IN MEMORIAM
FLOR1AN CAJOR1
SECOND COURSE IN ALGEBRA
A SERIES OF MATHEMATICAL TEXTS
EDITED BY
EARLE RAYMOND HEDRICK
THE CALCULUS
By ELLERT WILLIAMS DAVIS and WILLIAM CHARLES
BRENKE.
ANALYTIC GEOMETRY AND ALGEBRA
By ALEXANDER ZIWET and Louis ALLEN HOPKINS.
ELEMENTS OF ANALYTIC GEOMETRY
By ALEXANDER ZIWET and Louis ALLEN HOPKINS.
PLANE AND SPHERICAL TRIGONOMETRY WITH
COMPLETE TABLES
By ALFRED MONROE KENYON and Louis INGOLD.
PLANE AND SPHERICAL TRIGONOMETRY WITH
BRIEF TABLES
By ALFRED MONROE KENYON and Louis INGOLD.
ELEMENTARY MATHEMATICAL ANALYSIS
By JOHN WESLEY YOUNG and FRANK MILLETT MORGAN.
PLANE TRIGONOMETRY
By JOHN WESLEY YOUNG and FRANK MILLETT MORGAN.
COLLEGE ALGEBRA
By ERNEST BROWN SKINNER.
ELEMENTS OF PLANE TRIGONOMETRY WITH COM-
PLETE TABLES
By ALFRED MONROE KENYON and Louis INGOLD.
ELEMENTS OF PLANE TRIGONOMETRY WITH BRIEF
TABLES
By ALFRED MONROE KENYON and Louis INGOLD.
THE MACMILLAN TABLES
Prepared under the direction of EARLE RAYMOND HEDRICK.
PLANE GEOMETRY
By WALTER BURTON FORD and CHARLES AMMERMAN.
PLANE AND SOLID GEOMETRY
By WALTER BURTON FORD and CHARLES AMMERMAN.
SOLID GEOMETRY
By WALTER BURTON FORD and CHARLES AMMERMAN.
CONSTRUCTIVE GEOMETRY
Prepared under the direction of EARLE RAYMOND HEDRICK.
JUNIOR HIGH SCHOOL MATHEMATICS
By W. L. VOSBURGH and F. W. GENTLEMAN.
This book is issued in a form identical with that of the books announced above,
as is the First Course in Algebra by the same authors.
SECOND COURSE IN ALGEBRA
BY
WALTER BURTON FORD
n
PROFESSOR OF MATHEMATICS, THE UNIVERSITY OF MICHIGAN
AND
CHARLES AMMERMAN
THE WILLIAM McKINLEY HIGH SCHOOL, ST. LOUIS
gorfe
THE MACMILLAN COMPANY
1920
All rights reserved
COPYRIGHT, 1920,
BY THE MACMILLAN COMPANY.
Set up and electrotyped. Published January, 1920
Nortoooti
J. 8. Gushing Co. — Berwick & Smith Co.
Norwood, Mass., U.S.A.
PREFACE
IN the present volume, which is intended as a sequel to
the author's First Course in Algebra, the following features
may be noted :
(a) The first seven chapters furnish a systematic review
of the ordinary first course up to and including the subject
of simultaneous linear equations. In these opening chapters
the aim is to state briefly and concisely those fundamental
principles which are basal in all algebraic work and with
which the pupil already has some acquaintance. The
method of treatment is largely inductive, being based upon
solved problems and other illustrative material rather than
upon any attempt at proofs or formal demonstrations. The
various principles are explicitly stated, however, at all points
in the form of rules, which are clearly set off in italics. Upon
this plan the pupil is rapidly and effectively prepared for un-
dertaking the newer and more advanced topics which follow.
(6) Chapters VIII-IX (Square Root and Radicals) are
essentially a reproduction of the corresponding chapters in
the First Course, but all of the problems are new. This is
true also of the early parts of Chapter X (Quadratic Equa-
tions). These are topics which usually present more than
average difficulty ; hence, even for pupils who have studied
them carefully before, a complete treatment of them is de-
sirable in the second course.
The introduction and use of tables of square and cube
roots at this point (§ 43) is to be especially noted. It seems
vi PREFACE
clear that pupils should be made familiar with such tables
at an earlier date than formerly. One strong reason for this
is that a constantly increasing number of students pass
directly from the high schools into technical pursuits where
facility in the manipulation of tables of all kinds is especially
desirable.
(c) Part II, comprising Chapters XI-XX, presents the usual
topics of the advanced course. The order of arrangement
follows, so far as possible, that of the difficulty of the various
subjects, and the whole has been prepared with a view of
introducing a relatively large number of simple illustrative
examples drawn from nature and the arts. Throughout
the development, however, due emphasis has been given
to those fundamental disciplinary values which should be
preserved in any course in mathematics.
Among the unusual features, it may be observed that the
detailed consideration of exponents and radicals has been
delayed until logarithms are about to be taken up. These
topics in their extended sense have, in fact, but little to do
with algebra until that time.
Again, the chapter on logarithms is unusually full and com-
plete. All the essential features of this relatively difficult
but increasingly important subject are presented in detail.
In the past, much has ordinarily been left for the teacher to
explain.
(d) Functions, Mathematical Induction (including the
proof of the Binomial Theorem), and Determinants have
been grouped together under the title Supplementary Topics.
In fact, these subjects lie on the border line between the
second course and the college course. Only the elements of
each are taken up, but there is enough to show its im-
portant bearing in algebra and to pave the way for its
further development in the college course. For example,
PREFACE vii
the study of functions is so presented that it at once amplifies
material to be found in the earlier chapters of the book and
brings in new material which is connected with the graphical
study of the theory of equations. Thus it serves as an in-
troduction to the latter subject as presented in the usual
college texts.
As in the authors' other texts, a star (*) has been placed
against certain sections that may be omitted if desired with-
out destroying the continuity of the whole.
WALTER B. FORD.
CHARLES AMMERMAN.
TABLE OF CONTENTS
PART I. REVIEW TOPICS
CHAPTER PAGE
I. FUNDAMENTAL NOTIONS 1
II. SPECIAL PRODUCTS AND FACTORING ... 17
III. HIGHEST COMMON FACTOR AND LOWEST COMMON
MULTIPLE 26
IV. FRACTIONS 29
V. SIMPLE EQUATIONS . • 36
VI. GRAPHICAL STUDY OP EQUATIONS .... 41
VII. SIMULTANEOUS EQUATIONS SOLVED BY ELIMINATION 50
VIII. SQUARE ROOT 59
IX. RADICALS 67
X. QUADRATIC EQUATIONS 78
PART II. ADVANCED TOPICS
XI. LITERAL EQUATIONS AND FORMULAS ... 95
XII. GENERAL PROPERTIES OF QUADRATIC EQUATIONS . 108
XIII. IMAGINARY NUMBERS 116
XIV. SIMULTANEOUS QUADRATIC EQUATIONS . . . 122
XV. PROGRESSION 141
XVI. RATIO AND PROPORTION 166
XVII. VARIATION . . . . . . . .178
XVIII. EXPONENTS 193
XIX. RADICALS 205
XX. LOGARITHMS 216
viii
TABLE OF CONTENTS ix
PART III. SUPPLEMENTARY TOPICS
CHAPTER PAGE
XXI. FUNCTIONS . . .... 245
XXII. MATHEMATICAL INDUCTION — BINOMIAL THEOREM 255
XXIII. THE SOLUTION OF EQUATIONS BY DETERMINANTS 265
PART IV. TABLES
TABLE OF POWERS AND ROOTS 275
TABLE OF IMPORTANT NUMBERS 296
INDEX . . .297
LAPLACE
(Pierre Simon Laplace, 1749-1827)
Famous in mathematics for his researches, which were of a most advanced
kind, and especially famous in astronomy for his enunciation of the Nebular
Hypothesis. Interested also in physics and at various times held high polit-
ical offices under Napoleon.
SECOND COURSE IN ALGEBRA
PART I. REVIEW TOPICS f
CHAPTER I
FUNDAMENTAL NOTIONS
1. Negative Numbers. In the First Course in Algebra
it was shown how negative as well as positive numbers may
be used, the one being quite as common as the other in every-
day life.
Thus +15° (or simply 15°) means 15° above 0°, while -15°
means 15° below 0. Similarly, +$25 (or simply $25) means a gain
or asset of $25, while —$25 means a loss or debt of the same amount.
Negative numbers are compared with each other in much
the same way as positive numbers. Thus, just as in arith-
metic 4 is less than 5, 3 is less than 4, etc., until finally we
say that 0 is less than 1, so we continue this idea in algebra
by saying that — 1 is less than 0, —2 is less than — 1, etc.
The whole situation regarding the size of numbers is
vividly brought out to the eye in the figure below :
I I I I I I I I I I I I I I I I i I II I | I I I
-12-11-10-9-8—7-6-5-4-3-2-1 0 1 2 3 4~ ~S € 1 8 9 10 11 12
FIG. 1.
Here the positive (+) numbers are placed in their order to
the right of the point marked 0, while the negative ( — )
numbers are placed in their order to the left of that same
point. This figure shows all numbers (positive and negative)
arranged in their increasing order from left to right.
t Chapters I-X (pp. 1-94) furnish a review of the First Course,
The remaining chapters deal with more advanced topics.
B 1
2 SECOND COURSE IN ALGEBRA [I, §1
In Fig. 1, only the positive and negative integers and zero
are actually marked. A complete figure would show also
the positions of the fractions. Thus £ is located at the point
halfway between the 0 and the 1 ; 2^ is located at the point
one third the way from 2 to 3 ; — f is at the point f the dis-
tance from 0 to — 1 ; — 5£ is at the point f the distance from
— 5 to — 6 ; and so on for all fractions.
By the numerical value (or absolute value) of a negative
number is meant its corresponding positive value.
Thus the numerical, or absolute, value of —3 is +3, or simply 3.
NOTE. The numerical value of any positive number is the
number itself.
EXERCISES
In each of the following exercises, state which of the two
numbers is the larger. First locate the number at its proper
place on the line shown in § 1.
1. 7, 10. 6. |, f. 9. 3i, If.
2. 7, -10. 6. f,-f. 10. -3*, -If.
3. -7, 10. 7. -f, f. 11. -i 0.
4. -7, -10. 8. -f, -f. 12. -.3, -.05
2. Operations with Numbers. The following facts will
be recalled from the First Course in Algebra^
(a) To add two numbers having like signs, add their absolute
values (§1) and prefix the common sign.
Thus (-5)+(-6) = -ll.
(b) To add two numbers having unlike signs, find the dif-
ference between their absolute values and prefix the sign of the
one whose absolute value is the greater.
Thus (+3)+(-5) = -2.
t References to the authors' First Course in Algebra are given in
this book by page number.
I, §2] FUNDAMENTAL NOTIONS 3
NOTE. If we have more than two numbers to add together, as
for example (+4)+(-7)+(+5)+(-6)+(-l), the customary
way is to add all the positive parts together, then all the negative
parts, and finally to add the two results thus obtained. Thus, in
the example just mentioned, the sum of the positive parts is 4 +5 = 9,
while the sum of the negative parts is (— 7) + ( — 6) +( — 1) = — 14.
The final result sought is, therefore, (+9) +( -14) = -5.
(c) To subtract one number from another, change the sign
of the subtrahend and add the result to the minuend.
Thus (-7) - (-5) = (-7)+(+5) = -2.
(d) To multiply one number by another, find the product
of their absolute values, and take it positive if the two numbers
have the same sign, but negative if they have unlike signs.
Thus (+3) • (+2) = +6, and (-3) • (-2) = +6; but (-3) • (+2) =
-6 and (+3) • (-2) = -6.
(e) To divide one number by another, find the quotient of
their absolute values, and take it positive if the numbers have
the same sign, but negative if they have unlike signs.
Thus (+8H(+2) = +4, and (-8)-K-2) = +4; but (-8)^(+2) =
-4, and (+8)- ( -2) = -4.
EXERCISES
Determine the value of each of the following indicated ex-
pressions.
1. (+4) + (+7). 8. (+25) + (-32) + (-
2. (-4) + (+7). 9. (+7) -(+4).
3. (+4) + (-7). 10. (+7)-(-4).
4. (-4) + (-7). 11. (-7)-(+4).
5. (+9) + (-27). 12. (-7)-(-4).
6. (-32) + (+16). 13. (+34) -(-63).
7. (-3) + +2) + (-l). 14. -54--32.
SECOND COURSE IN ALGEBRA [I, §2
16.
[HINT. By (c) of § 2, this may first be changed into the form
(+6) +(+4) +(-2) +(+5).]
16.
17. (-23) + (+32)-(-
18. (+3) -(-4). 22. (+24) -5- (-6).
19. (-4) -(+3). 23. (-36) -5- (+6).
20. (+5) • (+4). 24. (-55) -(-11).
21. (-5) -(-4). 25. (f)-(-i).
3. Use of Letters in Algebra. Algebra is distinguished
from arithmetic not only because of its use of negative num-
bers, but also because of its general use of letters to represent
numbers. This is useful in many ways. In particular, it
enables us to solve problems in arithmetic which would other-
wise be very difficult. The following facts and definitions
will be recalled from the First Course in this connection.
The sum of any two numbers, as x and y, is represented by
x+y.
The difference between any two numbers, as x and y
(meaning the number which added to y gives x), is represented
bys-y.
The product of any two numbers, such as x and y, is written
in the form xy. It has the same meaning as xXy, or x - y.
Either of the numbers thus multiplied together is called a
factor of the product.
The quotient of x divided by y is expressed either by x -5- y,
or by -, or by x/y.
The product x • x is represented by x2 and is read x square;
similarly, x • x • x is represented by x3 and is read x cube.
More generally, x • x • x — to n factors is represented by
xn and is read x to the nth power. The letter n as thus
used in xn is called the exponent of x.
I, § 3] FUNDAMENTAL NOTIONS 5
The symbol Vx denotes that number which when squared
gives x. It is called the square root of x. Similarly, \/x is
ailed the cube root of x and denotes that number which
when cubed gives x. In general, ~\/x is called the nth root of
x, and denotes that number which when raised to the nth
power gives x. The letter n as thus used in Vx is called the
index of the root.
Whenever one or more letters are combined in such a way
as to require any of the processes just described, the result
is called an expression.
Thus 2 x +3 y, ax-bxy, 6 ran -3V^+2Vn, and 2 x+yz-x* +
xyz are expressions.
An expression is read from left to right in the order in
which the indicated processes occur. Indicated multiplica-
tions and divisions are to be carried out, in general, before
the indicated additions and subtractions.
Thus 2 x+y2— x3+xyz is read "Two x plus y square minus x
cube plus xyz'1
EXERCISES
Read each of the following expressions :
1. 2x2. 6.
6. ±_|_± ±£
y w 2
3. a»-2o&+6>. 7. VS+^. n. x3/£±2.
s-y
4. ms-n3. 8. v^+v^. 12. •\/aw+&r.
13. Express each of the following ideas in letters.
(a) The sum of the squares of x and y.
(b) The difference between m cube and n cube.
(c) Three times the product of mn diminished by twice
the quotient of x divided by the square root of y.
6 SECOND COURSE IN ALGEBRA [I, § 3
14. The fact that the area of any rectangle is equal to the
product of its two dimensions (length and breadth) is ex-
pressed by the formula A=ab. Express similarly in words
the meaning of each of the following familiar formulas :
(a) A =| bh. (Formula for the area of a triangle.)
(b) A =irr2. (Formula for the area of a circle.)
(c) C = 2irr. (Formula for the circumference of a
circle.)
(d) h2 = a?+b2. (Theorem of Pythagoras concerning any
right triangle.)
(e) V = % Tir3. (Formula for the volume of a sphere.)
(/) A = 4 TIT*. (Formula for the area of a sphere.)
4. Evaluation of Expressions. Whenever the values of
the letters in an expression are given, the expression itself
takes on a definite value. To obtain this value, we must
work out the values of the separate parts of the expression
and then combine them as indicated.
EXAMPLE. Find the value of the expression
o«+2-bc-c»
a+b+c
when a = 1, b = —2, and c = 3.
SOLUTION. Giving a, 6, and c their assigned values, the expres-
sion becomes
l2+2- (-2). 3-33_l+(-12)+(-27)_-38_
~~ :""
In evaluating expressions, it is useful to remember the
following general facts, which result from (d) of § 2 :
(a) The sign of the product of an even number of negative
factors is positive.
Ex. (-2). (-3)- (-1). (-4) = +24.
I, § 4] FUNDAMENTAL NOTIONS -f-
(b) The sign of the product of an odd number of negative
factors is negative.
Ex. (-2)- (-3)- (-l) = -6.
(c) A negative number raised to an even power gives a
positive result, but if raised to an odd power gives a negative
result.
Ex. ( -2)4 = +16, but ( -2)3 = -8.
(d) An odd root of a negative number is negative.
Ex. ^27 =-3; v^32=-2'; ^J=— 1.
EXERCISES
Evaluate each of the following expressions for the indi-
cated values of the letters.
1. a2+2a&+62, when a=l, b=-l.
2. 4xzy+4:xy*+xyz, when x=— 1, y = 2, z=— 3
3. ?™!±Z^whenm = 2,n = 6.
2 m—n
4. 2v/2x-3v/9Y, when a; =18, 2/= -3.
5.
, whenz=-l, 2/=-3, and a = l.
, when m = 2, n= -2, x = 5, y = 6.
8. By means of the formulas in Ex. 14, p. 6, find
(a) The area of the circle whose radius is 3 feet.
[HINT. Take7r=3i.]
(6) The circumference of the circle whose radius is 1^ feet.
(c) The volume of the sphere whose radius is 5 inches.
(d) The area of the sphere whose diameter is 1 yard.
8 SECOND COURSE IN ALGEBRA [I, § 5
5. Definitions. A monomial (or term) is an expression
not separated into parts by the signs + or — , as 5 xzy.
A binomial is an expression having two terms, as 3 x2 — 4 yx.
A trinomial is an expression having three terms, as
Any expression containing only powers of one or more
letters is called a polynomial.
A polynomial is said to be arranged in descending powers
of one of its letters if the term containing the highest power
of that letter is placed first, the term containing the next
lower power is placed second, and so on.
Thus, if we arrange \ x3 +-J- x5 — 1 -\-x — 3 x4 in descending powers
of x, it becomes -J x6 —3 x4 -f -J- x3 -fa; — 1.
Similarly, a polynomial is said to be arranged in ascending
powers of one of its letters if the term containing the lowest
power of that letter is placed first, the term containing the
next higher power is placed second, and so on.
Thus, if we arrange \ x3 -\-\ z6 — 1 -fa? — 3 x4 in ascending powers of
z, it becomes -l+x+%x3-3 z4+| x6.
Whenever a term is broken up into two factors, either
factor is known as the coefficient of the other one. Usually
the word is used to designate the factor written first.
Thus, in 4 xy, 4 is the coefficient of xy ; in ax, a is the coefficient
of x, etc.
A common factor of two or more terms is a factor that oc-
curs in each of them.
Thus 5 x, ax, x2, and x3 have the common factor x.
Whenever two or more terms have a common literal factor,
they are said to be like terms with respect to that factor.
Thus 5 x, ax, x, and — 2 x are like terms with respect to x ;
and 2a(x—y) and 3 b(x— y) are like terms with respect to x— y.
I, § 7] FUNDAMENTAL NOTIONS 9
6. Addition and Subtraction of Expressions. The follow-
ing rules will be recalled from the First Course, pp. 49-57.
(a) To add like terms, add the coefficients for a new coeffi-
cient and multiply the result by the common factor.
Thus 3 x +5 x -4 x = (3 +5 -4)z =4x.
Similarly, m(x—y)+n(x—y) = (m+ri)(x—y}.
(b) To add polynomials, write like terms in the same column,
find the sum of the terms in each column, and connect the results
with the proper signs.
Thus, in adding 3a+4b+2c, 5 a +3 b -2 c, and 7 a -9 6-5 c,
the work appears as follows :
3a+46+2c
5 a +3 6-2 c
7 a-9 b-5c
15 a— 26-5c. Ans.
(c) To subtract a term from another like term, change the
sign of the subtrahend and add the result to the minuend.
Thus 8 x*y - ( -3 x*y) =8 x*y +3 x*y = 11 x*y.
(d) To subtract one polynomial from another, change all
signs in the subtrahend and add the result to the minuend.
Thus, in subtracting 4 mn —2 nr +3 p from 5 ran — 4 nr — 4 p,
what we have to do is to add —4 ran +2 nr — 3 p to 5 ran —4 nr — 4 p.
Adding these (see the preceding rule for addition of polynomials)
gives ran —2 nr — 7 p. Ans.
NOTE. If two or more expressions can be arranged according to
the descending powers of some letter (§ 5), it is usually best to do
so before attempting to add, subtract, or perform other operations
upon them.
7. Parenthesis ( ), Bracket [ ], Brace J j, and Vincu-
lum . These are symbols for grouping terms that are to
be taken as one single number or expression.
Thus 4 x — (x +3 y — z) means that x +3 y — z as a whole is to be
subtracted from 4 x.
10 SECOND COURSE IN ALGEBRA [I, § 7
The following rules will be recalled :
(a) A parenthesis preceded by the sign + (either expressed
or understood] may always be removed without any other change.
(b) A parenthesis preceded by the sign -- may be removed
provided the sign of each term in the parenthesis be first changed.
Thus 2 a+3 6 + (z-3 y+z) = 2 a+3 b+x-3 y+z
but 2 a+3 b-(x-3y+z)=2a+3b-x+3y-z.
EXERCISES
1. State the common factors in each of the following
expressions.
(a) -4 x, -5 x, 6 x. (c) a(z+l)2, 6(z+l), c(;r+l)3.
(b) rs, 3 rs, - 10 r2s. (d) 2 mn(a+b), 4 m2n(a-b), 8 ran2.
2. Add 7a+66-3cand4a-76+4c.
3. Add2x+3y— 2xy, 7 xy— 4x — 9?/, andTx — 5xy — 4y.
4. Add z-S-T^+lSz3, 4+14 z3-!! x-x\ and
[HINT. See Note, § 6.]
5. Add 4(m+n)-3(g-r) and 4(m+n)+6(5-r).
6. Add 10(a+6)-ll(6+c), 3(a+6)-5(c+d), and
7. Add2mx+3nx — £qx, nx+lqx—ry, and py — qz-\-3w.
8. Subtract 3 x — 2 i/-fz from 5 x — y-\- z.
9. Subtract 4 x8 - 8 - 13 x2+ 15 a; from 6 x2+ 19 x3 - 4
+12*.
[HINT. See Note, § 6.]
10. From 13 a+5 6-4 c subtract 8 a+9 6+10 c.
11. From 2 a+3 c+d subtract a — 6+c.
12.' From 3 x2+7 x+ 10 subtract -x2-x-6.
13. Subtract l-a+a2-a3 from 1-a3.
14. From the sum of x2 — 4 xy+y* and 6 x2 — 2 xy+3 y2 sub-
tract 3 a;2 — 5xy+7 y2. Do it all in one operation if you can.
I, §8] FUNDAMENTAL NOTIONS 11
15. From the sum of 2 s+8 £—4 w and 3 s — 6 t+2 w take
the sum of 8 s+9 Z+6 w and 4 s-7 t-4 w.
Remove the parentheses and combine terms in each of the
following expressions.
16. a-(2a+4)-(5a+10).
17. 6x+(5x-\2x+ll).
[HINT. Remove the innermost group 'sign first.]
18. x-\x-(x-3x)\.
19. 2Qz-[(2z+7w)-(3z+5w)].
Find the values of the following when o = 4, 6 = 3, c= — 2,
and d= —I.
20. 10c2-(3a+6+d).
[HINT. Simplify the expression as far as possible before giving
to a, 6, c, and d their special values.]
21. 3d-\a-(c-b)\.
22. o-Hc-(3d-6)j+3Ka-c)-7(&+<f)!.
2c-(a2+b2)
'
24.
8. Multiplication The following formulas and rules
will be recalled from the First Course.
Formula I. xmxn = xm+n.
Thus 22-23=25; x2-x3=x6; (2 a)3 • (2 a)4 = (2 a)7;
(a +6)5. (a + &)8 = (a+6)13; etc.
Formula I leads to the following rule.
(a) To multiply one monomial by another, multiply the
coefficients for the new .coefficient and multiply the letters to-
gether, observing Formula I.
Thus, in multiplying —4 w2n3 by 2 m2n2 the new coefficient is
(— 4)X2, or -8, and the product of the letters is m2n3m2n2, or
w2m2n3n2, which reduces by Formula I to w4n5. The answer, there-
fore, is —8 m*n6.
12 SECOND COURSE IN ALGEBRA [I, § 8
Formula II. (xy)m = xmym.
Thus (2 • 3)2 = 22 • 32 ; (xy)* = x*y* ; (3 y)3 = 33 • y* = 27 y* ;
(2 mn)2 = (2 m • n)2 = (2 m)2 • n2 =22 • m2 • n2 =4 m2n2 ; etc.
Formula III. a(b-\-c)=ab-\-ac.
Thus 2(3+4)=2- 3+2- 4=6+8 = 14; x(y*+z*) =xy*+xz3',
ra2(mn2 — m2n) = m3n2 — m4n ; etc.
Formula III leads to the following rules.
(6) To multiply a polynomial by a monomial multiply each
term of the polynomial separately and combine the results.
Thus the process of multiplying the polynomial m— n+mn by
the monomial mn is as follows :
m —n+mn
_ mn
m2n — mn2+m2n2. Ans.
(c) To multiply one polynomial by another, multiply the mul-
tiplicand by each term of the multiplier and combine results.
Thus the process of multiplying x—y+3z by 2x+3y—z is as
follows :
x-y+3z
Multiplying by 2 x, 2x*-2xy+6xz
Multiplying by 3 y, 3 xy — 3 ?/2 + 9 yz
Multiplying by — z, _ — xz _ + yz— 3 z2
Combining results, 2 x- + xy +5 xz —3 y2 + 10 yz —3 z2. Ans.
EXERCISES
Find the product in each of the following indicated multi-
plications.
1. 10a5x6a2. 4. 3 a262c3 X - 2 aW.
2. -2a6x3a6. 5. 4 xyz2 X - 8 x2yz.
3. - 4 ra2™3 X2 w2n. 6. ( - 4 a26c2) X (2 a6) X ( - 3 ac) .
[HINT TO Ex. 6. Multiply the first two expressions together,
then multiply this product by the last expression.]
I, § 8] FUNDAMENTAL NOTIONS 13
Simplify each of the following expressions.
7. (3z)2. 8. (3a6)2.
[HINT TO Ex. 8. See fourth illustration under Formula II.]
9. (2 ran)3. 10. (8 abc)2.
[HINT TO Ex. 10. 8 abc may be written 8 a • be.]
11. (-2wn)4. 12. (-2z2?/2)3.
13. Show that if the side of one square is twice that of
another, its area is four times as great.
[HINT. Let a = a side of the small square. Then, a side of the
large square =2 a, and the area = (2 a)2. Now apply Formula II.]
14. Show that if the edge of one cube is twice that of an-
other, the volume is eight times as great.
15. Compare the areas of two circles, one of which has a
radius three times as great as the other. (See Ex. 14 (b) , p. 6.)
16. Compare the volumes of two spheres, one of which has
a radius twice as great as the other. (See Ex. 14 (e), p. 6.)
Find the product in each of the following multiplications.
17. (10a36+7a&4)X-2a26.
18. (2x2-3xy+5y2)x-2xy.
19. (a2- 10 a6+15 b2) x4 a262.
20. (a— 2a6+62)x(a-6).
21. (2z+7)x(3z+5).
22. (4a2-106+l)x(2a2-6+2).
23. Simplify the expression (2 x-\-y)2.
SOLUTION. ' (2 x +y)2 = (2x +y) • (2 x +y). Multiplying gives
4xz+4xy + y2. Ans.
24. If the side of a square is represented by 3 x — 2, what
represents its area?
[HINT. Simplify your answer as in Ex. 23.]
25. If the dimensions of a rectangle are represented by
x+2 and x—1, what represents its area?
14
SECOND COURSE IN ALGEBRA
[I, §8
26. What represents the area of the triangle whose base
is 2 x-\-3 and whose altitude is x — 5?
27. What represents the vol-
ume of the sphere whose radius
28. Explain how the figure to
the left illustrates the geometric
meaning of Formula III.
FIG, 2.
9. Division. The following formula and rules will be
recalled from the First Course.
Formula IV.
Thus =
(2o)4=24a4
Formula IV leads to the following rules.
(a) To divide one monomial by another, divide the coeffi-
cients for the new coefficient, observing the law of signs for
division (§2 (e)), and divide the literal factors, observing
Formula IV.
Thus, in dividing 28 a362 by -4 ab the process is carried out as
follows : _4 qb| 28 a3fr2
—7 a?b. Ans.
Here the division of 28 by —4 gives the new coefficient, — 7,
then the division of o3 by a gives a2 (by Formula IV), and finally the
division of 62 by b gives b.
(b) To divide a polynomial by a monomial, divide each
term of the polynomial separately, and combine results.
Thus the division of 8 x2y—4 x3y*+2 xy by 2 xy is carried out
as below : 2 xy \ 8 x*y-4 x*y*+2 xy
4z -2x2y +1. Ans.
I, §9] FUNDAMENTAL NOTIONS 15
(c) To divide one polynomial by another :
1. Arrange the dividend and divisor in the descending (or
ascending) powers of some common letter.
2. Divide the first term of the dividend by the first term of the
divisor, and write the result as the first term of the quotient.
3. Multiply the whole divisor by the first term of the quo-
tient; write the product under the dividend and subtract it from
the dividend.
4. Consider the remainder as a new dividend, and repeat
steps 1, 2, and 3, continuing in the same manner thereafter.
Thus the division of 17 x +20+3 x2 by 4+z is carried out as
follows :
3x2 + l2x 3z+5 Quotient. Ans.
5z+20
5s +20
0
In this example, a new dividend is finally obtained which is equal
to 0. Whenever this happens, the division is said to be exact. In
cases where the division does not come out in this way, there is a
remainder, as illustrated by the following example :
3 x* + 17 z+20 | x+3
3 x2 + 9 x 3 x +8 Quotient. Ans.
8z+20
8x+24
— 4 Remainder.
EXERCISES
1. How do you test (or check) to see whether an answer
in division is correct? Is the same method used for this in
both arithmetic and algebra? Check the correctness of the
results obtained for the last two examples in § 9.
Perform the following divisions, and check your answer.
16 SECOND COURSE IN ALGEBRA [I, §9
2. a* + a\ 3. (3 g)g+(3 g)\ 4.
5- (iP2<z)7-(iP2<?)3. 8. f7rr>-2 TTT.
6. -16zyz2-=-4z2/2z. 9. 3a6(a+6)2
7. 4a462c3^20a262c. 10. (9 m3np+18 wn3p)-=-3 mn.
11. (6 z2?/z+12 :n/2z-24 zi/z2) -5- (-3 xyz).
In each of the following divisions, find the quotient, also
the remainder if there is one. Check your answer for each.
12. (3z2-2z-l)-=-(z^l). 13. (15x2+x-2)-^-(3x-l).
14. (4</3+22/2-l)-K22/-l).
[HINT. Write the dividend in the form 4 y3 +2 yz +0 y — 1.]
15.
16.
17.
18.
19.
20.
MISCELLANEOUS EXERCISES
1. Add 4x*-2x*-7x + l, z3+3z2+5z-6, 4 x2-8
2x3-2x2+8x+4, and -2 x + 1+2 z3-3 x2.
2. Add 3(a+6)+6(6+c), 5(a+&)
3(&+c) -(a+6), 2(6+c) -10(a + &), and 3(a+6) -3(6+c).
3. From the sum of 1 +x and 1 — xz subtract 1 — x+x2 — x3.
4. Simplify the expression
ab -{5 +x - (b +c -ab +z)} + [x - (b - c -7)].
6. When x=3, m=6, n=2 find the value of the expression
(m +n +x)n — (m +n — x)n — (m - n +z)"( - m +n +x)n.
6. Simplify the expression
y3-[2x3-xy(x-y)-y*]+2(x-y)(x*+xy+y*).
7. Find the remainder in the following divisions
(a) (0^ + 1)^(0-1).
(6) [X3«-3 +2/3n+3] ^. [xn-l
CHAPTER II
SPECIAL PRODUCTS AND FACTORING
10. Special Products. Certain products occur so fre-
quently in algebra that it is desirable to study them with
especial care and to remember their forms. In this connec-
tion, the following formulas will be recalled.
Formula V. (x-y)(x+y = x2-yz.
Thus (x-8)(x+8)=x2-82=a;2-64;
(9 x-3 y)(9 x+3 y) =(9 z)2-(3 yY =81 x2-9 y\
Formula VI. (x+yY~ = x2+2 xy+y2.
Thus (r+6)2 = r2+2(r- 6) +62 =r2 + 12 r+36;
Formula VII. (x-y)z = x2-2 xy+y2.
Thus (r-6)2=r2-2(r- 6) +62 =r2-12 r+36;
Formula VIII. (x+m)(x+n) = x2 + (m+n)x+mn,
m and n being any (positive or negative) numbers.
Thus (x+4)(x+3)=x2 + (4+3)x+4- 3=z2+7z+12;
(-6
17
18
SECOND COURSE IN ALGEBRA
[II, § 10
ORAL EXERCISES
State under which formula each of the following products
comes, and read off the answer by inspection.
1. (z-3)(z+3). 3. (6+a)(6-o). 5. (5 x-2)(5z+2).
2. (r-5)(r+5). 4. (x-4)(x+4). 6. (1-
7. (xy-12)(xy+12).
8' (HXH
9. (z+8)2.
10. (z-8)2.
11. (2z+l)2.
12. (3z-4)2.
13. (a+26)2.
24. 4
14.
15.
16.
17. (3x-f)2.
18. (2
2). 26.
25. (3o-5|/)(3o+7y). 27.
28. }8-(r+«)H8+(r+«)}.
29.
30.
19.
20. (a+6)(a-8).
21. (a;-
22.
23. (a2+4)(a2-6).
WRITTEN EXERCISES
1. Show how Formulas VI and VII can be obtained as
special cases of Formula VIII.
2. Show how the following three figures illustrate re-
spectively the geometric meanings of Formulas V, VI, and VII.
-y—
(x-y)2
II, §11] SPECIAL PRODUCTS AND FACTORING 19
By use of Formulas V, VI, and VII write down (without
multiplying out) the simplest forms for the following prod-
ucts.
3. (a2+a-l)(a2-a+l).
[HINT. Write first as {a2 + (a-l)}{a2-(a-l)}, then apply For-
mula V, afterwards simplifying your answer by Formula VII. The
final result is a4 -a2 +2 a - 1.]
4. (a2+a6+62)(a2-a&+&2). 7. J(z+7/)-4(2.
6. (x2+x-2)(x*-x-2). 8. [7+(m-n)]2.
6. (a-b+m+n)(a-b-m-n). 9. [(x+y)-(m+ri)]2.
• 10. (a+6+c)2.
[HINT. Write as [(a +6) -fc]2 and apply Formula VI twice.]
11. The rule for finding the square of any polynomial is
as follows. The square of any polynomial is equal to the sum
of the squares of its terms plus twice the product of each term by
each term that follows it. For example,
a2+62+c2+d2+2 ab+2 ac+2 ad+2 bc+2 bd+2 cd.
Show that Formulas VI and VII conform to this general
rule ; also that your answer for Ex. 10 does so.
By means of the general rule in Ex. 11, write out the values
of each of the following expressions.
12. (a+6-c)2. 14. (2x+y-z)2.
13. (a-b-c)2. 15. (2x+2y-z+3w)2.
11. Type Forms of Factoring. Factoring is the reverse
of multiplication in the sense that in multiplication certain
factors are given and we are asked to find their product,
while in factoring a certain product is given and we are
asked to find its factors, that is, to find expressions which
multiplied together produce it. The following four types
of expressions are to be especially noted, as they can always
be readily factored.
20 SECOND COURSE IN ALGEBRA [II, § 11
(a) Expressions whose terms each contains a common factor.
Thus mx+my+mz = m(x+y+z); (Formula III, p. 12.)
a?x+ax*+a2xz=ax(a+x+ax) ;
ax — ay+bx-by=a(x-y) +b(x — y) = (x — y)(a+b) ;
2 x -y +4 x2 -2 xy = (2 x - y) +2 x(2 x - y} = (2 x - y) (1 +2 x).
Note that in all these examples the given expression has
finally been exhibited as a product. This is essential to every
example in factoring.
(b) Expressions which can be regarded as the difference of
two squares.
Thus 25x*-yz = (5x-y)(5x+y)', (Formula V, p. 17.)
a262 -c2d2 = (ab -cd)(ab +cd) ;
(c) Trinomials of the form x2 -\-px-\-q, where p and q have
such values that we can readily find two numbers whose sum is
p and whose product is q.
Thus, in factoring z2+7 a; + 12, we need only inquire whether we
can find two numbers whose sum is 7 and whose product is 12.
The numbers 3 and 4 are seen (by inspection) to do this. Hence we
know by Formula VIII that we may write
z2+7z+12 = (z+3)(z+4). Ans.
Similarly, z2-z-12 = (z-4)(z+3) ; Why?
z2-5 xy-36 y* = (x-9 7/)(z+4 y) ;
a2&2 -21 ab -72 = (ab -24) (ab +3).
(d) Trinomials of the form axz-\-bx+c which are perfect
squares, that is such that the coefficients a and c are perfect
squares while the other coefficient, namely 6, is equal in ab-
solute value (§ 1) to twice the product of the square roots of
a and c.
Thus 9 z2 + 12 z+4 is a perfect square because 12 =2 • A/9 • V4.
Hence, by Formula VI, we have
9z2 + 12z+4 = (3z+2)2 = (3z+2)(3:r+2). Ans.
Similarly, xz — l4x +49 is a perfect square, because
14=2- VI-
II, § 11] SPECIAL PRODUCTS AND FACTORING 21
Thus we have, using Formula VII,
x2 - 14 x +49 = (x -7)2 = (x -7) (x -7).
For a like reason, we see that z2 + 14 x+49 = (x+7)(x+7).
The following are two other examples that can be brought
under this case.
a2b2 -2 ab +4 = (ab -2)2 = (ab -2) (ab -2).
We could not, however, factor
by this method. Why ?
EXERCISES
Factor each of the following expressions :
1. (a) .z2+2 x. (c) 8 a2+24 a.
(6) x*y+xy*. (d)
(e)
— c)— g(a+&— c).
W pq — px — rq+rx.
(i) y*-±y+xy-±x. (j) 3 x*-15x+Wy-2x*y.
2. (a) 81 -x2. (/) 962-(a-x)2.
(6) a2-62c2. (g) 49 a2-(5 a-4 6)2.
(c) 144z2-4. (h) (2x+5)2-(5x-3)
(d) ixV-36. (i) (x+x2)2-(2x+2)2.
W ~- 0)
3. (a)
(6) x2-6o:+8. (g) 12+7 a+a2.
(c)
(e) x2-o;-110. 0')
22
SECOND COURSE IN ALGEBRA
[II, § 11
16-24(a-6)+9(a-6)2.
4. Test each of the following expressions to see whether
it is a trinomial square, and if so, factor it.
(a) z2-8z+16. (/)
(6) z2-12z+36. fa)
(c) 4z2+6z+l. (h)
(d) 81z2+18z+l. (i)
(e) 81-72r+16r2. (f)
6. The figure shows a square of side a within which lies
(in any manner) a smaller square of side b. Prove that the
area between the two (shaded in the figure)
is equal to (a +&) (a — b).
6. The result in Ex. 5 furnishes a rule
for determining quickly the area between
any two squares when the one lies within
the other. State the rule.
7. By means of Exs. 5 and 6 answer the
following : What is the area of pavement
in the street surrounding a city block one half mile on a
side, the street being 4 rods wide? (1 mile = 320 rods.)
8. Show that the area between a circle of radius R
and a smaller circle of radius r lying within it is equal to
ir(R+r)(R— r). Does it make any difference where the
smaller circle lies so long as it is within the large one?
9. Show that if a and b are the sides of a right triangle
whose hypotenuse is h, we shall have a2= (h+b)(h — b) ; also
V=(h+a)(h-a).
10. Formula V is frequently used to find the square of a
number quickly by mental arithmetic. Suppose, for example,
that we wish to know the value of 162. We first take 6 away
from the number, leaving 10, then we add 6 to it, giving 22.
II, § 12] SPECIAL PRODUCTS AND FACTORING 23
We multiply the 10 and the 22 thus obtained (as is easily
done mentally) , giving 220. Now all we have to do is to add
62, or 36, to the 220 to obtain the desired value of 162, giving
256 as the answer. The reason for these steps appears below.
(16-6)(16+6) = 162-62, (Formula V.)
whence (16-6)(16+6)+62 = 162.
Find (mentally) in this way the value of each of the fol-
lowing expressions.
(a) 152. (c) 172. (e) 312.
[HINT. Subtract 5 first, then add 5.]
(6) 142. (d) 22*. (/) 452.
* 12. Other Type Forms. Besides the type forms men-
tioned in § 11, the following may be noted :
(e) Trinomials of the form ax^+bx+c which are not per-
fect squares and hence do not fall under (d) of § 11. There is
no general rule in such cases, though we may frequently
discover by inspection whether a given trinomial of this form
is factorable readily, and if so, obtain its factors. This is
best understood from an example.
EXAMPLE . Factor 15 x2— 7 x — 2.
SOLUTION. For 15 x2, try 5 x and 3 x ; thus we begin by writing
(5 x ) (3 x ), where the open spaces are yet to be filled in.
For —2 try 1 and 2 with unlike signs, arranging the signs so that
the sum of the cross products shall give, as desired, the — 7 x of the
given expression; thus we now try (5 z + l)(3 x— 2). Here the
middle term of the product (cross product) is readily found to be
— 10 x+3 x, or — 7 x, as desired. The only other possibility would
be (5 x — 1)(3 x+2), but as the cross product term here becomes
10 x —3 x, or +7 x, this form cannot be the one we desire.
We have, therefore, 15 x2 -7 x -2 = (5 z + l)(3 x -2). Ans.
24 SECOND COURSE IN ALGEBRA [II, § 12
(/) The sum of two cubes. This form is factorable in ac-
cordance with the following formula.
Formula IX. x*+y*=(x+y)(x*-xy+y*).
Thus x3 +8 = x3 +23 = (x +2) (x2 -2 x +22)
= (x+2)(x2-2x+4). Ans.
Likewise,
W3n3 +64 p3 = (ran)3 + (4 p)3 = (ran +4 p) [(ran)2 - (ran) (4 p) + (4 p)2]
= (ran+4 p)(ra2n2— 4 ranp + 16 p2).
(g) The difference of two cubes. This form is factorable
in accordance with the following formula.
Formula X. x*-y*=(x-y)(x2+xy+y*).
-27=x3-33 = (x-3)(z2+3x-f32)
= (x-3)(x+3x+9). Ans.
8 x3 - 125 2/3 = (2 x)3 - (5 y)* = (2 x -5 y)[(2 x)2 + (2 x) (5 y) + (5 y)2]
= (2 x-5 2/)(4 x2 + 10 xy+25 i/2). Ans.
13. Complete Factoring. Each of the exercises on p. 21
concerns but a single one of the type forms mentioned in § 11,
but we often meet with problems in which two or more of the
types are concerned at the same time. Thus, in factoring
as£ _ a^} we first take oirt the common factor ab. This gives
a?b — abs = ab(a2-b*). But a2 — 62 is itself factorable, coming
under type (6) of § 11. The final answer, therefore, is
ab(a-b)(a+b).
Other illustrations of this idea occur below. Note that the
final answer in every case contains no factors which them-
selves can be still further broken up into other factors.
EXAMPLE 1. Factor completely x*-y2+x-y.
SOLUTION.
, §11.)
Ans.
(See (a) §11.)
11, § 13] SPECIAL PRODUCTS AND FACTORING 25
EXAMPLE 2. Factor completely a4 +3 a2b2— 4 64.
SOLUTION.
o4+3 a262-4 64 = (a2+4 62)(a2-62) (See (c), § 11.)
= (a2+4&2)(a-6)(a+6). (See (6), § 11.)
EXERCISES
Factor completely each of the following expressions. Those
preceded by the * involve the type forms mentioned in § 12.
1. x*-x2-x+l. 14.
[HINT. Write in the form. 15.
z2(z-l) -(*-!).] *16.
2. 2x*-8x2y+Sxy2. *17.
[HINT. Write in the form 18-
4xi/-[-4i/2).] *19.
3. x2+3ax-3a-x. *20. 20z2-6z-2.
4. a3+2a2-4a-8. 21- Sx*~
5. x4-lQ^2J-Qft 22> s4-1*
^ i . .4. r» —9.. 9 ^O.
.
24.
^JLuLOflfc 25t 9^4-30^2a+25a2.
26.
[HINT. 1 — a2 — 62 + 2 a6 = 1 — o i.-
(a-6)2.] ~
27. (a-x2)2-(6-i/2)2.
9. m2-4mn+4n2-16. *28
10. 2xy-x -y2+l. 29. x2-9 t/2+a:+3 y.
11. l+9c2+6c. 30. a3-2a26+4a62-863.
1O Q /y»3 ^_ Q /Y» I /I /y»4 /| /yi2 QO />»4 /1 7 /v*2
•Lv« O •*/ O »i/ \ A «|/ A «v • O*i» »i/ """" A I •*/
CHAPTER III
HIGHEST COMMON FACTOR AND LOWEST COMMON
MULTIPLE
14. . Prime Factor. A number which has no factor except
itself and unity is called in arithmetic a prime number.
Such a number when used as a factor is called a prime factor.
Thus the prime factors of 15 are 3 and 5.
The word prime factor is similarly used in algebra.
Thus we say that the prime factors of 3 abc are 3, a, 6, and c.
The prime factors of 18 x2y are 2, 3, 3, x, x, and y.
The prime factors of o26(a2 — b2) are a, a, b, a — 6, and a +b. (See
Formula V.)
15. Finding Common Factors. As soon as we have fac-
tored each of several expressions into its prime factors, we can
readily pick out their common factors (§5).
Thus, in finding the common factors of abc, a?b, a&2, and 3 ab, we
write
abc = a • b • c, azb = a • a • b, ab2 = a • b • b, 3 ab = 3 • a • b.
The common factors are, therefore, a and 6, since these occur in
each expression and they are the only factors thus appearing.
16. Highest Common Factor. The product of all the
common prime factors of two or more expressions is called
their highest common factor. It is called the highest be-
cause it contains all the common factors, and the usual ab-
breviation for it is H. C. F.
EXAMPLE 1. Find the H. C. F. of lOz2!/3, 2xy2, and 18sV-
SOLUTION. Resolving each into its prime factors,
10zV=2- 5- x- x- y y y,
2xy*=2- x- y y,
18zV =2- 3'3-x-X'X-yy.
The factors common to the three expressions are thus seen to be
2, x, y, y.
The H. C. F. is, therefore, 2 • x - y - y, or 2 xy*. Ans.
26
Ill, § 17] COMMON FACTORS AND MULTIPLES . 27
EXAMPLE 2. Find the H. C. F. of 3 z2+3 x-18,
6 z2+36 z+54, and 9 z2-81.
SOLUTION.
3z2+3z-18=3(z2-fz-6)=30c+3)(z-2). (See (c), §11.)
6z2+36z+54=6(z2+6-z+9)=2. 3(z+3)(z+3). (See (d), §11.)
9z2-81=9(z2-9)=3- 3(x+3)(x-3). (See (6), §11.)
The common factors being 3 and (x +3), the H. C. F. is 3(x +3.) Ans.
In general, to find the H. C. F. of two or more expressions :
1. Find the prime factors of each expression.
2. Pick out the different prime factors and give to each the
lowest exponent to which it occurs in any of the expressions.
3. Form the product of all the factors found in step 2.
NOTE. Since the H. C. F. of several expressions consists only
of factors common to them all, it is always an exact divisor of each
of the expressions. It is therefore called in arithmetic "the great-
est common divisor" and is represented by G. C. D.
EXERCISES
Find the H. C. F. of each of the following groups.
1. 12, 18. 7. a2+7a+12, a2-9.
2. 16,24,36. 8. x2-y2,(x-y)2,x2-Zxy+2y*.
3. x2y, xy2. 9. ra2+4 m+4, m2-6 m-16.
4. a2b, ab2, a2b2. 10. 3 ?/2-363, y2-7 T/-44.
6. 2 x*y, 6 x*y, 14 x*y2z4. 11. 2 a2+4 a, 4 a3+12 a2+8 a.
6. a2-62, a2-2a&+62. 12. 4-i/4, x*+xzy+xy2+y*.
13. 3 r5+9 r4-3 r3, 5 rV+15 rs2-5 s2, 7 ar2+21 ar-7 a.
*14. a3-63, a2-62, a-b. *15. x3+l, x2-x+l.
17. Common Multiple. In arithmetic a number which
is exactly divisible by two or more given numbers is called a
common multiple of them.
The word common multiple is similarly used in algebra.
Thus 4 x2y2 is a common multiple of x and y ; it is also a common
multiple of 4 and x. Similarly, a2 — b2 is a common multiple of a — b
and a +6.
28 SECOND COURSE IN ALGEBRA [III, § 18
18. Lowest Common Multiple. The lowest common
multiple of two or more numbers or expressions is that
multiple of them which contains the fewest possible prime
factors. Its abbreviation is L. C. M.
The following examples illustrate what the L. C. M. means
and how to obtain it.
EXAMPLE 1. Find the L. C. M. of 10 a2b, 16 a263, and
20 a364.
SOLUTION. Separate each expression into its prime factors. Thus
10a26=2- 5- a- a- 6,
16a263=2- 2- 2-2- a- a- 6- 6- 6,
20a364=2- 2-5-a-a-a.6-6.b-6.
The L. C. M. is 2 - 2 - 2 • 2 • 5 • a • a - a • 6 • 6 - 6 - 6, or 80 a364,
since this contains all the factors of each of the three expressions,
and at the same time is made up of fewer factors than any other
similar expression that can be found.
EXAMPLE 2. Find the L. C. M. of z4-10z2+9 and
SOLUTION.
Therefore the L. C. M. is (z-
In general, to find the L. C. M. of two or more expressions :
1. Find the prime factors of each expression.
2. Pick out the different prime factors, taking each the great-
est number of times it occurs in any one of the expressions.
3. Form the product of all the factors found in step 2.
NOTE. From the manner in which the L. C. M. is formed, it
must be exactly divisible by each of the given numbers, or expressions.
EXERCISES
Find the L. C. M. of each of the groups of expressions in
the exercises on p. 27.
CHAPTER IV
FRACTIONS
19. Definitions. Any expression of the form a/6 is called
a fraction. It means the number, or expression, which when
multiplied by b gives a. The part above the line, or a, is
called the numerator, while the part below the line, or 6, is
called the denominator. The numerator and denominator
taken together are called the terms of the fraction.
20. Equivalent Fractions. It is often desirable to change
the form of a fraction without changing its value. Such
changes all depend upon the following principle.
The numerator and denominator of a fraction may be multi-
plied or divided by the same number, or expression, without
changing the value of the fraction. Thus
3^3- 2^6.
4 4-2 8'
4 a_4 a • a_4 a2.
5 5 • a 5 a'
a+b
(a+6)2
21. Changes of Signs in Fractions. There are three signs
to be considered in a fraction ; the sign of the numerator, the
sign of the denominator, and the sign of the fraction itself.
_ o
Thus in H -- , the three signs in the order just mentioned are
4
-, +, +, while in --^7 they are +, -, -.
— D 0
29
30 SECOND COURSE IN ALGEBRA [IV, § 21
Since a fraction is merely an indicated division, the law
of signs for division (§ 2 (e), p. 3) must hold at all times, so
that we arrive at the following rule.
Any two of the three signs of a fraction may be changed with-
out altering the value of the fraction. Thus
_i_±3 = +:d* = -3= ±3
+4 -4 +4 -4*
Likewise
a _ — a _ — a _ a
b -b ~ 5 _&'
Care must be taken, however, in changing the sign of the
numerator or denominator of a fraction when polynomials
are present. For example, if the numerator is a polynomial,
we can change the sign of the whole numerator only by chang-
ing the sign of every term in it. A similar statement applies
when the denominator is a polynomial. Thus,
Q+26+c _ —a— 2b—c^ a +2 b+c
2a-3b-2c 2a-3b-2c -2o+36+2c*
Observe carefully the reason for every change of sign here.
22. Reduction of Fractions to Lowest Terms. A fraction
is reduced to its lowest terms when its numerator and de-
nominator have no common factor except 1.
To reduce a fraction to its lowest terms, factor numerator and
denominator, then divide each by all their common factors.
E
EXAMPLE 2. a2 -Ha +24 = (a-
a2-a+6 (a-
IV, §23] FRACTIONS 31
23. Lowest Common Denominator. The lowest common
denominator of two or more fractions is the lowest common
multiple (§ 18) of their denominators.
To reduce several fractions to their lowest common denomi-
nator :
1. Find the L. C. M. of the denominators.
2. For each fraction, divide this L. C. M. by the given de-
nominator, and multiply both numerator and denominator by the
quotient.
EXAMPLE. Reduce the following fractions to equivalent
fractions having their lowest common denominator :
and
SOLUTION. Factoring the denominators, the fractions may be
written
The L. C. M. of these denominators is (a; +3) (a; -3) (z +6). In
order to give the first fraction this L. C. M. as its denominator,
multiply its numerator and denominator by x+G (this being the
L. C. M. divided by the denominator of the first fraction).
In order to give the second fraction this L. C. M. as its denomina-
tor, multiply its numerator and denominator by x— 3 (this being
the L. C. M. divided by the denominator of the second fraction).
The desired forms are, therefore,
(x+2)(x+Q) d
(x +3) (x -3) (x +6) (x +3) (x -3) (x +6)
Observe that these fractions are respectively equivalent ( § 20) to
those with which we started, but these have denominators that are
alike, which was not the case with the original forms.
32 SECOND COURSE IN ALGEBRA [IV, § 23
EXERCISES
Write each of the following fractions in three other ways
without changing the value.
1 5 6 3 a-b_ . 3a-4
-6* ' l-x ' c-d ' (2z-l)i
Reduce each of the following expressions to lowest terms.
240
' * '
320 (a6+l)2
10 xy 4z2-2/2
O. * 11. — •
OA /y>2/)/2 ni O /y.
o\j ju y y — £i jc
„ 36 xr3 19 s2-6s+8
*• ^r — r~r* J-^.
72^^ s2-5s+6
-9a252c2 1Q r4-6r2+5
o. -3-: — r77~r~* !*•
54 aW r2-6r+5
28xy2-12x2y
Reduce all of the fractions in each of the following groups
to the lowest common denominator.
16 2f ^ £. go 2 -^ __ L,
' 4'6'2 ' 62'a2-62' a+b
17 5 ?. 21
' a 6 '
a r
y*>x-y ' (a+6)2' l'a2-62
1 1
23.
0. a;-2 rc-1 o:+3
Z4.
x2-2x-8' x2-3x-10'a;2-
IV, § 24] FRACTIONS 33
24. Addition and Subtraction of Fractions. The follow-
ing rule, which is the same in algebra as in arithmetic, will be
recalled from the First Course, p. 151.
To add, or subtract, fractions :
1. Reduce the fractions to equivalent fractions having their
lowest common denominator (L. C. D.).
2. Add, or subtract, each numerator according to the sign
before the fraction and write the result over the L. C. D.
3. Reduce the resulting fraction to its lowest terms.
EXAMPLE. Simplify
a-1 a+1 a2-!
SOLUTION. — Qn2+_3_a2_=4 a+4_a2-3 a+2+^g2_
a-1 a + 1 a2-! a2-! a2-! a2-!
_4a+4-(a2-3a+2)+3a2_2a2+7a-+2
a2-! a2-!
EXERCISES
Simplify each of the following expressions.
~3~ ~4~ 2' x y *' abz ab a?
2-2f+£?" 4-^-2J1^+I- 6-H+r
7 _ | _.. 12
2x-2 3(z+l) 3z-3 a2+2a6+62 a?-2ab+b2
l3-
9 _ . 14 _ .
a—b a+b n+4 1-n n2+3 n-4
10. ___ -+-~ 15. x-3+
r2-2r-8 r-4 r+2 x2+3 z+9
11 1 1 16 _x__ 1+a^2
' x-l x2+x+l
34 SECOND COURSE IN ALGEBRA [IV,
25. Multiplication and Division of Fractions. Fractions
are multiplied in algebra as in arithmetic by taking the
product of the numerators for the numerator, and the product
of the denominators for the denominator, canceling wherever
possible.
EXAMPLE 1.
4 xy* 9 ab3 4
Canceling like factors from numerator and denominator, this
reduces to
^- Ans.
662
EXAMPLE 2.
a2 +2 a -8 _ a2+5 a-6
a2 +6 a a2+a-12
.a(a-3)
In algebra we divide one fraction by another as we do in
arithmetic, by inverting the divisor and proceeding as in
multiplication.
EXAMPLE.
a2-b2 . a-b a2-b2
a-b
A
EXERCISES
Perform each of the following multiplications.
3., 2
5 rs2 1*8 m3n3
7 o 1^2 O
o « 5. A a k 2
1 6 c ' a2+&2
2/frl /d_ O'1^ P\ /•
CM/ *± •*/ M «J «
3 xy 5 a2 2 z2+4 Z7/+2 i/2 5
. 6 a \ 10 b8 8 r+2 g-1 4s-6
1 ' 8 a3' ' 2s-3 *2r+2* r+2"
IV, § 25] FRACTIONS 35
9.
10 . . - ___ -.
x — ?/ 6 — a £+?/ a— 6
Perform the following divisions.
4^2
' *
9*3 a(a+6) '2 0-6
12 10 & . 12 s r2-9s2
* ' 15<
13.
11 y y2 r2-4s2 r2-rs-6
a2+2a, (a+2)2
a2-2a* (a-2)2'
Perform the indicated operations and simplify each of the
following expressions.
16
\x+y x-yj \x+y x—yj
is
'
3,5
Li. 19.
3_4 V m m-3\m-2
x y
21. fi-
1 1 1
22 .
-X 1+05
CHAPTER V
SIMPLE EQUATIONS
26. Preliminary Considerations. Suppose we wish to
divide 64 into two parts such that if one part be divided by
5 and the other by 7 the sum of the quotients shall be 10.
Such a problem as this can be done only with some difficulty
by arithmetic, but it is a simple task by algebra.
SOLUTION. Let x represent one part.
Then 64— x will be the value of the other part.
From the statement of the problem, we are to have
Let us multiply both sides of this equality by 35 (as we may do
without destroying it), thus clearing it of fractions. This gives
7 x +320 -5 x =350.
Subtracting 320 from both sides of this last equality (as we may
do without destroying it), and replacing 7 x —5 x by its value 2 x,
we obtain 2 x= 30.
Hence (dividing both sides by 2) we have
z = 15.
The two parts sought are therefore 15 and 64 — 15, or 49. Ans.
CHECK. + = 3 +7 = 1°-
A statement of equality, like any of those above, wherein
a single unknown letter occurs, and occurs to no higher power
than the first, is called a simple equation. It is also known
as a linear equation, or an equation of the first degree.
The process of finding the value of the unknown letter is
called solving the equation.
The value of the unknown letter is called the solution,
or root, of the equation.
V, § 27] SIMPLE EQUATIONS 37
27. Principles Useful in the Solving of Equations. In
solving an equation we may at any point in the process add
the same amount to both sides, or subtract the same amount
from both, or we may multiply both by the same amount, or
divide both by the same amount, as was illustrated in § 26.
Derived from these are the following useful principles.
(a) A term may be transposed (carried over) from one side
(or member) of an equation to the other provided its sign is
changed.
Thus, in the equation 3 a: —4 =2 we may transpose the term — 4
to the second member, giving 3 x =2+4, or 3 x =6. This is equiva-
lent to adding 4 to both members of the given equation.
The solving of equations is greatly simplified by a free use
of this principle of transposing terms.
Thus, in solving 3 x— 4=x+2, we may transpose the —4 from
the first member to the second and at the same time transpose the
term x from the second member to the first, giving — z+3 x =2+4,
or 2 x =6. Therefore x =3. Ans.
(b) A term which appears in both members of an equation
may be canceled.
Thus, by canceling the 3 from both members of the equation
2 x +3 = 10 +3, we have simply 2 x = 10, and hence x =5.
Note that to cancel a term in this way merely amounts to
subtracting it from both members of the given equation.
(c) The signs of all the terms in an equation may be changed.
Thus -5z+3=z-9may be written 5 x -3= -x+9.
Note that to change all signs in this manner amounts
to multiplying both members of the given equation by —1.
(d) An equation may be cleared of fractions by multiplying
both members by the lowest common denominator of all the
fractions.
38 SECOND COURSE IN ALGEBRA [V, §27
EXAMPLE. Solve the equation - = 6
SOLUTION. The L. C. M. of the denominators is 12.
Multiplying both members by 12,
4(z-2) -3(3-3) =72-6(s-l).
Removing parentheses (§7),
4z-8-3z+9=72-6z+6.
Transposing, 6x+4:c_3 1=72+6+8_9>
7* =77.
Therefore
a: = 11. Ans.
EXERCISES
Solve the following, using the principles stated in § 27.
Check your answer for the first five.
2.
!+6-T-
33-1 2
3-3 , 6z+5
/v. 0 C
K * ° ° _Q
2 3
4 6
2z 12
1 3+1 2
4 5
7.
8.
Q
5 8
1 2
6' x x* 33
1
3+1 3(3+1)
1 1
12
4-6y 2-3i
1 ,r+l
r-2
r2_r_2 r-2 r+1
^ rv *k I -L *v " O
io- jn-ija
12
'
a;-4 x-8 z2-12z+32
V, § 27] SIMPLE EQUATIONS 39
13. If 10 be subtracted from a certain number, three
fourths of the remainder is 9. What is the number?
14. Divide 38 into two parts whose quotient is 122-.
15. Divide 96 into two parts such that f of the greater
shall exceed f of the smaller by 6.
16. A man started on a journey with a certain sum of
money. He spent -J of it for car fares and ^ of it for hotel
bills. When he returned home he found he had $9. How
much did he start with ?
17. I have $100 in one bank and $75 in another one. If I
have $45 more to deposit, how shall I divide it among the two
banks in order that they may have equal amounts?
18. A motor boat traveling at the rate of 12 miles an hour
crossed a lake in 10 minutes less tune than when traveling
at the rate of 10 miles an hour. What is the width of
the lake?
[HiNT. Time = Distance -5- Rate.]
19. A freight train goes 6 miles an hour less than a passen-
ger train. If it goes 80 miles in the same time that a passen-
ger train goes 112 miles, find the rate of each.
20. A tank can be filled by one pipe in 10 hours, or by an-
other pipe in 15 hours. How long will it take to fill the tank
if both pipes are open ?
[HINT. Let x = the number of hours. Then l/x = the part both
can fill in 1 hour. But, ^ = the part the first pipe can fill in 1 hour,
and T1g-=the part the second pipe can fill in 1 hour. Hence we must
have ! = _1_L_JL~|
x 10 15* J
21. How long will it take two pipes to fill a tank if one
alone can fill it in 5 hours and the other alone in 12 hours?
40 SECOND COURSE IN ALGEBRA [V, § 27
22. Two pipes are connected with a tank. The large one
can fill it in 3 hours ; the small one can empty it in 4 hours.
With both pipes open, how long before the tank will fill ?
23. A does a piece of work in 4 days, B in 6 days, and C in
8 days. How long will it take them working together?
24. A can do a piece of work in 16 hours, and B can do it
in 20 hours. If A works for 10 hours, how many hours must
B work to finish?
25. A's age is ^ that of his father's. 12 years ago he was
i as old as his father. How old is each now?
26. A boat goes at the rate of 12 miles an hour in still
water. If it takes as long to go 27 miles upstream as 45
miles downstream, what is the rate of the current?
27. An aviator made a trip of 95 miles. After flying 40
miles, he increased his speed by 15 miles an hour and made the
remaining distance in the same time it took him to fly the
first 40 miles. What was his rate over the first 40 miles?
28. A 5-gallon mixture of alcohol and water contains 80%
alcohol. How much water must be added to make it con-
tain only 50% alcohol?
[HINT. .50(z+5)=5X.80. Explain.]
29. How much water must be added to 65 pounds of a
10% salt solution to reduce it to an 8% solution ?
30. A train 660 feet long running at 15 miles an hour will
pass completely through the Simplon tunnel in Switzerland
in 49^ minutes. How long is the tunnel ?
DESCARTES
(Rent Descartes, 1596-1650)
Profound student and ranked as one of the greatest leaders of all time in
both mathematics and philosophy. He invented representation by graphs
and was thus led to the discovery and development of the branch of mathe-
matics called Analytic Geometry. He was also much interested in medicine
and surgery.
CHAPTER VI
GRAPHICAL STUDY OF EQUATIONS
28. Definitions. Let two lines XX' and Y Y' be drawn
on a sheet of squared (coordinate) paper, XX' being hori-
zontal and Y Y' vertical. Two such lines form a pair of
coordinate axes. The point 0 where they intersect is called
the origin.
Consider any point, as P, and draw the perpendiculars PA
and PB extending from P to the two axes Y Y' and XX'
respectively. PA is then called the abscissa of P and PB
is called the ordinate of P. The abscissa and ordinate
taken together are called the coordinates of P.
Y
A
A
P
i
I
(3
4)
(-2
,3)
n
1
X'
-
3
0
B
X
— J
4
-
2-
1
3 <
)
n
S
•*•
(3
)
R
-4
(
-3
-4)
Y'
FIG. 5.
Thus the point P in the figure has its abscissa equal to 3 and its
ordinate equal to 4.
All abscissas on the right of Y Y' are considered positive,
while all abscissas on the left of Y Y' are considered negative.
Thus the abscissa of Q is -2 ; that of R is -3 ; that of S is +3.
Similarly, all ordinates above XX' are considered positive,
while all ordinates below XX' are considered negative.
Thus the ordinate of Q is +3 ; that of R is -4 ; that of S is -2.
41
42 SECOND COURSE IN ALGEBRA [VI, § 28
In reading the coordinates of a point, the abscissa is always
read first and the ordinate second. Thus, in the figure, the
point P is briefly referred to as the point (3, 4) ; similarly, Q
is the point ( — 2, 3) ; R is the point ( — 3, — 4) ; and S is the
point (3, —2), etc.
In practice, XX' is commonly called the x-axis, and Y Y'
is called the y-axis.
EXERCISES '
[The pupil will find it convenient to use the prepared coordinate
paper such as usually may be secured at the stationery stores.]
1. Draw axes on a sheet of coordinate paper and then
locate (plot) the following points :
(2,4); (-3, -1); (2, -4); (2J, -3); (-2J, -2J);
(-4, +4); (0, -5); (4,0); (0,0).
2. The part of the plane within the angle XO Y (see figure
in § 28) is called the first quadrant, the part within the angle
YOX' is called the second quadrant, the part within X'O Y'
is called the third quadrant, etc. Hence, state in which quad-
rant a point lies when
(a) its abscissa is positive and its ordinate negative,
(&) its abscissa and ordinate are both negative,
(c) its abscissa is negative and ordinate positive.
3. What can be said of the position of a point whose ordi-
nate is positive ; whose abscissa is negative ?
4. A certain street runs due east and west. It is met by
another street which runs due north and south, thus form-
ing a " four corners." Taking the meeting place of the cen-
ter-lines of the two streets as origin, and the east and north
directions as positive, what are the coordinates of a flagpole
which stands due northwest from the origin at a distance
of 50 feet from the center-line of each road? Answer the
same when the pole is 45 feet due west of the crossing point.
VI, § 29] GRAPHICAL STUDY OP EQUATIONS 43
5. Plot the following three points and then see if it is
possible to draw a straight line that will pass through all of
them: (1,5); (0,3); (-1,1).
Do the same for the three points (2, 3) ; (—!,—!); (5, 0).
29. Graph of an Equation. We have seen in Chapter V
that if we have any linear equation containing a single un-
known letter, as for example the equation 2 x— l = 3(x— 1),
we can always solve it ; that is we can find the value of x.
Suppose now that we have a linear equation in which
two unknown letters, x and y, appear, that is an equation
in which no term contains both x and y nor any higher power
of either of them than the first, as for example
(1) x+y = 5.
The meaning of such an equation and the interesting
facts about it are best brought out by graphical methods in
ways which we shall now explain.
In the first place, it is to be observed that such an equa-
tion is satisfied by a great many pairs of values for x and y.
For example, the pair of values 0=1, y = 4) satisfies the
equation, because when we put these values for x and y
respectively in the equation, it becomes 1+4 = 5, which is
true. Again, the same is seen to be true of the pair (x = 2,
2/ = 3) (explain) ; and, similarly, the same is true of any one of
the pairs (x = ±, y = f), (x = Q, y=-l), (x = 8, y=-3), etc.
In fact, we can obtain as many such x, y pairs as we wish,
each pair having the property that the z-value and the
?/-value taken together satisfy the given equation.
If we place x =3 in the equation above, we have 3 +y =5 and this,
when solved for y, gives y =2. Thus (x =3, y =2) is a pair such as
mentioned above. Similarly, we can assign to x any value we wish
(positive or negative) and find from the equation the corresponding
value of y, thus forming a new pair of values of x and y.
44
SECOND COURSE IN ALGEBRA [VI, §29
Whenever an equation contains two (but no more) un-
known letters, such as x and y, any pair of values for x and
y that satisfy it is called a solution of the given equation.
It follows from what has been shown above that every such
equation has an indefinitely large number of solutions.
Returning to the equation x-\-y = 5, let us consider again
the special solutions which we noticed on page 43 :
(x = 2, 2/ = 3), (z = i, 2/ = f), (x=-l, i/ = 6), (x = 8, 2/=-3).
Following the ideas brought out in § 28, each of these may
now be plotted a*s a point, using x as abscissa and y as ordi-
nate. Upon locating these points carefully, it will be seen
that they all lie on one and the same straight line, as indi-
cated in the figure below.
\
Y
\
\<
i-
-)
\
(2
3)
•+
X
A
\
0
\
X
\
(6
-U
\
\
(
8,
3)
\
FIG. 6.
This line is called the graph of the equation x+y = 5. It
may be shown that every solution of the given equation gives
rise when plotted to some point upon this line, and vice versa,
every point upon this line has an a>value and a i/-value
which, when taken together, form a solution of the given
equation.
30. Graph Determined from Two Points. In practice,
the graph of a linear equation is drawn by locating two points
upon it, and connecting them by a straight line.
VI, § 30] GRAPHICAL STUDY OF EQUATIONS
45
EXAMPLE. Draw the graph of the equation 5 x— 4 y = 20.
SOLUTION. Placing x =0 in the equation gives y = — 5. Hence
(0, —5) is a point on the graph.
Placing y = 0 in the equation gives x = 4. Hence (4, 0) is a point
on the graph.
Plotting these two points (0, —5) and (4, 0) and drawing (with
ruler) the straight line through them, gives the required graph.
Y
V
A
f
/
0
(4
0)
/
x
/
/
^
/
«
4
/
/
(0
-5
)
b/
FIG. 7.
NOTE. As in the example just considered, it is often simplest
to select as the first point the one whose abscissa is 0, and as the
second point the one whose ordinate is'O. However, it is equally
correct to take any two points whose coordinates satisfy the given
equation. If the two points selected are too close to each other,
it is difficult to draw the line accurately ; if this happens, plot a third
point on the line at a considerable distance from the first two.
EXERCISES
1. State (orally) what is true of each point on the line in
the last figure.
[HINT. Its abscissa and ordinate, taken as a pair of numbers,
(x, y), form a . . . .]
Draw the graph of each of the following linear equations.
2. 2x-y = 4:. 4. 2x+3y=12. 6. 4x = 3y-7.
3. 2x+y = 2. 5. x-3y = 3. 1. 2x = 3y.
46 SECOND COURSE IN ALGEBRA [VI, § 30
8. If a person travels at the rate of 15 miles per hour, the
distance s which he will have traveled at the end of t hours
is given by the formula s = 15 t. Draw the graph of this
equation, using the ^-values as abscissas and the s-values as
ordinates. From your figure read off (approximately) how
far he will have traveled at the end of (a) 2 hours ; (6) 3
hours; (c) 3^- hours; (d) 4-J- hours.
[HINT. Take the Z-axis horizontal and the s-axis vertical, and
let the unit length on each be about half an inch. In order to get
the diagram into relatively compact form, allow each unit on the
s-axis to represent 15 miles, taking each unit on the 2-axis to repre-
sent 1 hour.]
9. A boy has $10 in the bank when he begins saving at the
rate of $3 a month, adding this amount month by month to
his account. Find graphically how many months must
elapse before his account will amount to $22.
[HINT. Let A represent the amount of the account at the end of
t months. Then, A =10 +3 L (Why?) Now draw the graph of
this equation, using ^-values as abscissas and A -values as ordinates,
and taking for convenience* one unit on the A -axis to represent $2,
while one unit on the t-axis represents 1 month. The problem then
calls for that abscissa which goes with the ordinate A =22.]
10. A boy has $30 in the bank when be begins spending it
at the rate of $4 a month. Find graphically how long it
will be before he has but $2 left.
[HINT. Use the same letters and units as in Ex. 9.]
11. A wheel is rotating at the rate of 10 revolutions a
second when the power is shut off. The wheel slows down
uniformly and comes to rest at the end of 30 seconds. Make
a diagram from which you can read off how many revolutions
the wheel was making at any given instant after the power
was shut off and use your diagram to determine how many
VI, § 31] GRAPHICAL STUDY OF EQUATIONS
47
revolutions per second were being made at the end of (a) 6
seconds ; (6) 9 seconds ; (c) 18 seconds ; (d) 26 seconds.
[HINT. Let r represent the number of revolutions per second at
the end of t seconds. Then the conditions of the problem tell us
that r = 10 when t=0, and that r=0 when £=30. Thus we have
two points on the graph, and we can draw the graph completely
without even getting its equation.]
12. The temperature at which water freezes is 32° on the
Fahrenheit scale, but it is 0° on the Centigrade scale. The
temperature at which water boils is 212° on the Fahrenheit
scale, but it is 100° on the Centigrade scale. Make a dia-
gram from which you can read off the Centigrade tempera-
ture that corresponds to any given Fahrenheit temperature.
[HINT. Let F represent the Fahrenheit reading and C the Centi-
grade reading. Then C=0 when F= 32 and C=100 when F=212.
This gives two points. The graph is the straight line joining these
two points.]
31. Simultaneous Equations. Suppose that, instead of
having a single linear equation containing the two unknown
letters x and y (as in § 29), we have two such equations;
for example
x+y = Q and 3 x-2 y= -2.
Of all the pairs of values (x, y) that
will satisfy the first equation and all
the pairs (x, y) that will satisfy the
second equation, is there a particu-
lar pair (x, y) that will satisfy them
both at the same time? We shall
consider this question graphically.
Draw the graphs of the two equa-
tions on the same sheet of coordinate paper, using the same
axes throughout. The lines thus obtained are seen to in-
tersect each other in the point (2, 4). This means that the
\
FIG. 8.
48
SECOND COURSE IN ALGEBRA [VI, § 31
pair (x = 2, 2/ = 4) satisfies both equations at once, since it
lies on both the graphs. This pair (x = 2, y = 4) is there-
fore the pair desired, and it is the only such pair because
two straight lines can intersect in but one point.
That this answer is correct is readily seen by substituting this
pair of values in the given equations. Thus, with x=2 and y = 4,
the equations become 2+4 = 6 and 6 — 8= —2, which are true.
The two equations above illustrate what is known as a
set of simultaneous equations, and the particular pair of
values (x = 2, i/ = 4) which we found would satisfy both the
equations at one time, illustrates what is called the solution
of the set. In general, two or more equations are said to be
simultaneous if they are considered at the same time. In
the present chapter we shall deal only with sets containing
two unknowns, as in the preceding example.
32. Inconsistent Equations. Although two linear simul-
taneous equations in x and y will in general have a solution
(as in § 31), there are cases where no
solution can be found, and indeed
none exists. For example, if. we draw
the graphs of the equations
x+y = 3, and x+y = 6,
we see that the lines do not intersect ;
in other words, they are parallel.
Thus, there is no pair of values (x, y)
that will satisfy both equations at
once ; that is, there is no solution. Such a pair of simul-
taneous equations is called inconsistent.
EXERCISES
Determine graphically which of the following sets of simul-
taneous equations has a solution arid which does not. In
\
Y
\
\
\
a
\
Xk
X
t
\
\
o3
\
\
\
X
0
\
\
\
"^
u> •> * *
VI, §32] GRAPHICAL STUDY OF EQUATIONS , 49
case a solution exists, determine it and check your result by
substituting it in the given equations.
f
9. A man starts at a given time and walks along a certain
road at the rate of 5 miles an hour. An hour later another
man starts from the same place and travels in the same
direction at the rate of 10 miles an hour. Find (graphically)
how far from the starting point they will meet.
[HINT. If s represents distance (in miles) traveled in t hours,
the first man's motion is described by the equation s =5 t, while the
second man's motion is described by the equation s = 10(Z — 1).
Now draw a pair of axes, and draw in the graphs of these two equa-
tions, using lvalues as abscissas. The problem then calls for that
value of s which belongs to the intersection of the two graphs.]
10. Use your diagram for Ex. 9 to answer the following
question : After how much time will the two men meet ?
11. A man starts and walks along a certain road at the
rate of 5 miles an hour. At the same instant another man
starts out at a point on the same road 15 miles distant and
travels toward the first man on a bicycle at the rate of 10
miles an hour. How far from the first man's starting point
will they meet? How long will it take them?
12. B and C start to save money. B has $10 when they
begin and saves at the rate of $3 a month, while C at the
start owes $6 and saves at the rate of $7 a month. Find
graphically how soon C will be able to cancel his debt and
have savings equal to B's, and how much each will then have.
CHAPTER VII
SIMULTANEOUS EQUATIONS SOLVED BY
ELIMINATION
33. Elimination by Substitution. The process of combin-
ing two equations in two unknowns in such a way as to
cause one of the unknowns to disappear is called elimina-
tion.
We shall consider first the method called elimination by
substitution. The process is illustrated by the following
example.
EXAMPLE. Solve the system
(1) 2x+3y = 2,
(2) 5z-4i/ = 28.
SOLUTION. From (1),
(3) 2x=2-3y.
Therefore
(4)
o o
Substituting ^ for x in (2) gives
(5)
Clearing (5) of fractions,
(6) 5(2 -3 y) -8 y = 56.
50
VII, § 34] ELIMINATION 51
Simplifying,
(7) W-15y-8y=5G.
Collecting,
(8) -23 y = 46.
Therefore
(9) y=-2.
Substituting -2 for y in (4) now gives
The required solution of the system (1), (2) is, therefore,
(z=4, i/=-2). Ans.
CHECK. Substituting x =4 and y = -2 in (1) gives
2(4)+3(-2),
which is equal to 8 — 6, or 2, as (1) requires.
Substituting x =4 and y = — 2 in (2) gives
5(4) -4( -2),
which is equal to 20+8, or 28, as (2) requires.
To solve two simultaneous equations by substitution :
1. Solve either equation for one of the unknown letters in
terms of the other one.
2. Place the result thus obtained in the other equation and
solve it.
3. Having thus found one of the unknown letters, substitute
its value in either of the given equations and solve for the other
unknown letter.
34. Elimination by Addition or Subtraction. The only
other method of elimination which we shall consider here is
called elimination by addition, or subtraction. The process is
illustrated by the following example.
52 SECOND COURSE IN ALGEBRA [VII, § 34
EXAMPLE. Solve the system
(1) 3s+4y=12,
(2) 2z-5y = 54.
SOLUTION. Multiplying (1) by 2,
(3) 6z+8'2/=24.
Multiplying (2) by 3,
(4) 6z-15 y = 162.
Subtracting (4) from (3) ,
(5) 232/=-138.
Therefore
2/=-6.
Substituting y = -6 in (1), 3 x -24 = 12, or, 3 x =36.
Therefore x = l2.
The required solution of the system (1), (2) is, therefore,
(z = 12, y= —6). Ans.
CHECK. Substituting 1? for x and —6 for y in (1), gives
3(12) +4( -6) =36 -24= 12, as (1) requires.
Substituting 12 for x and —6 for y in (2), gives
2(12) -5( -6) =24+30=54, as (2) requires.
NOTE. Instead of multiplying (1) by 2 and (2) by 3 and then
subtracting them, thus eliminating x, we might just as well have
multiplied (1) by 5 and (2) by 4 and added them, thus eliminating y.
Either plan leads to the same solution for the given system.
To solve two simultaneous equations by addition or subtrac-
tion:
1. Multiply one, or both, of the given equations by such num-
bers as will make the coefficients of one of the letters (say, y)
numerically equal.
2. Subtract (or add) the two equations thus obtained, thus
eliminating one of the unknown letters.
3. Solve the resulting equation for the letter it contains, and
obtain the value of the other letter by substituting the value of
the letter already found into either of the given equations.
VII, § 34]
ELIMINATION
EXERCISES
53
1.
Solve by substitution :
2x+Zy=12,
x+5y=13.
3.
4. <
[2x-3y=-W.
5. < _ , /
\ ^* ^ 4 *
Solve by addition or subtraction :
6.
[ A+B=-9, {10*-3V— .6,
\7A-3B = 7. \ 7x+42/ = 8.
Solve by either method :
x 4 y
3 x — y x-\-y
11.
9 Q J
2 15'
2~ 3 '
3z+i/ 3o:-2/_ 2
3
11 3
[HINT. First clear of fractions.]
12.
2H-6, s_8
»i. r '
M-n,
x y
10 12 4
12 L
— 4:.
HINT TO Ex. 16. Do not clear of fractions, but solve for - and --~|
2x-5?/=-l,
-+— = 5,
13.
x-2/ 3x+2y
7 23
15_30=_1
« V
14.
6x-8i/ 8x-20i/ ft
1 1 1 5
0 11 '
. 18'
x+V=|-
i r . ^ £>>
x-1 2/+1
2 , 3 _10
X—l y+l
54 SECOND COURSE IN ALGEBRA [VII, § 35
35. Simultaneous Equations in Three Unknown Letters.
We often meet with a system of three linear equations between
three unknown letters. Such a system, like those already
considered (§§ 33, 34), may be solved by elimination.
EXAMPLE. Solve the system
(1) x+y+z = 6,
(2) 2z-2/+3z = 9,
(3) x+2y-z = 2.
SOLUTION. Eliminate one of the unknowns, say y, from (1)
and (2). Thus
(4) 3 x +42 = 15. [(l)+(2)]
Eliminate the same unknown, y, from (2) and (3). Thus
(5) 4 x -2 y +6 z = 18. (2)X2
(6) x+2y- 2 = 2. (3)
5x +52=20, (5) + (6)
or
(7) x +2 = 4.
Equations (4) and (7) contain only x and z and hence may be
solved for these letters, as in §§ 33, 34. Thus
(8) 3 x +4 z = 15. (4)
(9) 3 x +3 z = 12. (7) X3
0=3. (8) -(9)
Substituting z =3 in (7), we obtain z+3 =4. Therefore x = 1.
Substituting 2=3 and z = 1 in (1), we find l+y+3 =6. There-
fore y=2.
The desired solution is, therefore, (x = l, y =2, 2=3). Ans.
To solve three simultaneous equations :
1. Eliminate one of the unknown letters from any pair of
the equations, then eliminate the same unknown from any other
pair of the equations.
2. Solve the two equations thus obtained, as in § 34.
3. This gives two of the letters, and the third may then be
found by substituting the letters already found in either of the
given equations.
- > a
VII, §35]
H
ELIMINATION
\\VA-- tA\-u
^vr- to
55
EXERCISES
Solve for x, y, and z each of the following sets of equations.
+3^ = 14, \x+y-z = Q,
1. 2z+?/+22 = 10, 4. z-?/ = 4,
-3^ = 2. z+z = 7.
2.
3.
5.
M+i-2,
x y z
2-i+i = 7,
x ?/ z
[HINT. See hint to Exer-
cise 16, p. 53.]
APPLIED PROBLEMS
1. The sum of two numbers is 75 and their difference is 5.
Find the numbers.
[HINT. Let x be one of the numbers and y the other, and form
two equations.]
2. One third of the sum of two numbers is 10, while one
sixth of their difference is 1. Find the numbers.
3. The perimeter of a certain rectangle is 10 inches less
than 3 times the base. If the base is 4|- times the height,
what are the base and height?
4. Each base angle of a certain isosceles triangle is 66°
more than the vertical angle. Find each angle.
5. A father's age is 1^- that of his son. Twenty years ago
his age was twice his son's. How old is each?
6. Four years ago A's age was 2^ B's age. Four years
hence A's age will be 1T8T B's age. What is the age of
each?
56 SECOND COURSE IN ALGEBRA [VII, § 35
7. A part of $2500 is invested at 6% and the remainder
at 5%. The yearly income from both is $141. Find the
amount in each investment.
8. One sum of money is invested at 5% and another at
6%. The total yearly income from both investments is
$53.75. If the rates should be reversed, the annual income
would be increased by $2.50. Find the sums of money in-
vested.
9. A and B together can do a piece of work in 12 days.
After A has worked alone for 5 days, B finishes the work in
26 days. In what time could each do the work alone?
[HINT. Let x = the time in which A can do it alone, y = the time
in which B can do it alone. Then the part A can do in one day is -,
and the part B can do in one day is -. So the equations become
i +1 = J_ and 5 +?§ = l. Now solve as in Ex. 16, p. 53.]
x y 12 x y
10. A and B can do a certain piece of work in 16 days.
They work together for 4 days, when B is left alone and
completes the work in 36 days. In what time could each
do it separately?
11. A laborer agreed to stay on a farm for 100 days.
For each day he worked he was to receive $2 and board,
but for each idle day he was to forfeit 75 cents for his board.
When the time expired, he received $180.75. How many
days did he work?
12. An errand boy went to the bank to deposit some bills,
some of them being $1 bills and the rest $2 bills. If there
were 38 bills in all and their combined value was $50, how
many of each were there?
[HINT. Let rr=the number of $1 bills, and y =the number of
$2 bills. Then their combined value was x +2 y dollars.]
VII, § 35] ELIMINATION 57
13. The receipts from the sale of 300 tickets for a musical
recital were $125. Adults were charged 50 cents each, and
children 25 cents each. How many tickets of each kind were
sold?
14. A grocer wishes to make 50 pounds of coffee worth
32 t;ents per pound by mixing two other grades, one of which
is worth 26 cents per pound and the other 35 cents per pound.
How much of each must he use ?
15. One cask contains 18 gallons of vinegar and 12 gallons
of water; another, 4 gallons of vinegar and 12 of water.
How many gallons must be taken from each so that when
mixed there may be 21 gallons, half vinegar and half water?
16. Two cities are 140 miles apart. To travel the distance
between them by automobile takes 3 hours less time than by
bicycle, but if the bicycle has a start of 42 miles, each takes
the same time. What is the rate of the automobile, and
what the rate of the bicycle ?
17. A boy rows 18 miles down a river and back in 12 hours.
He can row 3 miles downstream while he rows but 1 mile
upstream. What is his rate in still water, and what is the
rate of the stream ?
18. A motor boat can run r miles an hour in still water.
If it went downstream for s hours and took t hours to return,
what was the total distance traveled, and what was the rate
of the current?
19. The sum of three numbers is 20. The sum of the first
and second is 10 greater than the third, while the difference
between the second and third is 6 less than the first. Find
the numbers.
[HINT. Use the three letters x, y, z to represent the unknown
numbers, and form three equations. Solve as in § 35.]
58 SECOND COURSE IN ALGEBRA [VII, § 35
20. A, B, and C have certain sums of money. B would
have the same as A if A gave him $100 ; C would have four
times as much as B if B gave him $100 ; and C would have
twice as much as A if A gave him $100. How much has
each?
21. I have $90 on deposit in bank A, $51 in bank B, and
$75 in bank C. If I have $144 more to deposit, how shall I
distribute it among the three banks so as to make the three
deposits equal?
22. The perimeter of a certain rectangle is 16 feet. If
the length be increased by 3 feet and the breadth by 2 feet,
the area becomes increased by 25 square feet. What are
the length and breadth?
23. A barrel of vinegar is to be bottled for selling and it is
desired that some of the bottles be of pint size, others of
quart size and ojthers of gallon size. In order that there be
52 bottles in all, and twice as many of the pint as of the quart
size, how many of each will be necessary?
[HINT. 1 barrel =32 gallons.]
24. For any pulley block, the relation between the weight
to be raised and the pull necessary to raise it is
pull = x+yX weight,
where x and y are numbers that are different for different
pulleys.
In two experiments with a certain pulley block, a weight
of 100 pounds was raised by a pull of 22 pounds, and a weight
of 200 pounds was raised by a pull of 42 pounds. Find the
values of x and y for this pulley.
CHAPTER VIII
SQUARE ROOT
36. Definitions. The square root of a given number is
the number whose square equals that number.
Thus 2 is the square root of 4 because 22 =4. Likewise, 3 is the
square root of 9 because 32 =9, etc.
The square root of a number is denoted by the radical sign
V placed over it.
Thus V4=2, V9=3, Vl6=4, etc.
The process of finding the square root of a number is called
extracting its square root.
37. Extracting Square Roots in Arithmetic. Many times
we can pick out the square root of a number by inspection.
Thus, A/Ill is seen to be 12 because 122 = 144. Similarly,
Vl96 = 14. But in finding the square root of a large number,
such as 74,529, we cannot ordinarily determine the answer
by mere inspection. The process for such a case is illustrated
below, and is explained on the next page.
PROCESS.
7'45'29 | 273 Ans.
4
Trial divisor =2x20 =40
Complete divisor = 40 +7 =47
345
329
Trial divisor = 2 X270 = 540
Complete divisor = 540+ 3 = 543
59
1629
1629
60 SECOND COURSE IN ALGEBRA [VIII, § 37
EXPLANATION. First separate the number into periods of two
figures each, beginning at the right. That is, in the present case,
write the number in the form 7 '45 '29.
Find the greatest square in the left-hand period and write its
root for the first figure of the required root. This gives the 2 ap-
pearing in the answer.
Square this figure (giving 4), subtract the result from the left-
hand period and annex to the remainder the next period for new
dividend. This gives the 345 appearing in the process.
Double the root already found, with a 0 annexed (giving 40) for
a trial divisor and divide the last dividend (345) by it. The quo-
tient (or, in some cases, the quotient diminished) forms the second
figure, 7, of the required root. Add to the trial divisor the figure
last found (7), giving the complete divisor (47). Multiply this
complete divisor by the figure of the root last found (7), giving the
329 appearing in the process. Subtract this from the dividend, and
to the remainder annex the next period for the next dividend. This
gives the 1629 of the process.
Proceed as before, and continue until a new dividend equal to 0
is obtained. In the example above, this happens at once, giving
273 as the required root.
This process is the one commonly used in arithmetic, and
is stated here as a review. We shall see in § 38 that a sim-
ilar process may be used in extracting the square roots of
expressions in algebra.
In the example just solved, the root comes out exact be-
cause 74,529 (whose root is being extracted) is a perfect
square — that is, it is like one of the numbers 1, 4, 9, 16, 36,
etc. If we had started with a number which was not a per-
fect square, the process would be the same except that we
should not finally reach a new dividend which equals 0.
In such cases, in fact, the process continues indefinitely, but
if we stop it at any point, we have before us the desired root
correct (decimally) up to that point. For example, in find-
ing the square root of 550 correct to two decimal places, the
process is as follows.
VIII, § 37] SQUARE ROOT 61
PROCESS.
5'50.00'00 | 23.45 Ans. (correct to two
4 decimal places)
2X20 = 40
40+3 = 43
150
129
2X230 = 460
460+4 = 464
2100
1856
2X2340 = 4680
4680+5 = 4685
24400
23425
975
NOTE. In the process above we have first written 550 in the form
550.0000. If we had written it with six zeros, that is 550.000000,
and then carried the process forward until all these were used be-
low, we should have obtained the root correct to three decimal
places instead of two. In general, the root obtained would be
correct to a number of decimal places equal to half the number of
zeros added.
Square roots of decimal numbers, such as 334.796, are ob-
tained like those for whole numbers, except that in the begin-
ning the separation of the number into periods of two figures
each must be carried out both ways from the decimal point.
Thus 334.796 would be written 3 '34.79 '60. Similarly, 3.67893
would be written 3 '.67 '89 '30. The extraction of the root is then
carried out as in the process shown above.
EXERCISES
Find (by inspection or by the process shown in § 37) the
square root of each of the following numbers.
1. 49. 5. 576. 9. 8281. 13. f
= |Xf.]
2. 81.
6. 1444.
10. 15,876.
[HINT.
3. 64.
7. 4225.
11. 42,025.
14- M
4. 169.
8. 1681.
12. 95,481.
15. &
62 SECOND COURSE IN ALGEBRA [VIII, § 37
Find the square root of each of the following numbers
correct to two decimal places.
16. 567. 19. 17.76. 22. 3.
17. 633. 20. 13. 23. f.
[HINT. Write as [HINT. Write f as
13'.00'00.] .75.]
18. 1305. 21. 2. 24. .
38. Extracting Square Roots in Algebra.
(a) Monomials. The square root of a monomial can
usually be seen by inspection.
36m%2=6 mzn (because (6 m2n)2 =36 m4n2). Similarly,
(b) Trinomials. If a trinomial is a perfect square, its
square root can be obtained by inspection.
Thus suppose we wish to find the square root of 9 z2 + 12 xy +4 y*.
This trinomial is a perfect square because its terms 9 z2 and 4 y*
are squares and positive, while its remaining term, 12 xy, is equal to
2- A/9^2- V41/2. (See §11 (d), p. 20.) Hence the trinomial
can be expressed in the form (3 x +2 yY, whence the desired square
root of 9 x2 + 12 xy +4 yz is 3 x +2 y.
Similarly, V4s2-4s+l =2s-l because 4s2-4s+l is a perfect
trinomial square, and as such is factorable into
(2s-l)(2s-l) or (2s-l)2.
(c) Polynomials. To find the square root of a polynomial
of more than three terms we may follow a process much like
that employed for finding square roots in arithmetic. This
is illustrated in the following example.
VIII, § 38] SQUARE ROOT 63
EXAMPLE. Find the square root of 4x4-f-12x3-3x2-18x+9.
PROCESS.
4x4+12x3-3x2-18x+9| 2x2+3x-3
4 x4 Ans.
Trial divisor, 4 x2
Complete divisor, 4x2+3 x
12x3-3x2
12 x*+9 x2
Trial divisor, 4 x2-j-6 x
Complete divisor, 4x2+6 x — 3
-12x2-18x+9
-12x2-18x+9
EXPLANATION. Arrange the terms of the polynomial in the
descending (or ascending) powers of some letter. In the example,
the arrangement is in descending powers of x.
Extract the square root of the first term, write the result as the
first term of the root (giving the 2 x2 in the answer), and subtract
its square from the given polynomial (giving the 12 x3 —3 x2 in the
second line of the process).
Divide the first term of the remainder by twice the root already
found, used as a trial divisor. The quotient (3 x} is the next term
of the desired root. Write this term in the root, and annex it to the
trial divisor to form the complete divisor (the 4 x2 +3 x of the
process).
Multiply the complete divisor by this term of the root, and subtract
the product from the first remainder (giving the — 12 x2 — 18 x +9
of the process).
Find the next term of the root by dividing the first term of the
remainder by the first term of the new trial divisor. This gives the
—3 of the answer.
Form the second complete divisor and continue in the manner
above indicated until a remainder of 0 is obtained.
In the example just considered, only one letter, as x,
appears. A similar process may be employed, however, in
all cases by first arranging the expression in descending (or
ascending) powers of some one of the letters.
For example, 4 x4 +9 y* — 12 x^y3 + 16 x2 + 16 - 24 yz, when arranged
in descending powers of x, becomes
64 SECOND COURSE IN ALGEBRA [VIII, § 38
EXERCISES
Find (by inspection or by the process shown in § 38) the
square root of each of the following expressions. Check
each answer by squaring it to see if the result thus obtained
is the given expression.
1. 4 xY- 4. 81 a866c10. 7. 196 ploq12. 10. a?x2m.
2. 9a664. 5. 225 ra8n4. 8. m2nVr12. 11. 9 ra2r*4p.
3. 25z22/4-36. 6. 625 rW. 9. 529 r10s14. 12. m2*^.
13. z2+2z+l. 18. 9ra2-6raz-fz2.
14. z2-4z+4. 19. x2+xy+± y2.
15. 4m2+12mn+9n2. 20. 9 z2+66 z+121.
16. 4x2+4xy+y2. 21.
17. c2-4ac+4a2. 22.
23. 4
24. z6
25. a;4
[HINT. See remark at the close of § 38.]
26. z8+2a6z2-a4z4-2a2z6+a8.
[HINT. First arrange in descending powers of x.]
27. 9
28. 9
29. z8+4 x7-3 x4-20 x5-2 x6+4+4 x2- 16 x+32 x*.
39. The Double Sign of the Square Root. We know that
3 is the square root of 9 because 32 = 9. But we also have
( — 3)2 = 9. Therefore, —3 can also be regarded as a square
root of 9. In other words, 9 has two square roots, +3 and
— 3, which are opposite in sign but otherwise the same.
Similarly, 16 has the two square roots +4 and —4, and in
general, a2 has the two roots a and — a.
The double sign ± is sometimes used. Thus we say that the
square root of 9 is ±3. This is merely a brief way of saying that the
two roots are +3 and —3.
VIII, §40] SQUARE ROOT 65
In order to avoid all confusion, it is to be understood here-
after that the radical sign V~ when placed over a number
means the positive square root of that number. If it is de-
sired to indicate the negative square root, it is done by the
symbol — V .
Thus Vl6 means +4, while - Vl6 means -4. Similarly, Va
means +Va.
40. Equations Containing Radical Signs. Equations con-
taining radical signs may often be solved by squaring each
member. This is equivalent to multiplying each member by
the same amount, and hence is justified by § 27.
EXAMPLE 1. Solve the equation Vx — 2 = 6.
SOLUTION. Squaring both members gives
z-2=36,
whence x=38. Ans.
CHECK. V38-2 = V36~=6.
EXAMPLE 2. Solve the equation Vz— 1 — VE— 4 = 1.
SOLUTION. Transpose the Vx -4 to the right ; this gives
Square both members, using Formula VI, § 10 for finding
(1 + Vc-4)2. This gives
or
Canceling x from both sides and transposing the 1 and —4 to
the left, gives _
2 =2 Vc-4, or Vz-4 = l,
whence (squaring again)
z-4 = !2 = l.
Therefore x=5. Ans.
CHECK. V5^1 - V5^4 = V4- VI =2-1 =1.
66 SECOND COURSE IN ALGEBRA [VIII, § 40
NOTE. It is especially important to check all the answers ob-
tained for equations containing radical signs, since the process of
squaring both members sometimes leads to a new equation whose
roots do not all belong to the first one. Thus, if we square both
members of the equation x =5, we get x2 =25 and this last equation
has —5 as a root as well as 5.
EXERCISES
Solve each of the following equations and check each
answer.
3.
4. Vx+7-Vx =
[HINT. First transpose one of the radicals to the right side,
as in Example 2, § 40.]
9.
10. If 16 be added to 4 times a certain number, the square
root of the result is 6. What is the number?
11. If 9 be added to the square of a certain number, the
square root of the result is 5. What is the number?
12. The difference between the square root of a certain
number and the square root of 11 less than that number is 1.
Find the number.
13. Solve each of the following equations.
(a)
(6)
CHAPTER IX
RADICALS
41. Radicals. Suppose we have a square which we know
contains exactly 2 square feet. How long is each of its four
sides? In order to answer this, we naturally let x represent
the desired length. Then we must have
x -x = or
2 sq.ft.
Therefore x = A/2 ft. Ans.
This number A/2 cannot be determined
exactly because it is the square root of a pIG 10
number, 2, which is not a perfect square.
However, A/2 measures a perfectly definite length, as indi-
cated in the figure. Its value, correct to two decimal places
only, is 1.41.
Such a number as A/2 is called a quadratic radical. This
name is used in general to denote the indicated square root
of a number.
Thus V3, V7, Vzi, VI06, V213 are all radicals.
The word radical is also used in connection with other roots
than square roots. Thus VlO means the cube root of 10,
that is the number whose cube is 10. Similarly, v^6 means
the fourth root of 6, etc. All such numbers represent per-
fectly definite magnitudes, as did \/2 in the figure above,
even though we cannot express them exactly in decimal form.
67
68 SECOND COURSE IN ALGEBRA [IX, § 41
In general, the nth root of any number a is written A/a, and
this is known as a radical of the nth order. The number n is
here called the index of the root, and the number a is called
the radicand.
NOTE. When no index is expressed, the index 2 is to be under-
stood. Thus V3 means \/3.
The same definitions apply also to algebraic expressions.
Thus V3xy* and Vat+b2 are radicals.
42. Rational and Irrational Numbers. Surds. The posi-
tive and negative integers and fractions, and zero, are called
rational numbers. If an indicated root of a number cannot
be extracted exactly, that is, cannot be expressed exactly as
one of these rational numbers, it is called a sure?. Any posi-
tive or negative number that is not rational is called irrational
Thus_ V3, ^10, V6 are surds ; but V§, v^S, Vj^ are rational,
since Vg=3, ^8=2,
EXERCISES
Determine which of the following radicals are surds ; and
state the index and the radicand of each.
1. V7. 2. V8. 3. Vl6". 4. V^T 5. Vff . 6. V75.
7. V^S.
[HINT. -8 = (-2)3.]
8. Vis. 10. Vs. 12. V20. 14. Vs^y.
9. V27. 11. Vl6. 13. V32. 15. Vg m6(a+6)9.
43. Value of Radicals. Use of Table. To determine the
value of a radical correct to two or more decimal places
usually calls for a rather long process. (See § 37, p. 61.)
In order to save time and labor, the values of those radicals
which are needed most in ordinary life (the square and cube
IX, §43] RADICALS 69
roots) have been printed in a table and placed for convenient
reference at the end of this book. For the sake of complete-
ness, the second and third powers of numbers are also printed
in the table. Just how to use this table is described on page
275, which the pupil should now read carefully. Below are
a few illustrative examples.
EXAMPLE 1. Find \/7, using the table.
SOLUTION. The top number in the third column on page 290
(table) gives V? =2.64575. This value is correct up to the last
decimal figure given, that is to the fifth place. Thus the answer
may be written "^ = 2.64575+, the sign + indicating that this value
for V7 is correct up to the last decimal place stated.
EXAMPLE 2. Find vT, using the table.
SOLUTION. The top number in the sixth column on page 290
(table) gives ^7 = 1.91293+. Ans.
EXAMPLE 3. Find A/70.
SOLUTION. The top number in the fourth column on page 290
(table) gives V70 =8.36660+. Ans.
EXAMPLE 4. Find v/70.
SOLUTION. V70 = 4.12129+, from the seventh column, page 290.
EXAMPLE 5. Find ^700.
SOLUTION. v'TOO =8.87904+, from the eighth column, page 290.
EXERCISES
By means of the table, determine the approximate values
of the following radicals.
1. V6. 2. V60. 3. ^6. 4. v"60. 5. ^500.
6. VeT.
[HINT. 61=6.10X10. So use the information given for 6.10
in the table.]
70 SECOND COURSE IN ALGEBRA
7. S/6L 8. V92". 9. -v/920.
12. \/78^2 13. A/782.
SOLUTION. 782 = 7.82 X 100. Therefore V782 is the same as
V7.82 except that the decimal point in the root must be moved
one place farther to the right. (See p. 275.) Now, V7.82 = 2.79643
(table), so V782 =27.9643. Ans.
14. A/561.
[HINT. See Solution of Ex. 13.]
15. A/779. 16. A/895".
17. ^6120.
[HINT. 6120 =6.12 XlOOO. (See p. 276.)]
18.
19.
[HINT. .67 =^ X6.7 (Now see p. 276.)]
20. V.0676
APPLIED PROBLEMS
Use the tables in working the following problems.
1. If the sides of a right triangle are 3 inches and 2 inches
long, respectively, what is the length of the hypotenuse?
[HINT. If x be the hypotenuse, then x2 =32+22 = 13.]
2. A baseball diamond is a square 90 feet on a side. How
far is it from home plate to second base ?
3. If the diagonal of a square is 13 feet long, how long is
each side?
4. The dimensions of a certain rectangular field are 103
feet by 337 feet. In going from one corner to the opposite
corner, how much shorter is it to go by the diagonal than to
go around?
IX, §441 RADICALS 71
5. How long must the radius of a circle be in order
that the area be 12 square inches? (See Ex. 14 (6), p. 6.
6. What is the length of the edge of a cube if the volume
is 357 cubic inches?
7. If the volume of a sphere is 440 cubic inches, how long
is its radius ? (See Ex. 14 (e), p. 6.)
8. In the accompanying figure how
long should the radius of the inner semi-
circle be in order that the area inclosed ^ g • - Q D
may be 132 square feet? AB = CD= 2 feet
FIG. 11.
9. The area A of a triangle in terms
of its three sides, a, b, and c, is A = Vs (s — a) (s — b) (s — c) ,
where s = — The sides of a triangle are respectively
2i
6 inches, 7 inches, and 9 inches long. What is the area ?
10. Two circular cones have altitudes, h, which are the
same, but their bases have different radii. What is the
ratio of the longer radius to the shorter if the volume of the
one cone is three times that of the other?
[HINT. The formula for the volume of a cone is V = £irr2/&, where
h represents the altitude and r is the radius of the base.]
44. Simplification of Radicals. We know that the square
root of the product of two numbers is the same as the product
of their square roots. For example, V4x25 is the same as
A/4 X V25, because both are equal to 10 (the first being VIM,
or 10, and the second being 2 X5, or 10). In the same way,
v/8x3 = v/8XV/3, or simply 2^3. In fact, we have the
following general formula.
72 SECOND COURSE IN ALGEBRA [IX, §44
Formula I. Vafc = Va • V&.
Again, Vf is the same as — - because both are equal to f .
V9
(Explain.) Similarly, v^f may be written -— , or — — So
V 8 2
in general we have the following formula.
Formula II. V1/?
Formulas I and II enable us to simplify many radical ex-
pressions, as is illustrated in the following examples.
EXAMPLE 1. Simplify V63.
SOLUTION. Using Formula I, we have
V63 = V9X7 = V9 X V? =3 V?. Arcs.
EXAMPLE 2. Simplify \/32.
SOLUTION. ^32=4/8X4=^/8X^4=2^4. Ans.
EXAMPLE 3. Simplify
/8 Vg V4x2 Vix^/2 2\/2
SOLUTION. \~ = — — : = ^ = — — - — — = — — •
^27 V27 \/9x3 .V9XV5 3V3
EXAMPLE 4. Simplify V20 a6".
SOLUTION. V20 a6 = V4a6x5 = V4^ x V5 =2 a3 Vs. Ans.
EXAMPLE 5. Sunplify v/
SOLUTION.
) X (9 x2) = v'S
* Z* ^6 22 22
NOTE. It will be observed that in each of the examples above
the process of simplification consists in removing from under the
IX, § 44] RADICALS 73
radical sign the largest factor of the radicand which is a perfect
square, or perfect cube, as the case may be. Thus, in Example 1,
the radicand, 63, was first broken up into factors in such a way that
9 (which is a perfect square) appears clearly to the eye. Similarly,
in Example 2 (where we are dealing with a cube root) we first write
the radicand, 32, in a form which brings out conspicuously its
factor 8, which is a perfect cube. The first step in all such examples
is, therefore, to get the radicand properly broken up into factors.
This requires good judgment, but becomes very easy after slight
practice and experience.
EXERCISES
1. By Formula I, p. 72, we have A/20 = A/4x5 = A/4 X A/5
= 2A/5. Look up the values of A/20 and A/5 in the table
and thus prove that A/20 is the same as 2 A/5.
2. Show that Formula I, p. 72, gives A/54 = 3A/(T and
verify the correctness of this result by use of the table, as in
Ex. 1.
Simplify each of the following radicals. (See Note in § 44.)
3. A/18. 6. A/125. 9. v/54.
4. A/24. 7. A/108. 10. A^Sl.
6. A/Tl2i 8. A^32.
11. V||.
[HINT. First use Formula II, § 44.]
12- ^5S- 19. A^27 x*y*z\
20. \/16 M4.
o3/
6 a?bVab. Ans.
4(a2-62)
74 SECOND COURSE IN ALGEBRA [IX, § 44
Write each of the following in a form having no coefficient
outside the radical sign.
23. 2>/3.
SOLUTION. By Formula I, § 44, 2^3 = Vix \/3 = Vl2. Ans.
24. 2V2". 25. 5V5^ 26. 2v^
27. W2.
A/9 A/9
[HINT. Write W2 =-rr =-^, and apply Formula II, § 44.]
3 Vg
33. 2gVx-y. 38. fV3.
. Ans. 34. ra^Xn. 39. f^S.
35. 2aV«2-62. 40. fVf.
36. ^L- 41. (a-6)V2c".
42. o-
32. SmnVmn. 37. r\'--
45. Addition and Subtraction of Similar Radicals. When-
ever two radicals having the same index have also the same
radicand (or can be given the same radicand by simplifica-
tion) they are called similar radicals.
Thus 2 V2 and 3 V2 are similar radicals ; so also are V2 and V32
since the last of these may be simplified into 4V2, by § 44. Like-
wise, V3 a2x and V3 b2x are similar, being equal, respectively, to
a VWx and 6 V3 x, thus coming to have the same radicand.
Whenever similar radicals are added or subtracted, the
result may be expressed in a single term.
Thus 4V3+5V3 = (4+5)\/3=9\/3. Ans. _
Again, 3V32-2V8=3X4V2-2X2V2~=12V2-4V2
= (12-4)V2=8V2. Ans.
Likewise,
2 V4~cM> + V9 azb - Vl6 a*b =2 • 2 avT+3 a*N/6_-4 aV6
= (4 a+3 a-4 a)V6=3 aVF. Ans.
IX, §46]
RADICALS
75
NOTE. The pupil is especially warned that in general we cannot
write Va+6 = Va + Vb. Thus when a =4, 6=9 this would give
Vl3=2+3=5, which is clearly false. Similarly, we cannot write
V~a~—b = Va-Vb. Thus V25^9=Vl6=4, but V25-V9=5-3 = 2.
EXERCISES
Combine the following radicals.
1. V8+V18+V32. Check your answer by use of
the table; that is show that \/8+ VT8+V32, as computed
by the table, has the same value as your answer, similarly
computed.
2. V108+V27-V75.
3. v'm-h^lG-v^.
4. V72+V32-V50".
Check your answer as in Ex.1.
Check as in Exs. 1 and 2.
7. 32 a2-
8.
9.
46. Multiplication of Radicals. We may multiply one
radical, as Va, by another of the same index, as \/b, by For-
mula I of § 44. If n — 2, this gives as an important special
case
Va • Vb =
76 SECOND COURSE IN ALGEBRA [IX, §46
EXAMPLE 1. Find the product of V2^and Vl8.
SOLUTION. V% > Vl8 = V2 • 18 = V36 =6. Ans.
EXAMPLE 2. Multiply V3+A/5 by 2 A/3- \/5.
SOLUTION. V3+ V5
2V3- V5
2- 3+2VI5
- VI5-5
6+ 5-5 = l + Vl5. Ans.
EXAMPLE 3 . Multiply Va + Va — b by Va — Va — b.
SOLUTION. Va + Va — b
Va- Va-b
a + Va2—ab
— Va2— ab — (o-b)
a — (a — 6) =a— a+b = b. Ans.
EXERCISES
Find the following products, simplifying results as far as
possible.
1. V3 - \/27. 7. V7 . \/I
2. V8 - Vl2. 8. VI - VTOl.
3. V6-V4. 9. (V3-V^)(\/3+>/2).
4. \/7 • V9. 10. (2v/3-V2)(2\/3+v/2).
5. VI6 - V3 - V2. 11. (V6- V3)2.
6. \/f . V|. 12.
13. (3\/3+2V5)(V3-3\/5).
14. (V2+V3 + V5)(V2-V3).
15. Va-.Vtf. 18.
16. Vo6 • V^3. 19. (Va-\/6)2.
17. Vtf-V- 20.
IX, §47
RADICALS
77
21.
22. Find the value ofz2-4z-lifa: = 2+ \/5.
23. Find the value of z2+3z-2if
24. Does A/3+2 satisfy the equation x? — 4 z+1 = 0 ; that
is, is the equation true when x = A/3+2 ? Answer the same
question when x = A/3 + A/2.
47. Division of Radicals. We may divide one radical, as
Va, by another of the same index, as V6, by the Formula II
of § 44. If n = 2, this gives as an important special case
*-4
EXERCISES
Express each of the following quotients as a fraction under
one radical sign, and reduce your answer to simplest form.
A/15
Ans.
10. V81+9 a2+a4^- Va2-3 a+9.
CHAPTER X
QUADRATIC EQUATIONS
48. Quadratic Equation. An equation which contains
the unknown letter to the second (but no higher) power is
called a quadratic equation, or briefly, a quadratic.
Thus the equations 2 x2 — 4 x = I and ^ x2 +x — — 3 are quadratics,
but 2 x -3 =0 and 4 x3 -5 x2+x =2 are not.
49. Pure Quadratic. When the quadratic contains the
second power only of the unknown letter, it is called a pure
quadratic.
Thus 2 x2 —27 =0 and ax"2 = be are pure quadratics, but x2 — 4 x =2
and x2+bx+c=Q are not.
50. Affected Quadratic. When the quadratic contains
both the first and second powers of the unknown letter, it is
called an affected quadratic.
Thus z2+3z=7 and xz-\-2ax=az are affected quadratics, but
2 x2 -7 =0 and 5 x2 -16 o262 =c2 are not.
51. Solution of Pure Quadratics. The following example
will suffice to show how the solution of any pure quadratic
may be obtained.
EXAMPLE . Solve 2 x2 - 30 = 0.
SOLUTION. Transposing and dividing through by 2 gives^2 = 15.
Taking the square root of both members gives x = = Vl5. Ans.
To get the approximate value of Vl5, we may consult the table,
where we find Vl5 =3.87298+.
The answer may, therefore, be written in the form x = ±3.87298+.
CHECK. 2 (Vl5)2- 30 =2X15 -30 =30 -30=0, as required.
2( - Vl5)2 -30 =2 X 15 -30 =30 -30 =0, as required.
78
X, § 51] QUADRATIC EQUATIONS 79
NOTE. Strictly speaking, when we extract the square root of
both members of the equation x2 = 15 we get ±3 = ± VlS. But to
say that — x = =«= Vl5 means the same as +x = =±= Vl5, so it suffices
to write simply x = =±= Vl5 to cover all cases.
An examination of the example above shows that we have
the following rule.
To solve a pure quadratic, solve for x2, then take the square
root of the result. There will be two solutions, the one being
the negative of the other.
EXERCISES
Solve each of the following equations, checking your
answer for the first five. If you meet with a radical, find its
approximate value by use of the table.
1. z2-81 = 0. 11 3 _1 j • 1
2. 3*2-192 = 0. ' *+5 2*~5'
12 JL ! _Z_L 1
' '
3. 4z2+8 = 10*2-16.
4. 3x2-15 = 0. x+3 ,2x-l=Q
~
5. 3,2-!6 = 0.
„ _0
6. 3z2-17 = 0. -' —-
? ^_^ =19
x-2 x+3
[HINT. See § 40.]
17.
10. (o;+l)2-2(x+l)=4. 18. V25-6 x+ V25+6 a; = 8.
80 SECOND COURSE IN ALGEBRA [X, § 51
APPLIED PROBLEMS
1. What numbers are equal to their own reciprocals?
2. One side of a right triangle measures 3 inches and the
hypotenuse measures 7 inches. Find (approximately) the
length of the other side.
[HINT. Work by algebra, making use of the principle that in
any right triangle the square of the hypotenuse equals the sum
of the squares of the two sides.]
3. What is the length of the longest umbrella than can
be placed in the bottom of a trunk the inside of which is 33
inches long by 21 inches wide?
4. A certain square has a side which is three times as long
as the side of another square. If the difference of their areas
is 72 square feet, how long is the side of each ?
5. Find the mean proportional between 25 and 9; also
that between 17 and 21. In what particular is the latter one
essentially different from the first one ?
6. It is proved in geometry that whenever a perpen-
dicular is drawn from a point on a semicircle to the base,
as PQ in Fig. 12, its length is a mean Q
proportional between the segments AP
and PC of the base ; that is,
AP = PQ
PQ PC
IfAP = 8 inches and PC = 10 inches,
how long is PQ ?
7. Determine the formula for
(a) The side of the square whose area is a.
(b) The radius of the circle whose area is a.
X, § 52] QUADRATIC EQUATIONS 81
(c) The radius of the sphere whose area is a.
[HINT. See Ex. 14, p. 6.]
(d) The diameter of the base of a circular cone whose
volume is v and whose altitude is h.
(HINT. The volume of a circular cone is equal to the area of its
base multiplied by -^ its altitude, i.e., V = ^Trr2h.]
8. The distance s, measured in feet, through which an
object falls in t seconds when dropped vertically downward is
given by the formula s = %gt*, where 0 = 32 (approximately).
Hence, determine (approximately) how long it will take a
stone to drop to the bottom of a mine 300 feet deep.
9. One of the sides of a certain triangle is m units long.
What is the formula for the corresponding side of a similar
triangle whose area is n times as great?
[HINT. It is shown in geometry that if two geometric figures are
similar ; that is, have the same shape but are of different sizes, then
the square of any line in the first figure is to the square of the cor-
responding line in the second figure as the area of the first figure is
to the area of the second.]
10. A map of the United States is uniformly enlarged in
such a way as to cover twice as much area on the paper as
before. By what factor should the scale of the map be now
multiplied?
11. Find three consecutive integers such that the square
of the second plus the product of the other two equals 31.
[HINT. Let x be the second integer. Then the first and last
integers will be x — 1 and x + 1 respectively.]
52. Solution of Affected Quadratics by Factoring. It is a
familiar principle of arithmetic that the product of two
numbers is zero if either of the numbers is zero ; that is, if
either factor is zero.
For example 2X0=0, 0X4=0, (-3) X0=0, etc.
G
82 SECOND COURSE IN ALGEBRA [X, § 52
This principle is frequently used to solve affected quadratic
equations.
EXAMPLE. Solve by factoring the equation #2+8 x = 48.
SOLUTION. Transposing all terms to the left, we have
Factoring,
z2+8z-48 = (z-4)(z + 12). [§ 11 (c)]
Thus the given equation becomes
This equation will be satisfied, according to the principle men-
tioned above, whenever the factor x— 4 equals zero or the factor
z + 12 equals zero, that is in case x— 4=0 or a; + 12=0. Solving
these two simple equations gives z=4 and x= — 12, which must
therefore be the desired solutions.
CHECK. When x = 4 the left side of the original equation becomes
42+8x4, which reduces to 16+32 =48, as the equation demands.
When x = — 12 we have in like manner
(-12)2+8x(-12) = 144-96=48.
We thus have the following rule.
To solve quadratics by factoring :
1. Transpose all terms to the left so as to have 0 on the right.
2. Factor the left member of the resulting equation.
3. Place each factor equal to 0 and solve the resulting simple
equations. The two results are the solutions required.
EXERCISES
Solve each of the following equations by factoring, checking
your answer in the first five.
1. z2-7z+10 = 0. 3. z2+8z=-15.
2. z2-5z=-6. 4. z2+7z-30 = 0.
X, §53] QUADRATIC EQUATIONS 83
5. z2
6. l-
7.
[HINT. Write as
8. x2-l = 3
5 then apply the principle in § 52.]
~z+4 = 13. 3(z+l)(z-3)+4(z-3)=0.
10. j£- = __ i __ 5. 14- (z+l)2+3(z+l)+2 = 0.
x — 2 x — 2 [HINT. Solve first for x+L]
53. Solution of Any Quadratic by Completing the Square.
We often meet with a quadratic, such as xz-\-7 x — 5 = 0,
which we cannot solve as in § 52 by factoring. The difficulty
here is that we cannot factor readily x2+7 x — 5. However,
this quadratic and all others (whether solvable by factoring
or not) can be solved by a certain process known as completing
the square. How this is done will be best understood from a
careful study of the following examples.
EXAMPLE 1. Solve z2+ 6 x = 16.
SOLUTION. The first member of this equation, or x2 +6 x, would
become' a trinomial square [§ ll(d)] if 9 were added to it. Our
first step, therefore, is to add 9 to both members of the given equa-
tion, thus "completing the square" in the first member and giving
us the equation X2 _j_6 x _jig = 25,
or (x +3)2=25.
Taking the square root of both members of the last equation is
now an easy process and gives
x +3 = ±5.
Therefore we must either have #+3 =5, or x+3 = —5.
Solving the last two equations gives as the desired solutions
x=2andx=— 8. Ans.
CHECK. Substituting 2 for x in the first member of the given
equation gives 22+6X2, which reduces to 4 + 12 = 16, as desired.
Similarly, with x equal to —8, the first member of the given equa-
tion becomes (-8)2+6X(-8), or 64—48, which reduces to 16 as
required.
84 SECOND COURSE IN ALGEBRA [X, § 53
EXAMPLE 2. Solve x2- 8 x+ 14 = 0.
SOLUTION. Transposing, x2 — 8 x = — 14.
Completing the square by adding 16 to both sides gives
z2-8z + 16=2, or (z-4)2=2.
Taking the square root of both members,
x-4=±V2.
Solving the last two equations,
z=4 + V2andz=4- V2. Ans.
CHECK. With x=4 + V2, the first member of the given equation
becomes (4 + V2)2-8(4 + V2) +14. By Formula VI of §10 this
may be written
(16+8V2+2)-8(4 + V2)+14, or 16+8V2~+2-32-8V2 + 14.
Here the 8>/2 and the -8^2 cancel, while the rest of the expres-
sion (namely 16+2—32 + 14) reduces to 0, as required.
Likewise, when x has its other value, namely x =4 — V2, the first
member may be shown to become 0.
NOTE. Since the solutions obtained above for Example 2 con-
tain the surd V2, they cannot be expressed exactly (see § 37), but
we_can express their values approximately. Thus, the table gives
V2 = 1.41421+ so that the two solutions become 4 + 1.41421+ and
4-1.41421+, which reduce to 5.41421+ and 2.58579+. Ans.
EXAMPLE 3. Solve 3 z2+8 x = 15.
SOLUTION. Dividing through by 3 so as to have +1 as the co-
efficient of x2, the equation becomes
z2+fz=5.
Completing the square by adding (-|)2 (or -^-) to both sides gives
or,
Taking the square root of both members,
x + * = ±
Therefore, the two solutions are
z=-f+iV61 and z
These two answers may be written together in the condensed
form x =-^( — 4± V61) and by looking up the value of V61 in the
tables, these values of x may be determined approximately, as in-
dicated in the Note to Example 2.
X, § 54] QUADRATIC EQUATIONS 85
54. Summary and Rule. It is now to be observed care-
fully that in each of the three examples just considered
(§ 53) the first step in the solution consists in reducing the
given equation to the type form
where p and q are given numbers.
Thus, in Example 3, we first put the equation 3 x2 +8 x = 15 into
the form z2+f z=5. Here p=f, and q =5.
The next step is to complete the square. This is done in
each case by adding to both members the square of half the
coefficient of x, that is we add (p/2)2 to both members.
Thus, in Example 3, we had p=f, so we added (|-)2 to both
members.
After this, the equation is such that we can extract the
square root of the left member, and when we do so and
equate results, we obtain two simple equations, each
yielding a solution of the given quadratic.
This may now be summarized in the following rule.
To solve any quadratic :
1. Reduce the equation to the form
2. Complete the square by adding (p/2)2 to both members.
3. Extract the square root of both members of the new equa-
tion and equate results. This yields the two solutions desired.
EXERCISES
Solve each of the following equations, checking your answer
in the first five.
1. z2-5z = 14. 6. 8z = z2-180.
2. z2-20z = 21. 7. z2+22z=-120.
3. z2-12z+20 = 0/ 8. 2/2 = 10-3i/.
4. x2-2x = ll. 9. z2-
6. z2-3z-5 = 0. 10. 6z2
86 SECOND COURSE IN ALGEBRA [X, § 54
11. 2z2+
12. 1-3 * = 2*2.
17. ^—
13. 2 £(z-f-4)=42. x
14. (3x-2)2 = 6z+ll, 18. -^
15. x+— = 16.
.
z+5 z-2
55. Solution of Quadratics by the Hindu Method. A
simple way preferred by many for completing the square in
any quadratic is the one called the Hindu method. It con-
sists of two steps :
1. Multiply both members by four times the coefficient of x2.
2. Add to both members of the new equation the square of
the original coefficient of x.
EXAMPLE. Solve 2 x2 - 3 x = 2.
SOLUTION. Multiplying through by 4 times the coefficient of
x\ that is by 8, gives 16 x2 -24 x = 16.
Adding the square of the original coefficient of x to both sides,
that is adding ( — 3)2, or 9, to both sides, gives
16z2-24z+9=25.
The first member is now a perfect square, being equal to (4 x — 3)2.
Therefore, extracting square roots, we obtain
4 x -3 =5 and 4 x -3 =-5.
Solving the last two equations gives x=2 and x = — J. Ans.
EXERCISES
Solve each of the following quadratics by any method.
1. 9z2+6z = 35. 4. 16z2-7o;-123 = 0.
2. 4z2-12z = 27. 6. 1
3. 4z2-z-3 = 0. 6. 2
X, § 56] QUADRATIC EQUATIONS 87
7. 2z2+6z = f 17. 3z2+z-200
[HINT. First multiply both ^ r
numbers by 2.] 18. x+- = --
8. 3z2-2z = 5.
9. 3z2+7z-110 = 0. 19. — -- ^ =
r» o <-i f\ 3C O X
10. 5z2-7z=-2.
11. l-3z = 2z2. 20.
12. 4*2-3*-2 = 0.
13. 2a:2+3x = 27. 21«
14. 3x2-7x+2 = 0 [HINT. Proceed as in § 40.]
15. 4 x2- 17 z =-4. 22- Vx+1 — Va;-2 = A/2 a;- 5.
16. 8z = z2-180. 23. Vx-l+VlQ-x=3.
56. Solution by Formula. Every quadratic is an equation
of the type form ax2+bx+c = 0>
where a, b, and c are given numbers. We may solve this
equation as it stands by the process of § 54. Thus,
ax?-\-bx= —c.
Dividing through by a,
*M*--4
a a
Adding 6/(2 a)2 to both sides (§ 54) gives
x* \bx I fb b* c
a \2aJ \2aJ a 4 a2 a 4 o2
Extracting the square root of both members,
x+ — ==«= /fr2-4ac^ ±Vb2-4ac
2 a "V 4 a2 2 a
Transposing the term 6/(2 a), we thus have the following
formulas for the two roots :
x.-»+vy=«a and x=-b-v^=47c.
2a 2 o
88 SECOND COURSE IN ALGEBRA [X, § 56
NOTE. Observe that these formulas express the values of x
in terms of the known letters a, 6, and c, as should be the case. Thus,
in any example, we have only to put the given values of a, 6, and c
into the formulas in order to have at once the desired values of the
two roots.
EXAMPLE. Solve by formula 2 x2 — 3 x = 2.
SOLUTION. This equation may be written 2z2— 3#— 2=0.
Hence the values of a, b, and c in this case are as follows: a =2,
b = — 3, c = — 2. Placing these values in the formulas gives as the
roots _
-3)2 -4(2) ( -2)
2-2
and a,-(-3)-V(-3)«-4(2)(-2).
2 • 2
Simplifying, * =
and ,-
The two roots are, therefore, z— 2 and x= — ^. Ans. (Com-
pare solution in § 55.)
EXERCISES
1. Solve by the formula Exs. 13-17, p. 87.
2. Solve by the formula the equation 3 x2— 6 x+2 = 0.
[HINT. The roots are found to be x = 1 +^ V3 and x = 1 -| Vs.
This quadratic thus has roots which necessarily contain radicals.
From the table of square roots at the end of the book, we find
V3 = 1.73205. Hence, the roots, correct to five decimal places, are
1+ 1.73205 and 1_1.73205> which reduce respectively to 1.57735
3 3
and 0.42265. Ans.]
3. Solve by the formula the equation x2 — 5 z+3 = 0, and
use the tables if necessary to determine the roots decimally.
4. Solve by the formula the equation 4 x2— 3 x— 2 = 0, ex-
pressing the roots decimally correct to five places.
X, § 56] QUADRATIC EQUATIONS 89
APPLIED PROBLEMS
[The method of solving quadratics by formula (§ 56) is usually
the most direct.]
1. The square of a certain number is 4 less than five
times the number. Find the number.
[HINT. Remember that there should be two solutions.]
2. Divide 20 into two parts whose product is 96.
3. One side of a right triangle is 2 inches longer than the
other. If the hypotenuse is 10 inches long, how long are the
sides?
[HINT. Letting x represent the shorter side, the equation here
becomes x2 + (x+2)2 = 100, and in solving this we find that one of
the solutions is negative. But a negative solution can have no mean-
ing in such an example as this, so we keep only the positive solution.
This frequently happens in applied problems involving quadratics,
so the pupil must always be on his guard to keep only such solu-
tions as can actually fit a given problem.]
4. A gardener spades a bed 30 feet long by 20 feet wide.
He then decides to double its size by adding a border of uni-
form width throughout. How wide must ,_.
the border be made?
5. In Example 4 suppose that instead of
doubling the area, the gardener wishes merely
to add 200 square feet to it. Show that the
strip added around the outside must then be made a little
over 1.86 feet wide.
6. A circular swimming pool is surrounded by a walk 4
feet wide. If the area of the walk is one fourth that of the
pool, find (approximately) the radius of the pool. (Take
90
SECOND COURSE IN ALGEBRA
[X, § 56
FIG. 14.
is marked P
A G
7. If a train had its speed diminished by 10 miles an hour,
it would take it 1 hour longer to travel 200 miles. What is
p x B the speed ?
-c — ^ . L
8. A circular curbing touches the line of
\ the street curbing on each of two streets
/ that meet at right angles. In Fig. 14, the
middle point of the circular curbing is
marked M. The point at which the
straight curbings would meet if extended
If PM = 5 ft., find the radius (x in Fig. 14) of
the circular curbing, f
9. The figure represents a pattern fre-
quently used in window designs, consisting of
a square A BCD with a semicircle EFG
mounted upon it, the diameter GE of the
semicircle being slightly less than one of the
sides of the square. If the shoulders AG and
DE are each 1 foot long, how long must each
side of the square be made in order that the total lighting
surface shall be 88 square feet?
10. A soap bubble of radius r is blown out until the area
of its outer surface becomes double its original value. Show
that the radius has thus been increased by an amount h given
by the formula h = r(V2-l). [HINT. See Ex. 14 (/), p. 6.]
57. Graphical Solution of Quadratics. Consider the
quadratic x2 — 3 x — 4 = 0. Let us represent the left member
by the letter y ; that is, let us place
FIG. 15.
Now, if we give to x any value, this equation determines a
t This problem suggests a practical plan for finding the radius
of circular curbings when the center 0 of the circle cannot be reached.
X, § 57]
QUADRATIC EQUATIONS
91
corresponding value for y. For example, if x = Q, then
?/ = 02-3xO-4=-4. Again, if x=l, then i/=l2-3xl-4
= —6. The table below shows a number of x-values with
their corresponding ^/-values, determined in this way.
When x =
0
1
2
3
4
5
. 6
-1
-2
-3
then y =
4
-6
-6
-4
0
6
14
0
6
14
The graph is now obtained by draw-
ing an z-axis and a 2/-axis, as in § 28,
then plotting each of the points x, y
which the table contains, and finally
drawing the smooth curve passing
through all such points. The form of
the graph thus obtained is indicated in
the adjoining figure. Observe that this
graph is not a straight line and is there-
fore different in character from the graph
of a linear equation. (See § 29.) And
it is especially important to notice that
it cuts the x-axis in two points whose
^-values are —1 and 4, respectively.
These two values of x determined
in this purely graphical way are the
two solutions of the given quadratic, x2 — 3 x — 4 =0, for they
are those values of x that make y — 0, that is that make
z2-3z-4 = 0.
The graphical study which we have just made of the
special quadratic x2 — 3 x — 4 = 0 leads at once to the following
more general statements.
Every quadratic has a graph which -is obtained by first placing
y equal to the left member of the equation (it being understood
that the right member is 0) , then letting x take a series of values
FIG. 16.
92 SECOND COURSE IN ALGEBRA [X, § 57
and determining their corresponding y-values, plotting the
points, x, y, thus obtained and finally drawing the smooth curve
through them.
The x-values of the two points where the graph cuts the x-axis
will be the roots of the given quadratic.
EXERCISES
Draw the graphs of each of the following quadratics, and
note where each cuts the z-axis. In this way determine
graphically the solutions, and check the correctness of your
answer by actually solving by one of the methods explained
in 54-56.
1. z2-z-2 = 0. 2. x2-7x+l2 = Q. 3. xz+
4. z2-5z=-6.
[HiNT. Remember to write first as x2 —5 x +6 =0.]
5. x*+3x=W. 6. 2x2+3x = 9.
58. Quadratics Having Imaginary Solutions. Consider
the quadratic x2= — 1. This is a pure quadratic (§ 49) and
hence can be solved immediately by merely taking the
square root of each member. This gives as the required solu-
tions x=-\-V— 1 and x=—\^—\. But V— 1 means the
number whose square is —1, and there is no such number
among all those (positive or negative) which we have thus far met.
In fact, we know that the square of any number, whether the
number be positive or negative, is positive [§ 2(d)]. There-
fore, in any such case as this, we say that the solutions are
imaginary, and we speak of the numbers themselves which,
like V — 1, enter into algebra in this way, as imaginary
numbers. They are imaginary, however, only in the sense
that they have not been encountered before.
X, § 59] QUADRATIC EQUATIONS 93
As an example of an affected quadratic having imaginary
roots, let us consider the equation x2 — 6x+15 = 0. When
we proceed to solve this by the method of completing the
square, as in § 54, the work is as follows.
Transposing, we have
x2-6x= -15.
Adding 9 to both sides to complete the square,
z2-6z+9= -6, or (z-3)2= -6.
Extracting the square root of both sides,
x-3= ±V^6.
Therefore the solutions are
=3- V^Q. Ans.
Both of these solutions_are seen to be imaginary because they
contain the expression V —Q.
59. Definitions. A number like 3+v^6 or 3-V-6 is
frequently called a complex number in distinction to such a
number as V — 6, which is called a pure imaginary. Thus, a
complex number is a combination of a positive or negative
number with a pure imaginary.
All numbers considered in the chapters preceding this
(including irrationals) are called real numbers in distinction
from the imaginary numbers just described. Thus, the solu-
tions of all quadratics considered in §§ 54-56 are real instead
of imaginary.
EXERCISES
Find (by solving) whether the solutions of the following
quadratics are real or imaginary.
1. z2+9 = 0. 3. 2z2+2z+3 = 0. 6.
2. z2-6z+10 = 0. 4. 3z2+2z = 4. 6.
94
SECOND COURSE IN ALGEBRA
[X, §60
60. Determining Graphically Whether Solutions are Real
or Imaginary. It was shown in § 58 that the solutions of
the quadratic x2 — 6 x= — 15 are imaginary. Let us now see
what corresponds to this fact in the graph.
The table below shows several values of x and their corre-
sponding ^/-values, as determined from the given equation
When x =
-1
0
1
w_
2
3
4
5
6
then y =
22
15
7
6
7
10
15
FIG. 17.
Plotting the various points (x, y) thus
obtained and drawing the curve through
them gives the graph indicated in the
accompanying figure. This graph is es-
sentially different from those met with
in § 57 in one particular, namely it does
not cut the x-axis.
A similar result holds for the graph of
every quadratic whose solutions are
imaginary. Therefore, in order to tell
whether the solutions of any given quad-
ratic are real or imaginary, we need only
draw its graph and note whether or not
it cuts the z-axis. If it does, the solu-
tions are real ; if it does not, the solutions
are imaginary.
EXERCISES
Find by drawing the graph whether the roots of each of the
following quadratics are real or imaginary.
1. z2+2z+3 = 0. 3. z2-2z+3 = 0. 5. 6z2+5z+l=0.
2. z2+2z-3 = 0. 4. 3z2+4z+l = 0. 6. 2z2-3z+4 = 0.
LAGRANGE
(Joseph Louis Lagrange, 1736-1813)
Famous for his discoveries in all branches of mathematics and regarded as
the greatest mathematician of the 18th century. In algebra he gave much
attention to the study of equations and determinants, extending and unify-
ing the work of previous mathematicians in these fields.
PART II. ADVANCED TOPICS
CHAPTER XI
LITERAL EQUATIONS AND FORMULAS
61. Literal Equations. Equations in which some, or all,
of the known numbers are represented by letters are called
literal equations. The known letters are generally repre-
sented by the first letters of the alphabet, as a, b, c, etc.
Literal equations are solved by the same processes as numeri-
cal equations,.
EXAMPLE. Solve the following literal equation for x :
ax = bx+7 c.
SOLUTION. Transposing,
ax — bx=7 c.
Combining like terms,
(a-6)z=7c.
Dividing by (a— 6),
*=-?X Ans.
a—b
CHECK. Substituting the answer for x in the given equation,
Multiplying by (a — b),
7 ac =7 be +7 c(a -b) = 7 be +7 ac -7 be.
Transposing,
7 ac — 7 ac=7 be -7 be.
Simplifying, 0=0, which is a correct result.
95
96 SECOND COURSE IN ALGEBRA [XI, § 61
It is to be carefully observed that a literal equation is said
to be solved for the unknown letter, as x, only when that
letter has been expressed in terms of the other (known)
letters. Thus, in the example above, we obtained x in terms
of a, 6, and c. This when once done, is what we mean by the
solution.
NOTE. If a literal equation is satisfied no matter what values
be given to the letters appearing in it, it is called an identity. Thus
x2 — a2 = (x — a) (x +a) is an identity. This fact is often expressed by
means of the symbol =. Thus, x2— a?=(x— o)(x+o). Likewise,
(z-a)2 = z2-2 ax+a2, etc.
EXERCISES
Solve for x in the following, checking your answer in the
first five.
1. x-a = b. 9 *+b = *+a.
2. ax-l = b. a
3. ax+bx = c. - a , b _
4. 3z+6 = z-3&. cx
5. 4(3&-z) = 3(26+z).
6. (x-a)(x-b)=x(x+c).
7. -J-- a. 12.
1+x x-3
a, 6_o 13. Divide a into two parts
# # whose quotient is m.
14. If A can do a piece of work in a days, and B can do it
in b days, how long will it take them working together? (See
Exs. 23, 24, p. 40.)
[HINT. These are simultaneous equations, to be solved for the
two unknowns x and y in terms of a and b.]
XI, § 61] LITERAL EQUATIONS AND FORMULAS 97
16.
I ax-by = 2,
»•{
3 ax-f2 by = ab,
ax — by = ab.
18.
x y a
1-1=1.
x y b
[HINT. Solve first for l/x and l/y. See Ex. 16, p. 53.]
a_b _ _.,
x y~ lj
a
19.
x y
20. axz-c=l.
[HINT. See § 51. Ans. x = * >/— •]
21. az2-fa3 = 5a3-3az2.
22. !±+± = i.
x a x
23.
24. (x+a)(x+b)+4(x+a)=Q.
[HINT. Solve by factoring.]
25 /£2 ax = 2 a2
[HINT. See § 54, p. 85. Ans. x =2 a, or x = -a.]
29. x = 4 ax — 2 a .
30. Vx — a+Vb — x = Vb — a.
H
98 SECOND COURSE IN ALGEBRA [XI, § 62
62. Formulas. If a person travels for 10 hours at the
rate of 15 miles an hour, the distance he travels is 15 XlO =
150 miles. Stated in general (algebraic) language, we can
say in the same way that if a person travels for t hours at the
rate of r miles an hour, the distance s he travels is
s = rt.
This is a literal equation expressing the value of s in terms
of r and t. If we wish, we can solve it for t, giving t = s/r, and
what we now have is t expressed in terms of s and r. Or, we
can solve the original equation for r, giving r = s/t, and this
expresses r in terms of s and t.
These examples illustrate the important fact -that in
nearly all branches of knowledge, especially in engineering,
geometry, physics, and the like, there are general laws which
are expressed by means of mathematical formulas. Such
formulas are merely literal equations in which two or more
letters appear, and it is often desirable to solve them for some
one letter in order to express its value in terms of the others.
EXERCISES
1. The area A of a rectangle whose dimensions (length
and breadth) are a and b is given by the formula A=ab.
Solve this for a ; also for b. In each case state in terms of
what letters your answer is written.
2. The formula for the area A of a triangle whose height
(altitude) is h and whose base is a is A = % ah. Solve for a ;
also for h.
3. Solve for b in the formula
(Formula for the area A of a trapezoid whose
FIG. is. bases are B and b and whose altitude is h.)
XI, §62] LITERAL EQUATIONS AND FORMULAS 99
4. Solve for r in the formula A=irr2. (Formula for the
area of a circle whose radius is r.)
5. The interest / which a principal of p dollars will yield
in t years at r % is determined by the formula
100
Solve this for r and use your result to answer the following
question : What rate of interest is necessary in order that $50
may yield $6 interest in 2 years' time ?
[HiNT. Solving for r gives at once
r = 1007
pt '
Now see what the right member of this equation becomes when
7=6, p=50, and t=2.]
6. Using tne interest formula of Ex. 5, solve it for t and
use your result to answer the following question : How long
will it take $600 to yield $63 interest if invested at 6%?
7. The velocity of sound v, in feet per second, is given
by the formula v= 1090+1. 140-32), where t is the tem-
perature of the air in Fahrenheit degrees. Find
(a) The velocity of sound when the temperature is 75°.
(6) The temperature when sound travels 1120 ft. per sec.
8. Derive formulas for each of the following statements.
(a) The number N of turns made by a wagon wheel d feet
in diameter in traveling s miles.
(6) The number N of dimes in m dollars, n quarter dollars,
and q cents.
9. An automobile travels for T hours at the rate of v
miles per hour. By how much must this rate be increased
in order to make the same journey in t minutes less time?
10. A has $a and B has $6. Between them they give $c
to a certain charity, after which the amounts of money they
have are equal. How much does each contribute?
100
SECOND COURSE IN ALGEBRA [XI, § 63
63. Law of the Lever. If two weights are balanced at the
ends of any (uniform) bar, as shown in the figure, we have an
example of a lever. The point of
support, F, is called the fulcrum.
If we let W and w be the values
(in pounds or ounces or any
FIG 19 other convenient unit) of the two
weights, while D and d stand for
the distances respectively of W and w from F, then, when-
ever the balance is perfect, we have the formula
D F d
A
fw
W
w D
Sometimes a single weight W is balanced by a force, p,
usually called a power. This may happen in several ways, as
indicated by the following figures. In all such cases, if we
I— D-
(2)
let W represent the weight, p the power, D the distance from
W to the fulcrum, there exists the following formula when-
ever the balance is perfect :
W=d
P D
XI, §64] LITERAL EQUATIONS AND FORMULAS 101
This is called the general law of the lever. By clearing the
equation of fractions, it may be written in the form
WD=pd.
Translated into words, this last relation means that the
weight times the weight arm equals the power times the power
arm. It is in this form that the law is usually remembered by
engineers.
EXERCISES ON THE LEVER
1. If the fulcrum of a 5-foot crowbar is placed 1 foot from
the end, what weight can be lifted by a man weighing 180
pounds ?
[HINT. Here we have Fig. 19 with W = ?, w (or p) = 180 pounds,
D=lfoot, d =4 feet.]
2. The figure represents a simple
form of pump. If the pump handle
AF is 16 inches long, while the
piston-arm FC is 3 inches long, what
will be the upward pull at C when
there is a 9-pound downward push
at A?
3. A certain lever, after being balanced, has r pounds
added to the weight W. Determine (in terms of W, p, D, d,
and r) how much the power, p, must be increased to keep the
balance perfect.
64. Gear Wheel Law. Whenever a gear wheel having T
teeth revolves at the rate of N revolutions per minute and
turns another similar wheel having t
teeth at the rate of n revolutions per
minute, there exists at all times dur-
ing the motion the formula
T=n.
Fro. 22. t N
FIG. 21.
102 SECOND COURSE IN ALGEBRA [XI, § 64
This is the gear wheel law. Clearing this equation of frac-
tions, it becomes
TN=tn.
Translated into words, this last relation means that the
number of teeth in one wheel multiplied by its rate of turning is
equal to the number of teeth in the other wheel multiplied by its
rate of turning.
EXERCISES ON GEAR WHEELS
1. If in Fig. 22 the small wheel has 30 teeth and is making
96 revolutions per minute, how many teeth must the large
wheel have in order to revolve 16 times per minute?
2. In order that the large wheel revolve three fourths as
fast as the small one, how must the wheels be made ?
3. If the large wheel in § 64 be made larger by the addi-
tion of r teeth to its run, determine the amount by which the
speed of the smaller wheel will be thereby increased.
[HINT. Let x represent the unknown amount and find a formula
for x in terms of T, t, N, n, and r.]
65. Other Useful Formulas. In addition to the formulas
already mentioned, the following from plane and solid
geometry and from elementary physics are
often used.
1. The area A of an equilateral triangle
of side a (Fig. 23) is
_V3 2
FIG. 23. :~^~fl'
where A/3 = 1.732 approximately.
2. The area A of a regular hexagon of
side a (Fig. 24) is
3\/3
fl2.
2 FIG. 24.
XI, § 65] LITERAL EQUATIONS AND FORMULAS 103
3. The area A of any triangle in
terms of its three sides a, 6, and c
(Fig. 25) is
A = Vs(s-a)(s-b)(s-c),
where s =
a
FIG. 25.
4. The area A of the sector of a circle when
the intercepted arc is a and the radius is r
(Fig. 26), is
FIG. 26.
5. The length of the diagonal d of a
rectangular block whose dimensions b\
are a, 6, and c (Fig. 27) is
6. The volume V of a circular cylinder of altitude
h and radius of base r (Fig. 28) is
F=irr2/i.
7. The volume V of a circular
cone of altitude h and radius of base r (see
Fig. 29) is
FIG. 28.
V=±Trr*h.
8. The volume V of a pyramid
of altitude h and base B (Fig. 30) is
9. The volume V of a spheri-
cal segment (or slice of a sphere
between two parallel cutting planes), where
h is the altitude, and a and b the radii of the
two bases (Fig. 31) is
Fio. 29.
FIG. 30.
FIG. 31.
104
SECOND COURSE IN ALGEBRA
[XI, § 65
10. The surface S of the zone (or portion of the surface of
a sphere lying between two parallel cutting planes), where
h is the distance between the cutting planes and r the radius
of the sphere (Fig. 31), is
S = 2*117/1.
11. The length of belt I required to go around two wheels
whose diameters are D and d and whose centers are at the
distance a apart is
FIG. 32.
(a) In case the belt does not cross itself,
(6) In case the belt crosses itself once,
/ =
2
12. The force F, measured in pounds, with which a body
weighing W pounds pulls outward (centrifugal force) when
traveling with a velocity of v feet per
second in a circle of radius r (Fig. 33)
is determined by the formula
FIG. 33.
„
r
32
13. The pressure P exerted by a letter
press (Fig. 34) is determined by the
formula D 2irrF
where F is the value of the force applied at FIG. 34.
XI, § 65] LITERAL EQUATIONS AND FORMULAS 105
the wheel, r is the radius of the wheel, and h is the distance
from one thread of the screw to the next one.
14. The weight W which can be raised
by means of a toothed wheel and screw
such as indicated in Fig. 35 is deter-
mined by the formula
where P represents the pressure applied
at the handle and where R, r, d, and I are the dimensions
indicated in the figure.
Other important formulas from elementary geometry and
from physics may be found in Chapter XVII.
EXERCISES
1. Find by Formula 3 of § 65 the area of the triangle whose
three sides are respectively 5 inches, 5 inches, and 8 inches
long.
2. Show that in the case of an equilateral triangle of side
a the expression for A in Formula 3, § 65, reduces, as it should,
to that for A in Formula 1, § 65.
3. Show that in case the two wheels in Fig. 32 have the
same diameter D, the formula for the length of belt re-
duces in case (a) to the simple form l=TrD+2 a, and in case
(6) to Z = 2VD2+a2+7rD.
4. How much leather (surface measure) will it take for a
belt 6 inches wide to connect two pulleys whose diameters
are 5 feet and 1 foot, respectively, the distance between
centers being 10 feet ?
[HINT. Viol = 10.2, approximately.]
106 SECOND COURSE IN ALGEBRA [XI, § 65
5. A pail of water weighing 5 pounds is swung round at
arm's length at a speed of 10 feet per second. If the length
of the arm is 2 feet, find (a) the pull at the shoulder when
the pail is at the uppermost point of its course, (b) when at
the lowest point of its course. Also, find the least velocity
which the pail can have without the water dropping out at
the top point of the course.
6. What pressure is exerted by a letter press in case the
force applied at the wheel is 10 pounds, the diameter of
the wheel is 1^ feet, and the threads of the screw are % inch
apart ?
7. In the device shown in Fig. 35, show that if the distance
d between two adjacent threads be halved and the number
of teeth on the wheel be correspondingly doubled to fit the
new gear, other parts remaining the same, the weight W that
can be raised with a given pressure P will be doubled.
8. The volume V of the frustum of a cone or pyramid
made by a plane parallel to the base is given by the formula
where B and b denote the areas of the lower and upper bases,
and h denotes the altitude.
FIG. 36.
Determine the volume of the frustum of a circular cone the
radii of whose bases are 4 inches and 3 inches, the altitude
being 5 inches.
XI, § 65] LITERAL EQUATIONS AND FORMULAS 107
Show that in case the upper and lower bases are equilateral
triangles whose sides are a and b respectively, the formula
becomes
[HINT. See Formula 1, § 65.]
9. Show by means of Formula 3, p. 103, that if the
three sides of any triangle be extended to twice their original
length, the area will become quadrupled.
10. If, in the first of the two cases represented in Fig. 32,
the diameter of each wheel be increased by the same amount,
say a, show that the length of belt will thereby become in-
creased by the amount ira.
11. If, in the second of the cases represented in Fig. 32,
the diameter of the large wheel be increased by any given
amount and at the same time the diameter of the small
wheel be diminished by the same amount, show that pre-
cisely the same length of belt as before will fit over them
tightly.
12. A body is moving along a circle with a velocity of
3' feet per second. Show that in order for the centrifugal
force (12, p. 104) which it is exerting to be doubled, the
velocity must be increased by about 1J feet per second.
By how much would the velocity have to be diminished
in order that the centrifugal force become halved ?
13. By means of Formulas 9 and 10 (§ 65) obtain the
formulas for the area and volume of a whole sphere. Com-
pare with Ex. 14, (e), (/), p. 6.
CHAPTER XII
GENERAL PROPERTIES OF QUADRATIC EQUATIONS
66. The Classification of Numbers. A real number is
one whose expression does not require the square root of a
negative quantity, while an imaginary number is one whose
expression does require such a square root. (See § 59, p. 93.)
Thus 1, 3, -7, £, f, -f , \/2, 1+ \/3 are real numbers, while
-S, V- i, 2 + A/-3, are imaginary numbers.
In case a real number can be expressed in the particular
form ^ where p and q are integers (positive or negative) it is
q
called a rational number. The number zero is also included
among the rational numbers. (See § 42, p. 56.)
Thus l, f , — f , 5, 73, —10, — ^J, are rational numbers.
In case a real number cannot be expressed in the particu-
lar form just mentioned, it is called an irrational number.
Thus V2, V3, Vf , ^2, -vXJ", V^ 1 + A/6, are irrational num-
bers.
Imaginary numbers are either pure imaginaries, such as
V^3, or complex numbers (§ 59, p. 93), such as 1 + V — 3.
108
XII, § 67] GENERAL PROPERTIES OF QUADRATICS 109
These divisions and subdivisions may be summarized into
a table as below :
n , f Rational
Real{ T .. ,
»r » f ., Irrational
Numbers of Algebra T
. . j Pure Imaginanes
Imaginary { „ 1 ,yr ,
\ Complex Numbers
NOTE. The rational numbers are themselves subdivided into
three sub-classes : the single number zero, the integers (positive or
negative), and the rational fractions. The latter are those rational
numbers which, like f , cannot be expressed as integers.
All the numbers used in arithmetic are positive real numbers.
The negative real numbers together with the imaginary numbers
owe their existence to algebra.
67. Determining the Character of the Roots of a Quad-
ratic. It is often desirable to determine the character of
the roots of a given quadratic, that is, whether the roots are
real or imaginary ; and if real, whether they are rational or
irrational, etc.
Thus the roots of 2 z2-7 x+l = 0 are (by §56, p. 87), 7=t=V^.
4
Since 41 is positive, these roots are real numbers (§ 66).
Since 41 is not a perfect square, the roots are irrational (§ 66).
Since VH is added to 7 in the one root and subtracted from 7 in
the other root, the roots are unequal.
Thus the character of the roots in this case is described by saying
that they are real, irrational, and unequal.
There is, however, a much shorter method than the one
illustrated above for determining the character of the roots
of a given quadratic. Thus we know (§ 56) that the two
roots of any quadratic, namely, any equation of the form
ax2-\-bx-\-c = Q,
_ and _
2a 2a
110 SECOND COURSE IN ALGEBRA [XII, § 67
An examination of the form of these expressions gives
at once the following rule.
RULE. For any given quadratic, ax*+bx+c = Q, in which
the coefficients a, b, c are real numbers, the two roots will be
(1) Real and unequal if b2 — 4 ac is positive.
(2) Real and equal if b2 — 4 ac = 0. (Both roots then
reduce to — 6/2 a.)
(3) Imaginary ifb2 — 4acis negative.
Moreover, if the coefficients a, b} c are rational numbers, the
two roots will be
(4) Rational, if 62 — 4 ac is a perfect square; irrational if
b2—4ac is not a perfect square.
Because of the manner in which the character of the roots
thus comes to depend upon the value of b2 — 4 ac, this expres-
sion is called the discriminant of the given quadratic.
EXAMPLE 1. Determine the character of the roots of
2z2-3z-9 = 0.
SOLUTION. Here a =2, b =—3, c= — 9. Hence the value of
the discriminant, or 62-4ac, is ( -3)2-4(2)( -9) =9+72=81 =92.
Hence, by (1) and (4) of the rule, the roots are real, unequal,
and rational.
EXAMPLE 2. Determine the character of the roots of
SOLUTION. Here a =3, b=2, c = l. Hence 62— 4 ac=4 — 12= -8.
Hence, by (3) of the rule, the two roots must be imaginary.
EXAMPLE 3. Determine the character of the roots of
4z2-20z+25 = 0.
SOLUTION, a =4, 6= -20,c=25. Hence 62- 4 ac= 400 -400=0.
Therefore, by (2) of the rule, the roots are real and equal.
The common value which the two roots have may be found if
desired by actually solving the equation. It turns out to be f .
XII, §68] GENERAL PROPERTIES OF QUADRATICS 111
EXERCISES
Determine (without solving) the character of the roots of
each of the following equations.
1. 2z2-3z+l=0. 7.
x*+x=-l.
8.
9.
10.
11.
12.
*68. Character of Roots Considered Geometrically. We
have seen in § 57 that whenever a quadratic has its two roots
real and unequal, its graph will cut the #-axis in two distinct
points. On the other hand, if the roots are imaginary, the
graph of the equation will not cut the a>axis at all (§ 60).
Suppose now that we have a quadratic
whose two roots are equal to each other,
for example
(1) 4z2-12z+9 = 0.
Here the discriminant is
(-12)2- 4(4) (9) = 144 -144 = 0,
so that the roots must be equal (§ 67).
If we now proceed to draw the graph
of (1) in the usual way by placing
2/ = 4z2— 12 z+9, then forming a table
of x, y values, etc., it appears that the .
graph corresponding to (1) just touches
the x-axis instead of cutting through it.
This, in fact, is what we should expect,
since the equality of the roots virtually
means that there is but one root, and this can be possible
only when the graph merely touches (is tangent to) the x-axis.
FIG. 37.
112
SECOND COURSE IN ALGEBRA [XII, § 68
In order to illustrate in one single diagram all the facts
mentioned thus far about the graph, it is instructive to take
such an equation as
(2) z2-2z+c = 0,
and let c take different values, thus obtaining various quad-
ratics. For example, if we choose c— — 2, the quadratic
equation becomes
and if we draw the graph of this, we find that it cuts through
the z-axis at two points, thus in-
dicating that its solutions are real
and unequal.
Likewise, we find a similar result
when we let c=— 1, and when
c = 0, though the various graphs are
themselves different.
But if we choose c = l and pro-
ceed as before, the graph of the
corresponding quadratic no longer
cuts through the o>axis, but merely
touches it, thus indicating real
and equal roots.
Finally, for such values of c as 2,
3, or 4, the quadratics (2) come
to have graphs which do not cut
the x-axis, thus indicating that they have imaginary roots.
The effect of changing c is thus merely to slide one and
the same curve vertically up and down the coordinate paper.
The figure shows the positions of the curve corresponding
to c = — 2, c = 1 , and c = 4. The pupil is advised to draw
in for himself the positions of the curve for c = — 1, c = 0, c = 2,
and c = 3.
FIG. 38.
XII, §69] GENERAL PROPERTIES OF QUADRATICS 113
69. The Sum and Product of the Roots. We have seen
that every quadratic is an equation of the form
(1)
and we have also seen (§ 56) that the roots, or solutions, of
this equation are
and
-6-V&2-4
oc
2a
It is now to be observed that if we add these solutions
together the radical cancels and we obtain the simple result
-2b_ b
2 a a
Again, if we multiply the two solutions together, we obtain
a -final result which is very simple in form. Thus
_ (-fr)2- (Vb2-4 ac)2_62- (62-4 ac) _4 ac_ c
4 a2 4 a2 ~4~a?~a'
These results may be summarized in the following rule.
RULE. In the general quadratic equation ax2-\-bx-\-c = Q,
(1) The sum of the two solutions is — b/a.
(2) The product of the two solutions is c/a.
EXAMPLE. Find the sum and the product of the solutions
of the equation 3x2 — 2 z+6 = 0.
SOLUTION. Here a = 3, 6= —2, c = 6.
r>
Hence the sum of the solutions is , or f, and the product
of the solutions is % = 2.
l
114 SECOND COURSE IN ALGEBRA [XII, § 69
EXERCISES
State (by inspection) the sum and the product of the solu-
tions of each of the following equations. Check your answer
in Exs. 1, 2, 3, 4 by actually solving the equations and de-
termining the sum and the product of the solutions.
1. 2z2+5z-7 = 0. 5. 6z2+7z = 42.
2. 3x*-7x+2 = Q. 6.
3. 5z2-2z = 16. 7.
4. 3z = 200-z2. 8. x2+px = q.
9. Show that in the quadratic x2 -\-rnx +n = 0 the sum of the
roots is — m and the product of the roots is n. This general
result is important and may be stated in words as follows :
// in a quadratic the coefficient of x2 is 1, the sum of the solu-
tions will be the coefficient of x with its sign changed, while the
product of the solutions will be the remaining term. Explain
and illustrate by means of the equation z2— 10 z+12 = 0.
10. Apply the result stated in Ex. 9 to determine the sum
and the product of the solutions of the following equations:
(a) z2-5z+7 = 0. (e) x2-(a+b)x+ab = Q.
(b) z2-4z = 10. (/) 2z2+3z-4 = 0.
(c) x2 — %x = 2. [HINT. First divide through
(d) z2-V2z+V3 = 0. by 2.]
70. Formation of Quadratics Having Given Solutions.
EXAMPLE 1. Form the quadratic -whose solutions are
1 and -i-.
SOLUTION. If x = l, then x -1=0; if x = -J-, thenz+£ = 0.
Hence the equation (x-l)(z+£) =0, or z2-^ x-±=Q, will be
satisfied when either z = lorz= — ^-(§ 52).
The desired quadratic is therefore
z2 — 1- x — J- =0, or 2 x2 -x — 1 =0. Ans.
XII, §70] GENERAL PROPERTIES OF QUADRATICS 115
Similarly, if the given solutions are any numbers, as a
and 6, the equation having these as solutions is obtained by
subtracting a from x and 6 from x, then multiplying the two
factors thus obtained together and placing the result equal
to 0 ; that is, the desired equation is (x — a)(x — b) = 0»
EXERCISES
Form the quadratics whose roots are
1. 2, 3. 8.
2. -2, -3. 9. 3m, -5m.
3. 4, f 10. 2a-6, 2 a+b.
4. 12, -5. 11. 3+V2, 3-V2.
5. i, -£. 12. 2-V5, 2+ V5.
6. V2, V3. 13. 2±V3.
[HINT. Write answer in the 14. — \ (3 ± V§) .
form *2-(V2 + V3)z + V6 =0.] lg i(_i±V2).
7. V8, -\/2. 16. m(2±2V5).
17. Show that if in any quadratic, as az2+6z+c = 0, one
root is double the other, then the relation 2 b2 = 9 ac must
exist among the coefficients, a, 6, and c.
[HINT. Let r be one root. Then, by what the problem assumes,
the other root is 2 r. Now form the quadratic having r and 2 r as
roots, and examine its coefficients.]
18. Show that if in any quadratic, as ax2 -\-bx-\- c = 0, one
root is three times the other, then we must have 4 ac = 62 — 9 a2.
19. Find the relation which must hold between a, b, and c
in order that the roots of the quadratic ax2+bx+c = Q may
be to each other in the ratio 2:3.
[HINT. Use the Rule of § 69.]
CHAPTER XIII
IMAGINARY NUMBERS
71. Preliminary Statement. Just as V2 means the num-
ber whose square is 2 ; that is, (V2)2 = 2, so V— 2 means the
number whose square is —2; that is, (V — 2)2= —2. The
latter case differs essentially, however, from the former, be-
cause we cannot conceive of any number, positive or nega-
tive, whose square gives a negative result, like —2. In fact,
this would seem to contradict the law of signs [§ 2 (d)],
according to which the square of either a positive or negative
quantity is always positive. The explanation is that V— 2
belongs to an altogether new class of numbers, called im-
aginary numbers, so that we cannot expect to think of them
in the way just mentioned ; namely, as though they were real
numbers. Imaginary numbers first came to our notice in
§ 58, where we found that the very simple quadratic x2 = — 1
has the two imaginary roots x = V— 1 and x— — V— 1.
Imaginary numbers have certain definite properties.
They may be added, subtracted, multiplied, divided, etc.,
in ways which will be explained in the present chapter.
72. The Imaginary Unit. Every pure imaginary number
(§ 66) may be expressed as the product of a certain real num-
ber multiplied by V— 1. For this reason, V— 1 is called
116
XIII, § 72]
IMAGINARY NUMBERS
117
the imaginary unit, and for convenience is represented by
the letter i.
Thus V^
V27( - 1) = V27 V^I =
A pure imaginary number when thus written as a real
number multiplied by i is said to be expressed in terms of i.
EXERCISES
Write each of the following expressions in terms of i.
5. V^
1. V-IQ.
2. V-25.
3. V— 18.
SOLUTION OF Ex. 9.
10. v^J".
11. V^T.
12.
13. -
14.
SOLUTION OF Ex. 14.
ll^ J?7 V^
4 ' 4
i.
Ans.
16. =
118
SECOND COURSE IN ALGEBRA [XIII, § 73
73. Addition and Subtraction of Pure Imaginary Num-
bers.
EXAMPLE. Add V^9 and V-25 and express the result
in terms of i.
SOLUTION.
EXERCISES
Simplify each of the following expressions, obtaining the
result in terms of i.
-81 + -64+-100.
-16a2+ ~-100a2- V-81a2.
9. V - 25 zy + v - 225 x*y2+ V - 625 x2y2-
10. V-32m2-hV-20m2-V-27m2.
11. V-24 /i2/
12. V
74. Simplification of Complex Numbers.
EXAMPLE. Simplify 2+V-9.
SOLUTION. 24-^^9=2+3^. ^Ins.
In this exercise we have a real number, 2, to which is added the
pure imaginary number, V — 9, thus giving a complex number ( § 59).
EXERCISES
Simplify each of the following complex numbers.
1.
2. 6-2V
3-
XIII, §75] IMAGINARY NUMBERS 119
7. Show that the quadratic equation x2 — 4z+13 = 0 has
as its solutions x = 2 ± 3 i.
[HINT. Solve as in § 56.]
8. Find the solutions of the equation 4(2 x — 5) =x2.
75. Multiplication of Imaginary Numbers.
EXAMPLE 1. Multiply V^3 by V^4.
SOLUTION. V^3 • V^4 = V$i • V±i = Vl2- i*= Vl2(V-l)2
= -Vl2=-2V3. Ans.
Note that the process of multiplication consists in first
expressing each number in terms of the unit i, then making
use of the fact that (since i = V — 1) we may write — 1 for i2.
EXAMPLE 2. Multiply 2+^3 by 4-~^3.
SOLUTION. (2 + V^3) (4 - V~^3) = (2 + Vs i) (4 - VJTt).
8+4 V3i
8+2 V3i-3(-l)= 11+2 V3i. Ans.
EXERCISES
Find each of the following indicated products.
9.
10.
11.
12.
13.
14.
15. a
16.
120 SECOND COURSE IN ALGEBRA [XIII, § 76
76. Division of Imaginary Numbers.
EXAMPLE 1. Find the value of V^27-r- V^3.
so— -- --- -*
EXAMPLE 2. Find the value of 6 -^V^3.
SOLUTION. _^=-^_ =_^_ =^_=^M= _2Vs i.
^ V V2 -V
EXAMPLE 3. Find the value of 3 -5- (1 + V^~2) .
SOLUTION. —2 --- L— - 3(1 -V2i)
. ^
l-2i2 1+2
Note that in Example 2 we rationalized the denominator
of - %— by multiplying both numerator and denominator
-Va
by the value A/3. Likewise, in Example 3 we rationalized
Q
the denominator of - — — - by multiplying both numerator
and denominator by 1 — V2t. In general, any fraction hav-
ing a denominator of the form a-\- v — b may be rationalized
by multiplying both numerator and denominator by a— V — 6,
which is called the conjugate imaginary of a+ V— 6.
EXERCISES
Find the following indicated quotients, rationalizing the
denominator in each.
4. 2-i-V^6.
5. V% -r- V^2.
IMAGINARY NUMBERS
121
11.
*77. Geometric Representation of Complex Numbers. All real
numbers (positive or negative) can be represented as points on a line,
as explained in § 1. Similarly, all complex numbers may be repre-
sented as points in a plane, and it is convenient for many purposes
to regard them in this way. Thus,
the complex number 5 +4 i may be
looked upon as lying at the point
(5, 4), that is at the point whose
abscissa is 5 and whose ordinate is 4.
(See § 28.) Likewise, the complex
number -2+3 i lies at (-2, 3) ; the
number 3 — 2 i lies at (3, —2) ; and, in
general, the number x+yi, where x
and y are any (real) numbers, lies at
the point (x, y). Whenever a plane
is used in this way to represent com-
plex numbers, it is called a complex
plane. The re-axis is called the axis
.
Y
&
1
e
f
X4
y
8
3
1
5-
-4i
-±5
i
\
.Q-
/
/
/
,/
<
/
/
\
I/
Ax
S 0
f r«
als
0
^
Sw
X
^
3-2i
FIG. 39.
of reals and the ?/-axis is called the axis of pure imaginaries (because
the pure imaginary part of each complex number is to be measured
parallel to it). The straight line joining any point x+iy to the
origin is called the radius vector of that point, or number.
Observe that from this point of view all real numbers become
represented by the points on the axis of reals, while all pure imaginary
numbers become represented by the points on the axis of pure
imaginaries, the other points of the plane being then taken up by
what are properly the complex numbers.
CHAPTER XIV
SIMULTANEOUS QUADRATIC EQUATIONS
I. ONE EQUATION LINEAR AND THE OTHER QUADRATIC
78. Graphical Solution. In Chapter VI we have seen how
we may determine graphically the solution of two linear
equations e&ch of which contains the two unknown letters
x, y. The method consists in first drawing the graph of each
equation, then observing the x and the y of the point where
the two graphs intersect (cut each other). The particular
pair of values (x, y) thus obtained constitutes the solution.
We often meet with simultaneous equations which are not
both linear, as for example the two equations
(1) x-y = l,
(2) z2+2/2 = 25.
Here the first equation is linear (§ 26) but the second is not.
In order to solve them, that is in order to find that pair (or
pairs) of values of x and y which satisfy both equations, we
may proceed graphically as follows.
The graph of (1) is found (as in § 29) to be the straight
line shown in Fig. 40.
To draw the graph of (2), we first solve this equation for y.
Thus y2 = 25 - x2. Therefore
(3) 2/=±
122
XIV, § 78] SIMULTANEOUS QUADRATICS
123
We now work out a table of pairs values of x and y that will
satisfy (3), that is we give x various values .in (3) and solve
for the corresponding y value. (Compare § 60.) The result
is shown below. Observe that to the value x = 0 there cor-
Whenz =
0
+ 1
+2
+3
+4
+5
then y =
±V25
±5
W24
±4.8
±V21
±4.5
±VlQ
±4
±V9
±3
± Vo
0
respond the £wo values T/ = ± 5 ; similarly to x = 1 correspond
the two values y= ±4.8 (approximately), etc.
Moreover, if we assign to x the negative value, x=—l,
we find in the same way that corresponding to it y has the
two values y = ±4.8. Likewise, for x = — 2 we find y = ±4.5,
etc., the values of y for any negative value of x being the
same each time as for the corresponding positive value of x.
FIG. 40.
Plotting all the points (x, y) thus obtained and drawing
the smooth curve through them, the graph is as shown in
Fig. 40. This curve is a circle, as appears more and more
clearly as we plot more and more points (x, y) belonging to
the equation (3).
124 SECOND COURSE IN ALGEBRA [XIV, § 78
NOTE. The form of (3) shows that there can be no points in the
graph having x values greater than 5, for as soon as x exceeds 5 the
expression 25 — x2 becomes negative and hence V25 — x2 becomes
imaginary, and there is no point that we can plot corresponding to
such a result. Similarly, it appears from (3) that x cannot take
values less than —5.
Thus the graph can contain no points lying outside the circle
already drawn.
Returning now to the problem of solving (1) and (2), we
know (§31) that wherever the one graph cuts the other we
shall have a point whose x and y form a solution of (1) and
(2), that is we shall have a pair of values (x, y) that will
satisfy both equations at once. From the figure it appears
that there are in the present case two such points, namely
(z = 4, 2/ = 3) and (x=-3, y=-4). Equations (1) and (2)
therefore have the two solutions (z = 4, 2/ = 3) and (x=— 3,
y—— 4). Ans.
CHECK. For the solution (x = 4, y=3) we have x— y=4 — 3 = 1,
and x2 +yz = 16 +9 = 25, as required.
For the solution (x = —3, y = —4) we have x -y = -3 - ( -4) = 1,
and z2 +?/2= 9 +16 = 25, as required.
The following are other examples of the graphical study of
non-linear simultaneous equations.
EXAMPLE 1. Solve the system
(4)
(5)
SOLUTION. The straight line representing the graph of (4) is
drawn readily.
To obtain the graph of (5), we have 9 y2 = 100 —4 x2. Hence
y2 =£(100 -4 z2) =|(25 -x2),
and therefore
(6)
XIV, § 78] SIMULTANEOUS QUADRATICS
Corresponding to (6), we find the following table :
125
When z=
0
+ 1
+2
+3
+4
+5
greater than
+5
then y=
ifV/25
±§(5)
±3.3
±3^/24
±1(4.8)
±3.2
±|V/21
±§(4.5)
±3.0
±|X/16
±§(4)
±2.6
±|V^
±§(3)
±2
±1^0
±0
0
imaginary
imaginary
imaginary
For any negative value of x, the y- values are the same as for the
corresponding positive value of x. (See the solution of (1 ) and (2) .)
The graph thus obtained for (6), or (5), is an oval shaped curve.
It belongs to a general class of curves called ellipses.
\
FIG. 41.
The two graphs are seen to intersect at the points
(x = 4, y =2) and (x =-5,y = 0).
Therefore the desired solutions of (4) and (5) are (x=4, y=2)
and (x= — 5, 2/=0). Ans.
• EXAMPLE 2. Solve the system
(7) 2x-y=-2,
(8) Z2/ = 4.
SOLUTION. The graph of (7) is the straight line shown in Fig. 42.
To obtain the graph of (8), we have
(9) y=±,
126
SECOND COURSE IN ALGEBRA [XIV, § 78
from which we obtain the following table :
When x =
8
7
6
5
T
4
3
t
2
1
*
i
ST
i
T
16
i
5"
then y =
i
*
t
1
2
4
8
12
20
This table concerns only positive values of x, but it appears
from (9) that for any negative value of x the appropriate y value
is the negative of that for the corresponding positive value of x.
The graph thus obtained for (9), or (8), consists of two open
curves, each indefinitely long, situated as in Fig. 42. These taken
together (that is, regarded as one curve) form what is known as a
hyperbola (pronounced hy-per'bo-la). The part (branch) lying to
the right of the i/-axis corresponds to the above table, while the
other branch corresponds to the negative x values.
/\
FIG. 42.
The two graphs are seen to intersect in the points (a? = l, y=4)
and (x= -2, y= -2).
Therefore the desired solutions of (7) and (8) are (x = l, j/=4)
and (x = —2, y = —2). Ans.
NOTE. Ellipses and hyperbolas are extensively considered in
the branch of mathematics called analytic geometry — a study which
may be pursued after a course in plane geometry and a course in
algebra equivalent to that in this book. It is usually taught during
the first year of college mathematics.
XIV, § 78] SIMULTANEOUS QUADRATICS
127
EXAMPLE 3. Consider graphically the system
(10)
(11)
SOLUTION. The graph of (10) is found in the usual manner, and
is represented by the straight line in Fig. 43. The graph of (11)
has already been worked out (see discussion of (2)), being a circle
of radius 5 with center at the origin. The peculiarity to be es-
pecially observed here is that these two graphs do not intersect.
This means (as it naturally must) that there are no real solutions
to the system (10) and (11) ; in other words, the only possible solu-
tions are imaginary.
FIG. 43.
Likewise, whenever any two graphs fail to intersect, we may be
assured at once that the only solutions their equations can have
are imaginary. The system (10) and (11) and other such systems
will be considered further in the next article.
EXERCISES
Draw the graphs for the following systems and use your
result to determine the solutions whenever they are real.
4 z2+9 i/2 = 100.
[HINT. See Example 1, § 78.
128 SECOND COURSE IN ALGEBRA [XIV, § 78
3. J*-20=-l, 6
\y = x*.
[HINT. See Example 2, § 78.]
O.
5~ q „, n
Jj — o u — u.
79. Solution by Elimination. Let us consider again the
system (1) and (2) of § 78.
(1) x-y=l,
Instead of solving this system graphically, we may solve
it by elimination, that is by the process employed with two
linear equations in § 33.
Thus we have from (1)
(3) y = x-l.
Substituting this value of y in (2), thus eliminating y from
(2), we obtain
or, dividing through by 2,
(4) z2-z-12 = 0.
Solving (4) by formula (§ 56), gives as the two roots
1+7
= 4,
and
(-l)-V(-l)«-4(l)(-12)_l-Vl+48 = l-7^_q
2 22
When x has the first of these values, namely 4, we see
from (3) that y must have the value ?/ = 4— 1, or 3.
XIV, § 79] SIMULTANEOUS QUADRATICS 129
Similarly, when x takes on its other value, namely — 3, we
see that y has the value y= — 3 — 1, or —4.
The solutions of the system (1) and (2) are, therefore,
(^ = 4, ^ = 3) and (x= -3, y= -4). Ans.
Observe that these results agree with those obtained
graphically for (1) and (2) in § 78.
Further applications of this method are made in the
examples that follow.
EXAMPLE 1. Solve the system
(5) 2x-\-y = 4,
(6) z2+2/2 = 12.
SOLUTION. From (5),
(7) y=±-2x.
Substituting this expression for y in (6), we find
or
(8) 5z2-16z+4 = 0.
The two roots of (8), as determined by formula (§ 56), are
= - ( - 16) =fc V ( - 16)2 -4(5) (4) = 16 ± V256 -80= 16 =*= vT76
2(5) 10 10
_16^4VII^8=>=2Vn
10 5
The first of these values, namely x= (8+2vTF)/5, when sub-
stituted in (7), gives as its corresponding value of y,
j,.4_ia+4vH_4-|V]
The second value, namely x = (8— 2VTT)/5, when substituted
in (7), gives as its corresponding value of y,
16-4VT1 4+4VTT
,
5
Hence the desired solutions are
x 8+2VTT
8-2VTI
5 and
4-4VH
5 '
4+4^11
y 5 '
y" 5
130 SECOND COURSE IN ALGEBRA [XIV, § 79
To obtain the approximate values of the numbers thus obtained
we have VTI =3.31662 (tables), and hence the above solutions re-
duce to the forms
x= 2.9266, (x =0.2734,
y=- 1.8533, \y = 3.4533.
These are the solutions of the system (5), (6), correct to four places
of decimals, which is sufficient for ordinary work.
NOTE. It may be remarked that, while the graphical method of
solution described in § 78 is very instructive in showing how many
solutions a given system will have, and what their geometric
significance is, it does not usually afford a ready means of deter-
mining the exact values of the solutions. This is illustrated in the
example just solved, where, if the graphs of (5) and (6) be drawn,
they will intersect at points whose x and y contain the surd Vll
(as the above solution shows), and it would be difficult to measure
off any such values accurately on the scale of the diagram. In
fact, it would be practically impossible to determine graphically
the solutions of (5) and (6) correct to three or even two places of
decimals, yet this degree of approximation was easily obtained above
by the method of elimination. For such reasons, it is preferable,
whenever one is concerned only with finding the values of solutions,
to proceed from the beginning by the method of elimination.
EXAMPLE 2. Solve the system
(9)
(10)
SOLUTION. From (9),
(11) ^ = 10-z.
Substituting this expression in (10),
x2 + (100 -20 z +z2) = 25,
or
(12) 2 x2- 20x+75=0.
Solving (12) by formula, we find its solutions to be
and
Since these z-values contain the square root of the negative
quantity -200, they are imaginary (§58). The ^-values are also
XIV, § 79] SIMULTANEOUS QUADRATICS
131
imaginary, as appears by substituting the z-values just found into
(9), which gives the results
The desired solutions of the systems (9), (10) are therefore
x=-
10
-=- and
10
10
y =
10
This result should now be contrasted with what we saw in
Example 3 of § 78 regarding this same system (9) and (10). There
we found graphically that the solutions must be imaginary be-
cause the graphs failed to intersect, but we could not find the actual
imaginary numbers which form the solutions. This we have now
been able to do, however, by the method of elimination. The
method of elimination enables one to determine imaginary as well
as real solutions in all similar cases.
EXERCISES
Solve each of the following systems by the method of elimi-
nation, and, in case surds are present, find each solution
correct to two places of decimals by use of the tables.
f x+y=-l,
1.
2.
f x-2y=-l,
6.
7.
f x-2y = 2,
O« i <> * A 9 Of
8 i
9.
x-2y = 3.
*-2y*=-S,
x-2y=-3.
f 3o;2-a-5 =
15'
xy=-Q.
2x+y = 2,
xy=-Q.
f
10. I x-\-y x — y 6'
132
SECOND COURSE IN ALGEBRA [XIV, § 80
II. NEITHER EQUATION LINEAR
*80. Two Quadratic Equations. In each of the systems
considered in §§ 78, 79 one of the two given equations was
linear. However, the same methods of solving may often
be employed in case neither equation is linear. In such cases
four solutions may be present instead of two.
EXAMPLE 1. Solve the system
(1)
(2)
z2-?2=15.
SOLUTION. Here only xz and y2 appear and we begin by finding
their values. Thus, multiplying (2) through by 16 and adding the
result to (1), we eliminate y2 and find that 25 z2 =400, or
(3) z2 = 16.
Substituting this value of x in (2), we find
(4)
From (3) and (4) we now obtain
(5)
x=
and y ==*=!.
Forming all the pairs of values
x, y that can come from (5), we
obtain as our desired solutions
(x = 4, y = — 1) ; and
(x= — 4, y= — 1). Ans.
CHECK. Each of these pairs of
values of x and y is immediately
seen to satisfy both (1) and (2).
Let the pupil thus check each pair.
When considered graphically,
equation (1) gives rise to an
ellipse (compare § 78, Ex. 1), while
(2) gives a hyperbola situated as shown in the diagram. These
two curves intersect in four points which correspond to the four
solutions just obtained.
FIG. 44.
XIV, § 80] SIMULTANEOUS QUADRATICS
133
EXAMPLE 2. Solve the system
-l2.
(7)
(8) xy
SOLUTION. Here we cannot proceed as in Example 1 because
we cannot find readily the values of x2 and y*. But if we multiply
(8) by 2 and add the result to (7), we obtain
(9) x2 +2 xy +y* = l.
Taking the square root of both members of (9) gives
(10)
Similarly, multiplying (8)
by 2 and subtracting the result
from (7),
x2 - 2 xy +?/2 = 49,
and hence
(11) X-y=±7.
Taking account of the two
choices of sign in (10) and
(11), we see that they give rise
to the four simple (linear)
systems
(a) x+y = l, x-y=7;
(6) x+y=-l, x-y=7;
(c) x+y = l, x-y=-7;
(d) x+y=-l,x-y=-7',
Thus we have replaced the original system (7) and (8) by the
four simple systems (a), (6), (c), and (d), each of which may be
immediately solved by elimination, as in § § 33, 34. Since the solu-
tions of (a), (6), (c), (d) are respectively (x=4, y= -3), (x=3,
y = —4), (x = -3, y =4), and (x = —4, y =3), we conclude that these
are the desired solutions of (7) and (8). Ans.
The graphical significance of these solutions is shown in Fig. 45,
where the circle z2+?/2 = 25 is cut by the hyperbola xy= — 12 in
four points that correspond to the four solutions just found.
CHECK. That these four solutions each satisfy (7) and (8)
appears at once by trial.
FIG. 45.
134 SECOND COURSE IN ALGEBRA [XIV, § 80
While no general rule can be stated for solving two equations
neither of which is linear, the following observation may be
made. Unless the equations can be solved readily for x2
and y2 (as in Example 1), the system should first be carefully
examined with a view to making such combinations of the
given equations as will yield one or more new systems each
of which can be solved (as in Example 2) by methods already
familiar. All solutions obtained in this way should be
checked, since false combinations of the x- and ^-values are
frequently made by beginners when the work becomes at all
complicated.
EXERCISES
Solve each of the following systems, and draw a diagram
for each of the first three to show the geometric meaning of
your solutions.
= 25, f
= 7. \
f
[HINT FOR Ex. 4. First add, then subtract the two equations,
thus showing that the given system is equivalent to two others,
namely
Now solve each of these systems as in § 80.]
z2+92/2 = 85,
,
( x-
y+2xy=-2Q.
CARDAN
(Girolamo Cardan, 1501-1576)
An equation of the third degree (cubic equation) has three roots and
these can be found only by methods which are more powerful than those
employed in the study of quadratics. Cardan was the first to obtain and
publish a method for solving such equations. His methods are also sufficient
to solve any pair of simultaneous quadratics, but are too advanced to be
given in this book.
XIV, § 81J SIMULTANEOUS QUADRATICS 135
*81. Systems Having Special Forms. The systems of equations
considered in §^ 79, 80 illustrate the usual and more simple types
such as one commonly meets in practice. It is possible, however,
to solve more complicated systems provided they are of certain
prescribed forms. We shall here consider only two such type forms.
I. When one (or both) of the given equations is of the form
where the coefficients o, 6, c are such that the expression ax2+bxy+cy2
can be factored into two linear factors.
EXAMPLE. Solve the system
(1) x*+2x-y = 7,
(2) x2-xy-2y*=0.
SOLUTION. Here we see that (2) is of the form mentioned above,
since x2—xy—2 yz can be factored (as in §12 (e)} into (x— 2 y)(x+y}.
(2) may thus be written in the form
(3) (x-2y)(x+y)=Q.
It follows (§ 52) that either
x-2y=Q, or x+y=Q.
Hence the system (1), (2) may be replaced by the two following
systems :
and
fz2+2o;-2/ = 7,
I x+y=0.
Each of these two systems may now be solved as in § 80, and we
thus find that the solutions of the first system are
(x = 2, y = l) and (x=-%,y=— J)
while the solutions of the second system are
and
_
y=i(3+V37).
The desired solutions of (1) and (2) consist, therefore, of these four
solutions just obtained. Ans.
136 SECOND COURSE IN ALGEBRA [XIV, § 81
II. When both the given equations are of the form
axz+bxy+cy*=d,
where a, 6, c, and d have any given values (0 included).
EXAMPLE. Solve the system
(4) x*-xy+y* = 3,
(5) x*+2xy=5.
SOLUTION. Let v stand for the ratio x/y ; that is, let us set
y
Then
(6) x=vy.
Substituting in (4),
(7) 0V
Substituting in (5),
(8)
Solving (7) for y*,
(9)
Solving (8) for y2,
(10) ?/2
Equating the values of yz given by (9) and (10),
v2+2v v2-
Clearing of fractions,
(11) 2vz-llv+5 =
Solving (11) by formula (§56),
^H=faVl21-40
4 44
Therefore v = 5, or v =%. Substituting 5 for v in (9), or (10),
Hence
= i or L
Vf C V7
Substituting ^ for v in (9) or (10), yz =4. Hence y = +2, or -2.
XIV, § 82] SIMULTANEOUS QUADRATICS 137
The only values that y can have are, therefore, 1/V?, -1/V?,
2, and -2.
Since x_=vy (see (6)), the value of x to go with y = l/V7 is
x=5(l/v/7)=5/V7. _Similarly, when y= -1/V? we have
x =5( — I/VT) = — 5/V?. Likewise, when ?/=2 (in which case
v = -^, as shown on p. 136) then z =^ • 2 = 1, and when y — — 2, then
z=i(-2) = -l.
Therefore the only solutions which the system (4), (5) can have
are(x=5/V7, y = l/V7) ; (x = -5/S/7, »= -1/V?) ; (x = 1, y = 2) ;
OT=— 1, i/=— 2); and it is easily seen by checking that each of these
is a solution. Ans.
82. Conclusion. Every system of equations considered
in this chapter has been such that we could solve it by finally
solving one or more simple quadratic equations. We have
examined only special types, however, and the student should
not conclude that all pairs of simultaneous quadratics can be
solved so simply.
MISCELLANEOUS EXERCISES
Solve the following simultaneous quadratics. The star
(*) indicates that the exercise depends upon § 81.
3.
* y
2-- =
[HINT TO Ex. 4. First eliminate xy between the two equa-
tions so as to obtain a linear equation between x and y.]
z2+2z-2/ = 5, I
2 z2-3 x+2 y = S. \
,
lx2-i/2 = 9. i
[HINT TO Ex. 6. Divide the first equation by the second.]
138
SECOND COURSE IN ALGEBRA - [XIV, § 82
10
xy-y=l2.
:15.
APPLIED PROBLEMS
In working the following problems, let x and y represent the two
unknown quantities, then form two simultaneous equations and solve
them. If surds occur, find their approximate values by the tables.
1. The sum of two numbers is 13 and the difference of
their squares is 91. Find the numbers.
2. A piece of wire 48 inches long is bent into the form
of a right triangle whose hypotenuse is 20 inches long. What
are the lengths of the sides? (See Ex. 14 (d), p. 6.)
3. If it takes 26 rods of fence to inclose a rectangular garden
containing ^ of an acre, what are the length and breadth ?
4. Figure 46 shows two circles just
touching (tangent to) each other, the
smaller one being outside the larger one.
If their combined area is 15f square feet
and the distance CC' between the two
centers is 3 feet, find the radius of each
circle.
FIG. 46.
5. Work Ex. 4 in case the circles touch
on the inside of the larger one, taking the
shaded area to be 110 square feet and
CC' to be 5 feet.
FIG. 47.
XIV, § 82] SIMULTANEOUS QUADRATICS 139
6. Do positive integers exist differing by 5 and such that
the difference of their squares is 45 ? If so, find them.
7. Answer the question in Ex. 6 in case the differ-
ence of the squares is taken to be 10, other conditions re-
maining the same.
8. The area of a certain triangle is 160 square feet, and
its altitude is twice as long as its base. Find, correct to
three decimal places, the base and the altitude. (See Ex.
14 (a), p. 6.)
9. The area of a rectangular lot is 2400 square feet, and
the diagonal across it measures 100 feet. Find, correct to
three decimal places, the length and breadth.
10. The mean proportional between two numbers is 2
and the sum of their squares is 10. What are the numbers ?
(See Ex. 6, page 80.)
11. The dimensions of a rectangle are 5 feet by 2 feet.
Find the amounts (correct to two decimal places) by which
each dimension must be changed, and how, in order that
both the area and the perimeter shall be doubled.
12. Two men working together can complete a piece of
work in 6 days. If it would take one man 5 days longer
than the other to do the work alone, in how many days
can each do it alone? (Compare Ex. 9, p. 56.)
13. The fore wheel of a carriage makes 28 revolutions
more than the rear wheel in going 560 yards, but if the
circumference of each wheel be increased by 2 feet, the
difference would be only 20 revolutions. What is the
circumference of each wheel?
14. A sum of money on interest for one year at a certain
rate brought $7.50 interest. If the rate had been 1% less
and the principal $25 more, the interest would have been the
same. Find the principal and the rate.
140 SECOND COURSE IN ALGEBRA [XIV, § 82
15. A man traveled 30 miles. If his rate had been 5
miles more per hour, he could have made the journey in 1
hour less time. Find his time and rate.
16. Figure 48 shows a circle within
which a diameter A B has been drawn.
At a certain point P on AB the perpen-
B dicular PG measures 4 inches, while at
the point Q, which is 1 unit from P,
the perpendicular QF measures 3 inches.
How long is the diameter AB1
[HINT. Let x = AP,y = PB. Find x and y, using the fact stated
in Ex. 6, page 80, then take their sum.]
17. Show that the formulas for the length I and the
width w of the rectangle whose perimeter is a and whose
area is b are
18. Find the formulas for the radii of two circles in order
that the difference of the areas of the circles shall be d and
the sum of their circumferences shall be s.
and fi..«?=i£i Ans.
4 7TS 4 TTS
19. Find two fractions whose sum is f , and whose differ-
ence is equal to their product.
20. The diagonal and the longer side of a rectangle are
together five times the shorter side, and the longer side
exceeds the shorter side by 35 yards. What is the area of
the rectangle?
CHAPTER XV
PROGRESSIONS
I. ARITHMETIC PROGRESSION
83. Definition. An arithmetic progression is a sequence
of numbers, called terms, each of which is derived from the
preceding by adding to it a fixed amount, called the common
difference. An arithmetic progression is denoted by the
abbreviation A. P.
Thus 1, 3, 5, 7, ••• is an A. P. Each term is derived from the
preceding by adding 2, which is therefore the common difference.
The dots indicate that the sequence may be extended as far as one
pleases. Thus the first term after 7 would be 9, the next one would
be 11, etc.
Again, 5, 1, —3, — 7, — 11, ••• is an A. P. Here the common
difference is —4.
EXERCISES
Determine which of the following are arithmetic progres-
sions ; determine the common difference and the next two
terms of each of the arithmetic progressions.
1. 3, 6, 9, 12, ••-. 4. 30, 25, 20, 15, 10, •••.
2. 3, 5, 8, 12, .... 6. -1, -1^, -2, -2£, •••.
3. 6, 4, 2, 0, -2, -4, •••. 6. a, 2 a, 4 a, 5 a, •••.
7. a, a+3, a+6, a+9, •••.
8. a, a-\-d, a+2 d, a+3 d, a+4 d, •••.
9. x-4y, x-2y, x-y, •••.
10. 3x+3y, Qx+2y, 9 x+y, •••.
[HINT. The common difference may always be determined by
subtracting any term from the term immediately preceding.]
141
142 SECOND COURSE IN ALGEBRA [XV, § 83
11. Write the first five, terms of the A. P. in which
(a) The first term is 5 and the common difference is 2.
(6) The first term is —3 and the common difference 1.
(c) The first term is 3 a and the common difference is — b.
84. The nth Term of an Arithmetic Progression. From
the definition (§ 83) it follows that every arithmetic progres-
sion is of the form a, a+d, a+2 d, a+3 d, a+4 d, •••. Here
a is the first term and d the common difference.
Observe that the coefficient of d in any one term is 1 less
than the number of that term. Thus 2 is the coefficient of
d in the third term; 3 is the coefficient of d in the fourth
term, etc. Therefore the coefficient of d in the nth term
must be (n— 1). Hence, if we let I stand for the nth term,
we have the formula
EXAMPLE. Find the llth term of the A. P. 1, 3, 5, 7, • • •.
SOLUTION. We have a = l, d =2, n = ll, 1= ?
The formula gives Z = a + (n-l)d = l+10X2 = 1+20 = 21. Ans.
EXERCISES
1. Find the llth term of 3, 6, 9, 12, •••.
2. Find the 13th term of 6, 10, 14, 18, •••.
3. Find the 20th term of 4, 2, 0, -2, -4, •••.
%4. Find the 15th term of -1, -1£, -2, -2J, •••.
5. Find the 10th term of x-y, 2 x — 2 y, 3 x-3 y, • • •.
6. When a small heavy body (like a bullet) drops to the
ground it passes over 16.1 feet the first second, 3 times as far
the second second, 5 times as far the third second, etc. How
far does it go in the 12th second?
7. If you save 5 cents during the first week in January,
10 cents the second week, 15 cents the third week and so on,
how much will you save during the last week of the year?
XV, § 85] PROGRESSIONS 143
8. What term of the progression 2, 6, 10, 14, ••• is equal
to 98?
[HINT, a = 2, d=4, n=?, /=98.]
9. What term of 3, 7, 11, 15, ••• is equal to 59?
10. The first term of an A. P. is 8 and the 14th term is 47.
What is the common difference?
85. The Sum of the First n Terms of an Arithmetic Pro-
gression. Let a be the first term of an A. P., d the common
difference, I the nth term. Then the sum of the first n
terms, which we will call S, is
(1) s = a+(a+d) + (a+2d) + (a+Zd)+''.+ (l-d)+l.
This value for S may be written in a very much simpler
form, as we shall now show.
Write the terms of (1) in their reverse order. This gives
(2) S = l+(l-d) + (l-2d) + (l-3d)+---+(a+d)+a.
Now add (1) and (2), noting the cancellation of d with —dt
2 d with -2 d, etc. The result is
or 2S = n(a+l).
Therefore S = -(a+/).
2
This is the simple form for S mentioned above. If we re-
place I by its value a+(n—l)d (§84), this result takes the
form
2
EXAMPLE. Find the sum of the first 12 terms of the A. P.
2,6, 10, 14, —.
SOLUTION, a = 2, d=4, n = 12.
Therefore, by the second form for S in § 85, we have
4} = 6(4+44} = 6X48 =288. Ans.
144 SECOND COURSE IN ALGEBRA [XV, § 85
EXERCISES
Find the sum of each of the following arithmetic progres-
sions.
1. The first ten terms of 3, 6, 9, 12, •••.
2. The first fifteen terms of -2, 0, 2, 4, •••.
3. The first thirteen terms of 1, 3^, 6, ••-.
4. The first ten terms of 1, -1, -3, -5, •••.
5. The first n terms of 1, 8, 15, •••.
6. How many strokes does a common clock, striking
hours, make in 12 hours?
7. A body falls 16.1 feet the first second, 3 times as far
the second second, 5 times as far the third second, etc. How
far does it fall during the first 12 seconds?
8. Find the sum of all odd integers, beginning with 1 and
ending with 99.
9. If you save 5 cents during the first week in January,
10 cents during the second week, 15 cents the third week,
and so on, how much will you save in a year?
10. The first term of an A. P. is 4 and the 10th term is 31.
What is the sum of the 10 terms?
[HINT, a = 4, ft = 10, Z=31. Now use the first of the formulas
in § 85.]
11. The first term of an A. P. is £ and the 12th term is
11-J-. What is the sum of the 12 terms?
12. Figure 49 shows a series of 16
dotted lines which are equally distant
from each other. If the highest one is
6 inches long and the lowest one is 3 feet
long, what is the sum of all their lengths? Fl°- 49-
[HINT. The lines form an' A. P. since their lengths increase uni-
formly.]
XV, § 85]
PROGRESSIONS
145
13. The rungs of a ladder diminish uniformly from 2 feet
4 inches long at the base to 1 foot 3 inches long. at the top.
If there are 24 rungs, what is the total length of wood in
them?
14. Find the sum of the circumferences
of 10 concentric circles if the radius of the
innermost one is ^ inch and the radius of
the outermost one is 4 inches, it being under-
stood that the circles are equally spaced from
each other.
FIG. 50.
15. Figure 51 shows a coil of rope in the
ordinary circular form, containing 12 com-
plete turns, or layers. If the ength of the
innermost turn is 4 inches and the length of
the outermost turn is 37 inches, how long is
the rope?
FIG. 51.
[HINT. Regard each turn as a circle, thus neglecting the slight
effect due to the overlapping at the beginning of each turn after the
first.]
16. If in Fig. 51 the length of the innermost turn is a
inches and that of the outermost turn is b
inches, and the number of turns is n, what
represents the total length of the rope?
17. A small rope is wound tightly round a
cone, the number of complete turns being 24.
Upon unwinding the rope from the top, the
lengths of the first and second turns are
found to be 2J inches and 3^ inches respec-
tively. How long (approximately) is the rope?
FIG. 62.
146 SECOND COURSE IN ALGEBRA [XV, § 86
86. Arithmetic Means. The terms of an arithmetic pro-
gression lying between any two given terms are called the
arithmetic means between those two terms.
Thus, the three arithmetic means between 1 and 9 are 3, 5, 7,
since 1, 3, 5, 7, 9 form an arithmetic progression.
Whenever a single term is inserted in this way between two
numbers, it is briefly called the arithmetic mean of those two
numbers.
Thus, the arithmetic mean of 2 and 10 is 6, because 2, 6, 10 form
an arithmetic' progression.
A formula for the arithmetic mean of any two numbers,
as a and b, is easily obtained. Thus, if x is the mean, then
a, x, b forms an A. P. Therefore, we must have x — a = b—x.
Solving for x, this gives
x_a±b
2
Thus we have the following theorem : The arithmetic
mean of two numbers is equal to half their sum.
NOTE. The arithmetic mean of two numbers is also called their
average.
EXAMPLE. Insert five arithmetic means between 3 and 33.
SOLUTION. We are to have an A. P. of 7 terms in which a =3,
I =33, and n =7. We begin by finding d. Thus,
Z = a + (n-l)d (§ 84) so that 33=3+6d. Solving, d=5.
The progression is therefore 3, 8, 13, 18, 23, 28, 33 and hence
the desired means are 8, 13, 18, 23, 28. Ans.
EXERCISES
1. Insert three arithmetic means between 7 and 23.
2. Insert four arithmetic means between —5 and 10.
3. Insert seven arithmetic means between ^ and 25f .
4. What is the arithmetic mean of 8 and 30 ?
6. What is the arithmetic mean of J and — -J-?
XV, § 86]
PROGRESSIONS
147
/
FIG. 53.
B
6. Show that the first formula for S obtained in § 85
may be stated as follows : " The sum of n terms of an arith-
metic progression is equal to n multiplied by the arithmetic
mean of the first and nth terms."
7. A BCD is any trapezoid (that is, any four-sided figure
having its bases A B and DC parallel to each other). The
line EF, called the median, joins the
middle point of the side AD to the
middle point of the side BC, and it is
shown in geometry that the length of ^
this line EF will always be the arith- A
metic mean of the lengths of the bases
AB and CD. Hence answer the following questions.
(a) If the bases are 10 inches long and 2 inches long,
respectively, what is the length of the median?
(b) If the lower base is 14 inches long and the median 8
inches long, -how long is the upper base?
(c) If the upper base is 3 feet long and the median 4 feet
long, how long is the lower base?
(d) If the bases are a inches long and b inches long, re-
spectively, what represents the length of the median?
8. The figure shows the frustum of a cone and the frus-
tum of a pyramid, and in each case the " mid-section " has
FIG. 54.
been drawn in (that is, the section made by a plane which
passes midway between the bases AB and BC). It is shown
in solid geometry that in all such cases the perimeter of the
148 SECOND COURSE IN ALGEBRA [XV, § 86
mid-section will always be the arithmetic mean of the
perimeters of the two bases. Hence answer the following
questions.
(a) If the perimeters of the two bases are 30 inches and 10
inches respectively, what is the perimeter of the mid-section ?
(6) In the frustum of a cone, the radius of the upper base
is 2 inches and that of the lower base 8 inches. What is the
perimeter of the mid-section?
87. The Five Elements of an Arithmetic Progression. In
any arithmetic progression there are the five elements, a, d,
I, n, Sj defined in §§ 84, 85. If any three of these are given,
we can always find the other two by means of the formulas
in §§ 84, 85.
EXAMPLE 1. Given a =-•§•, ft = 30, S = 2l±. Find d
and I.
SOLUTION. From § 85, we have S =(a +Z).
Hence 21i = 15(
Solving, I = lii. Now, I = a + (n - l)d. Hence 1£J = -i +29 d.
Solving, d=^.
EXAMPLE 2. Given a = 3, d = 4, £ = 300. Find n and I
SOLUTION. S=%{2 a + (n-l)d}. Hence 300=S6 + O-1)- 4}.
2 2i
Therefore
600 = n{4n+2}; 4 n2+2 n-600=0; 2n2+n-300=0.
Solving the last (quadratic) equation by formula (§56), gives
or _
•4 44
Since n is the number of terms and therefore a positive integer,
it follows that n = 12. (See Hint to Ex. 3, p. 89.)
To find Z, we now use the formula I = a + (n — l)d. Thus
1=3+11 -4=3+44 = 47.
XV, § 87] PROGRESSIONS 149
EXERCISES
1. Given a = 3, n = 25, £ = 675, find d and /.
2. Given a= -9, n = 23, Z = 57, find d and S.
3. Given £ = 275, 1 = 4:5, n = ll, find a and d.
4. Find w and d when a = - 5, I = 15, S = 105.
5. Find a and n when 1 = 1, d=%, £=-20.
6. How many terms are there in the arithmetic pro-
gression 2, 6, 10, —70?
7. Given a, I, and n, derive a formula for d.
8. Given a, d, and Z, derive a formula for n.
9. Given a, n, and S, derive a formula for Z.
10. Given d, I, and S, derive a formula for a.
11. Find an A. P. of 14 terms having 13 for its 6th term
and 25 for its 10th term.
12. Find an A. P. of 16 terms such that the sum of the 6th,
7th and 8th terms is — 16J-, and the sum of its last two terms
is -28.
13. Find three integers in arithmetic progression such that
their sum is 24 and their product 384.
14. The figure represents one of the four
sides of a steel tower such as is commonly seen
at wireless telegraph stations. It is desired to
make one of these towers so that each girder,
such as A B, will be 2 feet longer than the one
just above it, as CD. How many girders will
the tower have (counting all four sides) in case
the total amount of girder steel used is to be
only 864 feet and the lowest girders are each to
FIG. 55. haye a lenth of 20 feet?
150 SECOND COURSE IN ALGEBRA [XV, § 88
II. GEOMETRIC PROGRESSION
88. Definitions. A geometric progression is a sequence
of numbers, called terms, each of which is derived from the
preceding by multiplying it by a fixed amount, called the
common ratio. A geometric progression is denoted by the
abbreviation G. P.
Thus 2, 4, 8, 16, 32, • • • is a G. P. Each term is derived from the
preceding by multiplying it by 2, which is therefore the common
ratio.
Again, 10, —5, +-JJ-, •—•§-, ••• is a G. P. whose common ratio is — -g-
The next two terms are +f , -y5¥.
EXERCISES
Determine which of the following are geometric progres-
sions, and find the common ratio and the next two terms of
each geometric progression.
1. 3,6, 12,24,48, •••.
2. 4, 12, 48, 75,---.
3. *,i,i,A, •;•••
4. -1,2, -4,8, -16,"..
5. a, a2, a3, a4, ••-.
6. 2x, 4z3, 8z5, 16 z7, ••«.
7. a, ar, ar2, or3, or4, •••.
8. a, aV2, a3r4, a r6, •••.
9. (0+6), (a+6)3, (a+6)5, (o+&)7, -.
in m2 m4 ra6 m8
n3' n4' n^ n«J '
11. Write the first five terms of the G. P. in which
(a) The first term is 4 and the common ratio is 4.
(6) The first term is —3 and the common ratio is —2.
(c) The first term is a and the common ratio is r.
XV, § 89] PROGRESSIONS 151
89. The nth Term of a Geometric Progression. From the
definition in § 88 it follows that every geometric progression
is of the form
a, ar, ar2, ar3, ar4, ••-.
Here a is the first term, and r the common ratio.
Observe that the exponent of r in any one term is 1 less
than the number of that term. Thus 2 is the exponent of r
in the third term ; 3 is the exponent of r in the fourth term,
etc. Therefore, the exponent of r in the nth term must be
(n — l). Hence, if we let Z stand for the nth term, we have
the formula
EXAMPLE. Find the 7th term of the G. P. 6, 4, f,
SOLUTION. We have a =6, r=§, n=7, l=?
The formula gives I = or»-i = 6 X (|) ' =2 X3 x|° =| = Ans.
EXERCISES
1. Find the ninth term of 2, 4, 8, 16, •••.
2. Find the eighth term of ^, ^, 1, •••.
3. Find the ninth term of — 1, 2, — 4, 8, •••.
4. Find the tenth term of 4, 2, 1, -J, •••.
5. Find the eighth term of -*-, £, f ,
6. Find the eleventh term of ax, a?x2, a3z3, a4#4, •••.
7. Find the tenth term of 2, A/2, 1, ••-.
8. What term of the G. P. 3, 6, 12, 24 is equal to 384?
9. What term of the progression 6, 4, f is equal to |4 ?
10. For every person there has lived two parents, four
grandparents, eight great grandparents, etc. How many
ancestors does a person have belonging to the 7th genera-
tion before himself, assuming that there is no duplication?
Answer also for the 10th generation.
152 SECOND COURSE IN ALGEBRA [XV, § 89
11. If you save 50 cents during the first three months of
the year and double the amount of your savings every three
months afterward, how much will you save during the last
three months of the second year ?
12. From a grain of corn there grew a stalk that produced
an ear of 100 grains. These grains were planted and each
produced an ear of 100 grains. This was repeated until
there were 5 harvestings. If 75 ears of corn make a bushel,
how many bushels were there the fifth year?
90. The Sum of the First n Terms of a Geometric Progres-
sion. Let a be the first term of a geometric progression, r
the common ratio, I the nth term. Then the sum of the first
n terms, which we will call S} is
(1) /S = a+ar+ar2+arH ----- \-arn-2+arn~l.
This value for S may be written in a very much simpler
form, as we shall now show.
Multiply both members of (1) by r. This gives
(2) rS = ar+ar2+arH ----- \-arn~l-\-arn.
Now subtract equation (2) from equation (1), noting the
cancelation of terms. This gives
S-rS = a-arn.
Solving this equation for S gives
l-r
This is the simple form for S mentioned above.
It is to be observed also that since I = arn~l, we may write
rl = arn. Putting this value of arn into the form just found
for S, we obtain as a second expression for S the following
formula.
fl-r?
XV, § 90] PROGRESSIONS 153
EXAMPLE. Find the sum of the first six terms of the G. P.
3, 6, 9, 12, .-.
SOLUTION, a =3, r = 2, n=6. To find S.
<y a -ar*_3-3.26_3 -3-64 _3 -192 _ -189 _1pn ,
=T^7=:~r=2~= ^r~ ^r ^r=
EXERCISES
Find the sum of the first
1. Eight terms of 2, 4, 8, •••.
2. Six terms of 1, 5, 25, •••.
3. Five terms of 1, !£, 2£, — .
4. Six terms of 2, — f , f, •-.
5. Ten terms of -£, i, -£, •••.
6. Six terms of 1, 2 a, 4 a2, •••.
7. Ten terms of 1, a2, a4, ....
8. What is the sum of the series 3, 6, 12, ••-, 384?
9. What is the sum of the series 8, 4, 2, •••, -j^?
10. Find the sum of the first ten powers of 2.
11. Find the sum of the first seven powers of 3.
12. A series of five squares are drawn such that a side of
the second one is twice as long as a side of the first one, a
side of the third one is twice as long as a side of the second,
etc. If a side of the first one is 2 inches long, find (by § 90)
the sum of the areas of all the squares.
13. What is the' combined volume of five spheres if the
radius of the first one is 16 inches, the radius of the second
one is half that of the first one, the radius of the third one is
half that of the second one, and so on to the fifth one ? (See
Ex. 14 (e), p. 6.)
14. Half the air in a certain corked empty jug is removed
by each stroke of an air pump. What fraction of the original
volume of air has been removed by the end of the seventh
stroke ?
154 SECOND COURSE IN ALGEBRA [XV, § 90
HISTORICAL NOTE. It is related that when Sessa, the inventor
of chess, presented his game to Scheran, an Indian prince, the latter
asked him to name his reward. Sessa begged that the prince would
give him 1 grain of wheat for the first square of the chess board, 2
for the second, 4 for the third, 8 for the fourth, and so on to the sixty-
fourth. Delighted with the inventor's modesty, the prince ordered
his ministers to make immediate payment. The number of grains
of wheat thus called for was (see § 90)
1-1.2"=2*-1=2<U 1
1-2 1
But the value of 264 is the enormous number 18,446,744,073,709,
551,616, so the number of grains of wheat owing was but 1 less than
this. This amount is greater than the world's annual supply at pres-
ent. History does not relate how the claim was settled. (From
Godfrey and Siddons' Elementary Algebra, Vol. II, pp. 336, 337.)
91. Geometric Means. The terms of a geometric progres-
sion lying between any two given terms are called the geo-
metric means of those two terms.
Thus the three geometric means of 2 and 32 are 4, 8, 16, since
2, 4, 8, 16, 32 form a geometric progression.
Whenever a single term is inserted in this way between two
numbers, it is briefly called the geometric mean of those two
numbers.
Thus the geometric mean of 2 and 32 is 8, since 2, 8, 32 form a
geometric progression.
A formula for the geometric mean of any two numbers,
as a and b, is easily obtained. Thus, if x is the mean, then a,
x, b forms a G. P. Therefore we must have x/a = b/x. Solv-
ing, we have z2 = ab, and hence
x = Vab.
Thus, we have the following theorem : The geometric mean
of two numbers is equal to the square root of their product.
NOTE. The geometric mean of two numbers is thus the same as
their mean proportional. See Ex. 6, p. 80.
XV, § 91]
PROGRESSIONS
155
EXAMPLE. Insert four geometric means between 3 and 96.
SOLUTION. We are to have a G. P. of six terms in which a =3,
I = 96, and n = 6. We begin by finding r. Thus •
I = arn-l(§ 89) so that 96 = 3 • r5, or r5 = 32. Hence r = 2.
The progression is therefore 3, 6, 12, 24, 48, 96, and hence the
desired means are 6, 12, 24, 48. Ans.
EXERCISES
1. Insert four geometric means between 2 and 486.
2. Insert three geometric means between 1 and 625.
3. Insert five geometric means between 4J and y^.
4. What is the geometric mean of 2 and 18?
5. What is the geometric mean of 8 and 50 ?
6. What is the geometric mean of \ and 3f .
7. Find, correct to four decimal places, the geometric
mean of 6 and 27, using the tables of square roots.
8. Find, correct to four decimal places, the geometric
mean of 2J and 3^.
9. Insert two geometric means between 5 and 9, express-
ing each correct to four decimal places.
10. Show that the number of units in a side of the square
is the geometric mean of the number of units in the two un-
equal sides of a rectangle that has the same area.
11. Figure 56 shows a square within which
is placed (in any manner) another square
whose side is half as long as that of the
first square. Show that the area between
the squares is equa1 to three halves of the
mean proportional between the' areas of the
squares themselves. FIG. 56.
12. Show that the result stated in Ex. 11 holds true also
in the case of the area between two circles, the smaller circle
lying within the larger and having its radius half as long as
that of the larger circle. Draw a figure.
156 SECOND COURSE IN ALGEBRA [XV, § 92
92. Infinite Geometric Progression. Consider the geo-
metric progression
(i) !,*,*,*, A, -•
Here o = l, r=%, and hence, by § 90, the sum of n terms is
-ar
1-r
Now, if the value selected for n is very large, the expres-
sion (l/2)n, which here appears, is very small, being the frac-
tion £ multiplied into itself n times. In fact, as n is selected
larger and larger, this expression (l/2)w comes to be as small
as we please, so that the value for S, as given above, comes as
near as we please to
1-0
which is the same as 2. So we say that 2 is the sum to in-
finity of the geometric progression above, meaning thereby
simply that as we sum up the terms, taking more and more
of them, we come as near as we please to 2.
The meaning of this result is seen in the figure below.
FIG. 67.
Here, beginning at the point marked 0, we first measure
off 1 unit of length, then, continuing to the right, we measure
off % unit, then -i unit, then -J unit, etc., each time going to
the right just one half the amount we went the time before.
As this is kept up indefinitely, we evidently come as near as
we please to the point marked 2, which is 2 units from 0.
This corresponds exactly to what we are doing when we add
more and more of the terms of the given progression
XV, § 92] PROGRESSIONS 157
A progression like the one just considered, in which the
value of n is not stated but may be taken as large as one
pleases, is called an infinite geometric progression.
Having thus considered the sum to infinity of the special
infinite geometric progression (1), let us now suppose that we
have any infinite geometric progression, as
a, ar, ar2, ar3, •••,
and (as before) that r has some value numerically (§1) less
than 1. Then the sum of the first n terms is
o a — arn
=T^7'
and, as n is taken larger and larger, the expression rn which
appears here becomes as small as we please, since we have
supposed r to be less than 1. • Hence, as n increases indefi-
nitely, the value of S comes as near as we please to
a-a -0
1-r '
or
a
1-r'
We have therefore the following theorem: The sum to
infinity of any geometric progression whose common ratio r
is numerically less than 1 is given by the formula
1-r
EXAMPLE. Find the sum to infinity of the progression
3> 1> -g-j ^j -fr, '"•
SOLUTION, a =3, r = ^. Since r is numerically less than 1, we
have by the formula of § 92,
< a 3 39.,
158
SECOND COURSE IN ALGEBRA [XV, § 92
EXERCISES
Find the sum to infinity of each of the following progres-
sions, and state in each case what your answer means, draw-
ing a diagram similar to Fig. 57 to illustrate.
2 Q 3 3 3
• O, -I". T'B't ~a~Ti .
[HINT. r=— i and hence is numerically less than 1. The
formula of § 92 therefore applies.]
4. 5, .5, .05, .005, •••.
6. 1 — x+x2 — x3+ •••whenx=f.
7. V2, 1,4=' I ..,
I 2 _2\/2 4
1 3' 3vr 9' '"*
10. A pendulum starts at A and swings
to B, then it swings back as far as C,
then forward as far as D, etc. If the
first swing (that is, the circular arc from
A to B) is 6 inches long and each suc-
ceeding swing is five sixths as long as the
one just preceding it, how far will the
pendulum bob travel before coming to
FIG. 58. rest?
11. At what time after 3 o'clock do the hands of a watch
pass each other?
[HINT. We may look at this as follows : The large (minute)
hand first moves down to where the small (hour) hand is at the be-
XV, § 93] PROGRESSIONS 159
ginning, that is through 15 of the minute spaces along the dial.
Meanwhile the small hand advances T^ as far or |f of a minute space.
This brings the small hand to the position indicated by the dotted
line in the figure. The large hand next passes
over this ^f of a minute space. Meanwhile
the small hand again advances ^ as far, which
is -£f± of a minute space. The large hand next
covers this ^f^ of a minute space, but the small
hand meanwhile advances ^ as far, or yyf-g-
of a minute space, etc. Thus, the successive
moves of the large hand, counting from the first
one, form the G. P. 15, if, ^, T^, ....
The sum of this to infinity will be the total distance passed over
by the large hand before the hands pass.]
93. Variable. Limit. We have seen (§ 92), in connection
with the geometric progression 1, -g-, -J-, -J-, ••'•, that the sum of
its first n terms is a quantity which, as n increases indefinitely,
comes and remains as near as we please to the exact value 2.
The usual way of stating this is to say that as n increases,
the sum of the first n terms approaches 2 as a limit. The sum
of the first n terms is here called a variable since it varies, or
changes, in the discussion. A similar remark applies to all
the infinite geometric progressions which we have consid-
ered. In every case the sum to infinity is the limit which
the sum of the first n terms, considered as a variable quantity,
is approaching.
NOTE. It may be asked whether the sum of the first n terms of
the G. P. 1, \, \, -|, ••• could ever actually reach its limit 2. The
answer is that it may or it may not, depending upon circumstances.
Thus, if we think of the terms, beginning with the second, as being
added on at the rate of one a minute we could never reach the end of
the adding process, since the number of the terms is inexhaustible and
hence the minutes required would have no end. In other words,
the sum of the first n terms could never reach its limit on this plan.
160 SECOND COURSE IN ALGEBRA [XV, § 93
But suppose that instead of this we were to add on the terms with
increasing speed as we went forward. For example, suppose we
added on the ^ in ^ a minute, then the -J in \ of a minute, then the
i in -1- of a minute, etc. On this plan we would actually reach the
limit 2 in 2 minutes of time. Here the constantly increasing speed
of the adding process exactly counterbalances the fact that we have
an indefinitely large number of terms to add, with the result that we
reach the end of the process in the definite time of 2 minutes. This
idea is practically illustrated in Ex. 11, p. 159, where the hands of the
watch would never pass each other at all except for the fact that the
successive moves of the large hand, which constitute the terms of
the progression 15, ^-|, -^£±, xrls"' ' * * are added on in less and less time
as the process goes on, each being added on in ^ the time occupied
by the one just before it.
The question of whether a variable can reach its limit is inti-
mately connected with the famous problem considered by the
Schoolmen in the Middle Ages and known as the problem of Achilles
and the tortoise. In this problem, Achilles, who was a celebrated
runner and athlete, starts out from some point, as A, to overtake
a tortoise which is at some point, as T, the tortoise being famous for
the slow rate at which it crawls along. Both start at the same in-
stant and go in the same direction, as indicated in the figure.
A T
FIG. 60.
Achilles soon arrives at the point T, from which the tortoise started,
but in the meantime the tortoise has gone some distance ahead.
Achilles now covers this last distance, but this leaves the tortoise
still ahead, having again gained some additional distance. This
continues indefinitely. How, therefore, can Achilles ever overtake
the tortoise? The Schoolmen never quite answered this question
satisfactorily to themselves. The secret of the difficulty lies in the
fact that, as in the other problems mentioned above, the successive
moves which Achilles makes are done in shorter and shorter inter-
vals of time, with the result that, although the number of moves
necessary is indefinitely great, they can all be accomplished in a
definite time.
XV, § 94] PROGRESSIONS 161
94. Repeating Decimals. If we express the fraction £|
decimally by dividing 12 by 33 in the usual way, we find that
the quotient is .363636 •••, the dots indicating that the divi-
sion process never stops (or is never exact) but leads to a
never-ending decimal. However, the figures appearing in
this decimal are seen to repeat themselves in a regular order,
since they are made up of 36 repeated again and again.
Such a decimal is called a repeating decimal. More generally,
a repeating decimal is one in which the figures repeat them-
selves after a certain point. Thus, .12343434 • • -, 1.653653653
•••, are repeating decimals.
Let us now turn the question around. Thus, suppose
that a certain repeating decimal is given, as for example
.272727 •••, and let us ask what fraction when divided out
gives this decimal. This kind of question is usually too
difficult to answer in arithmetic, but it can be easily answered
as follows by use of the formula in § 92.
Thus the decimal .272727 • • • may be written in the form
100 o o ~i~ ioooooo~f" *"•
This is an infinite geometric progression in which a =
r=TJTr. The sum of this progression to infinity must be the
value of the given decimal. Hence, the desired value is
= 27 100 = 27=3 A
100 99 99 11
This answer may be checked by dividing 3 by 11, the re-
sult being .272727 • • -, which is the given decimal.
NOTE. It is shown in higher mathematics that every rational
fraction in its lowest terms (that is, every number of the form a/6,
where a and 6 are integers prime to each other) gives rise when
divided out to a never-ending, repeating decimal, while every irra-
tional number (such as V2) gives rise when expressed decimally to
a never-ending non-repeating decimal.
M
162
SECOND COURSE IN ALGEBRA [XV, § 94
EXERCISES
Find the values of the following repeating decimals and
check your answer for each of the first six.
1. .414141 •••.
4. .3414141 -..
2. .898989 •••.
3. .543543543
SOLUTION. .3414141 • • • = .3 + .0414141 • • •
= .3 +^(.414141-)
6. .6535353 -
6. 1.212121 •
7. 3.2151515
_ 3 _,_'! ^ 41 ^100
~10+10X100X99
= A_i_^L=338 = 169 A
10 990 990 495'
8. 5.032032032
9. 6.008008008
10. 34.5767676 •
MISCELLANEOUS PROBLEMS
I. ARITHMETIC PROGRESSION
1. What will be the cost of digging a 20-foot well if the
digging costs 50 cents for the first foot and increases by 25
cents for each succeeding foot ?
2. Fifty-five logs are to be piled so that the top layer
shall consist of 1 log, the next layer of 2 logs, the next layer
of 3 logs, etc. How many logs will lie on the bottom
layer ?
3. In a potato race 30 potatoes are placed at the dis-
tances 6 feet, 9 feet, 12 feet, etc., from a basket. A player
starts from the basket, picks up the potatoes and carries them,
one at a time, to the basket. How far does he go altogether
in doing this?
XV, § 94] PROGRESSIONS 163
4. A row of numbers in arithmetic progression is written
down and afterwards all erased except the 7th and the 12th,
which are found to be — 10 and 15, respectively. What was
the 20th number?
5. If your father gives you as many dimes on each of
your birthdays as you are years old on that day, how old
will you be when the total amount he has given you in this
way amounts to $12?
6. How many arithmetic means must be inserted be-
tween the numbers 4 and 25 in order that their sum may
amount to 87 ?
7. Prove that equal multiples of the terms of an arith-
metic progression are in arithmetic progression.
8. Prove that the sum of n consecutive odd integers, be-
ginning with 1, is n .
9. The sum of three numbers in arithmetic progression
is 30 and the sum of their squares is 462. What are the
numbers ?
[HINT. The numbers may be represented as x— y, x, x+y.
Form two equations and solve for x and y.}
10. If a person saves $20 the first month and $10 each
month thereafter, how long before his total savings will
amount to $1700?
11. Divide 80 into four parts which are in arithmetic
progression and which are such that the product of the first
and fourth is to the product of the second and third as 2:3.
12. Find the sum of the first 40 terms of an A.P. in which
the ninth term is 136 and the sum of the first nineteen terms
is 2527.
13. lid = 2, n = 21, and 5 = 147, find a and I
14. Show that if, in any A. P., the values of d, I, and S are
given, then the formula for a is
164 SECOND COURSE IN ALGEBRA [XV, § 94
II. GEOMETRIC PROGRESSION
15. A wheel in a certain piece of machinery is making
32 revolutions per second when the steam is turned off and
the wheel begins to slow down, making one half as many
revolutions each second as it did the preceding second. How
long before it will be making only 2 revolutions per second?
16. Show that if a principal of $p be invested at r %
compound interest, the sum of money accumulating at the
ends of successive years will form a geometric progression,
while if the investment be made at simple interest, the sums
accumulating will form an arithmetic progression.
17. From a cask of vinegar ^ the contents is drawn off
and the cask then filled by pouring in water. Show that if
this is done 6 times, the cask will then contain more than
90% water.
[HINT. Call the original amount of vinegar 1, then express (as
a proper fraction) the amount of water in the cask after the first
refilling, second refilling, etc.]
18. A set of concentric circles is drawn, each having a
radius half that of the circle just outside it. Show that the
limit toward which the sum of their circumferences is ap-
proaching is equal to twice the circumference of the largest
circle.
19. A dipper when hung on a wall often swings back and
forth for a time, the swings gradually dying out. If the first
swing occupies 1 second, and each succeeding swing takes
.9 as long as the one before it, how long before the dipper
comes to rest?
20. It is found by experiment that the number of bacteria
in a sample of milk doubles every 3 hours. What increase
will there be in 24 hours, assuming that all outside conditions
remain the same?
XV, § 94] PROGRESSIONS 165
21. In Fig. 61 a series of ordinates equally spaced from
each other has been drawn, the first one being laid off 1 unit
long, the second one being laid off equal
to the first one increased by % its length,
the third being equal to the second in-
creased by ^ its length, etc. Show that
these ordinates represent the successive
terms of the G. P. whose first term is 1
and whose common ratio is 14-. In this
sense, the figure may be called the dia-
gram for the G. P. in which a = l, r=l-J.
22. Draw the diagram for the G. P. in which
(a) a = l, r = H; (6) a = 2, r = l£; (c) a = 4, r = J.
[HINT. Use 8 ordinates only, spacing them at any convenient
but equal distance apart.]
23. Prove that any series of numbers formed by writing
down the reciprocals of the successive terms of a geometric
progression is itself a geometric progression.
24. Three numbers whose sum is 24 are in arithmetic pro-
gression, but if 3, 4, and 7 be added to them respectively, the
results form a geometric progression. Find the numbers.
25. If a series of numbers are in G. P., are their squares
likewise in G. P. ? Answer the same for their cubes ; also
for their square roots and their cube roots.
Answer the same questions for an A. P.
[HiNT. See that your reasoning is general, that is, do not base it
upon an examination of some special cases.]
CHAPTER XVI
RATIO AND PROPORTION
95. Ratio. The quotient of one number divided by another
of the same kind is called their ratio.
Thus the ratio of 6 inches to 3 inches is |-, or 2 ; the ratio of 5 Ib.
to 3 Ib. is |> etc. Note that in each of these cases the ratio is simply
a fraction of the kind studied in arithmetic.
The first number, or dividend, is called the antecedent;
the second number, or divisor, is called the consequent.
Thus, in the ratio -J, the antecedent is 3 and the consequent is 4.
EXERCISES
1. What is the ratio of. 10 yards to 2 yards? of 7 yards
to 3 yards?
2. State (as a fraction in its simplest form) the value of
each of the following ratios.
(a) 5 to 25. (c) i to |. (e) 1 to 3. (g) 18 xz to 4 z2.
(6) 16 to 12. (d) 2toi. (/) 3 a to 6 b. (h) x*-y*tox-y.
3. State which is the antecedent and which the conse-
quent in each of the parts of Ex. 2.
4. What is the ratio of 10 inches to 2 feet?
[HINT. First reduce the 2 feet to inches so that we may com-
pare like numbers, that is numbers measured in the same unit.]
5. The dimensions of a certain grain bin are 3 feet by
6 feet by 7 feet. What is the ratio of its cubical contents to
that of a bin whose dimensions are 3 feet 6 inches by 5 feet
by 1£ yards?
166
XVI, § 96] RATIO AND PROPORTION 167
6. If one square has its sides each twice as long as the
sides of another square, what is the ratio of the area of the
first square to that of the second?
[HINT. Let a =a side of the smaller square.]
7. If one cube has its edges each twice as long as the edges
of another cube, what is the ratio of the volume of the first
cube to that of the second ?
8. Show that if a cylinder and a cone
have the same circular base and the same
height, the ratio of the volume of the cylinder
to that of the cone is 3 : 1.
FIG. 62.
9. When we sharpen a lead pencil a cer-
tain part of the cylindrical lead is exposed. What part of
the exposed lead is cut off when a smooth conical point is
made?
96. Proportion. A proportion is an expression of equality
between two ratios, or fractions.
For example, since j- is the same as f-, we have the proportion ^ = -|.
Likewise, we may write f = £, f=if> -f = -f, etc. ; hence
all these are true proportions. But f = -J- is not a proportion since
these two fractions are unequal.
Every proportion is thus seen to be an equality of the form
a/b = c/d, where a, 6, c, and d are certain numbers. These
four numbers are called the terms of the proportion. The
first and fourth (that is, a and d) are called the extremes
of the proportion, while the second and third (b and c) are
called the means.
Besides writing a proportion in the form a/6 = c/d, it may
be written in the form a : b : : c : d, or also in the form a : d =
c:d. In all cases it is read " a is to b as c is to d," and it
means that the fraction a/6 equals the fraction c/d.
168 SECOND COURSE IN ALGEBRA [XVI, § 96
EXERCISES
1. Using the language of proportion, read each of the
following statements.
(a) i = f. (c) 2: -1=8: -4.
(6) 1:4 = 3:12. (d) £:£::4:3.
2. State which are. the extremes and which the means
in each part of Ex. 1.
3. State such proportions as you can make out of the
following four quantities : 3 inches, 6 inches, 12 inches, 24
inches.
[HINT. 3 inches is to 6 inches as • • • . Make other proportions also.]
4. State such proportions as you can make out of the
following four quantities : 1 inch, 3 inches, 1 foot, 1 yard.
[HINT. First express all quantities in inches.]
5. State such proportions as you can make out of the
following : 1 pint, 1 quart, 1 gallon, 2 gallons.
6. Do as in Ex. 5 for the following : 2 seconds, 1 minute,
1 hour, a day and a quarter.
7. Do as in Ex. 5 for the following : 1 cent, 1 dollar, 1
centimeter, 1 meter.
[HINT. Compare money ratio with distance ratio.]
8. Do as in Ex. 5 for the following : 4 ounces, 1 pound,
1 gallon, 1 quart.
97. Algebraic Proportions. If we consider the algebraic
fraction (a26)/(a&2), we see (upon dividing both numerator
and denominator by ab) that it reduces to a/6. In other
words, we have
o26 = a
ab2 b
This is an example of an algebraic proportion. Similarly,
2x2y_ x
4 xyz 2 z
XVI, § 98] RATIO AND PROPORTION 169
is an algebraic proportion and may be written if desired in
the form
2 x2y : 4 xyz = x : 2 z.
Likewise, since
a2-62
__ _
we have
98. Fundamental Theorem. Let a/b = c/d be any pro-
portion. By multiplying both sides of this equality by bd,
we obtain
or
ad = be.
This result may be stated in the following theorem.
THEOREM A. In any proportion, the product of the means
is equal to the product of the extremes.
This theorem is useful in testing the correctness of a
proportion. Thus 6:9=14:21 is a correct proportion be-
cause the product of the means, which is 9 X 14, is equal to
the product of the extremes, which is 6 X21 ; but 6 : 9 = 8 : 15
is not correct because 9 X8 is not equal to 6 Xl5. Similarly,
x3 : x2y = x:y, because x2y - x = xz • y.
EXERCISES
By means of the theorem of § 98, test the correctness of
the following proportions.
1. 5:6 = 15:18. 5. 17:19 = 21:23.
2. 3:2 = 5:6. 6. 2 a : ab = W x : 5 bx.
3. 4: -1 = 8: -2. 7. 3 m2 : (a-b) =6 m : 2 m(a-b).
4. i:| = 8:4. 8.
9. a2-62:
170 SECOND COURSE IN ALGEBRA [XVI, § 98
By means of the theorem of § 98, determine the value
which x must have in the following proportions.
10. z:4 =3:2.
[HINT. The theorem gives 4 • 3 = x • 2, or 12 = 2 x.]
11. 10:z = 2:5. 13. (z-5) : 4: : 2: 3.
12. 25/32 = 8/z. 14. (x-3)/(x-4) = 5/6.
15. What number bears the same ratio to 4 as 16 does to
6?
[HINT. Let x represent the unknown number and form a pro-
portion.]
16. Divide 35 into two parts whose ratio shall be f .
[HINT. Let x be one part. Then 35 — x will be the other part.]
17. Divide 35 into two parts such that the lesser dimin-
ished by 4 is to the larger increased by 9 as 1 : 3.
18. A man's income from two investments is $980. The
two investments bear interest rates which are in the ratio
of 5 to 6. What income does he receive from each?
19. Concrete for sidewalks is a mixture made of two parts
sand to one part cement. How much of each is required to
make a walk containing 1500 cubic feet?
20. Prove that no four consecutive numbers, as n, ft+1,
ft +2, ft +3, can form a proportion in the order given.
99. Application to Similar Figures. When two geometric
figures have the same shape, though not necessarily the same
size, they are called similar figures. Thus any two circles
are similar figures; likewise, any two squares, or any two
cubes, or any two spheres.
Two triangles may be similar, as
, illustrated in Fig. 63.
The following facts are shown in
/ , geometry regarding any two similar
figures.
XVI, §99] RATIO AND PROPORTION 171
(a) Corresponding lines are proportional.
Thus, in the two similar triangles of Fig. 63, if the side AB of the
one is twice as long as the corresponding side A'B' of the other, then
BC is twice as long as B'C'. That is,
AB = BC
A'B' B'C''
In the same way, we have also
AB CA
A'B' C'A'
(6) Areas are proportional to the squares of corresponding
lines.
Thus, if one circle has a radius of length R and another circle has
a radius of length r, the area, A, of the first is to the area, a, of the
second as R2 is to r2. That is we have the proportion A /a = R2/r2.
(c) Volumes are proportional to the cubes of corresponding
lines.
Thus, if one sphere has the radius R and another has the radius r,
the volumes V and v of the spheres are such that V/v = R3/r3.
EXERCISES ON SIMILAR FIGURES
1. In the two similar triangles shown in § 99 suppose
AB = 2 feet, A'B' = l foot 4 inches, and BC = 3 feet. How
long will B'C' be?
2. If a tree casts a shadow 40 feet long when a post 3£
feet high casts a shadow 4 feet long, how high is the tree ?
3. Compare the areas of two city lots of the same shape
if a side of the one is three times as long as the corresponding
side of the other. Does it matter what the shape of each is ?
4. If a certain bottle holds -J pint, how much will a bottle
of the same shape but only half as high hold ?
172 SECOND COURSE IN ALGEBRA [XVI, § 100
100. Mean Proportional. If the means of a proportion
are equal, either is called the mean proportional between the
extremes.
Thus 2 is the mean proportional between 1 and 4 because
•£•=•}. Likewise, 2 x is the mean proportional between x2 and 4,
because x2/2x =2 z/4.
NOTE. The mean proportional between a and 6 is always equal
to Vo6, for we must have a/x=x/b. Hence, clearing of fractions,
x2 = ofc/andjtheref ore x = Vo6. This will be a surd (§ 42) unless the
product ab is a perfect square. For example, the mean proportional
between 2 and 3 is the surd V2 • 3, or V6 =2.44949 (table).
101. Third and Fourth Proportionals. The third propor-
tional to two numbers a and b is that number x such that
a: b = b:x.
Thus the third proportional to 2 and 3 is the value of x in the
equation ^ ^
o = ~- Solving, £=•§• = 4 \. Arts.
O X
The fourth proportional to three numbers a, b, and c is that
number x such that a:b = c:x.
Thus the fourth proportional to 2, 3, and 4 is the value of x in
the equation 2/3 = 4/x. Solving, x = 6. Ans.
EXERCISES
1. Find the mean proportional between 8 and 18.
[HINT. Let* be the desired mean. Then8/x =x/lS. Solveforz.]
Find the mean proportional between each of the follow-
ing pairs of numbers. In cases where the answer is a numer-
ical surd, use the table to find its approximate value.
2. 9 and 81. 6. | and |f.
3. 6 and 7. 7. 2£ and 3^.
4. 5 and 20. 8. 2 x2y and 32 xy2.
a2-2a-3
K K A 10 Q - ^
6. 5 and 19. 9. - and
a+1 a— 3
XVI, § 101]
RATIO AND PROPORTION
173
Find the third proportional to each of the following pairs.
10. 3 and 4. 12. 2| and 3^. 14. z2-9 and z-3.
11. 18 and 50. 13. 2 x and x. 15. 2 and 6.
Find the fourth proportional to each of the following sets.
16. 3, 4, and 5. 19. V2, \/6, and A/12.
17. 5, 4, and 2. 20. 3 a, 2 6, and c.
18. 2, 3^, 4i. 21. x, y, and xy.
NOTE. In Exs. 16-21, the numbers must be placed in the pro-
portion in the order in which they are given, as in the illustrative
examples of § 101.
22. In the semicircle ABC suppose
DE drawn perpendicular to AB. Then
(as shown in geometry) the length of
DE will be the mean proportional A
between the lengths of AE and EB. Fl<*- 64.
If A E = 4 inches and EB = 16 inches, find DE.
23. The figure shows a circle and a point P outside it
from which are drawn two lines PS and PT. The first of
p these lines (called a secant) cuts through the
circle at two points R and S while the
second line (called a tangent) just touches
the circle at the point T. In all such cases,
the tangent length, PT, is the mean pro-
portional between the whole secant, PS, and
its external segment, PR (as shown in
geometry).
Find the length of PT if PR = 4 and RS=11.
24. If a, b, c, d are unequal numbers such that a:b = c:dt
show that no number x can be found such that
a-\-x\ b+x = c-\-x \d-\-x.
FIG. 65.
174 SECOND COURSE IN ALGEBRA [XVI, § 102
102. Second Fundamental Theorem. THEOREM B. //
the product of two numbers is equal to the product of two other
numbers, either pair may be made the means of a proportion
in which the other two are taken as the extremes.
PROOF. Suppose mn = xy. Dividing both members by
nx gives m/x = y/n, or m : x = y : n, which is one of the possible
proportions mentioned in the theorem.
Similarly, if we divide both members of mn = xy by ny we
obtain m/y = x/n, or m:y = x:n, which is another of the
possible proportions mentioned in the theorem.
The other possible proportions are x:m = n:y and n:x =
y : m. The proof of these is left to the pupil.
For example, the equality 2 • 9=3 • 6 gives rise to the propor-
tions 2 : 3=6: 9, 2:6 = 3:9, 9:3=6:2, and 9: 6=3: 2.
103. Inversion in a Proportion. THEOREM C. If four
quantities are in proportion, they are in proportion by inversion;
that is the second term is to the first as the fourth is to the
third.
PROOF. We are to show that if a/b = c/d, then b/a = d/c.
Since a/b = c/d, we have, by Theorem A, ad = bc.
Therefore, by Theorem B, we may write b/a = d/c.
For example, f = f gives by inversion the new proportion f- = f .
104. Alternation in a Proportion. THEOREM D. // four
quantities are in proportion, they are in proportion by alterna-
tion; that is the first term is to the third as the second is to
the fourth.
The proof is left to the pupil. First use Theorem A. See
proof of Theorem C.
For example, f = f gives by alternation the new proportion £ =f •
105. Composition in a Proportion. THEOREM E. //
four quantities are in proportion, they are in proportion by
XVI, § 107] RATIO AND PROPORTION 175
composition; that is the sum of the first two terms is to the
second term as the sum of the last two terms is to the last
term.
PROOF. We are to show that if a/b = c/d, then (a+b)/b =
(c+d)/d.
Since a/b = c/d, we may add 1 to each member of this equa-
tion, thus giving
«+!=< + !, which reduces to
,
b a o a
For example, f = f gives by composition the new proportion
(2+3)/3 = (6+9)/9, orf=V5.
106. Division in a Proportion. THEOREM F. // four
quantities are in proportion, they are in proportion by division;
that is, the difference between the first two terms is to the
second term as the difference between the last two terms is to
the last term.
PROOF. We are to show that if a/b = c/d, then (a — &)/& =
(c-d)/d.
Since a/b = c/d, -we may subtract 1 from each side of this
equation, thus obtaining
f- 1=^-1, which reduces to *LZ&=^.
b a o d
For example, f =f gives by division the new proportion
(3-2)/2 = (9-6)/6, or|=f.
107. Composition and Division. THEOREM G. // four
quantities are in proportion, they are in proportion by composi-
tion and division; that is the sum of the first two terms is to
their difference as the sum of the last two terms is to their
difference.
PROOF. We are to show that if a/b = c/d, then
a—b c—d
176 SECOND COURSE IN ALGEBRA [XVI, § 107
By Theorems E and F, we have
a-j-b_c-\-d , a—b_c — d
~l d~' ! ~b 5"
By dividing the first of these equations by the second,
member by member, we obtain the desired result, namely
a— b c—d
For example, f = |- gives by composition and division the-new pro-
portion (3 +2) /(3-2) = (9 +6) /(9-6), or f =-^.
108. Several Equal Ratios. THEOREM H. In a series of
equal ratios, the sum of the antecedents is to the sum of the conse-
quents as any antecedent is to its consequent.
PROOF. We are to show that if a/b = c/d = e/f=g/h= •••,
then
a+c+e+g-\ — _a_^c _e_g _
b+d+f+h+~- b d f h
Let k be the value of any one of the equal ratios, so that
JL_« £_c r._e k_g
6'* d'* / -A'
= fc&, c = kd, e = kf, g = kh, •••.
Hence
b+d+f+h+-
or
a+c+e+g-\ — _^a_c ' _e_Q [_
b+d+f+h+- b d f h
For example, the three equal ratios f = f = f give the new pro-
portions
2+4+6^2^.4,6
3+6+9 369*
or
12 = 2=4=6
18 3 6 9*
XVI, § 108] RATIO AND PROPORTION 177
EXERCISES
1. Given the proportion f =J^-. Write down the vari-
ous proportions to be obtained from this by (1) inversion,
(2) alternation, (3) composition, (4) division, (5) composi-
tion and division. Note that each new proportion thus ob-
tained is a true one.
2. Show that if a/b = c/d, then a2/b2 = c2/d2', in other
words, if four numbers are in proportion, their squares are
also in proportion.
3. Show that if four numbers are in proportion, their
cubes are also in proportion; likewise their square roots;
likewise their cube roots.
4. Show by means of Theorem A (§ 98) that a/b = a2/b2
is not a true proportion, unless a = b. In other words, the
ratio of two numbers is not in general the same as the ratio
of their squares. Prove similarly that the ratio of two num-
bers is not in general the same as the ratio of their cubes,
or of their square roots, or of their cube roots.
5. If a : b = c : d, establish the following proportions.
(a) a2:b2c2 = l:d2.
SOLUTION. It suffices to show here that the product of the
extremes is equal to the product of the means, for if these two
products are the same, the proportion in question is true by Theorem
B. Thus we are to prove that azdz = 62c2.
Now, we know (by hypothesis) that a:b=c:d, or ad = bc.
Squaring gives, as desired, a?d2 = bzc2, thus completing the proof.
(b) ac:bd = c2:d2.
[HINT. Remember to use the hypothesis, namely that a : b = c : d.]
(c) Vad:Vb = Vc:l. (d) a:a+b = a+c: a+b+c+d..
(e) a+b: c+d=Va2+b2: Vc2+d2.
(/) a+b+c+d: a—b+c—d=a+b—c—d: a—b — c+d.
(g) If a : b = c : d, and x:y = z:w, show that ax : cz = by : dw.
CHAPTER XVII
VARIATION
109. Direct Variation. One quantity is said to vary
directly as another when the two are so related that, though
the quantities themselves may change, their ratio never
changes.
Thus the amount of work a man does varies directly as the
number of hours he works. For example, if it takes him 4 hours to
draw 10 loads of sand, we can say it will take him 8 hours to draw 20
loads. Here the first ratio is ^ and the second is -£$ and the two
are seen to be equal, though the numbers in the second have been
changed from what they were in the first. In general, if the man
works twice as long, he will draw twice as much ; if he works three
times as long, he will draw three times as much, etc. ; all of which
implies that the ratio of the time he works to the amount he draws
in that time never changes.
EXERCISES
Determine which of the following statements are true and
which false, giving your reason in each instance.
1. The amount of electricity used in lighting a room
varies directly as the number of lights turned on.
2. The amount of water in a cylindrical pail varies
directly as the height to which the water stands in the
pail.
178
XVII, § 110] VARIATION 179
3. The amount of gasoline used by an automobile in any
given time (one week, say) varies directly as the amount
of driving done.
4. The time it takes to walk from one place to another
at any given rate (3 miles an hour, say) varies directly as the
distance between the two places.
5. The time it takes to walk any given distance (5 miles,
say) varies directly as the rate of walking.
6. The perimeter of a square varies directly as the length
of one side.
7. The circumference of a circle varies directly as the
length of the radius.
8. The area of a square varies directly as the length of
one side.
9. x varies directly as 10 x.
10. x varies directly as 10 x2.
110. Inverse Variation One quantity, or number, is said
to vary inversely as another when the two are so related that,
though the quantities themselves may change, their product
never changes.
Thus the time occupied in doing any given piece of work varies
inversely as the number of men employed to do it. For example,
if it takes 2 men 6 days, it will take 4 men only 3 days. The point
to be observed here is that the first product, 2X6, equals the second
product, 4X3. In general, if twice as many men are employed it
will take half as long ; if three times as many men are employed, it
will take one third as long, etc. In all these cases, the number of
men employed multiplied by the corresponding time required to do
the work remains the same.
NOTE. The term varies inversely as is due to the fact that in
case xy never changes (as required by the above definition), it
follows that x + (\/y) never changes, since xy=x + (\/y). That
is, x varies directly as the reciprocal, or inverse, of y (§ 109).
180 SECOND COURSE IN ALGEBRA [XVII, § 110
EXERCISES
Determine which of the following statements are true and
which false, giving your reason in each instance.
1. The time it takes water to drain off a roof varies in-
versely as the number of (equal sized) conductor pipes.
2. The time it takes to walk any given distance (5 miles,
say) varies inversely as the rate of walking. (Compare
Ex. 5, p. 179.)
3. The weight of a pail of water varies inversely as the
amount of water that has been poured out of it.
4. x varies inversely as 10/z.
6. x varies inversely as 10/z2.
111. Joint Variation. One quantity, or number, is said to
vary jointly as two others when it varies directly as their
product.
Thus the area of a triangle varies jointly as its base and altitude,
for if A be the area of any triangle and b its base and h its altitude,
we have A=^bh, which may be written A/bh=%. Whence A
varies directly as the product bh (§ 109), that is the ratio of A to
bh is always the same, namely ^ in this instance.
EXERCISES
I
Determine whether the following statements are true,
giving your reason in each instance.
1. The area of a rectangle varies jointly as its two
dimensions, that is as its length and breadth.
2. The pay received by a workman varies jointly as his
daily wage and the number of days he works.
3. The amount of reading matter in a book varies jointly
as the thickness of the book and the distance between the
lines of print on the page.
XVII, § 113] VARIATION 181
4. The interest received in one year from an investment
varies jointly as the principal and rate.
5. The volume of a' rectangular parallelepiped (such as
an ordinary rectangular shaped box) varies jointly as its
length, breadth, and height.
[HINT. Here we have one quantity varying jointly as three
others. First make a definition yourself of what such variation
means.]
112. Variables and Constants. When we say that the
amount of work a man does varies directly as the number of
hours he works (see § 109), we are dealing with two quanti-
ties, namely the amount of work done and the time used in
doing it. But it is to be observed that these are not being
regarded as fixed quantities, but rather as changeable ones,
the only essential idea being that their ratio never changes.
In general, quantities which are thus changeable throughout
any discussion or problem are called variables, while quan-
tities which do not change are called constants.
113. The Different Types of Variation Stated as Equa-
tions. We may now state very briefly and concisely what is
meant by the different types of variation mentioned in
§§ 109-111 and certain other important types also. To do
this, let us think of x, y, and z as being certain variables
and k as being some constant. Then
(1) To say that x varies directly as y means (by § 109) that
T
- = fc, or x = ky.
y
(2) To say that x varies inversely as y means (by § 110) that
xy — k, or x = -.
y
(3) To say that x varies jointly as y and z means (by § 1 11) that
1*
— = k, or x — kyz.
yz
182 SECOND COURSE IN ALGEBRA [XVII, § 113
Two other important types of variation are described
below :
(4) To say that x varies directly as the square of y means that
- = k, or x = ky2.
(5) To say that x varies inversely as the square of y means that
xy2 = k, or x = — •
y2
In all these types of variation it is important to observe
that the value which must be given to the constant k depends
upon the particular statement or problem in hand. For
example, consider the statement that " The area of a rec-
tangle varies jointly as its two dimensions." This means
(see (3)) that if we let A be the variable area and a and b
the variable dimensions, then A = kab. But in this case we
know by arithmetic that A = ab, so the value of k here must
be 1. On the other hand, consider the statement that " The
area of a triangle varies jointly as its base and altitude."
Letting A be the variable area and b and h the variable base
and altitude, respectively, this means that A = kbh. But
here, as we know from arithmetic, k = \.
EXERCISES
Convert each of the following statements into equations,
supplying for each the proper value for the constant k
mentioned in § 113.
1. The circumference of a circle varies directly as the
radius.
[HINT. Let C stand for circumference and r for radius.]
2. The circumference of a circle varies directly as the
diameter.
3. The area of a circle varies directly as the square of the
radius.
XVIT, § 113] VARIATION 183
4. The area of a circle varies directly as the square of the
diameter.
5. The area of a sphere varies directly as the square of
the radius. (See § 14, (/).)
6. The volume of a rectangular parallelepiped varies
jointly as its length, breadth, and height. (See Ex. 5, p. 181.)
7. Interest varies jointly as the principal, rate, and time.
8. The volume of a sphere varies directly as the cube
of the radius.
[HINT. First supply for yourself the definition of what this
type of variation means.]
9. The volume of a circular cone varies jointly as the alti-
tude and the square of the radius of the base. (See p. 103.)
10. The distance, measured in feet, through which a body
falls if dropped vertically downward from a position of rest
(as from a window ledge) varies directly as the square of the
number of seconds it has been falling.
[HINT. It is found by experiments in physics that the value of
the constant A; is in this case 32 (approximately).]
11. The following, like Ex. 10, are statements of well-
known physical laws. Convert each into an equation with-
out, however, attempting to supply the proper value of k,
since to do so requires a study of physics.
(a) If a body is tied to a string and swung round and round
in a circle (as in swinging a pail of water at arm's length
from the shoulder) , the force, F, with which it pulls outward
from the center (called centrifugal force) varies directly as the
square of the velocity of the motion.
(6) The intensity of the illumination due to any small
source of light (such as a candle) varies inversely as the square
of the distance of the object illuminated from the source of
light.
184
SECOND COURSE IN ALGEBRA [XVII, § 113
(c) When an elastic string is stretched out, as represented
in Fig. 66, the tension (force tending to pull it apart at any
point) varies directly as the length to which the string has
been stretched (Hooke's Law).
FIG. 66.
(d) The pressure per square inch which a given amount
of gas (such as air, or hydrogen, or oxygen, or illuminating
gas) exerts upon the sides of the receptacle which holds
the gas (such as a bag) varies
inversely as the volume of the
receptacle (Boyle's Law).
FIG. 67.
For example, whenever air is con-
fined in a rubber balloon, as in the first
drawing in Fig. 67, it exerts a certain
pressure upon each square inch of the
interior surface. If the balloon be
squeezed, as in the second drawing
in Fig. 67 (no air being allowed to escape), until its volume is half
of what it was before, this pressure will be exactly doubled.
114. Problems in Variation. The problems naturally
arising in the study of variation fall into two general classes
as follows :
(1) Those in which the value of the constant k mentioned
in § 113 can be determined from the statement of the problem
and forms an essential part in the solution. This kind of
problem is illustrated by Exs. 1-10 on pages 185-187. The
solution given for Ex. 1 should be well understood before the
pupil undertakes Exs. 2-10.
XVII, § 114] VARIATION 185
(2) Those in which it is not necessary to know the value of
k. Such problems are illustrated in Exs. 11-20, pp. 187-189.
The pupil is advised to work several problems from each
group rather than to confine his attention to either.
EXERCISES
I. ILLUSTRATIONS OF CASE (1)
1. In a fleet of ships all made from the same model (that
is, of the same shape, but of different sizes) the area of the
deck varies directly as the square of the length of the ship.
If the ship whose length is 200 feet has 5000 square feet of
deck, how many square feet in the deck of the ship which
is 300 feet long?
SOLUTION. Let A represent the area of deck on the ship whose
length is I. Then the given law of variation, expressed as an
equation (§ 113), is
(1) A=klz. (k = some constant)
Since the ship which is 200 feet long has 5000 square feet of deck,
it follows from (1) that we must have
= /c(200)2.
This equation tells us that the value of k in the present problem
must be
.5000 = 5000 =1
(200)2 200X200 8*
Placing this value of k in (1), gives us an equation which deter-
mines completely the relation between A and I in the present problem,
that is
(2) A = iZ*.
Now the problem asks how many square feet of deck there are
in the ship whose length is 300 feet. This can be found by simply
placing £ = 300 in (2) and solving for A. Thus
A =4 X(30Q)2=300*3QP:= 11,250 square feet. Ans.
O O
186 SECOND COURSE IN ALGEBRA [XVII, § 114
NOTE. Observe that the first step in the above solution is to
express as an equation the law of variation belonging to the problem.
Next, the constant k is determined. After this, the first equation
is rewritten in its more exact form obtained by assigning to k its
value. The answer is then readily obtained.
These steps should be followed in working each of the Exs. 2-10.
2. In a fleet of ships all of the same model, the ship whose
length is 200 feet contains 6000 square feet in its deck. How
long must a similar ship be made if its deck is to contain
13,500 square feet?
3. To make a suit of clothes for a man who is 5 feet
8 inches high requires 6 square yards of cloth. How much
cloth will be required to make a suit for a man of similar build,
whose height is 6 feet 2 inches?
[HINT. The areas of any two similar figures vary directly as
the squares of their heights.]
4. If 10 men can do a piece of work in 20 days, how long
will it take 25 men to do it?
[HINT. The time required varies inversely as the number of
men employed.]
5. The horsepower required to propel a ship varies di-
rectly as the cube of the speed. If the horsepower is 2000
at a speed of 10 knots, what will it be at a speed of 15
knots?
6. A silver loving-cup (such as is sometimes given as a
prize in athletic contests) is to be made, and a model is first
prepared out of wood. The model is 8 inches high and
weighs 12 ounces. What will the loving-cup cost if made
10 inches high, it being given that silver is 17 times as heavy
as wood and costs $2.20 an ounce?
[HINT. The volumes and hence the weights of any two similar
figures vary directly as the cubes of their heights. See § 99 (c).]
XVII, § 114] VARIATION 187
7. When electricity flows through a wire, the wire offers
a certain resistance to its passage. The unit of this resist-
ance is called the ohm, and for a given length of wire the
resistance varies inversely as the square of the diameter. If
.a certain length of wire whose diameter is \ inch offers a
resistance of 3 ohms, what will be the resistance of a similar
wire (same length and material) ^ of an inch in diameter?
8. Three spheres of lead whose radii are 6 inches, 8 inches,
and 10 inches respectively are melted and made into one.
What is the radius of the resulting sphere ?
9. On board a ship at sea the distance of the horizon
varies directly as the square root of one's height above the
water. If, at a height of 20 feet, the horizon is 5.5 miles dis-
tant, what is its distance as seen from a light-house 80 feet
above sea-level?
10. The horsepower that a shaft can safely transmit
varies jointly as its speed in revolutions per minute and the
cube of its diameter. A 3-inch steel shaft making 100 revo-
lutions per minute can transmit 85 horsepower. How many
horsepower can a 4-inch shaft transmit at a speed of 150
revolutions per minute ?
II. ILLUSTRATIONS OF CASE (2)
11. Knowing that the force of gravitation due to the
earth varies inversely as the square of the distance from the
earth's center (Newton's Law of Gravitation), find how far
above the earth's surface a body must be taken in order to
lose half its weight.
SOLUTION. Letting W represent the weight of a given body at the
distance d from the earth's center, the law stated above, when ex-
pressed as an equation, becomes
( 1 ) W = - - (k = some constant)
a*
188 SECOND COURSE IN ALGEBRA [XVII, § 114
Now let Wi represent the weight of the body when on the surface.
Remembering that the earth's radius is 4000 miles (approximately),
equation (1) gives
(2) k
40002
Next, let x represent the desired distance, namely the distance
above the surface at which the same body loses half its weight.
At this distance its weight will consequently be %Wi, while its dis-
tance from the earth's center is now 4000 +x. So (1) gives
(3) IKi = , &
2 (4000 +z)2
Dividing equation (3) by equation (2), noting the cancelation of
W\ on the left and of the (unknown) k on the right, we obtain
1 = 400Q2
2 (4000 +xY
It remains only to solve this equation for x.
Clearing of fractions, (4000 +z)2 = 2 - 40002 = 40002 • 2.
Extracting the square root of both members, 4000 +x =4000 \/2.
Solving, x =4000 V2 -4000 =4000( V2 -1) miles. Ans.
To find the approximate value of this answer, we have (see table)
V2 = 1.41421
so that x =4000(1.41421 -1) = 4000 X. 41421 = 1656.84 miles. Ans.
NOTE. Observe that the first step in the above solution (as
also in the preceding exercises) is to. express as an equation the law
of variation belonging to the problem. Then write down the two
special equations which express the particular conditions given in
the problem and divide one of these equations by the other to
eliminate the unknown k. The answer is then readily obtained.
A similar process should be followed in working the remaining
exercises of this list.
12. Show that the earth's attraction at a point on the sur-
face is over 5000 times as strong as at the distance of the
moon, that is at the (approximate) distance of 280,000
miles.
[HINT. Call W\ the weight of a given body on the surface, and
let W2 represent the weight of the same body at the distance of the
moon from the earth's center. Then use the law expressed in (1)
of the solution of Ex. 11.1
XVII, § 114] VARIATION 189
13. A book is being held at a distance of 2 feet from an
incandescent lamp. How much nearer must it be brought
in order that the illumination on the page shall be doubled?
(See Ex. 11 (6), p. 183.)
14. If two like coins (such as quarter dollars) were melted
and made into a single coin of the same thickness as the origi-
nal, show that its diameter would be \/2 times as great.
[HINT. Call D the diameter of the given coins and A the area
of each. Note that the area of the new coin will then be 2 A. Use
the result stated in Ex. 3, p. 182.]
15. Find the result in Ex. 14 when four equal-sized coins
are used.
16. Show that a falling body will pass over the second
3 feet of its descent in about .4 of the time it takes it to
pass over the first 3 feet. (See Ex. 10, p. 183.)
17. The time required for a pendulum to make a complete
oscillation (swing forward and back) varies directly as the
square root of its length. By how much must a 2-foot pendu-
lum be shortened in order that its time of complete oscilla-
tion may be halved ?
18. If the diameter of a sphere be increased by 10%, by
what per cent will the volume be increased ?
19. Show that if a city is receiving its water supply by
means of a main (large pipe) from a reservoir, the supply
can be increased 25% by increasing the diameter of the main
by about 12%.
20. It is desired to build a ship similar in shape to one
already in use but having a 40% greater cargo space (or hold).
By what per cent must the beam (width of the ship) be in-
creased. (See § 99 (c).)
190
SECOND COURSE IN ALGEBRA [XVII, § 115
115. Variation Geometrically Considered. If a variable
y varies directly as another variable x, we know (§ 113) that
this is equivalent to having the equation y = kx, where k is
some constant. If the value of & is 1, this equation takes the
definite form y = x, and we may now draw its graph, the
result being a certain straight line. If, on the other hand,
k = 2, we have y = 2 x} and this again is an equation whose
234
FIG. 68. — DIRECT VARIATION.
graph may be drawn, leading to a straight line, but a differ-
ent one. In general, whatever the value of jfe, the corre-
sponding equation has a straight-line graph. The fact that in
all cases the graph is a straight line characterizes this type
of variation, that is, characterizes the type in which one
variable varies directly as another. The figure shows the
lines corresponding to several different values of k.
XVII, § 115]
VARIATION
191
In case a variable y varies inversely as another variable x,
we know (§ 113) that there exists an equation of the form
y = k/x, where k is some constant. If we let fc = l, this be-
comes y=l/x. By letting x take a series of values and
determining the corresponding values of y from this equa-
tion (thus forming a table as in § 57) we obtain the graph.
Similarly, corresponding to the value A; = 2 we have y = 2/x,
1 2 3 4 X
FIG. 69. — INVERSE VARIATION.
and this equation has a definite graph which is different from
the one just mentioned. In general, whatever the value of fc,
the corresponding equation has a graph, but it is now to be
noted that these graphs are not straight lines; they are
hyperbolas. (See Ex. 2, § 78.) The figure shows the curves
corresponding to several different values of k.
NOTE. Though these curves differ in form, they have the follow-
ing feature in common : Through the origin draw any two straight
192 SECOND COURSE IN ALGEBRA [XVII, § 115
lines (dotted in figure). Then the intercepted arcs AB, CD, EF,
GH, etc., are similar, that is the smallest arc when simply magnified
by the proper amount produces one of the others.
EXERCISES
Draw diagrams to represent the geometric meaning of
each of the following statements.
1. y varies directly as the square of x.
2. y varies inversely as the square of x.
3. y varies as the cube of x.
4. y varies directly as x, and y = 6 when x = 2.
[HINT. The diagram here consists of a single line.]
5. y varies inversely as x, and y = 6 when x = 2.
6. The cost of n pounds of butter at 40j£ per pound is
c = 40 n.
7. The amount of the extension, e, of a stretched string
is proportional to the tension, t, and e = 2 in. when t = 10 Ib.
(See Ex. 11 (c), p. 184.)
8. The pressure, p, of a gas on the walls of a retaining
vessel varies inversely as the volume, v ; and p = 40 Ib. per
square foot when v = 10 cu. ft.
9. The length, L, of any object in centimeters is propor-
tional to its length, I, expressed in inches; and L = 2.54
when 1=1.
NEWTON
(Sir Isaac Newton, 1642-1727)
Discoverer of the law of gravitation and famous in algebra for his discov-
ery of the binomial theorem. Inventor of the branch of higher mathematics
called the Calculus, wherein rates of motion and other changing, or variable,
quantities are extensively studied.
CHAPTER XVIII
EXPONENTS
I. POSITIVE INTEGRAL EXPONENTS
116, Powers. Involution. Just as a2 = u - a ; a3 = a -a • a ;
etc., so we define the nth power of a, where n is any posi-
tive integer, as follows :
an = a'Ci'a'a'a--a (n factors) .
The process of finding the power of a number, or expression,
is called involution.
117. Laws of Exponents. There are five fundamental
laws of exponents which are as follows, it being understood
that m and n everywhere stand for positive integers :
I. MULTIPLICATION LAW. This law for multiplying two
powers of the same quantity is
Qm . ^n=-^7n+n<
PROOF.
am = a° a - a • a • a ••• a (m factors). (§ 116)
an — a- a- a- a • a ••• a (n factors).
Hence
am . an={a • a- a • a ••• a (m factors)} • {a • a • a • a ••• a (n factors)}
= a- a- a - a--- a (m-\-n) f actors = am+n. (§ 116)
Therefore
am - an = am+n.
ILLUSTRATIONS.
194 SECOND COURSE IN ALGEBRA [XVIII, § 117
II. DIVISION LAW. The law for dividing one power by
another power of the same quantity is
am -^- an = am~n .
PROOF. am -=- a» = — = * ' * ' * ' **'" a ' a ' a'" a (m factor3)
an tf • tf • fa • • • JL (n factors)
= a • a • a • a ••• a (m—ri) factors = am~n. (§ 116)
Therefore
am -j-on = am~n.
ILLUSTRATIONS.
36 ^.32 =36-2 =34 . ( _2)5 + ( _2)3 = ( -2)2 ;
x8 -r-z5 =x3 • (a +b)7 -7- (a +6)3 = (a +6)4.
III. LAW FOR THE POWER OF A POWER. The law for
raising a power of a quantity to a new power is
mn
a
PROOF. (am)n = am • am • am • am -• a™ (n factors) (§ 116)
= am+m+m+m + "- +m (n terms)^ (La\V I)
Therefore (am}n = amn since m +m -\-m + • • • +m to n terms = ran.
ILLUSTRATIONS.
(42)3=42X3=46; £(_2)3}3 = (_2)9;
(X4)5=:C20; {(a +6)3)4 = (a +6)12.
IV. LAW FOR THE POWER OF A PRODUCT. The law for
raising to a power a product of two quantities is
(ab)n = anbn.
PROOF.
(aZ>) • (afc) ••• (06) (n factors) (§116)
{a • a • a ••• a (n factors)} • [b • b • b • b ••• b (n factors)}
anbn.
Therefore
ILLUSTRATIONS.
(2 X3)4 = 24 X34 ; {( -3) ( -2)}3 = ( -3)3( -2)3 ;
{(a +6) (c +d)}3 = (a +6)3
XVI11, § 117] EXPONENTS 195
V. LAW FOR THE POWER OF A QUOTIENT. The law for
raising to a power the quotient of two quantities is
Jb,
PROOF. Mn»/|j . M . /^ ... ^\ (n factors) (§ 116)
a (n factors) a
b - b - b ••• b (n factors) bn
Therefore
ILLUSTRATIONS. (|V~fJ; f-^V
\o/ o \ o / o"
(x\7_x^. /a -|-b\5 _ (q-j-b)5
W Z/7' \a — b) (a— 6)5
EXERCISES
Find the results of the indicated operations in the following
cases, using one (or more) of the five laws in § 117.
1. 25 • 23. 8. I2a • ta. 15. xl° + x2.
2( 1 "\3 / 1 "\2 Q ?r— 1 /yr+1 1C -»>jl2 • ^yj6
. ^ — i) • \ — L) . o. Z • Z . lu. 7/fr "!~7fl .
4/y.lO .7.2 -11 /7P+g— 1 . /^l+r 1 Q nm,^_rA
. ju JL , XJ.. ;/ y . xo. ly . y .
5. m12 -m13. 12. 83-^82. 19. i2a^-^.
6. yb-yn. 13. (-3)5^(-3)3. 20. z^+z*-1.
22. 0w+p-1-7-01+'\ 25. S(-8)3J4.
23. (a+b)2r+(a+b)r-1. 26. (x6)4. 28. (m4)8.
24. (25)3. 27. (?/3)7.
29. (a26)3.
[HINT ro Ex. 29. First use Law IV, then Law III.]
30. (z3!/2)2. 32. (a263c)4. 34. S(a+6)2(c+d)3|4.
31. (a6c)3. 33. (ra2n3w4)3. 35. (xY)2m- 36. (r2<
o\3 /»>?5\4 /rn2
37. f ? ) - 38. ^ ) . 39.
n J
. .
V n J \s
196 SECOND COURSE IN ALGEBRA [XVIII, § 117
40. f*-aY- . 42. c^y Y^y. 44. f-^v.
V2/y V2/V \ «V
43. f-Y-^C-9Y- 45. (- —
W \y*J V 2/3w
118. Roots. Evolution. Just as VcT means the number
whose square gives a, and Vo means the number whose cube
gives a, etc., so we define the nth root of a, Va, to be the
number whose nth power gives a, that is we agree that
Thus Vxl2=x4, because (z4)3=z12. (§ 117, Law III.) Similarly
Va1668 = a462, because (ct462)4 = ct16b8.
NOTE. In case n = 2, we write simply V" instead of v/ .
The number n is called the incfex of the root.
The number under the sign V~, as a, is called the radicand.
The process of finding the root of a number or expression
is called evolution.
119. Rule for Finding the nth Root of an Expression. The
nth root of an expression may be obtained readily in case the
expression itself is an exact nth power. This is illustrated
in the following examples.
EXAMPLE 1. To find the value of \^m*n9.
SOLUTION. The expression men9 may be written as an exact
cube, namely (m2n3)3. (Laws IV and III of § 117.)
Therefore \/m6n9= V(m2n3)3 = m2w3. Ans. (§118)
EXAMPLE 2. To find the value of
SOLUTION. The expression x V may be written as an exact 5th
z15
power, namely f ^) - (Laws IV and III of § 117.)
Therefore A/^-6==A/(^V=^- Ans. (§118)
XVIII, § 119]
EXPONENTS
197
Observe that the answer to Example 1 (namely ra2n3) is
the result of simply dividing each exponent of the radicand
(namely w6n9) by the index of the desired root (namely 3).
Similarly, in Ex. 2 if we simply divide each exponent in
f- by 5 we get the answer immediately. Thus, in practice,
&
we use the following rule.
To find the nth root of an exact nth power, divide the expo-
nent of each factor of the radicand by n.
Thus
NOTE. It jwill be recalled (§ 39) that, unless otherwise stated,
the symbol Va means the positive number whose square is a. Thus
V9=+3, the other root, —3, being represented by — V9. This
agreement is made in order to bring about perfect definiteness in the
use of the symbol V .
EXERCISES
Determine (from the definition in § 1 18) the value of :
_
a
625.
2. V-27.
3. -v/SL 6. \/A. 8.
Determine by means of the Rule in § 119 the value of:
10. ^64 a666. [HINT. Write 26 for 64.]
23.
24.
11. A/625 a864.
12. \/-27m?n«.
13. -
14. -
15.
22.
198 SECOND COURSE IN ALGEBRA [XVIII, § 120
II. FRACTIONAL, ZERO, AND NEGATIVE EXPONENTS
120. Introduction of General Exponents. Thus far we
have considered only positive integral exponents. Such
symbols as a3/4 and or2 thus have no meaning for us as yet
since there can be no such thing as taking a as a factor three
fourths times, or minus two times. However, we shall now see
that by extending our definitions we can assign perfectly
definite meanings to these symbols as well as to all others
wherein fractional, zero, or negative exponents occur.
121. Meaning of a Fractional Exponent. If a374 is to
obey the multiplication law (§ 117) then
a . a . a . a = ^
That is
(a3/4)4 = a3,
so that we must have
= 3/4+3/4+3/4+3/4 _
Thus we naturally take vV to be the meaning of a3/4.
Similarly (if the multiplication law is to hold true), the
meaning of a273 is v'a2, while that of a475 is x/a4, etc.
So, in all cases am/n means the nth root of am, that is,
Qinln _ ^/Qm.
Thus
82/3 = ^82 = ^64 = 4.
Similarly, _ _
(2^4)8/4 = #(x8g4)3 = ^24012 =X*y3. ($66 Rule 111 § 119.)
EXERCISES
Express with radical sign and find the value of :
1. 8173. 6. 27273. 11. G/10)175-
2. 41/2. 7. 32275. 12. (i/10)275.
3. 91/2. 8. 81374. 13.
4. 27173. 9. 64176. 14.
6. (-8)173. 10. (x6)173. 15. S(a+6)3i273.
XVIII, § 123] EXPONENTS 199
Express with radical signs :
16. 22/3. 18. 43/2. 20. (a2)4/3. 22. 2 xl/\ 24. m2/3n3/4.
17. 32/3. 19. a473. 21. (3z)3/4. 23. (2 a&)3/4. 25. (z+t/)5/6.
Express with fractional exponents :
26. V^4. 29. 2v/z*~. 32. Sv^n2". 35. V(a+6)3.
27. V^5. 30. ^(-a)2. 33.
28. v^S. 31. Vabc. 34.
122. Meaning of a Zero Exponent. If a° is to obey the
multiplication law (§117), then am-a° = am+°, that is
am •a° = am. Dividing both members of the last equality
by am gives a° = a™ -f- am = 1 . That is,
This means that the zero power of any number a (except 0)
must always be taken equal to 1.
Thus 3° = 1; (-32)° = l;(£)o = l;xo = i; (mn)° = l; (o+6)°=l;
etc.
123. Meaning of Negative Exponents. If a~m is to obey
the multiplication law (§117), then am • a~m = am-n = a°,
that is am • a~m=l (§ 122). Dividing both members of the
last equation by am gives
This means that a negative power of any number a must al-
ways be taken equal to 1 divided by the corresponding positive
power of a.
2' 8' (-4)' 1
Similarly, („+!>)-"* = — L— —±=.
200 SECOND COURSE IN ALGEBRA [XVIII, § 123
EXERCISES
Express with positive exponents and find the values of
each of the following expressions.
1. S-2. 5. 2-1 • 3-2. 9. 82 • 4~*. 13. 81~1/4.
2. 4~2. 6. 4° - S-3. 10. 8~1/3. 14. 64~1/6.
3. 2~4. 7. 7-4-4. 11. (-8)-1/3. 15. (-125)~1/3.
4. 8°. 8. 2~3 -8 •4-». 12. 27~1/3. 16. (-32)-1/5.
Write with positive exponents each of the following ex-
pressions.
17. x3y~*. 20. (2 a)"3?)3. 23. §~lmin~*.
18. x~2y2<r3. 21. 2~3a-363. 24. (a26c)"2.
19. 2a~363. 22. (-m)-\-ri)-\ 25. \a?(m-n) \~\
124. Negative Exponents in Fractions. This is best
understood from an example.
EXAMPLE. Write a o with positive exponents only.
cr2
1. 53
fl O d O O Cr u C? A /c i oo\
SOLUTION. !^f-™ — i — = —^ 4 ' T= ~T* ^ns. (§ 126)
<?
It is to be observed that the answer here results directly
by transferring the factor cr4 to the denominator by changing
the sign of its exponent, and transferring the factor c~2 to
the numerator by likewise changing the sign of its exponent.
Thus we have the following important principle.
A factor may be transferred from either term (numerator
or denominator) of a fraction to the other provided the sign of
its exponent be changed.
Thus we may write x ^~_^— = -^ — -.
Similarly, 4 q'^c"a= 4 a3b-3c-^d*e-6. Here we have written the
fraction in a form having no denominator, that is as a product.
XVIII, § 125J EXPONENTS 201
EXERCISES
Write each of the following expressions with positive
exponents only.
a-*b 3 a-3
c ' (26)
y-l '
-2
2(c+d)e-4
Change each of the following expressions to the form of a
product.
8
' "
11 JL2. 13 3r3g'2 15
' ' '
125. The Fundamental Laws for Any Rational Exponent.
The five fundamental laws stated in § 117 were there proved
true only for positive integral exponents, but it can be shown
that they hold equally well for fractional, zero, or negative
exponents. As the proof of this fact is long, it will be
omitted from this text. The following illustrations of the
meaning of the laws in such cases should, however, be care-
fully examined.
1 . a6 • a-3 • a4/3 • a273 = a*-3+473+2/3 = a6-^2 = a5. (Law I)
2. = a5/6-(-3) = a5/6+3 = a¥ = a3|> (Law II)
cr3
3. (a-372)475 = a'-372) ' 4/5 = a-675. (Law III)
4. (a-^62)-1/4 = a(-3)(-174) • 62<-1/4> = a374&-172. (Law IV)
202 SECOND COURSE IN ALGEBRA [XVIII, § 125
HISTORICAL NOTE. The idea of using exponents to mark the
power to which a quantity is raised is due to the French mathe-
matician and philosopher Descartes (1596-1650) ; see the picture fac-
ing p. 41), but he used only positive integral exponents, as in a1,
a2, a3, a4, •••. The English mathematician Wallis (1616-1703) en-
couraged the use of fractional and negative exponents and caused
them to be brought into general use.
EXERCISES
In the following exercises, assume that the laws of § 117
hold true for all rational exponents.
LAW I
Multiply
1. a2 by a-1. 8. a1/261/3 by a1/262/3,
2. a3 by a~2. 9. ra1/2rc1/3 by m3/2rc2/3.
3. a3 by or3. 10. p~1/4 q by 4 p5/4.
4. a by cr4. 11. n2 by 6n~3.
5. cr2 by a-3. 12. xm~n by xm+n.
6. a173 by a2/3. 13. z(m+n)/2 by x(m~n)/2.
7. ar1/2byz3/2. 14. a1/2+61/2 by a1/261/2.
16. a2/3+a1/361/3+&2/3 by a1/3-61/3.
[HINT. Follow the rule for multiplying one polynomial by
another, as given in § 8.]
16. m1/3+m1/6n1/6+n1/3 by m1/3-m1/6n1/6+n1/3.
Carry out the following indicated operations.
17. (a1/2+61/2)(a1/2-61/2). 20.
18. (x2/3+2/2/3)(x2/3-i/2/3). 21.
19. (a1/2+61/2)2. 22.
23. (z2
24. (x2
25. (2z2/3-3z1/3+4)(2+3ar1/3).
26. x2-
XVIII, § 125] EXPONENTS 203
LAW II
Divide
27. a4 by a5. 29. z2 by or2. 31. (ran)2/3 by (mnY/s
28. a3bya°. 30. z372 by z~1/2. 32. xl/ Y/2 by y~l/2.
33. x*-\-x2y2-\-y* by z2?/2.
SOLUTION. x ~*~x ^ +3r=_g — |_?J£. _| — ^_=^_.
X2^2 X2^2 X2^2 X2^2 ^/2
34. a3+a2+abya4.
35. a~4+a~26+62 by a~2b.
36. x4-|-2 az3-|-5 z"1?/ — Q?]r^~}~y* by x—12/~2.
37. a — 6 by a1/2+61/2.
SOLUTION, a —61
a+a1/2b1/2 a1/2-61/2.
— a1/261/2 — 6
38. a-6bya1/2-61/2. 41. z-1 by z2/3+z1/3+l.
39. a-f6 by a1/3+61/3. 42. x - y by z174 - 1/1/4.
40. z2+2/2 by z2/3+2/2/3. 43. m2-n3 by w1/3+n1/2.
LAW III
Simplify
44. (a1/2)3.
SOLUTION. As in Illustration 3 of § 125, the answer is a1/2X3,
or a3/2.
45. (a1/2)2. 46. (or173)6. 47. (or5)2. 48."(8-1/3)2. 49. (16~1/2)3.
50. Var2. [HINT. V^ = (a~2)1/2 by § 121.]
61. VaF"2. 52. v^z372.
204 SECOND COURSE IN ALGEBRA [XVIII, § 125
LAWS IV AND V
Simplify
63.
SOLUTION. As in Illustration 4 of § 125, we have (a~462/3)-1/2 =
4)(-l/2) . 52/3(-l/2) = a25-l/3.
64. (al73&-172)6.
65. (z4732T3)-173.
66. vV1726-3.
57. Vz473?/-3.
58.
59.
60.
61.
[HINT to Ex. 62. See Illustration 5 in § 125.]
63. . 64.
* MISCELLANEOUS EXERCISES
Expand, by use of Formulas VI and VII of § 10.
1. (a:172-i/172)2. 3. (a173+6173)2.
2. (a^+Zr1)2. 4. (1+2 x172)2.
Simplify, expressing results with positive exponents.
6.
-VWxVa-*
z y
^T^/W" 10.
Solve forz and check each result in the following equations.
11. z3/4 = 8. [HiNT. Write z3'4 in the form Or1/4)3.]
12. z275 = 9. 14. £z372 = 72. 16. 25ar273 = l.
13. z473 = 16. 16. la;273 = 25. 17. or372 -27 = 0.
CHAPTER XIX
RADICALS
126. Important Formulas. In § 3 we defined VcTas mean-
ing that number or expression which, when raised to the nth
power, would give a ; that is
(1) (W = a.
Unless a is an exact nth power of some number or expres-
sion, we agreed (§ 41) to call Va a radical of the nth order.
Thus V5, V23, Vj, V.05, Vx+y, Vm2+n2 are radicals of the
second order; #5~, v^ ^f' v/x+y are radicals of the third order,
etc.
Moreover, we saw (§ 44) that there exist two general
formulas as follows :
(2)
_ Va
(3)
And, in connection with the study of fractional exponents,
we havejseen (§ 121) that the meaning of a1/n must be taken
to be Va, that is we have the formula
(4) a1/n=Va.
The four formulas (1), (2), (3), (4) contain all that is
essential in the study of radicals. In fact, we have already
seen in Chapter IX how (1), (2), and (3) are thus used. In
the present chapter we shall review and extend those studies,
making use now of (4) also.
205
206
SECOND COURSE IN ALGEBRA [XIX, § 127
127. Simplification of Radicals.
EXAMPLE 1. Simplify V75.
SOLUTION. V75 = V25X3 = V25 X Vjj = 5 Vs. Ans.
(Formula (2), § 126)
Ay-J A fit *T
EXAMPLE 2. Simplify
SOLUTION.
Ans.
EXERCISES
Before undertaking the following exercises, review § 44, including
the note on page 72. These resemble the exercises on pages 73, 74,
but in some instances are more difficult because they refer to radicals
of as high orders as the fifth, sixth, and seventh.
Simplify each of the following expressions.
3/3. ^3
\«* 9
29.
30. A —
31.
27
32.
33.
34.
8/ 4
\125
[HINT. See Note in § 45.]
35.
81
37.
XIX, § 128] RADICALS 207
EXERCISES
Write each of the following in a form having no coefficient
outside the radical sign. First review the similar exercises
on page 74.
1. 3^2.
SOLUTION. 3^2 = ^27X^2 = v/27><2 = ^54. Ans.
2. 2\/3. 7. 4oV2a. 12.
3. 2\/2. 8. GsVjfi?. 2r
13 ^
4. 3A/3. 9.
5. 2"V/f. 10. mv^n. 2x
6. 3^6. 11. 2a^. T 2
128. Reduction of Radicals of Different Orders to Equiva-
lent Radicals of the Same Order.
EXAMPLE. Reduce V% v^, and ^5 to equivalent radi-
cals of the same order.
SOLUTION. By use of Formula (4), § 126, and the laws of expo-
nents (§ 117) we may write
V2"= 21/2 = 26/12 = v^» = V64,
#3= 31/3 = 34/12 = v^3* = v/81,
^5 =51/4 =53/12 = ^5?= 1^
NOTE. As now expressed, the given radicals may be compared
as to their magnitudes. Thus we see V5 is the greatest of the
three since (the orders of the radicals being now the same) it has the
largest radicand, namely, 125.
An examination of the process just followed leads to the
following rule.
To reduce radicals to equivalent radicals of the same order :
1. Express the radicals with fractional exponents.
2. Reduce the exponents to a common denominator.
3. Rewrite the results thus obtained in radical form.
208
SECOND COURSE IN ALGEBRA [XIX, § 128
EXERCISES
Reduce each of the following groups to equivalent radicals
of the same order.
1. V3andV?. 3. Vjfand V<T
2. V2 and Vs. 4. V5, V6, and
5. Vo~and VS. Va3 and VP". Ans.
6. V2~i and \^3x.
7. Vo-?/, V?/2, and Vxz.
8.
9.
Arrange the following in order of magnitude. (See § 128.)
11. V3, -^4. 14. VI, Vl3.
12. V2, V3. 16. V2, V4, V6.
13. VlO, V5. 16. Vl3, V7, vT74.
129. Multiplication of Radicals. We have already seen
in § 46 how to multiply two or more radicals of the second
order. Radicals of higher order than the second may be
multiplied by means of the general formula (2) of § 126.
EXAMPLE 1. VI- VIO = V40. (Formula (2), §126.) Sim-
plifying (§ 127), V40 = 2VB. Ans.
EXAMPLE 2. VTx -
But
= V(4z)3 • (2z2)2. (Formula (2), § 126)
V(4 a
2)2= V43 • 22 • a;3 • x*= V26 • 22 • x1
= V(2 x)6 - (4 x) = 2 x V4z. Ans.
Note that in Ex. 2 the given radicals are- of different orders,
in which case the first step is to reduce them to equivalent radicals
of the same order (§ 128). Then apply. (2) of § 126.
XIX, § 130] RADICALS 209
In general, we have the following rule.
To multiply radicals of any orders :
1. Reduce the radicals, if necessary, to equivalent radicals
of the same order (§ 128).
2. Multiply the resulting radicals by use of Formula (2),
§126.
3. Simplify the result as in §§ 44 and 127.
EXERCISES
[Compare with the exercises on page 76.]
Find each of the following indicated products.
We have already seen in § 47
how one radical of the second order may be divided by another
of that order. If the radicals are of higher order than the
second, we may divide them by means of the general formula
(3) of § 126.
EXAMPLE 1. Divide vT6 by
SOLUTION. = = (Formula (3), §126)
The simplest and most desirable form in which to leave this
result is that in which no fraction appears except outside the
sign. Thus
210 SECOND COURSE IN ALGEBRA [XIX, § 130
EXAMPLE 2. Divide V3 by \/9.
\i« (Formula (3), § 126)
But ^^E-iSLl*
EXAMPLE 3. Divide \^3xyby \/3 X2y*.
SOLUTION. -E = ^^ = ^^= $•
V3r.2?y3 -v/s T2,;3 *3 o;2-?;3 "ti
But
In general, we have the following rule.
To divide one radical by another :
1. Reduce the radicals, if necessary, to equivalent radicals
of the same order (§ 128).
2. Divide the resulting radicals by use of Formula (3) (§ 126).
3. Simplify the result in such a way that no fraction appears
except outside the radical sign.
EXERCISES
[Compare with the exercises on page 77.]
Find each of the following indicated quotients.
1. V7 + V2. 4. 2 + \/2. 7. \/l2x2+V2~x.
5. 3-^3. 8. v^81 xzy+\/9~xy.
-r- ^32 m.
9.
13.
14.
15.
XIX, § 132] RADICALS 211
131. Involution and Evolution of Radicals. By use of
Formula (4) of § 126 together with the general laws of expo-
nents (§ 117), we may raise a radical to a power or extract
a root of it.
EXAMPLE 1. Find the square of \/8.
SOLUTION. ( v'S)2 = (81/4)2 =82/4 =81/2 = V8 =2 V2. Ans.
EXAMPLE 2. Find the fourth root of V2 x.
SOLUTION. ^ V2x = [(2 x)1/2]1/4 = (2 zW* = \/2aT. Arts.
EXERCISES
Perform each of the following indicated involutions.
1. (v/3)2. 6. (2V3)3.
2. (v/S)2. 7. (3V2)3.
3. (3Vi)2. 8.
4. (2^3l02. 9.
5. x2vT2. 10.
Perform each of the following indicated evolutions.
16. ^V2.
17. V^f.
is. v^p;
19. V</|9.
132. Rationalizing the Denominator of a Fraction. If the
denominator of a fraction consists of a single quadratic radi-
cal, or is a binomial containing quadratic radicals, the frac-
tion may be changed into one which has radicals only in its
numerator. The process of doing this is called rationalizing
the denominator.
212 SECOND COURSE IN ALGEBRA [XIX, § 132
EXAMPLE 1. Rationalize the denominator in the fraction
Vf
VB
SOLUTION. Here the denominator contains the single surd V5.
To rationalize this denominator it is merely necessary to multiply
both numerator and denominator by V5^ giving Vl5/5. Ans.
EXAMPLE 2. Rationalize the denominator in the fraction
V3+V2
SOLUTION. Multiply both numerator and denominator by
V3 — V2, giving
V3-V2 = V3-V2 ^ V3- V2 = V3- V2
(V3-V2)(V3 + V2) (V3)2-(V2)2 3-2 1
= V3-V2. Ans.
EXAMPLE 3. Rationalize the denominator in the fraction
3V5+2V2
V5-V2
SOLUTION. Multiplying both numerator and denominator by
V5 + V2,- we have
= 3- 5+3V10+2V10+2 • 2
5-2
= 19+|V10 AnSf
We may then state the following rule.
To rationalize the denominator of a fraction :
If the denominator contains a single radical, multiply both
numerator and denominator by that radical.
If the denominator has either of the binomial forms Va+ Vb
or Va— VS, multiply both numerator and denominator by
Va — \/b, or Va-j-VS according as we have the first or second
of these two cases.
XIX, § 133] RADICALS 213
EXERCISES'
Rationalize the denominators in each of the following
fractions.
5 V3+V2 9 3V3-2V2
VI V3-A/2 ' 2V3+3\/2
6- -7F— ' 10.
2\/5
7. - — -. - 2\/q-3\/b
s. j
133. Finding the Value of Fractions Containing Radicals.
Suppose we wish to find the value of
V3+V2
correct to five places of decimals. It is well to begin by
rationalizing the denominator, thus making the fraction take
the form (see Ex. 2 worked in § 132) \/3 - \/2. All we now
need to do is to look up in the table the values of V3 and V2
so as to work out the value of A/3 — \/2. That is, we have
=V3-\2 = 1.73205+-1.41421+ = 0.31784+. Ans.
V3+V2
If, in this example, we had not first rationalized the denomi-
nator, we should have. had to find the value of
1
or
1.73205++ 1.41421+' 3.14626+'
which would compel us to divide 1 by 3.14626. Note how
much more difficult this is than the above, where virtually
all we need to do is to subtract 1.41421 from 1.73205.
214 SECOND COURSE IN ALGEBRA [XIX, § 133
This illustrates the general fact that to find the value of a
fraction, its denominator (if it contains radicals) should first
be rationalized whenever possible.
EXERCISES
Find (by first rationalizing the denominator and then using
the table) the approximate values of the following fractions.
2 , 1_ , 1
1. — — . 3. — — — • 5.
V3 V3-\/2 V250
2\/5-4
3V5 2-V3 ' 3V3-2
*134. Binomial Surd. A binomial, one or both of whose terms
are surds (§ 42), is called a binomial surd.
Thus 2 + A/5, A/2 + A/5, A/3-1 are binomial surds.
*135. To Find the Square Root of a Binomial Surd. The famil-
iar formula for (a +6)2 (§ 10, Formula VI) may be put into the form
(1) a2+62+2 ab = (a+&)2.
Since this relation holds true for any values of a and 6, let us sup-
pose in particular that^both are positive, in which case we may
write a = A/Z and b = Vy, where x and y are properly chosen positive
values. The equation just written then takes the form
Extracting the square root of both members now gives
(2) V
This formula, having been thus derived from (1), must therefore
hold true for any positive values of x and y.
Similarly, by starting with the familiar formula for (a — 6)2, we
arrive at the formula
(3)
Formulas (2) and (3) are frequently used to obtain the square
root of a binomial surd (§ 134) as illustrated in the following
example.
XIX, § 135] RADICALS 215
EXAMPLE. Find the square root of the binomial surd 11 +4 V7-
SOLUTION. We are to find V'n +4 V?.
This may be written Vn +2 V28, and it is now in the form of the
first member of Formula (2), provided we choose x and y so that
x +y = 11 while xy = 28.
The values of x and y which satisfy these last two equations are
seen (by inspection) to be x = 4, y = 7.
Substituting these values of x and y in the second member of
(2) gives V4 + V7=2 + V7.
Therefore V^H +4 V? =2 + V?. Ans.
CHECK.
(2 + V7)2=22+2- 2
The pupil will observe that all that is essential in working the
above example is to write the given surd term (4V7) so that it has
the coefficient 2 instead of 4. Similarly, all such problems may be
brought under Formulas (2) or (3) as soon as the coefficient of the
surd term has been reduced to 2.
* EXERCISES
Find the square root of each of the following expressions, and
check your answer.
1. 6+2V8.
[HINT. Herez+?/=6, xy =8.]
2. 6-2V8". 4. ll-2V3a 6. 6 + V32. 8. 8+4>/3.
3. 7 +4 Vs. 5. 6-V2a 7. 7-V|a 9. 20-6VIT.
10. Establish Formula (3) of § 135 by a process similar to that
used in establishing Formula (2).
CHAPTER XX
LOGARITHMS
I. GENERAL CONSIDERATIONS f
136. Definition of Logarithms. If we ask what power
of 10 must be used to give a result of 100, the answer is 2
because 102=100. Another common way of stating this is
to say that " the logarithm of 100 is 2." In the same way,
the power of 10 needed to give 1000 is 3 because 103 = 1000,
and this is briefly stated by saying that " the logarithm of
1000 is 3." Similarly, the power of 10 that gives .1 is —1
because 10~1=11jy, or .1 (§ 123), and this is equivalent to
saying that " the logarithm of .1 is — 1." Likewise, the loga-
rithm of .01 is -2. Why?
From these illustrations we readily see what is meant by
the logarithm of a number. It may be denned as follows :
The logarithm of a number is the power of 10 required to
give that number.
NOTE. A more general definition will be given in § 151, but this
is the one commonly used in practice.
The fact that the logarithm of 100 is 2 is written log 100 = 2.
Similarly, we have log 1000 = 3, log .1 = - 1, log .01 = -2, etc.
t Parts I and II give definitions and essential theorems which
should be well understood before Part III, which describes the im-
portant applications, is taken up.
216
XX, § 137] LOGARITHMS 217
EXERCISES
1. What is the meaning of log 10000 ? What is its
value f
» 2. What is the value of log .001? Why?
3. What is the value of log .00001 ? Why?
4. What is the value of log 10?
5. What is the value of log 1 ? (See § 122.)
6. As a number increases from 100 to 1000 how does its
logarithm change?
7. As a number decreases from .1 to .01 how does its
logarithm change? Answer the same as the number goes
from .01 to .001 ; from 1 to 10 ; from 1 to 1000.
8. Explain why the following are true statements :
(a) log 100000 = 5. (6) log .0001 =-4.
(c) logv^0 = i.
[HINT. Remember VlO = 101/2.
(d) log >^IO = i.
(e) log ^100 = 1.
[HINT. Remember \/100 = v/102 = 102'3. (§ 121.)]
137. Logarithm of Any Number. Suppose we ask what
the value is of log 236. What we are asking for (see defini-
tion in § 136) is that value which, when used as an exponent
to 10, will give 236 ; that is we wish the value of x which
will satisfy the equation 10X = 236. This question resembles
those in § 136, but is different because we cannot immediately
arrive at the desired value of x by mere inspection. All we
can say here at the beginning is that x must lie somewhere
between 2 and 3, because 102=100 and 103=1000, and 236
lies between these two numbers. In order to find z to a finer
degree of accuracy, it is now natural to try for it such values
218 SECOND COURSE IN ALGEBRA [XX, § 137
as 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, and 2.9, all of which lie
between 2 and 3. The result (which for brevity we shall
here state without proof) is that when x = 2.3 the value of 10*
is slightly less than our given number 236, while if we take
x= 2.4 the value of 10* is slightly greater than 236. Thus x
lies somewhere between 2.3 and 2.4. In other words, the
value of log 236 correct to the first decimal place (compare
§ 37) is 2.3.
It is now natural, if we wish to obtain x to still greater
accuracy, to try for it such values as 2.31, 2.32, 2.33, 2.34,
2.35, 2.36, 2.37, 2.38, and 2.39, all of which lie between 2.3
and 2.4. The result (which again is here stated without
proof) is that when x = 2.37 the value of 10X is slightly less
than our number 236, while if we take # = 2.38 the value of
10* is slightly greater than 236. This means that the second
figure of the decimal is 7, after which we may say that the
value of log 236 correct to two places of decimals is 2.37.
Proceeding farther in the same manner, it can be shown
that when x = 2.372 the value of 10* is slightly less than 236,
while for x = 2.373 the value of 10* is slightly greater than 236.
Thus the value of log 236 correct, to three places of decimals is
2.372. Similarly, it can be shown that the number in the
fourth decimal place is 9, and this is as far as it is necessary
to carry out the process, since the result is then sufficiently
accurate for all ordinary purposes.
In summary, then, we have log 236 = 2.3729, this value
being correct to four places of decimals.
NOTE. It thus appears that logarithms do not in general come
out exact, though they do so for such exceptional numbers as
100, 1000, 10,000, .1, .01, etc. (Compare § 37.) They can be ex-
pressed only approximately, yet as accurately as one pleases by
carrying out the decimal far enough. In this respect they resemble
such numbers as V2. \/2, V3, etc.
XX, § 138] LOGARITHMS 219
Other examples of logarithms are given below. Note
especially the decimal part of each, which is correct to four
places.
log 283 = 2.4518 log 196 = 2.2923 log 17 = 1.2304
log 6 = 0.7782 log 3.410 = 0.5328 log 5.75 = 0.7597
138. Characteristic. Mantissa. We have seen that the
logarithm of a number consists (in general) of an integral
part and a decimal part.
Thus log 236 =2.3729. Here the integral part is 2 and the deci-
mal part is .3729. Similarly, in log 6 = 0.7782 the integral part is 0,
while the decimal part is .7782.
These two parts of every logarithm are given special names
as follows :
The integral part of a logarithm is called the characteristic
of the logarithm.
The decimal part of a logarithm is called the mantissa of
the logarithm.
Thus the characteristic of log 236 is 2, while its mantissa is
.3729. (See above.) Similarly, the characteristic of log 6 is 0, while
its mantissa is .7782.
EXERCISES
1. What is the characteristic of log 100? What the
mantissa? Answer the same questions for log 1000, log 10,
and log 1.
2. What is the characteristic of log 156 ?
[HINT. Note that 156 lies between 102 and 103.]
3. What is the characteristic of log 276? of log 1376? of
log 97? of log 18? of log 5? of log 11? of log 14798-?
4. For what kind of number can one tell by inspection
both the characteristic and the mantissa of its logarithm?
(See § 136.)
220 SECOND COURSE IN ALGEBRA [XX, § 1I19
139. Further Study of Characteristic and Mantissa. We
have seen (§ 138) that log 236 = 2.3729, which is the same as
saying that
(1) 102-3729 = 236.
Let us now multiply both members of (1) by 10. The
left side becomes 102-3729+1 or 103-3729 (§ 117, Law I) while
the right side becomes 2360. That is, we have 103-3729 =
2360, which is the same as saying that
log 2360 = 3.3729
If, instead of multiplying both sides of (1) by 10, we divide
both by 10, we obtain in like manner lO2-3729"1 = 23.6 (§ 117,
Law II). That is, we have 101-3729 = 23.6, which is the same
as saying that
log 23.6 = 1.3729
Finally, if we divide both sides of (1) by 102, or 100, we
obtain 102-3729~2 = 2.36. That is, we have 10°-3729 = 2.36 which
is the same as saying that
log 2.36 = 0.3729
What we now wish to do is to compare the results which we
have just been obtaining, and for this purpose they are ar-
ranged side by side in a column below.
log 2360 = 3.3729
log 236 = 2.3729
log 23.6=1.3729
log 2.36 = 0.3729
Note that the mantissas here appearing on the right are
all the same, namely .3729, while the numbers appearing
on the left (that is, 2360, 236, 23.6, and 2.36) are alike except
for the position of the decimal point, that is they contain
the same significant figures. This illustrates the following
important rule.
(2)
XX, § 141] LOGARITHMS 221
RULE I. // two or more numbers have the same significant
figures (that is, differ only in the location of the decimal point) ,
their logarithms will have the same mantissas, that is their
logarithms can differ only in their characteristics.
Thus log 243, log 2430, log 24.3, log 2.43, log .243, and log .0243
all have the same mantissas. It is only their characteristics that
can be different.
EXERCISES
Apply Rule I, § 139, to tell which of the following loga-
rithms have the same mantissas.
log .167 log 8100 log 16.7 log 81 log .0072
log .081 log 7.2 log 720 log 1670 log 16700
II. To DETERMINE THE LOGARITHM OF ANY NUMBER
140. Purpose of This Part. When we wish to determine
the value of a logarithm, as, for example, to find log 236, we
can work out the characteristic and mantissa as explained
in § 137, but this requires considerable time. What we do
in practice is to use certain simple rules for determining the
characteristic, and we determine the mantissa directly from
certain tables which have been carefully prepared for the
purpose. We shall now state these rules (§§ 141-143)
and explain the tables and how to use them (§§ 144-146).
141. Characteristics for Numbers Greater than 1. If we
look again at the results in (2) of § 139, we see that the
characteristic of log 2360 is 3. Thus the characteristic is 1
less than the number of figures to the left of the decimal
point.
NOTE. 2360 is the same as 2360., so that there are four figures
here to the left of the decimal point.
SECOND COURSE IN ALGEBRA [XX, § 141
Again, we see from (2) of § 139 that the characteristic of
log 236 is 2 and this, as in the case already examined, is 1 less
than the number of figures to the left of the decimal point.
NOTE. 236 is the same as 236., so there are three figures here to
the left of the decimal point.
Similarly, since the characteristic of log 23.6 is 1 (see (2)
of § 139) this again obeys the same law as just observed in
the other two cases, that is, the characteristic is 1 less than the
number of figures to the left of the decimal point.
Finally, since the characteristic of log 2.36 is 0, the same
law is again present here. Explain.
The law which we have just observed can be shown in like
manner to hold good for the characteristic of the loga-
rithm of any number greater than 1 ; hence we may state
the following general rule.
RULE II. The characteristic of the logarithm of a number
greater than 1 is one less than the number of figures to the left
of the decimal point.
Thus the characteristic of log 385.9 is 2 ; that of log 8.679 is 0.
EXERCISES
State, by Rule II, § 141, the characteristic of the logarithm
of each of the following numbers.
1. 476.5 5. 89.65 9. 500.005
2. 325. 6. 105,000. 10. 3076.8
3. 8976. 7. 17.694 11. 41.
4. 1.6 8. 2.0815 12. 3.25679
State how many figures precede the decimal point of a
number if the characteristic of its logarithm is
13. 3. 14. 2. 15. 0. 16. 1. 17. 4. 18. 5.
XX, § 142] LOGARITHMS 223
142. Characteristics for Positive Numbers Less Than 1.
We have seen (see (2) in § 139) that log 2.36 = 0.3729, which
is the same as saying that
(1) 10°-3729 = 2.36
Let us now divide both members of this relation by 10.
We thus obtain (§ 117, Law II)
10o.3729-i = .236 (or 10-1+0-3729 = .236) ,
which gives us (by § 136)
log .236 =-1+0.3729
Observe that — 1 +0^3729 is really a negative quantity, being
equal to -(1-0.3729) which reduces to -0.6271. However, it is
more convenient for our present purposes to keep the longer form
— 1+0.3729. Note that this cannot be written as —1.3729 be-
cause this last is equal to -1-0.3729 instead of -1+0.3729.
If, instead of dividing both members of (1) by 10, we
divide both by 102, or 100, we obtain
JQO.3729-2 = <0236 (or lQ-2+0.3729 = .Q236),
which means that
log .0236 =-2+ 0.3729
Similarly, by dividing (1) by 103, or 1000, we find that
log .00236= -3+0.3729
Finally, if we divide (1) by 104, or 10000, we find that
log .000236= -4+0.3729
Let us now compare the four results just obtained. Be-
ginning with the last result, we see that in the number
.000236 there are three zeros immediately to the right of the
decimal point, that is, between the decimal point and the
first significant figure. Corresponding to this, the charac-
teristic on the right is minus four. Hence the characteristic
is negative and 1 more numerically than the number of zeros
between the decimal point and the first significant figure.
224 SECOND COURSE IN ALGEBRA [XX, § 142
Similarly, in the number .00236 there are two zeros between
the decimal point and the first significant figure, and corre-
sponding to this there is a characteristic on the right of
minus three. Hence, as before, the characteristic here is
negative and numerically 1 more than the number of zeros
between the decimal point and the first significant figure.
This statement, which is true in all cases mentioned above, can
be proved for the characteristic of the logarithm of any
positive number less than 1. Hence we have the following
rule.
RULE III. The characteristic of the logarithm of a (positive)
number less than 1, is negative, and is numerically 1 greater
than the number of zeros between the decimal point and the first
significant figure.
Thus the characteristic of log .0076 is -3 ; that of log .28 is -1.
NOTE. The logarithm of a negative number is an imaginary
quantity (as shown in higher mathematics), and hence we shall con-
sider here the logarithms of positive numbers only.
143. Usual Method of Writing a Negative Characteristic.
In § 142 we saw that log .236= -1+0.3729. If we add 10
to this quantity and at the same time subtract 10 from it,
we do not change its value, but we give it the new form
9+0.3729-10, which is the same as 9.3729-10. That is,
we may write.
log .236 = 9.3729 -10.
This is the form used in practice.
Likewise, instead of writing log .0236= -2+0.3729 (see
§ 142) we write in practice
log .0236 = 8.3729-10,
and similarly we write
log .00236 = 7.3729 -10.
XX, § 144] LOGARITHMS 225
Thus the usual method of expressing the characteristic
of — 1 is to write 9—10 for it ; if it is — 2, we write 8 — 10 for
it ; if it is —3, we write 7 — 10 for it, etc.
For example, log .0076 has the characteristic 7 — 10.
EXERCISES
State, by Rule III, § 142, the value of the characteristic
of the logarithm of each of the following ; state how it would
be written if expressed in the usual form described in § 143.
1. .06 -2, or 8-10. Ans.
2. .0087 5. .0835 8. .00978
3. .75 6. .835 9. .12345
4. .00067 7. .33764
How many zeros lie between the decimal point and the first
significant figure of a number when the characteristic of its
logarithm is
10. -3. 11. 9-10. 12. -5. 13. 8-10. 14. 7-10.
144. Determination of Mantissas. Use of Tables. Sup-
pose we wish to determine completely the value of log 187.
By Rule II, § 141, we know that the characteristic is 2.
To find the mantissa, we turn to the tables (p. 226) and
look in the column headed N for the first two figures of the
given number, that is, for 18. The desired mantissa is then
to be found on the horizontal line with these two figures and
in the column headed by the third figure of the given num-
ber, that is, in the column headed by 7. Thus in the present
case the mantissa is found to be .2718.
NOTE. For brevity, the decimal point preceding each mantissa
is omitted from the tables. It must be supplied as soon as the
mantissa is used.
The complete value (correct to four decimal places) of log
187 is therefore 2.2718.
226
SECOND COURSE IN ALGEBRA
N
O
1
2
3
4
5
6
7
8
9
1O
11
12
13
14
0000
0414
0792
1139
1461
0043
0453
0828
1173
1492
0086
0492
0864
1206
1523
0128
0531
0899
1239
1553
0170
0569
0934
1271
1584
0212
0607
0969
1303
1614
0253
0645
1004
1335
1644
0294
0682
1038
1367
1673
0334
0719
1072
1399
1703
0374
0755
1106
1430
1732
15
16
17
18
19
1761
2041
2304
2553
2788
1790
2068
2330
2577
2810
1818
2095
2355
2601
2833
1847
2122
2380
2625
2856
1875
2148
2405
2648
2878
1903
2175
2430
2672
2900
1931
2201
2455
2695
2923
1959
2227
2480
2718
2945
1987
2253
2504
2742
2967
2014
2279
2529
2765
2989
2O
21
22
23
24
3010
3222
3424
3617
3802
3032
3243
3444
3636
3820
3054
3263
3464
3655
3838
3075
3284
3483
3674
3856
3096
3304
3502
.3692
3874
3118
3324
3522
3711
3892
3139
3345
3541
3729
3909
3160
3365
3560
3747
3927
3181
3385
3579
3766
3945
3201
3404
3598
3784
3962
25
26
27
28
29
3979
4150
4314
4472
4624
3997
4166
4330
4487
4639
4014
4183
4346
4502
4654
4031
4200
4362
4518
4669
4048
4216
4378
4533
4683
4065
4232
4393
4548
4698
4082
4249
4409
4564
4713
4099
4265
4425
4579
4728
4116
4281
4440
4594
4742
4133
4298
4456
4609
4757
30
31
32
33
34
4771
4914
5051
5185
5315
4786
4928
5065
5198
5328
4800
4942
5079
5211
5340
4814
4955
5092
5224
5353
4829
4969
5105
5237
5366
4843
4983
5119
5250
5378
4857
4997
5132
5263
5391
4871
5011
5145
5276
5403
4886
5024
5159
5289
5416
4900
5038
5172
5302
5428
35
36
37
38
39
5441
5563
5682
5798
5911
5453
5575
5694
5809
5922
5465
5587
5705
5821
5933
5478
5599
5717
5832
5944
5490
5611
5729
5843
5955
5502
5623
5740
5855
5966
5514
5635
5752
5866
5977
5527
5647
5763
5877
5988
5539
5658
5775
5888
5999
5551
5670
5786
5899
6010
40
41
42
43
44
6021
6128
6232
6335
6435
6031
6138
6243
6345
6444
6042
6149
6253
6355
6454
6053
6160
6263
6365
6464
6064
6170
6274
6375
6474
6075
6180
6284
6385
6484
6085
6191
6294
6395
6493
6096
6201
6304
6405
6503
6107
6212
6314
6415
6513
6117
6222
6325
6425
6522
45
46
47
48
49
6532
6628
6721
6812
6902
6542
6637
6730
6821
6911
6551
6646
6739
6830
6920
6561
6656
6749
6839
6928
6571
6665
6758
6848
6937
6580
6675
6767
6857
6946
6590
6684
6776
6866
6955
6599
6693
6785
6875
6964
6609
6702
6794
6884
6972
6618
6712
6803
6893
6981
50
51
52
53
54
6990
7076
7160
7243
7324
6998
7084
7168
7251
7332
7007
7093
7177
7259
7340
7016
7101
7185
7267
7348
7024
7110
7193
7275
7356
7033
7118
7202
7284
7364
7042
7126
7210
7292
7372
7050
7135
7218
7300
7380
7059
7143
7226
7308
7388
7067
7152
71':;.-,
7316
7396
LOGARITHMS
227
N
0
1
2.
3
4
5
6
7
8
9
55
56
57
58
59
7404
7482
7559
7634
7709
7412
7490
7566
7642
7716
7419
7497
7574
7649
7723
7427
7505
7582
7657
7731
7435
7513
7589
7664
7738
7443
7520
7597
7672
7745
7451
7528
7604
7679
7752
7459
7536
7612
7686
7760
7466
7543
7619
7694
7767
7474
7551
7627
7701
7774
6O
61
62
63
64
7782
7853
7924
7993
8062
7789
7860
7931
8000
8069
7796
7868
7938
8007
8075
7803
7875
7945
8014
8082
7810
7882
7952
8021
8089
7818
7889
7959
8028
8096
7825
7896
7966
8035
8102
7832
7903
7973
8041
8109
7839
7910
7980
8048
8116
7846
7917
7987
8055
8122
65
66
67
68
69
8129
8195
8261
8325
8388
8136
8202
8267
8331
8395
8142
8209
8274
8338
8401
8149
8215
8280
8344
8407
8156
8222
8287
8351
8414
8162
8228
8293
8357
8420
8169
8235
8299
8363
8426
8176
8241
8306
8370
8432
8182
8248
8312
8376
8439
8189
8254
8319
8382
8445
70
71
72
73
74
8451
8513
8573
8633
8692
8457
8519
8579
8639
8698
8463
8525
8585
8645
8704
8470
8531
8591
8651
8710
8476
8537
8597
8657
8716
8482
8543
8603
8663
8722
8488
8549
8609
8669
8727
8494
8555
8615
8675
8733
8500
8561
8621
8681
8739
8506
8567
8627
8686
8745
75
76
77
78
79
8751
8808
8865
8921
8976
8756
8814
8871
8927
8982
8762
8820
8876
8932
8987
8768
8825
8882
8938
8993
8774
8831
8887
8943
8998
8779
8837
8893
8949
9004
8785
8842
8899
8954
9009
8791
8848
8904
8960
9015
8797
8854
8910
8965
9020
8802
8859
8915
8971
9025
8O
81
82
83
84
9031
9085
9138
9191
9243
9036
9090
9143
9196
9248
9042
9096
9149
9201
9253
9047
9101
9154
9206
9258
9053
9106
9159
9212
9263
9058
9112
9165
9217
9269
9063
9117
9170
9222
9274
9069
9122
9175
9227
9279
9074
9128
9180
9232
9284
9079
9133
9186
9238
9289
85
86
87
88
89
9294
9345
9395
9445
9494
9299
9350
9400
9450
9499
9304
9355
9405
9455
9504
9309
9360
9410
9460
9509
9315
9365
9415
9465
9513
9320
9370
9420
9469
9518
9325
9375
9425
9474
9523
9330
9380
9430
9479
9528
9335
9385
9435
9484
9533
9340
9390
9440
9489
9538
90
91
92
93
94
9542
9590
9638
9685
9731
9547
9595
9643
9689
9736
9552
9600
9647
9694
9741
9557
9605
9652
9699
9745
9562
9609
9657
9703
9750
9566
9614
9661
9708
9754
9571
9619
9666
9713
9759
9576
9624
9671
9717
9763
9581
9628
9675
9722
9768
9586
9633
9680
9727
9773
95
96
97
98
99
9777
9823
9868
9912
9956
9782
9827
9872
9917
9961
9786
9832
9877
9921
9965
9791
9836
9881
9926
9969
9795
9841
9886
9930
9974
9800
9845
9890
9934
9978
9805
9850
9894
9939
9983
9809
9854
9899
9943
9987
9814
9859
9903
9948
9991
9818
9863
9908
9952
9996
228 SECOND COURSE IN ALGEBRA [XX, § 144
Again, suppose we wish to determine log 27.6. The char-
acteristic (by § 141) is 1. The mantissa, by Rule I, § 139,
is the same as that of log 276 ; and the latter, as given in the
tables, is .4409. Therefore, log 27.6 = 1.4409. Ans.
As a last example, suppose we wish to determine log .0173.
The characteristic (by § 142) is — 2, or 8-10. The mantissa,
by the rule in § 139, is the same as that of log 173 and
the latter, as obtained from the tables, is .2380. Therefore,
log .0173 = 8.2380 -10. Ans.
These examples illustrate how the tables together with
Rules II and III, §§ 141, 142, enable us to determine com-
pletely the logarithm of any number provided it contains
no more than three significant figures. We may now sum-
marize our results in the following rule.
RULE IV. To find the logarithm of a number of three signifi-
cant figures :
1. Look in the column headed N for the first two figures of the
given number. The mantissa will then be found on the hori-
zontal line opposite these two figures and in the column headed
by the third figure of the given number.
2. Prefix the characteristic according to Rules II and III,
§'§ 141, 142.
EXERCISES
Determine the logarithm of each of the following numbers,
expressing all negative characteristics as explained in § 143.
1. 451. 2. 318. 3. 861. 4. 900.
5. 72.5 [HINT. Note how log 27.6 was obtained in § 144.]
6. 7.25 7. 93. [HINT. Write as 93.0]
8. 9. [HINT. Write as 9.00] 9. .0136
10. .936 11. .0036 [HINT. Write as .00360]
XX, § 145] LOGARITHMS 229
12. 8560. 15. .45 18. .000235
13. .081 16. 61.7 19. i
14. .8 17. 23,500. 20. f.
145. To Find the Logarithm of a Number of More Than
Three Significant Figures. Stippose we wish to determine
log 286.7. Here we have four significant figures while our
tables only tell us the mantissas of numbers having three (or
less) significant figures (as in § 144 and in the preceding ex-
ercises). In such cases we proceed as follows :
From the tables
log 286 = 2.4564 |
log 286.7 - ? Difference = 2.4597 - 2.4564 = .0033
log 287 = 2.4597 J
Since 286.7 lies between 286 and 287, its logarithm must
lie between their logarithms. Now, an increase of one unit
in the number (in going from 286 to 287) produces an increase
of .0033 in the mantissa. It is therefore assumed that an
increase of .7 in the number (in going from 286 to 286.7) pro-
duces an increase of .7 of .0033, or .00231, in the mantissa.
Therefore log 286.7 = 2.4564+.7 of .0033 = 2.4564+.00231
= 2.45871, so that
log 286.7 = 2.4587 (approximately). Ans.
In practice the answer is quickly obtained as follows :
The difference between any mantissa and the next higher
one in the table (neglecting the decimal point) is called a
tabular difference. The tabular difference in this example is
4597^564, or 33. Taking .7 of this gives 23.1, which (keeping
only the first two figures) we call 23, and adding this to 4564
gives 4587. This, therefore, is the required mantissa of log
286.7, so that log 286.7 = 2.4587 (approximately) . Ans.
230 SECOND COURSE IN ALGEBRA [XX, § 145
Similarly, § in finding log 286.75 the tabular difference (as before)
is 33. Taking .75 of 33 gives 24.75, which (keeping only two figures)
has the approximate value 25.
Hence the mantissa of log 286.75 is 4564+25 =4589. Therefore
log 286.75 =2.4589, Ans.
Below are two examples further illustrating how the above
processes are quickly carried out in practice. The student
should form the habit of writing the work in this form.
EXAMPLE 1. Determine the value of log 48. 731
SOLUTION. Mantissa of log 487 = 6875 1m,, -, . «.
Mantissa of log 488 =6884 ) Tabular dlff^nce =9
.31 X9 = 2.79 = 3 (approximately).
Hence
mantissa of log 48.731 =6875+3 = 6878.
Therefore
log 48.731 = 1.6878 Ans.
EXAMPLE 2. Determine the value of log .013403
SOLUTION. Mantissa of 134 = 1271
Mantissa of 135 = 1303 ( Tabular difference =32.
.03 X32 = .096 = 1 (approximately).
Hence
mantissa of log .013403 = 1271+1 = 1272.
Therefore
log .013403= -2 +.1272 = 8.1272 -10.
Ans.
NOTE. The process which we have employed for determining
a mantissa when it does not actually occur in the tables is called
interpolation. When examined carefully, it will be seen that the pro-
cess is based upon the assumption that if a number is increased by
any fractional amount of itself, the logarithm of the number will like-
wise be increased by the same fractional amount of itself. Thus, in
finding the mantissa of log 286.7 at the middle of p. 229, we assumed
that the increase of .7 in going from 286 to 286.7 would be accom-
panied by like increase of .7 in the logarithm. Such an assumption,
though not exactly correct, is very nearly so in most cases and is
therefore sufficiently accurate for all ordinary purposes.
XX, § 146] LOGARITHMS 231
Tables of logarithms much more extensive than those on pages
226, 227 have been prepared and are commonly used. See, for ex-
ample, The Macmillan Tables. By means of these, any desired
mantissa may usually be obtained as accurately as is necessary,
directly, that is without interpolation.
EXERCISES
Obtain the logarithm of each of the following n'umbers.
1. 678.4 8. 4.806 15. 62.856
2. 231.3 9. 1.508 16. 541.07
3. 785.4 10. 3.276 17. 6.3478
4. 492.6 11. .4567 18. 3.1416
6. 856.8 12. .08346 19- 1.7096
6. 42.17 13. 856.34 20. .15786
7. 9.567 14. 243.47 21. .085679
146. To Find the Number Corresponding to a Given Loga-
rithm. Thus far we have considered how to determine the
logarithm of a given number, but frequently the problem is
reversed, that is, it is the logarithm that is given and we wish
to find the number having that logarithm. The method of
doing this is the reverse of the method of §§ 144-145, and is
illustrated in the following examples.
EXAMPLE 1. Find the number whose logarithm is 1.9547
SOLUTION. Locate 9547 among the mantissas in the table.
Having done so, we find in the column N on the line with 9547 the
figures 90. These form the first two figures of the desired number.
At the head of the column containing 9547 is 1, which is therefore
the third figure of the desired number.
Hence the number sought is made up of the figures 901.
The given characteristic being 1, the number just found must
be pointed off so as to have two figures to the left of its decimal
point (Rule II, § 141).
Therefore the number is 90.1. Ans.
232 SECOND COURSE IN ALGEBRA [XX, § 146
EXAMPLE 2. Find the number whose logarithm is 0.6341
SOLUTION. As in Example 1, we look among the mantissas of
the table to find 6341. In this case we do not find exactly this man-
tissa, but we see that the next less mantissa appearing is 6335,
while the one next greater is 6345.
The numbers corresponding to these last two mantissas are seen
to be 430 and 431 respectively. Whence, if x represents the num-
ber sought, we have
Mantissa of log 430 = 6335 j D-ff =6 1
Mantissa of log x = 6341 J > Tabular difference = 10.
Mantissa of log 431 =6345
Since an increase of 10 in the mantissa produces an increase of 1 in
the number, we assume that an increase of 6 in the mantissa will
produce an increase of ^, or .6, in the number.
Hence the number sought has the figures 4306.
Since the given characteristic is 0, the number must be 4.306
(§141). Ans.
NOTE 1. The pupil will observe that in Example 1 the given
mantissa actually occurs in the tables, while in Example 2 it does
not, thus making it necessary in this last case to interpolate. (See
the Note in § 145.)
NOTE 2. The number whose logarithm is a given quantity is
called the antilogarithm of that quantity. Thus 100 is the anti-
logarithm of 2, 1000 is the antilogarithm of 3, etc.
EXERCISES
Find the numbers whose logarithms are given below.
1. 1.8751 9. 1.4893
2. 2.9405 10. 2.8588
3. 0.3856 11. 3.7430
4. 3.5866 12. 0.5240
6. 9.6955-10 13. 0.6970
6. 8.7152-10 14. 9.7400-10
7. 7.4900-10 15. 8.3090-10
8. 6.8519-10 16. 7.5308-10
NAPIER
(John Napier, 1550-1617)
Famous as the inventor of logarithms and first to show the advantage of
using them in reducing the labor of ordinary computations. Interested and
active also in the political and religious controversies of his day.
XX, § 147] LOGARITHMS 233
III. THE USE OF LOGAKITHMS IN COMPUTATION
147. To Find the Product of Several Numbers. The pro-
cesses of multiplication, division, raising to powers, and ex-
traction of roots, as carried out in arithmetic, may be greatly
shortened by the use of logarithms, as we shall now show.
Let us take any two numbers, for example 25 and 37, and
determine their logarithms. We find that log 25 = 1.3979
and log 37 = 1.5682. T^iis means (§ 136) that
25 = 101-3979 and 37 = 101-5682
Multiplying, we thus have
25X37 = l01-«»»+i.B682 (§ 117) Law j)
The last equality means (§ 136) that
log (25X37) = 1.3979+1.5682,
or log (25X37) =log 25+log 37.
Similarly, if we start with the three numbers 25, 37, and 18
we can show that
log (25X37X18)= log 25+log 37+log 18.
Thus we arrive at the following important rule.
RULE V. The logarithm of a product is equal to the sum
of the logarithms of its factors.
Thus log (13 X. 0156X99.8) =log 13+log .0156+log 99.8.
The way in which this rule is used to find the value of the
product of several numbers is shown below.
EXAMPLE 1. To find the value of 13 X. 0156X99.8
SOLUTION. log 13= 1.1139
log .0156= 8.1931-10
log 99.8= 1.9991
Adding, 11.3061 -10, or 1.3061
Hence, by Rule V, the logarithm of the desired product is 1.3061.
It follows that the product itself is the number whose logarithm
is 1.3061. When we look up this number (as in § 146) we find it to
be 20.23. Hence 13 X. 0156X99.8 =20.23 (approximately). Arcs.
234 SECOND COURSE IN ALGEBRA [XX, § 147
EXAMPLE 2. To find the value of
8.45X.678X.0015X956X.111
SOLUTION. log 8.45= 0.9269
log .678= 9.8312-10
log .0015= 7.1761-10
log 956= 2.9805
log .111= 9.0453-10
/Adding, 29.9600-30=9.9600-10.
f
Hence, by Rule V, the logarithm of the* desired product is seen to
be 9.9600 -10.
Therefore the product itself is found (as in § 146) to be -912
(approximately). Ans.
These examples lead to the following rule.
RULE VI. To multiply several numbers :
1. Add the logarithms of the several factors.
2. The sum thus obtained is the logarithm of the product.
3. The product itself can then be determined as in § 146.
NOTE. It may happen (as in Example 2) that the sum of several
logarithms is negative. In such cases it is best to write the sum in
such a form that it will end with — 10, thus conforming always to
§ 143.
EXERCISES
Find, by Rule V, § 147, the value of each of the following
logarithms.
1. log (35.1X7.29). 3. log (145.7 X 8.35 X. 00456).
2. log (5X3.17X.0016). 4. log (3.456 X. 001798 XI. 456).
Find (by Rule VI, § 147) the value of
5. 56.8X3.47 X. 735
Check your answer by multiplying out the long way as in arith-
metic. Compare the two results and see how great was the error
committed by following the short (logarithmic) method. Compare
also the time required for the two methods.
XX, § 148] LOGARITHMS 235
6. .975X42.8X3.72
7. 896X40.8X3.75X.00489
8. 34.56X18.16X.0157
[HINT. See § 145.]
9. 576.8X43.25X3.576X.0576
10. 60.573X8.087X.008915X1.2387
11. 23X23X23X23X23X23X23, (or 237)
12. 1.2X2.3X3.4X4.5X5.6X6.7X7.8
13. .31X5.198X6.831X2.584X.00312X.07568
14. Since 25 X 15 = 375 we know by Rule V, § 147, that the
logarithm of 25 added to the logarithm of 15 is equal to the
logarithm of 375. Show that the values given in the tables
for log 25, log 15, and log 375 confirm this result. Invent
and try out several other similar problems for yourself.
148. To Find the Quotient of Two Numbers. Let us take
any two numbers, for example 41 and 29, and look up their
logarithms. We find
log 41 = 1.6128
log 29=1.4624
These mean that
41==101.6128
and 29 = 101-4624
Whence, dividing the first of these equalities by the second,
we obtain
101'6128~M624 (§ 117> Law n)
The last equality means that
log (41^-29) = 1.6128-1.4624 = log 41-log 29.
236 SECOND COURSE IN ALGEBRA [XX, § 148
This result illustrates the following general rule.
RULE VII. The logarithm of a quotient is equal to the
logarithm of the dividend minus the logarithm of the divisor.
Thus log (467.3 -J-.00149) =log 467.3 -log .00149
The way in which this rule is used to find the value of the
quotient of two numbers is shown below.
EXAMPLE 1. To find the value of 236 -i-4. 15
SOLUTION. log 236=2.3729
log 4.15 =0.6180
Subtracting, 1.7549
Hence the logarithm of the desired quotient is 1.7549 (Rule VII)
The number whose logarithm is 1.7549 is found (as in § 146)
to be 56.875
Therefore 236-^4.15=56.875 (approximately). Ans.
EXAMPLE 2. To find the value of 1 . 46 -^ .00576
SOLUTION. log 1.46 =0.1644 = 10.1644 - 10 (See Note below.)
log .00576 = 7.7619-10
Subtracting, 2.4025
The number whose logarithm is 2.4025 is found to be 252.64
Therefore 1 .46 -T- .00576 = 252. 64 (approximately) . Ans.
Thus we have the following rule.
RULE VIII. To find the quotient of two numbers :
1. Subtract the logarithm of the divisor from the logarithm
of the dividend.
2. The difference thus obtained is the logarithm of the quo-
tient.
3. The quotient itself can then be determined as in § 146.
NOTE. To subtract a negative logarithm from a positive one,
or to subtract a greater logarithm from a less, increase the charac-
teristic of the minuend by 10, writing —10 after the mantissa to
compensate. Thus, in Example 2, we wished to subtract the nega-
tive logarithm 7.7619-10 from the positive one 0.1644. There-
fore 6.1644 was written in the form 10.1644-10, after which the
subtraction was easily performed.
XX, § 149] LOGARITHMS 237
EXERCISES
Find, by Rule VII, § 148, the value of each of the follow-
ing logarithms.
1. log(13-f-9). 3. log (38.76 -J-. 0017).
2. log (217-5-8.16). 4. log (8.764 -f- 114.3).
Find, by Rule VIII, § 148, the value of each of the follow-
ing quotients.
5. 246 -i- 15.7
Check your answer by dividing out the long way as in arith-
metic. Compare the two results and see how great was the error
committed by following the short (logarithmic) method.
6. 34.7^-5.34 8. 45.67^38.01
7. 389.7^-4.353 9. 3.25 -f-. 00876
[HINT. See § 145.] [HINT. See Note in § 148.]
10. 49.6 -i- 87.3
n 40.3X6.35
3.72
[HINT. Find the logarithm of the numerator by Rule V, § 147.]
12 .0036X2.36 24.3 X. 695 X. 0831
.0084 8.40 X. 216
14. Since 27-:-9 = 3 we know, by Rule VII, § 147, that the
logarithm of 9 subtracted from the logarithm of 27 is equal to
the logarithm of 3. Show that the values given in the tables
for log 9, log 27, and log 3 confirm this result. Invent and
try out several other similar problems for yourself.
149. To Raise a Number to a Power. Let us take any
number, for example 25, and raise it to any power, say the
fourth. We then have 254, which means 25X25X25X25.
Hence, by Rule V, § 147, we must have
log 254 = log 25-flog 25+log 25+log 25, or log 254 = 4 log 25.
238 SECOND COURSE IN ALGEBRA [XX, § 149
This illustrates the following rule.
RULE IX. The logarithm of any power of a number is
equal to the logarithm of the number multiplied by the exponent
indicating the power.
Thus log 3.1710 = 10 log 3.17 ; similarly, log. 001746 = 6 log .00174.
The way in which this principle is used to raise a number
to a power is shown below.
EXAMPLE 1. To find the value of 2.37*
SOLUTION. log 2.37= 0.3747
4
Multiplying, 1.4988
Hence
Iog2.374 = 1.4988 (Rule IX)
The number whose logarithm is 1.4988 is found to be 31.535
Therefore
2.374 =31.525 (approximately). Ans.
EXAMPLE 2. To find the value of .8565
SOLUTION. log .856 = 9.9325 - 10
Multiplying, 49.6625-50 = 9.6625 - 10
The number whose logarithm is 9.6625-10 is .4597 (§ 146)
Therefore
.8565= . 4597 (approximately). Ans.
Thus we have the following rule.
RULE X. To raise a number to a power :
1. Multiply the logarithm of the number by the exponent
indicating the power.
2. The result thus obtained is the logarithm of the answer.
3. The answer itself can then be determined as in § 146.
EXERCISES
Find, by Rule IX, § 149, the value of each of the following
logarithms.
1. log 166 2. log 3.123 3. log .01762 4. log 36.644
XX, § 150] LOGARITHMS 239
Find, by Rule X, § 149, the value of each of the following
expressions.
5. 8.823
Check your answer by raising 8.82 to the third power as in
arithmetic. Compare the two results and see how great was the
error committed by following the short (logarithmic) method.
6. 4'.124 7. 4.1234
8. .1755 [HINT. See Ex. 2 in § 149.]
9. 813X.0152 [HINT. Combine the rules of §§ 147 and 149.]
10. 43X8.92X.0753
8.76X53.9X4.53
' 2.32X3.15X5.143
[HINT. Use Rules VI, VIII, X.]
12. Since 93 = 729 we know, by Rule IX, § 149, that three
times the logarithm of 9 is equal to the logarithm of 729.
Show that the values given in the tables for log 9 and log 729
confirm this result. Invent and try out several other similar
problems for yourself.
150. To Extract Any Root of a Number. Let us take any
number, for example 36, and consider any root of it, say the
fifth, that is, let us consider \/36l
Supposing x to be the value of the desired root, we have
z5 = 36. (§ 118)
Now the logarithm of the first member of this equality is
equal to 5 log x by Rule IX.
Hence 5 log x = \og 36, or log x = ^ log 36.
This illustrates the following rule.
RULE XI. The logarithm of the root of a number is equal
to the logarithm of the radicand divided by the index of the root.
Thus log ^73=^ log 2.73 ; similarly, log -v/.01685=| log .01685.
The way in which this principle is used to extract the roots
of numbers in arithmetic will now be shown.
240 SECOND COURSE IN ALGEBRA [XX, § 150
EXAMPLE 1. To find the value of 'V/85.2
SOLUTION. log 85.2 = 1 .9304,
so that -J- of log 85.2=0.4826.
Hence log ^85^2" =0.4826. (Rule XI)
The number whose logarithm is 0.4826 is 3.038 (§ 146)
Therefore ^85.2=3.038 (approximately). Ans.
EXAMPLE 2. To find the value of v^.0875
SOLUTION. log .0875 = 8.9420 - 10,
so that i of log .0875 = ^8.9420 - 10) = i(48.9420 -50)
= 9.7884 - 10. (See Note below.)
The number whose logarithm is 9.7884-10 is .6143 (§ 146)
Therefore ^.0875 = . 6143 (approximately). Ans.
These examples lead to the following rule.
RULE XII. To find any root of any number.
1. Divide the logarithm of the number by the index of the root.
2. The quotient thus obtained is the logarithm of the desired
root.
3. The root itself can then be determined as in § 146.
NOTE. To divide a negative logarithm, write it in a form where
the negative part of the characteristic may be divided exactly by
the divisor giving — 10 as quotient. Thus, in Example 2, we wrote
8.9420-10 in the form 48.9420-50 after which the division by 5
was easily done and resulted in a form ending in — 10.
EXERCISES
Find, by Rule XI, § 150, the value of each of the following
logarithms.
1. log vT6 2. log v^332 3. log ^0175 4. log v/38^6"
Find, by Rule XII, § 150, the value of each of the following.
5.
Check your answer by extracting the square root of 315 (correct
to three decimal places) as in arithmetic. Compare the two results
and see how great was the error committed by following the short
(logarithmic) method.
XX, § 150] LOGARITHMS 241
8.
[HINT. See Example 2 in § 150.]
9. -^8.76 X. 0153
[HINT. Use Rules IX and XI.] _
1576 X 9. 132
11.
3.8X5.323
12. Since A/49 = 7 we know, by Rule XI, § 150, that one
half the logarithm of 49 is equal to the logarithm of 7. Show
that the values given in the tables for log 49 and log 7 con-
firm this result. Invent and try out several other similar
problems for yourself.
APPLIED PROBLEMS
Solve the following exercises by logarithms.
1. How many cubic feet of air are there in a schoolroom
whose dimensions are 50.5 ft. by 25.3 ft. by 10.4 ft. ?
2. How many gallons will a rectangular tank hold whose
dimensions are 8 ft. 10 in. by 9 ft. 3 in. by 10 ft 1 in. ?
3. How much wheat will a cylindrical bin hold if the
diameter of the base is 9 ft. 5 in. and the height is 40 ft. 4 in. ?
4. How much would a sphere of solid cork weigh if its
diameter was 4 ft. 3 in., it being known that the specific
gravity of cork is .24? (See Example 14 (e), page 6.)
[HINT. To say that the specific gravity of cork is .24 means that
any volume of cork weighs .24 times as much as an equal volume of
water. Water weighs 62.5 pounds per cubic foot.]
5. The diameter d in inches of a wrought-iron shaft re-
quired to transmit h horse power at a speed of n revolutions
per minute is given by the formula d = \ Find the
* n
diameter required when 135 horse power is to be transmitted
at a speed of 130 revolutions per minute.
242 SECOND COURSE IN ALGEBRA [XX, § 150
6. The amount to which P dollars will accumulate at
r% compound interest in n years is given by the formula
Find A if P = $500, r = 5, and n = W.
Find A if P = $100, r = 3.5, and n=15.
7. By means of Formula 3 of § 65, find the area of the
triangle whose sides are 3.15 in., 4.87 in., and 2.68 in.
8. The height H of a mountain in feet is given by the
formula
H = 49,00(/^— rVl +—\
\R+rJ\ 900 /
where R, r are the observed heights of the barometer in inches
at the foot and at the summit of the mountain, and where T,
t are the observed Fahrenheit temperatures at the foot and
summit.
Find the height of a mountain if the height of the barom-
eter at the foot is 29.6 inches and at the summit 25.35
inches, while the temperature at the foot is 67° and at the
summit 32°.
9. By means of the formula in Ex. 6 answer the follow-
ing question : How long will it take a sum of money to
double itself if placed at compound interest at 5 % ?
14.2 years. Ans.
GENERAL LOGARITHMS
*161. Logarithms to Any Base. In § 136 we defined the loga-
rithm of a number as the power to which 10 must be raised to obtain
that number. Thus, from such equalities as 102 = 100, 103 = 1000,
etc., we had log 100 = 2, log 1000 = 3, etc. Strictly speaking, this
defines the logarithm of a number to the base 10, or, as it is usually
called, a common logarithm.
We may and frequently do use some other base than 10. For
example, since 32=9, 33=27, 34=81, etc., we can say that the
logarithm of 9 to the base 3 is 2, the logarithm of 27 to the base 8 is 3,
XX, § 154] LOGARITHMS 243
the logarithm of 81 to the base 3 is 4, etc. The usual way of denot-
ing this is to write Iog39 =2, Iog327 =3, Iog381 =4, etc. Observe that
the number being used as the base is thus placed to the right and
just below the symbol log.
Similarly, we have Iog216=4, Iog864 = 2, Iog6125=3, etc.
Thus we have the following general definition. The logarithm of
any number x to a given base a is the power of a required to give x. It is
written \ogax. Any positive number except 1 may be used as the base.
NOTE. When the base a is taken equal to 10 (that is, in the usual
case) we write simply log x instead of
EXERCISES
State first the meaning and then the value of
1. Iog24. 2. Iog28. 3. Iog416. 4. logsf
6. Iog2i. 6. log^. 7. Iog5.2 8. Iog832.
*152. Logarithm of a Product. We can now show that Rule V,
§ 147, holds true whatever the base. That is, if M and N are any two
numbers, and a the base, then
logaMN=logaM+logaN.
PROOF. Let x = logaM and y = logaN. Then ax = M and a" = N
(§151). Hence a*-a« = MN, or ax+« = MN (§117, Law I).
But the last equality means that
logaMN=x+y=\ogaM+logaN. (§ 151)
*153. Logarithm of a Quotient. Rule VII, § 148, holds true
whatever the base. That is, if M and N are any two numbers, then
lQg.(M+N) = \OgaM-logaN.
PROOF. Let x = log0M and y = \ogaN. Then ax = M and a« = N.
(§ 151). Hence, a* -=- a" = M -=- N, OTax~v = M + N (§ 117, Law II).
But the last equality means that
loga(M + N)=x-y=logaM-logaN. (§ 151)
*164. Logarithm of a Power of a Number. Rule IX, § 149, holds
true whatever the base. That is, if M is any number and n any (posi-
tive integral) power, then
log0Mre = n logaM.
PROOF. Letz = logaAf. Then a* = M (§ 151) and hence anx = Mn
(§ 117, Law III). But the last equality means that
log0Mn = nx=n log0M. ( § 151 )
244 SECOND COURSE IN ALGEBRA [XX, § 155
*155. Logarithm of a Root of a Number. Rule XI, | 150, holds
true whatever the base. That is, if M is any number and n any (posi-
tive integral) root, then
= log0M.
n
PROOF. Let x=logaM Then ax = M (§151) and hence
(0*)i/» = M1/n, ora*/n = VM" (§ 121). But the last equality means
that
n n
*166. Summary. From the results established in §§ 151-155 it
appears that Rules V-XII, §§ 147-150, are not only true when the
base is 10 (as was there taken) but they are true for any base.
Complete tables have been worked out for various bases other than
10, but we shall not consider them further here.
NOTE. The reason why 1 cannot be used as a base is that 1 to
any power is equal to 1, that is, we cannot get different numbers by
raising 1 to different powers.
*157. Historical Note. Logarithms were first introduced and
employed for shortening computation by JOHN NAPIER (1550-1617),
a Scotchman. (See the picture facing p. 233.) However, he did
not use the base 10, this being first done by the English mathemati-
cian BRIGGS (1556-1631), who computed the first table of common
logarithms and did much to bring logarithms into general use.
*158. Calculating Machines. The Slide-Rule. Machines have
been invented and are now coming into very general use, especially
by engineers, by which the processes of multiplication, division,
involution, and evolution can be immediately performed. The
construction of these machines depends upon the principles of
logarithms, but to describe the machines and their methods of
working would take us beyond the scope of this text. The simplest
machine of this kind is the slide rule, the use of which is easily
understood. A simple slide rule with directions is inexpensive
and may ordinarily be secured from booksellers.
IA
A
I c ((
l,I|p!.l ^111!
i""l" ^ ! ^
D
FIG. 70. THE SLIDE RULE.
PAET III. SUPPLEMENTARY TOPICS
CHAPTER XXI
FUNCTIONS
159. The Function Idea. In ordinary speech we make
such statements as the folio whig :
1. The area of a circle depends upon the length of its
radius.
2. The time it takes to go from one place to another de-
pends upon the distance between them.
3. The power which an engine can exert depends upon the
pressure per square inch of the steam in the boiler.
Another way of stating these facts is as follows :
1. The area of a circle is a function of the length of its
radius.
2. The time it takes to go from one place to another is a
function of the distance between them.
3. The power which an engine can exert is a function of
the pressure per square inch of the steam in the boiler.
The idea thus conveyed by the word function is that we
have one magnitude whose value is determined as soon as we
know the value of some other one (or more) magnitudes upon
which the first one depends. This idea is at once seen to be
universal in everyday experience and for that reason it be-
comes of great importance in mathematics, f In the present
t The extended formal study of the function idea enters into
that branch of mathematics known as the Calculus.
245
246 SECOND COURSE IN ALGEBRA [XXI, § 159
chapter we shall indicate briefly how it is related to some of
the subjects treated in the preceding chapters, noting es-
pecially the significance of the idea when considered graphi-
cally.
160. Types of Algebraic Functions. An expression of the
form
(1) «wr+fli,
where the coefficients aQ and ai have any given values (except
a0 must not be 0) is called a linear function of x. Observe
that every such expression depends for its value upon the
value assigned to #, and is determined as soon as x is known.
Hence it is a function of x in the sense explained in § 159.
It is called a linear function since it is of the first degree in x.
(Compare § 26.)
For example, 2 x +3 is a linear function of x. Here we have the
form (1) inwhicha0 = 2 and ai=3. Similarly, 3x— 2, x— 4,-z+i
and 3x are linear functions of x. (Why?)
Likewise, 3^+2 is a linear function of t, while — r+5 is a linear
function of r, etc.
As an example of a linear function in everyday experience, sup-
pose that in Fig. 71 a person starts from the point P and moves to
the right at the rate of 15 miles per hour, and let Q be the point 10
S E
h— 10— >!
FIG. 71.
miles to the left of P. Then we may say that the distance of the
traveler from Q is a linear function of the time he has been traveling,
for if t represent the number of hours he has been traveling, his dis-
tance from P is 15 1 (see § 62) and hence his distance from Q is
15 £ + 10. This is seen to be a linear function of t, being of the form
(1) in which a0 = 15 and 01 = 10.
. Likewise, the interest which a given principal, P, will yield in one
year is a linear function of the rate, for, if r be the rate, the interest
XXI, § 160] FUNCTIONS 247
in question is given by the formula P X-^:,or— — r, and this is seen
, -LvJO J.LHJ
p
to be of the form (1) in which «o = rrr:, and ai=0.
J_ \j\J .___ -
An expression of the form
(2) a0x2+aix+a2,
where the coefficients a0, «i, and a2 have any given values
(except that a0 must not be 0) is called a quadratic function
of x.
For example, 2rr2+3 x — 1 is a quadratic function of x because it
is of the form (2) in which a0 = 2, a\ = 3, 02 = — 1. Likewise, x2 +-|- x ;
x2+i; — z2+3 x ; 5 x2 ; z2 are quadratic functions of x. (Why?)
Again, we may say that the area of a square is a quadratic func-
tion of the length of one side, for if x be the length of side, the area
is xz and this is of the form (2) in which a0 = 1, a\ = a2 =0.
Similarly, the area of a circle is a quadratic function of the
radius. (Why?)
An expression of the form
(3) flo*3+ai*2+a2Jt+a3,
where the coefficients a0, «i, «2 and a3 have any given values
(except that a0 must not be 0) is called a cubic function of x.
For example, 3 x3-x*+±x-l ; 4z3-z; z3-2x2 + l; 5x3; x3,
etc. (Why?)
Again, we may say that the volume of a cube is a cubic function
of the length of one edge. (Why?) ! Also, the volume of a sphere
is a cubic function of the radius. (Why?)
It may now be observed that the expressions (1), (2), and
(3) are but special forms of the more general expression
(4) a0xn+ai^n~1+fl2Xn-2H \-an_iX+an
where it is understood that n can be any positive integer,
while the coefficients OQ, ai, a2, ••• a,, have any given values
(except that a0 must not be 0). This is called the general
248 SECOND COURSE IN ALGEBRA [XXI, § 160
integral rational function of x, or, more simply, a polynomial
in x. It reduces to the linear function (1) when n=l;
to the quadratic function (2) when n = 2 ; etc.
Expressions such as
Vx,
and all others composed merely of powers or roots (or both)
of x are classed under the name of algebraic functions.
Since all functions of the form (4) are composed of integral
powers only of x, they are but special cases of the algebraic
functions just mentioned.
EXERCISES
1. Show that the thickness of a book is a linear function
of the number of its pages.
[HINT. Let x be the number of pages, d be the thickness of each
page, and D the thickness of each cover. Now build up the formula
for the thickness of the book and note which of the functional
types in § 160 is present.]
2. The supply of gasoline in a tank was very low, its
depth being but 1 inch all over the bottom, when it was re-
plenished from a pipe which delivered 3 gallons per minute.
Show that the amount in the tank at any moment during the
filling was a linear function of the time since the filling began.
3. Show that the force which a steam engine has at any
moment at its cylinder is a linear function of the area of the
piston ; also that it is a linear function of the boiler pressure
of the steam per square inch.
4. A certain room contains a number of 16-candle-power
electric lights and a number of Welsbach gas-burners. Show
that the amount of illumination at any time is a linear
function of the number of electric lights turned on. Is this
true regardless of the number of gas-burners already lighted ?
XXI, § 160] FUNCTIONS 249
5. Show that the perimeter of a square is a linear func-
tion of the length of one side ; also that the circumference
of a circle is a linear function of its radius.
6. Show that if each side of a square be increased by x,
the corresponding increase in the area will be a quadratic
function of x.
[HINT. Let a = the length of one side of the original square.
Then the area is a2 and the area of the new square is (a +z)2. Now
formulate the expression for the increase in area.]
7. Show that if the radius of a circle be increased by x,
the corresponding increase in area will be a quadratic func-
tion of x.
8. Show that if the edge of a cube be increased by x
the corresponding increase in volume will be a cubic function
of x. State and prove the corresponding statement for a
sphere.
9. Show that if y varies directly as x (see § 113), then y
is a linear function of x. Is the converse of this statement
necessarily true, namely if y is a linear function of x, then y
varies directly as x ?
10. When y varies as the square of x, to which one of the
functional types mentioned in § 160 does y belong? Answer
the same question when y varies inversely as x ; when y varies
inversely as the square of x.
11. A certain linear function of x takes the value 5 when
x = l and takes the value 8 when x = 2. Determine com-
pletely the form of the function.
SOLUTION. Since the function is linear, it is of the form aox +«i.
Since this expression must (by hypothesis) be equal to 5 when x = 1,
we have a0 • l+ai=5. Likewise, placing x=2, gives a0 • 2+ai=8.
Solving these two equations for a0 and ai we obtain oo=3, ai=2.
The desired function is therefore 3 x +2. Ans.
250
SECOND COURSE IN ALGEBRA [XXI, § 160
12. A certain linear function of x takes the value 14
when x = 3, and takes the value —6 when x= — 1. Deter-
mine completely the form of the function.
13. A certain quadratic function takes the value 0 when
x = 1, and the value 1 when x = 2, and the value 4 when x = 3.
Determine completely the form of the function.
14. Show that the area of any triangle is an algebraic
function of the sum of its three sides. (See Formula 3 in
§65.)
161. Functions Considered Graphically. By the graph
of a function is meant the line or curve which results when
some letter, as y, is placed equal to the function and the graph
is drawn of the equation thus obtained. The purpose of
the graph is to bring out clearly and quickly to the eye the
relation between the given function and the quantity (vari-
able) upon which it depends for its values.
The method of drawing such graphs is precisely the same
as that given in § 29, p. 43 for equations of the first degree,
and in § 57, p. 90, for quadratic equations.
Thus, in order to obtain the graph of the function x*, we place
y=*xz and proceed to draw the graph of this equation in the way
explained in § 29, that is, we assign various values to x and compute
(from this equation) the corresponding values of y, then we plot
each point thus obtained and finally draw the smooth curve passing
through all such points.
Below is a table of several values of x and y thus computed;
and the graph is shown in Fig. 72.
When x =
-2
-1
0
1
2
3
4
then y =
-8
-1
0
1
8
27
64
The portion of the curve lying to the right of the y-a,xis extends up-
ward indefinitely, while the portion to the left of the same axis ex-
tends downward indefinitely. Note that, from the way this curve has
XXI, § 161]
FUNCTIONS
251
been drawn, it at once brings out to the eye
the value of the given function x3 for any
value of the letter x upon which this func-
tion depends, the function values being the
ordinates (§ 28) of the points on the curve.
For example, at x = 2 the corresponding
ordinate measures 8, which is the function
value then present.
This curve may be used as a graphical
table of cubes of numbers. Thus, if x = 1.5,
2/ = 3.4, approximately, etc. Likewise, if y
is given first, the curve shows the cube root
of y; for example, if y=4, x is about 1.6.
The figure may be drawn by the student on
a much larger scale ; the values of x and y
can be read much more accurately from
such a figure than from the small figure on ,
this page.
Another means of improving the accu-
racy of the figure is to take a longer dis-
tance on the horizontal line to represent one
unit than is taken to represent one unit on
the vertical scale.
FIG. 72.
The graph of every linear function is a straight line. The
graph of every other algebraic function is a curved line.
-B
FIG. 73.
For example, in considering the graph
of the linear function \x— 5, we place
y=\ x— 5. But this is an equation of the
first degree between x and y and hence
(§ 29) its graph is a straight line. Fig. 73
shows the result.
Note that the graph cuts the z-axis
in one point. The abscissa of this par-
ticular point is 4, which indicates that 4
is the root, or solution, of the equation
f x— 5=0, for it is this value of x that
makes y=Q.
252
SECOND COURSE IN ALGEBRA [XXI, § 161
The graph of every quadratic function belongs to the class
of curves known as parabolas. A parabola resembles in form
an oval, open at one end. It never cuts the z-axis in more
than two points.
Fig. 74 shows the graph of the quadratic function xz+x-2.
Note that the curve cuts the z-axis at two points whose abscis-
sas are — 2 and 1, respectively. This indicates that — 2 and 1 are
the roots of the quadratic equation x2+x —2=0.
1-0
FIG. 74.
FIG. 75.
The general form of the graph of a cubic function is that
of an indefinitely long smooth curve which cuts the x-axis
in no more than three points.
Fig. 75 shows the graph of the cubic function x3 — 3 z2 — x +3. It
cuts the rr-axis at three points whose abscissas are respectively —1,
1, and 3. These values, therefore, are the roots of the cubic equa-
tion z8-3z2-z+3=0.
XXI, § 161]
FUNCTIONS
253
Similarly, the general form of the graph of the rational
integral function of the fourth degree is that of an indefinitely
long smooth curve which cuts the x-axis in no more than
four points. And it may be said likewise that the graph
of the general integral function of degree n (see (4), § 160)
is an indefinitely long smooth curve which cuts the z-axis
in no more than n points.
Fig. 76 shows, for example, the graph of 2 x4 —5 x3 +5 x - 2, this
being a function of the fourth degree. The four points where the
curve cuts the z-axis have abscissas which are equal respectively
to —1, -1-, 1, and 2. These values, therefore, are the roots of the
equation 2 x4 — 5 z3+5 x — 2 = 0.
FIG. 76.
FIG. 77.
Fractional expressions give rise to more complex graphs, which
may have more than one piece. Fig. 77 shows, for example, the
graph of 1/rc. If we let y = l/x, y varies inversely as x (§ 110).
The -curve is therefore similar to those drawn in § 115, Fig. 69.
The graph consists of two branches and belongs to the class of
curves known as hyperbolas. These we have already met in § 78.
254 SECOND COURSE IN ALGEBRA [XXI, § 161
EXERCISES
Draw the graphs of the following functions by plotting
several points on each and drawing the curve through them.
Try to plot enough points so that the form and location of
the various waves, or arches, of the curve will be brought
out clearly, as in the figures of § 161. Note how many
times the curve cuts the x-axis and make such inferences as
you can regarding the roots of the corresponding equation.
[HINT. When the graph of a quadratic function fails to cut the
x-axis, this indicates that the roots of the corresponding quadratic
equation are imaginary. (See §§ 57, 60.) Similarly, when the
graph of a cubic function cuts the z-axis in but one point, this indi-
cates that there is but one real root to the corresponding equation,
the other two roots being imaginary. In general, the number of
times the graph cuts the x-axis indicates the number of real roots
of the corresponding equation, the number of imaginary roots being
the degree of the equation minus the number of real roots.]
1. 3x+4. 2. x. 3. xz-x-2. 4. z2-4.
5. x2+l. 6. x*-3x*-x+3. 7.
CHAPTER XXII
MATHEMATICAL INDUCTION — BINOMIAL THEOREM
162. Mathematical Induction. The three following
purely arithmetic relations are easily seen to be true :
l+2+3 =
We might at once infer from these that if n be any positive
integer, there exists the algebraic relation
(1) 1+2+3+4+ •••+n = £(n+l),
«
the dots indicating that the addition of the terms on the right
continues up to and including the number n.
For example, if n = 8, this would mean that
l+2+3+4+5+6+7+8=f(8 + l). -
Again, if n = 10, it would mean that
1+2+3+4+5+6+7+8+9 + 10=^(10 + 1).
That these are indeed true relations is discovered as soon as we
simplify them. Let the pupil convince himself on this point.
It is now to be carefully observed that the inference just
made, namely that (1) is true- for any n, is not yet justified,
strictly speaking, from anything we have done, for we have
only shown that (1) holds good for certain special values of n,
and we could never hope to do more than this however long
we continued to try out the formula in this way.
Something more than a knowledge of special cases must always
be known before any perfectly certain general inference can be made.
For example, the fact that Saturday was cloudy for 38 weeks in suc-
cession gives no certain information that it will be so on the 39th
week.
255
256 SECOND COURSE IN ALGEBRA [XXII, § 162
We shall now show how the general formula (1) may be
established free from all objection, that is in a way that
leaves no possible question as to its truth in all cases.
Let r represent any one of the special values of n for which
we know (1) to be true. Then
(2) 1+2+3+4+ ."+r=|(r+l).
Let us add (r+1) to both sides. The result is
1+2+3+4+.- H-r+(r+l) = £(r+l) + (r+l).
2i
In the second member of the last equation we may write
while the first member has the same meaning as
Thus, (2) being given us, it follows that we may write
(3)
But (3) is seen to be precisely the same as (2) except that
r+1 now replaces r throughout. Stated in words, this re-
sult means that if (1) is true when n = r, as we have supposed,
then it holds true necessarily for the next greater value of n,
which is r+1.
The original fact which we wished to establish (namely,
that (1) is true for any n) now follows without difficulty.
In fact, we know (see beginning of this section) that (1) is
true when n = 4, from which it now follows that it must be
true also when n = 5. Being true when n = 5, the same
reasoning says it must be true also when n = 6. Being true
when n = 6, it must be likewise true when n = 7. Proceeding
in this way, we may reach any integer n we may mention,
however large it may be. Hence (1) is true [for any such
value of n.
XXII, § 162] BINOMIAL THEOREM 257
This method of reasoning illustrates what is termed
mathematical induction. Another example of the process
will now be given, the steps being arranged, however, in a
more condensed form.
EXAMPLE. Prove by mathematical induction that
(1) 1+3+5+7'H ----- \-(2n— l)=n2. (n = any positive integer)
SOLUTION. When n = 1, the formula gives 1 = I2 ; when n = 2, it
gives l+3=22; when n=3, it gives l+3+5=32, all of which
arithmetical relations are seen to be correct.
Let r represent any value of n for which the formula has been
proved. Then
(2) 1+3+5+7+.. . +(2 r-l) = r2.
Adding (2 r+1) to each member gives
(3) 1+3+5+7+.- -+(2 r+1) =r2 + (2 r + l)=r2+2 rfl = (r + l)2.
But (3) is the same as (2) except that r has been replaced through-
out by r + 1. Hence, if (1) is true for any value of n, such as r, it
is necessarily true also for that value of n increased by 1.
Now, we know (1) to be true when n=3. (See above.) Hence it
must be true when n = 4. Being true when n = 4, it must be true
when w=5, etc., and in this way we now know that (1) is true for
any value (positive integral) of n whatever.
EXERCISES
Prove the correctness of each of the following formulas
by mathematical induction, n always being understood to
be any positive integer.
[HINT. First try out f orn = 1, n = 2, and n = 3. Let r represent
a number for which the formula holds. Add 2(r + l) to both mem-
bers of the resulting equation and compare results.]
2. 3+6+9+12+ •••+3 rc= — (n+1).
3.
4. 22+42+62+---+(2n)2 =£n(n+l)(2ri
5.
s
258 SECOND COURSE IN ALGEBRA [XXII, § 162
6 _1 ,_1 i__l I [_ 1 = n
1-2 2-3 3-4 n(w+l) n+l'
7. 2+22+23+24H r-2» = 2(2»-l).
8. Prove that if n is any positive integer, an—bn is divisi-
ble by a — b.
[HINT. Since or+1-6r+1 =a(ar -br)+br(a-b), it follows that
ar+1— br+l will be divisible by a — 6 whenever ar — br is divisible by
0-6.]
9. Prove that a2n — b2n is divisible by a +b.
163. The Binomial Theorem. If we raise the binomial
(a -\-x) to the second power, that is find (a+x)2, the result
is a2-}- 2 ax+x2 (§10). Similarly, by repeated multiplica-
tion of (a+x) into itself, we can find the expanded forms for
(a+z)3, (a+x)4, (a+x)5, etc. The results which we find in
this way have been placed for reference in a table below :
> = a2+2ax+x2.
' = a3+3 a2z+3 az2+z3.
(ct-hx)4 = tt4-|-4 a3x-\^6 a2x2-f-4 axs-}-x*.
(a-f-x)5 = tt5-|-6 a4x-j-10 a3x2-\-10 a2xs-\-Q ax*-\-x5, etc.
Upon comparing these, we see that the expansion of (a+x)n,
where n is any positive integer, has the following properties :
1. The exponent of a in the first term is n, and it decreases
by 1 in each succeeding term.
The last term, or xn, maybe regarded as a°xn. (See § 122.)
2. The first term does not contain x. The exponent of x in
the second term is 1 and it increases by 1 in each succeeding
term until it becomes n in the last term.
3. The coefficient of the first term is 1 ; that of the second
term is n.
4. // the coefficient of any term be multiplied by the exponent
of a in that term, and the product be divided by the number of
the term, the quotient is the coefficient of the next term.
XXII, § 163] BINOMIAL THEOREM 259
For example, the term 6 o2x2, which is the third term in the ex-
pansion of (o+z)4 (see p. 258) has a coefficient, namely 6, which
may be derived by multiplying the coefficient of the preceding term
(which is 4) by the exponent of a in that term (which is 3) and
dividing the product thus obtained by the number of that term
(which is 2).
5. The total number of terms in the expansion is n-\- 1.
The results just observed regarding the expansion of
(a-\-x)n, where n is any positive integer, may be summarized
and condensed int.o a single formula as follows :
(a+x)n =
1 • 2
1-2-3
the dots indicating that the terms are to be supplied in the
manner indicated up to the last one, or (w+l)st.
This formula is called the binomial theorem. By means
of it, one may write down at once the expansion of any
binomial raised to any positive integral power. That the
formula is true in all cases, when n is a positive integer, will
be proved in detail in § 165. We assume its truth here for
those small values of n for which its correctness is easily
tested.
NOTE. The formula is generally attributed to Sir Isaac Newton
(1642-1727) ; see the picture facing p. 193.
EXAMPLE 1. Expand (a+z)6.
SOLUTION. Here n = 6, so the formula gives
'''
.6-5-4-3-2 5 .6-5-4-3.2- 1 x6
^l- 2- 3-4- 5 ^1.2.3-4.5.6
Simplifying the various coefficients by performing the possible
cancelations in each, we obtain
5-Fz6. Ans.
260 SECOND COURSE IN ALGEBRA [XXII, § 163
NOTE. It may be observed that the coefficients of the first and
last terms turn out to be the same ; likewise the coefficients of the
second and next to the last terms are the same, and so on symmetri-
cally as we read the expansion from its two ends. This feature is
true of the expansion of (a +#) to any power. (Note the expansions
of (o+x)2, (a-f-z)3, (a+rc)4, etc., as given at the beginning of § 163.)
EXAMPLE 2. Expand (2 — m)5.
SOLUTION. Here a =2, x=—m, and n=5. The formula thus
gives
S' *' j* • 22(-m)3
1 • ^ • o
Simplifying the coefficients (as in Example 1) this becomes
(2-m)5=25+5- 24(-m)+10- 23(-m)2 + 10- 22(-ra)3
Making further simplifications, we obtain
(2-m)5 = 32-80m+80ra2-40ra3 + 10ra4-m5. Ans.
NOTE. The result for (2— x)5 is the same as that for (2+x)6
except that the signs of the terms are alternately positive and
negative instead of all positive. A similar remark applies to the
expansion of every binomial of the form (a— x)n as compared to
that of (a+x)n.
EXERCISES
Expand each of the following powers.
1. (x+y)*. 9. (a2-z2)4. 1? (I , IV.
2. (a+6)4. 10. (2a+l)4. (x yj
3. (x — vY. 11. (x— 3 vY.
40 / u, a/
4. (a-6)4. 12. (1+x2)6.
5. (2+r)5. 13. (l-o:)8.
6. (a+x)7. 14. (x — ^-)5.
7. (fif-3)6. 15. (3 a2- 1)4. 2a
8. (a2+x)6. 16. (a+x)
'»
XXII, § 1653 BINOMIAL THEOREM 261
*164. The General Term of (a+x)n. The third term in the
expansion of (a+x)n, as given by the formula in § 163, is
(n-l
Observe that the exponent of x is 1 less than the number of the
term ; the exponent of a is n minus the exponent of x ; the last
factor of the denominator equals the exponent of x ; in the numerator
there are as many factors as in the denominator.
Precisely the same statements can be made as regards the fourth
term, or
1 • 2 • 3
In the same way, it appears that the above statements can be
made of any term, such as the rth, so that the formula for the rth
term is
rth
L ' 2i ' O • • • (r — 1)
EXAMPLE. Find the 7th term of (2 6 -c)10.
SOLUTION. Here a = 2 b, x = (—c), n = 10, and r = 7. Therefore
(using the formula), the desired 7th term is
*!• 2- 3- L 56 65 ' (26)4(-c)6 = 210(2&)4(-c)6 = 3360^c«. Ans.
EXERCISES
Find each of the following indicated terms.
1. 5th term of («+*)-. ?. 6th term of tx+Vfl.
2. 6th term of (x-y}*.
3. 7th term of (2+ z)9. 8. 9th term of ^-
4. 10th term of (w-n)14. /2 2x12
6. 6th term of (*-&.)» 9' 5th term of (f -f J '_
6. 20th term of (1 +x)24. 10. 4th term of (2 \/2 - v/3) 6.
165. Proof of the Binomial Theorem. The way in which
the binomial formula was established in § 163 is, strictly
speaking, open to objection because we there made sure of
its correctness only for certain special values of n, such as
ft = 2, n = 3, n = 4, and n==5. Though the formula holds
262 SECOND COURSE IN ALGEBRA [XXII, § 165
true, as we saw, in these cases, it does not follow necessarily
that it is true in every case, that is for eveiy positive inte-
gral value of n. We can now establish this fact, however,
by the process of mathematical induction, when n is a positive
integer.
Let ra represent any special value of n for which the formula
has been established (as, for example, 2, 3, 4, or 5). Then we
have
(1) m(m-l) - (m-r+2)
1-2-3- (r-1)
Let us now multiply both members of this equation by
a+z. On the left we obtain (a+x)m+l. On the right we
shall have the sum of the two results obtained by multiply-
ing the right side of (1) first by a and then by x, that is
we shall have the sum of the two following expressions :
m(m-l) ... (m-r+2)
1-2.3-. (r-1)
and
Adding these, and making the natural simplifications in
the resulting coefficients of amx, am~lx2, etc., and equating
the final result to its equal on the left (namely (a+x)m+1, as
noted above) gives
(2)
1 • & • o ••• (T — i)
XXII, § 167] BINOMIAL THEOREM 263
But (2) is precisely (1) except for the substitution of ra+1
for m throughout. Hence, if the binomial formula holds
for any special value of n, as m, it necessarily holds for the
next larger value, namely m-j-1. But we have already ob-
served that it holds when n = 5. It must, therefore, hold when
n = 5+l, or 6. But if it holds when n = Q, it must likewise
hold when n = 6+l, or 7. Thus we may proceed until we
arrive at any chosen value of n whatever. That is, the for-
mula must be true for any positive integral value of n.
*166. The Binomial Formula for Fractional and Negative Ex-
ponents. In case the exponent n is not a positive integer but is
fractional or negative, we may still write the expansion of (a-\-x)n
by the formula of § 163, but it will now contain indefinitely many
terms instead of coming to an end at some definite point, that is
we meet with an infinite series. (Compare § 92.)
For example, the formula gives
1-2 1 • 2 • 3
= a1/2 +4- a-1/2.? +n a~3/2x2 -f~~ a-5/2x3 +•
1-2 1-2-3
= a^+±a-^x-±a-*<
Here we have written only the first four terms of the expansion,
but we could obtain the 5th term in the same way and as many
others in their order as might be desired.
*167. Application. If in (a+x}n the value of x is small in com-
parison to that of a (more exactly, if the numerical value of x/a is
less than 1) then the first few terms of the expansion furnish a
close approximation to the value of (a+z)n. This fact is often
used to find approximate values for the roots of numbers in the
manner illustrated below.
EXAMPLE. Find the approximate value of VlO.
SOLUTION. Write VlO = V9 + l = V(32 + l) and expand this
last form by the binomial formula. Thus (using the final result
264 SECOND COURSE IN ALGEBRA [XXII, § 167
in the worked example of § 166), we have
1/2- 1-- J-(32)-3/2- I2
o ,_L_ 1.1
2-3 8-33 16-35
= 3+.166666-.004629+.000257 =3.162288 (approximately).
Observe that the value of VlO as given in the tables is 3.16228,
thus agreeing with that just found so far as the first five places of
decimals are concerned.
Whenever extracting roots by this process we use the following
general rule.
Separate the given number into two parts, the first of which is the
nearest perfect power of the same degree as the required root, and ex-
pand the result by the binomial theorem.
*EXERCISES
Write the first four terms in the expansion of each of the following
expressions.
1. (o+aO»/». 6. (2 o+6)3/4.
2. (a+z)-2. 6. (a3-z2)-3/4.
3. (l+x)V3. 7.
4. (2- a;)-1/4. 8.
9. Find the 6th term in the expansion of
[HINT. Use the formula in § 164, with n =-J and r = 6.]
Find the
10. 5th term of (a+rr)1/2. 13. 9th term of (a -
11. 7th term of (a+x)~2/3. 14. 10th term of
12. 8th term of (1+z)1/3. 15. 6th term of \/2a+6.
Find the approximate values of the following to six decimal
places and compare your results for the first three examples with
those given in the tables.
16. VT7. 17. V27. 18. \/9
19. ^11. 20.
[HINT. Write 14 = 16-2=24-2.]
CHAPTER XXIII
THE SOLUTION OF EQUATIONS BY DETERMINANTS
168. Definitions. The symbol
a b
c d
is called a determinant of the second order, and is defined
as follows :
a b
Thus
8 3
2 4
7
2
-10
Q
= ad—bc.
8 -4-2 -3 = 32-6 = 26.
= 7 -4-(-2) -3 = 28+6 = 34.
6] = 4[-50+18]
= 4(-32) = -128.
The numbers a, b, c, and d are called the elements of the
determinant.
The elements a and d (which lie along the diagonal through
the upper left-hand corner of the determinant) form the
principal diagonal The letters b and c (which lie along the
diagonal through the upper right-hand corner) form the
minor diagonal.
265
266
SECOND COURSE IN ALGEBRA [XXIII, § 168
From these definitions, we have the following rule.
To evaluate any determinant of the second order, subtract
the product of the elements in the minor diagonal from the
product of the elements in the principal diagonal.
EXERCISES
Evaluate each of the following determinants.
1.
3.
7 1
4 -8
8 7
-2 3
-2 6
7 -3
6. 5
7. f
8.
2a
3a
7a
36
56
0
66
x+y 3
x — y 4
169. Solution of Two Linear Equations. Let us consider
a system of two linear equations between two unknown
letters, x and y. Any such system is of the form
(1)
(2)
where ai, 61, Ci, etc., represent known numbers (coefficients).
This system may be solved for x and y by elimination, as
in Chapter VII. Thus, multiplying (1) by 62 and (2) by 61
and then subtracting the resulting equations from each other,
the letter y is eliminated and we reach the equation
(«i62 — 0^61)0; = 62Ci — bic%.
Therefore
(3) x
Likewise, we may eliminate x by multiplying (1) by 02 and
XXIII, § 169]
DETERMINANTS
267
(2) by «i and subtracting the resulting equations from each
other. This gives
(ai&2 — 0261) y = aid —
Therefore
(4)
y
It is now clear, by § 168, that the numerators and denomi-
nators in (3) and (4) are all determinants of the second order ;
and by the definition of § 168, (3) and (4) may be written
respectively in the forms
61
(5)
x =
y=
These forms for the solution of (1) and (2) are easily re-
membered. In particular, observe that :
1. The determinant for the denominator is the same for
both x and y.
2. The determinant for the numerator of the z-value is the
same as that for the denominator except that the numbers
Ci and c% replace the ai and a^ which occur in the first column
of the denominator determinant.
3. The determinant for the numerator of the ^/-value is
the same as that for the denominator except that the num-
bers Ci and 02 replace the bi and 62 which occur in the second
column of the denominator determinant.
The usefulness of the forms (5) lies in the fact that they
express the solution of a system of two linear equations in
condensed form, enabling us to write down the desired values
of x and y immediately, without the usual process of elimina-
tion. This will now be illustrated.
268 SECOND COURSE IN ALGEBRA [XXIII, § 169
EXAMPLE. Solve by determinants the system
(6)
(7) x-7y=-S.
SOLUTION. Using the forms (5), we have at once
x =
18
3
-8
-7
_18. (_7)-(-8)-
3 -126+24 -102 R
2
3
2(-7)-l • 3
-14-3 -17
1
-7
2
18
1
—8
2- (-8)-!- 18
-16-18 -34 o
2
3
2(— 7)— 1 • 3
-14-3 -17
1
-7
The solution desired is therefore (x = 6, y=2). Ans.
CHECK. Substituting 6 'for # and 2 for y in (6) and (7) gives
12 +6 = 18 and 6 - 14 = -8, which are true results.
EXERCISES
Solve each of the following pairs of equations by determi-
nants, checking your answers for each of the first three.
1.
2.
3.
4.
Sx+3y=-7.
3x-7 y=-&,
= 14.
7.
8.
I 2x-y=lS.
I ax-\-by = r,
\bx — ay = s.
j .2a4-.56 = 30,
5* \. 4 a-. 8 6= -16.
XXIII, § 170]
DETERMINANTS
269
170. Determinants of the Third Order. The symbol
(1)
is called a determinant of the third order.
Its value is defined as follows :
This expression, as we shall see presently, is important in the
study of equations.
The expression (2) is called the expanded form of the determinant
(1). It is important to observe that this expanded form may be
written down at once as follows.
Write the determinant with the first two columns repeated at
the right and first note the three diagonals which then run down
from left to right (marked +). The
product of the elements in the first of
these diagonals is ai 62 c3, and this is seen to
be the first term of the expanded form (2).
Similarly, the product of the elements in
the second of these diagonals is 6iC2«3,
which forms the second term of (2) ; and
likewise the third diagonal furnishes at
once the third term of (2).
Next consider the three diagonals which run up from left to right
(marked with dotted lines) . The product of the elements in the first
of these is o3 62 Ci, and this is the fourth term of (2), provided it be
taken negatively, that is preceded by the sign — . Similarly, the
other two dotted diagonals of (3) furnish the last two terms of (2),
provided they be taken negatively.
NOTE. Every determinant of the third order when expanded
contains a total of six terms.
EXAMPLE. Expand and find the value of the determinant
379
214
632
270
SECOND COURSE IN ALGEBRA [XXIII, § 170
SOLUTION. Repeating the first and second columns at the right,
we have
379
214
632
3 7
2 1
6 3
The diagonals running down from left to right give the three
products
3-1-2, 7-4-6, 9-2-3,
which form the first three terms of the expansion.
The diagonals running up from left to right give the products
6-1-9, 3-4-3, 2-2-7,
which, when taken negatively, form the three remaining terms of the
determinant.
The complete expanded form of (3) is, therefore,
3- 1- 2+7-4- 6+9-2- 3-6- -1-9-3-4. 3-2- 2- 7,
which reduces to
6+168+54-54-36-28 = 110. Ans.
* EXERCISES
Expand and find the value of the following determinants.
3.
4.
13 7
24 6
35-4
-7
3
2 2
-4 6
8
-5 -3
8
2
3
6
0
5
.
3
0
7
2a 3 66
3 a 2 -56
a. 0 -26
5.
6.
7.
8.
x 7 1
23-4
4 2 1
a 6 2
-453
210
a 6 c
d e f
x y z
1 0 0
0 x-y 0
0 0 x+y
XXIII, § 171]
DETERMINANTS
271
*171. Solution of Three Linear Equations. Let us consider a sys-
tem of three linear equations between three unknown letters, such
as x, y, and z. Any such system is of the form
(1)
where 01, 61, ci, di, a2, b2, etc., represent known numbers (coefficients).
This system may be solved for x, y, and z by elimination, as in § 35,
but the process is long. We shall here state merely the results,
which are as follows (compare with (3) and (4) of § 169) :
_dibzC3-\-d2b3c\-}-dzbiCz — dsbzCi
_
gib-ids -\-dzb3di -\-d3bidz—d3bzdi — dib3dz—a3bidzw
(2)
It is clear by § 170 that in these values for x, y, and z, each nu-
merator and denominator is the expanded form of a determinant of
the third order. In fact, it appears from the definition in § 170,
that we may now express these values of x, y, and z in the following
condensed (determinant) forms :
i 61 (
(3) x =
61
b3
61
d3
The importance of these expressions for x, y, and z lies in the fact
that they give at once the solution of any system such as (1) in
very compact and easily remembered forms. The following features
should be especially noted :
1. The denominator determinant is the same in all three cases.
(Compare statement 1 of § 169.)
2. The determinant for the numerator of the z-value is the same
as that for the denominator determinant except that the numbers
di, dz, dz replace the 01, a2, 03 which occur in the first column of the
denominator determinant.
272
SECOND COURSE IN ALGEBRA [XXIII, § 171
3. Similarly, the numerator of the y-v&lue is formed from that
of the denominator determinant by replacing the second column by
the elements di, d2, d3 ; while the numerator of the 2-value is formed
from that of the denominator determinant by replacing the third
column by the elements dif d2, d3. (Compare statements 2 and 3
of § 169.)
The readiness with which (3) may be used in practice to solve a
system of three linear equations is illustrated by the following
EXAMPLE. Solve the system
2x-y+3 0 = 35,
z+37/-15=-2z,
SOLUTION. Arranging the equations as in (1) of § 171, the given
system is
2 x-y +3 z =35,
Therefore, using (3) of § 171, we have at once
0+180-2-9-280-0_-lll_
: 0 + 12-6-27-16-0 ~~^
35
-1
3
15
3
2
1
4
0
2
-1
3
1
3
2
3
4
0
2
35
3
1
15
2
3
1
0
2
-1
3
1
3
2
3
4
0
2
-1
35
1
3
15
3
4
1
2
-1
3
1
3
2
3
4
0
_o ,R
~3'
_0+3+210-135-4-Q^ 74
-37 -37
= -2,
_6 + 140-45-315-120+l^-333
-37 " -37
= 9.
XXIII, § 172] DETERMINANTS 273
The desired solution is, therefore, (3=3, 0= —2, 2 = 9). Ans.
CHECK. With x = 3, y = —2, z = 9, it is readily seen that the three
given equations are satisfied.
EXERCISES
Solve each of the following systems by determinants.
3-20+2=5, f 3+20=0,
1. { 33+60-42=3, 6. 3+2= -3,
83-100+32=34. [40+3-22 = 4.
f 23-3 0-42 = 25,
3. 3 +0-2 =-4,
4.
-9» f 3+0 = 3 a,
33+20-2 = 2, 7. <|z+2=56,
23-0+2=^.
* 172. Determinants of Higher Order. Determinants of the
fourth order exist and are studied in higher algebra, as are deter-
minants of the fifth order, sixth order, etc. Moreover, determi-
nants of the fourth order bear a similar relation to the solving of
four linear equations between four unknown letters, as determinants
of the third order bear to the solving of three linear equations be-
tween three unknown letters ; and a similar remark may be made
regarding determinants of the fifth order, sixth order, etc. In all
cases, the solutions of such systems of equations can be expressed
very simply by means of determinants.
APPENDIX
TABLE OF POWERS AND ROOTS
EXPLANATION
1. Square Roots. The way to find square roots from the
Table is best understood from an example. Thus, suppose
we wish to find Vl.48. To do this we first locate 1.48 in
the column headed by the letter n. We find it near the
bottom of this column (next to the last number). Now
we go across on that level until we get into the column
headed by Vn. We find at that place the number 1.21655.
This is .our answer. That is, Vl.48 =1.21655 (approxi-
mately) .
If we had wanted V14.8 instead of Vl.48 the work would
have been the same except that we would have gone over
into the column headed VlO n (because 14.8=10X1.48).
The number thus located is seen to be 3.84708, which is,
therefore, the desired value of V14.8.
Again, if we had wished to find Vl48 the work would take
us back again to the column headed Vn, but now instead
of the answer being 1.21655 it would be 12.1655. In other
words, the order of the digits in Vl48 is the same as for
Vl.48, but the decimal point in the answer is one place
farther to the right.
Similarly, if we desired V1480 the work would be the same
as before except that we must now use the column headed
VlO n and move the decimal point there occurring one place
farther to the right. This is seen to give 38.4708.
Thus we see how to get the square root of 1.48 or any
power of 10 times that number.
275
276 APPENDIX
In the same way, if we wish to find V.148, or V.0148, or
V.00148, or the square root of any number obtained by
dividing 1.48 by any power of 10 we can get the answers
from the column headed Vn or VlOn by merely placing
the decimal point properly. Thus, we find that V.148 =
.384708, V^148 = .121655, VML48 = .0384708, etc.
What we have seen in regard to the square root of 1.48
or of that number multiplied or divided by any power of 10
holds true in a similar way for any number that occurs in
the column headed n, so that the tables thus give us the
square roots of a great many numbers.
2. Cube Roots. Cube roots are located in the tables
in much the same way as that just described for square
roots, but we have here three columns to select from instead
of two, namely the columns headed Vn, VlO n, VlOO n.
Illustration.
•v/1.48 occurs in the column headed %/n and is seen to be 1.13960.
-v/14.8 occurs in the column headed -^10 n and is seen to be 2.4552.
\/148 occurs in the column headed v'lOO n and is seen to be
5.28957.
To get vO48 we observe that .148 = ^M^ = ^jjj: = ^ ^148.
* .LvJ JLLHJU -L\J
Thus, we look up \/148 and divide it by 10. The result is instantly
seen to be .528957. Similarly, to get \/.0148 we observe that
^0148 = -vS| = A/J^ = i 3/Ul8. Thus, we look up \/14^Fand
10U • 1UUU 10
divide it by 10, giving the result .24552.
To get ^.00148 we observe that V/.00148 = A/T?H = A V^8' so
-LvJv/vJ JLU
that we must divide Vl.48 by 10. This gives .11396.
Similarly the cube root of any number occurring in the
column headed n may be found, as well as the cube root of
any number obtained by multiplying or dividing such a
number by any power of 10.
TABLE OF POWERS AND ROOTS 277
3. Squares and Cubes. To find the square of 1.48 we
naturally look at the proper level in the column headed n2.
Here we find 2.1904, which is the answer. If we wished the
square of 14.8 the result would be the same except that
the decimal point must be moved two places to the right,
giving 219.04 as the answer. Similarly the value of (148)2
is 21904.0 etc.
On the other hand, the value of (.148)2 is found by moving
the decimal place two places to the left, thus giving .021904.
Similarly, (.0148)2 = .00021904, etc.
To find (1.48)3 we look at the proper level in the column
headed n3 where we find 3.24179. The value of (14.8)3 is
the same except that we must move the decimal point three
places to the right, giving 3241.79. Similarly, in finding
(.148)3 we must move the decimal place three places to the
left, giving .00324179.
Further illustrations of the way to use the tables will be
found in § 43.
EXERCISES
Read off from the tables the values of each of the following ex-
pressions.
1. \/4l 4. v'GTO 7. V93/7 10. v/,00154
2. V8.9 5. V^9 8. V93.7 11. V.000143
3. >/67 Q. VjOlG 9. V.00154 12. v^.000143
278
Table I — Powers and Roots
n
n2
V^
vio™
n*
*£
VWn
^ioow
1.00
1.0000
1.00000
3.16228
1.00000
1.00000
2.15443
4.64159
1.01
1.02
1.03
1.0201
1.0404
1.0609
1.00499
• 1.00995
1.01489
3.17805
3.19374
3.20936
1.03030
1.06121
1.09273
1.00332
1.00662
1.00990
2.16159
2.16870
2.17577
4.65701
4.67233
4.68755
1.04
1.05
1.06
1.0816
1.1025
1.1236
1.01980
1.02470
1.02956
3.22490
3.24037
3.25576
1.12486
1.15762
1.19102
1.01316
1.01640
1.01961
2.18279
2.18976
2.19669
4.70267
4.71769
4.73262
1.07
1.08
1.09
1.1449
1.1664
1.1881
1.03441
1.03923
1.04403
3.27109
3.28634
3.30151
1.22504
1.25971
1.29503
1.02281
1.02599
1.02914
2.20358
2.21042
2.21722
4.74746
4.76220
4.77686
1.10
1.2100
1.04881
3.31662
1.33100
1.03228
2.22398
4.79142
1.11
1.12
1.13
1.2321
1.2544
1.2769
1.05357
1.05830
1.06301
3.33167
3.34664
3.36155
1.36763
1.40493
1.44290
1.03540
1.03850
1.04158
2.23070
2.23738
2.24402
4.80590
4.82028
4.83459
1.14
.15
.16
1.2996
1.3225
1.3456
1.06771
1.07238
1.07703
3.37639
3.39116
3.40588
1.48154
1.52088
1.56090
1.04464
1.04769
1.05072
2.25062
2.25718
2.26370
4.84881
4.86294
4.87700
.17
.18
.19
1.3689
1.3924
1.4161
1.08167
1.08628
1.09087
3.42053
3.43511
3.44964
1.60161
1.64303
1.68516
1.05373
1.05672
1.05970
2.27019
2.27664
2.28305
4.89097
4.90487
4.91868
1.20
1.4400
1.09545
3.46410
1.72800
1.06266
2.28943
4.93242
1.21
1.22
1.23
1.4641
1.4884
1.5129
1.10000
1.10454
1.10905
3.47851
3.49285
3.50714
1.77156
1.81585
1.86087
1.06560
1.06853
1.07144
2.29577
2.30208
2.30835
4.94609
4.95968
4.97319
1.24
1.25
1.26
1.5376
1.5625
1.5876
1.11355
1.11803
1.12250
3.52136
3.53553
3.54965
1.90662
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2.00038
1.07434
1.07722
1.08008
2.31459
2.32079
2.32697
4.98663
5.00000
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1.27
1.28
1.29
1.6129
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1.6641
1.12694
1.13137
1.13578
3.56371
3.57771
3.59166
2.04838
2.09715
2.14669
1.08293
1.08577
1.08859
2.33311
2.33921
2.34529
5.02653
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1.30
1.6900
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3.60555
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5.06580
1.31
1.32
1.33
1.7161
1.7424
1.7689
1.14455
1.14891
1.15326
3.61939
3.63318
3.64692
2.24809
2.29997
2.35264
1.09418
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2.35735
2.36333
2.36928
5.07875
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1.34
1.35
1.36
1.7956
1.8225
1.8496
1.15758
1.16190
1.16619
3.66060
3.67423
3.68782
2.40610
2.46038
2.51546
1.10247
1.10521
1.10793
2.37521
2.38110
2.38697
5.11723
5.12993
5.14256
1.37
1.38
1.39
1.8769
1.9044
1.9321
1.17047
1.17473
1.17898
3.70135
3.71484
3.72827
2.57135
2.62807
2.68562
1.11064
1.11334
1.11602
2.39280
2.39861
2.40439
5.15514
5.16765
5.18010
1.40
1.9600
1.18322
3.74166
2.74400
1.11869
2.41014
5.19249
1.41
1.42
1.43
1.9881
2.0164
2.0449
1.18743
1.19164
1.19583
3.75500
3.76829
3.78153
2.80322
2.86329
2.92421
1.12135
1.12399
1.12662
2.41587
2.42156
2.42724
5.20483
5.21710
5.22932
1.44
1.45
1.46
2.0736
2.1025
2.1316
1.20000
1.20416
1.20830
3.79473
3.80789
3.82099
2.98598
3.04862
3.11214
1.12924
1.13185
1.13445
2.43288
2.43850
2.44409
6.24148
5.26359
6.26564
1.47
1.48
1.49
2.1609
2.1904
2.2201
1.21244
1.21655
1.22066
3.83406
3.84708
3.86005
3.17652
3.24179
3.30795
1.13703
1.13960
1.14216
2.44966
2.46620
±•111071!
6.27763
5.28957
6.30146
Powers and Roots
279
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1.50
2.2500
1.22474
3.87298
3.37500
1.14471
2.46621
5.31329
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1.52
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2.2801
2.3104
2.3409
1.22882
1.23288
1.23693
3.88587
3.89872
3.91152
3.44295
3.51181
3.58158
1.14725
1.14978
1.15230
2.47168
2.47712
2.48255
5.32507
5.33680
5.34848
1.54
1.55 •
1.56
2.3716
2.4025
2.4336
1.24097
1.24499
1.24900
3.92428
3.93700
3.94968
3.65226
3.72388
3.79642
1.15480
1.15729
1.15978
2.48794
2.49332
2.49867
5.36011
5.37169
5.38321
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1.59
2.4649
2.4964
2.5281
1.25300
1.25698
1.26095
3.96232
3.97492
3.98748
3.86989
3.94431
4.01968
1.16225
1.16471
1.16717
2.50399
2.50930
2.51458
5.39469
5.40612
5.41750
1.60
2.5600
1.26491
4.00000
4.09600
1.16961
2.51984
5.42884
1.61
1.62
1.63
2.5921
2.6244
2.6569
1.26886
1.27279
1.27671
4.01248
4.02492
4.03733
4.17328
4.25153
4.33075
1.17204
1.17446
1.17687
2.52508
2.53030
2.53549
5.44012
5.45136
5.46256
1.64
1.65
1.66
2.6896
2.7225
2.7556
1.28062
1.28452
1.28841
4.04969
4.06202
4.07431
4.41094
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4.57430
1.17927
1.18167
1.18405
2.54067
2.54582
2.55095
5.47370
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1.67
1.68
1.69
2.7889
2.8224
2.8561
1.29228
1.29615
1.30000
4.08656
4.09878
4.11096
4.65746
4.74163
4.82681
1.18642
1.18878
1.19114
2.55607
2.56116
2.56623
5.50688
5.51785
5.52877
1.70
2.8900
1.30384
4.12311
4.91300
1.19348
2.57128
5.53966
1.71
1.72
1.73
2.9241
2.9584
2.9929
1.30767
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1.31529
4.13521
4.14729
4.15933
5.00021
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5.17772
1.19582
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2.57631
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5.55050
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1.74
1.75
1.76
3.0276
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1.31909
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4.17133
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5.26802
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1.77
1.78
1.79
3.1329
3.1684
3.2041
1.33041
1.33417
1.33791
4.20714
4.21900
4.23084
5.54523
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2.60610
2.61100
2.61588
5.61467
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1.80
3.2400
1.34164
4.24264
5.83200
1.21644
2.62074
5.64622
1.81
1.82
1.83
3.2761
3.3124
3.3489
1.34536
1.34907
1.35277
4.25441
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4.27785
5.92974
6.02857
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1.21869
1.22093
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2.62559
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2.63522
5.65665
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1.84
1.85
1.86
3.3856
3.4225
3.4596
1.35647
1.36015
1.36382
4.28952
4.30116
4.31277
6.22950
6.33162
6.43486
1.22539
1.22760
1.22981
2.64001
2.64479
2.64954
5.68773
5.69802
5.70827
1.87
1.88
1.89
3.4969
3.5344
3.5721
1.36748
1.37113
1.37477
4.32435
4.33590
4.34741
6.53920
6.64467
6.75127
1.23201
1.23420
1.23639
2.65428
2.65901
2.66371
5.71848
5.72865
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1.90
3.6100
1.37840
4.35890
6.85900
1.23856
2.66840
5.74890
1.91
1.92
1.93
3.6481
3.6864
3.7249
1.38203
1.38564
1.38924
4.37035
4.38178
4.39318
6.96787
7.07789
7.18906
1.24073
1.24289
1.24505
2.67307
2.67773
2.68237
5.75897
5.76900
5.77900
1.94
1.95
1.96
3.7636
3.8025
3.8416
1.39284
1.39642
1.40000
4.40454
4.41588
4.42719
7.30138
7.41488
7.52954
1.24719
1.24933
1.25146
2.68700
2.69161
2.69620
5.78896
5.79889
5.80879
1.97
1.98
1.99
3.8809
3.9204
3.9601
1.40357
1.40712
1.41067
4.43847
4.44972
4.46094
7.64537
7.76239
7.88060
1.25359
1.25571
1.25782
2.70078
2.70534
2.70989
5.81865
5.82848
5.83827
280
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1.41421
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2.02
2.03
4.0401
4.0804
4.1209
1.41774
1.42127
1.42478
448330
4.49444
4.50555
8.12060
8.24241
8.36543
1.26202
1.26411
1.26619
2.71893
2.72344
2.72792
5.85777
5.86746
5.87713
2.04
2.05
2.06
4.1616
4.2025
4.2436
1.42829
1.43178
1.43527
4.51664
4.52769
4.53872
8.48966
8.61512
8.74182
1.26827
1.27033
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2.73239
2.73685
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5.88677
6.89637
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2.07
2.08
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4.2849
4.3264
4.3681
1.43875
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1.44568
4.54973
4.56070
4.57165
8.86974
8.99891
9.12933
1.27445
1.27650
1.27854
2.74572
2.75014
2.75454
5.91548
5.92499
5.93447
2.10
4.4100
1.44914
4.58258
9.26100
1.28058
2.75892
5.94392
2.11
2.12
2.13
4.4521
4.4944
4.5369
1.45258
1.45602
1.45945
4.59347
4.60435
4.61519
9.39393
9.52813
9.66360
1.28261
1.28463
1.28665
2.76330
2.76766
2.77200
5.95334
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5.97209
2.14
2.15
2.16
4.5796
4.6225
4.6656
1.46287
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1.46969
4.62601
4.63681
4.64758
9.80034
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10.0777
1.28866
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2.78495
5.98142
5.99073
6.00000
2.17
2.18
2.19
4.7089
4.7524
4.7961
1.47309
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4.65833
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4.67974
10.2183
10.3602
10.5035
1.29465
1.29664
1.29862
2.78924
2.79352
2.79779
6.00925
6.01846
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2.20
4.8400
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10.6480
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5.0176
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5.1529
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4.83735
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12.8129
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2.37
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2.39
5.6169
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1.53948
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4.86826
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13.3121
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2.40
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13.8240
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6.21447
2.41
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5.8081
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1.55242
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4.90918
4.91935
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13.9975
14.1725
14.3489
1.34072
1.34257
1.34442
2.88850
2.89249
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6.22308
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6.24025
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4.93964
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14.5268
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1.34993
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15.0692
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15.4:382
1.35176
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2.91227
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6.27431
6.28276
6.29119
Powers and Roots
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6.3001
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1.58430
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15.8133
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1.35902
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6.30799
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6.4516
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6.33303
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6.6049
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1.60312
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5.06952
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16.9746
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17.3740
1.36976
1.37153
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6.35786
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2.60
6.7600
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24.8971
25.1538
1.42768
1.42931
1.43094
3.07584
3.07936
3.08287
6.62671
6.63429
6.64185
2.94
2.95
2.96
8.6436
8.7025
8.7616
1.71464
1.71756
1.72047
5.42218
5.43139
5.44059
25.4122
25.6724
25.9343
1.43257
1.43*19
1.43581
3.08638
3.08987
3.09336
6.64940
6.65693
6.66444
2.97
2.98
2.99
8.8209
8.8804
8.9401
1.72337
1.72627
1.72916
5.44977
5.45894
5.46809
26.1981
26.4636
26.7309
1.43743
1.43904
1.44065
3.09684
3.10031
3.10378
6.67194
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6.68688
282
Powers and Roots
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1.73205
5.47723
27.0000
1.44225
3.10723
6.69433
3.01
3.02
3.03
9.0601
9.1204
9.1809
1.73494
1.73781
1.74069
5.486:35
5.49545
5.50454
27.2709
27.5436
27.8181
1.44385
1.44545
1.44704
3.11068
3.11412
3.11756
6.70176
6.70917
6.71657
3.04
3.05
3.06
9.2416
9.3025
9.3636
1.74356
1.74642
1.74929
5.51362
5.52268
5.53173
28.0945
28.3726
28.6526
1.44863
1.45022
1.45180
3.12098
3.12440
3.12781
6.72395
6.73132
6.73866
3.07
3.08
3.09
9.4249
9.4864
9.5481
1.75214
1.75499
1.75784
5.54076
5.54977
5.55878
28.9344
29.2181
29.5036
1.45338
1.45496
1.45653
3.13121
3.13461
3.13800
6.74600
6.75331
6.76061
3.10
9.6100
1.76068
5.56776
29.7910
1.45810
3.14138
6.76790
3.11
3.12
3.13
9.6721
9.7344
9.7969
1.76352
1.76635
1.76918
5.57674
5.58570
5.59464
30.0802
30.3713
30.6643
1.45967
1.46123
1.46279
3.14475
3.14812
3.15148
6.77517
6.78242
6.78966
3.14
3.15
3.16
9.8596
9.9225
9.9856
1.77200
1.77482
1.77764
5.60357
5.61249
5.62139
30.9591
31.2559
31.5545
1.46434
1.46590
1.46745
3.15483
3.15818
3.16152
6.79688
6.80409
6.81128
3.17
3.18
3.19
10.0489
10.1124
10.1761
1.78045
1.78326
1.78606
5.63028
5.63915
5.64801
31.8550
32.1574
32.4618
1.46899
1.47054
1.47208
3.16485
3.16817
3.17149
6.81846
6.82562
6.83277
3.20
10.2400
1.78885
5.65685
32.7680
1.47361
3.17480
6.83990
3.21
3.22
3.23
10.3041
10.3684
10.4329
1.79165
1.79444
1.79722
5.66569
5.67450
5.68331
33.0762
33.3862
33.6983
1.47515
1.47668
1.47820
3.17811
3.18140
3.18469
6.84702
6.85412
6.86121
3.24
3.25
3.26
10.4976
10.5625
10.6276
1.80000
1.80278
1.80555
5.69210
5.70088
5.70964
34.0122
34.3281
34.6460
1.47973
1.48125
1.48277
3.18798
3.19125
3.19452
6.86829
6.87534
6.88239
3.27
3.28
3.29
10.6929
10.7584
10.8241
1.80831
1.81108
1.81384
5.71839
5.72713
5.73585
34.9658
35.2876
35.6113
1.48428
1.48579
1.48730
3.19778
3.20104
3.20429
6.88942
6.89643
6.90344
3.30
10.8900
1.81659
5.74456
35.9370
1.48881
3.20753
6.91042
3.31
3.32
3.33
10.9561
11.0224
11.0889
1.81934
1.82209
1.82483
5.75326
5.76194
5.77062
36.2647
36.5944
36.9260
1.49031
1.49181
1.49330
3.21077
3.21400
3.21722
6.91740
6.92436
6.93130
3.34
3.35
3.36
11.1556
11.2225
11.2896
1.82757
1.83030
1.83303
5.77927
5.78792
5.79655
37.2597
37.5954
37.9331
1.49480
1.49629
1.49777
3.22044
3.22365
3.22686
6.93823
6.94515
6.95205
3.37
3.38
3.39
11.3569
11.4244
11.4921
1.83576
1.83848
1.84120
5.80517
5.81378
5.82237
38.2728
38.6145
38.9582
1.49926
1.50074
1.50222
3.23006
3.23325
3.23643
6.95894
6.96582
6.97268
3.40
11.5600
1.84391
5.83095
39.3040
1.50369
3.23961
6.97953
3.41
3.42
3.43
11.6281
11.6964
11.7649
1.84662
1.84932
1.85203
5.83952
5.84808
6.85662
39.6518
40.0017
40.3536
1.50517
1.50664
1.50810
3.24278
3.24595
3.24911
6.98637
6.99319
7.00000
3.44
3.45
3.46
11.8336
11.9025
11.9716
1.85472
1.8.7742
1.86011
5.86515
5.87367
5.88218
40.7076
41.0636
41.4217
1.50957
1.51103
1.51249
3.25227
3.25542
3.25856
7.00680
7.01358
7.02035
3.47
3.48
3.49
12.0409
12.1104
12.1801
1.86279
1.86548
1.86815
5.89067
5.89915
5.90762
41.7819
42.1442
42.5085
1.51394
1.51540
1.51685
3.26169
3.26482
3.26795
7.02711
7.03385
7.04058
Powers and Roots
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3.50
12.2500
1.87083
5.91608
42.8750
1.51829
3.27107
7.04730
3.51
3.52
3.53
12.3201
12.3904
12.4609
1.87350
1.87617
1.87883
5.92453
5.93296
5.94138
43.2436
43.6142
43.9870
1.51974
1.52118
1.52262
3.27418
3.27729
3.28039
7.05400
7.06070
7.06738
3.54
3.55
3.56
12.5316
12.6025
12.6736
1.88149
1.88414
1.88680
5.94979
5.95819
5.96657
44.3619
44.7389
45.1180
1.52406
1.52549
1.52692
3.28348
3.28657
3.28965
7.07404
7.08070
7.08734
3.57
3.58
3.59
12.7449
12.8164
12.8881
1.88944
1.89209
1.89473
5.97495
5.98331
5.99166
45.4993
45.8827
46.2683
1.52835
1.52978
1.53120
3.29273
3.29580
3.29887
7.09397
7.10059
7.10719
3.60
12.9600
1.89737
6.00000
46.6560
1.53262
3.30193
7.11379
3.61
3.62
3.63
13.0321
13.1044
13.1769
1.90000
1.90263
1.90526
6.00833
6.01664
6.02495
47.0459
47.4379
47.8321
1.53404
1.53545
1.53686
3.30498
3.30803
3.31107
7.12037
7.12694
7.13349
3.64
3.65
3.66
13.2496
13.3225
13.3956
1.90788
1.91050
1.91311
6.03324
6.04152
6.04979
48.2285
48.6271
49.0279
1.53827
1.53968
1.54109
3.31411
3.31714
3.32017
7.14004
7.14657
7.15309
3.67
3.68
3.69
13.4689
13.5424
13.6161
1.91572
1.91833
1.92094
6.05805
6.06630
6.07454
49.4309
49.8360
50.2434
1.54249
1.54389
1.54529
3.32319
3.32621
3.32922
7.15960
7.16610
7.17258
3.70
13.6900
1.92354
6.08276
50.6530
1.54668
3.33222
7.17905
3.71
3.72
3.73
13.7641
13.8384
13.9129
1.92614
1.92873
1.93132
6.09098
6.09918
6.10737
51.0648
51.4788
51.8951
1.54807
1.54946
1.55085
3.33522
3.33822
3.34120
7.18552
7.19197
7.19840
3.74
3.75
3.76
13.9876
14.0625
14.1376
1.93391
1.93649
1.93907
6.11555
6.12372
6.13188
52.3136
52.7344
53.1574
1.55223
1.55362
1.55500
3.34419
3.34716
3.35014
7.20483
7.21125
7.21765
3.77
3.78
3.79
14.2129
14.2884
14.3641
1.94165
1.94422
1.94679
6.14003
6.14817
6.15630
53.5826
54.0102
54.4399
1.55637
1.55775
1.55912
3.35310
3.35607
3.35902
7.22405
7.23043
7.23680
3.80
14.4400
1.94936
6.16441
54.8720
1.56049
3.36198
7.24316
3.81
3.82
3.83
14.5161
14.5924
14.6689
1.95192
1.95448
1.95704
6.17252
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6.18870
55.3063
55.7430
56.1819
1.56186
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7.24950
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3.84
3.85
3.86
14.7456
14.8225
14.8996
1.95959
1.96214
1.96469
6.19677
6.20484
6.21289
56.6231
57.0666
57.5125
1.56595
1.56731
1.56866
3.37373
3.37666
3.37958
7.26848
7.27479
7.28108
3.87
3.88
3.89
14.9769
15.0544
15.1321
1.96723
1.96977
1.97231
6.22093
6.22896
6.23699
57.9606
58.4111
58.8639
1.57001
1.57137
1.57271
3.38249
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7.28736
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3.90
15.2100
1.97484
6.24500
59.3190
1.57406
3.39121
7.30614
3.91
3.92
3.93
15.2881
15.3664
15.4449
1.97737
1.97990
1.98242
6.25300
6.26099
6.26897
59.7765
60.2363
60.6985
1.57541
1.57675
1.57809
3.39411
3.39700
3.39988
7.31238
7.31861
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3.94
3.95
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15.5236
15.6025
15.6816
1.98494
1.98746
1.98997
6.27694
6.28490
6.29285
61.1630
61.6299
62.0991
1.57942
1.58076
1.58209
3.40277
3.40564
3.40851
7.33104
7.33723
7.34342
3.97
3.98
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15.7609
15.8404
15.9201
1.99249
1.99499
1.99750
6.30079
6.30872
6.31664
62.5708
63.0448
63.5212
1.58342
1.58475
1.58608
3.41138
3.41424
3.41710
7.34960
7.35576
7.36192
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Powers and Roots
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6.32456
64.0000
1.58740
3.41995
7.36806
4.01
4.02
4.03
16.0801
16.1604
16.2409
2.00250
2.00499
2.00749
6.33246
6.34035
6.34823
64.4812
64.9648
65.4508
1.58872
1.59004
1.59136
3.42280
3.42564
3.42848
7.37420
7.38032
7.38644
4.04
4.05
4.06
16.3216
16.4025
16.4836
2.00998
2.01246
2.01494
6.35610
6.36396
6.37181
65.9393
66.4301
66.9234
1.59267
1.59399
1.59530
3.43131
3.43414
3.43697
7.39254
7.39864
7.40472
4.07
4.08
4.09
16.5649
16.6464
16.7281
2.01742
2.01990
2.02237
6.37966
6.38749
6.39531
67.4191
67.9173
68.4179
1.59661
1.59791
1.59922
3.43979
3.44260
3.44541
7.41080
7.41686
7.42291
4.10
16.8100
2.02485
6.40312
68.9210
1.60052
3.44822
7.42896
4.11
4.12
4.13
16.8921
16.9744
17.0569
2.02731
2.02978
2.03224
6.41093
6.41872
6.42651
69.4265
69.9345
70.4450
1.60182
1.60312
1.60441
3.45102
3.45382
3.45661
7.43499
7.44102
7.44703
4.14
4.15
4.16
17.1396
17.2225
17.3056
2.03470
2.03715
2.03961
6.43428
6.44205
6.44981
70.9579
71.4734
71.9913
1.60571
1.60700
1.60829
3.45939
3.46218
3.46496
7.45304
7.45904
7.46502
4.17
4.18
4.19
17.3889
17.4724
17.5561
2.04206
2.04450
2.04695
6.45755
6.46529
6.47302
72.5117
73.0346
73.5601
1.60958
1.61086
1.61215
3.46773
3.47050
3.47327
7.47100
7.47697
7.48292
4.20
17.6400
2.04939
6.48074
74.0880
1.61343
3.47603
7.48887
4.21
4.22
4.23
17.7241
17.8084
17.8929
2.05183
2.05426
2.05670
6.48845
6.49615
6.50384
74.6185
75.1514
75.6870
1.61471
1.61599
1.61726
3.47878
3.48154
3.48428
7.49481
7.50074
7.50666
4.24
4.25
4.26
17.9776
18.0625
18.1476
2.05913
2.06155
2.06398
6.51153
6.51920
6.52687
76.2250
76.7656
77.3088
1.61853
1.61981
1.62108
3.48703
3.48977
3.49250
7.51257
7.51847
7.52437
4.27
4.28
4.29
18.2329
18.3184
18.4041
2.06640
2.06882
2.07123
6.53452
6.54217
6.54981
77.8545
78.4028
78.9536
1.62234
1.62361
1.62487
3.49523
3.49796
3.50068
7.53025
7.53612
7.54199
4.30
18.4900
2.07364
6.55744
79.5070
1.62613
3.50340
7.54784
4.31
4.32
4.33
18.5761
18.6624
18.7489
2.07605
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2.08087
6.56506
6.57267
6.58027
80.0630
80.6216
81.1827
1.62739
1.62865
1.62991
3.50611
3.50882
3.51153
7.55369
7.55953
7.56535
4.34
4.35
4.36
18.8356
18.9225
19.0096
2.08327
2.08567
2.08806
6.58787
6.59545
6.60303
81.7465
82.3129
82.8819
1.63116
1.63241
1.63366
3.51423
3.51692
3.51962
7.57117
7.57698
7.58279
4.37
4.38
4.39
19.0969
19.1844
19.2721
2.09045
2.09284
2.09523
6.61060
6.61816
6.62571
83.4535
84.0277
84.6045
1.63491
1.63619
1.63740
3.52231
3.52499
3.52767
7.58858
7.59436
7.60014
4.40
19.3600
2.09762
6.63325
85.1840
1.63864
3.53035
7.60590
4.41
4.42
4.43
19.4481
19.5364
19.6249
2.10000
2.10238
2.10476
6.64078
6.64831
6.65582
85.7661
86.3509
86.9383
1.63988
1.64112
1.64236
3.53302
3.53569
3.53835
7.61166
7.61741
7.62316
4.44
4.45
4.46
19.7136
19.8025
19.8916
2.10713
2.10950
2.11187
6.66333
6.67083
6.67832
87.5284
88.1211
88.7165
1.64359
1.64483
1.64606
3.54101
3.54367
3.54632
7.62888
7.63461
7.64032
4.47
4.48
4.49
19.9809
20.0704
20.1601
2.11424
2.11660
2.11896
6.68581
6.69328
6.70075
89.3146
89.9154
90.5188
1.64729
1.64851
1.64974
3.54897
3.55162
3.55426
7.64603
7.65172
7.65741
Powers and Roots
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4.50
20.2500
2.12132
6.70820
91.1250
1.65096
3.55689
7.66309
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4.52
4.53
20.3401
20.4304
20.5209
2.12368
2.12603
2.12838
6.71565
6.72309
6.73053
91.7339
92.3454
92.9597
1.65219
1.65341
1.65462
3.55953
3.56215
3.56478
7.66877
7.67443
7,68009
4.54
4.55
4.56
20.6116
20.7025
20.7936
2.13073
2.13307
2.13542
6.73795
6.74537
6.75278
93.5767
94.1964
94.8188
1.65584
1.65706
1.65827
3.56740
3.57002
3.57263
7.68573
7.69137
7.69700
4.57
4.58
4.59
20.8849
20.9764
21.0681
2.13776
2.14009
2.14243
6.76018
6.76757
6.77495
95.4440
96.0719
96.7026
1.65948
1.66069
1.66190
3.57524
3.57785
3.58045
7.70262
7.70824
7.71384
4.60
21.1600
2.14476
6.78233
97.3360
1.66310
3.58305
7.71944
4.61
4.62
4.63
21.2521
21.3444
21.4369
2.14709
2.14942
2.15174
6.78970
6.79706
6.80441
97.9722
98.6111
99.2528
1.66431
1.66551
1.66671
3.58564
3.58823
3.59082
7.72503
7.73061
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4.64
4.65
4.66
21.5296
21.6225
21.7156
2.15407
2.15639
2.15870
6.81175
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6.82642
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21.8089
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6.84105
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101.848
102.503
103.162
1.67150
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1.67388
3.60113
3.60370
3.60626
7.75840
7.76394
7.76946
4.70
22.0900
2.16795
6.85565
103.823
1.67507
3.60883
7.77498
4.7?
4.72
4.73
22.1841
22.2784
22.3729
2.17025
2.17256
2.17486
6.86294
6.87023
6.87750
104.487
105.154
105.824
1.67626
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1.67863
3.61138
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7.78049
7.78599
7.79149
4.74
4.75
4.76
22.4676
22.5625
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6.88477
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106.496
107.172
107.850
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1.68217
3.61903
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4.79
22.7529
22.8484
22.9441
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6.90652
6.91375
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108.531
109.215
109.902
1.68334
1.68452
1.68569
3.62665
3.62919
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7.81339
7.81885
7.82429
4.80
23.0400
2.19089
6.92820
110.592
1.68687
3.63424
7.82974
4.81
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4.83
23.1361
23.2324
23.3289
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6.93542
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111.285
111.980
112.679
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3.64180
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7.84601
4.84
4.85
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23.4256
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23.6196
2.20000
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6.95701
6.96419
6.97137
113.380
114.084
114.791
1.69154
1.69270
1.69386
3.64431
3.64682
3.64932
7.85142
7.85683
7.86222
4.87
4.88
4.89
23.7169
23.8144
23.9121
2.20681
2.20907
2.21133
6.97854
6.98570
6.99285
115.501
116.214
116.930
1.69503
1.69619
1.69734
3.65182
3.65432
3.65681
7.86761
7.87299
7.87837
4.90
24.0100
2.21359
7.00000
117.649
1.69850
3.65931
7.88374
4.91
4.92
4.93
24.1081 •
24.2064
24.3049
2.21585
2.21811
2.22036
7.00714
7.01427
7.02140
118.371
119.095
119.823
1.69965
1.70081
1.70196
3.66179
3.66428
3.66676
7.88909
7.89445
7.89979
4.94
4.95
4.96
24.4036
24.5025
24.6016
2.22261
2.22486
2.22711
7.02851
7.03562
7.04273
120.554
121.287
122.024
1.70311
1.70426
1.70540
3.66924
3.67171
3.67418
7.90513
7.91046
7.91578
4.97
4.98
4.99
24.7009
24.8004
24.9001
2.22935
2.23159
2.23383
7.04982
7.05691
7.06399
122.763
123.506
124.251
1.70655
1.70769
1.70884
3.67665
3.67911
3.68157
7.92110
7.92641
7.93171
286
Powers and Roots
n
W2
Vn
VlOw
W8
Vn
vwn
^iooli
5.00
25.0000
2.23607
7.07107
125.000
1.70998
3.68403
7.93701
5.01
5.02
5.03
25.1001
25.2004
25.3009
2.23830
2.24054
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7.07814
7.08520
7.09225
125.752
126.506
127.264
1.71112
1.71225
1.71339
3.68649
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3.69138
7.94229
7.94757
7.95285
5.04
5.05
5.06
25.4016
25.5025
25.6036
2.24499
2.24722
2.24944
7.09930
7.10634
7.11337
128.024
128.788
129.554
1.71452
1.71566
1.71679
3.69383
3.69627
3.69871
7.95811
7.96337
7.96863
5.07
5.08
5.09
25.7049
25.8064
25.9081
2.25167
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7.12039
7.12741
7.13442
130.324
131.097
131.872
1.71792
1.71905
1.72017
3.70114
3.70357
3.70600
7.97387
7.97911
7.98434
5.10
26.0100
2.25832
7.14143
132.651
1.72130
3.70843
7.98957
5.11
5.12
5.13
26.1121
26.2144
26.3169
2.26053
2.20274
2.26495
7.14843
7.15542
7.16240
133.433
134.218
135.006
1.72242
1,72355
1.72467
3.71085
3.71327
3.71569
7.99479
8.00000
8.00520
5.14
5.15
5.16
26.4196
26.5225
26.6256
2.26716
2.26936
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7.16938
7.17635
7.18331
135.797
136.591
137.388
1.72579
1.72691
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3.71810
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3.72292
8.01040
8.01559
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5.1T
5.18
5.19
26.7289
26.8324
26.9361
2.27376
2.27596
2.27816
7.19027
7.19722
7.20417
138.188
138.992
139.798
1.72914
1.73025
1.73137
3.72532
3,72772
3.73012
8.02596
8.03113
8.03629
5.20
27.0400
2.28035
7.21110
140.608
1.73248
3.73251
8.04145
5.21
5.22
5.23
27.1441
27.2484
27.3529
2.28254
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7.21803
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141.421
142.237
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1.73359
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8.04660
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5.24
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27.4576
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7.23878
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143.878
144.703
145.532
1.73691
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5.27
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27.7729
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2.29565
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7.25948
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146.363
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5.30
28.0900
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148.877
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5.31
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28.1961
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1.74789
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1.75007
3.76571
3.76806
3.77041
8.11298
8.11804
8.12310
5.37
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5.39
28.8369
28.9444
29.0521
2.31733
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7.32803
7.33485
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154.854
155.721
156.591
1.75116
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8.12814
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5.40
29.1600
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29.2681
29.3764
29.4849
2.32594
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7.35527
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158.340
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8.14828
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5.44
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29.5936
29.7025
29.8116
2.33238
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7.37564
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160.989
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5.47
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29.9209
30.0304
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2.33880
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163.667
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8.17829
8.18327
8.18824
Powers and Roots
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5.50
30.2500
2.34521
7.41620
166.375
1.76517
3.80295
8.19321
5.51
5.52
5.53
30.3601
30.4704
30.5809
2.34734
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2.35160
7.42294
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7.43640
167.284
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1.76624
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3.80526
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8.19818
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5.54
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30.6916
30.8025
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7.44312
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170.031
170.954
171.880
1.76944
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5.57
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31.0249
31.1364
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2.36008
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7.46324
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172.809
173.741
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3.81902
3.82130
3.82358
8.22783
8.23275
8.23766
5.60
31.3600
2.36643
7.48331
175.616
1.77581
3.82586
8.24257
5.61
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31.4721
31.5844
31.6969
2.36854
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7.48999
7.49667
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176.558
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178.454
1.77686
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3.82814
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8.24747
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179.406
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5.67
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32.1489
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7.52994
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182.284
183.250
184.220
1.78318
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8.27677
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5.70
32.4900
2.38747
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185.193
1.78632
3.84850
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5.71
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32.6041
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186.169
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32.9476
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189.119
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1.79048
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3.85748
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33.2929
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192.100
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33.7561
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34.1056
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7.64199
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199.177
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201.230
1.80082
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8.35868
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5.87
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34.4569
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2.42281
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202.262
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34.8100
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34.9281
35.0464
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206.425
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1.80799
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5.94
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35.2836
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7.72010
209.585
210.645
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5.97
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35.6409
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212.776
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214.922
1.81409
1.81510
1.81611
3.90833
3.91051
3.91269
8.42025
8.42494
8.42964
288
Powers and Boots
n
w2
Vw
VWJi
W3
va
VWn
^100™
6.00
36.0000
2.44949
7.74597
216.000
1.81712
3.91487
8.43433
6.01
36.1201
2.45153
7.75242
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1.81813
3.91704
8.43901
6.02
36.2404
2.45357
7.75887
218.167
1.81914
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8.44369
6.03
36.3609
2.45561
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219.256
1.82014
3.92138
8.44836
6.04
36.4816
2.45764
7.77174
220.349
1.82115
3.92355
8.45303
6.05
86.6025
2.45967
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221.445
1.82215
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8.45769
6.06
36.7236
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1.82316
3.92787
8.46235
6.07
36.8449
2.46374
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223.649
1.82416
3.93003
8.46700
6.08
36.9664
2.46577
7.79744
224.756
1.82516
3.93219
8.47165
6.09
37.0881
2.46779
7.80385
225.867
1.82616
3.93434
8.47629
6.10
37.2100
2.46982
7.81025
226.981
1.82716
3.93650
8.48093
6.11
37.3321
2.47184
7.81665
228.099
1.82816
3.93865
8.48556
6.12
37.4544
2.47386
7.82304
229.221
1.82915
3.94079
8.49018
6.13
37.5769
2.47588
7.82943
230.346
1.83015
3.94294
8.49481
6.14
37.6996
2.47790
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8.49942
6.15
37.8225
2.47992
7.84219
232.608
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8.50403
6.16
37.9456
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1.83313
3.94936
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6.17
38.0689
2.48395
7.85493
234.885
1.83412
3.95150
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6.18
38.1924
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7.86130
236.029
1.83511
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6.19
38.3161
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237.177
1.83610
3.95576
8.52243
6.20
38.4400
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1.83709
3.95789
8.52702
6.21
38.5641
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7.88036
239.483
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8.54532
6.25
39.0625
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244.141
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6.26
39.1876
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39.3129
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8.58620
6.34
40.1956
2.51794
7.96241
254.840
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3.98746
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256.048
1.85179
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6.41
41.0881
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8.00625
263.375
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267.090
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41.6025
2.53969
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269.586
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42.3801
42.5104
42.6409
2.55147
2.55343
2.55539
8.06846
8.07465
8.08084
275.894
277.168
278.445
1.86721
1.86817
1.86912
4.02279
4.02485
4.02690
8.66683
8.67127
8.67570
6.54
6.55
6.56
42.7716
42.9025
43.0336
2.55734
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2.56125
8.08703
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279.726
281.011
282.300
1.87008
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4.02896
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8.68012
8.68455
8.68896
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43.1649
43.2964
43.4281
2.56320
2.56515
2.56710
8.10555
8.11172
8.11788
283.593
284.890
286.191
1.87293
1.87388
1.87483
4.03511
4.03715
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8.69778
8.70219
6.60
43.5600
2.56905
8.12404
287.496
1.87578
4.04124
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6.61
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43.9569
2.57099
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8.13019
8.13634
8.14248
288.805
290.118
291.434
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8.14862
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292.755
294.080
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302.112
303.464
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306.182
307.547
308.916
1.88895
1.88988
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8.76772
8.77205
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46.1041
2.60192
2.60384
2.60576
8.22800
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8.24015
310.289
311.666
313.047
1.89175
1.89268
1.89361
4.07564
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8.78071
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46.2400
2.60768
8.24621
314.432
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46.3761
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8.25227
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315.821
317.215
318.612
1.89546
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320.014
321.419
322.829
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47.3344
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324.243
325.661
327.083
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329.939
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334.255
335.702
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338.609
340.068
341.532
1.91019
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351.896
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353.393
354.895
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1.91929
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1.92109
4.13498
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8.90854
8.91274
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50.4100
2.66458
8.42615
357.911
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4.14082
8.92112
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50.5521
50.6944
50.8369
2.66646
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8.43208
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359.425
360.944
362.467
1.92290
1.92380
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4.14276
4.14470
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8.92531
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8.44985
8.45577
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363.994
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8.93784
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51.5524
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8.46759
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370.146
371.695
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51.8400
2.68328
8.48528
373.248
1.93098
4.16017
8.96281
7.21
7.22
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51.9841
52.1284
52.2729
2.68514
2.68701
2.68887
8.49117
8.49706
8.50294
374.805
376.367
377.933
1.93187
1.93277
1.93366
4.16209
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52.4176
52.5625
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8.50882
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379.503
381.078
382.657
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4.16786
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62.8529
52.9984
53.1441
2.69629
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2.70000
8.52643
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384.241
385.828
387.420
1.93722
1.93810
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53.4361
53.5824
53.7289
2.70370
2.70555
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390.618
392.223
393.833
1.94076
1.94165
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4.18125
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9.00822
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8.56738
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395.447
397.065
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8.68332
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428.661
430.369
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4.22465
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57.7600
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8.71780
438.976
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58.8289
58.9824
59.1361
2.76948
2.77128
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8.75785
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451.218
452.985
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4.25248
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59.2900
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59.4441
59.5984
59.7529
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2.78029
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458.314
460.100
461.890
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2.78388
2.78568
8.79773
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463.685
465.484
467.289
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9.18150
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60.3729
60.5284
60.6841
2.78747
2.78927
2.79106
8.81476
8.82043
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469.097
470.911
472.729
1.98065
1.98150
1.98234
4.26717
4.26900
4.27083
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2.79285
8.83176
474.552
1.98319
4.27266
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7.81
7.82
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61.1524
61.3089
2.79464
2.79643
2.79821
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8.84308
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476.380
478.212
480.049
1,98404
1.98489
1.98573
4.27448
4.27631
4.27813
9.20910
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7.84
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61.4656
61.6225
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2.80000
2.80179
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8.85438
8.86002
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481.890
483.737
485.588
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1.98742
1.98826
4.27995
4.28177
4.28359
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62.2521
2.80535
2.80713
2.80891
8.87130
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487.443
489.304
491.169
1.98911
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4.28722
4.28903
9.23262
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2.81069
8.88819
493.039
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2.81603
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498.677
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1.99415
4.29265
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4.29627
9.24823
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63.0436
63.2025
63.3616
2.81780
2.81957
2.82135
8.91067
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500.566
502.460
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8.97775
519.718
521.660
523.607
2.00333
2.00416
2.00499
4.31604
4.31783
4.31961
9.29862
9.30248
9.30633
8.07
8.08
8.09
65.1249
65.2864
65.4481
2.84077
2.84253
2.84429
8.98332
8.98888
8.99444
525.558
527.514
529.475
2.00582
2.00664
2.00747
4.32140
4.32318
4.32497
9.31018
9.31402
9.31786
8.10
65.6100
2.84605
9.00000
531.441
2.00830
4.32675
9.32170
8.11
8.12
8.13
65.7721
65.9344
66.0969
2-84781
2.84956
2.85132
9.00555
9.01110
9.01665
533.412
535.387
537.368
2.00912
2.00995
2.01078
4.32853
4.33031
4.33208
9.32553
9.32936
9.33319
8.14
8.15
8.16
66.2596
66.4225
66.5856
2.85307
2.85482
2.85657
9.02219
9.02774
9.03327
539.353
541.343
543.338
2.01160
2.01242
2.01325
4.33386
4.33563
4.33741
9.33702
9.34084
9.34466
8.17
8.18
8.19
66.7489
66.9124
67.0761
2.85832
2.86007
2.86182
9.03881
9.04434
9.04986
545.339
547.343
549.353
2.01407
2.01489
2.01571
4.33918
4.34095
4.34271
9.34847
9.35229
9.35610
8.20
67.2400
2.86356
9.05539
551.368
2.01653
4.34448
9.35990
8.21
8.22
8.23
67.4041
67.5684
67.7329
2.86531
2.86705
2.86880
9.06091
9.06642
9.07193
553.388
555.412
557.442
2.01735
2.01817
2.01899
4.34625
4.34801
4.34977
9.36370
9.36751
9.37130
8.24
8.25
8.26
67.8976
68.0625
68.2276
2.87054
2.87228
2.87402
9.07744
9.08295
9.08845
559.476
561.516
563.560
2.01980
2.02062
2.02144
4.35153
4.35329
4.35505
9.37510
9.37889
9.38268
8.27
8.28
8.29
68.3929
68.5584
68.7241
2.87576
2.87750
2.87924
9.09395
9.09945
9.10494
565.609
567.664
569.723
2.02225
2.02307
2.02388
4.35681
4.35856
4.36032
9.38646
9.39024
9.39402
8.30
68.8900
2.88097
9.11043
571.787
2.02469
4.36207
9.39780
8.31
8.32
8.33
69.0561
69.2224
69.3889
2.88271
2.88444
2.88617
9.11592
9.12140
9.12688
573.856
575.930
578.010
2.02551
2.02632
2.02713
4.36382
4.36557
4.36732
9.40157
9.40534
9.40911
8.34
8.35
8.36
69.5556
69.7225
69.8896
2.88791
2.88964
2.89137
9.13236
9.13783
9.14330
580.094
582.183
584.277
2.02794
2.02875
2.02956
4.36907
4.37081
4.37256
9.41287
9.41663
9.42039
8.37
8.38
8.39
70.0569
70.2244
70.3921
2.89310
2.89482
2.89655
9.14877
9.15423
9.15969
586.376
588.480
590.590
2.03037
2.03118
2.03199
4.37430
4.37604
4.37778
9.42414
0.42789
9.43164
8.40
70.5600
2.89828
9.16515
592.704
2.03279
4.37952
9.43539
8.41
8.42
8.43
70.7281
70.8964
71.0649
2.90000
2.90172
2.90345
9.17061
9.17606
9.18150
594.823
596.948
599.077
2.03360
2.03440
2.03521
4.38126
4.38299
4.38473
9.43913
9.44287
9.44661
8.44
8.45
8.46
71.2336
71.4025
71.5716
2.90517
2.90689
2.90861
9.18695
9.19239
9.19783
601.212
603.351
605.496
2.03601
2.03(582
2.03762
4.38646
4.38819
4.38992
9.45034
9.45407
9.45780
8.47
8.48
8.49
71.7409
71.9104
72.0801
2.91033
2.91204
2.91376
9.20326
9.20869
9.21412
607.645
609.800
611.960
2.03842
2.039L':i
2.04003
4.30166
4.39338
4.: s'.>r> 10
9.46152
9.46525
9.46897
Powers and Boots
293
n
w2
Vti-
VWn
n8
3/n
VWn
^100 n
8.50
72.2500
2.91548
9.21954
614.125
2.04083
4.39683
9.47268
8.51
8.52
8.53
72.4201
72.5904
72.7609
2.91719
2.91890
2.92062
9.22497
9.23038
9.23580
616.295
618.470
620.650
2.04163
2.04243
2.04323
4.39855
4.40028
4.40200
9.47640
9.48011
9.48381
8.54
8.55
8.56
72.9316
73.1025
73.2736
2.92233
2.92404
2.92575
9.24121
9.24662
9.25203
622.836
625.026
627.222
2.04402
2.04482
2.04562
4.40372
4.40543
4.40715
9.48752
9.49122
9.49492
.8.57
8.58
8.59
73.4449
73.6164
73.7881
2.92746
2.92916
2.93087
9.25743
9.26283
9.26823
629.423
631.629
633.840
2.04641
2.04721
2.04801
4.40887
4.41058
4.41229
9.498(51
9.50231
9.50600
860
73.9600
2.93258
9.27362
636.056
2.04880
4.41400
9.50969
8.61
8.62
8.63
74.1321
74.3044
74.4769
2.93428
2.93598
2.93769
9.27901
9.28440
9.28978
638.277
640.504
642.736
2.04959
2.03039
2.05118
4.41571
4.41742
4.41913
9.51337
9.51705
9.52073
8.G4
8.65
8.66
74.6496
74.8225
74.9956
2.93939
2.94109
2.94279
9.29516
9.30054
9.30591
644.973
647.215
649.462
2.05197
2.05276
2.05355
4.42084
4.42254
4.42425
9.52441
9.52808
9.53175
8.67
8.68
8.69
75.1689
75.3424
75.5161
2.94449
2.94618
2.94788
9.31128
9.31665
9.32202
651.714
653.972
656.235
2.05434
2.05513
2.05592
4.42595
4.42765
4.42935
9.53542
9.53908
9.54274
8.70
75.0900
2.<)4958
9.32738
658.503
2.05671
4.43105
9.54640
8.71
8.72
8.73
75.8641
76.0384
76.2129
2.95127
2.95296
2.9546(5
9.33274
9.33H09
9.34345
660.776
663.055
665.339
2.05750
2.05828
2.05907
4.43274
4.43444
4.43613
9.55006
9.55371
9.55736
8.74
8.75
8.76
76.3876
76.5625
76.7376
2.95635
2.95804
2.95973
9.34880
9.35414
9.35949
667.628
669.922
672.221
2.05986
2.06064
2.06143
4.43783
4.43952
4.44121
9.56101
9.56466
9.56830
8.77
8.78
8.79
76.9129
77.0884
77.2641
2.96142
2.96311
2.96479
9.36483
9.37017
9.37550
674.526
676.836
679.151
2.06221
2.06299
2.06378
4.44290
4.44459
4.44627
9.57194
9.57557
9.57921
8 80
77.4400
2.96648
9.38083
681.472
2.06456
4.44796
9.58284
8.81
8.82
8.83
77.6161
77.7924
77.9689
2.96816
2.96985
2.97153
9.38616
9.39149
9.39681
683.798
686.129
688.465
2.06534
2.06612
2.06690
4.44964
4.45133
4.45301
9.58647
9.59009
9.59372
8.84
8.85
8.86
78.1456
78.3225
78.4996
2.97321
2.97489
2.97658
9.40213
9.40744
9.41276
690.807
693.154
695.506
2.06768
2.06846
2.06924
4.45469
4.45637
4.45805
9.59734
9.60095
9.60457
8.87
8.88
8.89
78.6769
78.8544
79.0321
2.97825
2.97993
2.98161
9.41807
9.42338
9.42868
697.864
700.227
702.595
2.07002
2.07080
2.07157
4.45972
4.46140
4.46307
9.60818
9.61179
9.6154C
890
79.2100
2.98329
9.43398
704.969
2.07235
4.46475
9.61900
8.91
8.92
8.93
79.3881
79.5664
79.7449
2.98496
2.98664
2.98831
9.43928
9.44458
9.44987
707.348
709.732
712.122
2.07313
2.07390
2.07468
4.46642
4.46809
4.46976
9.62260
9.62620
9.62980
8.94
8.95
8.96
79.9236
80.1025
80.2816
2.98998
2.99166
2.99333
9.45516
9.46044
9.46573
714.517
716.917
719.323
2.07545
2.07622
2.07700
4.47142
4.47309
4.47476
9.63339
9.63698
9.64057
8.97
8.98
8.99
80.4609
80.6404
80.8201
2.99500
2.99666
2.99833
9.47101
9.47629
9.48156
721.734
724.151
726.573
2.07777
2.07854
2.07931
4.47642
4.47808
4.47974
9.64415
9.64774
9.65132
294
Powers and Boots
n
n*
Vti
VWn
n*
V*
VWn
VJMn
9.00
81.0000
3.00000
9.48683
729.000
2.08008
4.48140
9.65489
9.01
9.02
9.03
81.1801
81.3604
81.5409
3.00167
3.00333
3.00500
9.49210
9.49737
9.50263
731.433
733.871
736.314
2.08085
2.08162
2.08239
4.48306
4.48472
4.48638
9.65847
9.66204
9.66561
9.04
9.05
9.06
81.7216
81.9025
82.0836
3.00666
3.00832
3.00998
9.50789
9.51315
9.51840
738.763
741.218
743.677
2.08316
2.08393
2.08470
4.48803
4.48969
4.49134
9.66918
9.67274
9.67630
9.07
9.08
9.09
82.2649
82.4464
82.6281
3.01164
3.01330
3.01496
9.523(35
9.52890
9.53415
746.143
748.613
751.089
2.08546
2.08623
2.08699
4.49299
4.49464
4.49629
9.67986
9.68342
9.68697
9.10
82.8100
3.01662
9.53939
753.571
2.08776
4.49794
9.69052
9.11
9.12
9.13
82.9921
83.1744
83.3569
3.01828
3.01993
3.02159
9.54463
9.54987
9.55510
756.058
758.551
761.048
2.08852
2.08929
2.09005
4.49959
4.50123
4.50288
9.69407
9.69762
9.70116
9.14
9.15
9.16
83.5396
83.7225
83.9056
3.02324
3.02490
3.02655
9.56033
9.56556
9.57079
763.552
766.061
768.575
2.09081
2.09158
2.09234
4.50452
4.50616
4.50781
9.70470
9.70824
9.71177
9.17
8.18
9.19
84.0889
84.2724
84.4561
3.02820
3.02985
3.03150
9.57601
9.58123
9.58645
771.095
773.621
776.152
2.09310
2.09386
2.09462
4.50945
4.51108
4.51272
9.71531
9.71884
9.72236
9.20
84.6400
3.03315
9.59166
778.688
2.09538
4.51436
9.72589
9.21
9.22
9.23
84.8241
85.0084
85.1929
3.03480
3.03645
3.03809
9.59687
9.60208
9.60729
781.230
783.777
786.330
2.09614
2.09690
2.09765
4.51599
4.51763
4.51926
9.72941
9.73293
9.73645
9.24
9.25
9.26
85.3776
85.5625
85.7476
3.03974
3.04138
3.04302
9.61249
9.61769
9.62289
788.889
791.453
794.023
2.09841
2.09917
2.09992
4.52089
4.52252
4.52415
9.73996
9.74348
9.74699
9.27
9.28
9.29
85.9329
86.1184
86.3041
3.04467
3.04631
3.04795
9.62808
9.63328
9.63846
796.598
799.179
801.765
2.10068
2.10144
2.10219
4.52578
4.52740
4.52903
9.75049
9.75400
9.75750
9.30
86.4900
3.04959
9.64365
804.357
2.10294
4.53065
9.76100
9.31
9.32
9.33
86.6761
86.8624
87.0489
3.05123
3.05287
3.05450
9.64883
9.65401
9.65919
806.954
809.558
812.166
2.10370
2.10445
2.10520
4.53228
4.53390
4.53552
9.76450
9.76799
9.77148
9.34
9.35
9.36
87.2356
87.4225
87.6096
3.05614
3.05778
3.05941
9.66437
9.66954
9.67471
814.781
817.400
820.026
2.10595
2.10671
2.10746
4.53714
4.53876
4.54038
9.77497
9.77846
9.78195
9.37
9.38
9.39
87.7969
87.9844
88.1721
3.06105
3.06268
3.06431
9.67988
9.68504
9.69020
822.657
825.294
827.936
2.10821
2.10896
2.10971
4.54199
4.54361
4.54522
9.78543
9.78891
9.79239
9.40
88.3600
3.06594
9.69536
830.584
2.11045
4.54684
9.79586
9.41
9.42
9.43
88.5481
88.7364
88.9249
3.06757
3.06920
3.07083
9.70052
9.70567
9.71082
833.238
835.897
838.562
2.11120
2.11195
2.11270
4.54845
4.55006
4.55167
9.79933
9.80280
9.80627
9.44
9.45
9.46
89.1136
89.3025
89.4916
3.07246
3.07409
3.07571
9.71597
9.72111
9.72625
841.232
843.909
846.591
2.11344
2.11419
2.11494
4.55328
4.55488
4.55649
9.80974
9.81320
9.81666
9.47
9.48
9.49
89.6809
89.8704
90.0601
3.07734
3.07896
3.08058
9.73139
9.73653
9.74166
849.278
851.971
854.670
2.11568
2.11642
2.11717
4.55809
4.55970
4.56130
9.82012
9.82357
9.82703
Powers and Roots
295
n
n2
Vn
VWn
n8
^n
^10 n
^100 n
9.50
90.2500
3.08221
9.74679
857.375
2.11791
4.56290
9.83048
9.51
9.52
9.53
90.4401
90.6304
90.8209
3.08383
3.08545
3.08707
9.75192
9.75705
9.76217
860.085
862.801
865.523
2.11865
2.11940
2.12014
4.56450
4.56610
4.56770
9.83392
9.83737
9.84081
9.54
9.55
9.56
91.0116
91.2025
91.3936
3.08869
3.09031
3.09192
9.76729
9.77241
9.77753
868.251
870.984
873.723
2.12088
2.12162
2.12236
4.56930
4.57089
4.57249
9.84425
9.84769
9.85113
9.57
9.58
9.59
91.5849
91.7764
91.9681
3.09354
3.09516
3.09677
9.78264
9.78775
9.79285
876.467
879.218
881.974
2.12310
2.12384
2.12458
4.57408
4.57567
4.57727
9.85456
9.85799
9.86142
9.60
92.1600
3.09839
9.79796
884.736
2.12532
4.57886
9.86485
9.61
9.62
9.63
92.3521
92.5444
92.7369
3.10000
3.10161
3.10322
9.80306
9.80816
9.81326
887.504
890.277
893.056
2.12605
2.12679
2.12753
4.58045
4.58204
4.58362
9.86827
9.87169
9.87511
9.64
9.65
9.66
92.9296
93.1225
93.3156
3.10483
3.10644
3.10805
9.81835
9.82344
9.82853
895.841
898.632
901.429
2.12826
2.12900
2.12974
4.58521
4.58679
4.58838
9.87853
9.88195
9:88536
9.67
9.68
9.69
93.5089
93.7024
93.8961
3.10966
3.11127
3.11288
9.83362
9.83870
9.84378
904.231
907.039
909.853
2.13047
2.13120
2.13194
4.58996
4.59154
4.59312
9.88877
9.89217
9.89558
9.70
94.0900
3.11448
9.84886
912.673
2.13267
4.59470
9.89898
9.71
9.72
9.73
94.2841
94.4784
94.6729
3.11609
3.11769
3.11929
9.85393
9.85901
9.86408
915.499
918.330
921.167
2.13340
2.13414
2.13487
4.59628
4.59786
4.59943
9.90238
9.90578
9.90918
9.74
9.75
9.76
94.8676
95.0625
95.2576
3.12090
3.12250
3.12410
9.86914
9.87421
9.87927
924.010
926.859
929.714
2.13560
2.13633
2.13706
4.60101
4.60258
4.60416
9.91257
9.91596
9.91935
9.77
9.78
9.79
95.4529
95.6484
95.8441
3.12570
3.12730
3.12890
9.88433
9.88939
9.89444
932.575
935.441
938.314
2.13779
2.13852
2.13925
4.60573
4.60730
4.60887
9.92274
9.92612
9.92950
9.80
96.0400
3.13050
9.89949
941.192
2.13997
4.61044
9.93288
9.81
9.82
9.83
96.2361
96.4324
96.6289
3.13209
3.13369
3.13528
9.90454
9.90959
9.91464
944.076
946.966
949.862
2.14070
2.14143
2.14216
4.61200
4.61357
4.61514
9.93626
9.93964
9.94301
9.84
9.85
9.86
96.8256
97.0225
97.2196
3.13688
3.13847
3.14006
9.91968
9.92472
9.92975
952.764
955.672
958.585
2.14288
2.14361
2.14433
4.61670
4.61826
4.61983
9.94638
9.94975
9.95311
9.87
9.88
9.89
97.4169
97. (5144
97.8121
3.14166
3.14325
3.14484
9.93479
9.93982
9.94485
961.505
964.430
967.362
2.14506
2.14578
2.14651
4.62139
4.62295
4.62451
9.95648
9.95984
9.96320
9.90
98.0100
3.14643
9.94987
970.299
2.14723
4.62607
9.96655
9.91
9.92
9.93
98.2081
98.4064
98.6049
3.14802
3.14960
3.15119
9.95490
9.95992
9.96494
973.242
976.191
979.147
2.14795
2.14867
2.14940
4.62762
4.62918
4.63073
9.96991
9.97326
9.97661
9.94
9.95
9.96
98.8036
99.0025
99.2016
3.15278
3.15436
3.15595
9.96995
9.97497
9.97998
982.108
985.075
988.048
2.15012
2.15084
2.15156
4.63229
4.63384
4.63539
9.97996
9.98331
9.98665
9.97
9.98
9.99
99.4009
99.6004
99.8001
3.15753
3.15911
3.16070
9.98499
9.98999
9.99500
991.027
994.012
997.003
2.15228
2.15300
2.15372
4.63694
4.63849
4.64004
9.98999
9.99333
9.99667
TABLE II — IMPORTANT NUMBERS
A. Units of Length
ENGLISH UNITS METRIC UNITS
12 inches (in.) = 1 foot (ft.) 10 millimeters = 1 centimeter (cm.)
3 feet = 1 yard (yd.) (mm.)
6£ yards = 1 rod (rd.) 10 centimeters = 1 decimeter (dm.)
320 rods = 1 mile (mi.) 10 decimeters = 1 meter (m.)
10 meters = 1 dekameter (Dm.)
1000 meters = 1 kilometer (Km.)
ENGLISH TO METRIC METRIC TO ENGLISH
1 in. = 2.5400 cm. 1 cm. = 0.3937 in.
1 ft. = 30.480 cm. 1m. = 39.37 in. = 3.2808 ft
1 mi. = 1.6093 Km. 1 Km. = 0.6214 mi.
B. Units of Area or Surface
1 square yard = 9 square feet = 1296 square inches
1 acre (A.) = 160 square rods = 4840 square yards
1 square mile = 640 acres = 102400 square rods
C. Units of Measurement of Capacity
DRY MEASURE LIQUID MEASURE
2 pints (pt.) = 1 quart (qt.) 4 gills (gi.) = 1 pint (pt.)
8 quarts = 1 peck (pk.) 2 pints = 1 quart (qt.)
4 pecks = 1 bushel (bu.) 4 quarts = 1 gallon (gal.)
1 gallon = 231 cu. in.
D. Metric Units to English Units
1 liter = 1000 cu. cm. = 61.02 cu. in. = 1.0567 liquid quarts
1 quart = .94636 liter = 946.36 cu. cm.
1000 grams = 1 kilogram (Kg.) = 2.2046 pounds (Ib.)
1 pound = .453593 kilogram = 453.59 grams
E. Other Numbers
T = ratio of circumference to diameter of a circle
= 3.14159265
1 radian = angle subtended by an arc equal to the radius
= 57° 17' 44".8 = 67°.2957795 = 180°/ir
1 degree = 0.01745329 radian, or w/lSQ radians
Weight of 1 cu. ft. of water = 62.425 Jb.
296
INDEX
Abscissa, 41.
Absolute value, 2.
Addition, of expressions, 9; of frac-
tions, 32 ; of radicals, 74.
Antecedent, 166.
Arithmetic progression, 141 ; means,
146.
Ascending powers, 8.
Axes, coordinate, 41.
Base, logarithm to any, 242.
Brace, 9.
Bracket, 9.
Binomial, 8; theorem, 258; proof
of theorem, 261.
Calculating machines, 244.
Change of signs in fractions, 29.
Characteristic, 219, 221, 223.
Coefficient, 8.
Common difference, 141.
Common logarithm, 242.
Complex numbers, 93.
Consequent, 166.
Constants, 181.
Coordinates of a point, 41.
Cube root, 5.
Decimals, repeating, 161.
Denominator, 29.
Descending powers, 8.
Determinant, of the second order,
264; of the third order, 269; of
higher order, 273.
Difference, tabular, 230.
Discriminant, 110.
Division, formulas and rules, 14.
Elements of a determinant, 265.
Elimination by substitution, 50;
by addition or subtraction, 51.
Ellipse, 125.
Equation, simple, 36 ; linear, 36 ;
of the first degree, 36; solution
of, 36; root of, 36; principles
useful in the solution of, 37, 38;
containing radicals, 65; literal,
95; quadratic, see Quadratic
equation.
Equations, simultaneous, 47; in-
consistent, 48; simultaneous in
three unknowns, 54.
Evolution, 196.
Exponent, 4; fractional, 198; zero,
199 ; fundamental laws for any
rational exponent, 201.
Exponents, laws of, 193 ; introduc-
tion of general, 198; negative in
fractions, 200.
Extremes of a proportion, 167.
Factor, prime, 26.
Factoring, type forms of, 19, 23.
Factors, common, 26 ; highest com-
mon, 26.
Formulas, 97, 102, 205.
Fractions, definition, 28 ; equiva-
lent, 29 ; change of signs in, 29 ;
reduction to lowest terms, 30;
reduction to lowest common de-
nominator, 31 ; addition and sub-
traction of, 32 ; multiplication
and division of, 36.
Functions, idea of, 245 ; types of alge-
braic functions, 246; considered
graphically, 250.
Gear wheel law, 101.
Geometric progression, 150; means,
154; infinite, 156.
Graph, of an equation, 43 ; deter-
mined from two points, 44 ; of a
quadratic, 90 ; of a function, 250.
297
298
INDEX
Hyperbola, 126.
Imaginary numbers, 116; pure,
93; unit, 116; addition and sub-
traction of, 118; multiplication
of, 119; division of, 120; geo-
metric representation of, 121.
Inconsistent equations, 47.
Index, 5, 68, 196.
Involution, 193.
Irrational numbers, 68.
Letters, use of in algebra, 4.
Lever, law of, 100.
Limit, variable, 159.
Linear equation, 36 ; graph of, 43, 44.
Logarithm, 217 ; of any number,
217 ; number corresponding to,
231; of a power, 235; general,
242; common, 242; of a root,
244; tables, 291.
Mantissa, 219; determination of,
225.
Mathematical induction, 255.
Mean proportional, 172.
Means of a proportion, 167.
Monomial, 8 ; square root of, 62.
Multiple, common, 27 ; lowest com-
mon, 28.
Multiplication, formulas and rules,
11.
Negative exponents, 199.
Negative numbers, 1 ; operations
with, 2.
Numbers, negative, 1 ; positive, 1 ;
operations with, 2; rational, 68,
108; irrational, 68; complex, 93;
real, 93, 108; summary, 109;
imaginary, see Imaginary number.
Numerator, 29.
Order of a determinant, 265, 269.
Ordinate, 41.
Origin, 41.
Parenthesis, 9.
Perfect square trinomial, 20.
Periods in square root, 60.
Polynomial, 8; arranging a, 8;
square root of, 62.
Power, 4; tables, 274.
Powers, 195.
Prime factor, 27.
Progression, arithmetic, 141 ; geo-
metric, 150.
Proportion, 167; terms of a, 167;
extremes of a, 167 ; means of a,
167; algebraic, 168; fundamental
principles of, 168, 174; inversion
in a, 174 ; alternation in a, 174 ;
composition in a, 174 ; division
in a, 175; composition and divi-
sion in a, 175 ; several equal ratios
in a, 176.
Proportional, mean, 172; third and
fourth, 172.
Quadratic equation, 78; pure, 78;
affected, 78; solution of pure,
78 ; solution of affected by fac-
toring, 81 ; solution by complet-
ing the square, 83, 85; solution
by the Hindu method, 86 ; solu-
tion by formula, 87 ; graphical
solution of, 90, 122 ; having imag-
inary solutions, 92; character of
the roots of, 109 ; character of roots
considered geometrically, 111; for-
mation of from given solutions,
114; solution by elimination,
128; simultaneous, 137.
Radical, or quadratic radical, 67;
of the nth order, 68 ; value of,
68.
Radicals, simplification! of, 71, 206;
similar, 74; addition and sub-
traction of, 74; multiplication
of, 75; division of, 77.
Radicand, 68, 196.
Ratio, 166; of geometric progres-
sion, 150.
Rational number, 68.
Rationalizing the denominator, 2J.1.
Root, square, 5; nth, 5, 196.
Root of an equation, 36.
INDEX
299
Roots, 196; imaginary of a quad-
ratic, 108; character of, 111
Simultaneous equations, 47.
Slide rule, 244.
Solution of an equation, 44.
Special products, formulas of, 17.
Square root, 5 ; of a number, 59 ;
in arithmetic, 59 ; in algebra, 62 ;
of trinomials, 62 ; double sign
of, 64 ; tables, 274.
Subtraction, 8 ; of expressions, 9 ;
of fractions, 32; of radicals, 74.
Surd, 68; binomial, 214.
Table, use of, 68.
Term, 8.
Terms, like, 8 ; of a proportion, 167.
Theorem, binomial, 258.
Trinomial, 8; perfect square, 20;
square root of, 62.
Unit, imaginary, 116.
Variables, 181.
Variation, direct, 178; inverse, 179;
joint, 180; problems in, 185;
geometrically considered, 190.
Vinculum, 9.
Printed in the United States of America.
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