + 2
wVre „=x(/i + l)(2-4-6---27i)
115. Ratio test. The test which in many cases is the most power-
ful and at the same time is the simplest to apply is described in the
following
Theorem. Let n^ + ?/., + u,^ -\ \- ?^„ -\ he an infinite series
of positive tei^ns which is to he tested.
I. Jfi "'^ _i±l < Ij the series is convergent.
II. If ^^^n -!L±i ^ 1, the series is divergent.
n=co 11^
III. If I'^^n _iL±i = \^ the test fails to give us any inj^ormation.
200 HIGHER ALGEBRA
I. Suppose that _^ _^_+i_ _ ^^ where t is a constant the precise
value of which we need not specify further than to say that it is
positive and is less than 1. From the definition of the limit of a
variable (p. 186) we know that for all valiies of n equal to or greater
than some integer vi, the variable ratio -1±1 differs from its limit
n
by as little as we please. Let r be a number greater than t but less
than 1. Then we can find an m so large that for all ?i's equal to or
0 t r I
1 1 1 1 —
greater than m the variable ratio """"^ will differ from its limit t
by less than the quantity r — t; that is, each ratio, for values of
the subscript greater than m, will be less than r. Symbolically
expressed, -iL±i <; ,.^ when n = m. Letting n take on the values
n
m, m + 1, m- + 2, • • •, we have, then,
—-—<r, or ^(,n + l<ru^,
'^'m + 1
<r, or y,n + o<ru„, + i<i-'u„,
--^ < r, or w,„ + 3 < ?•»,„+, < I'hi^.
"'m + 2
Adding the inequalities on the right, we have
< w„ (r + r' + '•'+•••)•
The expression inside the p)arenthesis is a geometrical series, and
since r < 1, it converges to some limit, say, L. Then we have
Hence the original series converges (p. 192).
II. If t is greater than 1, we may take r less than f but greater
than 1, and by the definition of the limit of a variable we can find
a value, w, of n so great' that for it and all greater values the ratio
INFINITE SERIES 201
in question will (lirCcr fioni t by less than tlit; quimtity t — r and
7/
hence be greater thun /•; that is, "'"'"'>?•, where r>l.
0 \ r t
\ 1
It follows that 7/,,, ^ 1 > ?■//„ ; that is, each term is greater than the
preceding one. Hence the series must diverge (§ 110).
III. When t = l we are unable^ t(j determine by this method
whether the series converges or diverges.
The present test fails, for example, for the Harmonic Series, though we have
proved that the series diverges. It also fails for the series in exercise 5,
p. 11)4, although the series is convergent.
EXAMPLE
2- 3"^ 4"'^
Test the series 1 + 91 + q^ + TT + ""-
o , . «^ (« + 1)2
Solution. H„ = , Un + i = — ,
n ! (h + 1) !
where u„ + i is obtained from h„ by replacing n by n + 1.
Un + i ^ (n + 1)2 111 ^ n+ 1
u„ (ji + 1) ! ' n- n-
since (n + 1) ! = (?i + 1) • n !,
lim "» + ^ ^ lim n + 1 _ ]i,„ /1_^ 1\ q^j_
» = °° w„ " = " n'^ » = »\n JiV
Hence the series is convergent.
EXERCISES
DeterniiuL' whutlier thci'olluwing series are convergent or divergent;
111 1 '^ S 4-
^' ^"^27'^3!"^r! + "*- ^' 3^3^"^3'«"^3^ + ■■■•
123£ 2-^3^£
^' 2"^2- 2«"*"2*'^'"' ^' "'"2!"^3!"'"4!"^'"''
2! 3! 4! 2i« 3'" 4'" 5"^
3- 1+^ + ^ + 7^+. ••. 7. - + -;^ + -:;j + -;^ + ---.
^ fj '-t W «^ Arf «rf
1 1 1 «1_l2,3,
*• ^+2V2+3V3+4V4+-- ^•2T3 + 3:i + 4.5 + ---
202 HIGHER ALGEBRA
43 "*" ^3 + "1^23 16^ ^2*""
10. 1+2-3 + 3, + ^,+ .... 12. X 3.6.9...3.
^ " n=l
A 1 1-4 1.4-7
"^ 10 10 . 20 "^ 10 • 20 . 30 "^ " ■'
,. . 1,1.3 1.3.5
^^- ^ + 2^^ + 2!T2^ + ^!T2^+-'-
116. Alternating series. Up to the present all of our tests for
convergence hold only for series with positive terms. The simplest
case where some of the terms of a series are negative is that of an
alternating series ; that is, a series in which the signs of the terms
alternate. For this case we have the
Theorem. If the absolute value of each term of an alternating
series is less than that of the 'preceding term, and if the limit of the
nth term is zero as n becomes infinite, the series converges.
Given the series
^ = "x - "1 + "3 - "4 + "5 -",+ •••»
where the a's are positive, o^^^ < r/„, and lim r/„ = 0.
Consider the two following methods of adding up the terms of the
series to obtain the sum of the first n and 71 -\- 1 terms respectively.
Assuming first that n is an even integer, we may write
Sn = («x - <',) + (% - «4) + («5 - %) +■■■+ ("n-1 - ^n), (1)
and
Since the a's decrease in value as n increases, each of the parentheses
in (2) contains a positive number. But since they are all subtracted
from a^, the sum of ?i + 1 terms of S cannot exceed a^, however
great n may be.
An inspection of (1) shows that S^ is the sum of positive terms,
and since it dilfers from 5'„ + i by «„+i, which can be made arbi-
trarily small, it follows that .S^„ is less than a positive constant, and
hence approaches a limit by the assumption on page 192.
INFINITE SERIES 203
Now the difference between S„ and its limit can be made less
than any assigned value for a sufficiently large even value of n.
But since „ ,, .
•^n + 1 — '^n ~r "„ + 1>
it appears that lim .S„ ^ j = lim .S",, -f lim r/„ ^ j.
Hence since limr/^^^ = 0, the series S converges when n becomes
intinite by taking on all integral values in succession.
In coniputiiii;- the approximate value of the sum of an alternating series by
adding together tlic lirst few terms, the foregoing method of proof assures us
that the error in stopping the computation with any term, as ui, does not
exceed the value of the next term, a^ + i. For the part which is disregarded is
an alternating series, and an inspection of (2) shows that the sum of a conver-
gent alternatiui'- series cannot exceed its first term.
t>
117. Series with positive and negative terms. "When the signs in
a series are not alternately plus and minus, we may often settle the
question of convergence by
TuEORE.Ar I. A series, w^ + m^ + ^'., + • • ••, some of ivJiose terms
are negative, is convergent if the series formed hy the absolute
values of the terms is convergent.
If the series w^ + ttg + Mg + f- »„ + • • • is given, where some of the terms
are positive and others arc negative, the series formed by the absolute values
of these terms may be denoted by | m^ | + | m., | + | M3 | + • • • + ] it,, | + • • • . All of
these terms are positive quantities.
Thus the series formed by the absolute values of the terms of the series
1-l + i-i + l isi + ^ + i + j + i-i-....
Proof. Let the series of absolute values be a^ + rr., + "3 + • • • , and
let it be convergent to the number .1. It is certain that if in this
series some of the signs are changed to minus, the resulting series
will converge to some value not greater than A ; for the sum of
the absolute values of such terms must have a definite value. Even
if we should change the signs of all the a's, the resulting series
would converge to the value — .1. But the ?<-series may be obtained
by changing the signs of properly selected terms of the series of
absolute values ; hence the ?/-series converges.
We can now extend the proof of the ratio test for convergence so
that it will ap})ly to the case where some of the terms of the series
are negative.
204
HIGHER ALGEBRA
Theokem II. An infinite series u^ + w ,+ m^ + • • •, consisting of
positive and negative terms, converges if
lim
n = CO
V
n+\
ti„
<1.
Since
u.
n+l
«.,
w.
= p^, it follows that if
n„
v.
n + l
n„
< 1, then
\u.
n+W
(1)
<1.
W.
This latter inequality is the condition that the series IwJ + h'al + i'^'sl
+ • • • converges ; and if this series converges, then, by Theorem I,
the original series converges.
Similarly, we may prove that if
u
n+l
w„
> 1, the series ?/^ + ii_^ +
Mg + • • • diverges.
118. Power series. The infinite series
«Q + a^x -\- a^"^ + rtgCK^ + • • • + fn^"" + • • •,
where the o's do not contain cc, is called a power series in x. When
the a's are numerically given, such a series may converge for certain
values of x and diverge for others. For example, when the a's are
each equal to 1, we have the geometrical series, which converges
when ja'l < 1 and diverges for other values of x. The ratio test may
be used to determine for what values of a:; a given series converges.
EXAMPLE
^ "^ 7
„ . , . X X X
Eor what values of x is the series x — — + —— — +
6 0 «
vergent ? divergent ?
Solution.
Mn+l
a;2n'-i
2n-l'
a;2n + i 2n-l
w» + i = (-l)"
2n-l
2(n + l)-l'
lim
Mn
M« + l
2 n + l x2«-i
«„
,. 2n-l
; lim ;
« = =0 2 n + 1
2-
2?i + 1
1
= lim
7i= 00
n
2 + -
n
con-
Hence the series will be convergent when x^<\^ that is, when — l<x<l ;
and divergent when x2>l, that is, when x< — 1 or >1.
The ratio test gives no information concerning the convergence of the series
when X = 1 or — 1. A separate investigation is necessary to determine what
happens for these values of x.
TXFINITK SKIMES . 205
When X — I the series becomes
i_I+'_l+. .. + (-,)-, _!_ +
This is ail altenuiting series and
iiin I u,, I = liin = 0.
ji =z JO n = 00 Z 71 — 1
Therefore the series eon verges by § 11*>.
When X =— 1 the series becomes
-o-ui -} + ■■■)■
Tliis series is also convergent, as it is tlie negative of tlie preceding series.
Hence the original series is convergent wlien — l^^Sl ai>il divergent for all
other values of x.
EXERCISES
For what values of x are the following series convergent?
divergent ?
£ •>^_^, 6. 1 -./■ + ^-'-.'-'+ ■•••
A • -L ^ t J ft I ' ■ * •
2 4 6
7.1 + .'■ + 2 ! X- -f- 3 ! a-« ! +
'^' •' 2^3 4 ^ 8. 1-- + -2--3 +
iC^ . X^ X' . ^.2 _^,3
^' "-* ~ 3! + 57 " 7! + ■ ■ • ^- ^ + •'■ + 2l + 3! +
x'^ a;« X* a;'' x^ a-*
4. a; j^-\ pr 7^+ ••••10. —x
V2 Va V4 2 3 4
x^ a;^ a;« , ,, , a;^ , x^ , a-' ,
^•^-2!+4!-6-! + --- ^^-^+3 + 5 + 7 +
1 -a-^ 1 . 3 a-^ 1 • 3 • 5 a-^
^'^^ ^ + 2 . 3 "^ 2 . 4 . 0 "^ 2 . 4 . G . 7 "^ ■ ■ ■•
13. ^(-l)'""^^^;)". 15. 2t"
Fl=l H=l-'^
+ a;"
:-\??V' ,_ -^ ??''.)■"
14. X-;^^- 16. 2^ — p , where /> IS any nnite
"=^ " ri=i • number.
206
HIGHER ALGEBRA
119. Important special series. In the calculus a general method
will be derived for expressing any ordinary function in terms of a
series. The rational integral functions are really series with a finite
number of terms, but rational nonintegral functions and the func-
tions of trigonometry give rise to infinite series. The most important
of these series are the following :
loge(l + a-) = a-
log, (!-»■) = -
loo'
2 ^
3
—
i-
x"
x^
x'
'-- 2
3
4
x'
"+3
+
x^
5
+ •
^3 ^,5 ^j I
smx = x-^ + --- +
lyt^ rw*^ /ytO
J »A' tt' i/y
cosa: = l-- + --- +
sin~^a:; =■ x -\-
x°
+
1.3x5
2-3 2.4.5
+
tan ^x = X
x" x"
3 + 5
^7 _^^
7 "^ 9
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
In the trigonometric functions the unit of angle is the radian in
each case. Each of these series has been shown to be convergent for
certain values of x in the preceding exercises.
120. Approxipiate computation of the sum of a series. It is, in
general, impossible to find the value of the sum of a power series for
a particular value of x in terms of rational numbers, or even radicals.
Fortunately it is unnecessary, for we can usually compute the value
of the sum to as many places of decimals as we please, and deter-
mine the limit of error of the computation. We may find the limit
of error by determining a convergent series whose sum is greater
than the sum of the series which we neglect by breaking off our
computation with any particular term.
INFINITE SERIES 207
Tliis is illustrated in the computation of natural logarithms, that
is, of logarithms for the base e, by the use of series (4) of the pre-
ceding section ; namely,
+
This series is used for the computation of logarithms instead of (2) or (3) be-
cause it contains only terms with odd exponents, and hence converges much
more rapidly than they do.
In this series we make the substitution
1 + ,r n + 1.
1 — X n
where n is a positive integer.
This gives n + «-*' = n — no- + 1 — cc,
2nx -\- X = 1,
_ 1
Hence log, (^^-^j = 1"^",. \^~^) ^ ^*^^'<' (^ + 1) " '^^Scfi,
and the series becomes
log,(/^ + l)
= ^"^'^" + ' [2^1 + 3(2 u + ly + 5(2 n-^ 1)^ + • • •]• (^>
By means of this series we may compute the natural logarithm of
a.'iy positive integer, n -\-l, if we know the logarithm of n for the
base e.
In order to find the logarithm of a number for the base 10 from
the logarithm for the base e, we apply the theorem on the change of
base, § lOi) ; namely, log x = - — ^^ •
log,. 10
We shall tind in exercise 14, p. 210, that log, 10 = 2.303. Hence
to find the logarithm of any number for the base 10 we divide its
logarithm for the base e by 2.303. Since it is simpler to multiply
than it is to divide, it is customary to multijily the natural log-arithm
by :; 77: = .4343, which is called the modulus of the system of
•^ log, 10 -^
common logarithms.
208 HIGHER ALGEBRA
EXAMPLES
1. Find the sum of the series
rw*^ rv*^ /y>*
^"^1.2'^2.3'^3.4'^"""^(n-l)ri"^""'
for the value x = i, correct to four decimal places.
Solution. Replacing x in eacli of tiie first four terms by ^, we obtain
> =.02,
25-2
1
125 • 6
1
625 • 12
= .001.3.33 . • •,
= .0001.33 • • •
.221466 . . •
This gives us .2215 as the sum of the first four terms.
We will now determine an upper limit to the sum of the infinite series
475 + 5:6+ ■■■'
which the foregoing computation neglects, in order to find out whether the
series neglected would affect the fourth place in the sum .2215. We use the
principle that decreasing the denominator of a fraction increases the fraction.
Hence c „« _,
H h — + ••■ = x^{ \ 1 1-
4.55-6 6-7 \4.5 5-6 6-7
<^{l + x + x'^ + •■■)
4 • 5
by §10, <^(^) (if |a:|<l)
4 • 5 \1 — a;/
1 5
subs;ituti;i:i i for x, < = .00002.
° ^ ' 3125.20 4
Hence the series neglected does not affect the fourth decimal place, and the
result .2215 is correct to the required four places.
2. In computing \ogg(n + 1) by means of series (Z) find the limit
of error if only the first r terms of the series in the brackets are used.
Solution. The series in brackets is
1 +,,,J-^^,.,,J_^+...^ ' ^..,
2n+l 3(2n + 1)3 ' 5(2n+ 1)5 (2r - 1) (2 n + l)-""- ^
where the last term written is the rth term.
INFINITE SERIES 209
The remainder of the series after the rth term is
'■'^ {2r + \){2n + iy^'-+^ (2r + 3) (2ji + l)2'- + « (2 r + r,)(2,t + l)2r + 6 "^ " ' "
1 f 1 1 1
*^{2r+ l)(2n + 1)2'-+ i l "^ (2 n + 1)^"^ (2ji + 1)* "^ ' ' '/ "
The geometrloal series in the Ijnicos lias a ratio which is less
^ (2 n + 1)2
tlian 1, since n is a positive integer. Hence its sum is (§ 10)
1
S^ =
'
(2n + l)2
Now, since n = 1, we have 2 u + 1 ^ 3,
1 ^ 1
1-
(2n + 1)2- 9'
1 ^8
(2 n + 1)2 ~ 9 '
1 ^0
1
1 -8
(2n+l)^
9
s_ =
Hence Rr <
8(2/- + l)(2n + l)2'- + i
Now multiplying by 2, the coefficient of the bracket in (L), w-e find fur the
limit of error, n
E= -^
4(2r+ l)(2ji+ 1)2'- + 1
In computing the value of a logarithm by means of series (L), the error in
using only the first r terms of the series in brackets is always less than E.
3. Compute the value of log^2 correct to three decimal places and
show that the remainder of the series after three terms does not
affect the third decimal place.
Solution. Substituting ji = 1 in series (L), we have, since log 1 = 0,
Now - = .3333
3
- = .0123
3*
^ = .0008
5.3»
.34()4
2
lege 2 = .093
210 HIGHER ALGEBRA
Substituting n = 1, r = 3, in tlie expression for E, we find
E = = — ^ = .00014.
4 • 7 . 37 61236
Thus we see tliat the error in neglecting all terms of the series after the third
is not greater than .0002. Hence this remainder does not affect the third deci-
mal place of the result, and log^ 2 = .693 correct to three decimal places.
EXERCISES
In the following exercises make use of the series in § 119. In
each case show that the terms of tlie series which are neglected in
the computation do not affect the result :
1. Find the value of e correct to three decimal j)laces.
2. Find the value of e*^^ correct to four decimal places.
3. Find the value of logg2 correct to five decimal places.
4. Find the value of log^ 3 correct to four decimal places.
5. Find the value of cos 1° correct to four decimal places.
Hint. 1° = — radian = — = .0175 correct to four places, or radians.
^ 180 180 ^ ' 400
Let X = -— in series (6).
400
6. Find the value of sin 1° correct to four decimal places.
7. Find the value of sin 5° correct to four decimal places.
8. Find the value of cos 5° correct to four decimal places.
9. Using the results of exercises 7 and 8, find by division the
value of tan 5° to three decimal places.
10. Find the value of tt correct to two decimal places by letting
X = ^ in series (7).
11. Compute correct to five decimal places the value of tan~^ (.1).
12. The series of exercise 14, p. 202, gives the value of V2. Check
this to three decimal places.
13. Compute the logarithms for the base e of the positive integers
up to and including ten, correct to three decimal places.
14. Compute the logarithms of the first nine positive integers for
the base 10, correct to three decimal places.
TABLES
Table of Logakithms
Rule for determining the characteristic of the logarithm of a
number.
I. The characteristic of a luonhcr (freater than 1 in one lesa than
the number of digits to the left of the decimal poitit.
II, TJie characteristic of a number less than 1 is negative and
numerically one greater than the number of zeros between the decimal
point and the first significant figure.
Rule for determining the mantissa of the logarithm of a number.
Prefix the proper characteristic to the mantissa of the first three
significant figures of the given number.
Then )nnJtiply the difference bettveen this mantissa and the next
greater mantissa in the table (^called the tabular difference, column
D of the table^ by the remaining figures of the number preceded by
a decimal point.
Add the product to the logarithm of the first three figures, taking
the nearest decimal in the fourth place.
Rule for finding the antilogarithm.
Write the number of three figiires corresponding to (he lesser of
the tivo mantissas betiveen which the given mantissa lies.
Subtract the lesser mantissa from the given mantissa and divide
the remainder by the tabular difference to one decimal place.
Annex this figure to the three already found and 2)1 ace the decimal
point where indicated by the characteristic.
211
212
HIGHER ALGEBRA
N
0
1
2
3
4
5
6
7
8
9
D
10
GOOD
0043
0080
0128
0170
0212
0253
0294
0334
0374
42
11
0414
0453
0492
0531
0509
0607
0645
0682
0719
0755
38
12
0792
0828
0864
0899
0934
0969
1004
1038
1072
1106
35
13
1139
1173
1206
1239
1271
1303
1335
1367
1399
1430
32
14
1461
1492
1523
1553
1584
1614
1644
1673
1703
1732
30
15
1761
1790
1818
1847
1875
1903
1931
1959
1987
2014
28
16
2041
2068
2095
2122
2148
2175
2201
2227
2253
2279
26
17
2304
2330
2355
2380
2405
2430
2455
2480
2504
2529
25
18
2553
2577
2601
2625
2648
2672
2695
2718
2742
2765
24
19
2788
2810
2833
2856
2878
2900
2923
2945
2967
2989
22
20
3010
3032
3054
3075
3096
3118
3139
3160
3181
3201
21
21
3222
3243
3263
3284
3304
3324
3345
3365
3385
3404
20
22
3424
3444
3464
3483
3502
3522
3541
3560
3579
3598
19
23
3617
3636
3655
3674
3692
3711
3729
3747
3766
3784
18
24
3802
3820
3838
3856
3874
3892
3909
3927
3945
3962
18
25
3979
3997
4014
4031
4048
4065
4082
4099
4116
4133
17
26
4150
4166
4183
4200
4210
4232
4249
4265
4281
4298
16
27
4314
4330
4346
4362
4378
4393
4409
4425
4440
4456
16
28
4472
4487
4502
4518
4533
4548
4564
4579
4594
4609
15
29
4624
4639
4654
4669
4683
4698
4713
4728
4742
4757
15
30
4771
4786
4800
4814
4829
4843
4857
4871
4886
4900
14
31
4914
4928
4942
4955
4969
4983
4997
5011
5024
5038
14
32
5051
5065
5079
5092
5105
5119
5132
5145
5159
5172
13
33
5185
5198
5211
5224
5237
5250
5263
5276
5289
5302
13
34
5315
5328
5340
5353
5366
5378
5391
5403
5416
5428
13
35
5441
5453
5465
5478
5400
5502
5514
5527
5539
5551
12
36
5563
5575
5587
5599
5611
5623
5635
5647
5658
5670
12
37
5682
5694
5705
5717
5729
5740
5752
5763
5775
5786
12
38
5798
5809
5821
5832
5843
6855
5866
5877
5888
5899
11
39
5911
5922
5933
5944
5955
5966
5977
5988
5999
6010
11
40
6021
6031
6042
6053
6064
6075
6085
6096
6107
6117
11
41
6128
6138
6149
6160
6170
6180
6191
6201
6212
6222
10
42
6232
6243
6253
6263
6274
6284
6294
6304
6314
6325
10
43
6335
6345
6355
6365
6375
6385
6395
6405
6415
6425
10
44
6435
6444
6454
6464
6474
6484
6493
6503
6513
6522
10
45
6532
6542
6551
6561
6571
6580
6590
6599
6609
6618
10
46
6628
6637
6646
6656
6665
6675
6684
6693
6702
6712
9
47
6721
6730
6739
6749
6758
6767
6776
6785
6794
6803
9
48
6812
6821
6830
0839
6848
6857
6866
6875
6884
6893
9
49
0902
6911
6920
6928
6937
6946
6955
6964
6972
6981
9
50
6990
6998
7007
7016
7024
7033
7042
7050
7059
7067
9
51
7076
7084
7093
7101
7110
7118
7126
7135
7143
7152
8
52
7160
7168
7177
7185
7103
7202
7210
7218
7226
7235
8
53
7243
7251
7259
7267
7275
7284
7292
7300
7308
7316
8
54
7324
7332
7340
7348
7356
7364
7372
7380
7388
7396
8
TAl'.LES
213
N
0
1
2
3
4
5
6
7
8
9
D
65
7404
7412
7419
7427
7435
7443
7451
7459
7466
7474
8
50
7482
7490
7497
7505
7513
7520
7528
7536
754:5
7551
8
67
7559
7566
7574
7582
7589
7597
7604
7012
7619
7027
8
68
7634
7642
7649
7657
7664
7672
7679
7686
7694
7701
7
59
7709
7716
7723
7731
7738
7745
7752
7760
7767
7774
7
60
7782
7789
7790
7803
7810
7818
7825
7832
7839
7846
I
61
7853
78(50
78(58
7875
7882
7889
7896
7903
7910
791 7
t
62
7924
7931
7938
7945
7952
7959
7966
7973
7980
7987
7
(53
7993
8000
8007
8014
8021
8028
8035
8041
8048
8055
7
64
8062
8069
8075
8082
8089
8096
8102
8109
8116
8122
7
65
8129
8136
8142
8149
8156
81(52
81(59
8176
8182
8189
7
66
8195
8202
8209
8215
8222
8228
8235
8241
8248
8254
(
67
8201
8267
8274
8280
8287
8293
8299
830(5
8312
8319
(5
68
8325
8331
8338
8344
8351
8357
83(53
8370
8376
8382
6
6!)
8388
8395
8401
8407
8414
8420
8426
8432
8439
8445
(5
70
8451
8457
8463
8470
8476
8482
8488
8494
8500
8506
(5
71
8513
8519
8525
8531
8537
8543
8549
8555
85(51
8567
6
72
8573
8579
8585
8591
8597
8603
8609
8015
8621
8627
6
73
8(533
8(539
8045
8(551
8657
8663
8669
8675
8681
8086
6
74
8692
8698
8704
8710
8716
8722
8727
8733
8739
8745
6
75
8751
8756
8762
8768
8774
8779
8785
8791
8797
8802
6
76
8808
8814
8820
8825
8831
8837
8842
8848
8854
8859
6
77
88(55
8871
8876
8882
8887
8893
8899
8904
8910
8915
6
78
8921
8927
8932
8938
8943
8949
8954
89(50
89(55
8971
6
70
8976
8982
8987
8993
8998
9004
9009
9015
9020
9025
5
80
9031
9036
9042
9047
9053
9058
9063
9069
9074
9079
5
81
9085
9090
9096
9101
9100
9112
9117
9122
9128
9133
5
82
9138
9143
9149
9154
9159
9165
9170
9175
9180
9186
5
83
9191
9196
9201
9206
9212
9217
9222
9227
9232
9238
5
84
9243
9248
9253
9258
9263
9269
9274
9279
9284
9289
5
85
9294
9299
9304
9309
9315
9320
9325
9330
9335
9340
5
86
9345
9350
9355
9360
9365
9370
9375
9380
9385
9390
5
87
9395
9400
9405
9410
9415
9420
9425
9430
9435
9440
5
88
9445
9450
9455
94(J0
9465
94(59
9474
9479
9484
9489
5
89
9494
9499
9504
9509
9513
9518
9523
9528
9533
9538
5
90
9542
9547
9552
9557
9562
9566
9571
9576
9581
9586
5
ill
9590
9595
9(500
9605
9(509
9614
9619
9(524
9(528
9633
6
i)2
9638
9643
9647
9652
9(557
9661
96(56
9671
9675
9680
5
93
9685
9689
9(594
9(599
9703
9708
9713
9717
9722
9727
5
94
9731
9736
9741
9745
9750
9754
9759
9763
97(58
9773
5
95
9777
9782
9786
9791
9795
9800
9805
9809
9814
9818
5
9(5
9823
9827
9832
9836
9841
9845
9850
9854
9859
98(53
5
97
9868
9872
9877
9881
9886
9890
9894
9899
9903
9908
4
98
9912
9917
9921
9926
9930
9934
9939
9943
9948
9952
4
99
9956
99(51
99(55
9969
9974
9978
9983
9987
9991
9996
4
214 HIGHER ALGEBRA
Square Root
To find the square root of a number Avith an even number of digits
to the left of the decimal point, use Table I.
To find the square root of a number with an odd number of digits
to the left of the decimal point, use Table II. If the number con-
tains three significant figures, interpolate in order to make the cor-
rection for the third place as in the use of the logarithmic tables.
If the decimal point of a number is so placed that by shifting it
to the right or to the left over two (or over any multiple of two)
digits it comes in one of the places mentioned above, use the table
which corresponds to that case.
Thus ^^ = 4.899 ; a/2400 = 48.99 ; \/724 = .4899 ; V24..3 = 4.929. Each
of the foregoing is fi'om Table I. The last requires interpolation.
Each of the following is from Table II : V2A - 1.549 ; V240 = 15.49 ;
Vy024 = .1549 ; V24300 = 155.9.
TABLES
21i
TABLE I
0.
1.
2.
3.
4.
5.
6.
7.
8.
9.
0
0.000
1.000
1.414
1.732
2.000
2.236
2.449
2.646
2.828
3.000
1
3.162
3.317
3.464
3.606
3.742
3.873
4.000
4.123
4.243
4.359
2
4.472
4.583
4.690
4.706
4.899
5.000
5.099
6.196
5.292
5.385
3
5.477
5.508
5.657
5.745
5.831
5.910
6.000
6.083
6.164
6.245
4
6.325
6.403
6.481
6.557
6.633
6.708
6.782
6.856
6.928
7.000
5
7.071
7.141
7.211
7.280
7.348
7.416
7.483
7.550
7.616
7.681
6
7.746
7.810
7.874
7.937
8.000
8.002
8.124
8.185
8.246
8.307
7
S.367
8.420
8.485
8.544
8.002
8.000
8.718
8.775
8.8.32
8.888
8
».'J44
9.000
9.055
9.110
9.165
9.220
9.274
9.327
9.381
9.434
9
9.487
9.539
9.592
9.644
9.695
9.747
9.798
9.849
9.899
9.950
TABLE II
.0
.1
.2
.3
.4
.5
.6
.7
.8
.9
0
0.000
.316
.447
.548
.0.32
.707
.775
.837
.894
.949
1
1.000
1.049
1.095
1.140
1.183
1.225
1.265
1.304
1.342
1.378
2
1.414
1.449
1.483
1.517
1.549
1.581
1.612
1.643
1.673
1.703
3
1.732
1.761
1.789
1.817
1.844
1.871
1.897
1.924
1.949
1.975
4
2.000
2.025
2.049
2.074
2.098
2.121
2.145
2.168
2.191
2.214
5
2.230
2.258
2.280
2.. 302
2.324
2.345
2.366
2.387
2.408
2.429
6
2.449
2.470
2.490
2.510
2.530
2.550
2.569
2.588
2.608
2.627
7
2.646
2.665
2.683
2.702
2.720
2.739
2.757
2.775
2.793
2.811
8
2.828
2.846
2.804
2.881
2.898
2.915
2.933
2.950
2.966
2.983
9
3.000
3.017
3.033
3.050
3.066
3.082
3.098
3.114
3.130
3.146
216 HIGHER ALGEBRA
Cube Root
If the cube root of a number with two digits to the left of the
decimal point is wanted, use Table I. If the cube root of a num-
ber with one digit to the left of the decimal point is wanted, use
Table II. If the cube root of a number with three digits to the left
of the decimal point is wanted, use Table III.
If the decimal point of a number is so placed that by shifting it
to the right or to the left over three (or any multiple of three)
digits it comes into one of the places mentioned above, use the table
which corresponds to that case.
Thus V22 = 2.802 ; V22000 = 28.02 ; a .000022 = .02802 ; \ 22.6 = 2.827.
Each of the foregoing i.s from Table I. The hist requires interpolation.
From Table II we obtain -^/2j = 1.301 ; y/2206 = 13.01 ; a .00226 = .1312.
The last requires intei-polation.
rrom Table III we obtain V^ = .604 ; ^220 = 6.04 ; -y 226000 = 60.9.
TAliLES
217
TABLE I
0.
1.
2.
3.
4.
5.
6.
7.
8.
9.
0
0.000
1.000
1.200
1.442
1.587
1.710
1.817
1.913
2.000
2.080
1
2.154
2.224
2.289
2.351
2.410
2.466
2.520
2.571
2.621
2.668
2
2.714
2.759
2.802
2.844
2.884
2.924
2.962
3.000
3.037
3.072
3
3.107
3.141
3.175
3.208
3.240
3.271
3.302
3.332
3.362
3.391
4
3.420
3.448
3.470
3.503
3.530
3.557
3.583
3.609
3.0.34
3.059
6
3.()84
3.708
3.733
3.750
3.780
3.803
3.82(i
3.849
3.871
3.893
6
3.015
3.930
3.958
3.979
4.000
4.021
4.041
4.062
4.082
4.102
7
4.121
4.141
4.160
4.179
4.198
4.217
4.236
4.254
4.273
4.291
8
4.309
4.327
4.344
4.362
4..S80
4.397
4.414
4.431
4.448
4.465
9
4.481
4.498
4.514
4.531
4.547
4.563
4.579
4.595
4.610
4.626
TABLE II
.0
.1
.2
.3
.4
.5
.6
.7
.8
.9
0
0.000
.464
.585
.669
.737
.794
.843
.888
.928
.965
1
1.000
1.032
1.063
1.091
1.119
1.145
1.170
1.193
1.210
1.239
2
1.260
1.281
1.301
1.320
1.339
1.357
1.375
1.392
1.409
1.426
3
1.442
1.4,58
1.474
1.489
1.504
1.518
1.5.33
1.547
1.560
1.574
4
1.587
1.601
1.613
1.626
1.6.39
1.651
1.663
1.675
1.687
1.698
5
1.710
1.721
1.732
1.744
1.754
1.765
1.776
1.786
1.797
1.807
6
1.817
1.827
1.837
1.847
1.8.57
1.866
1.876
1.885
1.895
1.904
7
1.913
1.922
1.931
1.940
1.949
1.957
1.966
1.975
1.983
1.992
8
2.000
2.008
2.017
2.025
2.0.33
2.041
2.049
2.057
2.065
2.072
9
2.080
2.088
2.095
2.103
2.110
2.118
2.125
2.133
2.140
2.147
TABLE III
0
1
2
3
4
5
6
7
8
9
.0
0.000
.215
.271
.311
.342
.368
.391
.412
.431
.448
.1
.404
.479
.493
.507
.519
.531
.543
.554
.565
.575
.2
.585
.694
.604
.613
.621
.630
.638
.646
.6.54
.062
.3
.669
.677
.684
.691
.698
.705
.711
.718
.724
.731
.4
.737
.743
.749
.755
.761
.766
.772
.777
.783
.788
.6
.794
.799
.804
.809
.814
.819
.824
.829
.834
.839
.6
.843
.848
.853
.857
.862
.866
.871
.875
.879
.884
.7
.888
.892
.896
.900
.905
.909
.913
.917
.921
.924
.8
.928
.932
.936
.940
.944
.947
.951
.955
.958
.962
.9
.965
.969
.973
.970
.980
.083
.986
.990
.993
.997
INDEX
(TIic iiiunhcrs refer to i>:i};e.s.)
Sllll-
Aliscissa, 32
Ace-i'lcnitt'd motion, 30
A('(.'ur;icy of results, 24
^\dilitioii, proportion by, 9; ami
traction, proportion by, '.»
(/-form of pent-ral equation, 87
Alternating series, 202
Alternation, ])roportion by, 0
Angle (complex numbers), 77; of a
power, 7!); of a product, 7'.t; of a
quotient, 80 ; of a root, 82
Antecedent, 9
Antilogaritlim, 211
Aritliuietical ])rogression, 12
Axes, coordinate, 32
Axis, of imaginaries, 72, 74; of reals,
72, 74
Base, 176; change of, 183
Hinomial conjugate surd, 104
Hinoniiai ijuadralic surd, 104
Binomial Theorem, 6
Cardan's Fornndas for the solution nf
cubics, 132
Change, of base, 183; of sign, 1 Hi
Check, on solution of (juadratics, 41,
40; on solution of cubic, 132
Classification of numbers, 52
Coefficients and roots, relations be-
tween, i|uadratic, 48; general, lOo
Combinations, 145
Connnon difference, 12
Comparison lest, convergent series,
102; divergent series, 105
Complex nund)ers, 52, 72 ; angle <if
77 ; conjugate, 75 : etjuality betwet'U.
72; graphical representation of. 72.
73,77; in jxilar forn),78, 70. 80 ; mod-
ulus of, 77; operations iqton. 73.
75 : ])olar representation of, 77 ; roots
of, 82
Complex roots, general etination, 10(t;
quadratic equation, 53, 50
Compound interest, 181
Condition, equations of, 14
Conditional inequalities, 04 ; linear,
07 ; (juadratic, 07
Conjugate binomial surds, 104
(.'onjugate complex nundiers, 75
Conse(iuent, 9
Constant, gravitational, 30; multiiili-
cation of roots by a, 115
Contiiuiation oi sign, IKi
Convergence, 100 ; comjiarison test for,
102; ratio test for, 100
Convergent scries, 1!)1, 102; sum of,
191 ; approximate computation, 200
Coordinate axes, 32
Coordinates, 32
f'orrespoudence of points to numbers,
72
Cube roots, tables of, 217
Cubic equation, graphical .solution,
134 ; solution by Cardan's Formula.s,
130
Cubic function, derivative of, 135
Curves, families of, 58 ; maximum ami
miinnmm ])oints, 00. ^Ste a/so Graph,
Kijuatiou
Decimal places, convention for last
significant figure, 178
Degenerate quadratic, 50
De Moivre's theorem, 79
Dependent eijuations, 10
Dependent variable, 20
Derivative, of the cubic function, 135 ;
of the polynomial, 130
Descartes's rule of signs, 110; for
jiegative roots. 118
Determinant, development, 151, 15G;
development, by minors, 100; ele-
ments of, 151 ; evaluation, 102 ;
nnnor of, 160; of nth order, 152;
of second order, 151 ; of third order,
152; priniijial diagonal of, 151;
properties of, 157, 158, 159, 1(50;
transposition, 159
Determinants, .solution of .system of
liiu'ar eipiations bv means of, 154,
105, 107
219
220
HIGHER ALGEBRA
Development of a determinant, lol,
156 ; by minors, 1(50
Diauonal, principal, 151
Diminution of roots of an equation,
111) ; i;rapliical interpretation of, 1:J2
Discriminant, 53
Divergence, lUO ; comparison test tor,
195; ratio test for, 199
Divergent series, 191
Division, by zero, 14, 16, 57; syn-
thetic, 91 ; rule for synthetic. 93
Double roots of an equation, 102, 137
Elements of a determinant, 151
Equal roots of an eijuation, 53, 55 ; de-
termination t)f maxima and minima
by means <if, 60
Equality between complex numbers,
72
Equation, 14 ; exponential, 180 ; graph
of, 24. 32; homogeneous, 1(37; in
quadratic form, 44 ; of condition,
14 ; with known roots, formation of,
107. See also Quadratic, General,
Linear; and Cubic Equation, Koot,
Curve, Graph, Identity
Equations, dependent. 19 ; families of,
53 ; incompatible, 19 ; system of, 19,
34, 154, 165, 167
Error in computation, 139
Exponent, fractional, 4 ; irrational,
175; negative, 5; zero, 5
Exponential and radical notation, 4
Exponential ecjuation, 180
Exponents, laws of, 4
Extent of table of values, 95
Extraneous roots, 39
Extremes, 9
Factor, rationalizing, 5
Factor Theorem, general equation, 91 ;
quadratic ecjuation, 47
Factoring, 1 ; solution of equations by,
37 ; type forms for, 1
Falling body, formula for, from rest,
29; with initial velocity, 30
Families, of curves, 58 ; of equations, 53
Fractional exponent, 4
Fractions, simplification of, 3
Function, 24 ; integral, 87 ; rational,
87 ; representation of, 25
General equation, 87 ; a-form. 87 ;
Factor Theorem for. 91 ; j>-form. 88 ;
relation between coefficients and
roots, 105. See also Equation, Hoot
Geometrical progression, 12
Geometrical series. 12. 192. 196
Graph, 24; of equation. 32; of func-
tion, 94 ; of incompatible equation.s,
34; of linear equation, 34 ; of quad-
ratic e(iuation, 59; of system of
linear equations, 34
Graphical interpretation, of dinnnu-
tioii of roots, 122 ; of linear con-
ditional inequality, 67
Graphical representation of complex
numbers, 72, 73, 77
(jraphical solution, of cubic equation.
134 ; of ijuadratic equation, 59, 133 ;
of system of two equations, 34
Gravitational con.stant, 30
Harmonic series, 195
IbMuogeneous equation, 167
Homogeneous equations, solution of
system of, by determinants, 167
Horner's Method of approximating ir-
rational roots, 124
Identity. 14
Imaginaries, axis of, 72, 74
Imaginary numbers, 52 ; necessity for,
09 ; operations with, 70, 71
Imaginary unit, 69
Inaccessible limit, 187
Incompatible equations, 19; graph of,
34
Independent variable, 29 ; choice of, 33
Indeterminate forms, 197
Induction, method of complete. 106
Ine(iualities, operations with, 64
Inequality, 64; conditional, 64; linear
conditional, 67 ; graphical interpre-
tation of, 67 ; quadratic conditional,
67 ; unconditional, 64
Infinite root, 57
Infinite series, .see Series
Infinity, 37, 185
Integral function. 87
Integral term, 87
Interest, compound. 181
Inversion, 156; proportion by, 9
Irrational exponent, 175
Irrational number, 52 ; necessity for, 69
Irrational roots, Horner's Method of
approximation, 124
Limit, 12, 186, 187 ; inaccessible, 187
Linear conditional inequality, 67
Linear equation, graph of. 34 ; in one
variable. 14 ; in two variables, 19
Linear equations, system of, 19 : graph-
ical solution of system of, 34 ; solu-
tion of sy.stem of, by determinants,
154, 165.'l67
INDEX
221
Location iniiiciplc, 123
J.ofiariLliiii, 175; iiiodiilus of. :^(ii ; of
a power, 177; of a prodiu-l, 177; of
a (jiiotieiit, 178; of a root, 178
Lofiaiitlims, operations witii, 177;
tables of, 212
Maxima aii<I iiiiniiiia. iM); dftiTiniiia-
tioii of, by ciiual roots, GO
Means, 9
.Minima, (iO
Minor, KiO
Minors, development by. 1(30
Moihiiius (eoiiiplex nuniliers), 77; of a
power, 7l>; of a proihict, 7'.»; of a
(luotieiit. 80; of a root, 82
Modulus (ioixaritliins), 207
.Motion, accelerated. ;50 : unifoiin. 28
Multiple roots, 104, lo7
Multiplication, by the unknown. 4H ;
of roots bj' a constant, 115
Nefrative and positive terms, series
with, 202, 203
Negative exponent, 5
Negative roots, Uescartes's rule of
signs for, 118
Number of roots, 4(5, 07
Numbers, cla.ssiticalion, 52: complex,
52 ; correspondence of, to points, 72 ;
irrational, 52; pure imaginary, 52;
rational, 52 ; real, 52. Hee also above
headlnyx iiidivkhidUi/
One-to-one corres])ondence of points
to numbers, 72
Operations, upon complex numliers,
73, 75; with inecjualilies, 04; witli
logarithms, 177
( >rdinate, 32
( )rigin, 32
O.scillating series. 101
Parallel lines, 34
Tarameter, 53
Partial remainders, sign of, as indica-
tion of roots. 0(i
rernuitations, 142; of things all <Uf-
ferent, 143; with repetitions, 143
j)-form of general eiiuati()n, 88
Points and numbers, correspondeni'e
between, 72
polar form of ct)inplex nund)er.s. 78
Polar representation, of complex num
bers. 77 ; of division of comi)lex num
bers,80: of mull iiihca I ion of complex
innnbers, 70; oi powers of ci niplex
numbers, 70
Polynomial, 87; general, of jHli de-
gree, 87; derivative of, \:Hi
Pcjsitive and negative terms, series
with, 202, 203
Positive root indicated by radical sign,
5
Power, angle of a. 70; logarithm of a,
177 ; modulus of a. 7'.»
Power si'ries, 204
Prime expression, 1
Principal diagonal, 151
Probaliility. 148
Product, angle of. 70; logarithm of a,
177 ; modulus of a. 70
Progression, arithmetical, 12; geomet-
rical, 12
Proportion, 0; liy addilion, 0; by ad-
dition and subtraction, 0; by al-
ternation, 0; l)y inversion, 0; liy
subtraction. 0
Pure imaginary nund)er, 52
Quadratic conditional ine(|uality, 07
Qiiatlratic e(|uation, character of roots,
52, 55; check on solution, 41 ; com-
plex roots, 53, 50 ; degenerate, 5(5 ;
determination of maxima and min-
ima l)y t'i|ual roots. 00 ; discriminant,
53 ; ilistinctroots. 53 ; etjual roots. 53,
55. (iO ; extraneous roots, 80 ; Factor
Tlieorem for, 47; fornuda for solu-
tion, 30; graph, 50; graphical solu-
tion, 50, 133; inlinite root, 57;
number of roots, 4(»; rational roots,
53; real roots. 53; reduced form,
48; relation between roots and co-
efficients, 48; roots. 38, 30, 4(!, 48,
52, 53, 55, 57, 50, GO; solution by
factoring, 37; solution by formula,
38 ; special cases, 55 ; zero root, 55
Quadratic form. 44
Quantity, 24
Quotient, angle of . 80 ; logarithm of,
178 ; modulus of, 80
Padical, 4; and exponential notation,
4
Uailical siirn. convi-ntion of, 5
Patio. 0. 12
Paiio test for infinite series. 100
Rational function. 87
national luunlH-r, 52
Palional roots. 53: deieciion of. Ill
Pational term. 87
Pationalization. 5
Patioiiali/.ing farior, 5
Peal nundx-r, 52
Peal roots, 53
222
HIGHEK ALGEBRA
Heals, axis of. 72. 74
Retlucedforiaot I juadi'atic equation, 48
Keinainder, partial, 9(3
Remainder Theorem, 90
Root, angle of, 82 ; cube, tables of, 217 ;
logarithm of a,178 : modulus of a. 82 ;
positive, denoted by radical sign, 5 ;
square, tables of, 215
Roots, of a complex number, 82 ; of an
e(juation, 14; of a number, 4
Roots of the general equation, bino-
mial quadratic surd, 104 ; complex,
100;(liuiinutioiiof, 119, 122; double,
102, i;:!7 ; existence indicated by sign
of partial remainder, 90 ; irrational,
Horner's Method of approximation,
124 ; multiple, 102, 104. 137 ; nuilti-
jilication of, by a cou8tant,115 ; num-
ber, 97; rational, detection of. 111 ;
relation between coefficients and,
105; threefold, 102
lioots of the quadratic equation, see
Quadratic equation
.Series, 188; alternating, 202; com-
parison test for, 192, 195 ; conver-
gent, 191, 192; convergent, sum of,
191,20(i; divergent. 191; geometrical,
192, 19(); harmonic, 195; oscillat-
ing, 191 ; power, 204 ; ratio test for,
199; special, list of, 206; with posi-
tive and .legative terms, 202, 203
Sign, change of, 11(5: continuation of,
116
Sign, of partial remainder as indica-
tion of root, 96; radical, conven-
tion for, 5
Significant figure, convention for last,
178
Signs, Descartes's rule of, 116; for
negative roots, 118
Sinniltaneous system of equations, 19,
34, 154, 165, 167
Solution, of cubic equation by Cardan's
Fonnulas,130 ; of quadraticequation
by factoring, 37 ; of quadratic etjua-
tion by formula, 38; of quadratic
equation, check on, 40, 42 ; of system
of equations by determinants, 154,
165, 167. See also Graphical solution
Square roots, tables of, 215
Straight line, 34
Subtraction, pnjportion by, 9
Sum of a series, 191 ; approximate
computation, 206
Surd, binomial quadratic, 104; con-
jugate binomial, 104
Synthetic division, 91 ; rule for, 93
System of ec^uations, 19, 34, 154, l(i5,
167
Table of values, extent of, 95
Tables, of cube roots, 217; of loga-
rithms, 212; of square roots, 215
Term, integral, 87; rational, 87
Test for convergence, comparison.
192 ; ratio, 199
Test for divergence, comparison, 195;
ratio, 199
Threefold roots, 102
Transposition in determinants, 159
Unconditional inecjuality, 64
Uniform motion, 23
Values, extent of table of. 95
Variable, 29. 53. 185; dependent,
29 ; independent, 29 ; independent,
choice of, 33
Variation, 9
Zero, division by, 14 ; expression equal
to, 37 ; root e(iual to, 55
Zero exjjonent, 5
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