OURSE i
ALGEBRA
HAWKES -lUBY-TOUTON
J
GIFT OF
MATHEMATICAL TEXTS
FOR SCHOOLS
EDITED BY
PERCEY F. SMITH, Ph.D.
PROFESSOR OF MATHEMATICS IN THE SHEFFIELD
SCIENTIFIC SCHOOL OF YALE UNIVERSITY
SECOND COURSE IN ALGEBRA
BY
HERBERT E. HAWKES, Ph.D.
PROFESSOR OF MATHEMATICS IX COLUMBIA UNIVERSITY
WILLIAM A. LUBY, A.B.
HEAD OF THE DEPARTMENT OF MATHEMATICS
KAKSAS CITY POLYTECHNIC INSTITUTE
AND
FRANK C. TOUTON, A.M.
FORMERLY PRINCIPAL OF CENTRAL HIGH SCHOOL
ST. JOSEPH, MISSOURI
REVISED EDITION
GINN AND COMPANY
BOSTON • NEW YORK • CHICAGO • LONDON
ATLANTA • DALLAS • COLUMBUS • SAN FRANCISCO
tt33
ENTERED AT STATIONERS' HALL
COPYRIGHT, 1911, 1918, BY
HERBERT E. HAWKES, WILLIAM A. LUBY
AND FRANK C. TOUTON
ALL BIGHTS RESERVED
921.3
gbc satftengum 3pre<<
GINN AND COMPANY • PRO-
PRIETORS • BOSTON • U.S.A.
x^
PREFACE
This revision of the " Second Course in Algebra " has been
carried out in the same spirit as was the recent revision of the
" First Course in Algebra " by the same authors. The exercises
and problems, mainly new, have been carefully graded, and
the exposition has been wholly rewritten. Some advantageous
changes have been made in the order of topics, and some
chapters for which no well-grounded demand exists have been
omitted. Such simplifications of exercises and exposition have
been made as are consistent with a standard course in third-
semester algebra.
In the chapters devoted to a review of first-year algebra the
fact was borne constantly in mind that the material would be
handled by students who had not pursued the study of algebra
during the preceding year. It was consequently thought desira^
ble to have work in equations come much earlier than before.
The subject of fractions, the topic usually most in need of
review, has received full and careful treatment. Linear sys-
tems have been presented without the use of determinants.
Instead of treating square root, radicals, and exponents in
one chapter, the work under these topics has been made more
accessible by giving a separate chapter to each. The needs of
classes, even under almost identical conditions, differ widely,
one class needing more review on a certain topic than does
another. Consequently the review material has been expanded
so as to afford ample work for any class. It is not intended,
however, that all the exercises and problems should be solved
by any one student.
V
459908
SECOND COURSE IN ALGEBRA
CHAPTER I
REVIEW OF FUNDAMENTAL OPERATIONS
1. Order of fundamental Operations. The numerical value
of an arithmetical or an algebraic expression involving
signs of addition, subtraction, multiplication, and division
depends on the order in which the indicated operations
are performed. It is understood that
In a series of operations involving addition^ subtraction^
multiplication^ and division, first the midtiplications and' divi-
sions shall be performed in the order in which they occur.
Then the additions and subtractions shall be performed in
the order in which they occur or in any other order.
Within any parenthesis the preceding rule applies.
If the multiplication of two numbers or number symbols is indi-
cated by juxtaposition without any symbol of multiplication, it is
customary to think of the multiplication as already having been
9 a
performed. Thus 9 a ^ 7 a^ =
EXERCISES
Simplify :
1. 6 - 2 + 8 - 7. 5. 36 -- 4 . 3 - 8 + 2.
2. 8 + 12-3-5. 6. (19-3.5)(8-5)-(14-f-7).
3. 8-6-4. 7. 50 - 4(20 - 4 • 2)- 3 + 2 -7.
4. 24 - 2 -r- 3. 8. 12 - 8 -f- 2 + 10 . 3 - 6 + 8 . 2.
9. (28 - 14 . 24 - 8 - 3 + 10) (30 - 3 - 5 -^ 2).
10. (16 - 32 X 48 H- 8 - 8 -h 3) -f- (42 - 6 . 7 - 42 - 6) . 6.
1
9
2 • SEGO-JSTB COURSE IN. ALGEBRA
' ' Fmd'tiie name^ieai v.alue of :
11. 9ic-7 itx = 3.
12. 2x^-5x + 3iix = 4..
13. t^-St''-^St-lift = 2.
• 14. Does 4(2;:c-5) + 15 = 3(£t;4-10) if cc = 7?.
15. Does (t -\-4:)(t-{-3)-(t-\- 1) (z^ + 2)= 42 if ^ =
ic T^ 3 • 2x 3 + 5^-2.^2. „
16. Does — - = — iix = 2?
X — 1 x -{-1 x^ — 1
2. Similar terms. Terms which are ahke iii every respect
except their coefficients are called similar.
3. Addition. In algebra, addition involves the uniting
of similar terms which have the same or opposite signs
into one term. For this we have the following rules:
/. To add two or more positive numbers^ find the arith-
metical sum of their absolute values and prefix to "this sum
the plus sign,
II. To add two or more negative numbers^ find the arith-
metical sum of their absolute values arid prefix to this sum
the minus sign.
III. To add a positive and a negative number, find the dif-
ference of their absolute values and prefix to this difference
the sign of the number which has the greater absolute value.
Obviously 2 + 4+7=2+7+4 = 7+24-4, etc. Even
if we have a series of positive and negative numbers,
the order in which they occur does not affect their sum.
This principle of addition is called the Commutative Law
for Addition.
For the addition of polynomials we have the
Rule. Write sitnilar terms iri the same column.
Find the algebraic sum of the terms in each column and
write the results ift succession with their proper signs.
EEVIEW OF FUNDAMENTAL OPERATIONS 3
4. Subtraction. For the subtraction of polynomials we
have the
Rule, Write the subtrahend under the minuend so that
similar terms are in the same column.
Then change mentally/ the sign of each term of the sub-
trahend and apply the rule for addition to each column.
EXERCISES
Add:
1. 12, - 8, 4- 4, - 3, and 6.
2. 4 a, 3 a, — 7a, 6 a, and — 2 a.
3. 5 a - 3 c 4- 6, 2 a - 6 c 4- 11, and 4 a — c - 9.
4. 3s-4i^ + 6, -7!f + 6s-8, and t-\-s^l^.
b. X - 2 f - ?> z, ^ f - ^x -{- 2 z, d.n& 4cz - f.
6. ^2 - 3 a -f 1, - 2 «2 _ 7f^ _|- 6, and 3 cv" - 4 + 5 a.
Write so that x, y, or t shall have a polynomial coefficient;
1. ax — 2x. 11. at — aH — 2 st.
Solution. ax — 2x = {a — 2) x. 12. ay -\- by -{- y.
8. ace + bx -\- ex. IZ. ^ ax — A:bx -{- ^ x.
9. 5 at — Abt — 2t. 14. 4 a; — abx — x.
10. 4 2^ — 3 a?^ — S2^. 15. 7x — 3ax~ 4:a^x.
Write the following so that the binomial will have a binomial
coefficient :
16. (a-3)x-c(a-3). 18. S(a -{- b)- G(a -^ b).
17. A(a-x)-^c(-x-{-a). 19. 6a(^c - 2 c)- S(x - 2 r).
20. 3a(x-l)-2b(x-l).
21. 4:b(3x-2)~Sc(3x-2).
22. 4 7w,(5 a — 3 c) — 6 /t(— 3 c + 5 «).
Subtract the second expression from the first :
23. 6a, 4a. 25. 4x4-3, 8.T4-6.
24. 8 a\ 15 a^ 26. Tor^ - 10, 5x^-h 20.
SECOND COURSE IN ALGEBRA
27. 5x-6, 20" + 8.
28. x^-5x-{-6, 2cc2_|_3^_io
29. a'-4:ac-3c',4:a'-^10ac-e'.
30. Sa--2b-6c,4:a-h6b-7c-2.
31. 3a^-2c''-6ac,5a' + 4:c'-Sac.
32. a^-Sa^c + 6 ac\ 7c^-h4:a^c-2 a\
S3. X — Sf -\- z — 4:ac -\- 7 ax, 4:X — f + 8 — 5ax -h 9 ac.
34. a^—c-\-Sx — a^m — 8 ac, 4 a* + m — 8 x — 10 ac + 4
Find the expression which added to the first will give the
second:
35. Sx^ - 5x -\- 2, 6x^ -llx -^ S.
36. 4:X^^Scx + c%10x^-j-Scx-9 g\
Find the expression which subtracted from the second will
give the first :
.37. 4.a'-2ab-\-b^, 7 a^ -lOab -h 6b\
38. c'-ecx -i-Sx% 9x^ - lOcx + 4 + c^.
39. From the sum of 2^^ — 4 # — 9 and 3 ^^^ _ g ^^ _|_ ^1^ ^^i^^, ^.^^g
sum of 3 2^ - 6 + 4 ?J2 ^nd 5 - 8 2^2 + 4 1.
40. From the sum oi ax — ac — ^ ^ and 4 c^ — 3 ac take the
sum of 4 c^ — 8 ace -|- a^ and Aac -\- Sax — 5 c\
ORAL EXERCISES
1. What name is given to each 3 in a^ -\-Sa? Define both.
2. Distinguish between an exponent and a power. What is
the meaning of 4 in cc* ? of 2 in 3^ ? of a in x^ ?
3. What is the coefficient of x in 3 a'^bx ? ofa^? ofb?
4. What is the coefficient of x in the expressions ax -\-x?
4:X — ax? ex — ax — X? What is the meaning of 3 in 3 ic ? of
10 in lOic? of a in OK?
KEVIEW OF FUNDAMENTAL OPERATIONS 5
5. What is meant by the absolute vahie of a number?
Illustrate.
6. What is a literal exponent ? Illustrate.
8. x^^x'^? x« -^ a;' = ? a^^^+i -^x^ = ?
9. How can the correctness of the result of addition be
checked ? of subtraction ?
10. What is meant by arrangement of an algebraic expres-
sion with respect to a certain letter ?
11. Arrange a^ -\- b* — 4: a^b — 6 a%^ + 4 ab^ according to the
descending powers of b.
12. Arrange t^ — Sf^ — 5 -\-t — 2t^ according to the ascend-
ing powers of t ; according to the descending powers of t.
13. Is arrangement of divisor and dividend in the same
order, desirable ? Why ?
5. Multiplication. In multiplying one term by another
the sign of the product, the coefficient of the product, and
the exponent of any letter in the product are obtained as
follows :
7. 7^e sign of the product is plus if the multiplier and the
multiplicand have like signs, and minus if they have unlike
II, The coefficient of the product is the product of the coeffi-
cients of the factors.
III, The exponent of each letter in the product is determined
hy the general law n°xnf>= n«+ K
For the multiplication of polynomials we have the
Rule. Multiply the midtiplicand by each term of the multi-
plier in tu/m, and add the partial products.
SECOND COURSE m ALGEBEA
EXERCISES
Perform, the indicated multiplication-:
1. (6x^-3x-h4.)Sx. 5. Qix^ - 2x + 6)(7 - 3x^ - x).
2. (3x-5)(4:X + 3). 6. (2x^ - 5x-i-3)(x^- 5x-{- 6).
3. (2s-St)(4:S + 5t). 7. (a*-}-2a''-4)(a^-2a-3).
^. (x^-.x-^2)(3x-4:). 8. (t^-2t-\-6)(t*-3t''-ty
9. (2s^-3st-\-f-)(s^-{-5st-4:t^.
10. (2(4 _|_ 8 4- 4 2^2) (f^ + 8 - 4 1^).
11. (6^2 - ttc + c') («' + c" + ac).
12. (a'* -^2ab-[- h^) {a} + b- ab).
13. {t^ — t^ + t) (at^ + a-\- at).
14. (4 A2 _j_ 6 Jc^ + 9 AA)) (4 A^ 4. 6 A;^ - 9 M).
15. (3 c/.z^2 _ 2 a?^^ 4. 5 ^) (6 a?^' - 2 at - 4. af).
16. (2 ^2 _ 3 c« 4. 4 ac8) (2 a^ - 3 c« - 4 ac«).
,^ /a 2a« «^/» 2^2 a^
^®* V2~T""4A2~ 3 ~ 4.J'
19. (?^^ + 1 + f) {t'-\-l- f) (f' + 1-0-
20. (x^ -\^ 9f -{- l(y -h Sxij - 4:x -\- 12y)(x - 3y + 4).
Find the value of :
21. 3x^-^2x-\-r) if ic=:5.
22. 3t^-it-\-Siit=-3.
23. 9 - 8 jJ + 5 ?{' - 3 ^» if ^ =5 2.
24. 2 «« - 3 a^c 4- 4 ac2 - 3 c2 if a = 3 and c =^ -- 2.
25. Does 15(x - a)- 6(x -i- a)= 3(5 a - 3x) itx = 2a?
^6. Does ax(a -\- 3) -\- a(10 - a^)^ x + 3 if x =^ a - 3?
^7. Does T -T = 5 — ; ifa; = 0? ifx = -2?
REVIEW OF FUNDAMENTAL OPERATIONS 7
6. Division. In dividing one term by another the sign of
tlie quotient, the coefficient of the quotient, and the expo-
nent of each letter in the quotient are obtained as follows :
/. The sign of the quotient is plus when the dividend mid the
divisor have like signs, and minus when they have unlike signs.
II. The coefficient of the quotient is obtained hy dividing the
coefficient of the dividend hy that of the divisor.
III. The exponent of each letter in the quotient is determined
hy the law tv^ -i- n^ = n°- *.
The method of dividing one polynomial by another is
stated in the
Rule. Arrange the dividend and the divisor according to
the descending powers of some common letter, called the letter
of arrangement.
Divide the first term of the dividend hy the first term of the
divisor and ivrite the result for the first term of the quotient.
Multiply the entire divisor hy the first term of the quotient,
write the result under the dividend, and suhtract, heing careful
to write the terms of the remainder in the same order as those
of the divisor.
Divide the first term of the remainder hy the first term of
the divisor to get the second term of the quotient, and proceed
as before until there is no remainder, or until the remainder is
of lower degree in the letter of arrangement than the divisor.
EXERCISES
Perform the indicated division :
1. (2cc2-5a;-f 3)--(2x-3).
2. (6a^2-13£c + 6)-(3-2£c).
3. {^x^ -[-Bxy-2y'')^(x-\-2if).
4. (6a2 4-23o-!^-55?^2)-f-(3ci-5)5).
5. (6a« + 6a2-28-26a)-(2c^ + 4). .'
8 SECOND COURSE IN ALGEBRA
6. (6x^ -5x' + 25cc» - 17x^)-^(5x^ - 2a^«).
7. (6-«-7s^-8)^(6'2_^4 + 2s).
8. (x^-5 a'x + 2 a«) -^ (o? -\- 2 ax - a").'
9. (2 ^* - 12 jJ^ _ 2 _|. 11 25 _ 7 jj8^ ^ (^j^ _ 3 ^ _ 2 ^2>)^
10. (23s2-13s^4-2s''-60-s)--(5 4-35-6-2).
11. (32a;*- 60- 2£c- 104x^ + 92 ic2)--(5 + 6x-4a;2).
12. (a?-2ah-\-h^-^x')^{^x-)rh-a).
13. (4^»c4-l-4c2-^»2)^(2c-^»-l).
14. (a;^ + i»' + 8ic2+8)H-(£c2-2x + 4).
15. (27 a - 18 a^'^ _ 3 ^9 ^ 3 ^J4^) ^ (3 _ i^ + ^4>^.
16. (3£c^ + 9cc2 + 2 - 5aj - 8a;« - cc*)h- (x - 1)1
18. (4.0" -{- ^ - ^ah + h^ - Qb + 12 a)^(h - ?> - 2 a).
19. (£c« - 4icy + lQy^)^(x' + 2;rV + 2x^/4- 4ic/ + 4/).
20. (x8 + 8/ + 12o-30«2/)^(a; + 2?/ + 5).
21. (a?« + 2/' + ^' - ^xyz)^{x + 2/ + ^)-
7. Parentheses. The removal of a parenthesis preceded
by a plus sign is governed by the
Principle, A parenthesis and the sign before it, if plus, may
be removed from an expression without changing the signs of
the terms which were inclosed by the parenthesis.
The removal of a parenthesis preceded by a minus sign
is governed by the
Principle. A parenthesis and the sign before it, if minus,
may be removed from an expression, provided the sign of each
term which was inclosed by the parenthesis be changed.
The principle just stated is equivalent to the principle
which holds in the subtraction of one polynomial from
another.
EEVIEW OF FUNDAMEJSTTAL OPERATIONS 9
When one parenthesis incloses another, either the outer
or the mner parenthesis may be removed first. Usually it
is best to use the
Rule. Rewrite the expression^ omitting the innermost paren-
thesis, changing the signs of the terms which it inclosed if the
sign preceding it he minus aiid leaving them unchanged if it
he plus.
Comhine like terms that may occur within the new inner-
most parenthesis.
Repeat these processes until all the parentheses are removed.
EXERCISES
E/emove parentheses and simplify :
1. a + {a-h)-{a-Zh). 3. ^b -\_{a - h)-{c - a)'].
2. ^_(a-6') + (2c-3a). 4. 2c - 2(a - c)-f 3(c - a).
b. x-2j^-^(x-y)-A.(2x-y).
6. 4ic-a-f[-(3c-a')-(2a-3cc)].
7. a_[_(a_3) + (3c-2a)-5«]+6c.
8. X - 3 - (ci - 2 ic) - [2 (a - X - 5) - 3 (6 - 2 a)].
9. 2t^-^t-2t{^ + t).
10. x'-^-(x-l){x-2).
11. ^x" - ^ a^ -{a - 2x)(2>x - a).
12. 6cc+(3c- 8x + 2)-(c-a;-2). ' .
13. 6a;-[-(a-c)4-(3c-4a)].
14. 7c-[(3c-4)-6-(4a^-3a-c)].
15. 4.T. -2(£c-3)-3[x-3(4-2a;)+ 8].
16. 6cc-4(3-5cc)-4[2(x-4)+3(2ic-l)-(ic-7)].
17. 3x - 2[1 - 3(2 £c - 3 - a)- 5{a -(3cc - 2 a) - 4}].
18. 2t''-lt-(2t-l)(t-{-l).
19. (x - 4>(£c - 3) - (x - 3) (;r + 2).
RE
10 SECOND COUESE IK ALGEBRA
20. (a + b)a -(a — b)b-\-3 ab.
21. 3a(a - b)-(a + b)(a - b).
22. (x -3)(x-4:)-{x- 5) (x + 3).
8. Important special products. Certain products are of
frequent occurrence. These forms should be memorized
so that one can write or state the result without the labor
of actual multiplication.
I. For the square of the sum of two terms we have
the formula . (a+bf = a" + 2ab+Ir'.
This expressed verbally is :
The square of the sum of two terms is the square of the first
term plus twice the product of the two terms plus the square
of the second terin.
II. For the square of the difference of two terms we
have the formula , ,.„ 9 « i. , 1.0
(a —hy = a^ — 2ah-\- b".
This expressed verbally is:
The square of the difference of two terms is the square of
the first term, minus twice the product of the two terms, plus
the square of the second term.
III. For the product of the sum and tlie difference of
two terms we have the formula
, ' {a + b)(a-b) = a'-bi'.
This expressed verbally is:
T^e product of the sum and the difference of two terms
equals the difference of their squares taken in the same order
as the difference of the terms.
IV. For the product of two binomials having a common
term we have the formula
{x+a){x^-b) = jc^ -f {a^b)x-\-ab.
REVIEW OF FUNDAMENTAL OPERATIONS 11
This expressed verbally is: -
The product of two binomials having a common term equals
the square of the common term^ plus the algebraic sum of the
unlike terms multiplied by the common term^ plus the algebraic
product of the unlike terms,
V. The square of the polynomial a-\-b — c gives the
formula ^^^^_^y =: ^2 ^ ^ _^ c'+2ab~2ac-2hc.
This expressed verbally is:
TJie square of any polynomial is equal to the sum of the
squares of each of the terms plus twice the algebraic product
of each term by every term that follows it in the polynomial,
VI. The cube of the binomial a + b gives the formula
{a +by = (^ + Sa'b+3ab^ + b".
This expressed verbally is:
The cube of the sum of two numbers equals the cube of the
first, plus three times the square of the fivst times the second,
plus three times the first times the square of the second, plus
the cube of the second.
VII. Similarly,
{<i-bf = (^-Za^b+^ali'-l^.
This can be expressed verbally in a manner similar to VI.
ORAL EXERCISES
State the result of the indicated multiplication :
1. (£c + 5)1 6. (2 £c - If. 11. {x" - xf.
2. {x + 7)^ 7. {^x- 5)1 12. {a" + 3 af.
3. (2x + lf. 8. {ax -If. 13. {x-a){x + a).
4. (3 cc + 2)-. 9. {x + 7 a)\ 14. {x - 5) (x + 5).
5. {x^^^f. 10. (2x-^ijf. 15. (2x-l)(2a;+l).
12
SECOND COUKSE IN ALGEBRA
16. {ax + 3) {ax - 3).
17. (4£c-3c)(4£c + 3c).
18. {ax — c) {ax -{- c).
19. (4a + c)(c-4a).
20. {Za^2x){2x-2>a).
21. (a^ + 3)(.^ + 4).
22. {x + 2){x + b).
23. {x + l){x + l).
24. (x-3)(x-5).
25. (cc - 2) (ic - 7).
26. {x - 3) (a; + 4).
27. (a; H- 6) (ic - 7).
28. (£c H- 2) (£c - 9).
29. (x - 5) (£c + 10).
30. (o^cc — 3) {ax -\- 5).
31. (2a;-l)(2^ + 3).
32. (3cc + l)(3a; + 4).
33. (4x-2)(4ajH-3).
34. {a^-3a){a''^4.a).
35. (a + ^>4-c)2
36. {a^-G + xf
37. (c* + c - cc)*'^.
38. (a-c + ic/
39. {a + e-\- Vf
40. (a-c + l)2
41. (a + c + 2)2
42. (a + c -h 6)-^.
43. (2 a -I- c + 1)2.
44. (a + 20- -3)2.
45. (a -3 a; 4- 2)2.
46.
47.
48.
49.
50.
51.
52.
53.
54.
55.
56.
57.
58.
59.
60.
61.
62.
63.
64.
65.
66.
67.
68.
69.
70.
71.
72.
(,, _ 3 ^ - cf.
a:^ + b^ + c' + </2
-\-2ab + 2ac + 2ad
+ 2hc + 2hd
+ 2cd.
+ c + x^ df.
-J- ^ -j_ c 4- xf.
a -\-h -\-c — xy,
-\-b-c-xy.
-y -c- af.
+ 2a + c + yf.
[3 X -^ a -\- y — c-y.
« + 2 z» + 3 c + dy.
a — 2c-\-x — Z yy.
2a — e — Sx -\- yy,
a + cy.
a + 2)1
[a-\-sy.
a — xy.
a - 2y.
a + 1)=.
1 - ay.
a - sy.
a-{-2xy.
3 a - x)\
?ya + 2xy.
'^^a-Sxy.
a^ - ay.
x' + 2 .t)«.
[2 a2 _ ay.
EEVIEW OF FUNDAMENTAL OPERATIONS 13
EXERCISES
Find the following products and expand the results :
1. [(^ + y)+l][(^4-2/)-l].
2. l(x-^a) +3][(x + ^)-3].
3. [(a^-r/)+3][(x-a)-3J
4. [(:r + 4)4-c][(^ + 4)-4
5. [(2 a-b) + c^[(2a-b)- c].
6. lx-\-(b + c)-][x-(h + c)l
7. [x +(b - c)Xx -(b - c)^.
8. l3+(x-^J)X^-(x-^J)y
9. [10 -(a- 5)] [10 +(a- 5)].
10. [4.x + (2y-x)X^x-(2y-x)'].
11. From each corner of a square piece of tin of side a inches
a square of side b inches is cut. By turning up the sides an
open box is formed. Show that a^ — 4:b^ is the area of the
inside of the box in square inches.
12. Express the area a^ — 4:b^ of Exercise 11 as the product
of two binomials.
13. Using the results of Exercise 12, find by a short method
the area of the inside of the box if a = 12 and 6 = 3; if a = 95
and b = 5.
14. The dimensions of a rectangular box are d, d + 3, and
6^ + 6. Express (a) the sum of the edges, (b) the total outer
surface, and (c) the volume of the box.
15. Assume each dimension of the box of Exercise 14 equal
to n inches and solve as before.
16. If two equal boxes of dimensions n, n — 4:, and n + 5
are placed end to end, find (a) the sum of the outer edges,
(b) the outer surface, and (c) the combined volume.
17. The area of a circle is given by the formula tt)^, in
which TT = ^- and r equals the radius of the circle. What is
14 SECOND COURSE IN ALGEBRA
the area of a circular ring left by cutting a circle of radius r
from the center of a circle of radius R ?
18. The surface of a cylinder is 2 7rr(h -]-r), in which r
is the radius of the base and h is the height of the cylinder.
Find the amount of tin in 1250 cylindrical cans which have a
circular base of radius 2 inches and a height of 5 inches.
19. How much less tin is used in making a cylindrical can
of height h inches and radius r inches than in making one of
height h -\- 2 inches and radius r + 2 inches ?
20. Would more tin be used in constructing a cylindrical
can of height h and radius r -^ 2 or of height A + 2 and
radius r ? How much more ?
21. Find the total surface of a cylinder the radius of whose
base is r + 8 and whose height is r — 5.
22. The formula for the volume of a cylinder is Trr^/i, in
which r equals the radius of the circular base and h the height.
Find the number of gallons contained in the 1250 cans of
Exercise 18, given that 1 gallon contains 231 cubic inches.
23. Find the volume of a cylinder the radius -of whose base
is r — 5 and whose height is r -\- 5.
24. The formula for the surface of a sphere is 4 tt/^, in
which r is the radius of the sphere. Compare the sum of the
surfaces of 1728 balls of radius one-half inch with the surface
of a ball of radius 9 inches.
25. Compare the surface of a sphere of radius 3 n with the
surface of a sphere of radius n.
26. The formula for the volume of a sphere is |- tt;**. How
many spherical balls of radius 2 inches can be made from a
spherical ball of lead of radius 12 inches ?
27. Write the formula for the volume of a sphere whose
radius is r — 6 inches.
28. Compare the volume of two spheres of radii r and 2r
respectively.
CHAPTER II
LINEAR EQUATIONS IN ONE UNKNOWN
9. Definition of an equation. An equation is a statement
of the equaKty between two equal numbers or number
symbols.
Thus a (a — 2) = a^ — 2 a and a: + 5 = 7 are equations.
Equations are of two kinds — identities and equations of
condition.
10. Identities. An arithmetical or an algebraic identity-
is an equation in which either the two members are alike,
term for term, or become so if indicated operations be
performed.
Thus 15 - 8 = 6 + 1 and (a + Z>) (a - &) = a^ _ j2 ^re identities,
for in each, if the indicated operations be performed, the two mem-
bers become precisely alike.
An identity involving letters is true for any set of
numerical values of the letters in it.
Thus the identity a(h — c) = ah — ac becomes 2 (9 — 5) = 18 — 10,
or 8 = 8, when, for example, a = 2, & = 9, and c = 5.
11. Equation of condition. An equation which is true
only for certain values of a letter in it, or for certain sets
of related values of two or more of its letters, is an
equation of condition, or simply an equation.
Thus 2 a; + 5 = 17 is true for a: = 6 only; and a: + 2 ?/ = 10 is true
for a: = 8 and y = 1 and for many other pairs of values for x and y,
but it is not true for x = ^ and y = 2 and for many other (though not
for all) pairs of values for x and y.
15
16 SECOND COUKSE IN ALGEBRA
J.2. Satisfying an equation. A number or literal expres-
sion which, being substituted for the unknown letter in
an equation, changes it to an identity, is said to satisfy the
equation.
Thus X = Q a satisfies the equation 2x + 5 a = 17 a, for on sub-
stituting 6 a for X we have 2-Qa + 5a = 17a, or 17 a = 17 a, which
is an identity.
After the substitution is made it is usually necessary to
simplify each member before the identity becomes apparent.
13. Root of an equation. A root of an equation is any
number or number symbol which satisfies the equation.
Thus 8 is the root of the equation 3 x + 2 = 26, for it satisfies the
equation.
14. Axioms. An axiom is a statement the truth of which
is accepted without proof. Some of the axioms most fre-
quently used are
Axiom I. If the same number is added to each member of
an equation^ the result is an equation.
Axiom II. If the same number is subtracted from each
member of an equation, the residt is an equation.
Axiom III. If each member of an equation is multiplied by
the same number, the result is an equation.
Axiom IV. If each member of an equation is divided by
the same number (iiot zero), the result is an equation.
Each of the foregoing axioms is used in the solution of the
EXAMPLE
Solve^-i = x + 7.
Solution. -=z X + 7.
(1)
Multiplying (1) by 2,
ox-l = 2x-\-U.
(Ax. ITT)
(2)
LINEAR EQUATIONS IN ONE UNKNOWN 17
Adding 1 to each member of (2), (Ax. I)
5x = 2x + 15. (3)
Subtracting 2x from each member of (3), (Ax. II)
3 a; = 15. (4)
Dividing (4) by 3, (Ax. IV)
X = 5. (5)
Check. Substituting 5 for x in (1), we have
2^5 _ i := 5 _,_ 7^ or 12 = 12.
Since substituting 5. for x satisfies (1), 5 is the root of (1).
15. Transposition. In the solution of the foregoing sec-
tion by the appKcation of Axiom I to (2), the term — 1 is
omitted from the first member and -f- 1 is combined with the
second member. Again, by applying Axiom II to (3), the
term + 2 :?:; is omitted from the second member and — 2 2^ is
combined with the first member.
It thus appears that a term may he omitted from one
member of an equation., provided the same term with its
sign changed from -{- to — or from — to -\- is written 4n or
combined with the other member. This process is called
transposition.
Hereafter, in order to simplify an equation, instead of
subtracting a number from each member or adding a
number to each member, as illustrated in the foregoing
example, the student should use transposition, since it is
usually more rapid and convenient. He should, however,
always remember that the transposition of a term is really
the subtraction of that term from each member of the equatio7i.
16. Equivalent equations. Two or more equations in one
unknown, even if of very different form, are equivalent if
all are satisfied by every value of the unknown which
satisfies any one of them.
18 SECOND COURSE IN ALGEBRA
Equations (2), (3), (4), and (5) of section 14 are each
equivalent to equation (1) and to each other, for all are
satisfied by the same value of the unknown.
Of the four axioms or assumptions of section 14 we shall
make constant use. If the '' same number " referred to in
each is expressed arithmetically, the result is always an
equation equivalent to the original one. Further, if identical
expressions involving the unknown be added to or subtracted
from each member of an equation, the resulting equation
is equivalent to the first. If, however, both members of
an equation be multiplied by identical expressions con-
taining the unknown, the resulting equation mag not be
equivalent to the original one.
Multiplying each member of the equation x — 2 = 3 by a; — 1, we
get x"^ — Q X -\- 2 = 'd X — '3, or x^ — Q X + = 0. Now this last equa-
tion has the roots 1 and 5, whereas the given equation has the root 5
only. Here the root 1 was introduced by multiplying the given
equation by a: — 1. Results obtained from the use of Axiom III with
multipliers which contain an unknown should always be carefully
checked. When a root is obtained which does not satisfy the original
equation, this root should be rejected.
The use of Axiom IV when the divisor contains the
unknown may result in the loss of a root which the process
of checking will not discover. If an equation is divided
hy a factor containing an unknown^ this factor should be set
equal to zero. The root thus obtained is a root of the
given equation.
For example, if each member of a:^ — 4 = 3 a; -I- 6 is divided by
a: + 2, the result is a; — 2 = 3, whence x- = 5. But a; = — 2 satisfies
a;2 — 4 = 3 a; + 6. This root was lost by dividing by a; -H 2.
With these and with certain other rare exceptions which
will be noted later, the application of the axioms will
produce an equation equivalent to the given one.
LINEAR EQUATIONS IN ONE UNKNOWN 19
For solving equations in one unknown which do not
involve fractions we have the
Rule. Free the equation of any parentheses it may contain.
Transpose and solve for the unknown involved.
Reject all values for the unknown which do not satisfy the
original equation.
Checking the solution of an equation is often called testing
or verifying the result. For this we have the
Rule. Substitute the value of the unknown obtained from
the solution in place of the letter which represents the unknown
in the original equation. Then simplify each member of the
resulting identity until the two members are seen to be identical.
If the correct substitution of the root for the unknown
does not transform the equation into an identity, an error
has been made in the solution.
EXERCISES
Solve, and check the results as directed by the teacher:
1. hx-\-\ = 2x-\-l. 3. l + £c + 5£c + 17 = 0.
2. 6cc + 10 = 10x4-2. 4. 15cc = 3(4£c-5).
5. 2(£c + 2)-(a; + 5)=0.
6. 3(2£c-l)-(5a^-l)=0.
7. 3(2a;-7)-2(5-2^)-f-l = 0.
8. 3(£c4-2)-5<2x-3)-:0.
9. 6(cc + 4)-4(a; + 2)- 0.
10. 4(6j^4-2)-3(7 25 + 3)=0.
11. 6(4z^-5)-ll(2^-3)=0.
12. 6(2a^-l)+4(3-4a;)=0.
13. 4(4ic-l)+3-2(34-^)=0.
14. 5 7i-9(2 7i-f 4)-2(7z,-9)=0.
15. 4(x-2)+3(2-x)-3* = 6(ir + l). •
20 SECOND COUESE IN ALGEBRA
16. Sn - 5(4: - 71)= 5 - S(l-\- n).
17. (x-\-l)(x-2)=x^-[-3.
18. (x -{-5)(x + l) = (x- 3) (x-2)-^ 10.
19. (4cc-3)(2x-5)-(4a;-7)(2x-l)=0.
20. (x + 2)2 -f- 48 = (x - 4)2.
21. (a^ + 3)2 + 40=(cc + 5)2.
22. (2x-\-Sy = 4.(l-xy.
23. (x + 4)2 - (2 - £c)2 = 84.
24. (x-^ 3) (6 £c 4- 5) - (2 a: + 4) (3 £c - 8) = 38.
25. (1 - x) (x + 2) + (x + 3) (x + 4) = 0.
26. (2/_4)(6-2/) + (2/ + 2)(2/-4)=0.
27. (x + 4)(x4-3) = (x + 2)(x + l)+42.
28. (2v-S)(3v + 2)-(4.-6v)(l-v)=0.
29. (5x-3)(4-6x) + (3x + 4)(10x-21)-9 = 0.
30. 3(x + 2)(x-4)+5(cc-l)(x + 3) =
4(2x-l)(.T-2)+l.
31. X — 2a = 4:a — X.
32. c — ic = X — c.
33. 8s — x = X — 4r.
34. ax — 2 ah = A: ah — ax.
35. 2t ax — 5 ac = 2f ac — ax.
36. 3c(ic- 2a)=2c(a-ic).
37. m (x — 5 tt) — 3 m (^i — ic) = 0.
38. x(h-\-a)= ah-{-a^. •
39. x(a — g)= a
2 6'2
4c2.
41. 2ax-4a2-f 4a = l4-a^.
42. ax-a2 + 5a = e-j-3x.
LINEAE EQUATIONS IN ONE UNKNOWN 21
43. ax + 1 — a^ = x.
44. 2mx-\-5m-3 = 3x-\-2m^.
45. (x -j- a)(x -{- b)= x" -\- 2a^ -^ Sab,
, 46. (a — c){x — m) = (m — c) (x — a).
47. J ax - Slab = l^a? + lbh^ - bhx.
48. a^x — a^-\-Sax = S — 10a-{-x.
49. c(l + a:)+m(ic + l)-ic(m4-c + l)=0.
50. m (m — 2 x) + 2 am = a (2 £c — a).
17. Solution of problems. In the solution of problems lead-
ing to simple equations the following steps are necessary :
/. Read the problem carefully aiid find the facts which will
later be expressed by the equation.
II. Represent the unknown number by a letter and express
any other unknown involved in terms of this letter.
III. Express the conditions stated in the problem as an
equation involving this letter.
IV. Solve the equation.
V. Check by substituting in the problem the value found
for the unknown.
In the preceding sentence the words " in the problem "
are of importance, for substituting the value found m the
equation would not detect any errors made in translating
the words of the problem into the equation.
PROBLEMS
1. The sum of two consecutive numbers is 975. Find the
numbers.
2. One number is five times another, and their sum is 102.
What are the numbers ?
3. The sum of two numbers is 54. Twice one of them
equals ten times the other. What are the numbers ?
22 SECOND COURSE IN ALGEBRA
4. The sum of three consecutive even numbers is 1044
Find the numbers.
5. The product of two consecutive even numbers is 1416
less than the product of the next two consecutive even num-
bers. Find the numbers.
6. Of four consecutive numbers the product of the sec-
ond and fourth exceeds the product of the first and third by
201. Find the numbers.
7. One pupil is four years older than another. Eight years
ago the first was twice as old as the second. Find their ages now.
8. Two men are 48 and 18 years of age respectively. How
many years hence will the older be twice as old as the younger ?
9. One man is three times as old as another. Fifteen years
ago the first was six times as old as the second. Find their
ages now.
10. A's age is double B's. Twelve years ago B's age was
one fourth of A's. How old is each ?
11. A is twice as old as B, and C is three times as old as D.
B is 8 years older than D. In ten years the sum of their ages
will be 113. How old is each ?
12. If each side of a square is increased 7 feet, its area will
be increased 329 square feet. Find the side of the square.
13. A certain rectangle is 8 feet longer than it is broad. If
it were 2 feet shorter and 5 feet broader, its area would be
60 square feet greater. What are its length and breadth?
14. A certain, rectangular plot of ground is 15 yards longer
than it is wide. If it were 20 yards shorter, it would have to
be 40 yards wider in order to have the same area. What are
its dimensions ?
15. A certain square plot is surrounded by a border 6 feet
wide. The area of this border is 816 square feet. What is the
side of the square ?
LINEAR EQUATIONS IN ONE UNKNOWN 23
16. A certain picture is 4 inches longer than it is wide, and
the frame is 2 inches wide. The area of the framed picture
is 192 square inches greater than that of the picture alone.
What are the dimensions of the picture ?
17. Can a rectangle of perimeter 112 inches be drawn which
has a length 5 inches greater than twice the width? If so,
give its dimensions.
18. A sum of |15.75 consists of dollars, quarters, and dimes.
If there are 6 more quarters than dollars, and twice as many
dimes as quarters, find the number of coins of each kind.
19. A certain sum consisting of quarters, dimes, and pennies
amounts to $8.62. The number of dimes equals twice the num-
ber of quarters, while the number of dimes and quarters
together is 2 greater than the number of pennies. Find the
number of coins of each kind.
20. A collection of 109 coins is made up of quarters, dimes,
and nickels. There are 7 fewer dimes than quarters, and 3
less than five times as many nickels as dimes. Find the
amount of the collection.
21. The sum of the digits of a certain two-digit number
is 14. If the order of the digits is reversed, the number is
decreased by 36. Find the number.
22. Change the word "decreased" to "increased" in Prob-
lem 21 and solve.
23. The digits of a certain three-digit number are consecutive
odd numbers. If the sum of the digits is 15, find the number.
Facts from Geometry.* The area of a circle is the square of
the radius multiplied by 7r(7r = ^Y^- approximately). This is
expressed by the formula A — irR^.
The circumference of a circle equals the diameter times ir.
The usual formula is C = 2 ttR.
24. If the radius of a given circle is increased 14 inches, the
area is increased 1232 square inches. Find the first radius.
24 .SECOND COURSE IN ALGEBRA
25. By adding 2 inches to the radius of a circle whose
radius is 8 inches, how much is the circumference increased?
the area ?
26. Substitute R inches for 8 inches in Problem 25 and
solve. Interpret your results.
27. Imagine that a circular hoop 1 foot longer than the cir-
cumference of the earth is placed about the earth so that it
is everywhere equidistant from the equator and lies in its plane.
How far from the equator will the hoop be ?
28. Compare the result of Problem 21 with the one obtained
when a similar process is carried out with a sphere 18 inches
in diameter, instead of with the earth.
29. If the height of a square is increased 4 feet and its
length is increased twice that amount, the area of the figure
will be increased 248 square feet. Find the side of the square
and the area of the rectangle.
30. The rope attached to the top of a flagpole is 5 feet
longer than the pole. The lower end of the rope just reaches
the ground when taken to a point 25 feet from the base of the
pole. Find the height of the pole.
31. The length of a given rectangle is three times its width.
A second rectangle is 9 inches shorter and 1 inch wider than
the first, and has a perimeter one half as great. Find the
dimensions of each rectangle.
CHAPTER III
FACTORING
18. Definition of factoring. Factoring is the process of
finding the two or more algebraic expressions whose
product is equal to a given expression.
In multiplication we have two factors given and are required to
find their product. In division we have the product and one factor
given and are required to find the other factor. In factoring, how-
ever, the problem is a little more diJfficult, for we have only the
product given, and our experience in multiplication and division is
called upon to enable us to determine the factors.
19. Rational expressions. A rational algebraic expression
is one which can be written without the use of indicated
roots of the letters involved.
Thus 2, ox, 3?/— v2, and a^ are rational expressions. In this
chapter factors which involve radicals will not be sought.
20. Integral expressions. If a rational expression can
be written so as not to mvolve an mdicated division in
which an unknown letter occurs in a denominator, it is
said to be integral.
Thus 3, 7 a, -> and 4 a; — 3 are integral expressions. In this chap-
o 1
ter factors which involve fractions will not be sought.
21. Prime factors. An integral expression is prime when
it is the product of no two rational integral expressions
except itself and 1.
«^ 25
26 SECOND COURSE IN ALGEBEA
It must be remembered that to factor an integral expres-
sion means to resolve it into its prime factors.
The methods of this chapter enable one to factor integral rational
expressions in one letter which are not prime, as well as some of
the simpler expressions in two letters. No attempt is made even to
define what is meant by prime factors of expressions which are not
rational and integral.
There is no simple operation the performance of wliich
makes us sure that we have found the prime factors of a
given expression. Only insight and experience enable us
to find prime factors with certainty.
A partial check that may be applied to all the exer-
cises in factoring consists in actually multiplying together
the factors that have been found. If the result is the
original expression, correct factors have been found, though
they may not be prime factors.
22. Polynomials with a common monomial factor. The
type form is ab+ac-ad.
Factoring, ab -\- ac — ad = a(lb + c — tl),
ORAL EXERCISES
Factor :
1. 3a + 6. 10. bax — 2ax^.
2. 5£c + 15. 11. 2c + 4c--2cc?.
3. a^+a. 12. 4.a-10a^-2a\
4. 2c-Qc\ 13. 6ac-3^c4-3c.
5. 9x^-3cc. 14. 10cc+15a;«-5«?/.
6. cd^c^d\ , 15. 14a*-7«« + Ta*-7rt'.
7. ax" - A. 16. 3 c^ - 6 c^ -h 9 c» - 15 c\
8. 4fx-8c2. 17. 6rV-3r8«2 + 3/'V-3rV.
9. 14 A - 21 A^A;. 18. 10 xY - 2 a-/ -f 2 a^/ - 2 xy.
FACTORING 27
23. Polynomials which may be factored by grouping
terms and taking out a common binomial factor. The
type form is ax+ay+bx^hy.
Factoring,
ax -{- ay -{- hx -\- hy = (ax + ay') + (hx + hy)
= a(x-\-y')-]-b(x-^y)
= (_x + y)(a-hh').
EXERCISES
Separate into polynomial factors :
1. 2(a-\-b)-{-x(a-['b). 5. a(c - d) -b(c - d).
2. S(b-^5)-{-a(b-j-5). 6. 2(x - y)- x{x - y).
3. 5x(a — cy-^y(a — c). 7. h(m +Sn) — 2k(m -{-3n).
^.2a(x-2y)-\-b(x-2y). 8. 2r(5x-4:2j)-9s(5x-4.tjy
9. -2a(Sh-k)—b(Sh-k).
10. x(r — s)-\- y(s — r).
Hint. Write in the form x{r — s)— y{r — s).
11. 2a(c-Sd)-\-b'(3d-c).
. 12. 5r(Sm-2n)-2s(2n-Sm).
13. ac -{- ad -}- be -\- bd.
14. ac -\- 2 ex -\- 3 ay -{- ^ xy.
15. mx — my -\- nx — ny.
16. cd-3cf-\-2d-^f.
17. «./z,r -|- aki' — ahs — (7A;s.
18. r^s + 2rs- 37^t-^ rt.
19. 4 /^m + 8 Ati - 6 A^m - 12 A;/i.
20. 2 a^ — 2 ax — ac + ca?.
21. x^-^xy^-^x^y + 'dy^.
22. aa? + ay -\- bx -{- by -\- ex -\- cy.
23. mr — 2 r + ms — 2 s + m^ — 2 ^.
28 SECOND COURSE IN ALGEBRA
24. Trinomials which are perfect squares. The type
^^^^ ^^ a^±2ab+lf.
Factoring, a^± 2ab -{-b'^=(a ± by.
ORAL EXERCISES
Separate into binomial factors :
1. a^-\-2ax-^x\ 7. r^ - 10 r*- + 25 si
2. x^-2xt/-\-f. 8. 9 + 6a-\-a\
3. m^ - 2 7nn + n^ 9. 16 - 8 ao; + aV.
4. a;2_|_4^_^4 IQ, 9(r2-12x?/4-4/.
5. 4_4a;H-ic2 11^ 1 4- 10 a^> + 25 a^^A
6. a^-4: ah + 4 Z-'l 12. {a + 5)^.- 2(a + 6) + 1.
13. (r-.s)2-6(r-.9)+9.
14. 4 (a + 2)2 - 12 c {a + 2) + 9 <?.
15. 9 + 6(a + x) + (a4-ic)l
16. 16-8(a-2a:)4-(^-2ic)'l
17. (^ + ^»)2 + 4 (^ -h />) (c 4- 2) + 4 (6' + 2)1
18." {a + Z;)2 - 6(a + ^) (c - ^) 4- 9(c - 6/)l
19. a2n_i2a^^ + 36.
20. x?"" - 14 x^/* + 49 ?/^
25. A binomial the difference of two squares. The type
form is cfi — W
Factoring, cfi—l)^={a-\- b) (a — b).
More generally,
«2^2aft4-^2_^4-2cc?-c?2
= a^+2ah + b^-((^-2cd+ d^)
= (a-hby-(c-dY
= (a-}-b-\'C-d){a-^b-c-\-d).
FACTORING
29
ORAL EXERCISES
Factor :
1. a^-o'^
10. 25^2 -36 5V.
18.
x^-1.
2. m^ - 'n?.
11. 36a2^»2_49c2c^l
19.
x'-l. '
3. a^-4.
12. a^-25 6^
20.
x'' - 81.
4. x" - 9.
13. xy-9.
21.
16 - a\
5. 16 -xl
14. xy-64^^
22.
625 - x\
6. a^-16h\
15. 81a2_l00^»V.
23.
81 - c\
7. l-25cl
16. a* -16.
24.
^2m yin
8. 9a;y-l.
Hint. Find three factors.
25.
^2m _ j2n^
9. 160^2-25
2/^.
17. a^'-h'.
26.
C''* - d""^.
EXERCISES
Factor :
1. a'-x\ 7. a?-{b + cf.
2. x'if-z\ 8. 4r'^-(r-s)*.
3. (tt-2f-c2. 9. 9m2- 4(71 + 3)1
4. 4 (a^ + 3)^ - 2/1 10. (a - c)^ - (cZ + ef.
b. 16{x-yf-z\ II. a (/• + 2 s)^ - e^ (a^ - 2/)^
6. 9(x-^yf-16z\ 12. x2(2^-A:)2-cc2(77^-27^)*.
13. 4 a{a - cc)^ - 9 a(c - 2 d)\
14. a2cc(cR-2/)2-Z'2£c(3c-^)2.
15. m2 + 2m7i + 7i2-(£c2-2x?/ + ^).
16. x'-2xij + 2f
2 ah- y.
17. cc^ + 6a3 + 9 - ^2 4. 2 ^/^ - z^.
18. a^ - 22 a 4- 121 - &2 + 20 &c - 100 cl
19. 2-8£c2 + 8.x^-22/'-8?/^- 8.-;l
20. «2^<2 + 2a& -f 1 - c^ + lOccZ - 25 (Z'^.
21. ^c^-a^-2ab-l}'.
22. 50 6^%2 _ 8 ^,2 _^ 8 Z>c - 2 c^.
80 SECOND COURSE IN ALGEBEA
23. 49 x'- 49 2/2_ 14 ^_ 1. 28. r" + rs - {i^ - s^.
24. 121 ic»- 1-18 1/^-81 2/^. 29. nv" - n- - m - n.
25. (^2 - ^2 - (a - 6). ' 30. m-{-n- m^ + nl
*26. cc^ - / 4- («^a; + a^/). 31. x^-4:f-\-x-2 y.
27. r'-rs-(7^-s'). 32. 7^ - r - S s - 9 s\
26. The quadratic trinomial. The type form is
x^ + bx+c.
Since (x -{- h) {x -\- k^ = x'^ + (li + ¥)x-\- hk,
it follows that a^ -\-hx-{- c can be factored into the bino-
mial factors (x -\-}i)(x-\- k^ if two numbers h and k can
be found which have the sum h and the product c. The
method of determining these factors is illustrated in the
EXAMPLE
Factor x^ -2x-15.
Solution. Here — 15 = 1 • — 15 or — 1 • 15 or + 5 • — 3 or + 3 • — 5.
Of these pairs of factors of — 15 only the pair — 5 and + 3 give the
sum — 2.
Hence z^ - 2x -15 = (x - d)(x + 3).
For factoring expressions of the type x^ -{-hx + e we have
the
Rule. Find two numbers whose algebraic product is + c and
whose algebraic sum is + b.
Write for the factors two binomials both of which have x
for their first terms and these numbers for the second terms.
EXERCISES
Separate into binomial factors :
1. x^-h5x-\- 6. 4. a'' + 8 ^ + 12.
7. d^-Sd-10.
2. x' + 7x + 12. 5. d"" + 11 r/ + 18.
8. ^2_2^_35.
3. x' + 7a; + 6. 6. a^ -\- 10 a -\- 25
9. 0^-4:0-12.
. FACTOEING 31
10. r" -r- 90. 18. a" -^ 9 a - 10.
11. m^ -3m- 18. 19. (a -{- bf - 2 (a -{- h) - 8.
12. a^2_2ic-24. 20. (x-yy + 4.(x-y)-{-3.
13. 9-lOic + a^l 21. a^n _|_ 12 ct^ + 35.
14. 7-2 - 4rs + 35^. 22. c^^ - 7 c« - 18.
15. m^ - 7 mn + 10 Til 23. m'"' - 11 m^ - 12.
16. 1 - 5x + 6x\ 24. a^^ - 5 a^^ - 6.
17. 1 _|_ 2 Tfc - 24 Til 25. yy -2h^y - 35.
27. The general quadratic trinomial. The type form is
For many trinomials of this type two binomial factors
of -the form Qix + It) (rnx + n) may be found. The method
of factormg such trinomials is illustrated in the
EXAMPLE
Factor 2 x^ + 7 a:; — 15. ? a; + ? (\\
Solution. 2a:2+7x-15 = (?a: + ?)(?a: + ?). ?a: + ? (2)
To find the proper factors we must supply ^x ■{■ , x
such numbers for the interrogation points in + ?a: —15
(1) and (2) as will give 2 2'^ + 7 a: - 15 (3)
2 3^ for the. product of the first two terms of the binomials,
— 15 for the product of the last two terms of the binomials,
-\- 1 X for the sum of the cross products.
Now 2x^ = 2x-x, (4)
and - 15 = - 1 • + 15 ; 1 • - 15 ; + 3 . - 5 ; - 3 • + 5. (5)
The factors of 2 and — 15 from (4) and (5) may be substituted
for the interrogation points in (1) and (2) to form the following
pairs of binomials, each having a product containing the first and
last terms of the trinomial :
2x-l 2x+15 2i;+l 2x-15 2a: + 3 2ar-5 2x-Z 2a: + 5
a:+15 x—1 a; — 15 x-\-l x~5 x + 3 x + 6 x — d
32 SECOND COUESE IN ALGEBRA
By trial we find that only the seventh pair has +7 x for the sum
of its cross products, which gives the middle term of the trinomial.
Therefore 2 a;^ + 7 a; - 15 = (2 a; - 3) (a; + 5).
After a little practice it will usually be found unnecessary to
write down all of the pairs of binomials that do not produce the
required product.
If none of the pairs gives the required product, the
given trinomial is prime.
If an expression of the form ax^ + bx+c is not prime,
it can be factored by applying the
Rule. Find two binomials, such that
I. The product of the first terms is ax^ ;
//. The product of the last terms is + c\ *
///. The sum of the cross products is + bx.
EXERCISES
Factor :
1. 2oc^ + ^x + 2. 4. 4^2 4- 7a + 3. 7. 4;r" + 8.T + 3.
2. 2a;2 + 7a^ + 6. 5. 3 .t^ + 13 .r + 12. 8. 6 ^-^ + 7 r -}- 2.
3. 2a2 + 9a + 10. 6. ^x' + llx + W. 9. 2b^-5h^2.
10. ?>x?-^x-\-^. 21. 9r7.2 + 3a-2.
ir. 6(-2^7c4-2. 22. 12 7'2 + 10r-12.
12. 3 .r^ - 11 a: + 6. 23. 10 /-^^ - 19 r.s- - 15 s\
13. 3.x' - 11 .T 4- 8. 24. iSa' -lla'^h-ld^bK
14. 4 .r' - 13 a; + 10. 25. 6 x» + 10 xhj - 4 xf.
15.10x^-29x4-10. • 26. 2x2_5^y_3y2
16. 12x''-llxy + 2f. 21.-2x^^Bxy -12y\
17. 2a'^ + ^a-2. 28. ^a^-2a%-% ah\
18. S7^ + r-2. 29. 6 a^"" - 7 a"" + 2.
19. 2a^-a-15. 30. 3 a1»» - 10 a" - 8.
20. 8 .s-2 -6s -9. 31. 10 a»" - a,'' - 3.
FACTORING 33
28. Expressions reducible to the difference of two squares.
The type form is ^4 _|_ ^^^2^^ ^ ^^
If k has siach a value that the trinomial is not a perfect
square, a trinomial of this type can often be written as the
difference of two squares. Thus, if ^ = 1, the addition and
subtraction of aW accomplishes this result.
EXAMPLES
1. Factor a" + (1%'' -\-h\
Solution. «4 + 0.%"- + &* = a^ + 2 a^U^ + h^ - a^l)^
= (a2 + V^y - (aby
= (a2 + &2 4. ab) (a2 + &2 _ ab).
2. Factor 49 h"" + 34 h^k^ + 25 k\
Solution. If 36 A^p is added, the expression becomes a perfect
trinomial square. Adding and subtracting 36 h^ki^, we have
49 h^ + 34 AU-2 + 25 ^4 = 49 ¥ + 70 /j^F + 25 yl'* - 36 ^2^^2
= (7h'' + Dky -(6hky
= (7A2 + 5 ^'2 4. 6 7<^-) (7A2 + 5 ^2 _ g J^J.y
EXERCISES
Factor :
1. x^ 4- xy -f ?/. 11. 25 X* - 19 £c2 4- 9.
2. c* + c2(^2 + d\ 12. 9 ax« - 28 axY + 4 r/7/l
3. a' -\- a:'b' -\- b\ 13. 4 A + 3 c^^^^x + 9 ^>«ic.
4. a'' + 3a%^-i-4.b\ 14. 4a* + l.
5. m* + m^ + 1. Hint. 4a* + 1 =
6. x^^5x'^9. 4a* + 4a2 + l_4a2.
7. c*H- 4c2 + J6. 15. c^ + 4i7^
8. 25a*-19a2 + i. le. 64aV + icl
9. 25cc*-lla^2 4-l. 17. 4a*^ + ^^^
10. 47'*-44r^52_j_49^.4 ^^ x""^ -\- 4.y^'\
34 SECOND COUESE IN ALGEBKA
29. A binomial the sum or the difference of two cubes.
The type form is c^zkW
a^ + h^ divided hj a-\-h gives the quotient a?' — ah -{- b%
and a^ — h^ divided hj a— b gives the quotient a? -\- ah-\- b\
Therefore a^ + 5^ = (a + 5) (^2 - ab -^ b^), (1)
and a^-b^ = (a-b}(a^-i-ab-hl^'). (2)
Formulas (1) and (2) above may be apphed as in the
EXAMPLES
1. Factor (^^ + 27.
Solution, a^ + 27 = a^ + S^ = (a + 3) (a- - a • 8 + B^) '
= (a + 3) (a2 - 3 a + 0).
Solution. 8 - 2:8 =
: 23 - a:8 = (2 - a-) (2^
f 2x + a:2)
= (2 - a;)
(4 + 2x4-2:2).
EXERCISES
Factor :
1. x' + 7,A
9. x' - 2^
17. ««+(/>2/.
2. 6t^4-^^'.
10. a^-27.
18. c« + ^«.
3. a^4-2«.
11. m^ - 64.
19. m« - nl
4. c' + 51
12. m8-(2 7i)».
20. (2ay-(3by.
5. c^'^ + S.
13. 8;r«-7/«.
21. 8a:«-27v/«.
6. 6/« + 27.
14. 7^-27s\
22. 125x« + 8//.
7. a-« + 125.
15. 64-a;».-
23. (a -{- by + c^
8. x' - y\
16. 125 -a;«.
24. ^?8 4- />» + or. -h ^>.
^5. x^-y''-\-x
-?/•
28.
,3_8.s.8 4.r-25.
26. m^ — n^ — m -f- w.
29.
^8» ^ /,8n^
27. ic» - 8 / +
^; — 2 ?/.
30.
^8m _ pn^
FACTOEIKG 35
30. The Remainder Theorem. If any rational integral
expression in x be divided by x — n, the remainder is the
same as the original expression with n substituted for x.
This fact is illustrated in the
EXAMPLE
Divide x^ — 5x -\- 6 hj x — n.
• Solution. x^ — 5 X + 6
X + (n — 5^
(n-5)x + 6
(n — o)x ~ n^ -{- 5 n
n^ — 5 n + 6 = Remainder
Here the remainder n^—3n + 6is the same as a:^ — 5 .r + 6,
the given expression, when n is substituted for x.
EXERCISES
1. Divide x^ -\- bx -\- c hj x — n and show that the remain-
der is n^ -{- bn -{- c.
Z. Divide x'^ -{- bx -{- c hj x — a and find the remainder.
3. Divide x^-j- ax^-\- bx + chj x — n and find the, remainder.
4. Ill (x* -{- x'^ — 5 x + S) -i- (x — 2) find the remainder (a) by
division, (b) by the Remainder Theorem.
Solution (b). 2^ + 2^ - 5 • 2 + 3 = 5.
5. In (x^ — X -\- 5) -^ (^x — 3) find the remainder (a) by divi-
sion, (b) by the Remainder Theorem.
By use of the Remainder Theorem find the remainders in
the following :
6. (x^ -\- x^ - 5x -\- S)-ir(x - 3).
. 7. (x^-3x-15)-^(x-^4.).
8. (x^-2x''-100)^(x-5).
9. (x^ - 2x^ -2x- 3)^(x - 3).
10. (x^-2x^-^x- 2)^(x - 2).
36 SECOND COURSE IN ALGEBEA
31 . Factor Theorem. By substituting 2ioTxmx^—5x-\-6
we obtain 4 — 10 + 6, or 0. Hence a; — 2 is an exact divisor
(or factor) ot a^ — 3 x -\- 6. Again, if 3 is substituted for x
m x^ — 5 X -\- 6, the expression equals zero. Hence a; — 3 is
a factor oi x^ — 3 x -\- 6. These examples illustrate the
Theorem. If any rational integral expression in x becomes
zero when a number n is substituted for x^ then x—n is a
factor of the expression.
The Factor Theorem may be used to factor some of the
preceding exercises and, in addition, many others which are
very difficult to factor by previous methods.
Note. By means of the Factor Theorem we are able to solve
cubic and higher equations when the roots are integers. The solu-
tion of the general cubic equation is one of the famous problems of
mathematics and one which is accompanied by many interesting
applications. This problem was first solved by the Italian, Tartaglia,
about 1530, but was published by, Cardan, to whom Tartaglia ex-
plained his solution on the pledge that he would not divulge it. For
many years the credit for the discovery was given to Cardan, and to
this day it is usually called Cardan's Solution.
When searching for the values of x which will make an
expression zero, only integral divisors of the last term of the
expression (arranged according to the descending powers
of x') need be tried, for the last term of the factor must
be an integral divisor of the last term of the expression.
EXAMPLE
Factor cc» + 2cc-3.
Solution. If X — n is a factor oi x^ ■\- 2x — 3, then n must be an
integral divisor of 3. Now the factors of — 3 are 1, — 1, 3, and — 3.
If 1 is put for X, then x^ -I- 2 x — 3 equals zero, hence a: — 1 is a factor
of a:^ + 2 a: — 3. Dividing a,*^ -f- 2 a; — 3 by a; — 1, we obtain the quotient
a;^ + a: + 3. Since a;^ -f- a: + 8 is prime, the factors of x* -H 2 a; — 3
are a: — 1 and ar^ + a: + 3.
FACTORING 37
ORAL EXERCISES
1. Is X — 1 a factor of ic^ + 3 cc — 4 ?
2. Is X - 2 a factor of 2x^-^x' - 20?
3. Is a - 2 a factor of a^ - 3 (^ + 2 ?
4. Is X — 1 a factor of cc^ + 3 x^ — 4 ?
5. Is r 4- 1 a factor of r'^ — 4 r^ — 4 r + 1 ?
6. Is r — 3 a factor of 2 r^ — 7^ + 5 ?
7. Is 5 +1 a factor of 3s^ - 5^^ + 8 ?
8. Is A: - 3 a factor of 2 A:^ - 5 ^b^ - 9 ?
EXERCISES
Factor :
1. x^-{-x-2. 8. y^-^f -^y-{- 9.
2. x^+-2xH-3. ' 9. .T'^-Tx'-ex.
3. tt«4.«2_36. 10. x^ -7x^4- 4x4-12.
4. ^3 ^ .^ _io. 11. 2x^ - 2x2 - X - 6.
5. ^^ + ^"-12. 12. x3-x2-4.
6. x^-2x2-5x + 6. 13. 3x^-2x2 + 2x- 3.
7. ^8_x24-4x-4. 14. a^-Q>a^ +lla'-^a.
32. The sum or difference of two like powers. The type
form is a" zfc 6".
The cases m which a^ ± If' is divisible by a 4- ^ or a — h
can be determined by the Factor Theorem.
Thus in a^ — If^ n being either an odd or an even in-"
teger, substitute h for a. Then a^ — h^ becomes b'^ — h^ = 0.
Therefore a— b is always a factor of a^ — h^.
In a" — 6^, n being even, put — b for a. Then d!^ — b^
becomes b^ — b'^ = 0, since (— 5)^ is positive when n is even.
Therefore when n is even a 4- ^ as well as a — 6 is an exact
divisor of a^ — b^.
38 SECOND COURSE IN ALGEBRA
In a'* + h''\ n being even, put either -\-h or —h for a.
Then a^ + V^ becomes If^ + h^^ which is not zero. Therefore
^i _|_ yti ^g i^eyer divisible hy a-\-b oy a — h when n is even.
In a** + h''\ n being odd, put — b for a. Then a" + 6"
becomes (~ by^ + ^'^ = 0, since (— 5)" is negative when ?i
is 06?(i. Therefore when n is odd a-\-b is a divisor of a^. + 5".
Summing up:
I. a" — b" is always divisible by a — 6.
II. a" — &", when n is e^^e/i, is divisible both by a + &
and by a — &.
III. a" + &" is never divisible by a—b.
IV. a" + &", when n is oc7c?, is divisible by c + 6.
ORAL. EXERCISES
I
For each of the following, state a binomial factor :
1. x'-f.
6. x''-2f. 11. 7?^'^+2l
16. 1-7-1
2. x^-5\
7. o:^ - 2^. 12. a' + 8.
17. l-r»
3. 27 - a\
8. 2^cc^-7/l 13. iK^+7/.
18. 1 + 7-^
4. cc^-2^
9. a'^ - c-". 14. r^ + 2'.
19. l + /«
5. 32^^-/.
10. a«-^A 15. a^+32.
20. 5« + l.
21. 8^-1.
23. Is 10' -f 1 divisible by 11?
22. 10^-1.
24. Is 10^-1 divisible
EXAMPLE
by9?
Factor x^ + y^
Solution. By division,
£!
-i-^ = x*- xh + xhr - xy^ + v^.
Hence a;'^ + y* = (a: + ?y) (.r^ — ar*?/ + a-V — 2:?/' + y^).
Note that the signs of the second factor are alternately plus and
minus. Also note the order in which the exponents occur.
FACTORING
39
EXERCISES
Factor :
1. x^-^z\
8.
a^-32af.
18. 1 - r^.
2. x'' + 1.
9.
(2x)^-2432/^
Hint. Write only the
first five terms and the
3. a^-^2\
10.
a} - x\
last term of the poly-
4. x' + 32.
11.
1 - r^.
nomial factor.
5. (ay + (hy
'.
19. 1 + r^
6. cc^-«^
12.
x^ - 128.
20. x^ — if.
Hint. Find the ,
see-
13.
x^' + y'K
Hint. Factor first
ond factor by division
and observe the signs
14.
o}'^Z2x^K
as the difference of two
squares.
of the terms and
order in which the
the
ex-
15.
x' 4- a\
21. x« - y\
ponents occur.
16.
\^t\
2^2. a}''-}?-\
7. a' - 2K
17.
128:r^-M.
23. a"-"--}?^..
33. General directions for factoring. The following sug-
gestions will prove helpful in factoring :
/. First look for a common monomial factor^ and if there
is one (other than i), separate the expression into its greatest
monomial factor and the corresponding polynomial factor.
II. Then from the form of the polynomial factor determine
with which of the following types it should he classed^ and
use the methods of factoring applicable to that type.
1. ax+ay+bx+by. 5. ax'^ + bx+c.
2. a'±2ab-\-1y', 6. a^ + kaW + b\
3. a'-b'. . 7. a^zkb^.
4. x' + bx+c. 8. fl"±6".
///. 'Proceed again as in II with each polynomial factor
obtained^ until the original expression has been separated into
its prime factors.
IV. If the preceding steps fail, try the Factor Theorem.
40 SECOND COURSE IN ALGEBRA
REVIEW EXERCISES
Factor :
1. 6x^ + 2x^-{-2x\ 29. x^ - Sx^-4.x-^12.
2. 5a^ + 2a^-15a- 6. 30. x - x^ - x^ -^ x\
3. a^-\-4.ab-\-4: h\ 31. (a + xf + 10(a + x)-\- 25.
4. 2cH-%cd\ 32. {x + rf-^{x + r)-l^.
5. 3m«-3m2-18m. 33. x^ - Sx"" - x -[- ^.
6. 2^2 + 3 aa; + al 34. a^ + 27 6i"'.
7. x'-lxhf-^-^y^' 35. 64c^8-|-2^«.
8. a^c - ac^. 36. a^'^ - a^»V + a^^c - a6c».
9. 2x\j-2xy\ 37. (ic + 2/)2- 6(a; 4-2^)^ + 9;5;2.
10. cc^ - 2 cc^ - 9 a;2. 38. (a - xf - 16(m - ti)^
11. ac^2hG-ad-2hd. 39. 2 ic^ - 2 ic^ - 12 cc.
12. 18 r^ - 24 r^s + 8 7-/. 40. 10 a?c - 15 aV - 70 ac».
13. 45xV-20x/. 41. a* -11^2 + 1.
14. 2 A% 4- 4 «2 _ 30 ;j,3 42. 6t^ + aVK
15. 3a2-10a/> + 3^;^. 43. a^ + ct'i^
16. ic^ + 7 .T^ + 16. 44. x^{x - 1)2 - .T(a; - 1).
17. aH-%ad\ 45. 4(a-^')2-12(a-^>)^ + 9<'2.
18. 2 6i^ + 64 dK 46. a^ - 10 aZ» + 25 ^'^ _ c^.
19. m^n-\-mn\ 47. cc2_2ic(«_^,)_35(a-^)l
20. x^ -\-^x- 5. 48. 6^2 - 13a: 4- 6.
21. ax'-^a^Zx^-Vl. 49. ««- 8 ar» + 17 a; - 10.
22. a%'' - 4 a2^»2 _|_ 4 a6. 50. ic* - ^if + a;» - a^V-
23. aV-c2-a2^_i. 51. ^2_i2,r + 36-a*.
24. xy - 13 x\f - 14 a;y. 52. a"" - b^ -\-(a- bf.
25. 27» -b't^s-Srs". 53. c2-2c(Z + rf''-2(c-r/)-85.
54. 2{a-\-by-^^e{a + b)-2c'.
55. r??/* - 7 wV + 71*.
56. a8 + ^<8 + 3a'''^ + 3a^'^.
26.
x'-
-l^x^
+ 36.
27.
a^-
- a2/>8 -
- a»^»'-^
^^'^
28.
x"^
-x\
FACTORING 41
57. ^2x^-xh/\ 69. 9ic'-4i/2-3x-2y.
58. b{a-bf-a + b. 70. x" -b(2x- 5).
59. c2 + 4^2_^2_4^^ rj^ a^_S-7a^ + Ua.
60. (a-2x}c'-\-(2x-a)d:\ 72. 3a2-14a(^>-c)4-8(*-c)l
61. x^-20-\-x\ 73. aV4-4.
62. 6a^-\-Sa^-3 al 74. ccV + x'z^
63. 16a* + 7a^ + l. 75. cc^ - lOx - 3.
64. ccy - 64. 76. x' - xS/ - a-* (cc^ - ij").
6b. x' -lSx^-21x\ 77. a«-5a* + 4.
66. a' - a^Z*^ + ^2^ - a^ 78. ti* - Qa^ - « + 3.
67. a* + 8 fi^^ + a' + 2 «^>. 79. x - 1 + a-^ - x\
68. a2 + 2a + l-Z'' + 2^»c-cl 80. x^ + x - y - f.
%l.,x' + 2xij + f-2^ a^ -lOa-1.
82. a''-b^-2ah{a'-b'').
83. 2 ir^ + 3 x^ + a;. 92. 7^'"^ - 2 A"^A:" + A:^^
84. m^ - 8 m - 7 m*. 93. a^"^ - ^'^^^
85. 3 (^^ - 9 a (2 a - 3). 94. x^'^ + (r + s)^^'' 4- ^'s.
86. 2x^-10x^-\-4:X. 95. /^^-x'*^ + hrx'' + Ascc^ + As.
87. 10 a -7 a" -6 a\ 96. a*"* + a^'^b^'' + &*'\
88. a^-^-l-^-Sa' + Sa. 97. a«"^ + ^'^
89. 12x^-Sx-^x^-6x\ 98. a«"* - b^"".
90. a^x^ + 2 (^»ic + a« 99. a^"* + ^>s^
91. mr — ms — m- + ws. 100. a^^ — ^^^
101. Solve for x, ax -\-bx — 3a = 3b.
Solution. Rewriting, ax + bx = ^ a -^ ^b.^
Factoring in each member, a:(a + &) = 3 (a + 6).
Dividing each member by a + &, a; = 3.
102. Solve for x, ex -\- Ux = (^ -\- ed.
103. Solve for m, wa — ?w^ -f Jc = ai.
42 SECOND COURSE IN ALGEBRA
104. Solve for y, 5 ay — S bi/ — 5 a^ -\- S ab = 0.
105. Solve for z, az — 3 ad = bz — S bd.
" 106. Solve for ic, cic — 2 dx = c^ — 4 c?l
107. Solve for m, am — m -\- 1 — a^ = 0.
108. Solve for y, by -^S dy - b^ + 9d^ = 0.
109. Solve for r, r(2a -7c)= 4.a^ - 28 ac + 49 c^.
110. Solve for m, m(Sa — c)— 9 ad = 2 ec — 6ae — Scd.
111. Solve for a;, <xx — 3 Z»x — a^ = 3 Z»^ — 4 ab.
112. Solve for s, 2 CIS - 7 s + 13 a = 2 «2 -h 21.
113. Solve ioTt,2te-2e-}-15-Se'-{-3t=0.
114. Solve for 2/, 2/ 4- 21 6^2 + 4 c^ = 7c?y + 1.
115. Solve for x, d(l — 3a)-{-x-\-Sac = c-\~S ax.
116. Solve for -z, az -\- ae — 2 ec -\-2cd = 2cz -\- ad.
117. Solve for 2/, a - 3 == a?/ - 3 ?/ - 2(a - 3).
118. Solve for £c,5cc-2cx-a(5-2c) = 5a + 5-2c-2«c.
119. Solve for x, o?x — a^ + x — a^ — 1 = 0.
120. Solve for x, c» - c^x. - 2cdx - Sd^ = 4: d^x.
121. Solve for m, m(a - 2) (a^ + 4) = a* - 16.
122. Solve for s, 16 s - 8 as + 4 a^s - 2 ah + a^s - 32 = a^
34. Solution of equations by factoring. The methods
of factoring enable us to solve many equations in one
unknown. In the solution of equations by factoring use
is made of the
Principle. If the product of two or more factor's is zero^
one of the factors must he zero.
Each of the factors may be zefo, but the vanishing of
one factor is sufficient to make the product zero.
FACTORING 48
EXAMPLE
Solve the equation x^ — x^ = ^ r.
Solution. Transposing 6 x, we have
a,.3 _ ^2 _ g 2^^ ^ 0.
Factoring, x (x + 2) (x - 3) = 0.
Solving the equations resulting from setting each factor separately
equal to zero, we have x = 0, — 2, and +3.
Substituting these values in the original equation, we find that
0, — 2, and + 3 are roots of the equation.
The method of solving the above example is stated in the
Rule. If necessary, rewrite the equation so that the second
member is zero. Then factor the first member, set each factor
which contains an unknown equal to zero, and solve the result-
ing equations.
Check as usual.
In the above example the division of each member oi x^ — x^ = Qx
by x gives x^ — x = 6. This last equation has only the roots — 2 and
3, while the original equation had, in addition, the root 0.
Again, if the equation a;^ — 5 x + 6 = a; — 2 is solved by the rule, the
roots are 2 and 4. If, however, each member of a:^ — 5 a: + 6 = a; — 2
is divided by x — 2, the resulting equation a- — 3 = 1 has the root 4
only. In each of these cases the root would not have been lost if
the factor used as a divisor had been set equal to zero and the equa-
tion thus obtained had been solved. The solution of an equation
is often simplified by the use of this method.
EXERCISES
Solve by factoring :
1. x2-4 = 0. 5. 4x^ = 25x.
2. ar^-5x = 0. 6. y^=^A:y.
3. a;2 - aa; = 0. 7. a;^ + 5 = 6x.
8. 2x' + x = Q.
44 SECOND COUESE IN ALGEBRA
9. m:^-a7n-2a^=0. 25. Sx^ -\-71x^ - 9x' = 0.
10. x^-2ax-Sa^ = 0. 26. (2x-Sy -(5x-\- 6y = 0.
11. f-y + 2 = 2y\ 27. (x-r)2-(x-s)2=0.
12. £c2-9«^2_^_^3^^()_ 28. (£c-3)2-2(a;-3)=8.
13. ic« - ax^ - 12 a^ic = 0. 29. cc« + 50-25x-2«2=o.
14. cc»-a2£c + 2aa;2_2a8=0. 30. 2^/^ - 2^^ _ g y + 8 = 0.
15. x« + £c2- A-ax = 0. 31. x^-ax^-4:a^x-\-4:a'^=0.
16. 4y 4- 19/ -52/^ = 0. 32. cc^ + 8 + 6ar» + 12ir = 0.
17. /_77/-6 = 0. 33. 7/-^Sf-\-Sy-\-l = 0.
18. 7/^-13 7/2 4-36 = 0. 34. acx^ -{-bcx -{- adx +bd=0.
19. x®-5a:«=-4x. 35. .x^ - 7x - 8 = a^ 4- 1.
20. x^-7x^ = -6x. 36. 40^^ _^ cc -, 1 = 2a; - 1.
21. 3r«4-24/-2 4-48r=:0. 37. 3.^2 _ j-j^^ _j_ g ^ 3^ _ 2
22. 2;» + 12^- 6«2_8 = 0. 38. a^x^ - 2 ax - S = ax - S.
23. 3a;^+72ic 4- 33x^ = 0. 39. x^ -2x^ -15x = x^ -5x.
24. 5.s2-f 12s-3s^ = 0. 40. Gx^ + 7a; - 3 = 3a; - 1.
35. The highest common factor. The highest common factor
(H.C.F.) of two or more monomials or polynomials is the
expression of highest degree, with the greatest numerical
coeificient, which is an exact divisor of each.
Thus the H.C.F. of 28 a%^ and 42 aV)'' is 14 a^^ The H.C.F. of
x^ — 4:Z and x^ — 5 a:^ 4 6 a; is a; (a: — 2), or a:^ — 2 x.
EXAMPLE
Find the H. C. F. of 9 a;^ - 36 x^ and 3 a-' - 12 a;« 4- 12 x*.
Solution. Factoring, we have
9 a;4 - 36 a:2 = S^x^ (x + 2) (x - 2),
3 r' - 12 a:6 4 12 a:6 = 3 x^ (x - 2)2.
Therefore the H.C.F. is 8 x'^(x - 2), which equals 3 .r» - 6 x%
FACTORING 45
The method used in the preceding solutions for finding
the H.C.F. of two or more monomials or polynomials is
stated in the
Rule. Separate each expression into its prime factors. Then
find the product of such factors as occur in each expression^
using each prime factor the least number of times it occurs in
any one expression.
If two or more polynomials have no common factor
other than 1, then 1 is their H.C.F., and the polynomials
are said to be prime to each other.
EXERCISES
Find the H.C.F. of the following:
1. 12, 18, 24. 5. 30 c^^, 45 c^^ l^cH\
2. 15, 25, 40. 6. 2Sa%\ A2 ab', 70a%\
3. 24, 60, 72. 7. 66 c% 132 c'x^, 165 cV.
4. 12 a^, 30 a', 36 a\ 8. a^ ^ 2 ab + b% a^ - b\
9. 3 a^ - 3 b'', 9(a - bf, 3 a^ - 3 b\
10. ax^ — 2 axy + ay^^ a^x^ — a'^y'^, 2 ax^ — 2 ay^.
11. 2 aV — 2 A^, 4 am^ — 12 amn + 8 an^, 10 am — 10 an.
12. 25 x^ - 25 x\ 10 x^ - 20 xhj - 30 xtf, 5x'-5 xy\
13. 3 X* - 6 x% 6x^-24t xSf, 12 x"" - 96 xhf.
, 14. 24 a« - 6 a%'', 48 a« + 24 a%, 48 a'' - 48 a% - 36 a%''.
15. 5 x^ - 160 x^, 15 cc^ - 60 x\ 25 x' - 200a;^
16. 18 a* - 2 a%% 12 a« - 8 a'b - 4 c^^^^^ 30 a'^P + 10 a^6l
17. 4 X* - 4 xy, 5 x^- 5 a^y, 8 x^i - 8 xy.
18. 6^5-3a*5 + 2a3^^ a^ - 2 a*6 - a^^^2 + 2 a^^^.
Note. The most famous, and in some respects the most perfect,
treatise on elementary mathematics ever written is Euclid's "Ele-
ments." About one third of the material of the thirteen books treats
46 SECOKD COURSE IN ALGEBRA
topics which to-day would be considered arithmetical in character.
In appearance and language, however, they are all geometrical, for
Euclid represents quantities not by numerals, as we do in arithmetic,
or by letters, as we do in algebra, but by lines. Book VII contains
the earliest statement of a general method for finding the G.C.D. of
two numbers. This method, though never necessary in elementary
mathematical work, is so perfect and beautiful from a scientific point
of view that until recently it remained in elementary treatises on
algebra and arithmetic by force of tradition. It is a great- tribute to
Euclid's genius that he was able to devise so perfect a method for
the process that all the efforts of two thousand years have been
unable to improve it essentially. It 'is of fundamental importance
in advanced portions of algebra.
CHAPTER IV
FRACTIONS
36. Operations on fractions. The change of a fraction to
higher or to lower terms, and the addition and the sub-
traction of fractions in both arithmetic and algebra, depend
on the
Principle. Tlie numerator and the denominator of a fraction
may he multiplied hy the' same expression or divided hy the
same expression without changing the value of the fraction.
Thus
3 3.4 12
4 "4.4 ~ 16'
^18 18 ^ 6 3
^"^ 30 30-^6 5
Similarly,
a a • n an
b b • n bn
, a a ^ n n
and - = = -.
b -^ n b
It should be noted that by the application of this principle a
fraction is changed in form but not in value.
ORAL EXERCISES
Divide both numerator and denominator of
^•if^^'S. 5.gby3.^ 8.^by. + 2.
47
48
SECOND COUESE IN ALGEBRA
Reduce to lower terms :
9
10.
11.
12.
^ ax
~Za'
10 xy
2{x-
5)
^x{x-by
13.
14.
15.
3(a--2)
^2-4*
5a;+10
2£c-h4 *
x(x + ^)'
EXERCISES
Reduce to lowest terms :
2a«&2
19.
10 6^2^,8
15 x]/^
24.5 x^y'
x''-9
2x^-{-6x'
5x^ -\- 5xy
25x^-25xhf'
2a^-{-S d'b + 8 ah^
5 rt.^ - 20 (^
48.T^-6x/
32x*-32icV+8.Ty
3a^»-375
2ic2+10£c + 50'
^,8 + 6 a^^ 4- 9 </^»'^
^a^b + ^ah''
17.
18.
x2-6a' + 9
9.
ro.
11.
12.
13.
14.
15.
16.
4a2
16.
1 Tl
x'
X^
' + 4
-16
a;«4-8
x^
+ 4a; + 4
1 Q
4
-2.^ + J--
iC^ +
2c^ + 2c^"-4cZ^
2c^-^6Acd'
16c*-40c^^+16cV-^'
ac? — 3aa; + 2cd — 6cx
ad — Sax — cd -\- 3 ex
2x^-Sx'' + Sx '
'4a;^ + 6x-40
X —15 a -\- 6 ax — 10
5x*-4:0x
Sx^-96 '
12^^+10^^^ -12 <?.
4:-\-9e(^-12a
Sx*y-\-SxY-\-Sy^
5 x^y -\- 5 xxf + 5 £cy
ic2-f-4«a-4-4a2- 9
32^^^^-.^'
32+16icH-4£c»
2 . -I
^ + 1
21.
(f)'+(i)' •
FRACTIONS 49
37. Changes of sign in a fraction. In the various opera-
tions on fractions three signs must be considered : the sign
of the numerator, the sign of the denominator, and the
sign before the fraction. Since a fraction is an indicated
quotient, the law of signs in division, when considered in
connection with the sign of a fraction, gives the following
identity:
+ a_ — «_ — a _ -\-a ^
+ :^ - "^ i:^ ~ ~ IjTi ~ " - ^ *
From the above we have the
Principle. In a fraction the signs of both numerator and
denominator, or the sign of the numerator and the sign before
the fraction, or the sign of the denominator and the sign be-
fore the fraction, may be changed without alte7'ing the value
of the fraction.
In applying this principle to a polynomial denominator
or numerator, like Sx^ — 3x -{- 2, we change its sign by
changing the signs of every term of the polynomial. If
the polynomial denominator is in factored form, like
(a; — 3) (2 a? — 1) (3 a; 4- 1), its sign is changed by changing
the sign of ang one factor or of an odd number of factors.
ORAL EXERCISES
Read the following fractions in three additional ways, using
the above principle :
1.^. 2.-?-. 3.^- 4.-/^.
X — X I — X \— a
Change the sign of the following indicated products :
5. (£t;-3)(£c + 3). 6. (2x4- 5)(3a^- 5). 7. {x-a){x-b).
Change in two ways the sign of the following :
%. {x ^-V){x -V){x} ^-V). 10. (2x-3)(3a:-4)(4a^-5).
9. {x-a){x-b){c-x). 11. (cc-2)2(£c-l)(.T + 2).
50 SECOND COURSE IN ALGEBRA
12. Find the indicated product in Exercise 8. Then change
the sign of two factors and find the product. Compare the
results.
13. Make a general statement of which the result of Exer-
cise 12 is an illustration.
Change the form of the following fractions so that each will
contain the factor cc — 1 in its denominator :
14. A:- 16. -f^- 18.+. '"
1— X 1— X 2 — X — X^
^^' 1-x'' x(l-xy ^ -.i^2x-x^
By proper changes of sign make the denominators of the
following fractions as nearly alike as possible :
3 x-\-S
20.
21.
(x -2)(x-j- 2) (2 -x)(2 + x)
a h
(a — h){c — a) (b — c) (b — a) (c — a)(b — c)
c
(a — h){a — c) (c — b)
38. Lowest common multiple. The lowest common multiple
(L.C.M.) of two or more rational integral expressions is
the expression of lowest degree, with the least numerical
coefficient, which will exactly contain each.
Thus the L.C.M. of 6, 8, and 12 is 24, and tlie L.*..M. of a-c,
ac^, and 2 abc^ is 2 a^bc^.
EXAMPLE
Eind the L. CM. of a;« - 4 cc, 2 a-^ - 4 x^, and 4 .r* - 8 x^
Solution. Factoring, x^ — 4:x = x(x + 2) (x— 2),
2x^-4:X^ = 2x^(x-2),
4x*- 8x^ = ix'^(x-2).
The L.C.M. iHix\x + 2)(x- 2).
FKACTIONS 51
For such expressions as can be readily factored the
method of finding the L.C.M. is stated in the
Rule. Separate eacJi expression into its prime factors. Then
find the product of all the different prime factors^ using each
factor the greatest number of times it occurs in any one
expression.
ORAL EXERCISES
Find the L.C.M. of the following:
1. 4, 8, 12. 5. x-2, x^ - 4, 2(ic -f-'2).
2. 25, 50, 75. 6. x^ - 8, 2{x - 2), o^^ + 2a; + 4.
3. 4a;, 6£c2, 8:r. 7. ^^2 _ 9^ ^2 _ g^ _l_ 9
4. 2aly', 4:a% 6ab^ S. x - 5, x^ - x - 20.
9. a^-4.,a^-^a-\-4., S(a + 2).
10. a^-2a-\-l,l-a, a.
11. x^ -'125, x^-^5x + 25,5-x.
12. ax(a^ - x^), a^(x^ - a^), x\a -\- x).
EXERCISES
Find the L. C. M. of the following :
1. a^- ab^, a^-2ab-\- b^, a" + ab.
2. ax -\- ay -\- bx -\- by, o? — V^, x^ — y'^.
3. c'-^cd-^ 16 d"", c" - 16 d"", 6^ + 4 cd.
4. ^^ _ 4^5 _ 21^2^ 7^ - 49 s^ r^ - 95^.
5. 2m' -\-mn- 10n% 4^^ - 25n^, m« - Sn\
e. Sr'-Sr\l-2r-^7^,r'-l.
7. 125 - n% n^ - 25/1, 50n-i- IOti^ + 2?i«.
8. a^ - 32, 4 - a^, 5a^-\- 10a, 5a -10.
9. «' - Tec + 6, a;2 _ 3^ 4. 2, ic^ -f 2ic - 3.
10. x^-^4.x^-4.x-16,x^-4:,x'-\-2x-S.
11. a?^ — £c, x^ — x'*, x^ -^ x^ -\- X, x^ -- X.
12. 4 a^ - 20 ^ + 25, 25 - 4 a^, 2 a^ -f- 15 a + 25.
52 SECOND COURSE IN ALGEBRA
39. Equivalent fractions. Two fractions are equivalent
if one can be obtained from the other either by multiply-
ing or by dividing both numerator and denominator by the
same expression.
Two fractions having unlike denominators cannot be
added or subtracted until they have been reduced to re-
spectively equivalent fractions having like denominators.
EXAMPLE
Reduce to respectively equivalent fractions having the lowest
common denominator (L.C.D.):
2 5 . 2x
and
3 X x^ — 4 x^ — 2x
Solution. Rewriting with denominators in factored form, we have
2 5 J 2a.-
and
Then
and
3x (a: + 2)(x-2) x{x-2)
TheL.C.D. is 3 x(x + 2) (a; - 2).
2__ 2 (a: + 2) (a: - 2)
3x~3a:(a; + 2)(a;-2)'
5 ^ 5.3a;
(a: + 2)(a;-2) 3a;(a; + 2)(a;-2)'
2a: _ 2 _ 2 • 3 a: (a; + 2)
a;(a; - 2) ~ X - 2 ~ 3 ar(a; + 2) (a: - 2) '
To change two or more fractions (in their lowest terms)
to respectively equivalent fractions having the L.C.D. we
have the
Rule. Rewrite the fractions with their denominators in
factored form.
Find the L.O.M. of the denorhinators of the fractions.
Multiply the numerator and the denoming,tor of each frac-
tion hy those factors of this L. C. M. which are not found in
the denominator of the fraction.
FRACTIONS 53
EXERCISES
Change to respectively equivalent fractions having the L. C. D. :
25 ^ 2x — 1 —x-\- 5
1. TT-^ 7^^' 5.
3ic 2cc' x2-6£c + 9 -2x + 6
^ 2x 3 ^ cc + l — a?
2. ;;>— ^- 6. -7i z r^J
a;
—
2'
x + 2
3
2c^
a^
—
9'
'a-3
3
a
2 a'
7.
8.
x^ — 5ic + 6 — x^ + 4
2 a ' 5 «.
2a2 + 3a + l'2«^-hl'
2a ^4-3 2^2 + 5
(^-2)2^-2 a + 2 3a + 13a2 + 7a-f2
a 2 a 5a
9.
10.
2^ 3x x'-\-2x
2x4-1 5
a;2-2ic + 4 a;^ + 8 x -^ 2
2x-9 -2
11. -^ ^»
12
^5_32 _a;8 + 2x2 x'-{-2x^ + 4.x^+Sx'-\-16x
ax — hx — ar -\- br x' — i^ o^ — IP"
ax -\- hx -\- ar -\- hv a? — ly^ x? — r^
2x 5x 4 3ic + 2
13. -;; -y- J
x^-S 2-x 4:-x' x^ + 2x' + 4.x
40. Addition and subtraction of fractions. To find the
algebraic sum of two or more fractions in their lowest
terms we proceed as in the
EXAMPLE
Find the algebraic sum of — H — r r-
° a a'- ah
Solution. TheL.C.D. isa%
Rewriting with common denominators, and adding numerators,
^® ^^^® 2^ . ?i! _ 5l^ = 2 «&' + 3 ?>2 - 5 g
64 SECOND COUKSE IN ALGEBRA
Check. Substituting 2 for a and 3 for b, we have
6 9 5 _ 36 + 27-10
2 4 6" 12
53^53
12" 12'
For finding the algebraic sum of two or more fractions
we have the
Rule. Reduce the fractions to respectively equivalent frac-
tions having the lowest common denominator. Write in succes-
sion over the lowest common denominator the numerators of the
equivalent fractions^ inclosing each polynomial numerator vn a
parenthesis preceded hy the sign of the corresponding fraction.
Rewrite the fraction just obtained^ removing the parentheses
in the numerator.
Then combine like terms in the numerator and^ if neces-
sary^ reduce the resulting fraction to its lowest terms.
Check. Set the original expression equal to the final result
and substitute in each member numerical values for the letters
involved. The equation should be an identity.
EXERCISES
Find the algebraic sum of :
2x 3^_2£r . c 2c
510 15 • "c-4c-f3
2c5c^c a 2a 1
^ a-b a-Zb Sa 3a;-l 3
ab a" y x'-2x^-Z x-2
2r8^-\-r r-Qs _ s^ 2g + l 4
r's ~ 3r^ ~2r x^-^ x-Z
5a 2 ,. «-3 a- 3 5
10.
3 3 a-^-4 2-« ' a + 2
FEACTIOKS 55
6^ + 2 c-5 ,, ^ x^
^''- c(c-2) 8-c^ ^-2
13.
^^ _ 1 _ ^m_-2^ 17. a' + ab + P- -^
m — 1 m m^ — 1 a — o
14.2 + -^- . ''■''-f^S-^-
X — 7
20. -i- ^^ ^^^
a; + 2 ' X ^x' + 2x-2
c-^2d 1 • 2
22.
23.
c^^2Gd + d'' d-G c + 2^
a_3 2-a 1
a,2_2o^-3 a2_3^_^2 l-a^
^^* 4-^2 a-2 "^ a + 2
a + 2 a-\-2
25.
a2_7^_,_12 6 + ^2- 5a 6a -a^- 8
2S.*±_«_2(*-^).
— a \a a — 01
28. ,. '.+ ' . + \*'. . , + " + •
(^ — c) (c — a) ac — a?" — be -\- ab (a — b)(b — c)
x^ — (m — ny m^ — {x — nY t? —{x — TYif
(x + ny — Tf? {x -\- KYif — n^ (rti + rtf — x^
i
•1 =
= M-
a
c
ac
b
d
-u'
5.1f =
-i-h'-
= -V-.
n
a
n a
na
h *
56 SECOND COUESE IN ALGEBRA
41. Multiplication of fractions. In algebra, as in arith-
metic, the product of two or more fractions is the prod-
uct of their numerators divided by the product of their
denominators.
Thus
Similarly,
and
In like manner
Integral and mixed expressions are reduced to fractional
form before the multiplication is performed. Factors com-
mon to any numerator and any denominator are canceled
the same number of times from each.
EXAMPLE
Multiply 20«^(l-i)(^^^j^).
Solution. Writing the above in fractional form, we have
20 g^ g^ - 4 6 a
1 ' a2 '5a3-10a2*
Factoring and canceling,
4
1 ^ -^rtJ^XjOr — -^ a
a
To find the product of two or more fractions or mixed
expressions we have the
Rule. If there are integral or mixed expressions, reduce
them to fractional form.
Separate each numerator and each denominator into its
prime factors.
FRACTIONS 5T
Cancel the factors (^factor for factor^ common to any
numerator and any denominator.
Write the product of the factors remaining in the numerator
over the product of the factors remaining in the denominator.
Check, Set the given expression equal to the final result
and substitute in each member numerical values for the let-
ters. Simplify each member. The result should be an identity.
42. Division of fractions. For division of fractions we
have the
Rule. Reduce all integral or mixed expressions to fractional
form.
Then invert the divisor or divisors and proceed as in mul-
tiplication of fractions.
Check as usual.
EXERCISER
Perform the indicated operations :
6 a 10 xip' ^ a^ — 4 4 a
25 xV 4 A 2^2
^ ^ 3(^ 4c ^ 3^2
2. 5c. TT^^-TT-,- 9.
lOc^^ ^d x + 3 ^x^
laV' ^^cU jQ a}-\ ^a? 5a-5
^cd"- Ua% 2a a'-2a-{-l 3(l-a)
^ Sah 21 c^ 20 a%^ ,, (^ , 1\/ a'' \( 1\
4
^ 2a^ _^ 12. A-tV
2 ' 3x'' l^a 13. If H-lf:
6. lOa.I^.^. 14. 1^^ ^^-
21 (T^ "• 15
ax"
4:m\ 6ns 2mV 120 e^d 100 cd
9 7iV 4.m^s 2mn^ *"' 42c^x * 147 c^ic^
RE
58 SECOISTD COUESE IN ALGEBRA
24 a;* _8^ ^ _48^ 22 mV . 52 m'n , 275 n-x
^<^xif' 22x'y '' \2\xY 35^3 ' 125 ar^ ' 39m-^x
_ 15 rs^ 300ri(« 1605^* ,^ 5x^ ^{x-yf 1
14^2^ 98^2^ 28rV x-y 2^^ x{x-y)
x + S a;^-4a;"H-4 , x'-2x
x^~4:' 2x-t-6^ ' x^ + 2x'
x^^2x + 4l ^ a;»-8 . x-^ ~ 2a;
a;'-2x + 4 ' ic2 + 2x * x^ + 8 *
„ 2a;^ + 9a; + 9 ^ x^ - 9 a;'-6x + 9
2a;2-9cc + 9 * 4x^-9x' 2x2- 3«
x^ — y^ . i^^ + i^y + ^z"* ^'^ — ^2/ + 3/^
*a7^ + y^' x + y a; — 2/
a^ + 32 , a« + 2a2 af" -^ 2a^ + Aa"" -^ Sa + 16
a" -32 ' a^-2a' a^-2a^ + 4.a^-Sa + 16'
25.
26,
x^-Qx' + llx-S , a;2-5a; + 6 a;'' + 1
a;*-l ' x^-\-2x-^l' x-\-l'
2x + 5 ^ 25-4a;2 ^ 2:^2 + a; - 15
^3_6^2^_18.^_27 • a;» + 27 ' x-3
29.
30
-©)"-Ki5)'©)«(if.)e)'
-•(— D(^)fei)•
1 4. «2 • \^-^ a« + 1/ V «« + 1 /
33
34
FEACTIONS 59
9
• '^;rri^\Sa'-{-la + 2}v''~l~V
35
— : — 12"
X- 5 +
a? -|- 3
Hint. An expression of this form is called a complex fraction. It is
simply another way of writing
\ x + 3/ \ x + 3/
X — 2 \^ — y / \x — y /
, 3
X-1-- 2 11
X
X + 1 +
38. — b a-\-b a — b
x__ 45_
X
39.
a -{- b a — b
a , b a-b~a+b
b a b
^ + 1 + ^^ 46.'^^ + ^^ « + ^
a
. a 1 — a
a -\- 1 a
41. ^
a 1 — a 48^
-^2
X
a —
- a
T
f 1
1 ^^-
a;"
•s^
2«/
a — i <x + ^
&Hc
a + b a — b 2 b
42. TT^rr—TTT:- 49.
2 a^/>" 4- 2 ^>^ A.__2 "M— -- i_-
^j'.^ — />^ 2 a \ c a c I
b^cll -
' 4a6
a
60 SECOND COUESE IN ALGEBRA
50. If ic = ■; > y = — ; — ) and z = -> find the value
6 + c a -\- c a -{-
n , ,.; when a = x— - and b = x -\- -
a/ -^b^ X X
51. What value has ^ , ,., ; when a = x— - and b —x-V-'i
52. 1 ^-. 53. 1 ?: — 54. 1 +
1 + ^ 1-
1 + ^ 1-| 1 + i
Z 6 a
55. Brouncker (1620-1684) proved that ir (the circumference
of a circle divided by its diameter) is four times the following
fraction :
1
1 +
2 +
2 + ?^
24- '^
2 + etc.
225
(cC) Rewrite the fraction, continuing it to
2 + etc.
(&) Stopping with 2 -f- -^^- , find the difference in value between
four times the value of this fraction and 3.1416, the approxi-
mate value of TT.
Note. William Brouncker, one of the brilliant mathematicians
of his time, was an intimate friend of John Wallis (see " First Course
in Algebra," Revised Edition, p. 48). These two scientists were
among the pioneers in the study of expressions with a countless
number of terms.
The complex fraction in the exercise, if continued indefinitely
according to the law which its form suggests, is called an infinite
continued fraction. Brouncker was the first to study the properties
of such expressions.
FRACTIOKS 61
43. Equations involving fractions. Equations involving
fractions are solved as in the
EXAMPLE
Solve ' 5-« = ^(^. (1)
X X +1
Solution. The L. CM. of the denominators is x(x + 1).
(1) . x(x + 1), 5a:(a; + 1) - 8(x + 1) = 5^2- 5a:. (2)
From (2), 5a;2 + 5a: - 8 a: - 8 = 5a;2 - 5a;, (3)
or . 2a: = 8.
Whence a: = 4.
Check. Substituting 4 for x in (1), 5 - 2 = 3, or 3 = 3.
For solving equations in one unknown which may or may
not ipivolve simple fractions we have the
Rule. Where polynomial denominators occur^ factor them
if possible.
Find the L.Q.M. of the denominators of the fractions and
multiply each fraction and each integral term of the equation
by it, using cancellation wherever possible.
Transpose and solve as usual.
Reject all values for the unknown which do not satisfy the
original equation.
EXERCISES
Solve the following for x and check as directed by the teacher :
1.
2 2 2^-
2.
¥-l = f + 3.
3.
2x , , 7
4.
llx 2
9 3 = "-
3x 16
2 3
25
= x--^
^x 21
6 5
^x-2
10
2(^-4)
3
3
2(0.-2)
2 + 5a;
9
62 SECOND COURSE IN ALGEBRA
^ 2x + l lzL?^_Qi 1^ a: + 5_10
^' ~4 3~~ ~ ^^- ^^=^ ~ 3 '
3 7 4' 14. ^ = — ^
4 a? 16 3 a;
1, 2x-7 3
• 3a; 5x 45 3x-2 3
12. 3(a; + 2)-4(2a;-3) + 2 = 0. a; + 7 "^ 5
17. (a; + 5)(a; + l)-(ir-3)(x-2) = 10.
1ft _A_ . __J___o 99 ^ + l _ x + 4
^** 2a;-5^2a;-l"^' ^'^* a; + 3~a; + 2*
2a;-5 ^ 3a;-14 a;+3 3a; + 4
* 2a;4-7~ 3a;-2* 2aj -f 1 ~ 6a; - 2*
„^ 5a; + 3 5a; + 7 ^^ 17-3a; 13-3a;
20. = • 24. =
2a;-3 2a; 3»-13 3a;-8
«^11 — ^/v «,.^ — 3h 5
21. 7i 5 = 0. 25. — ; = 1
a?6 — a;* a;4-7 x -\- 1
26.
3(1- 8a;) 2(1+ 8a;) _ 1- 34a;
5x 8a; —1 5x
27._1- + ^^±1==1_ ^^
28.
29.
2x+l l-2a; 4a;2-l
c -\- X _ x(c —1)
~ x-\-l " c (a; + 1) '
l+2a; _ 3 a; +1 _ 8 + 3a; -4a;''
2a;-l x+1 ~ 2a;2 + a;-l
30. a^- — h.
a
x^a x-\rh x -- x ^
31. :; — == — a. 33. tH —:r=^Za,
a b ' a —1 a +1
a X ^ <( -\- 2
o« 1 , 1 1 oA ^ a; +16 .
32. - + - = -• 34. — —rr ?r = 4a.
36.
FRACTIONS 63
2a;+l 3x-7 _ 9-Sx-a^
'x-{-32~x~x'^x-6^
X -\- cd X — cd _2 chl — 2 cd'^
c -\~ d c — d c^'— d^
37. f^-|^^ = 0.
Lx — a ox — c
5x-\- A 1.3 a; -.05 30.35 -8 a;
^^* '~~^~ + 4 - 2A •
5 — 23- r 4-2
39. 11.3-^-— ^ = 2.3 -(5- 7a;) + --^.
,' 2x-.3 .4x + 5 275 ^,
41. ;r = — 9.4.
Jx ' X .25 X
PROBLEMS
1. Separate the number 286 into two parts such that the
greater will be 2^ times the less.
2. Separate the number 1010 into three parts such that the
second will be -^- of the first and the third will be -^-^- of the
first.
3. By what number must 352 be divided so as to give a
partial quotient 15 and the remainder 7 ?
4. What number must be subtracted from both terms of
the fraction |-| to give a fraction equivalent to J- ?
5. Separate 133 into two parts such that their quotient is 2^.
6. Separate 96 into two parts such that 56 exceeds two
thirds of the one by as much as the other exceeds 16.
7. A boy is 12 years old and his sister 8 years old. In how
many years will the boy be f as old as his sister ?
8. Two thirds a man's age now equals |- his age 30 yearg
ago What is his age ?
64 SECOND COUKSE IN ALGEBRA
9. The square of a certain number is 4 greater than two
thirds of the product of the next two consecutive numbers.
Find the number.
10. The length of a certain rectangle is 2|- times the width..
If it were 10 yards shorter and Ij yards wider, its area would
be 1260 square feet less. Find the dimensions of the rectangle.
11. A square court has f the area of a rectangular court
whose length is 4 yards greater and whose width is 3 yards
less. Find the dimensions of the square court.
12. A can do a piece of work in 10 days and B in 12 days.
How many days will they both require working together ?
Hint. Let x = number of days required by A and B together. Then
- = fractional part of the piece of work that they can do in 1 day.
13. A can do a piece of work in 10 days and B in 15 days.
After they have worked together 5 days, how many days will
A require to finish the work ?
14. A tank has a supply pipe which fills it in 4 hours and
a waste pipe which empties it in 6 hours. If the tank is empty
and both pipes are opened, how much time must elapse before
the tank is filled ?
15. A tank has a supply pipe which fills it in 4 hours and
two waste pipes which empty it in 6 and 8 hours respectively.
If the tank is full and all three pipes are open, how much
time will be required to empty the tank ?
16. If all three pipes of Problem 15 were outlet pipes, how
long would be required to empty the tank ?
17. If in Problem 15 the supply pipe had been closed after
4 hours, how much more time would have been required to
empty the tank?
18. The diameter of the earth is 3| times that of the moon,
and the difference of the two diameters is 57 GO miles. Find
each diameter in miles.
FEACTIONS 65
19. The diameter of the sun is 3220 miles greater than 109
times the diameter of the earth, and the sum of the two diame-
ters is 874,420 miles. Find each diameter in miles.
20. The diameter of Jupiter is lOj-J times the diameter
of the earth, and the sum of their diameters is 94,320 miles.
Find each diameter in miles.
21. A man who can row^ miles per hour in still water
rows up a stream the rate of whose current is Ij- miles per
hour. After rowing back he finds that the entire journey re-
quired 10 hours. Find the time required for the trip upstream.
22. A man who can row 4 J miles per hour in still water
finds that it requires 6i hours to row upstream a distance
which it requires 2|- hours to row down. Find the rate of the
current.
23. A passenger train whose rate is 40 miles per hour leaves
a certain station 2 hours and 45 minutes after a freight train.
The passenger train overtakes the freight in 5 hours and 15
minutes. Find the rate of the freight train in miles per hour.
24. A man invests a part of $8000 at 5% and the remainder
at 4%. If the yearly interest on the whole investment is $345,
how much was invested at each rate ?
25. A man invests $6800 in two parts : the first part at 5%,
and the second part at 4%. If the average rate of interest is
4f %, find the amount of each investment.
26. Two thousand dollars of Mr. A's income is not taxed.
All of his income over that amount is taxed 2%, and all above
$10,000 is taxed 2% in addition. He pays a tax of $180.
What is his income?
27. How much water must be added to a gallon of alcohol
95% pure so as to make a mixture 10% pure ?
Hint. Let w = the number of gallons of water to be added.
Then nkll = i?-, etc.
1 + w 100
66 SECOND COURSE IN ALGEBRA
28. How many ounces of alloy must be added to 45 ounces
of silver to make a composition 60% silver ?
29. It is desired to mix coffee which sells for twenty-five
cents per pound with coffee which sells for thirty-five cents
per pound so as to obtain a 10-pound mixture which may be
sold at the same profit for thirty-two cents per pound. In what
ratio must the parts be taken f r^m each grade ?
30. Milk of a certain grade is known to test 20% cream.
How much water must be added to 25 gallons of this milk to
make a mixture 18% cream ?
31. Grun metal of a certain grade is composed of 16% tin
and 84% copper. How much tin must be added to 410 pounds*
of this gun metal to make a composition 18% tin ?
Hints. Since the composition is 16% tin, ^W • 410 = the number of
pounds of tin in the first composition.
Let X = the number of pounds of tin to be added.
. Then \- x = the number of pounds of tin in the sec-
ond composition,
and 410 + X = the number of pounds of both metals in
the second composition.
16.410
+ X
Therefore ■' = — , etc.
410 + X 100
32. How many "gallons of alcohol 90% pure must be mixed
with 12 gallons of alcohol 96% pure to make a mixture 93%
pure ?
33. A certain lot of pig iron contains 93% pure iron. How
much pure iron must be melted with 10 tons of pig iron to
make iron 98% pure ?
34. The arms of a lever are 5 feet and 6 feet in length
respectively. Excluding the weight of the beam, what weight
on the shorter arm will balance 70 pounds on the longer ?
Hint. The products of the weights by their respective arms are equal.
FKACTIONS 67
35. The arms of a balanced lever are 7 feet and 12 feet
respectively, the shorter arm carrying a load of 36 pounds.
Find the load on the longer arm.
36. If the load on the longer arm in Problem 35 be reduced
7 pounds, how many feet from the fulcrum must a 12-pound
weight be placed on the longer arm to restore the balance ?
37. A beam 12 feet long supported at each end carries a load
of 2 tons at a point 4 feet from one end. Find the load in tons
on each support.
38. At what time between 3 and 4 o'clock will the hands of
a clock be together ?
Solution. The minute hand moves twelve times as fast as the hour
hand. While the minute hand travels x spaces, the hour hand travels
— spaces. Hence a; — — equals the number of spaces gained by the
minute hand in any given time x.
In the time from 3 o'clock until the hands are together, the minute
hand must gain 15 minute spaces to overtake the hour hand.
X
Therefore x — — = 15.
Whence x = 16 y\.
Hence the hands are together 16 y\ minutes after 3 o'clock.
39. At what time between 9 and 10 o'clock are the hands of
a clock together ?
40. At what times between 3 and 4 o'clock are the hands of
a clock in a straight line ?
41. At what time between 5 and 6 o'clock is the minute hand
10 minute spaces ahead of the hour hand ? 10 minute spaces
behind the hour hand ?
42. At what times between 4 and 5 o'clock are the hands of
a clock at right angles ?
43. If the average height of n boys is x inches, what is the
sum of their heights in yards ?
68 SECOND COURSE IK ALGEBKA
44. If 3 m 4- 5 packages weigh j9 pounds, wliat is the weight
of n of them ?
45. If it takes a man t hours to do a piece of work, what
portion of the work can he do in 1 hour ? What portion of the
work would n men do in 1 hour ? What portion would k men
do in h hours ?
46. If it takes n men h hours to do a piece of work, how
long will it take x men to do it ?
47. A man buys bananas at d cents a dozen and sells them
for h cents each. What does he gain on n dozen ?
48. A man bought m articles for c cents per hundred. He
sold them all for |10. How many dollars did he lose ?
49. Itn yards of ribbon cost c cents, find the cost of x yards.
50. If «/ yards of ribbon cost x cents, how many yards can
be bought for d dollars ?
51. A man buys goods for b dollars and sells them for s
dollars. What is his per cent of gain ?
52. One man can do a piece of work in d days, another can
do the same work in n days. How many days will it take both,
working together ?
53. If it takes h hours to mow m acres, how many days of
10 hours each will it take to mow n acres ?
54. A train goes y yards in t seconds. If this equals m miles
per hour, write an equation involving y, #, and m.
55. If 71 men can do a piece of work in d days, how many
men would it be necessary to hire if the work had to be done
in t days ?
56. A transport plying between two ports is under fire for
/ feet of the way. If she steams k knots per hour, for how
many minutes is she under fire ?
Hint. 1 knot = 6080 feet.
CHAPTER V
LINEAR SYSTEMS
44. Graphical solution of a linear system. The construc-
tion of the graph of a single linear equation in two unknowns
or of a linear system in two unknowns depends on several
assumptions and definitions. It is agreed:
I. To have at right angles to each other two lines,
X^OX^ called the jr-axis, and F'OY, called the y-axis.
II. To have a line of definite length for a unit of distance.
Thus the number 2 will correspond to a distance of twice the
unit, the number 4| to a distance 4^ times the unit, etc.
III. That the distance (measured parallel to the a:-axis)
from the ?/-axis to any point in the paper be the j:-distance
(or abscissa) of the point, and the distance (measured
parallel to the ?/-axis) from the :r-axis to the point be the
y-distance (or ordinate) of the point.
IV. That the a;-distance of a point to the riglit of the
?/-axis be represented by a positive number, and the x-
distance of a point to the left by a negative number ; also
that the ^-distance of a point above the a:-axis be repre-
sented by a positive number, and the y-distance of a point
helow the a:-axis by a negative number. Briefly, distances
measured from the axis to the right or upward are positive-, to
the left or downward are negative.
V. That every point in the surface of the paper corre-
sponds to a pair of numbers, one or both of which may be
positive, negative, integral, or fractional.
69
70 SECOND COUKSE IN ALGEBRA
•
VI. That of a given pair of numbers the first be the
measure of the a^-distanee and the second the measure of
the ^-distance.
Thus the point (2, 3) is the point whose x-distance is 2 and whose
^-distance is 3.
The point of intersection of the axes is called the origin.
The values of the a;-distance and the ^-distance are often
called the coordinates of the point.
The relation between an equation and its graph may be
stated as follows :
The equation of a line is satisfied hy the values of the
x-distances and the y-distance of any point on that line.
Any pointy the values of whose x-distance and whose y-
distance satisfy the equation, is on the graph of the equation.
The graph of a linear equation in two unknowns is a
straight line. Therefore it is necessary in constructing
the graph of such an equation to locate only two points
whose coordinates satisfy the equation and then to draw
through the two points a straight line. It is usually most
convenient to locate the two points where the line cuts
the axes. If these two points are very close together,
however, the direction of the line will not be accurately
determined. This error can be avoided by selectmg two
points at a greater distance apart.
The graphical solution of a linear system in two unknowns
consists in plotting the two equations to the same scale
and on the same axes and obtaining from the graph the
values of x and y at the point of intersection of the lines.
Through the graphical study of equations we unite the
two subjects of geometry and algebra, which have hitherto
seemed quite separate, and learn to interpret problems of
the one in the language of tke other.
LINEAR SYSTEMS
Tl
EXAMPLE
Solve graphically the system
3^ _ 4?/ + 20 = 0,
Solution. Substituting zero tor x in S x — 4: y + 20 = 0, we obtain
y = b. Substituting zero for y, we obtain a:=— 6|. This may be
expressed in tabular form :
li x =
-6|
then y =
5
Similarly, for the equation
2x + 2^4-6 = we obtain
the following table :
If x =
-3
then y =
-6
Then, constructing the
graph of each equation as
indicated in the adjacent
figure, we obtain for the
coordinates of the point of
intersection of the two lines
a; = — 4 and y = 2.
Y
y
5 ^^
V ^y
5 -..^4
V ^^ *
\,yr
\T\
X ^ 3 X
/'-« -♦ A'' " '
j^ 3r 2
4 =^
IS -4
^
A;
^
3
r
r
Solve graphically :
2. aj - 2/ = 6,
5x + 4t/ = — 15.
3 aj 4-2^4-11 = 0,
' y-x = l.
EXERCISES
iK + 2y=4,
6y4-2x = 20.
^ + 4 = 10x,
2 - a; = 0.
2x4-42/ = 20,
22/-2 = x.
72 SECOND COURSE IN ALGEBRA
x + y = ^, x-y = b,
* 2/ + 2 = 0. 3a^-f-2t/ = 6.
0^ + 5 = -32/, 3a^-42/ = 33,
• 62/ + 2£c-ll = 0. 4« + 32/=-6.
«H-?/ = 4, 2x-4ty = %
' x + 2y = 7. x-2y=S.
13. In each of the first three exercises will the values of the
X- and ^/-distances of the point of intersection of the two lines,
as obtained from the graph, satisfy the equation obtained by-
adding the two given equations ?
14. Graph the equation x — 2 y = S. Then multiply both
members by 3 and graph the resulting equation. Compare the
two graphs. Th^n use — 2 as a multiplier and graph the result-
ing equation. Compare the three graphs. What conclusion
seems warranted?
15. What are the coordinates of the origin ?
16. Is a graphical solution of a linear system ever impossible?
Give an example.
17. What is the form of the equation of a line parallel to
the £c-axis ? the ^/-axis ?
18. The boiling point of water on a centigrade thermometer
is marked 100°, and on a Fahrenheit 212°. The freezing point
on the centigrade is zero and on the Fahrenheit 32°. Con-
sequently a degree on one is not equal to a degree on the other,
nor does a temperature of 60° Fahrenheit mean 60° centigrade.
Show that the correct relation is expressed by the equation
C = f (F — 32), where C represents the number of degrees
centigrade and F the number of degrees Fahrenheit.
19. Construct a graph of the equation in Exercise 18, using
C and F as abscissa and ordinate. Can you, by means of this
graph, express a centigrade reading in degrees Fahrenheit
and vice versa?
LINEAR SYSTEMS 73
20. By means of the graph drawn in Exercise 19 express
the following centigrade readings in Fahrenheit readings and
vice versa: (a)60°C.; (^) 150°F. ; (c)-20°C.; (d)-SO°¥.
21. From the graph determine what reading means the same
temperature on both scales.
45. Elimination. In order to find values of x and 2/
which satisfy the equation
32:+2y=20, (1)
when we know that
^ = 22: + 3, (2)
we may substitute for 7/ in the first equation the value of
y from the second, obtainmg the single equation in x,
Sx-h2(^2x-{-S)=20, or 72^ = 14.
The process by which we have obtained one equation
containing one unknown from the two equations (1) and
(2) each of which contains two unknowns illustrates one
method of elimination.
In general, the process of deriving from a system of n
equations a system of n — 1 equations, containing one
variable less than the original system, is called elimination.
For example, when n = 2, if we have a system of two equations
in two unknowns, the process of elimination leads to one equation
in one unknown. Since we can always solve such an equation, it
appears that we can solve a system of two equations in two unknowns
whenever it is possible to eliminate one of the unknowns. We shall
see that only in certain exceptional cases is elimination impossible.
This is either because more than one unknown is removed when
we try it or because the result of attempted elimination is not an
equation.
Only two methods of solution will be considered — that
involving elimination by substitution and that involving
elimination by addition or subtraction.
74 SECOND COUKSE IN ALGEBEA
46. Solution by substitution. The method of solving a
system of two linear equations by substitution is illustrated
in the
EXAMPLE
Solve the system/^ ~^^,^^^^;, W
^ 18 0- + 11 2/ = 18. (2)
Solution. From (1), a; = 13 ?/ + 81. (3)
Substituting this value for x in (2),
8 (13 2^ + 31) + 11^ = 18.
Simplifying, 104 z/ + 248 + 11 ?/ = 18.
Combining terms, 115 y = — 230.
Whence y =—2.
Substituting — 2 tor y in (3), and solving,
x = 6.
Check. Substituting 5 for x and — 2 for y in (1) and (2),
5-13(-2) = 31, or 31 = 81,
and 8-5 +11 (-2) = 18, or 18=18.
The method of the preceding solution is stated in the
following
Rule, Solve either equation for the value of one unJmown in
terms of the other.
Substitute this value for the unknown which it represents^ in
the equation from which it was not obtained^ and solve the
resulting equation.
In the simplest of the preceding equations which contains
both unknowns, substitute the definite value just founds and
solve, thus obtaining a definite value for the other unknown.
Check. Substitute for each unknown in both original equa-
tions its value as found. If the resulting equations are not
obvious identities, simplify them until they become so.
3x
-22/:
= 18,
42/
+ Sx.
= 0.
8m
-Sn
+ 6/
2
4 m
-1 =
-Sn.
LINEAR SYSTEMS 75
EXERCISES
Solve by substitution :
Sx-Sy = 20, 5.
X — 61/ = 0.
2x + 52/ = 8,
^ 2(x + y)+Stj = 4:, 4.t-2n = lS,
' 5==x-hy- ^- 202^ = 7/1 + 63.
^ 16x + 7 = 152/, 6a; + 38 = 12y,
• 4a:4-52/ = 0. ^' 4:X-Sy = 0.
47. Solution by addition or subtraction. The method of
solving a system of two linear equations by addition or
subtraction is illustrated in the
EXAMPLE
c 1 .1. . fl3x + 3y = U, (1)
Solve the system |^^_^^^^^^; (2)
Solution. First eliminate y, as follows :
(1) • 2, 26 a; + 6 ?/ = 28 (3)
(2). 3, 21a:-6.y:^66 (4)
(3) + (4), 47 a: =94 (5)
(5) ^47, a; = 2. (6)
Substituting 2 for x in (2), we obtain y = — 4.
Check. Substituting 2 for x and — 4 for y in (1) and (2),
26 - 12 = 14, or 14 = 14,
and 14 + 8 = 22, or 22 = 22.
Before starting to eliminate, the system should always
be inspected carefully to determine which unknown can
be removed most conveniently. In this case the elimina-
tion of y is the simpler, because it involves multiplication
•by smaller numbers than does the elimination of x.
76 SECOND COUESE IN ALGEBRA
The method of the preceding solution is stated in the
following
Rule. If necessary^ multiply each member of the first equa-
tion hy a number^ and each member of the second equation by
another number, such that the coefficients of the same unhnoum
in the resulting equations will be numerically equal.
If these coefficients have like signs, subtract one equation
from the other ; if they have unlike signs, add ; then solve the
equation thus obtained.
In the simplest of the preceding equations which contairis
both unknowns, substitute the value just found and solve for
the other u7iknown.
Check. As on page 74.
ORAL EXERCISES
Solve the following systems :
' x-y = 2, ' 2x + y = ^. *2ic-3?/ = 9.
g a^ + 27/ = 3, ^ 2aj+32/ = 8, ^hx-6y = 7,
' X -\- y = 2. ' x — y = 1. ' Ax — Sy = 2.
48. Special cases. The equation x + y = 10 has as roots
any set of two numbers whose sum is 10. li x-\- y = 5 is
taken as the other equation of a system, one can see im-
mediately that tlie two equations have no set of roots in
common, since the sum of two numbers cannot be 10 and
5 at the same time.
A system of equations which has a common set of
roots is called a simultaneous system.
A system of equations which does not have a common
set of roots is called inconsistent or incompatible.
LINEAE SYSTEMS 7T
The attempt to solve an incompatible system results in
getting rid not only of one but of both unknowns and
leads to a statement in the form of an equation which is
false.
Consider x -\- y = 10, (1)
x + y = 5. (2)
(1) - (2), = 5, which is false.
If, on the other hand, the equation x-\-i/ = 10 is taken
for one equation of a system and 2a: + 2?/=20 for the
other, it appears that any set of numbers which satisfies
one equation satisfies also the other, since if the sum of
two numbers is 10, the sum of twice those numbers is 20,
and any one of the countless sets of roots of one equa-
tion is a set of roots of the other. In fact, the second
equation may be obtained from the first by multiplying
each member by 2.
If one equation of a system can be obtained from one
or more of the other equations of the system by applica-
tion of one or more of the axioms, it is called a derived or
dependent equation. If it cannot be so obtained, it is called
independent.
Thus equations (1) and (2) in the example on page 75 are
independent, while equation (5) is derived from them.
An attempt to eliminate one unknown from a system
of two equations in two unknowns which are not inde-
pendent results in gettmg rid not only of both unknowns
but of the constant terms as well, so that only the identity
= remains.
Thus x + y = 10, (1)
2x + 2y = 20. (2)
(1).2, 2a; + 2^ = 20. (8)
(2) -(3), = 0.
78 SECOND COURSE IN ALGEBRA
ORAL EXERCISES
Which of the following systems are incompatible, which are
simultaneous, and which are dependent ?
x + y = 2, ^x-22j=4:,
'x-\-y = l. *5ic — 102/ = 4.
3x + 32/ = 6, ?>x-y = 2,
^' x + y = 2. • 9^^-32/ = 6.
2 a; + 2/ = 4, 3^ -2/ = 2,
"*• 4a^ + 23/ = l. * 9a: + 32/ = 6.
0^ + 2/ = 3, 2t/-^ + 6 = 0,
*• x-2/ = l. • 3a; -67/ -21 = 0.
2a;4-32/ = 4, 2a3 = 72/-5,
4a; + 62/ = 8. * 8a; = 282/ 4- 3.
EXERCISES
Solve the following systems :
2a; , _
11 1
3 +^= ^'
6. "^ ^ 6
2 3_4
5a; -3^/ = 105.
3r 7 s
a; 2/~3*
4 2 12'
1 1
r+8=-2s.
Hint : Solve first for - and -
X y
32/ + 1 ^ + 22
= 3
4 12
>
2r-h4s 38
«-2y = l.
2r-s ~ 3 '
^ ^ -
.5x + .73, = |,
7* S
.8a;-.22/ = 3f.
w -h 2 71 2m —
71
5
2m + Sn-2 4
__ ;::::
— ,
5 10
2
m+^+6 3
w, + 71 m — n
71
4 7
'2*
LINEAR SYSTEMS 79
x-^ y ' 3 -11
' X — ^y _2 5 5
5 '~n' 2x + y = l.
2.^h-\-l-.lk ^ X y 6j/ 2x
k-l(^ + h ' 5 "^2 2 "^ 5
.8A-2.2"^35-5.5A: ' ^^- 4 2
2h-Sk
= 3,
X
1
4-2y 4
n. ^-^^ 5^±^4.2= ^^
^— = 39. ^^-. + 7 17 + . _Q
3 « 5-«
Solve for ic and y :
jg 5a; + 4?/ = 10a + 4,
jc — 2 ay = 0.
7x + 5y = 21c,
2 acc — 3^*^ = 7,
* 5aic + 7Z>y = 3.
Zx y
2-2 = 6,
19. " "
^ + 1_ = 13
2a 3c Sax-6by = 5c.
20.
a Sa
X y
Za a 1
^ y~2'
21.
4a /
a
22.
X y a -\-h
ah ab
a^-Jy"
X — y = ;
^ ah
oo
2 ace ~ 4 6?/ = 7 c,
^0 SECOND COURSE IN ALGEBRA
^^ ax + hy = c, ^^ ax + hy = c,
kax + khy — ck. ' dx -\-fy = ff.
26. Show that if af—hd = 0, the equations in Exercise 25
are inconsistent, unless they form a dependent system.
27. Solve the system of Exercise 17 for a and h in terms
of X and y.
28. Solve the system of Exercise 23 for a and b in terms
of Xy y, and c.
49. Equations in several unknowns. We have already
seen that the equation x-{-y = 10 is satisfied by an un-
limited number of sets of roots, since there is an infinity
of pairs of numbers whose sum is 10.
An equation or a system of equations which is satisfied by
an infinite number of sets of roots is said to be indeterminate.
If a simultaneous system is satisfied by only a limited
number of set.* of roots, it is said to be determinate.
The system x + ?/ = 10, a; — y = 2, is determinate and has the set of
roots (6, 4). The system 2 x + 2 y = 20, x + y = 10, is indeterminate.
When we consider systems of equations in three un-
knowns, the question arises whether two such equations
form a determinate system. For example, the equation
x + y-^z=\0 (1)
is satisfied by an infinite number of sets of roots. If we
consider a system consisting of (1) and
^-(y + ^)=2, (2)
it appears from inspection that the system is satisfied if
x = Q and ?/ -f 2 = 4. But the equation y -\-z = ^ is satis-
fied by an infinite number of sets of roots. Hence equar
tions (1) and (2) form an indeterminate system.
LINEAR SYSTEMS 81
If, however, we adjoin a third equation to the system, as
y-z=% (3)
it becomes determinate, since ?/ + ^ = 4 and y — z=2 are
satisfied only by the set of numbers 3, 1. It is usually
true that three equations in three unknowns form a deter-
minate system.
In general, when the number of unknowns in a system
of linear equations exceeds the number of equations, the
system is indeterminate. If the number of equations
equals the number of unknowns, the system is usually
determinate and simultaneous. If the number of equations
exceeds the number of unknowns, the system is usually
inconsistent. There are many special cases wliich arise in
the study of hnear systems in n unknowns, corresponding
to those mentioned for two unknowns in section 48, but
they become very complicated for larger values of n^ and.
a thorough study of them is quite beyond the scope of
this text.
Note. It is not a little remarkable that the writings of the first
great algebraist, Diophantos of Alexandria (about a.d. 275), are de-
voted almost entirely to the solution of indeterminate equations ;
that is, to finding the sets of related values which satisfy an equa-
tion in two unknowns, or, perhaps, two equations in three unknowns.
We know practically nothing of Diophantos himself, except the
information contained in his epitaph, which reads as follows :
".Diophantos passed one sixth of his life in childhood, one twelfth
in youth, one seventh more as a bachelor; five years after his mar-
riage a son was born who died four years before his father, at half
his father's age." From this statement the reader was supposed
to be able to find at what age Diophantos died. As a mathemati-
cian Diophantos stood a;lone, without any prominent forerunner
or disciple, so far as we know. His solutions of the indetermi-
nate equations were exceedingly skillful, but his methods were so
obscure that his work had comparatively little influence upon later
mathematicians.
82 SECOND COURSE IN ALGEBRA
50. Determinate systems. The method of obtaining the
set of roots of a determinate system in three unknowns is
illustrated in the
EXAMPLE
(x-\-62j-5z = 21, (1)
Solve the system iSx — Sy-\-z=— 5, (2)
[5x-7y-^2z = 4:. (3)
Solution. First eUminate one unknown, say z, between (1) and (2) :
X + 6 3/ - 5 2 = 21. (1)
(2). 5, 15x-^0y + 5z=-2o. (4)
(l) + (4), lQx-Uij=-4:. (5)
Now eUminate z between (2) and (3) :
(2) -2, Qx-lQy + 2z=-10. (6)
5x-7y -\-2z = ^. (3)
(6)-(3), x-9y=-U. (7)
The equations (5) and (7) contain the same two unknowns x and y.
lQx-Uy=-^. (5)
(7) -16, 16 a: -144?/ =-224. (8)
(5) -(8), 110 y = 220.
y = 2.
Substituting in (7), a; = 4.
Substituting both these values in (1),
4 + 12 - 5 ^ = 21.
Whence z= — l.
Check. Substituting 4 for x, 2 for y, and — 1 for 2 in (1), (2),
and (3) respectively,
4 + 6-2-5(-l) = 21, or 21 = 21.
3.4-8.2 + (-l)=-5, or -5=-5.
5.4-7.2 + 2(-l) = 4, or 4 = 4.
LINEAR SYSTEMS 83
For the solution of a simultaneous system of equations
in three unknowns we have the
Rule. From an inspection of the coefficients decide which
unknown is most easily eliminated.
Using any two equations^ eliminate that unknown.
With one of the equations just u^ed and the third equa-
tion again eliminate the same unknown.
The last two operations give two equations in the same two
unknowns. Solve these equations.
Substitute in the simplest of the original equations the two
values found, and solve for the third unknown.
Check. Substitute the values found in the original equations
and simplify results.
. A system of four independent equations in four un-
knowns may be solved as follows :
Use the first and second equation, then the first and
third, and lastly the first and fourth, and eliminate the
same unknown each time. This gives a system of three
equations in the same three unknowns, which can be solved
by the rule given above.
EXERCISES
Solve for x, y, and z and check the results :
x + ^y — ^z = 2, x + y + z = l,
I. 2x — y — z = 1, ^. X -{- y — z = 2,
Sx-{- 5y-7z=-10. x-?/ + ^ = 3.
2x+Sy-i-4.-z=-14:, . x + 2y -\- z = 1,
2.x — y + 3z = 0, 5.2x-^y — z = 0,
5x-\-2y + z = 14:. x-\-2y -{-z = 0.
x+'2y-\-z = -l, 2x-y-^5z=0,
3. 2x-y-}-z=-20, 6. Sx -^7 y -\- z = SS,
— X — y — 5 z = IS. X — 6y — z = 7.
84 SECOND COUESE IN ALGEBEA '
X y z
X 1^ y
x-\-y £_2.
7/ 4- ^ X ^
""4 2 "12'
Sx + 2y=12-3z.
x-6y-\-Sz = ^
4x-32/ = «,
8.
s = £c -4- 2/,
2x = 32/ + l.
2x + 2/ = S + «,
9.
05-2^ = 6,
32/ + 2« = aj.
^_2y = 10,
10.
3^4-4^=-!,
5 X — ^ = 18.
^ = 3^ + 2,
11.
y = .x-7i,
^ = 6i/-l.
4x-22/ = 0,
12
. 6^-82/=-2,
x^z = \\.
18.
A« J[-hy — lz = 2 hk
19. A^i/ — /^ic + ^^ = 2 A;^,
hx — hy + 1^ = 2 hi.
20.
2a; + 2/ + ^ + ^ = ^»
jC _ 1/ _ ;2 + 2 ?^^ = 4,
X + 2 7/ — ^ — 'i^ = 0,
oj —'2/ + 2 ^ — i^; = 1.
i.
^3-- + ^ = t"' ^^•4x-22/ + ^ — =-9
^ + " = ^^' 2x-32/-2. + ^^ =
35c4.2i/ = 6a-2^,
14. x-5« + 62/ = 2a-ll^ x + 2/-« = ^j
6a; -82/ = 12 a + 8^. ^ _|- 2/ - w; = 2,
ic-?/ + « + w; = S)
ax + hy = 0, 2sc-3y-« + i^ = 7.
15. cx-6« = 2 6c,
bx -\- az — cy = b\ ^ ^y -\- z = l,
.3x + .22/ + .4. = 1.9, 23. ^"^"""^"i'
16. .02x = .l-.01//-.02^, x-z-w:^-5,
a; + 2/ + « = 6.
z
— « + !(; = 0.
LINEAR SYSTEMS . 85
PROBLEMS
Express the conditions of the following problems by means
of simultaneous systems, solve, and check the results :
1. The sum of two numbers is 109 and their difference
is 49. Find the numbers.
2. A workman is hired for 30 days. He is paid $3.50 per
day and board, but is credited with 80 cents for each day that
he does not receive board. At the end of the 30 days he re-
ceives $113. How many days did he receive board ?
3. Thirty -nine tons of material are to be moved by motor
trucks and drays. It is found that the work can be done in a
given time either by 10 trucks and 6 drays, or by 8 trucks and
10 drays. What is the capacity of a truck and of a dray ?
4. Two men travel from New York to the same station by
rail. It costs one of them three times as much for excess
baggage as it costs the other. One pays $7.40 in all, the other
$10.60. How much does each pay for his ticket?
5. A man and a boy can do in 18 days a piece of work
which 5 men and 9 boys can do in 3 days. In how many days
can 1 man do the work ? 1 boy ?
6. A and B together can do a piece of work in 5 days. If
they work together 3 days and A can then finish the job alone
in 4 days more, how many days does each require alone ?
7. Two sums are put at interest at 5% and 6% respectively.
The annual income from both together is $100. If the first
sum had yielded 1% more and the second 1% less, the annual
income would have decreased by $2. Find each sum.
8. If ax -{-by = 2 is satisfied by x = 2 and y = S and also
hj X = 6 and y = 5, what values must a and b have ?
9. li 2x-\-b't/ = c is satisfied when x = l and y = — l and
also when x = 5 and y = 4=, what values must b and c have ?
86 SECOND COUESE IN ALGEBRA
10. A sum of $4000 is invested, a part in 5% bonds at 90,
and the remainder in 6% bonds at 110. If the total annual
income is |220, find the sum invested at each rate.
11. The purity of gold is measured in carats, 18 carats
meaning that 18 parts out of 24 are pure gold. A goldsmith
has 20 ounces of pure gold which he wishes to use in making
16-carat and 10-carat alloys. How much pure gold can he use
for each alloy if he makes 39 ounces in all ?
12. A bag weighing 18 ounces contains two sizes of steel
balls — ounce balls and |^-ounce balls. There are 23 balls in all.
Find the number of balls of each size.
13. If the length of a rectangle is decreased by 7 feet and
the breadth is increased by 8 feet, the area is unchanged. If
the length is increased by 14 feet and the breadth is decreased
by 4 feet, the area is also unchanged. Find the dimensions of
the rectangle.
14. A man has |4.50 in dimes and quarters. If he has 36
coins in all, how many has he of each ?
15. A man has $6.00 in quarters, dimes, and nickels. He
has as many quarters as he has dimes, and three times as many
nickels as dimes. How many of each has he ?
16. A is half as old as B. Seven years ago A was one third
as old as B. How old is each now ?
17. A man can walk 4 miles per hour. He reaches a point
20 miles from his starting point in three hours, having been
taken part of the way by a stage traveling 12 miles per hour.
How far did the stage carry him ?
18. A man and his two sons can do a piece of work in
t of a day. The two boys together can do it in 1|- days and
one of them can do it in 1 day less than the other. What
portion of the work does each do when they work together?
LINEAR SYSTEMS 87
19. Two automobiles 25 miles apart travel toward each other
and meet in 1 hour. If they had both traveled in the same
direction, the faster would have overtaken the slower in 5 hours.
Find the rate of each.
20. An aeroplane travels a certain distance in 3 hours. If
the distance had been half again as great, the aeroplane would
have been forced to travel 50 miles per hour faster in order to
cover it in the same time. Find the distance and the speed of
the aeroplane.
21. One angle of a triangle is twice another, and their sum
equals the third. Find the number of degrees in each angle of
the triangle.
22. The sum of three numbers is 108. The sum of one third
the first, one fourth the second, and one sixth the third is 25.
Three times the first added to four times the second and six
times the third is 504. Find the numbers.
. 23. The sum of three numbers is 217. The quotient of the
first by the second is 5, which is also the quotient of the second
by the third. Find the numbers.
24. If the tens' and units' digits of a three-digit number be
interchanged, the resulting number is 27 less than the given
number. If the same interchange is made with the tens' and
hundreds' digits, the resulting number is 180 less than the
given number. The sum of the digits is 14. Find the number.
25. In 1 hour a tank which has three intake pipes is filled
seven-eighths full by all three together. The tank is filled in
1^ hours if the first and second pipes are open, and in 2 hours
and 40 minutes if the second and third pipes are open. Find
the time in hours required by each pipe to fill the tank.
26. The sum of two sides which meet at one of the vertices
of a quadrilateral is 20 feet. The sum of the two which meet
at the next vertex is 27 feet. The sums of the two pairs of
88 SECOND COUESE IN ALGEBRA
opposite, sides are 23 feet and 29 feet respectively. Find
each side. (Two solutions.)
27. Two chairs cost h dollars. The first cost m cents more
than the second. Find the cost of each in cents.
28. Find two numbers whose sum is a and whose difference
is h.
29. Find two numbers whose sum \s> a-\-h and whose differ-
ence \s, a — h. '
30. Two relays of messengers carry a message k miles. The
first relay travels c miles further than the second. How far
does each go?
31. A man has a dollars and b cents in dimes and quarters.
If he has c coins in all, how many of each kind has he ? ^
32. A man has a dollars in quarters and nickels, with b
more quarters than nickels. How many of each has he ?
33. A and B together can do a piece of work in m days.
B works G times as fast as A. How many days does each
require to do the work alone ?
34. A man rows m miles downstream in t hours and returns
in a houi's. Find his rate in still water and the rate of the
river.
35. Two contestants run over a 440-yard course. The first
wins by 4 seconds when given a start of 200 feet. They finish
together when the first is given a handicap of 40 yards. Find
the rate of each in feet per second.
36. A train leaves M two hours late and runs from M to P at
60 % more than its usual rate, arriving on time. If it had run
from M to P at 25 miles per hour, it would have been 48 min-
utes late. Find the usual rate and the distance from M to P.
37. A train leaves M 30 minutes late. It then runs to N at
a rate 20% greater than its usual rate, arriving 6 minutes late.
Had it run 15 miles of the distance from ilf to iV at the usual
LINEAE SYSTEMS 89
rate and the rest of the trip at the increased rate, it would
have been 12 minutes late. Find the distance from M to N
and the usual rate of the train.
38. It is desired to have a 10-gallon mixture of 45% alcohol.
Two mixtures, one of 95% alcohol and another of 15% alcohol,
are to be used. How many gallons of each will be required to
make the desired mixture ?
Hint. Let x and y = the number of gallons of 95% and 15% alcohol
respectively. Then —. '- — - = — , and x 4- y = 10.
^ 10 100 ^
39. A chemist has the same acid in two strengths. Eight
liters of one mixed with 12 liters of the other gives a mixture
84% pure, and 3 liters of the first mixed with 2 liters of the
second gives a mixture 86% pure. Find the per cent of purity
of each acid.
40. When weighed in water the crown of Hiero of Syracuse,
which was part gold and part silver, and which weighed 20
pounds in air, lost 1^ pounds. How much gold and how much
silver did it contain?
Hint. When weighed in water 19^ pounds of gold and 10^ pounds
of silver each lose 1 pound.
41. Find the positive integers which satisfy the equation
5x + 22/ = 42.
4.2 — 9 y
Solution. X = -•
5
42 — 2 y
If X is to be a positive integer, -^ — - must be integral ; that
o
is, 42 — 2 y must be a positive integral multiple of 5. Hence y can
only have the values 1, 6, 11, and 16. The corresponding values of
X are 8, 6, 4, and 2.
The various related sets of integTal values of the unknowns which
satisfy an equation may be effectively represented to the eye by the
graph of the equation. Since the equation 5 re + 2 y = 42 has the
integral sets of roots (8, 1), (6, 6), etc., the line of which this is
the equation passes through the points whose coordinates are these
RE
90 SECOND COURSE IN ALGEBEA
sets of integers. If the line does not enter the first quadrant, we
can see at a glance that the corresponding equation has no positive
integral sets of roots.
42. Solve in positive integers 7x -|- 2y = 36, and illustrate
the result graphically.
43. In hov^ many ways can a debt of ^73 be paid with 5-dollar
and 2-dollar bills ? Illustrate the result graphically.
44. A man buys calves at |6 each and pigs at |4 each,
spending |72. How many of each did he* buy?
45. In liow many ways can |1.75 be paid in quarters and
nickels ?
46. A farmer sells some calves at |6 each, pigs at $3 each,
and lambs at |4 each, receiving for all |126. In how many
ways, could he haye sold 32 animals at these prices for the
same sum ? Determine the number of animals in the various
groups.
Hint. Eorm two equations in three unknowns. Then eliminate one
of the unknowns.
47. In how many ways can a sum of $2.40 be made up with
nickels, dimes, and quarters, on the condition that the number
of nickels used shall equal the number of quarters and dimes
together ? Determine the various groups.
CHAPTER VI-
EXPONENTS
51. Fundamental laws of exponents. The four laws of
exponents used in the preceding chapters are:
I. Law of Multiplication,
If a and b are positive integers, we have
x'^ = X ' X ' X ' X ' ' • to a factors,
and x^ = X- ' X ' X • ' ' to h factors.
Hence x^ * x^ =(x • x * x - - -to a factors) x (^x - x > x - - »
to b factors)
= X'X'X'-' to a-^b factors
_ ^a+6 \yj ^i^Q definition of an exponent.
II. Law of Division,
Xa-^ X^ = X°-^.
Again, if a and b are positive integers, we have
. x^ X' X ' X • ' ' to a factors
X^^X^=—: — = —T
x'^ X' X ' X ' • ' to factors
If b is less than a, the b factors of* the denominator may
be canceled with b factors of the numerator, leaving a — b
factors in the numerator.
x^
Hence — = x^~^.
x^
• 91
92 SECOND COUESE IN ALGEBRA
If h is greater than a,
III. Law of Involution, or raising to a power,
As before, if a and h are positive integers, we have
(x^y = x^ ' x^ ' x^ ' ' 'to b factors
_ ^a+a + a+ ' ' ' . . . to 6 terms of the exponent
IV. Law of Evolution, or extraction of roots,
Law I may be stated more completely thus :
x"" 'X^ 'X' " . = 2:^+^ + «+--'.
Law
Ill includes the more general forms
(a)
Qx^yhy — ^ac^ftc^
(^)
((xf'yy ' . . = x^^''-'.
When Laws I, II, and III were used in previous work
in multiplication and division, we always assumed that a
and b were positive integers and, in Law II, that a was
greater than b. In the work on radicals (see "First Course
in Algebra," Revised Edition, pp. 251-252) the meaning
of an exponent was extended so as to include fractional
exponents, as defined by Law IV. Though Laws I-IV
have thus far been restricted to positive integers and frac-
tions, they hold, nevertheless, for any rational values of a
and b. This fact will be assumed without proof. We shall
now explain the meaning which, according to these laws,
must be given to a zero or to a negative exponent.
EXPONENTS 93
52. Meaning of zero as an exponent. From Law II,
But a^^a^ = - = l.
Therefore ;c** = 1.
More generally, 2f^ -h sf' = af-'^ = x^,
and, as before, aP =1.
That is, any number (except zero) whose exponent is zero is equal
to 1. Hence 4^ = (f )^ = (— 6)^, for each equals 1. Again, if x is not
zero, (5 xy = 1, and if a; is not 1, (x^ — 2 x + 1)*^ = 1.
53. Meaning of a negative exponent. From Law II,
a'
d>
Obviously, a^ _j_ ^5 _
Therefore ar'^ is another way of writing — •
Then 4-3^1_ ^ .'
43 64
Also a" t z= — z= — ^=^ .
1
In general terms, x~^ = —
Consequently = — = Jf°.
Similarly, we obtain the more general results
hx~ " = — and = hxf^.
94 SECOND COURSE IN ALGEBRA
Therefore, Any factor of the numerator of a fraction
may he omitted from the numerator and written as a factor
of the denominator, and vice versa, if the sign of the exponent
of the factor he changed.
It follows that any expression involving negative exponents may
be written as an expression involving only positive exponents. That
is to say, negative exponents are not a mathematical necessity but
merely a convenience. The extension of the laws of exponents
which brings with it the zero exponent and the negative exponent
is an illustration of what is called the Law of Permanence of Form.
It is to be understood that the part of the rule for
muItipHcation (p. 5) and of the rule for division (p. 7),
which determines the exponents in the product or m the
quotient, applies to all numbers, whether positive or nega-
tive, integral, fractional, or literal. Hence those rules need
not be restated here.
ORAL EXERCISES
Fractional, Negative, and Zero Exponents
Simplify :
1.
xKcci
10. x^-^x-\
19.
{a^-a^).aK
2.
cc^^l^
11. x^ -x-^.
20.
{al^a^j^l)a\.
3.
x^.xl
12. x^ 'X-^.
21.
^8a-l ^ x'^-'2a
4.
x^.xi.
13. x:'-^x-\
22.
x« - x\
5.
X" -T- X^.
14. x^-i-x-\
23.
x2«-i.xi
6.
x' . x\
15. {bxy-nx'.
24.
Vx 'X.
x^.x.
16. lx''x\
25.
^X -h x\
7.
8.
x" . x°.
17. x^ 'X^.
26.
vs.^i.
9.
aj« 'X.
18. ai . a^ ' a^.
27.
Vi*-- Vi.
EXPONENTS 95
28. ^x . Vx\ 33. {x'^y . x\ 37^ ^2 _^ J_
29. V^. V^. 34. (x^a-x-''-.
X
1
30. a^^-^-x^-^. 35. a^^^^- 38. ^ ^ : ^_,
x~^
31. x2«-^--cc^-2«. ^ • 39. ici-^*-T-x-2«.
32. (x2)3 . x\ ^^' ^^" ^ ^ ' 40. ax-^ -^ a-^x\
Kead the following with positive exponents and simplify the
results :
41.
42.
2a-\
49.
4c«
xy-^
54.
10- ^a
bc^
43.
44.
Sab--\
lxhj-\
50.
y-'
55.
45.
46.
x~'^y~^z.
51.
5-\aby
10- •■^^»'^
56.
4-V-V-
47.
3
52.
■3a^b-^c
57.
2 s-'
48.
4:X
53.
Vlxhj-'
2yx-^
58.
2e-: ^
Arrange terms so that the exponents of one letter occur in
the descending order :
59. a^-\-a-'-ea-{-Sa''.
60. a^-]-a~^^~-a^ -i-a + Sa'.
61. a + a^ + 2 + a-^ i-a-\
62. a^ + a~^ + a~^ -\- a -j- S.
63. a^ + -^ + « + - + 6^'.
67. Arrange the polynomials in Exercises 17 and 21 on
page 97 in descending order with respect to the letter a, and
Exercises 22 (p. 97) and 28 (p. 98) with respect to x.
64.
a^ a
65.
••4.+^&
66.
--S-?-¥-^
96 SECOND COURSE IN ALGEBRA
EXERCISES IN MULTIPLICATION
Perform the indicated multiplication and simplify :
1. {x + x^-^x-^)2xi. 13. (e^ + e-y.
2. \a^x-[-ax^ — a^x ^)ax^. ^ ^
3. (x^ — a^)x^a^. ^ /I 1\
4. (x^-5ax + 6a^)Jx-k ^^' ^^'" +.^"H^ " ^)*
5. (x^ + y^jx^yK ^ ^ ^
/I IX / 1 tx 18. (3a-^ + 2a*)'.
19. (x-i-3x-2ir-2)2.
7. (cc^ + y'^) (cc^ — 2/^). " / 1 ^ 1 ^ 3\2
^ ^^ -^ ^ 20. U-^4-2a;^-3x^).
8. (a-2 + 3)(a-2-5).
9. (.--3.«)(.^-2.). ^1- (-* + -^+l)U^--^+l)-
10. (a-i - a)l ^ o ,
11. (6i«-2«-y. 23. {x^-(^){x^-\-a^-J^aS).
12. (a- 1 _ 2 a + 3 a- 2)2. 24. (a^ - a^cc + 4 ic^) (a^ + 2 ic).
25. (x^ + 2 ?/^) (x3- - 2 x^?/^ + 4 2/^).
26. (a + a^^>^ + ^)U + ^^ - f^^^'^).
27. (a* - <i*^»* + h^) {h^ + «^ + o^^^'^).
28. {x 4- xV^ + y~^)^y~^ +.^ - ^^2/"^)-
29. (^/^^-5Va)(>/^^-5V^).
Hint. This is best written (ai — 6 a^) (a^ — 6 a^).
30. (5 V^ + -v^^ - a -v/^'f.
31. W-^^n^-v;^).
5.
EXPONENTS 97
EXERCISES m DIVISION
Perform the indicated division and simplify :
1. x*^x\ 6. (x«- 2x2^-1 + 3 ic^«-2)-f-x2«-i.
2.x^^x^. 7. (6 67.3+*^-9a^™-2 + 12a2-n^)_^3^»-2^
3, x^ ^ x^. 8. (x-y)-^ {x^ — y^).
4. axi-i-Jx^. 9. (x + y)^{x^ + yi).
ax-a^x\ 10. (a^-8 2/)-(aj*-2 2/^).
a^a:^ 11. (16a^^-A096f)-i-{2x^-^Syi),
12. (at + Z,)-^(V^-^^).
13. (a^ + c»6-i + ^*-2)^(^^ _ Jf^-h + ^-i).
14. (e2^4-e-2^ + 2)--(e-^ + e^).
16. (6 + e-^^ + e^o; _ 4 ^-20; _^ 4 e^^^)-- (e^ - e"^).
17. (a + 2 Z^ + 2 ah^ + a*^»*) - («^* + 2 5^).
18. (m' - 7m-2 + m"* + 7 m^ + 8) - (5 - 7/^-^ + m^).
19. (9a;^*^-^-.T3«-2_^2cc2«-i-)_j_^2ic'^-i + 3ic2"-2).
20. (I6 £c - 8 y~^ -\-x~^y-2x~ V) ^ (^~ ^y~^ -S xhj- 1).
21. (40 ah - 2^a-h^ - 16 aH^) - (- 4 a'^V^ + 5 a^h-^)
22. (2:r-2« - 28.T-« + 33 + lla^^^ + 38a:« + cc^^)
-^(4 + a;«-2cc-«).
Divide :
23. a^ - ah^ + a^^ - h^ by a^ - 5^
24. 3£c-i° + x«-4a^-«by 2ir-2 + a^2_^3a^-«
98 SECOND COUESE IN ALGEBRA
25. x^ — y^ by \_{x^ — y^)-i-{x^^ + 2/^)].
26. 9m + 4m-i-13 by 3w^-5 + 2m~i
27. ^2a ^ 4^-2a _ 29 by a!« _ 2cc-« - 5.
28. 9x2« + 25£c-*« - 19x-« by 5a:-2« + Sic*^ - Ix
29 (^ + ?I^
?9- ' 8 ^ 64
^ [/^^ -f ^"j (64 a; - 96 x^y^ + 144 y)l.
Note. To us, who use the notation of exponents every day, it
seems so simple and natural a method of expressing the product of
several factors that it is difficult to understand why such a long
time was necessary to develop it. But here, as in many other in-
stances, it required a great man to discover what to us seems the
most obvious relation. The man who brought the notation of
exponents to its modern form was John Wallis (1616-1703), an
Englishman.
Though the idea of using negative and fractional exponents had
occurred to writers before Wallis, it was he who showed their natu-
ralness, and who introduced them permanently. He also was the
first to use the ordinary sign oo to denote infinity.
MISCELLANEOUS EXERCISES ON EXPONENTS
Find the numerical value of the following :
1. 3-^
10. 1-
18. 16"^.
2. 4-^
3°
19. 8-1
3. 2-* • 3^
20. 16-1
4. 2--. 3-".
12. 5.2°-
.(5 . 2)^
21. 25 H
5. 7 . T'' . 0.
,o 4-^3-
-2
22. (- 64)-*.
6. a)-^
13. ,_.
~*
23. (- 32)i
7. (|)-^.4o.
14. 32-1
24. (32)-*.
8. (i)-^ •(¥)-'•
15. 0^ . h\
25. (-125)- i
2
16. 4"i
26. ■v^27-^
*-3--
17. 8-^.
27. -^8-^.
EXPONENTS
99
ay-
22 23
28. {^/^sy.
30. (I)" I ^^^'^- ^r-7. — ::r-.= . ' . ^ etc.
31. (.04)1
32. (.027)" i
33. (.064)- 1
34. (.00032)1
Write with positive exponents and simplify the results
36.
2-1
2-
-2 2-3
TTttx
rT.
2-1
2-^-2-3
37.
3-
-2 2-2
3-
■1-2-1
38.
2-
■1 + 3-1
2-
-3_^3-8
QO
3^
-3 9-3
40.
Hint.
44.
2
5-2 J. _ 1_
-2
46.
etc.
41.
42.
43.
+ ^'-
45.
a-%-
7.-4
-hb-
47.
a-^-b-'
a-^ + b-
-1
a
a-'~ - b-
-2
Bse"
s-- + e-
-2
18 '~'
+ ^»-l
*®- a-3
+ Z»-3
-27-1
-3-^
"V^rite without a denominator and simplify the results :
50.
51.
52.
53.
2xy
b' '
4 6-c-^
2-is-'
54.
55.
56.
57.
12 a%^
4.xf '
Ix-hf
2-y *
c{x-yf
58
59. ^(^-?/)-^
^6'iC (iC — y)
42m-«?i2"*
60.
61.
r~V(s — r)*
100 SECOND COURSE IN ALGEBRA
Simplify :
62. (28/. 71. (x-^yK 80. (a8)2^ • (a2)8^.
63. (28)-2. 72. (Sx-'^y. 81. (a^+i)^ • (ai-^)^
64. (2-')-\ 73. (5(^^)-^ 82. (««)^ + ^-(a'')^-^
65. [(f)-^^ 74. (0-^)^. 83. (a%y • (W)^
66. W- 75. (5o.2«.38)i 8^. (.^-.->3.
67. U^j. 76. (25a^^«)-^
68. {x^)\ 77. (2a?y . 8-4^.
^2^8\2a?
yn C2 71
88. ^^-;:2_ = 2
86. 2'^. 42 = 2?.
87. 2" .4^ = 2?.
69. (xy\ 78. (6i2^^)«(a + Z.). 2
88 —
70. (x-«)-2. 79. (a^)^'^ . a^^. * 43^^
89. 2'' .4^+^^ 2'' = 2?.
4.TC + 1 J^^+1
Oft : _J_ 5 — 9
2"('4'''~^V* * /4» + iyi-i "^
Solve for w:
^^ -95. 81.27^^=(9")2.
92. 3«.9^ = 812. g,„
93. 9'* . 3« = 27^\ ^^- (^^'')'' == (l25y^'
94. 2^'^"^^ • 4" + '^ = 8^". 97. 2^'"''^^ • 4^^+^ = ('8"')".
Solve for x :
98. a-^ = 8. 103. x-^ = 4. 108. ^x~^ = 2.
99. x-^ = 5. 104. x^ = 2. 109. (cc"*)"* = 49.
100. x~^=25e. 105. £c^ = 16. 110. (aici)"* = 27.
101. x"i = 2. 106. ic"^ = 32. V^ _ \/25
102. x^ = 4. 107. x^ = 343. * ^x^ -^
CHAPTER VII
SQUARE ROOT
54. Square root. The square root of any number is one
of the two equal factors whose product is the number.
From the law of signs m multiplication it follows that
Every positive number or algebraic expression has two square
roots which have the same absolute value but opposite signs,
55. Square root of a monomial. For extracting the
square roots of any monomial we have the
Rule. Write the square root of the numerical coefficient
preceded by the sign ± and followed by all the letters of the
monomial^ giving to each letter an exponent equal to one half
its exponent in the monomial.
A rule similar to the preceding one holds, for the fourth
root, the sixth root, and other even roots.
56. Cube root. The cube root of any number is one of
the three equa,l factors whose product is the number.
For extracting the cube root of a monomial we have the
Rule. Write the cube root of the numerical coefficient fol-
lowed by all the letters of the monomial^ giving to each letter
an exponent equal to one third of its exponent in the monomial.
A rule similar to the preceding one holds for the fifth
root, the seventh root, and other odd roots.
57. Principal root. For a given index the principal root
of a number is its one real root if it has but one, or its
positive real root if it has two real roots of that index.
101
102 sircc»:^rD course in algebra
Then the principal square root of 9 is + 3 ; the principal fourth
root of 16 is + 2, not — 2. The square root of — 4 or — 9 is not
real ; such numbers have no principal square root.
The principal cube root of 8 is 2, of — 27 is — 3. The principal
fifth root of 32 is +2, of - 32 is - 2.
Every number has more than one root of given odd index.
The number 8, for example, has two other cube roots besides
its principal cube root 2. What they are and how they are
obtained will be made clear in the chapter on Imaginaries,
where the consideration of the square roots of negative
numbers will also be taken up.
Unless otherwise specified, only the principal odd root
of a number will be considered.
ORAL EXERCISES
Find the principal square root of the following :
1. 16.
2. 25.
3. 4.a\
5. 36 a«.
6. 49 t\
7. 64 1\^
9.
10.
h xJ"
^' 13. x-\
15. 9a;V*.
16. 4x\
17. x^.
4. 9 a*.
8. 81a?«.
11.
18. x^.
Find the
principal cube root of the following
19. 8.
24. 27 rr*.
29. -125a«.
34. 343 a«.
20. 27.
25. 64 x\
30. - 27 a\
35. -512a«.
21. 64.
26. -8.
31. -64a\
36. - 27 a^.
22. 8 a".
27. -27.
32. -%a\
37. -^x-\
23. 8a«.
28. - 64.
33. 216 a\
38. -27ic-«.
Find the principal fourth root of the following :
39. 16. 42. a\ 45. 16 x\ 48. x^,
40. 81. 43. x\ 46. 625 a^ 49. 16 a;-*.
41. 266. 44. x"^. 47. 16x-\ 50. 81 x"^.
SQUARE BOOT 103
Give the principal root and one other root for the following :
51. The fourth root of 81; of w" \ of x~^.
52. The sixth root of 64; of a«; of x-\
53. What is the sign of the principal odd-Yoo^, of a positive
number ? The principal odd root of a negative number ?
54. What is the sign of the principal even root of a positive
number ?
55. State the rule for extracting the fourth root of a mono-
mial.
56. State the rule for extracting the fifth root of a monomial.
57. Can one obtain the fifth root of a monomial by extract-
ing the square root of its cube root ? by extracting the cube
root of its square root ? Explain.
58. Square root of polynomials. Extractmg the square
root of a number is essentially an undoing of the work
of multiplication. The square of any polynomial* may be
represented by
Qi^t^uy' = h^ + 2]it + t'^^-2hu + 2 tu 4- w2
A little study of this last form and a comparison with
the example which follows will make clear the reason for
each step of the process.
EXAMPLE
J^
First trial divisor, 2 h
+ 2]it-\-t^ =(2A+ 0^
First complete divisor, 2 A + ^
Second trial divisor, 2^ + 2^ |+2AM + 2m+u2 = (2^ + 2«+w)u
Second complete divisor, 2 ^ + 2 « + m \+2Tiu-\-2tu-\-u'^ = {2h-\-2t-\- u)u
Therefore th^e required square roots are ± (h -{■ t ■\- u).
104
SECOND COURSE IN ALGEBRA
EXERCISES
Extract the square root of the following :
1. x^-{-4.x^-2x^-12x-\-9.
2. a^-10d^-4.a^-\-25a^-{-20a-\-4..
3. x^ + 4lx^ + 16-4.x^-{-Sx^-16x.
4. t^ ^ 4:t^ -{- 9 + 4.t' - 6 1^ - 12 t\
5. 4 a* + 9 aH^ + t^ +*12 aH - 4 aH"" - 6 a^^-
6. 4 a« + 12 c^* - 7 - 24 c*"* + 16 «"«.
7. 49 c-« - 28 c-4 + 74 c-^ -20 + 25 e".
8. 9x2 + 4a^ + l + 12ir*4-6x + 4£A
9. 9i»*-6a^^ + a;^-66x^ + 22ic2_^121:r.
10. 25x^-10x''-\-90xi + x-lSxi-j-Slxk
11. 16 772.-^-
12. -^ + 7 + /-7, +
-* + 104 m - 26 m* + 169 m« + m-\
3 2/
iC
Solution. Arranging terms in descending powers of x and apply-
ing the method of page 103, we obtain the following :
(t)
4x2 12
y^ y
4a:2
X 4:X^
_ + 3-^
r
ff-)=
4x
y
i^ + 3
4a:
12 X
y
y
"(t-)^
+ 6
i£ + „_JL
V 2 a;
2_^ + JC
a: 4ar2
-2
a: 4
a:2 \?/ ^ 2a:/\ 2x1
Therefore the square roots are ±(— + 3 — ^j-
SQUARE ROOT 105
13. x« + 6a;'^-^ + 2x + -•
15. ^ ^g -2a;+ 3 +^4-
4 9 o
17. ^a' - 2a' -\-7^a' - -^^a -{- ^K
18. 9c*-12c^ + 4c2-- + i + 6.
19. ^'-18 + -, + 9(^2-2 + 2a2.
9 a^
20. ^ 1 + -^ + 4i.
21- I^ + ^ + ^'^' + 2^2 + 2 + 2^2.
x^ a^ X a
g^ 2a2 117 40 16
**'4c*^ ^25^* c^^Sac 5^2
^^- 9c2"^ a^ "*"3c"^ g "^3'
J[__3£ 9^ _a^_6^ J^
4a* a« ■*" a"- "^25:^2 5 "^5a;'
Find the first four terms in the square root of the following :
27. l + 2a;. 28. ^'^-a?.
KE
106 SECOND COUKSE IN ALGEBRA
59. Square root of arithmetical numbers. The abbrevi-
ated process of extracting the square roots of an arithmetical
number is as follows :
7^32^67^89 1 2706.8 +
4
47
332
329
5406
54128
36789
32436
435300
433024
2276
Therefore the square roots of 7326789 are ±2706.8+.
It follows from the preceding example that the work of
extracting the positive square root of a number may be a
never-ending process. The number 7,326,789 has no exact
square root, and no matter how far the work is carried,
there is no final digit. As the work stands we know that
the required root lies between 2706.8 and 2706.9.
The method just illustrated for extracting the positive
square root of a number is the one eommonly used. For
it we have the
Rule. Begin at the decimal point and point off as many
periods of two digits each as possible : to the left if the num-
ber is an integer^ to the right if it is a decimal^' to both the left
and the right if the number is part integral and part decimal.
Find the greatest integer whose square is equal to or less
than the left-hand period and write this integer for the first
digit of the root.
Square the first digit of the root^ subtract its square from
the first period and annex the second period to the remainder.
SQUARE KOOT 107
Double the part of the root already founds for a trial divisor,
divide it into the remainder (omitting from the latter the right-
hand digit^, and write the integral part of the quotient as the
next digit of the root.
Annex the root digit just found to the trial divisor to make
the complete divisor, multiply the complete divisor hy this root
digit, subtract the result from the dividend, and annex to the
remainder the next period, thus making a new dividend.
Double the part of the root already found, for a new trial
divisor, and proceed as before until the desired number of digits
of the root have been found.
After extracting the square root of a number involving
decimals, point off one decimal place in the root for every
decimal period in the number.
Check. If the root is exact, square it. The result should
be the original number. If the root is inexact, square it and
add to this result the remainder. The sum should be the
original number.
EXERCISES
Find the positive square root of the following :
1. 6241. ' 5. 53.29. 9. 2,932,900.
2. 9216. 6. 1.4641. 10. 7,049,025.
3. 15,129. 7. 216.09. 11. 3.9601.
4. 56,169. 8. 988,036. 12. .0061504.
Find to three decimal places the square root of the following:
13. 7. 15. .01235. 17. f. 19. ^K 21. 23^\.
14. .63. 16. .96384. 18. 4|. 20. |. 22. 89i.
23. Find the hypotenuse of a right triangle whose legs are
136 and 273 respectively.
108 SECOND COUESE IN ALGEBRA
24. A baseball diamond is a square 90 feet on each side.
Find the distance from the home plate to second base, correct
to .01 of a foot.
25. The hypotenuse of a right ttiangle is 207 feet, and one
leg is 83 feet. Find the other leg, correct to .01 of a foot.
26. The hypotenuse and one leg of a right triangle are
respectively 5471 and 4059. Find the other leg.
27. The side of an equilateral triangle is 17 inches. Find
its altitude, correct to .1 of an inch.
28. Find the side of an equilateral triangle whose altitude
is 15 inches, correct to .001 of an inch.
Fact from Geometry. If a, ^, and c represent the sides of a
triangle and s equals one half of a + & + c, the area of the
triangle equals Vs(s — a){s — b) (s — c).
29. Find the area of a triangle whose sides are 12, 27, and
35 inches respectively, correct to .001 of a square foot.
30. By the method of Exercise 29 find to .01 of a square
inch the area of a triangle each side of which is 22 inches.
31. Find to two decimals the sum of all of the diagonal
lines that can be drawn on the faces of a cube whose edge is
11 inches.
32. Find to two decimals the radius of a circle whose area
is 70 square feet.
33. Find to two decimals the diagonal of a room whose
dimensions in feet are 15, 22, and "28.
34. Find to two decimals the diagonal of a cube whose edge
is 8 feet.
35. A room is* 24 feet by 40 feet by 14 feet. What is the
length of the shortest broken line from one lower corner to
the diagonally opposite upper corner, the line to be on the
walls or the floor, but not through the air ?
CHAPTER VIII
RADICALS
60. Radical. A radical is an indicated root of an algebraic
or aritnmetical expression.
Thus V9, V5, V2x, and Va;^ — a; — 12 are radicals.
61. Index. The small figure like the 3 in v^7 is called
the index of the radical.
The index determines the order of the radical and
indicatefcj the root to be extracted.
For square root the index is usually omitted. Thus V 2 and ViS
mean v2 and vl8 respectively.
62. Radicand. The radicand is the number or expression
under the radical sign. In V5 and \^2ax the respective
radicands are 5 and 2 ax.
63. Fractional exponents. Radical expressions may be
written in either of two ways : with radical signs or with
fractional exponents.
Thus v/5 and 5^ have the same meaning, and Va^ equals as:, etc.
64. Rational numbers. A rational number is a positive or
a negative integer or any number which can be expressed
as the quotient of two such iutegers.
Thus 7, — 6, |, or 2.871 are rational numbers.
65. Irrational numbers. Any real number which is not
rational is irrational. (See section 67.)
109
110 SECOKB COURSE IN ALGEBRA
If a number under a radical sign is such that the root
indicated cannot be exactly obtained, the radical represents
an irrational number.
For example, V? and Vi are irrational. Approximate values for
these are given on page 274.
A repeating decimal, though endless, is not an irrational
number, ' for any repeating decimal can be expressed as a
common fraction, and is therefore rational.
Thus the repeating decimal .272727 • • • is not irrational, as it
exactly equals ^^. Similarly, .2857142857142 • • • exactly equals f , etc.
Note. The number f reduced to a decimal repeats the digits in
groups of six each, and the mere fact that a decimal does so repeat
is proof that it is a rational number. On the other hand, the num-
ber TT is known to be irrational, and its value has been computed to
707 decimals, showing, of course, no repetition. The fact that it
does not repeat in 700 digits is, however, no proof that tt is irrational,
for decimals with even more than that many digits do repeat. For
example, the fraction WgV^ equals the decimal 1.29 + , which repeats
in groups of 7698 digits each.
66. Imaginary. An indicated square root of a negative
number is called an imaginary number.
Thus V— 4, V— 7, and V— 12 are imaginary numbers. And
3 + V^^ is also imaginary, though, as will be seen later (Chapter XII),
such numbers are better called complex numbers.
67. Classification of numbers. All the numbers of alge-
bra then may be placed in one or the other of two classes :
real numbers and imaginary numbers.
Real numbers, as we have seen, are of two kinds, rational
numbers and irrational numbers.
68. Surd. A surd is an irrational number in which the
radicand is rational.
Thus Va, V^, etc., are surds. But v 2 + Vi} and Vtt are not surds.
EADICALS 111
69. The algebraic sign of a radical. The square root of
4 is both + 2 and — 2. The symbol Vl, however, sig-
nifies only + 2, the principal root (section 57). Similarly,
^81 is + 3 and V64 is + 2^ But -V9 is - 3, and -^l6
is — 2. The symbol ± V25 denotes both + 5 and — 5.
Further, +^27- +3, -^27 = - 3, and -^327== +3.
The foregoing remarks apply also to fractional expo-
nents. Thus 4^ = + 2 only, and 81^ = + 3 only, etc. It
should be noted that these statements really define the
meaning of such symbols as V~? ^? v^ » etc. Such an un-
derstanding as this avoids all the ambiguity which would
arise if Vl6 meant both + 4 and — 4. The distinctions
here made are especially needed in radical equations.
ORAL EXERCISES
Find the numerical value of the following :
1. -Vi.
7. -^81.
13. 8*
19. 16^
2. -V9.
/ —
8. v^32.
14. 36t
20. (i)^.
3. Vl6.
9. -^-125.
15. 49^.
21. (f)*.
4. Vs.
10. --v^-32.
16. 8^.
22. (- 64)*-
5. ^-8.
11. ^64.
17. 16*.
23. (49)1
6. -^/i6.
12. -^625.
18. 27^.
24. (121)1
Bead in radical form :
25. x\
30. Srx^.
35.
a(a-l)^.
26. x^.
31. 4.ax^.
36.
2c(2x-3)i.
27. (at)\
32. 2aV.
37.
x^'k
28. (3t)i
33. 5ux^.
38.
2ixk
29. St^.
34. 4Ai
39.
a 2 71
xH'^.
112 SECOND COURSE IN ALGEBRA
Read with fractional exponents :
40. V^^ 43. V'^K 46. -^{a + x)\
41. V^«. 44. 2V^. 47. -s/^^x-ay.
42. Va\ 45. ^v^. 48. -^x'^y - 1).
49. What are the two square roots of 36 ?
50. What are two fourth roots of 16 ? of 81 ? of 625 ?
51. What is the value of a/16 ? of "v^ ? of -v/625 ?
52. What are two sixth roots of 64 ?
53. What is the distinction between a rational number and
an irrational one ?
54. Which of the numbers 8, |, 343, VI, Vs, and ir
(it = 3.14159 + ) are rational ? Which are irrational ?
55. Give a geometrical illustration of an irrational number
by means of a right triangle.
56. Is a radical always a surd ? Illustrate.
57. Is a surd always a radical ? Illustrate.
58. Distinguish between a surd and a radical.
59. Which of the numbers Vs, VI, ^27, V^, \/2 + ^^3,
and V2 tt are surds ? Which are radicals ?
60. What is the principal root of ± Vi, Vs, and V— 8 ?
61. Name the order of V6 ; of a^; of VS; of c*; of Vm^.
62. Give an example of (a) a real number ; (b) an imaginary
number; (c) a rational number; (^Z) an irrational number;
(e) a radical; (/) a surd; («7) an index; (h) a radicand;
(i) the principal odd root of a positive number ; (j) the prin-
cipal even root of a positive number; (k) the principal odd
root of a negative number.
70. Simplification of radicals. The form of a radical
expression may be changed without altering its numerical
value. It is often desirable to change the form of a radical
RADICALS 113
so that its numerical value can be computed with the least
possible labor.
The simplification of a radical is based on the general
A radical is in its simplest form when the radicand
/. Is integral.
n. Contains no rational factor raised to a power which
is equal to, or greater than, the order of the radical.
m. Is not raised to a power, unless the exponent of the
power and the index of the root are prime to each other.
For the meaning of I, II, and III study carefully the
EXAMPLES
Examples of I :
- 1. Vf = Vf = vT^ = Vi^ = i^-
2. 6\4 = 6Vf = 6vTr3 = 6.iV3 = 2V3.
Examples of II :
2. 5 ■\/24.x' =5^Sx^-Sx^ = 5 V(2x)^ 'Sx^ = 10a; VS^.
3. Vl6 - 8 V2 = \/4(4 - 2 V2) = 2 V4 - 2 Vi
Examples of III :
1. ■^ = -v^ = 2* = 2^ = V2.
2. ^ = ^2_3* = 3^ = -v^3.
3.' -\/'^^ = Jb^ = ah--=b\^.
114 SECOND COURSE IN ALGEBEA
A radical of the second order is simplified by the use
of the
Rule. Separate the radicand into two factors one of which
is the greatest perfect square which it contains. Tiien take the
square root of this factor and write it as the coefficient of a
radical which has the other factor as radicand.
If the original radical has a coefficient other than the num-
ber i, multiply the result obtained above by this co&^cient.
A similar rule holds for simplifying radicals involving
the cube root and roots of higher orders.
EXERCISES
Simplify :
1. Vl8. 6. V52. 11. Vl92. 16. S Vu.
2. V20. 7. V63. 12. 2V45. 17. V^.
3. V28. 8. V68. . 13. Vi6. IS.
ax".
4. Vii. 9. V75. 14. 4V54. 19. VaV
5. V5O. 10. VT08. 15. -v^. 20. nVT^\
21. V|.
Solution. V^ = V5 = V^~^ = i V3.
22. x/f. 23. Vf 24. V\. 25. V|.
• 26. ^.
ya _
Solution. ^ ft = ^/£ = ^/i . « = - V^.
\a \a2 \a2 a
34. 6v
27. \ —
28.
«^ -35. VWl?.
IT 3|2^ 32. Vi_ 36. v.^^£(|y.
\^«' ^^- NT* 33. V-|. 37. x/r+(|y.
RADICALS 115
38.
44. \/a^ + aW3.
45. V16-8V3.
39. 4'+gJ- 46. V54-9Vi8.
42. Vrr^V!. 49. ^/J»-|V3.
Hint. V4-8V3 =
V4 (1 - 2 ^
43. V36 + I8V5.
51. -^.
v/3).
54.
55.
56.
50.
Soli
ition.
21 = 2^= V2.
52. -\/4.
53. -Va^t
«l4a«
Express entirely under the radical sign :
59. 2V7. 63. a-Va. 68. e^Ve^ + e"^.
Solution. 2V7= 64. 2c^. ^^ , . in^O_
60. 3V5. ee.x^^\ 70. (2^ + 1)^)^^
61. 4 Vs.
^„ a 3f9 „^ .x-3a 3| 125
62.2^8. S^a^' 5 \(a5-3a)2
Express in simplest form with one radical sign :
72. VV2. 73. VV^. 76. VV^»
Solution. VV2=V2^ 74. VVa. 77. V Vo-.
= 2i:=^. 75. vWl 78. V-?/^.
116
SECOND COUESE IN ALGEBRA
79.
82. VsVsVS. 85. VVV^.
80. V-\/Sd^x.
81. V3V3.
83.
Vs.
84. 2\/2V2.
Find by the formula of Exercise 28, page 108, the areas of
the triangles whose sides are
88. 6, 8, and 10. 90. 33, 56, and 65.
89. 7, 24, and 25. 91. 104, 153, and 185.
71. Addition and subtraction of radicals. Similar radicals
are radicals of the same order, with radicands which are
identical or which can be made so by simplification.
The sum or the difference of similar radicals can be
expressed as one term, while the sum or difference of
dissimilar radicals can only be mdicated.
Simplify and collect :
1. V8 + VI8.
Solution. Vs + V18 =
EXERCISES
>V^.
2 V2 + 3 V2
2. VI -f 3 V2.
3. V5O + V98-V32.
4. V12 + 5V75-2V27.
5. 3VI8-V98 + V128.
6. V75 + 3V147- V12.
7. 2V54 + V24-V96.
8. V45 - V20 + 5 V245.
9. 3V275 + 2V99-5V44.
10. -v/ie + "v/si - 3 -v^.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
+ V375.
V192 - 4 -V24 -,
V54 + -v/ie - VT28.
^625 + ^40 + Vi35.
loVf-V^ + V^-
3x/f + 3Vi-2V^.
a Vac^— -VA— 5 Va^.
\Sa . 13 X lax
[a fa |5x*
EADICALS 117
21. v^32 x^ 4- a/1250 x - 4^512 x - V2^.
22. V(a ^cf-c ^{a + c)2 + 2 c V {a + cf.
23. -^(ct _ cf + c ^^2 - 2 ac + c^ + (» + c) -v/^
24. \^--aJ-+aJ— -^— + 2--^J-
■'±^'_9
ac
3y
25. ^^ _|- ^(3 c^ + 9) (ct + 3)^ - V8i + c, Vg - 4 -V3,
26. 2 V9 a^ - 9 a^^* - 3 V9 aV" - 96* + V(a--^ - V') (a + &).
36a^^-36^^^
a-hh
27. (a-^»)-J^ + V25a^-25&^ + ^A|
^ ^ ya—h a—h >
72. Multiplication of real radicals. Real radicals of the
same order are multiplied as follows :
Example 1. Multiply 2 -\/x — 3 V« — 4 ^ax by 2 Vox.
Solution. 2 Vx_— 3 V« — 4 Vaa;
2 Vaa:
4 a: V« — 6 a Vx — 8 aa:
Real radicals of different order are multiplied as follows :
Example 2. Multiply Vn by -^x.
Solution. Vn = n^ = n5 = "V^n^
"v/x = x^ = xt = "v/^^.
Then Vn • -^a; = -v^ • V^ = xn^x^.
The method of multiplying real radicals is stated in the
Rule, If necessary, reduce the radicals to the same order.
Find the products of the coefficients of the radicals for the
CQefficient of the radical part of the result.
Multiple/ together the radicands and write the product under
the common radical sign.
Reduce the result to its simplest form.
The preceding rule does not hold for the multiplication of
imaginary numbers. This case is discussed in Chapter XII.
118 SECOND COURSE IN ALGEBRA
EXERCISES
Perform the indicated multiplication and simplify the
products :
1. V3V27. 5. VI- Vi- VI-
2. Vi2Vi8. 6. (V3-Va)V2.
3. Vf . Vf . 7. ( V2 - 3 VS) V5.
4. V|->/t- Vf- 8. (V3-2V2)(V2-V3).
9. (V5-3V2)(2V5-V3).
10. (Va — Vax)(Va + 2 Va^).
11. (V2 + V3)(V2- V3). 16. (V3^-V2^)^
12. (3V5- V2)(3V5 4-V2). 17. {-^x - 3)1
13. (V7-V5)(V7 + V5). 18. (2V3a:-l)'.
14. (2V3- V3)(2V3 + V3). 19. (V^ - V.X- - 2)'.
15. (4V5 + 2V7)(4V5-2V7). 20. 3Vaj-3V4x-8,
21. (V5 - V3 - V2)(V5 4- V3 - V2).
22. (3 V2 + 2 V3 + V30)(2 V2 + 2 V3 - 2 V5).
2J.(;,-fV3)(2« + ^V5).
Square :
26. V2 - Va: - 3. 29. Va- - 3 - Va: + 3.
27. -v^ _ Va^4-4. 30. 2 Vx — 3 V2 .r + 1.
28. Vrr - 3 + Vic + 5. 31. 3 V;r - 1 -f- 2 vT
a;.
RADICALS 119
Perform the indicated multiplication :
32. (a + Va + ^)(a - Va -t- b).
33. (Va — b — ^a){-y/a — b + Va).
34. (V2x - 3 - V3^)(V2x - 3 + Vs^).
Express as radicals of same order :
35. V2 and V^3. 37. Va^ and VS.
36. VS and V^. 38. V^ and VS.
Multiply the following :
39. V3, -v^. 43. ^, V2. 47. V^, v^.
40. VS, VS. 44. Vl2, x4. 48. xl^ , ^P •
41. V^, Vs. 45. Vc, VS. 49. ^y27^ V3^.
. 42. VS^, VS. 46. VS^, VS^. 50. VST^; VS+^.
73. Division of radicals. Direct division of radicals co-
efficient by coefficient and radicand by radicand is often
possible.
Thus 6V5-^3V3 = 2V| = §V15,
and 3 Vox -t- 2 VS = | Va.
Direct division of radicals when the divisor is a radical
expression with more than one term is usually very diffi-
cult. In sucji cases a rationalizing factor of the denomi-
nator is used. We then carry out the operation of division
indirectly by resorting to multiplication.
74. Rationalizing factor. One radical expression is a
rationalizing factor for another if the product of the two
is rational.
120 SECOND COURSE IN ALGEBRA
A rationalizing factor for Vz is Vt, since Vz • V? = 7. For
"v/5 a rationalizing factor is ■v^25, since "Vb • ■v^25 = 5. Similarly,
Vs — V3 is a rationalizing factor of V5 + Vs, as their product,
(Vs - V3) (V5 + V3), is equal to 5 - 3, or 2.
In like manner (3 Vt + 2 Vs) (3 V? - 2 V5 ) = 63 - 20 = 43.
Two important radical expressions are V^+Vi and
Va — Vb. Two such binomials are called conjugate radicals,
and either is a rationalizing factor for the other.
Rationalizing factors are used in division of radicals as
follows :
, , /;: /r V6V5 V30
Example 1. V 6 -^ V5 = \_ ,_ = -—- •
V5V5 5
Example 2. (6 V2 - 15 Vs) . 3 Vs = ^^^"l^^j'^
3V5.V5
_ 2VlO-5\^
5
Examples. (Vs +^)^(V5- Vii) = M±^^1M±2^
(V5-V3)(V5 + V3)
_ Vi5 + ViQ + 3 + Ve
5-3
= K^^ + ^^ + 3 + Ve)-
Therefore when direct division of radicals is impossible
use the
Rule. Write the dividend over the divisor in the form of a
fraction. Then multiply the numerator and denominator of
the fraction by a rationalizing factor for the denominator and
simplify the resulting fraction.
This rule applies in all cases, while the rule for direct division
fails when dividing a real radical by an imaginary number.
RADICALS
121
EXERCISES
Find a simple rationalizing factor for
1. Vt.
2. 3 Vs.
3. Vs.
4. Vi.
5. Vs.
6. Vs.
7. Vs-Vr.
8. V3 - 2. .
9. 3V6-2VII.
10. V3a-V2x.
Perform the indicated division :
16. V8^V2.
17. eVio^Vs.
18. Vl2^V3.
19. V8--V24.
20. Va^H-Va^.
21. V2a^-v-V3a^.
11. Vx-3-V3.
12. Vic - 3 - 2 V3^.
13. Va — ^ + Va -}- 6
14. V8+V2-V5.
15. V3H-Vi8-V^.
22. (Vi4-Vi0)--V2.
23. (2yiO-3V5)--2V5.
24. ( Va^ — Va^) -^ VS.
25. 8 --4 Vs.
26. 8 --2 Vs.
27. 24--3V3.
28. V3^V2.
„ , ^. V8 V3.\/4 V3.^4 ^27 • 16
^^^"'"" vi = v^:^ ^ -^ ^ -^—
29. V5-V3. 35. Vi-VJ.
30. V^--V2.
31. V8--V2.
32. Va-f-Va.
33. V2^^
4 -^ V 2
O. -f- C ViC.
34. V32-V2.
R£
37. V3^(V3-2).
38. V5--(V5+V2).
39. (2V3+V5)--(V3-V5)
40. (V7+V3)-(V7-V2).
122 SECOND CGHTESE m ALGEBRA
Change to respectively oqiaiYail^nt fractions having rational
denominators: ., , , ,
V5---^ ^. 2V5+"3V7 . ,„ Va!-3+V3
■'42: —^ ■%. y^. 48. ^^^^
V5;-V2, Vcc+Vc V2-V2
Perform the indicated division :
50. ( ViO - V5)-( VlO ^i'W):^^^ i>i^toiL/ii 61LI nnol-i..<!
51. (i;'^^)-^<^-^-S^^).^'S .SV-8V .01
53&'^( VW 4- c - ^^) V (:^i/a^c + V^). .0 V -- oIV l) ^.TI
53,n(^-^3,H7V2),^y3^ V2). .FiV-^ siV ,81
54. ( V5 + V7^V2l3( V| - V7). . ,v_ aV .ei
55. Is there any distinction between the meaning of the
direction before Eiercise il^lid of that bef of e Exercise 5<) ?
56. Does 3 -Vz satisfy a:?&- 6 a; + 2 = ? on ^jV .iSi
57. Does '^"^ satisfy 2x'-75x + 161 ^ ?8S
58. Does ^(B' ± Vto^)*satisfy ^3 i^^jc^^ ^ = ?
75. Square roQt of surd expressions. The square of a
binomial is usually a 'trinomial. However, the result"* of
squaring a binomial of the form Va + V^ is a binomial if
a and h ara rational number^. Thus ..^ /' \y
In 10-^2Y21^ lO^i^vtl^a sgp of 7 and 3, and ^1 is^the
product of 7 and 3. Tnese relations anii the lact tnat
thfe^ooeffi^eHiof/the'^kdieftl V2T is 2 eiiake Ue^to-ffiid
A>i<i.ivM. RADICALS azooa;^ i2$
the square root of many expressions of the fomTa ±V6
by writing each in the form of x±^^xy-\-y and thea
extracting the square root of the trinomial square as follows f
Example. Extract tne square foot of 9 — yoo.
, ^ __>/,,!, y>',miLy6'iq iloiiiw anoift89iqz9 \imm
Solution. 9-V66=J,^-^^A(^^ dfl-gjiodi od won v/ifn smiiq
We must now find two numbers whose sum is 9 and w;hose
product is 14. These are 7 and 2.' ' - " ^ ~ "' '" rwiiy
Then 9 - 2 Vii = 7 - i^^' ^f'^^^ -^ . ^'^^
Heiice th^ sqtmi^ rdot^^f ^i^^^^^i^ "tg (H^H)^. ^^^^^ «I
•• -iL^-j.Jjjn iO -- • ' iijrl) hoojaiohnn yIibsIo
.8ioio.Bl oipXERCiSES-t Qflj fii ginsioifiooo odd
ei Find the positive squafe roots in Exereise^^d[*-:l^iri^'89'i oT
^'\B^k V^- ^.11-3 Vs. ,; ' ^3 tMte /n-^Lim
4. 7-V40. 8. Il-Vi20. 4 .Soj^fd
12. 2 cc + 2 Vx2 - 49. 14. V^ + 3 VS = V? 4-^.
13. a + Va^ - 1. 15.' V15 - 5 VS = ?
'* 16. V« + V^^^^^^^ = ?
17. V ??i"^ + 7?^ H- 2, ?i -f- 2 ??i V m + 2 71 = ?
Note. In the writings of one of the later Hindu mathematicians
(about A. D. 1150-)'we find a method of extracting the square root of
suTd.s which is poetically the.ipame as that given in^the te?:t.§In
fact, the formula for the operation is given, apart froin the modern
symbols, as follows : Va + Vh = V « + Z> + 2 Vah. The study of ex-
pressitdTS of the t^i^" V Va i ^6 had been carried to a i^tto^tTeii^rk-
able degree of acQuracy by the Greek, Euclid. His researches on this
subject, if original with him, place him among the keenest mathe-
maticians of all time, but his work and all of his results are ex-
pressed in geometrical language which is very far reindved f^oirPthe
algebraijc aymboli^iXJi,of tp-d^ _ g _ - ^^ ^ .01
124 SECOND COURSE IN ALGEBRA
. 76. Factors involving radicals. In the chapter on Factor-
ing it was definitely stated that factors involving radicals
would not then be considered. This limitation on the
character of a factor is no longer necessary. Consequently
many expressions which previously have been regarded as
prime may now be thought of as factorable.
Thus 3 x2 - 1 = (a: V3 + l)(a: VS - 1),
and 4.x^-5 = (2x + V6)(2x-^^').
In this extension of our notion of a factor it must be
clearly understood that the use of radicals is limited to
the coefficients in the terms of the factors.
To restrict the use of radicals in the way just indicated is
necessary for the sake of definiteness. Otherwise it would
be impossible to obey a direction to factor even so simple
an expression as x^—y"^; for if the unknown is allowed
under a radical sign in a factor, x^ — y^ has countless
factors.
Thus x^ — y^ = (x ■\- y^(x -^ y)
= (x + y) (Vx + Vy) ( V.r — Vy)
= (x + y) (Vx + Vy) (Vx + V^) (Vx — Vy),
and so on indefinitely.
EXERCISES
Factor :
1. ar^-11. 3. a;« + 3. 5.3x^-27,
2. 3a^2-8. 4.0^^-12. 6. 6x«4-125.
Find the algebraic sum of the following :
2-Vb 2 x + c x^ -^c^
' a-b V«-f V^^* * V^-Vc ^-c
Solve by factoring, and check the results :
9. x'-b^Q. 11. (T* + 144 = 26x1
10. 2x2-3 = 0. 12. 4.x'-{-c = x^+A:Cx\
RADICALS 125
PROBLEMS
(Obtain answers in simplest radical form.)
1. The side of an equilateral triangle is 8. Find the altitude.
2. The side of an equilateral triangle is s. Find the altitude
and the area.
3. The altitude of an equilateral triangle is 24. Find one
side and the area.
4. Find the side of an equilateral triangle whose altitude is a.
5. Find the altitude on the longest side of the triangle
whose sides are 11, 13, and 20. Find the area of the triangle.
Hint. Let the altitude on the side 20 be x, and the two parts into
which the altitude divides side 20 be y and 20— y ; then set up two
equations involving x and y, and solve.
6. Find the altitude on the longest side of the triangle
whose sides are 10, 12, and 16.
' ' E n
Fact froTYi Geometry. A regu- / \
lar hexagon may be divided into / \
six equal equilateral triangles by / \
lines from its center to the vertices. fI 2 \fy
In the adjacent regular hexa- \
gon,AB = BC = CD, etc. is the \ 7
center and OK perpendicular to \/
^^ is the apothem of the hexagon. A ' K B
7. Find the apothem and the area of a regular hexagon
(a) whose side is 18; (h) whose side is s.
8. Find the side and the area of a regular hexagon
(a) whose apothem is 30; (h) whose apothem is h.
Facts from Geometry. The volume of a pyramid or cone is
— > where a is the altitude and h is the area of the base.
The altitudes of an equilateral triangle intersect at a point
which divides each altitude into two parts whose ratio is 2 to 1.
126 SECOND Ce^ESE' IN ALGEBRA
9. The base of a pyramid is a square, each side of which
is 10 feet. The other four edges are each 20 feet. Find the
altitude and the volume of the pyramid.
,3 [ , jjp^ r jTJ;ie: side pf an equilateral triangle ift ^^ , j|!in^ the two
parts into which each altitude is divided by the othey altitudes.
'>i 'A' regular- tetrahedron is '^^i^P'^ 'ifJ^ to oLiJ^jjUu siiT .£
pyramid whose four faces are .Bs^kaif.^ hua mFjih
eq^ual equilateral trianglesJ •rr.^tiiliirpo iir> io oMh dkM)[ii .
sl-The altitude of a regulajrf.j no obufi/B }\\\^1 .a
t%%ahedron , (>Z>^ in the adja-,os; bar, ,^/,tl ojias^H sao/lw
§^^t iigui-e) ^eets,the,base,^t ,,i, ,,0 9/jijiB ..iJ.li .\rH
<fc^e point where the.- altitudes os ahia/jLivib Qbiijjjifi j)j\[:)h[v/
of the base intersect. ..»7[oa brf^\y bits tftgriiJloini 8/ioiVxri>o
rsihedT^. If each, edge is 24,- =''''^N<^' ''■'<\^^)}^^''^>:yC'ff
find C/!^'VK'^a,ndt;-^ lastly, tlm^^yi A av^''^J?*v^'^^<>'\V. "^^'^
altituae Z)^. oJiii bobivib.;:Jt:l v(^i;iii>8<{AaKyiI ud
J%. Find the altitudeandvai ;:''^""":' '^"''""f" *" ^" . ' ,'
ui^e of a regdlar tetrahedron each of whose ed^esis ^6 iiiclies.
ik Show that the /ititude and the voluthe^of a»tfegulartetra^
hedron\j^hose-^4^is e are respectively % "V^ and tt: ^(2.
MISCELLANEOUS EXERCISES
f r* 1 iRediice! fc6 respectively equivalent fractions haviingf tatibnal
denominaltoaB i^i'MiJoqij o^.uii ^/ (A), -(jii rii tiirj(l,jo<j.B abOiiYV (^^ i
Find the positive square root" of the following :
7. 28 + 10V3...^ li_i2.a.-5I-=. l^yiS.
8. 30 + 12 Ve. J~V 10. c^ + c + 2 c^ V^.
11. Show that X — .. .■ , ^ . .; isa root.^^ii^^ — 7 a? + 1 = 0.
12. Show that X = ^^„ :^^rB ruuls o^f^x"^ + 6 a; = 1.
o
13. Show that x = 3 a ± 2 Vc are roots of a;^ — 6 ax + 9 a^
1. Qv. .T. . - ^^> + V60,^^ + a^^>^ . , . ,,
14. Show that x = ig^^ root of the
equation 3 ax^ + abx = 5 b. ~ ~
Simplify :
15. 2V18H-3V5- V8-*V32. 17, V3 - V6.
— : , — -. ^ ., .lo
16. sVf + 2\/^-^Vf H-'^^^^- 18- V5H-3V2.
20. (5 V7 - ^)i;^( v^ + V7)r ^^
21. sVi + Ss/S+f V96-V66|.
22. ; ^^ - ^, ^ ^^ <,-)
25. 03 . i (a^ - xYH- 2 ^) - («' - ^')^.
26.
(x^)^
his)
x'(S x^) - (x^ 4- 5) 4 x^ x^ + 20
(x*)2 • x^
128 SECOND COURSE IN ALGEBRA
(x^ - If
28.
29.
30.
X
x^
x\ax^-
x'-l
-\-l)2x
{xy
x^^'nx^-
x'
-1 - x'' .
2ax^^-^
(x^y .
,2 a
31.
32.
x-'(- 2x-^)-(x^-\-S)(- 5x-*)
(x-y
2 e"'^ + 1
33. ^^ ^pr-^
■\x
•^ + 1
CHAPTER IX
FUNCTIONS AND THEIR GRAPHS
77. Functions. One of the most important concepts of
mathematics is the notion of function.
As the term is used in mathematics the basic idea is the
dependence of one quantity upon another or upon several
others. Countless functional relations exist in everyday-
affairs. The velocity of a falling body is a function of
the time since it started to fall, the interest on a definite
sum of money is a function of the time and the rate, and
the quantity of water transported yearly by the Mississippi
is a function of the rainfall (and the snowfall) in its basin.
Relations such as these can often be expressed either exactly
or approximately by equations. For a body falling from
rest near the earth's surface, s = 16 t\ Here the distance in
feet, s, is expressed as a function of the time, ^, in seconds.
In y = 7? -\- 2> X -\-l the value of y is a function of x. Such
a relation is often represented more clearly and strikingly
by a graph of the equation than by the equation itself.
78. Names of functions. A function is called linear, quad-
ratic, or cubic according as its degree with respect to the
unknown or unknowns is first, second, or thirS respectively.
Thus 4 a: — 7 is a linear function of a: ; 3 x^ — 8 a; + 1 is a quadratic
function of x ; and a:^ — 3 a:^ + a: — 10 is a cubic function of x.
In the study of functions the unknown is often called the
variable, since from this point of view the problem is not so much
the finding of an unknown as it is the study of the changes of a
variable quantity.
129
130
SECOND COUESE IN ALGEBRA
79. Notations for a function. After a function of any
variable x has once been given it is usual to refer to it
later in the same discussion by the symbol /(.r), which is
read the function of Jr, or, more briefly, / of x.
80. Linear functions^ THd dxpresfeion 3 a; 4- 2 is a func-
tion of x^ and the value of this binomial varies with x.
The following table gives a partial view of the relative
^change of val,u^s,,J?ptw^l^|}5 a::,,an4, tl;^^ fung|4^^7§^
f i oij.mtr ^ t
If
n,i.
i?
theii/(i) = 8x + 2=i^l0! tt^ v4i4i im^' -iQ'! i'5 • 'i8-'Iiili(j)I
,•>.,
.?,
u.
This relation can, be xepo^^^anted graphically l;>yiTisuigtih,e
same ir-axis as beforQj
(section 44) and using
the.j/-axis as the func-
tion axis; that is, laying ;
off values of x horizour. ,
tally aad corresponding .j
values of the function^ \
^x + 2 vertically. Thev
graph resulting from, ^
the above t^J^le iqf y^U^k
ues is shown in the,
^CQmpanying figu^p,,,.
J^ic^ bja show^ th^^t, M
the graph ofr a Imea^ , , ,
function is , always „ a
straight Un^-,,j.,„^,^ ^l^^^f,y ^> ^j or . i • , j; - •'■•, I.jik ;x lu aolUnn^
Mift i)'>rii;') n'ttlii «f iiwoii /({EXERCISES 'i'fojiult ^o \Imi\<'. 'mH nl
il'UMU OK Jtur yi n!''iifo'iif -Mit /■'•'i • ',"■""! HfiM, rnn-fj -yiiiiij ^'tUlnhivf
1. Construct , the graph of the function 3 a: ~L., .. ,,
2. Construct the graph of the function/^ w,rln .3. .Mjiirijv
ru3srcTiONS axd their graphs
131
81. Quadratic functions. The function x^-{-x—6 may
be represented. graphically by proceeding a^ :^ollows:
Qd!
a
1;h^ii'J^(i9±i:?ar*'{P'dt-t^6t=it.q(i6f \>i^ : u.4
' i'iM k,M.[ l' ;! l!
^S
rhrrrrr
a^-l'.
^e
«'
-6
-4
^
s
■6!'
-^f' Plotting the\
*^'6ints corre-
sponding to the
numbers in thj^^'r
■-'..:
T-rr
•In
IT,
"\
I , 1 1 j
up-:
;t8y
I<Jif ir; \
■jM'/
• ^ ; i ,'
• ',
! '>
if
i ,
GRAPH
OF
ii
-lo
iiq:
1 /
table we obtain ^-
i'l'J
i'l'J
\i/
m-
^^f.
;aj-
B ;
tn '
; ,( if
• r i
/'
the accompany-' -
- "" ■\.
jil
A
::■:
:n
'
.1
!!''!
" li
/ '
ing graph* ^-^oUoi
8S
•o,.i
\
' r 1
I '/
(i V
!];;■
j~-Th;e - g
df a quae
ttmvtK
1
1 tip 11
.ratio
r
I, _
V-
- 1-^
/
i
\
/
fmnctionii
ijone 1
\
/
. . _ . 1
'•
\
/
. 1
y'
variable is a
3
c
.
^-'
J -i
i /2 3
,,:■■
[fiurve called K .
\
/
liurabola. Itmayl
/ ;.>
\
~
) 1 '
~^
/
:ib
l!<.>'
f'.' ' .
•■■■'-
■>■; -
' ; '»
\
b^ sha^
r ierj
--,
\
\y
'flatter thai
a the;
hV
;
^^
.■ : : !
■]'■
i ' '
' " j
accompanyijig^
1\
-h
'-J
!!)
p*
'
graphs jbd
+ ^ '
i '
^n..^
' .1
V'-i-
,-
^ y.L- : I \i
the same general shapfi^anfl-the opening may be upward
or downwfi^d^ .J'.i:...ii i ^no :ji[} nl
f^- j T j EXERCISES »i^^ 8088010 OYlir'o OflT
^1 ri^^T L „, « .:.. ^ •„ .:;^!ffiii 6^-nI} aixB--^;
Uonstruct tna graph of the lollowmg : ^ r
7. A body falling from rest near the earth's surface obeys
the law s = 16t^, where s is in feet and ^ is in seconds. Con-
stwt(&t the gi?^h 4ii fit) = 16 1^. ii ^ .xioirniiebxii alliil Ji 1 .
132
SECOND COUKSE IN ALGEBRA
- Note. In the study of analytical geometry one takes up system-
atically the curves which represent equations of the various degrees
beginning with the simplest. It turns out, as we have already seen,
that the linear equation is represented by a straight line.. Equations
of the second degree in x and y lead to the so-called "conic section."
One of the most interesting and important aspects of the graphi-
cal method is the fact that the simplest equations correspond to the
most useful curves both in pure science and in nature. The com-
monest curves in nature are the circle and the parabola. Their
equations are the. very simplest equations of the second degree.
82. Graph of a cubic' function. The graph of a cubic
function is obtained in the same general way as that of
a quadratic function. The function a^— 5x+^ may be
represented graphically by proceeding as follows:
If x =
- 4
-3
-2
-1
1
2
n
3
then/(a;)=a:«-5a: + 3 =
-41
-9
5
7
3
-1
1
6i
15
Plotting the points
corresponding to the
numbers in the table
(except the first and
last), we obtain the
points ^(-3, -9), B,
C, D, E, G, and II,
in the order named.
The curve crosses the
a:-axis three times :
once between 1 and
2, again between 0.
and 1, and a third
time between — 2 and
— 3. Above If the curve rises indefinitely, and below
A it falls indefinitely. In each case it becomes more
F
GR
WoF
1
f(3
f) =
x'-
5£C
+3
/
P
/
/
fN.
Ih-
Yn
/
/
/
\
/
\
^
X
X
f
^ V
I
i
-1
6
\
\
y
1
\
E
1
1
1
1
1
f
o
\
A
F'
FUNCTIONS AND THEIR GRAPHS 133
and more nearly straight as it recedes from the a:-axis,
never crossing either axis again.
In forming a table of values two pairs are sufficient for a linear
function, but more are needed for a quadratic function and still
more for a cubic function. Usually the higher the degree of the
function, the more points are needed in constructing its graph. It
should be noted that in making a good graph the number of points
is not so important as is their distribution, which should be such
as faithfully to outline the entire curve. Where the graph curves
rapidly or makes sharp turns the points should be close together.
Such places are difficult to locate before the graph is constructed ;
hence one should make a table of values which appears to be suffi-
cient and plot them. Then inspection of the plotted points will
usually show where sharp turns or rapid curvature exists. The
table of values should then be properly extended and the additional
points located. Repetition of this last step will enable one to draw
a graph which accurately pictures the variation of the function.
It should be observed here that scales on the two axes need not be
the same. Some experience is required to choose for the two axes
the scales which are best suited to bring out clearly the shape of the
curve. In general the graph should be drawn to as large a scale, in
both directions, as the size of the paper permits. What that will be
for each axis can be decided by inspecting the table of values. For
example, when the dimensions of the preceding graph are once deter-
mined one can see from the table that all values of x are easily repre-
sented but that it is undesirable to try t.o represent the values of the
function, — 41 and 15.
EXERCISES
Construct the graph of the following :
1. x^-Sx-^l. 3. x»-4cc-2.
2. x^-Sx + 2. 4. £c*-ll£c2 + 24.
Note. The notion of a function is one of the three or four most
fundamental ideas in modem mathematics. Only the simplest ex-
amples are given in this book, but many others involving expressions
of the utmost complexity have been studied by mathematicians for
many years. An important reason for the-study of functions is found
134 SECOND COURSE IN ALGEBRA
i|Q, the fact that all kinds of facts and principles which we meet in the
study of nature can be expressed symbolically, by means of , functions,
and the discovery of the properties of such functions helps us to
understand the meaning of the facts. A cohipl6te understaiidiiig of
the laws of falling bodies, light, electriditj^, o'l* 'teoiind ' could tLevei^%(e
reached Without the study of <te ' Blattfeiii^W^k-'fi^'^
these phenomena suggest. ^^ bsij^irr str p.:imo(i sioiu eiiJ .noitoiiii't
When the electrician, the architect, or the artilleri'st tnefefts a
problem, he frequently must represent quantities by letters. The x
and the y may represent the measures of objects in nature, but the^
solution has become merely an operation of algebra. As students
of algebra we are not concerned with the origin of the functdon or
expression, but merely with the iiumerical determinatloitucrf -eonael
unknown or the simplification of some expression. '.Iq hnn in-^ro
^_ _ , . - , . ■: i.- •■: 'VJ (flRl!^-; ')'i-)ii^// 7/Oii« {lli\lLi>.i(
83. Graphical solution of equations in one, unknown. Jn
of the preceding sections is their use in solving equations^
in one unknown. The ideas involved can be made clear
by questions? <bii -the graphs of- sections ^ 80, ' 81, and ^82.
'>il:l 10 '<i[iidr. od^ \hsif)h ]'■: . ' ifjiri 7«')d fyw ii;uiiw fcj>i^Or'- e»ni
m ^ofno?. r> ^v;h>\ ?,o o* a^OItAL EXERCISfifi^'' ^' ^'^* iBiJ-^no^ xiT .svtjjo
ndJIi;i,.ij;n:t JxMiV/' -^iuniKf v,(ns'f ^^. i k, '•^■l^'niitJ) .^.^\vyV^vuV AaoA
X. ¥rom the graph m sectibn 80 id^termine,the value fOi tna
function ^ 3 x -f- 2 at the 'ppmt yrhev^e it^ ,g;^'f!'P,^ iC?^^^s^^.^ i^h^ ,«?7^fi9f>
3. Does this value of aa satisfy the equation 3 a;-}- 2. *i)0t2n!
8. Solve 3a5 + 2 = without reference tb th^' gtaphl'^ f''^"»«
4. What point on the. gi;^ph represents" the root of
3a; + 2 = 0? ^'^^^''
5. Irom the graph in section 81 determine the vulues ui
the function x^ ^ x — 6 at tHe points where its graph crosWs
the a;-axis. .*^- 4- '^ - + ''-^ " ' ' -^
6. Do these values of « satisfy the equation'^4^'5'— 6ii/0?
7. Solve x^-{- x—6 =="(y without referving'to' tie graphl '. ^ ^
, ,i8,„TO^,,piQip^,pp,,tia^ ngxaph repre^Qftii tbft .^ootBiipf,
F^KCTIONS .iLasrE> THEIR GRAPHS 135
9. From >tbe giapiiiin action 82 determine tlie values of
the functioixiqff!^ 5.;?^ rb 3 at the points where its graph crosses
It). Do these values of oj mak^-the functioa i^-^&As^f B
equal zero ? .3ini gi eifia
11. What method could beHsi&d' to solve the ^qiuatloii
84. The process of graphical solution. From what pre-
cedes, it is apparent tlmt tfee steps in. the graphical solution
of an equation ip one unknown are: , m^
^,, Transpose the terms so that the right member 4$ ^^(%idBi}pe
^iilGrmjpJp ^HJkM^^9V( in the, left {m^mf^er, , : , : ; , , j r , ,,, . r... ; t
The values of xfor the points wlipretfi^ gi^ph- crp,^s&»,t^,
X-axis are the real roof$ ^f the eqpation.\Q imyi dirjji8i!00 erij
The algebraic solutions of a linear equation and a quad-
ratic equation in one unknown are so simple that except
for the purpose of illustration their graphical solution is
comparatively unimportant. The algebraic solution of cubic
and higher equations, except in simple eases, is much more
difficult than the solution of the quadratic and is never
iresented\ij^s^an elementary course. For this i»^asOil and
ijoT the insiglit it gives into equations in general, the
Graphical solution of the cubic and higher equations is
i|mportant Ti!q[d: - fflnminating. For: subfe' equations if the
method of factoring fails, the graphical method is the only
method open to the student at this point in bis. progl^ss.
I, x^-Tx + S, f^jO.xo' a = 't 8ort4.[^? -8^ = 0. , )(c. - 5.)
136
SECOND COURSE IN ALGEBRA
85. Imaginary roots. An equation of the second or
a higher degree often has imaginary roots. Such roots can-
not be obtained by the graphical methods so far considered.
A study of the graphs which follow will make clear why
this is true.
Consider the following equations:
a^-4:x-5 = 0, (1 )
2^_4:^; + 4 = 0, (2)
^_42; + 13 = 0. (3)
The graphs of the functions in the left members of
equations (1), (2), and (3) are given in the accompanying
figure. The three functions differ only in their constant
terms, for 9 added to
the constant term of
(1) gives the constant
term of (2), and 9
added to the constant
term of (2) gives the
constant term of (3).
Apparently, as the
constant term is in-
creased the graph rises
without change of
shape and without
motion to the left or
to the right.
From the graph the
roots of a? — 4 x — 5 = are seen to be 5 and — 1. These
results are obtained from factoring ; a^ — 4a: — 5 = 0, or
(^x — 5)(a; + 1) = 0. Whence x=5 or — 1.
If we imagine curve (1) to move upward, the two roots
change in value and become the single root of, curve (2),
F
V ~!
\
Vf- t1
Iv 71
\ i \ / ' /
\* ' '7
3aS ttt
Cv^ - U
SjuA. 9n Z-J-i-
\5^v 6 -/^Q
Xx^^.'X yJ-l-
yyj^i-^'^ztz
\ "^ 4 •'' /
X \ N (2) ..' / X
-2 -\ 1 2 3 A
^ y4 1 y' '
-8*^4-^
F 1
FUNCTIONS AND THEIR GRAPHS
137
which touches the a;-axis at a point where x equals 2. Solv-
ing 2j2 — 42^ + 4 = by factoring gives (x — 2) (a; — 2) = 0.
Whence x=2.
If we now imagine curve (1) to move still farther upward
from its position (2), it will no longer cut the a^-axis.
Further, when the curve reaches the position of (3) it
does not cut the a;-axis at all, and hence cannot show the
values of the roots of the equation a:^ _ 4^_j_13 _ 0^ as,
in fact, it does not. The graph does show, however, that
the value of a:^ — 4a; + 13 at the lowest point of the curve
is 9. This means that for every real value of a?, positive
or negative, x^ — 4:x+l'^ is never less than 9. The graph
of (3) makes clear that no real
number if substituted for x will
make a:^ — 4 a; + 13 equal zero.
It can be shown by the method
of section 87 that the roots of
a^-4ic + 13 = are 2 + 3V^
and 2-3^/^.
Note. It required the genius of no
less a man than Sir Isaac Newton first
to observe from, the graph of a function
that two of its roots become imaginary
simultaneously. He also saw that an
equation with two of its roots equal to
each other is, in a certain sense, the lim-
iting case between equations in which
the corresponding roots appear as two
real and distinct roots and those in
which they appear as imaginary roots.
86. Imaginary roots for a cubic equation. If we attempt
to solve 2;^ — 2^; — 4 = graphically, we obtain the graph
of the accompanying figure. The curve crosses the 2;-axis at
x=2. This is the only real root the equation has ; the other
RE
F
J
]
X
'
X
_9
3 1
/
/
/
^
1
/
^
J
/
-5
/
GRA
PH
OF
^
n
xS=
X'-
-Zx
^4
"r
_
F
138 SECOND COURSE IN ALGEBRA
two are imaginary. The roots can here be obtained by fac-
toring a^-2x-4: = 0; thus (x-2)(x^'{-2x + 2} = 0.
The roots of 2:24- 2 a:;— 2 = are the two imaginary roots
of the euhic equation.
EXERCISES
As far as possible solve graphically, finding results to one
decimal place :
1.
x^-5x-9 = 0.
6. x^-4:X-^5 = 0.
2.
x^ = 4.x-5.
7. X* - 7 ic + 4 = 0.
3.
a:^ — 3 X + 4 = 0.
8. ic*-4x» + 12 = 0.
4.
a;3 + ic - 4 = 0.
9. a;* = 10x^-9.
6.
x^-lx'-2x + S =
= 0.
10. x^- 2x^ = 0.
CHAPTER X
QUADRATIC EQUATIONS
87. Solution by completing the square. An equation of
the form a7? -{- 62; -f- c = 0, where a, 5, and c denote num-
bers or known literal expressions, is called a quadratic equa-
tion. Any such equation can be solved by the method of
completing the square. This method gets its name from
the fact that in the course of the solution there is added to
each member of the equation a number making one member
a perfect square.
ORAL EXERCISES
What terms should be added in order to make the following
expressions perfect squares ?
1. x2 + 2ic + ?
5. ir2+a^ + ?
9. cc2 + |x + ?
2. x^-2x^l
6. a;2_3^_^9
10. x^-\x+l
3. a;2_6^4-?
7. £c2_^fx + ?
11. cc^ + |x + ?
4. ic2 + 8x + ?
8. a^2_4^_|.?
EXAMPLE
12. x'-\-ax-\-'i
Solve5ar^-3x
— 2 = and check the result.
Solution. 5 :r2 - 3 a: - 2
«
Transposing, 5 a;^ — 3 a:
Dividing by the coefficient of a:^, x^ — \x
Adding (— ^-^y to each member,
= 0. (1)
= 2.
or
139
140 SECOND COURSE IN ALGEBRA
Extracting the square root of each member,
X
h
±/zT-
Whence ^ = i% ± tV = 1 ^^^ ~ §•
Check. Substituting 1 for x in (1),
5- 12 -3- 1-2 = 0,
5-3-2 = 0,
= 0.
Substituting — f for x in (1),
5(-f)^-3(-f)-2 = 0,
1 + 1-2-0,
= 0.
The method of solving a quadratic equation illustrated
in the preceding example' is stated in the
Rule, Transpose so that the terms containing x are in the
first member and those which do not contain x are in the
second.
Divide each member of the equation by the coefficient of x^
unless that coefficient is +1.
In the equation just obtained^ add to each member the
square of one half the coefficient of jc, thus mahing the first
member a perfect trinomial square.
Rewrite the equation^ expressing the first member as the
square of a binomial and the second member in its simplest
form.
Extract the square root of both members of the equation^
and write the sign ± before the square root of the second
member^ thus obtaining two linear equations.
Solve the equation in which the second member is taken
with the sign +, and then solve the equation in which the
second member is taken with the sign — . The results are the
roots of the quadratic.
QUADEATIO EQUATIONS 141
Check. Substitute each result separately in place of x in the
original equation. If the resulting equations are not obvious
identities^ simplify each until it becomes one,
EXERCISES
Solve by completing the square and check real results as
directed by the teacher :
1. £c2 _^ 4£c -h 3 = 0. 9. 6ic2 + ic - 35 = 0.
2. 0^-2^-8 = 0. 10. 12x2 -25a; + 12 = 0.
3. ^2 - 5 - 2 = 0. 11. 3 ^2 _^ 8 ^ + 4 = 0.
4. x2-h2=-3a:. 12. x-{-2 = ^x\
5. 2x2 + 5^2 + 3 = 0. 13. x-4 + x2 = 6-2x2 + 8x
6. 3x2 + 7x-6 = 0. ,^^+14
7. 2^2- 3x- 5 = 0. 4 '
8. 5x2-7a!-6 = 0. 15. (3x - 2)^ + (x - 1)^ = 1.
In Exercises 16-25 obtain results to three decimal places :
16. £c2-12x + 31 = 0.
Hints. By applying the rule we get
X = 6 + VS,
and X = 6 — Vs.
, From the table on page 274, V'5 = 2.236.
Hence we get the result, to three decimals,
a; = 8.236 and 3.764.
4~ 3
17.
x^-x-^l^^.
22.
Sx'
18.
n^ + 2 = 5n.
23.
x'-
19.
3x2_l2x + 9 = 2.
Q
20.
5a;2 + 8x + 2 = 0.
24.
21.
10-2x' = x^-Tx.
25.
(y-\
25. (2/+l)(2/ + 2) = 32/(y-4).
142 SECOND COUESE IN ALGEBRA
26. £c* - 3 cc^ + 2 = 0.
Note. This is not a quadratic equation, but many equations of
this form can be solved by the methods applicable to quadratics.
i-
Solution.
x4 - 3 a:2 + 2 = 0.
xi-Sx^=-2.
x^-3a:2+|=-2 + | =
x^-^=±h
a:2 = 2 or 1.
Whence
x = ±V2, ±1.
Check as
usual.
Note. It should be particularly observed that the equation of
Exercise 26 has four roots instead of two. In general an equation
has a number of roots equal to its degree. Thus the equations in
Exercises 29 and 31 have six and eight roots respectively, although
some of them are imaginary and the student at present should not
be required to find them at all.
27. ic*-5a;2 + 4 = 0. 32. 6 cc^ - 11 ic^ -j- 3 = 0.
28. x' - ISoc^ + 36 = 0. , 33. 4m« - 15 = 7m«.
29. x^-j-S = 9x\ 34.. x^-4.x^-\-S = 0.
30. x^ - 7x^ = 8. 35. 2/' + 7/ = 8.
31. a;«- 17 0-^ + 16 = 0. 36. m*-4 V2m2 + 7 = 0.
37. (x^ - 2xy- 7 (x'' - 2x) = -12.
Solution. Let x^ — 2 x = y.
Substituting y for x^ — 2xy we obtain
Solving, .y = 3 and 4.
Then a;^ - 2 a: = 3.
Whence a: = 3 and — 1.
Also a;'* — 2 .r = 4.
Whence • x = l± V5.
QUADRATIC EQUATIONS 143
In Exercises 38-41 do not expand, but solve as in Exercise 37 :
38. (x-iy-\-4.(x-l)=5.
39. (x^-4.xy~5(x'-4.x)-24: = 0.
40. S(2f-^Syy-7(f + S2j)=20.
41. (x - ^y+ Jx-^-5 = 0.
42. aa^ + 6x -f c = 0.
Solution, ax^ + bx + c = 0.
x^ ■{ — X = — -'
a a
.^b , / bV J2 ^ j2_
a:2 + -a;+ — - =—--- = — -
a \2 a/ 4 a^ rt 4
— iac
, b Vft^ — 4 ac
2a 2
___6_ \^b^ - '^ ac _ - b ±Vb^ - 4: ac
^~ 2a'^ 2 a ~ 2 a
43. x^-\-Sax-h2a^ = 0. 46. *cc2-a(J+l)a? + ^t2=0.
44. 2f + b2j=6b\ ,. h ^ b ,
45. bx^ + x = l+bx. 2a 2a
48. cx^ -{(^-itd)x + cd = 0.
49. a^ic^ — aj (a + ^) Va^ + aZ> = 0.
50. 3a.*-aV-2a^ = 0. „ 3 , ^ 1 . 9a^
53. 7cc^ — 2aa:-h-a2 = -- —
51. 6ay-7ar2/ + 2?-2 = 0. 4 2 4
52. (3ax + 4&)2 = (2ax-&)l 54.. aV4-3aicV^-10Z» = 0.
55. 9A:2-cc2 = 2;ta;V3.
56. (aa; + fey^4-3(r^r + i)+2 = 0.
57. («V - 3 ax)2 = 14(aV - 3 aa^) - 40.
144 SECOND COUESE IN ALGEBRA
88. Solution by formula. In Exercise 42, above, the
general quadratic ax^ -{-hx-\- c = Q has been solved and
the roots found to be
x = (F)
2a ^ ^
The expression (i^) is a general result and may be
used as a formula to solve any quadratic equation in the
standard form aa^ -\- bx -\- c = 0, in which a, b, and e may
represent numbers, single letters, binomials, or any other
form of algebraic expression not involving x.
If the numbers a, b, and c are such that the expression 6^ — 4 ac
is negative, the formula contains the square root of a negative num-
ber, which is a kind of number not yet fully considered in this text.
In the exercises that follow it will be assumed that only such numer-
ical values of the literal coefficients are involved as will not make
6^ — 4 ac negative. A discussion of the case here ruled out will be
found in Chapter XII.
EXERCISES
Solve for x by formula and check the results as directed
by the teacher :
1. 4:X^-^SX = S.
Solution. Writing in standard form,
4ar2 + 8a:-3=0.
Comparing with ax^ + bx + c = 0, evidently 4 corresponds to a,
8 to 6, and — 3 to c. Substituting the values in (F) gives
8±V64-4«4-(-3)
2-4
8±V64 -H48 -8±Vll2
Check as usual.
-8db4V7
8
8 8
QUADKATIC EQUATIONS 145
2. x"- ex -16 = 0. 9. 3£c + 4 = x\
3. x^-x-l = 0. 10. Ua^ -Wx = 26.
2 -^ 2"
5. 4a;2 + 5a^-3. 12. x + f = fa;2.
e.2x-\-4. = x\ 13. f -i£c2z=2x. •
7. X = 1 - xi 14. 6x == 1 + 2£c2
8. 4^2 - lliT - 60 = 0. 15. .^2 + 2cc = 8 - a-^
16. 3j9V^j9x + 4.
Solution. Writing in standard form,
Here a = 3 j?^, b= — p, and c=— 4.
Substituting these values in the formula (F),
^ - (-p)±^(-py - 4 • Sp^(- 4)
2.3jo2
_ p±^p^ + 48 jp^ _ JP =fc 7jp _ 4 -1
Check as usual.
17. x''-2kx-Sk'' = 0. „^ .^,, 13A;i 871-
25. 6 h^ H
18. 3a^ = 7ax + 6x^.
19. 4ic2 ^ A;a! - 14/^2 =0. «« ,2 . . . 1
26. ic^ _^ ^>£c + - = 0.
20. bx = 12x^-b\
.2^2
c
21. a'x^-\-4.abx-{-Sb^ = 0. 27. - + a^ + 3^ = 0.
22. 12pV-4j9ra;-?- = 0.
23. Sa^x^-\-Sabx-h4.b^=0. 28. — - - _ i = 0,
24. X - 3 VJ - — = 0.
2
ic 29. 13 a%^x^ = 9 ^/ -f 4 aV.
146 SECOND COURSE IN ALGEBRA
30. 2x^ + 5x = cx^-{-Scx-\-S.
Solution. (2 - c) a;2 + (5 - 3 c) a; - 3 = 0.
Here a = 2 — c, 6 = 5 — 3 c, and c = — 3.
Hpi
ace
3 c - 5 ± V(5 - 3 c)2 + 12 (2 - c)
XlCJ
2(2-c)
3c-5±V49-42c + 9c2
4-2c
_3c-5±(7-3c)
4-2c
2 1 ^ 6c-'l2
4-2 c' ^^ 2-c' "^ 4-2c
Check
as usual.
31.
P'
-\-x'' = 22)x-j-2x-2p.
32.
(X
-lf = a(x-x^).
33.
x' + 2x = kx^-\-kx-l.
34.
Sx + ax^ = 2(x^ -{-ax- 1).
35.
cc^
— ex — ^(ax — ac).
36.
4aV + 4c2 = 16aV + cV.
37.
x«
-a^x^=Sb'x^-Sa%\
38.
a^
-i-x'' = 2ax-\-b\
39.
12jpq = x^-}-4:qx-Spx.
40.
x^
- aV + a%' = ^>*a;l
or -3.
89. Comparison of the various methods. Four methods
have been given for the solution of tlie quacbatic equation :
(a) Solution by factoring.
(J) Solution by graphing,
(c) Solution by completing the square,
(c?) Solution by formula.
In practice, the first and the last of these methods are
most convenient.
QUADRATIC EQUATIONS 147
When an equation with integral coefficients can be
solved by factoring, as explained in Chapter III, the roots
are always rational numbers.
For example, 3x2 + 2a;-8 = factors into (3 a; - 4) (a: + 2) = 0.
Hence the roots are x = ^ and — 2.
An inspection of the formula
-h± Vb^ -4:ac
2a
shows that, if a, b, and c are integers, one always obtains
roots involving radicals unless the expression under the
radical sign, 5^ — 4 ac, is a perfect square. In this case,
however, the values of the roots are rational numbers.
For example, in the equation 3a;^ + 2a: — 8 = the value of the
expression 6^ — 4 ac is (2)^ — 4 • 3 (— 8) = 100, which is a perfect
square. Hence the roots of 3 a:^ + 2 a: — 8 = are rational num-
bers, since the radical term, in the roots can be expressed as a
rational number.
Hence to determine whether a quadratic equation of
the form aa^ -j-bx-^ c = can be solved by factoring into
ratienal factors, we have the
Rule. Compute the value of b^ — 4: ac for the equation.
If the result is the square of an integer, the left member
of the equation can be factored and the roots are rational.
ORAL EXERCISES
Determine which of the following can be solved by factoring :
1. x" + lOcc + 25 = 0. 6. x^ - 4x - 4 = 0.
2. x'-^x- 4.00 = 0. 7. 3 a;2 _ ic + 4 = 0.
3. x2_7^_3 = 0. ' 8. 3a^2-5ic + 8 = 0.
4. 2x2_f_3^_^-^^Q 9^ 2£c2-10£c-25 = 0.
5. 2cc2 + 3x + 2 = 0. 10. 3x^-h2ax-a^==0.
148 SECOND COUESE IN ALGEBRA
REVIEW EXERCISES
Solve the following by the method best adapted to each
1. ^x" -7x - 10 = 0. 6. x' - 7x2 + 12 = q
2. 5x2 + 14.r + 8 = 0. 7. 6x«-20 = 2a;l
3. 4x2 4-3x-2 = 0. S.x^-x = 0.
4. x^ -1.1 X- .84 = 0. 9. x^ = x.
5. 8x2-f-2V5x-15 = 0. 10. x^ + 64x = 0.
11. (a^-3x)'^-2(x2-3x)=8.
12. 5^2- 9^ + 3-0. ,_ x^-l x^ + 1 _
x' + l
14. x2 + xV2-V6 = xV3. 19 14 X - 1 ^ X + 1
a; + 4x-2 x + 2
13. .03x' + .01x = .l.
x2 + xV2-V
X — 1 X -\-l
^^' X - 6 ' ^^ 2ic -1 1-x x + 1
X — 1 cc — 2ic + 2
X — 6 _ X — 4:
xHi ~ 2a; + 6* ^1. pqx^ - rqx -\- psx = rs.
x±l x + 3^22 22.x'-2ax-\-a'~b' = 0.
24. (x2 + 5x + 2)2-6(«2-f5x + 2)=16.
3x-l x + 1 48
25.
5.^ + 1 x-1 (5x + l)(l-a:)
a;»-10a^ + l
27. 15^2 -1.95x4- .054 = 0.
28. 10-9x = 7x''.
29. x^ + 961 a^ = 62 ax.
30. (x8 + 4)^ = 4 + a:»:
31. (2x + 3)(x-4) = (3x-8)(4x-l).
32. (3x - 2)(x - 5) = (4x - 3)(x + 1).
QUADRATIC EQUATIONS 149
33.
34.
35.
36.
X -j- b X + o
38. 17a; = 6ic2-10.
39. 7x*-4t2x'-\-25 = 13x''-5x^-\-4.00.
40. 2k^x^ -h Skx = 5(3kx -1).
1 2
1
x-^7 5
3-a:
^
a: + 2
ar^ + Scc-
4"^x + 4
£C + 3
x-1
x-2
x^-3x-\-2 x-2
x-1
5x-ll
l-3a^
2x-l
x-1 '
cc + 2
0^ + 1
x + 12 ^
, x + 7
41.
37j9^a: = 210^2-pV.
42.
.x + 1 ic-1 2ic-f-l
x^-1 x^ + 1 (x^^x + l)(x^-
-0^ + 1)
43.
3x-2 2 5(^ + 1)
0,-1 '3~^6-4^_i
44.
cc*-a;2^4(9 + ^').
45.
^ , 4:k k-2x
e — Za3 c
46.
47
4-30^ 2x 2x-\-5
X 1-x 3-2ic
6cc2^x-5 12x=^H-2x-4
4ic2-£c-2 8a^2-2£c + 4
48. 8.4x2 4- .005 a: -.15 = 0.
49. 324 a;^ + 1936 = 1665x1
50. 3x2 + .7x=.2.
51. 3 aV _ 10 abx - 25b^ = 2(2abx - aV).
150 SECOND COURSE IN ALGEBRA
1 1 1
52. ^ -7^ +
53
54.
x'-Zx + 2 x^ + x-^ ?,-2x-x^
x + 2 2 3(2-0;)
(x + 4.)(x- 1) (x + 4)(3 -x)~ {x-'d){l-x)'
2(ar^-2)-l ^ x'' + 2
x^-1 ~(x'-2)-4.'
55. (3x - 7)(2x -^1) - (5x -^ 2)(2x - S) = 0.
. 56. (x - 2)2- (1 - 2x)(Sx-\- 5) = 5 - (1 - 2x)(3a3 + 2).
57. (a32-2cc-2)2+2(cc2-2x-2)-3 = 0.
58. (£c2 - 4ic + 1)2 _ 4(cc2 - 4x) - 16 = 0.
59. (2a;-^>)2 = ci(2£c-Z>) + 2a2.
60. « + - = w H
x m
PROBLEMS
1. Separate 42 into two parts such that the first shall be
the square of the second.
2. Find two consecutive odd numbers whose product is 143.
3. The sum of the reciprocals of two consecutive numbers
is \^. Find the numbers.
4. The time, in hours, required for a trip of 216 miles waa
6 greater than the rate, in miles per hour. Find the time and
the rate.
5. The altitude of a triangle is 6 feet less than the base.
The area is bQ> square feet. Find the base and altitude.
6. One leg of a right triangle is 9 feet shorter than the
other, and the area is 45 square feet. Find the three sides.
7. The area of a trapezoid is 180 square feet. One base is
4 feet greater than the altitude, the other is three times the
altitude. Find the two bases and the altitude.
8. A rectangle whose area is 27 square inches has a perim-
eter of 21 inches. Find the length of the shorter side.
QUADRATIC EQUATIONS 151
9. A polygon of n sides always has ^n(n — S) diagonals.
How many sides has a polygon with 119 diagonals ?
10. If AB, in the accompany-
ing figure, is a tangent to the
circle and BD is any secant, then
(ABf = BC . BD. li AB = 6 and
CD = 10, find BC.
11. A requires 4 more days than
B to do a piece of work. Work-
ing together they require 2|^ days.
How many days will each require
alone ?
12. In selling an article at $20, a merchant's per cent of
profit is 9 greater than the original cost of the article in
dollars. Find the original cost and the per cent of profit.
13. A 3-inch square is cut from each corner of a square
piece of tin. The sides are then turned up to form an open
box of volume 12 cubic inches. What was the side of the
original square ?
14. The number of lines on a certain printed page is 20 less
than the average number of letters per line. If the number of
letters per line is decreased by 16, the number of lines must
be increased by 15 in order that the new page may contain as
much matter as the old. How many lines and how many
letters were there on the original page ?
15. For what positive value or values of x will the product
of 0? — 5 and a? -f- 5 be 1 greater than their difference ?
16. What values of x will make the product of 4ic — 5 and
X — 2 equal in value to sc -f 4 ?
17. The length of a rectangular room exceeds its width by
2 feet. A rug placed in the middle of the floor leaves a mar-
gin of 1 foot all around. If the area of the rug is eight times
that of the margin, find the dimensions of the room.
152 SECOND COURSE IN ALGEBRA
18. A sum of |4000 is invested, and at the end of .each year
the year's interest, plus |400, is added to the investment. At
the beginning of the third year the investment afnounts to
$5230. What is the rate of interest?
19. The cost of an outing was $60. If there had been two
more in the party, the share of each would have been a dollar
less. How many were there ?
20. The dimensions of a rectangular box are expressed by
three consecutive numbers. Ics surface is 214 square inches.
Find its dimensions.
21. The radius of a circle is 28 inches. How much must this
be shortened in order to decrease the area of the circle by 1078
square inches ?
22. Two bodies, A and B, move on the sides of a right
triangle. A is now 5 feet from the vertex, and moving from
it at the rate of 10 feet per second. B is 35 feet from the ver-
tex, and moving toward it at 5 feet per second. At what time
(past or future) are they 75 feet apart ?
23. How high is a mountain which can just be seen from a
point on the surface of the sea, 40 miles distant ?
Hint. See Exercise 10.
24. The distance in feet, s, through which a body fallg from
rest in t seconds, is given by the formula s = I gf/^, where
<7 = 32, approximately. A bomb dropped from an aeroplane
struck the ground below 8 seconds later. How high was the
aeroplane at the time ?
25. If a stone be thrown vertically upward, with a velocity
of 100 feet per second, it is known that, neglecting the
resistance Of the air, its height after t seconds will be
100^ — 16^^ feet. After how many seconds will it be 136 feet
high? Explain the double answer.
26. After how many seconds will the stone of Exercise 25
return to its starting point ? Explain the zero root.
CHAPTER XI
IRRATIONAL EQUATIONS
90. Definitions and discussion. An irrational equation in
one unknown is an equation in which tlie unknown occurs
under a radical, or is affected by a fractional exponent.
Thus 3x -\-5Vx = l, x-x^ + 1 = 0, and Vx^ -3x4-2 = 6 are
irrational equations.
The chief difficulty involved in the solution of sucK
equations arises from the fact that sometimes results are
obtained which do not satisfy the given equation and
hence are not roots of that equation. A result of this
kind is called extraneous.
EXAMPLE
(a) Solve Vic — 6 — 4 =
:0.
(^)
Solve -Vcc-6-4:
= 0.
Solution. Transposing,
Vx - 6 = 4.
(1)
-Va:-6 =4.
(1)
Squaring, a; — 6 = 16.
(2)
a: - 6 = 16.
(2)
Solving, X = 22.
X = 22.
Check. V22 -6-4 = 0.
- V22 -6-4 = 0.
Vl6 -4 = 0.
-Vi6-4 = 0.
4-4 = 0,
-4-4 = 0,
lich is true.
which is not true.
It appears from a study of these solutions that state-
ments (1) differ only in the signs preceding their left
members. Consequently this distinction disappears after
»K 163
154 SECOND COUESE IN ALGEBRA
squaring, and equations (2) are identical. Since the re-
mainder of the work in both («) and (5) consists in the
solution of (2), the result obtained is really the root of
this equation. Whether the root obtained satisfies both
(a) and (5), or only one of them, can be determined only
by substitution. Hence it appears that (a) is an equation
and that (5) is not, but is merely a false statement in the
form of an equation.
In any case, all of the roots of the original equation are
sure to be among the results found, provided no factor
containing the unknown has been divided out. But no
result should be called a root unless it satisfies the origiual
equation. This means that all results must be checked.
In irrational equations, as m all the work up to the
present, it is understood that unless a radical or an ex-
pression affected by a fractional exponent is preceded by
the double sign ± it has only the one value, just like any
other number symbol.
Thus Vl6 means + 4, and not — 4.
Also 4^ means + 2, while — ^^ means — Vi, or — 2.
. If this fact is kept in mind, it is clear from an inspec-
tion of (5), above, that it could have no root, since the
sum of two negative numbers could not possibly be zero.
The method of solving equations containing radicals
is stated in the
Rule. Transpose the terms so that one radical expression
(the least simple one if there are two or more} is the only
term in one member of the equation.
Next raise both members of the resulting equation to the
same power as the index of this radical.
Combine like terms in each member and^ if radical ex-
pressions still remain^ repeat the two preceding operations
lERATIONAL EQUATIONS 155
until an equation is obtained which is free from radicals ;
then solve, this equation.
Check. Substitute in the original equation the values found
and reduce the resulting numerical equation to its simplest
form by extracting roots, but not by raising both members of
the equation to any power.
Finally, reject all extraneous roots.
EXERCISES
Solve the following for real roots and check results :
1. Vcc + 1 = 5. 7. Vic + l=V3x~5.
2.. -s/?>x-5 = 7. • 8. 2V8x = xVi.
3. ■\/2a^-h3 = 3. 9. 3 V2a: + 6 = VGx^ - 6.
4. 2 Vx + 4 = 4. 10. -\/4x + 3--v/4-3a^ = a
5. 3V2x-8-7 = 17. 11. V2x + 2=V2x-2.
6. 7 + 2a/2cc-1=:13. 12. ^iTl=V^^^.
13. 1+Vcc-2=V^.
Solution, Transposing,
Squaring,
Transposing and collecting,
Squaring,
Check. Substituting | for x in the original equation,
l + V|^=V|,
14. V9ic- 4 = Vcc.
15. V3ic-2 + 5 = Vcc4- 35.
16. V3 .'^5 + 4 — Vcc + 5 = Vd — X.
17. 3Va; + 4 + Vl-.T==3Vl-ic + 2Vic + 4.
18. Va; + 2 = Vi + V2. 19. a; -f 1 = V^ic^ + S;
156 SECOND COUKSE IN ALGEBKA
20. ir -h 4 + V^ = V^M^e.
21. V2x-4 4-Va; + 6 = VT^
22. -Wx -\- 2 -\- -Vx -1- ^3 X -\- 3 = 0.
23. V3cc+1- Vic+1= Vx-4.
24. x^- 9x^ + 20 = 0.
Note. The equation here given is not a quadratic equation, but
it is of the general type ax^^ + 6a;'» + c = 0. Here x occurs in but
two terms, and its exponent in one term is twice that in the other
term. Many equations in this form can be solved by completing the
square (compare Exercise 42, p. 143).
Solution. x^ - 9 xt + -\L = _ 20 + -S|^.
art = 5 or 4.
Whence a; = ± 5* or ±8.
Check. Substituting ± 5^ for x in the original equation,
(±5f)|_9(± 51)1 + 20 = 0,
or 52 - 9 . 5 + 20 = 0, or = 0.
Substituting ± 8 for a: in the original equation,
(± 8)i-9(± 8)^ + 20 = 0.
16 - 36 + 20 = 0, or = 0.
25. x^ + 5a;' -14 = 0. 30. 2a;^ - 7 -^sc + 6 = 0.
26. .'-10x1=11. 31...-! = 4x-i + 32.
, 32. aji - a;i - 6 = 0.
27. 6 = x^-{-x. OK 1
33 r-i — -4- i —
28. Sxi -{-5x^-\-2 = 0. SQxi 9~^'
29. x-''-17x-^-\-52 = 0. 34. 7ic^-6=2a;i
35. x^-^5x + aVa^^ + Sx - 54 = 0.
Hint. Let y = x^ ■}■ 6x.
lEEATIOKAL EQUATIONS 157
36. 3x2-4x-llV3ic^-
4a; +28=0.
37. 2x^-'^x-^--\/2x'-
-3a^-l+l = 0.
38. x^-lx-h-^x"- -2x-
-4 + 2 = 0.
39. V2x-4 + 5 = l.
40. <Ix^?>=^^\x-X
41. 12£c^-27ic'^ = 20£c3 _
-45.
42. 15-2Vir2_^9 = x^ + 9.
43. 27(8 + 27a;*-ic^) = 8xi
V2a^-V8
44. 4Vic + 2 = 3Vic + 4.
45.- ^x + 4 = Va^ - 8.
2V3 \/2a: + 12
V2x-3 3V3
48. Vx + 1 = Va; - 4 + 1.
49. 4(cc - 5) - 2(aj _ 5)* - 2 = 0.
50. ^x^'x = {x -\- 3) V9^.
^, Va; + 2- ^8-^ 8
51. , — , = o *
V8 - cc Vx + 2 3
52. 5x^ + «7x^ - 200 = 16 - 3icl
53. 4 V^- 21 ^^ + 27 = 0.
54. ix" - 5a; + 2)* - h{x? - 5cc + 2)^ + 6 = 0.
55. V 17 + 2 V3 + s + V7+?- 5 = 0.
56. 3cc2-a;-22 = 6V3ic'-cc- 6.
57. If # = TT-v -J solve for I and cr.
>^
58 . VtT = \x V3 and ^ = 2 7rr^, express A in terms of x.
59. If ^^•= 2 7'^ and C = 2 vrr, express C* in terms of a.
60. If ^ = — - V 3 and a; = -|- r V3, express A in terms of x.
61 . If i!L = 3 ?-^ and a = J ?■ v 3 + V2, express i^ in terms of a.
CHAPTER XII
IMAGINARIES
91. Definitions. When the square root of a negative
number arose in our previous work, it was called an
imaginary, but no attempt was then made to use it or to
explain its meaning. The treatment of imaginaries was
deferred because there were so jnany topics of more im-
portance to the beginner. It must not be supposed,
however, that imaginaries are not of great value in mathe-
matics. They are frequently used in certain branches of
applied science ; and it is unfortunate that symbols which
can be employed in numerical computations to obtain prac-
tical results should ever have been called imagmary. By
such a name something unreal and fanciful is suggested,
to obviate which it has been proposed to call imaginary num-
bers orthotomic numbers^ but this name has been httle used.
The equation a:^ + 1 = 0, or x^ = — l^ states that a:; is a
number whose square is —1. By defining a new number,
V— 1, as one whose square is —1, we obtain one root for
the equation x^ + l = 0.
Similarly, V— 5 is a number whose square is — 5. And,
in general, V— rv is a number whose square is — n. Obvi-
ously, V— 6 means something very different from V5,
and V— n from Vn.
The positive numbers are all multiples of the unit -f 1,
and the negative numbers are all multiples of the unit — 1.
Similarly, pure imaginary numbers are real multiples of
the imaginary unit V— 1, as 2V— 1, — 5V— 1," and 5V— 1.
168
IMAGINARIES 159
Furthermore v — 4 = Sv— 1 ; -y — a^ = a^/— 1 ; and
V^=V5.V^. __
The imaginary unit V— 1 is often denoted by the letter
I ; that is, 3 V— 1 = 3 z.
If a real number be united to a pure imaginary by a
plus sign or a minus sign, the expression thus obtained is
called a complex number.
Thus — 2 + V— 1 and 3 — 2V— 4 are complex numbers. The
general form of a complex number is a + hi, in which a and h may
be any real numbers.
Note. Up to the time of Gauss (1777-1855) complex numbers
were not clearly understood and were usually thought of as absurd.
The situation reminds one of the time when negative numbers were
similarly regarded, and the veil was removed from both in about the
same way. It was found that negative numbers rfeally had a sig-
nificance — that they could be used in problems that involve debt,
opposite directions, and many other everyday relations. The inter-
pretation of imaginary numbers is not quite so obvious, and is not
considered in this text. But as soon as it was seen that an interpre-
tation was possible the ice was broken, and it needed only the insight
and authority of a man like Gauss to give complex numbers their
proper place in mathematics.
ORAL EXERCISES
Express as multiples of V— 1, or i :
1. V-16.
5. 3 V- 6.
2. V-25.
6. 2^-c.
3. V-6^2_
7. V3.V-3.
4. 2V-3.
8. V2.V-5.
11. ^-x'-2x-l.
12. V-?/ + 62/-9.
92. Addition and subtraction of imaginaries. The funda-
mental operations of addition and subtraction are performed
on imaginary and complex numbers as they are performed
on rational numbers and ordinary radicals of the same form.
160 SECOND COURSE IN ALGEBKA
Thus
and 5V^-3V^=2V^.
Also (8 + 5V^)+(4-2V^) = 7+3V^.
Similarly, (a + bi) + (c + di) = a -{■ c ■\- (b + d) i
Simplify :
1. 3V^+2Viri.
2. 4V^4-V^.
EXERCISES
7. V-18 + V^.
3. V- 25 - V- 16.
4. V^ + V^.
5. V-4 4-V~16.
6. (_ 8)^ + (- 32)^.
8. 4V-25cc2_2V-36ar».
9. 24-3V^ + 6-5V^.
10. 7VI^2-5a + 4V^ir;^.
11. (Sa- 6ib)-]-(a-\-{h).
12. 4-8i + 16- 3V^.
13. 5 4-3V-49cc2_6V^6x2H-4.
14.^ 18 - 3(- 1)* + 6(- 25)* + 4.
15. 5 V^ + 3 V^ _ V- 27 + 2 V^s.
16. (8-5 V^Il6)-(7 + 3 V- 25).
17. 4 V- 9 a* - 6 a^ V- 16 + 3
18. (5x-6iy)-(Sx-\-2iy).
6 + 5 V- 54.
93. Multiplication of imaginaries. By the definition of
square root, the square of (— ny is — n.
Therefore {y/'^f = -l,
(V3i)3 = (V3i )VZi = _ V3T.
(v::i )^ = (vzi )2 ( vzT)2 = (_ 1 ) (_ 1) = 1.
IMAGIKARIES
161
'^)
To multiply V— 2 by V— 3, we first write
v32'^V2.viri,
and V^ = V3.V3l.
Then V32 . V^g = ( V2 . V:^) ( Vs • VI3)
== V6 . V3i . V^=-V6.
Similarly,
(2 Vi:5)(- 3 Vr2)= (2V5 . vC3) . (- 3V2 • a
= _6VlO(-l) = 6VlO.
In general, if V^^ and V— 5 are two imaginaries whose
product is desired, they should first be written in the form
Va . V— 1 and y/b • V— 1 and the multiplication should only
then be performed. This method will prevent many errors.
In this connection it must be clearly understood that one rule
followed in the multiplication of real radicals (see page 117) does
not apply to imaginary numbers.
In the case of ordinary radicals we have
But the product of two imaginaries like V— 2 • V— 3 does not
equal V(— 2)(— 3), for this equals Vc. We have seen above that
In multiplying two complex numbers, write each ex-
pression in the form a ± hi and proceed as in the following
EXAMPLE
Multiply 4 + V^ by 2 - V^.
Solution. 4 + V^ = 4 + V5 . V^,
2 _ VITe = 2 - Ve . V^.
(1)
12)
Multiplying (1) by (2), 8 + 2 Vs • V^ - 4 V6 • V^- V30 (- 1).
Rewriting, 8 + 2 V^ - 4 V^ + V30.
162
SECOND COUKSE IN ALGEBRA
EXERCISES
Perform the following Indicated multiplications and simplify
results :
17. Vm + n ' V— 7n — n.
18. (3+v:ri)(3-V:^).
19. (5+V^)(5-V^).
20, (3-4V2^)(3 + 2V2^).
21. (4+v:^)(5-V:r3)..
22. (5-3i)(6-5V2i).
23. {a + ib){c + id).
32. (2 + 2 aAITs)' - (2 - 2 V^
25. {a-\-ib)(a-ib).
26. (-i + iVirs)^
27.(-i-iV33;.
28. (-2 4-2V^)'.
29. (-2-2V^)'.
30. (x — iy)\
31. {a + ibf-{a-ibf.
Sf.
33. (a + ^ Vl - b^) {a - I Vl - h'^).
34. Determine wh ether the su m an d the product of the
numbers 5 + 6 V— 2 and 5 — 6 V— 2 are real.
35. Determine whether the sum and the product of the
numbers 2 + V— 3 and 2 — V— 3 are real.
94. Division of imaginaries. One complex number is
the conjugate of another if their product and their sum
are real. Thus a + hi and a — hi are conjugates. Conjugate
IMAGINAEIES
163
complex numbers are used in division of imaginary ex-
pressions as conjugate radicals are used in division of real
radicals.
In case either the numerator or the denominator of a
fraction is imaginary or complex, the division may be
performed as in the following
1. V^-v-V2.
EXAMPLES
V-6.V2 2V-3
OVIUUIUU.
V2 (V2)^ 2
2. V8--V^
^.
Solution.
V8 V8.V-2 4ti
Vi:2 (V^)' - 2
-21
3. V-6--^
7:^2.
Solution.
V- () V- 6 . V- 2 V6 •
'•^^•'_,/3
v:^ C^/^)'
-2 ^^•
4. 3-(2+-
s/33).
Ortll-l<-Trt«
3 _ 3 (2 - V- 3)
6-3V-3
2 +
V-3 (2 + V-3)(2-V-
6 - 3 V- 3
-3) 4 + 3
The method of the above examples is stated in the
Rule. Write the dividend over the divisor in the form of
a fraction.
Then multiply both numerator and denominator of this
fraction hy the simplest expression which will make the new
denominator real and rational.
Reduce the result to its simplest form.
164
SECOND COUESE IN ALGEBRA
1.
V- 12 .
-V2.
2.
Ve^v
^-2.
3.
2VE^
4V-1.
4.
V-9^
-V-1.
5.
i^Vi:
~3.
6.
4-^V-
"5.
7.
Vs^v
'-2.
8.
(- 49)*
^(-64)*.
9.
^by^^
^-y.
10.
V— m -
f-V— 7i.
21.
. (3 + 2i)(l
EXERCISES
Perform the indicated operations :
11. (- 6/>a;)*-f-(-5a;)t
12. [(-a^6■)i_(-_^2^^i]^(^_^)i
13. 3-(l+V^).
14. 2-(l-V^).
15. 2V^^(V^H-6).
16. 2V^-f-(3V^ + 3).
17. (_i_v^)^(-i+v:r3).
18. (l+2^)--(3-4^).
19. X -r-(£C + ^^/).
20. {(i-^ih)^{c-^id).
22. Is 4- 1 - V- 3 a cube root of - 8 ?
23. Does a;2_6ic4-12 = 0ific = 3± V^^s ?
24. Does X = f(V-lO), y=- |-(V-10), satisfy the sys-
tem x'-xy- 12 2/-2 = 8, x" + xy-10y^ = 20 ?
95. Equations with imaginary roots. The student should
now be able to check the solution of an equation which
has imaginary roots.
EXERCISES
Solve the equations which follow, and check the results :
1. x' + 4a; + 8 = 0. 6. cc^ + a; + 1 = 0.
2. 3,2 _ 8a; + 24 = 0. 7. oi^ - a; + 1 = 0.
3. ic2 + 3a; + 9 = 0. 8. 5a;=^ - 6a; + 14 = 0.
4. a;2-6a; +16 = 0. 9. 6 a;^ + 10 a; + 21 = 0.
5. 3a;2 4-2a; + 4 = 0. 10. 3ar^ + 16a; + 21 = 0.
IMAGINARIES 165
11. x^ = l.
Hints. If x^ = 1, x^ - 1 = 0.
Hence (x - 1) (x^ + x + 1) = 0.
Then . x - 1 = 0,
and x2 + X + 1 = 0.
12. ic^ + 1 = 0. 15. x^ = l. 18. x^ = 64.
13. x^ = 8. 16. x^ = 9. 19. x^ = 64.
U.,x^=-27. 17. £c' = l. 20. x^ = -125.
21. How many square roots has any real number? cube
roots ? fourth, roots ? sixth roots ?
22. What do the preceding exercises suggest regarding the
number of Tith roots which any real number has ?
23. 27 ic^- 8 = 0. 27. a^« + 7a;^-8 = 0.
24. 64a;«f 125 = 0. * 28. 3ic* + 16cc2 + 21 = 0.
25. x'' - 2ic2 -.8 = 0. 29. 27 o;^ -I2x''- 64. = 0.
26. x^-\-x^-2x-2 = 0. 30. 6x^ + 21x^ + 9 = 0.
31. 25£c^ + 40a^2_^64 = 0.
32. (x^ 4- 4)(.x2 4- 3 a; + 7) = 0.
33. (^2 + 20^)2 + 15 (x^ + 2a:) + 54 = 0.
34. (x^ + 5xy + 9(x^ + 5x)- 112 = 0.
35. Solve a: + ?/ = 4, x^ — 3 xi/ — y^ = — 39, and check the
results.
36. Solve X -\- 2 y = 4z, 'if — X = 0, and check the results.
37. Solve x^ -\- if = 4:, X — 1/ = 6, and check the results.
Note. Long before the time of Gauss, mathematicians had
performed the operations of multiplication and division on com-
plex numbers by the same rules that they used for real numbers.
As early as 1545 Cardan showed that the product of 5 -f V— 15
and 5 — V— 15 was the real number 40. However, he was not
always equally fortunate in obtaining correct results, for in another
place he sets -( — \ — -) = — — = = -•
4\ >J 4/ V64 8
166 SECOND COURSE IN ALGEBRA
Even the rather complicated formula for extracting any root of
a complex number was discovered in the early part of the eighteenth
century. But all these operations were purely formal, and seemed
to most mathematicians a mere juggling with symbols until Gauss
showed clearly the place and usefulness of such numbers.
96. Factors involving imaginaries. After studying radi-
cals we enlarged our previous notion of a factor and, with
certain limitations, employed radicals among the terms of
a factor. Now in a similar manner, with like restrictions,
we extend our notion of a factor still farther and use
imaginary numbers as coefficients or as terms in a factor.
For example, x^ + 1 may hereafter be regarded as factorable,
for a^-\-l=2^-(-l) = {x+V^){x-^'^).
Similarly,
4 :^^ + 9 = 4 2^2 _ (_ 9) _ (2 ^^ + 3 V^) {2x - 3 V^)
and a:2 4- 6 = a:2 - (- 6) = {x + V^^)(2: - V^).
Further, a^ -1= (^x -1} (^x^ -{- x -hi) . Hitherto the trino
mial a:^-{-x-\-l has been regarded as prime ; but the stu-
dent can easily prove that a^ + x+lis equal to the product
(2; + l4.jV33)(a: + | -^V^). Therefore a^-1 has
three factors, these two and x — 1.
Note on the use of imaginaries. We have explained the laws
of addition, subtraction, multiplication, and division for imaginary
(and complex) numbers and have made some use of them. It is
largely because imaginaries obey these laws that we call them
numbers, for it must be admitted that we cannot count objects
with imaginary numbers. Nor can we state by means of them our
age, our weight, or the area of the earth's surface. It should be
remembered, however, that we can do none of these things with
negative numbers. We may have a group of objects — books, for
example — whose number is 5 ; but no group of objects exists whose
number is — 5, or — 8, or any negative number -wliatever. If it is
asked, How, then, can negative numbers and imaginary numbers
IMAGINARIES 167
have any practical use ? the answer is this : They have a practical
use because when they enter into our calculations and we have
performed the necessary operations upon them and obtained our
final result, that result can frequently be interpreted as a concrete
number like those dealt with in ordinary arithmetic. Moreover, if
the result cannot be so interpreted, it is, in applied mathematics
at least, finally rejected.
In that part of electrical engineering where the theory and meas-
urement of alternating currents of electricity are treated, complex
numbers have had extensive use. Their employment in the difficult
problems which there arise has given a briefer, a more direct, and a
more general treatment than the earlier ones where such numbers are
not used.
In theoretical mathematics complex numbers have been of great
value in many ways. For example, numerous important theorems
aboat functions are more easily proved under the assumption that
the variable is complex. Then, by letting the imaginary part of the
complex number become zero, we obtain the proof of the theorem
for real values of the variable. Indeed, the student need not go very
far beyond this point in his mathematical work to learn that, if e is
2.7182 + (see page 253), e ""^l^ + e-"^/-i is equal to the real number
1.082 + • At the same time he will learn also how such a form arises,
and something of its importance. In a way which we cannot now-
explain, even so involved an expression as (a + 16)'' + *'' has in higher
work a meaning and a use. If the student pursues his mathematical
studies far enough, that meaning and use and a multitude of other
uses for complex numbers will become familiar to him. But the
numbers which we have learned in this book to use, namely fractions,
negative numbers, irrational numbers, and compl-ex numbers, com-
plete the number system of ordinary algebra, for it can be proved
that from the fundamental operations no other forms of number
can arise.
CHAPTER XIII
THEORY OF QUADRATIC EQUATIONS
97. Formation of equations with given roots. According
to section 34, the equation (x—2)(x—S)=0 has the roots
2 and 3. In general, the equation (a; — r^) (a; — rg) = has
the roots r^ and r^, because either of these numbers,
when substituted for x, satisfies the equation. Hence we
can always find an equation whose roots are two given
numbers r^ and r^ by setting the product of the binomials
x — Ti and x — r2 equal to zero.
For example, an equation whose roots are 3 and 4 is seen in
(a; - 3) (x - 4) = 0, or a;^ - 7a; + 12 = 0.
EXERCISES
Form an equation whose roots are the following :
1. 2,3. 5. 2,|. 9. 1 + V3, 1-V3
2. 3, 7. 6. 5, f 10. 3 ± V7.
3. 1, - 3. 7. - I, 1 . 11. 2 + V^, 2 - V^.
4.-2,-5. is. -|,-|. 12. -7+V:r5, -7-V^.
13. i+Vf,i-\/|. 17. 1,-1,2.
14. i+vr|,i_vri. i8..i,i,.i. _
15. 2, 3, 4. 19. Vir2, - V- 2, 2.
Hint, (x - 2) (x - 3) (x- 4) = 0. 20. 1 + V2, 1 - V2, 3.
16. 1, 3, 5. 21.1,- 1, V^, ~ VI^i.
22. V2 - V3, V2 + V3, - V2 - V3, - V2 + Vs.
168
THEORY OF QUADRATIC EQUATIONS 169
98. Relations between roots and coefficients. By direct
multiplication we obtain from
(x-r^)(x-r,)=0 (1)
the equation x^ — (r-^ + r^) x + r^r^ = 0. (2)
Since r-^ and r^ are the roots of (1), it appears from an
inspection of (2) that the quadratic equation
x^-\-hx-\-c=0
has the roots r-^ and r^^ provided h = — (r-^-\- r^) and c = r^r^.
For example, we may form at once the equation whose roots are
4 and 9, as follows :
x-2-(4 + 9)x + 4 • 9 = 0, or a:2 _ 13 a; + 36 = 0.
Similarly for the cubic equation x^ + bx^ + ex + d = 0, whose
roots are r^, r^, and r^, we have (x — r^)(x — r^) (x — r^) = 0.
Then b =-(r^ -{- r.-, -\- r^),
c = r^r^ + r^r^ + r^r^,
and d=— r^r^r^.
ORAL EXERCISES
Form equations whose roots are the following :
1.2,9. 4.-3,-5. 7. 2V2, -2V2.
• 2. 4,5. 5. -7,2. 8. 3 + Vt, 3 - Vz.
3. -1,6. 6. V3, -V3. 9. I+V2, I-V2.
We will now show the relations which exist between the
roots and the coefficients of the general quadratic equation
a2^+ hx-\- c=Q. By section 88 the roots of ax^ -\-hx+ c =
are
and r« = .
2a 2 2a
170 SECOND COUKSE IN ALGEBEA
Adding r^ and rg, we have
- b + Vja _ 4 ac - J - Vj2 _ 4 ae
-26
2a
h
'■i ' '^- 2«
a
Therefore — (r^ + ^2) ~ " *
Multiplying r^ by r^^ we have
-4ac
4 a2 4 ^2 a
Therefore r^^ = - •
These results may be expressed verbally as follows:
For the equation ax^ + ftjr + c = 0,
/. The sum of the roots with its sign changed is -•
c ^
II. The product of the roots is - •
These relations frequently afford the simplest means of
checking the result of solvmg a quadratic equation, as
illustrated in Exercises 18-31, below.
EXERCISES
Form the equation whose roots are the following :
solution. -:(l-i\=-L^^., (?\/i\=_5 = £.
\2 5/ 10 a \2/\ ;5/ 6 a
Hence the required equation is
a;2 - — a: - 5 = 0, or 10 a;2 - 7 a: - 12 = 0.
10 5
THEORY OF QUADRATIC EQUATIONS ITl
2.
hh
3.
h-h
4.
H, - H-
5.
4.41, 1.59.
6.
2 + 3V3, 2-
-3 Vs.
7.
4 + V- 2, 4
-V-2.
8.
f + Vt, 1—
V7.
9.
HiV-3,
i-iV:
11.
±V-1
12.
1
a, —-'
a
13.
S a 5 a
2' 2
1 + 1.
14. 1 + a, 1
a.
10. If ± V3.
a -{-b a — h
^^' T' — TT'
a — a -\-
16. 6, 8, f
17. - 4, 4, f
Solve the following equations by the use of the formula
and check the result by the use of I and II, above r
18. cc2_5ic + 6 = 0.
19. x'-x-^^O.
20. x2-2x-4 = 0.
21. x^- 9a; -10 = 0.
22. a;2 + 2a; + l = 0.
23. ^2 + 8a: + 16 = 0.
24. £c2 + 5a; + 5 = 0.
25. 2r»2_|_3x- 6 = 0.
26. 3cc2+3£c-5 = 0.
27. 5a;2-6a; + 10=0.
28. x^ 4- £c + 1 = 0.
1
29
- + - + -
3 ^4^5
0.
30. - + 4x-7 = 0.
31. Qx'-^.x-
0.
Find the value of the literal coefficient in the following :
32. a;^ + 2 i^ — c = 0, if one root is 2.
33. 0?^ — a? — c = 0, if one root is 6.
34. a^ — ca- — 70 = 0, if one root is 7.
35. ar^ + 2 ^>x + 20 = 0, if one root is - 4.
36. a?^ — 8 a; + c = 0,if one root is twice the other.
37. a;^ + 7 a: + c = 0, if one root exceeds the other by 2.
38. x^ + 11 X + ^ = 0, if the difference between the roots is 10.
_^,+V62-
- 4 ac
2^
-h-^h^-
- 4 ac
172 SECOND COURSE IN ALOEBEA
99. Character of the roots of a quadratic equation. It
is often desirable to determine whether the roots of a given
quadratic equation are real or imaginary, rational or irra-
tional, equal or unequal, without solvmg the equation.
This can be accomplished by use of the formulas for the
roots of the quadratic ax^ + 5a; + c = :
(1)
r,= ^^^^~^-^. (2)
These expressions are seen to differ from each other only
in the sign preceding the radical. The expression
ft2 - 4 ac,
which appears under the radical sign, is called the dis-
criminant of the quadratic. The only way in which r^
or ^2 can be a complex number is for the discriminant
to be negative. If all of the coefficients a, 5, and c a re
rational, r^ or r^ can be rational only when Vj2 — ^ac is
rational ; that is, when the discriminant is a perfect square.
Hence if a, 6, and c are rational, an inspection of (1)
and (2) shows that the following statements are true:
/. If y^— 4iac is positive and not a perfect square, the
roots are real, unequal, and irrational.
For example, ina:^ — 8a; + 2 = the discriminant h^ — 4:ac equals
(— 8)* — 4 • 1 • 2 = 56, which is not a perfect square. The roots of the
equation are the real, unequal, and irrational numbers 4 ± vl4.
//. If y^ — ^ac is positive and a perfect square, the roots
are real, unequal, and rational.
For example, in the equation 2 a;^ — 3a; — 9 = the discriminant
62 — 4 ar equals (— 3)^ — 4-2-(— 9)= 81, wfiich is a perfect square.
The roots are the real, unequal, rational numbers — J and + 3.
THEORY OF QUADRATIC EQUATIONS 173
///. i/ ^2 _ 4 ^^ {g 2^ro, the roots are equal.
For example, in the equation 4:x^ — 12x + 9 = 0, the discrimi-
nant P - 4 ac equals (- 12)2 _ 4 • 4 • 9 =^ 144 - 144 = 0. The only
number which satisfies this equation is f , so in one sense the equa-
tion has only one root. But since the left member has two identical
factors each of which affords the same root of the equation, it is
customary to say that the equation has equal roots.
IV. If IP' — 4: ac is negative^ the roots are imaginary.
For example, in the equation 2 x^ — b x -\- ^ = the discrimi-
nant 62 _ 4 ac equals (- 5)2 - 4 • 2 • 4 = 25 - 32 = - 7. The roots
of the equation are the conjugate imaginaries and
+ 5-V~7 ^
4
ORAL EXERCISES
Determine the character of the roots of the following
equations without solving:
1. ic'-f 5aj + 6 = 0. Q. x'-{-x+l=0.
2. 3a;2-4ic+l=0. 7. x^- 8a; +16 = 0.
3. 2x2- 6a;- 3 = 0. 8. 2x2h-3x + 5 = 0.
4. 2x'-3x-2 = Q. 9. 4x2-4x+l=0.
5. 5x2-5x + 4 = 0. 10. 3^2 -2x- 2 = 0.
EXERCISES
Determine the value of k which will make the roots of the
following equations equal :
1. x2-A;x-M6 = 0.
Solution. a = l,h=-k,c = lQ.
Hence ' h'^-^ac = k^- 64.
In order for the roots to be equal, U^ — ^ac must equal zero.
Therefore P - 64 = 0.
Whence 1- = ± 8.
174 SECOND COURSE IN ALGEBRA
Check. Substituting 8 for k in the original equation,
a;2 - 8 X + 16 = 0.
Whence x = 4, only.
Similarly, substituting — 8 for k,
x^ + 8 X + IQ = 0.
Whence x = — 4, only.
2. x^-^kx-j-16 = 0. 7. 9cc2 + 30£c + A; 4- 9 = 0.
3. x^ - lOcc -\-k = 0. S. 4.kx^- 60ic + 25 = 0.
4. 2x^-hSx + k = 0. 9. 9k^x^-S4:X-\-4:9 = 0.
5. x^-3kx + S6 = 0. 10. 49x2 - (A: + 3)a; + 4 = 0.
6. 3£c2 + 4A;x + 12 = 0. 11. (k"" -\- 5) x^ - SO x -{- 25 = 0.
Determine the values of a for which the following systems
will have two sets of equal roots :
12. ^' ^ ^ "^^ 13. ^' + ^' = "^'^ 14. ^2/ = «^'
'2/ = ic + a. *y = x + l. *a;H-2/ = l.
100. Number of roots of a quadratic. In section 87 we
found that every quadratic equation has two roots.
Up to the present we have assumed that a quadratic
equation can have no other root than the ones found by
the method of completing the square. This fact can be
proved as follows:
Proof. If we writ/e the equation ax^ + 62; + c = in the form
x2 ^^, ^ ^ E - and substitute therein from (I) and (II) on
a a
page 170, we get x^ — (r^ + r^) x + r^r^ = 0. This can be factored
and written as (x — r{) (x — r^) = 0. Now if any value of x different
from rj and rg, say r, be a root of this equation, such a value when
substituted for x must satisfy the equation (x — r^) (x — r^) = 0.
Hence (r — r^)(r — r^) must equal zero. By assumption, however,
r is different from r^ and r^. Consequently neither the factor r — r^
THEORY OF QUADRATIC EQUATIONS 175
nor r — r^ can equal zero, and therefore their product cannot equal
zero. This proves that no additional value r can satisfy the equation
x^ — (r^ + rg) a: + r^^ — 0. As this equation is but another form of
03?" + 62; + c = 0, the latter has only two roots.
101. Factors of quadratic expressions. Let r^ and r^ be
the roots of a^ + 5a: + c = ; then
I c
or a7^-\- hx -\- c = a(x — r-[) (x — r^.
Therefore the three factors a, x — r-^^ and x — r^ of any
quadratic expression can be found if we first set the ex-
pression equal to zero (see section 34) and solve the equa-
tion thus formed. Obviously the character of the roots so
obtained will determine the character of the factors. Hence
by the use of the discriminant l^— 4:ac we can decide
whether the factors of a quadratic expression are real or
imaginary, rational or irrational, without factoring it.
EXERCISES
Determine which of the following expressions have rational
factors :
1. a;2-3cc-40. 3. Tic^-Qx + lS. 5. 120? -11x^1.
2. 2a:2 + 5aj-7. 4. ^^.x'-x-lO. 6. 5ic2 + 3cc-20.
7. 3^2 - 9a- + 28. 9. x" - 2«cc + {a? - b^.
8. 33 A^ - 233 A, - 6. 10. ahx" - {W- + a')x + ah.
Separate into rational, irrational, or imaginary factors :
11. 2x2 + 5x-8.
Solution. Let 2 a:2 + 5 a: - 8 = 0.
Solving by formula, x = — ^^ ^ =
176 SECOND COURSE IN ALGEBRA
rr. -5+V89 ^ -5-V89
1 hen r, = and r„ =
4
Therefore 2x^-{-5x
= 2\x
-S+Vsoir _5-V89'
|(4x + 5-V89)(4a; + 5 +V89).
12. x'-lx-T. 21. £c2 + 7£f + 8.
13. x'-Ax- 1. 22. x^-\-x + l.
14. x^-}-2x-^2. 23. x^ + 1.
15. a^2 4.4^_9. 24. £c2 + 9.
16. 4a;2_l2£c-9. 25. cc^ - 2 ao: + a'^ - ^».
17. 25x2 + 20ic + 4. 26. a:^ + 6acc + 9'a2_4^
18. 6;z^2^14£c-40. 27. 4£c2 + 4aic + a^ _ 4c.
19. 10-9a;-9a;2. 28. x"" - ^ ax + 4: a^ -\- c.
20. 10x2 4-12-26cc. 29. a^t-^ _,_ ^^ _|_ ^
30. x^ — xy-j- 5x — 2y + 6.
Solution. Let x^ — xy •{■ 5 x — 2i/ +6 = 0.
Then x^ + (6 - y)x - 2 y + Q = 0.
Solving for x in terms of y by the formula,
(5 - y) ± V (5 - ?/)2 - 4 (- 2 y + 6)
_ -5 + y±-y/y^-2y + l
~ 2 ' '
Whence x =— 2 and y — 3.
Therefore x^ - xy -\- 5 x - 2 y -\- Q = (x -\- 2)(x - ?/ + 3).
31. Sx^ - 6xy -\-14:X - 4:y + S.
32. x^-xy-2f-\-Sx-6y.
33. ar^ - 4xy - ?/ + Sy- - 2 - cc.
CHAPTER XIV
GRAPHS OF QUADRATIC EQUATIONS IN TWO VARIABLES
102. Graph of a quadratic equation in two variables.
Before solving graphically a quadratic system, the method
of graphing one quadratic equation m two variables must
be clearly understood.
EXAMPLES
1. Construct the graph of ic^ = 3 y.
Solution. Solving the equation for y, y = — .
o
We now assign values to x and then compute the approximate
corresponding values of y. Tabulating the results gives :
If x =
6
4
3
2
1
- 1
— 2
-4
-6
then y =
12
^/
3""
1
i
h
t
¥-
12
Using an a;-axis and a y-SLxis
as in graphing linear equations,
plotting the points correspond-
ing to the real numbers in the
table, and drawing the curve
determined by these points, we
obtain the graph of the adja-
cent figure. The curve is called
a parabola.
The graph of any equation of
the form x^ = ay is a parabola.
Y
\
p
AR
AB
)Li
^
1
^
c
mi
IPHQ]
7
^
X
•i—
3y
Y
J
\)
q
f
V
J
A
/
X
r\
s
1
/
X
-2
j-
1
> I
\
Y'
1.77
178 SECOND COUESE IN ALGEBRA
2. Graph, the equation xy -{- S = 0.
8
Solution. Solving for y, y =
X
Assigning values to x as indicated in the following table, we
then compute the corresponding values of 2^ :
If x =
-6
-5
-4
-3
-2
-1
-i
\
1
2
3
4
5
6
8
then y —
1
f
2
1
4
8
16
-16
-8
-4
-!
-2
-1
-i
-1
Proceeding as before with the numbers in the table, we obtain
the two-branched curve of the figure below, which does not touch
either axis. The curve is called a hyperbola.
Y
y
Q
HI
fPE
RB
OL
A
A
Y
H
X
X
-7
-^
-.
l-X
]
.
I 3 ^
t
yA
H
U^
GRAPH OF
-2
X
^:2/H
■8-
=0
f
r
_J
The graph of any equation of the form xy = K is a
hyperbola. The curve for xy — K (Jr=any constant) is
always in the same general position ; that is, if K is posi-
tive, one branch of the curve lies in the first quadrant and
the other branch in the third. If A' is negative, one branch
lies in the second quadrant and the other in the fourth.
GRAPHS OF QUADRATIC EQUATIONS 179
3. Graph the equation x- -\- y^ — X^.
Solution. Solving for y, y = ± Vl6 — x^.
Assigning values to x as indicated, in the following table, we
obtain from page 274 the corresponding values of y :
If x =
-5
-4
-3
-2
-1
1
2
3
4
5
theni/ =
±3V^
±2.64
±3.46
±3.87
±4
±3.87
±3.46
±2.64
±3V^
For values of x numerically greater than 4 it appears that y is
imaginary. The points corresponding to the pairs of real numbers
in the table lie on the circle in the accompanying figure. The center
of the circle is at the origin,
and the radius is 4.
The graph of any equa-
tion of the form oP- -\- tf^^ r^
is a circle whose radius is r.
This can be proved from
the right triangle FKO,
If P represents any point
on the circle, OK equals
the 2;-distance of P, KF
equals the ^/-distance, and
OF equals the radius. Now ~0K^ ■\-KF'^ =0F'^ \ that is,
o?'-\-y^ — r^. It follows, then, that the graphs of 2^-|-?/2=9
and a^ + «/2 = 8 are circles whose centers are at the origin
and whose radii are 3 and V8 respectively. Hereafter,
when it is required to graph an equation of the form
2^ 4- ?/2 = 7^, the student may use compasses and, with the
origin as the center and with the proper radius (the square
root of the constant term), describe the circle at once.
In all the graphical work which follows, the student will save
time by obtaining from the table on page 274 the square roots or
cube roots which he may need.
Y
^A
M
'
Q
H
k.
/
r,s
•■ V
2
>
/
\
C
IRC
LE
X'
\
X
K
-
]
, I
> J
i
V
/
\
k
-2
J
/
N
K
^
l>
Y
GRAPH OJ
i'
^■'f2/i = 16
r
180
SECOND COURSE IN ALGEBRA
4. Graph the equation I6x^ -\-9y^ = 144.
Solution. Solving, y = ± ^ V9 — x^. Proceed as in Example 3
If x =
-4
-3
-2
-1
+ 1
+ 2
+ 3
+ 4
then y =
±^v^
±2.98
±3.77
'±4
±3.77
±2.98
±^V^
For any value of x numerically greater than 3, y is imaginary.
The points corresponding to the real numbers in the table lie on
the graph of the adjacent figure.
The curve is called an ellipse.
The graph of any equation
of the form of ax^ -\-by^ = c
in which a and b are unequal
and of the same sign as c is
an ellipse.
Note. These three curves —
the ellipse, the hyperbola, and the
parabola — were first studied by
ihe Greeks, who proved that they
are the sections which one obtains
by cutting a cone by a plane. Not
for hundreds of years afterwards did anyone imagine that these curves
actually appear in nature, for the Greeks regarded them merely as
geometrical figures, and not at all as curves that have anything to
do with our everyday life. One of the most important discoveries of
astronomy was made by Kepler (1571-1630), who showed that the
earth revolves around the sun in an ellipse, and stated the laws
which govern the motion. Those comets that return to our field of
vision periodically also have elliptic orbits, while those that appear
once, never to be seen again, describe parabolic or hyperbolic paths.
The path of a ball thrown through the air in any direction,
except vertically upward or downward, is a parabola. The approxi-
mate parabola which a projectile actually describes depends on the
elevation of the gun (the angle with the horizontal), the quality of
the powder, the amount of the charge, the direction of the wind, and
various other conditions. This makes gunnery a complex subject.
Y
y^' H^
Y- 3- "Si
^ 3 I
-: «> e
2 ^
t V \
t
tL Lx
-2 - 2
i.± m
, "^ ELLIPSE /
^ 2 t
\ -2 f-
\ J-
S. z
GRAPH OF >jv^ ^
16xH 91/^14'
III Y
GRAPHS OF QUADRATIC EQUATIONS 181
EXERCISES
Construct the graphs of the following equations and state
the name of each curve obtained:
1. x' = 4:y. 3, x'+i/ = 49. 5. x^ - f = 16.
2. y^-{-2x = 0. 4. x^-{-t/ = 18. 6. xij = 12.
7. xy = -6. 9. 1603^-9?/^ = 144.
8. 9x''-\-4.y^ = S6. 10. 25 x^ -\- 9 if = 225.
103. Graphical solution of a quadratic system in two
variables. That we may solve a system of two quadratic
equations by a method similar to that employed in section 44
for linear equations appears from the
EXAMPLES
1. Solve graphically [^^^_^^^ _ {j
(1)
(2)
Solution. Constructing the graphs of (1) and (2), we obtain the
straight line and the parabola shown in the adjacent figure. There
are two sets of roots correspond-
ing to two points of intersection,
which are
Note. If the straight line in
the adjacent figure were moved
to the right in such a way that
it always remained parallel to its
present position, the points A and
B would approach each other and
finally coincide. The line would
then be tangent to the parabola
at the point x = 4, y = 1.
Were the straight line moved still farther, it would neither touch
nor intersect the parabola and there would be no graphical solution
(see page 182).
Y
\
^^
\
A
^
v^
\
Q
N
\
o
N
.<2)
\
N
X
^
L
K
-^
~ 1
>
o\
L \
>
5
r
>
b\
>
^
y
1)
^
^
\
^
F'
182
SECOND COUESE IN ALGEBEA
2. Solve grapMcally i
20^ + 2/ = 12, (1)
y + 4:x = 19. (2)
Solution. The graphs of (1) and (2) are the straight line and
the parabola of the adjacent figure. These curves have no real
points of intersec-
tion. There are,
however, two pairs
of imaginary roots.
Solving (1) and
(2) by substitution,
X = JJ- +V^ or
^7^- — v — 1, and
y = l-2V^ or
1 + 2V-1.
The essential
point to be em-
phasized here is
that real roots
of a simultane-
ous system cor-
respond to real intersections, and imaginary roots correspond
to no intersections of real graphs.
ex
3. Solve graphically |
x^-f = 4,
£c - 6 = 0.
(1)
(2)
Solution. Constructing the graphs of (1) and (2), we obtain the
hyperbola and the parabola of the figure on the opposite page.
There are four sets of roots corresponding to the four points of
intersection, which are approximately
= -2.7.
1^ = 3.1. ^1^ = 1.8. ^
a: =-2.7,
y=-1.8.
3.7,
-3.1.
If the two curves had been so chosen as to intersect only twice,
their equations would have had only two sets of real roots.
GRAPHS OF QUADRATIC EQUATIOKS 188
Examples 1, 2, and 3 partially illustrate the truth of the
following statement:
If in a system of two equations in two variables one
equation is of the mth degree and one .of the nth, there
are usually mn sets of roots (real or imaginary) and never
more than mn such sets.
r
\
/^
\
^^'^
A
>■
/
\
s.
^^
Y_
m
—
.
\
^£
/
l/
^
\
/
X
'
y'
f
X
V
-^
]
\ ?
^
J
\
/
^
* —
-2
\
V
)
i!l
/
/
-\
^
—
■—
/
-4
\(
1)
y
Y'
EXERCISES
If possible, solve graphically each of the following systems
2/^ = 4 cc,
x" 4- tf = 25,
2. -^
2/
3.
2x = 10.
4.
5.
x2/= 6,
10.
11.
x^2y
= 17,
9.
^2^4y = 17,
x-}-22/ = 12.
«2 + 7/2 = 25,
x^ + 2/^ = 9,
a; 4- ?/ = 10.
0^2 + ^2 = 36,
x^-y' = 25.
12
13
a^2 + y^=9,
^- X^ - 7/2 = 16.
x?/ = 12,
^' 2a; + 2/ = 10.
0^2 = 4 7/,
x'-\-y = h,
if + x = Z.
ic_2/V8 = 0, .
x^ = 7/8 - 9 7/.
9.
184
SECOND COURSE IN ALGEBRA
104. Graphical presentation of numerical data. A great
variety of statistics can be presented graphically in a very
striking manner. Business and commercial houses have
during the past few years used the method extensively
not only to present facts but also to aid in interpreting
them and in indicating their tendencies.
The following exercises illustrate some of the possibili-
ties in the graphical presentation of numerical data.
EXAMPLE
For the year 1912 the total income and expenses .per mile
of line of all the railroads in the United States having a
yearly revenue of one million dollars or more was as follows :
Jan.
Feb.
Mar.
Apr.
May
June
July
Aug.
Sept
Oct.
Nov.
Dec.
Income in dollars .
Expenses in dollars
930
730
970
710
1050
750
980
720
1040
740
1080
750
1120
760
1220
780
1210
780
1320
830
1220
810
1170
800
Delia
rs
>
y
y'
V.
B
Nl«
^
"»^
^
V
^
uu
^
— -
^
-^
_
-1
_
i-i
PE
NS
E-
__
_.
=»
—
—
—
-
—
—
-
—
—
—
—
—
—
—
—
OO
RC
2
T
-
-
=-
•—
V
^
"^
^
•ja
n.
F<
b.
M
ar.
A
Dr.
M
ay
Ju
n.
Ji
1.
Ai
Iff.
s«
p.
k.
N
w.
D
^c.
-
GRAPHS OF QUADRATIC EQUATIONS 185
The foregoing graph represents the given data on the same
axes.
The lowest curve on the diagram shows the profits of the
railroads. It was obtained by plotting the differences of the
numbers in the table.
EXERCISES
1. The average incomes of 155 members of a certain college
class for the first ten years after their graduation is given in
the following table :
Years after graduation
1
2
3
4
5
6
7
8
9
10
Income in dollars . . .
706
902
1199
1651
2039
2408
2382
2709
3222
3804
Graph the data. What tendencies do you note '!
2. The number of inches of rainfall during the month of
July and the number of bushels of corn yielded per acre for a
term of years in a certain locality is given in the following
table :
Year
'89
'90
'91
'92
• '93
'94
'95
'96
'97
'98
'99
'00
'01
'02
Rainfall
Corn yield ....
5.4
32
2.6
23
5.1
27
3.7
27
3.4
24
1.9
18
4.8 5.4
30 38
3.7
25
4.3
26
4.6
28
4.7
30
1.2
19
6.0
32
Plot these data on the same axes.. How do you account for
the similarity of the curves ?
3. The numbers of hundreds of telephone calls in certain
business and residential sections of New York City are given
in the following table for various hours of the day. Plot both
sets of data on the same axes and explain the reason for differ-
ences in the shapes of the graphs.
Time of day
7
8
■9
10
11
Noon
1
2
3
4
5
6
7
8
9
10
11
12
Business district .
Residence district
1
.25
2
4
35
55
73
108
76
101
65
78
52
75
48
93
47
87
40
67
38
35
38
8
36
2
32
1.5
25
1
15
1
4
1
3
RE
186
SECOND COURSE IN ALGEBRA
4. Measurements of the breadth of the heads of a thousand
students in a certain school were as follows :
Head breadth in inches
Number of students . .
,5.5
3
5.6
12
5.7
43
6S
80
5.9
131
6.0
236
6.1
185
6,2
142
6.3
99
6.4
37
6.5
15
6.6
12
6.7
3
6.8
2
Construct a graph of the above data.
5. The chest measurements of 10,000 soldiers were tabulated
as follows :
Chest measure- -»
ment in inches /
Number of soldiers
322
38
1305
39
18G7
40
1882
41
1628
42
1148
38
48
Construct a graph of the above data.
It may appear accidental that the foregoing measure-
ments should group themselves with any regularity. But if
the number of measurements of this type is large and each
is made with care, they obey a law called the law of proba-
bility. In fact the graph of the data in Exercises 4 and 5 is
a close approximation to what is called the probability curve.
The equation of this curve is y = e"^* when eJ = 2.7
approximately.
6. Construct the graph of the equation y = e'^"^ between
the values cc = — 2 and x = 1.
Hint. Let x = — 2, — |,
\, 0, etc.
Note. It is well established that physical characteristics, such
as those illustrated by the graphs of Exercises 4 and 5, obey the
law of probability. If the graph of Exercise 4 is carefully consid-
ered, it may raise the question, Do mental characteristics also obey
the law? An interesting aspect of this is given by the fact that an
increasing number of high schools and colleges assume that such
is the case, and grade, or mark, their students by a system based
GEAPHS OF QUADRATIC EQUATIONS 187
on the law of probability. Such a system assumes that if one hun-
dred or more students in any subject are examined, the number of
students and their degree of mastery of the subject arrange them-
selves according to the probability "curve shown below. Obviously
between very many students the differences in grades attained will
be small, between many others they will be moderate, and between
only a few will they be great.
In the statistical study of problems which have their origin quite
remote from each other, this curve frequently occurs, and occupies a
central 'position in the mathematical theory of statistics.
CHAPTER XV
SYSTEMS SOLVABLE BY QUADRATICS
105. Introduction. The general equation of the second
degree in two variables is ax^ + hy^ + cxy -\-dx + ey -i-f = 0.
To solve a pair of such equations requires the solution of
an equation of the fourth degree. Even the solution of
x^-{- y = 3 and ^2 _j_ ^. _ 3 requires the solution of such an
equation. In fact only a limited number of systems of the
second degree in two variables is solvable by quadratics.
By the graphical methods of Chapter XIV the student can
solve graphically for real roots any system of quadratic
equations, provided the terms have numerical coefficients.
The algebraic solution of such systems will in many cases
be possible for him only after further study of algebra.
106. Linear and quadratic systems. Every system of
equations in two variables in which one equation is linear
and the other quadratic can be solved by the method of
substitution.
EXAMPLE
Solve the svstem i , * * ^ „ '
-^ la; + 2 7/ = 13.
(1)
(2)
Solution. Solving (2) for y in terms of x,
1^-x
y 2 '
Substituting — - — for y in (1),
(8)
.^-..f«r)=r.
(4)
From (4), 2 x" _ 13 x - 7 = 0.
188
(5)
SYSTEMS SOLVABLE BY QUADRATICS 189
Solving (5), a: = 7 or — ^.
Substituting 7 for a: in (3),
13-7
= 3.
Substituting — ^ iov x in (3),
y--
13 + Jj
2
= 6i
The two sets of roots
^vQx = l,y--
= 3, and ;
r =-
-i,y = 6|.
Check. Substituting
7 for X and 3 for y in
(1)
and (2),
49 - 42 =
= 7,
«
7+6 =
= 13.
Substituting - i for
X and 6f for
3/ in (1)
= 7,
= 13.
and
(2),
EXERCISES
Solve the following systems, pair results, and check each
set of roots :
1.
x-\-y = 5,
2.
2x-^y = l.
3.
x^ -\-xy = 15,
x + y = ^.
4.
x'' + 2xy = 21,
x + 2y = ^.
5.
x' + 2y = ll,
2x-y = 2.
A
^_10:r = 6.
x4-3y = 13.
8
2 ar2-a;y = 70,
o»
ic+-4y = 23.
Q
^2 + 25?^ = 108,
V •
4s-3^=6.
10.
5^-352^ = 22,
4s + 2^ = 2.
11.
5^ +. ^2 ^ 169,
2s + ?J = 22.
12.
s^ + 2 ^=^ = 27,
4^-25 = 18.
SECOND COURSE IN ALGEBRA
13.
s^ + Sts-\-t" = 44,
2s-t = 0.
22.i + | = a,
14.
s^-\-ts + t^ = 12,
s-\-t = 2.
1 X 5
• 2/ 2~4*
15.
16.
2s^-ts -{-t^ = 16,
2s-t = 5.
6 4_2
s t~ S'
4:X -\- y = 6.
o. ^ + 13
4.^ + ^^=28, 6- + 1 2
• 2a:2 + 3a^2/-98. ^ + 4_1
-4-^ = 2,
2/ .T 6
19. -4-^ = 2, 32/ + 2x = 2.
?/ a?
3^5 + 22/ = 5. 26 ^' + '^' + 2^^ = 40,
* s-\-t + 2 = 0.
on ^ 1 1
Sx-y = 2.
y — X = V2,
X- 2, 98 ^' + y' + 4x + 6./ = 40,
10_10^3 29^ + ^^^ = ^^
y X 2 • ^Z^ + x^zzzieo^.
107. Homogeneous equations. If both the equations of
a system are quadratic, an attempt to solve it by substi-
tution usually gives an equation of the fourth degree. In
most cases such an equation could not be solved by fac-
toring, and at the present time its solution by any other
method is beyond the student. With certain types of
SYSTEMS SOLVABLE BY QUADRATICS 191
systems, however, which occur more or less frequently, we
can employ special devices and avoid the solution of any
equation of higher degree than a quadratic. Among these
systems are the so-called " homogeneous " systems.
An equation is homogeneous if, on being written so that
one member is zero, the terms m the other member are of
tHe same degree with respect to the variables.
Thus x^ + y^ = xy and x^ — oxy-{-y^ = are homogeneous equa-
tions of the second degree. 2x^ -{• y^ = x'^y — 3 xy"^ is a homogeneous
equation of the third degree.
The system \ { ^ „ ' )► is a homogeneous system.
Systems like i ^ „ „ , ' ^ are often called homogeneous
1,2 a:^ — 2/2 4- 4 a;y = 6 J
systems, but strictly speaking they are not. As will be seen, the
method of solving such systems is about the same as the method
of solving a homogeneous system. Hence they are classed with
homogeneous systems.
108. Systems having both equations quadratic. Occasion-
ally when both equations are quadratic the terms which
occur in the two are so related that the elimination of
terms involving both variables or of the constant terms
can be performed. The following system is of such type:
EXAMPLES
x"
1. Solve the system r; , ^2/ " 2, (1)
Solution. First eliminate xy by addition (§ 47),
(1) -4, 4 a:2 - 4 a:y = 8. (3)
(2) + (3), 7x2 = 28.
Whence j: = ± 2.
192
SECOND COURSE IN ALGEBRA
Substituting + 2 for x in (1), 4 — 2 ?/ = 2, whence y = \.
Substituting — 2 for x in (1), 4 + 2 ?/ = 2, whence ^/ = — 1.
Therefore a: = 2, - 2,
and ' ?/ = 1, — 1.
These values may be checked as usual.
The method of solving a system in which every term in each
equation except the constant terms is of the second degree is
as follows :
-xy-\-'iif = ^, (1)
x^-\-y' = 10. (2)
2. Solve
Solution. First we combine the two equations to obtain a homo-
geneous equation :
(1) -5, 5 xy + 15 2/2 :== 30.
(2) -3, 3 a:2 + 3 2/2 = 30.
(3) - (4), - 3 x2 + 5 xy + 12 y2 = 0.
Solving (5) for x in terms of y,
and
Substituting ^y iox x in (2),
From (8),
Substituting from (9) in (6),
4
x^^y,
4«
9 ?/2 + ^,2 ^ 10.-
ar=±3.
16 7/2
Substituting - —^ for a; in (2), — -^ + y'^
From (10),
Substituting from (11) in (7),
10.
y=±^Vi0.
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
When X =
3
-3
+ tVio
-^VIo
then 2/ =
1
- 1
-^ VlO
+ '^Vio"
Each pair of values can be checked as usual.
SYSTEMS SOLVABLE BY QUADRATICS 193
A quadratic system in which one equation is homoge-
neous is easier to solve than the system of Example 2, as
can be seen from what follows:
3. Solve the system |^. _^ ^ _ 5 ^ ^ 3^ ^2)
Hints. Solving (1) for x, x = 2 </, , (3)
and ^ = \- (^)
o
We may now substitute from (3) in (2) and from (4) in (2), solve the
resulting equations, and then complete the solution as in Example 2.
EXERCISES
Solve, pair results, and check each set of real roots :
Zi/-\-x' = 2d>. 2x^-^xy + if=().
'^3.^ + 2.^ = 7, 2a^-.^ + 2^ = 12,
2x -^xy = -^. 8. 2x''+xy-{-2f=^.
^if-xy = 2,
^* 2 2/=^ + 3 0^1/ = 38. 9 20^2 + 2/2 = 3a;y,
x^ + ^xy = lQ.
y"^ — xy = 2 x^^
' 2x^-{-xy = 16. _ 52_3^^^4
5 x''-Sy-{-S = 0,
''']^-,y=,r
y^-\-x'-25. s^ + 2st-\-4.t^ = lS,
6. Solve Example 3 completely. " t^ + S — — 2st
109. Symmetric systems. A system of equations in
X and 1/ is symmetric if the system is not altered by sub-
stituting X for y and 1/ for x.
Thus x^ + ^2 _ g and x + y = 11 is a symmetric system, but
x^ — y^ = Q and z -{- y ~ 11 is not.
194 SECOND COURSE IN ALGEBRA
Certain symmetric systems or systems which are nearly
symmetric can be easily solved by the method of substi-
tution. Of such the following are types:
1x1/ = 4:. \x±y=2. l2a;±«/ = ll.
A few other systems which are symmetric or nearly so
are more easily solved by certain special methods. The
following list contains typical systems, and the methods
applicable are given in Exercises 1, 10, and 12.
EXERCISES
1. Solve <^ ^^^ )^{
\x2j = 6. (2)
Hints. These equations can be combined in such a way as to obtain
definite values ior x + y and x — ?/ as follows :
(2). 2, 2xy = 12. (3)
{l) + (3), a;2 + 2x^ + 2/2 = 49. (4)
From (4), x-\- y= ±7. (5)
(l)-(3), x^-2xy + y^ = 26. (6)
From (6), x-y = ±6. . (7)
(5) and (7) combined give four systems of equations :
rx + 2/ = 7, (8) fx-{.y=-7, (11)
-^\x-y = 6. (9) \x-y = 5. (9)
rx+7/ = 7, (8) fx^y=-7, (11)
^\x-2/=-6. (10) \x-?/=-5. (10)
The solution of systems A^ B, C, and D is left to the student.
The pairs of roots found for the four systems A, B, C, and
D will check in the original system.
.r*^ + ;//=:100, x'' + xy + 9f==19, ^ x^-\-4y^ = 20,
xy = 48. * xy = 6. ' o-y = 4.
SYSTEMS SOLVABLE BY QUADRATICS 195
9x' + t/ = ei,
xy = 10.
4.x''-{-xy + f = So,
xy = 12.
9.
6
x = -■
y
7
"3
4x^ + 252/^ = 41,
'• xy = 2.
10.
x' y'
1 = 6.
xy
x'-xy-\-y' = S,
Hint. To clear of fractions in Exercise 10 would merely increase the
difl&culty of solution. Instead we solve in a manner similar to that of
Exercise 1 (see Exercise 6, p. 78).
1 = 12.
xy
Then
1
X2
+
^ + i =
xy j/2
:25.
Whence
1
X
+ ' =
y
:±5.
In like manner
1
X
1_
±1.
— = 12. .
xy
13.
11 13
?>x^ 2y~ 6'
11 97
9ic2 ' 4y^"36*
11 5
~1 + ~5 = 7'
^^ x^ ^r 4
12.
1 1 _ 1
x y~ 2
14.
x^ x^^2/^ ''
i = 6.
^2/
XI 12 1
Hints. — 1 =
x2 xy ?/2
Then — =
xy
1
"4"
= 1,
etc
15.
x^ xy f
l + i = 7.
196 SECOND COUKSE IK ALGEBRA
110. Equivalent systems. Equivalent systems of equations
are systems which have tlie same set or sets of roots.
If the two systems
r^ ,^ = 12, (1) ^^^ r. , = 2, (8)
lx + y = ^. (2) Vx-\-y = Q. (1)
are solved, only one pair of roots, x — 4: and y = 2, is obtained
for A and the same pair for B. Systems A and B are
equivalent, though usually a system which consists of a
linear and a quadratic has two sets of roots and hence
cannot be equivalent to a linear system (see page 183,
second paragraph).
Sometimes an equation simpler than either of those given
in a system can be derived from this system by dividing
the left and right members of the first equation by the corre-
sponding members of the second. In systems like those of
the following Ust the equation so obtained taken with one of
the first two gives an equivalent system more easily solved
than the original one.
EXERCISES
Solve, using division where possible, pair results, and check
each set of real roots :
"" x + 2y^2.
4.
4a^^-y = 16,
2x + y = S.
Hint. Division gives the
equiv-
5.
R^h - 75 = 0,
alent system
Rh = 15.
x-22/ = 10,
X — y = 1.
6.
X y
^- 2y-x=-\l.
7.
x'-¥y' = 2^,
a; + 2/ = 4.
SYSTEMS SOLVABLE BY QUADRATICS 197
In the following figure, (1), (2), and (3) are the graphs of
x^ -{- y^ = 28, x^ — xy •\- i/ = 7, and x -\- y = ^ respectively.
These equations are all used in the solution of Exercise 7. The
graph makes clear in a striking way that the system (1) and (3)
is equivalent to the system (2) and (3).
JF 1
^
3)^^
S^
\
X 4^ ^
1^ ^
"^^^ =^ S
~^ "^--^^^
^'^.> l^^S
-7 ^ J-^-A
/ 1 "^
/ >\
X' ,: t ^^ :^
/-2 -1 U 1 2y 3 \
^ 4- 1 i^ ^
-yr -1 ^1 ^
^T ^ \ '3:^
4^^ y" \ %
^---^ ^
(1^.
J5
\
Y
a^«-8y^ = 35,
x — 2y = h.
X y
1,
y"^ x^
X y
152,
MISCELLANEOUS EXERCISES
Solve by any method and pair results. If any system can-
not be solved algebraically by the methods previously given,
solve it graphically.
a! 4- 2/ = 4.
• x-'dy = \,
' £C + «/ = 7.
198 SECOIS'D COURSE IN ALGEBRA
^^o A i« 0^^ + 2^2/ + 22/^ = 10,
• 5. y "^ ' 3x'-xi/-f = 51.
Q (x + yf = 9, ' x' + y = 11.
• (^_3/). = 49. ^^ ^. + ^,^ + ^. = 7,
^•.^
2aj2_^2/2^33, x2 + 2/' = 10.
-^ 072 4-0-?/ + 2/ = 0,
3 ^2 _ 8 A)2 = 40, * £c2 + ici/ + a; = 0.
5 ^2 + ^^2 ^ 81
— 4- — = 13
4i2f + 3 = 9i2|, - 0)-^^/
12i2f + i2|-^9^. 1 1
= 1.
^^' a;^ + y = 20. -^8-^=^
0^2 + 2/2 = 169, 23. ^ ^'^
a? ?/~ *
xy -\- X = 18,
ajy = 60,
24^ 40.2 + 2/2 = 289,
£C2/ = 60.
25 0^^ + 2/^ = 9,
' a; V = 20.
• x + 2/ = 5-
27. 9x -^2/=lS = ^2/-
28.4. + J = 46 = ?|^-i.
5 Tr,2- 6.8 >F| = 99.55,
TF2-Tr| = 20. _i_ + — !— = -,
20 ^-2 2/-2 4
0^2^ + 22/^ = 2, l_l = i.
^ ' 3iC2^ + 62/^ = 2, X y 12
12.
x^ = y,
xy = S.
13
x — xy = 5,
AO.
2y-\-xy = 6.
14.
x'-y' = 19,
x'-\-xy-\-7/ = 19.
4 7^2 + 7 ?/i2 = 9,
15.
2 7^2 _ 9 ^ ^,,2^
SYSTEMS SOLVABLE BY QUADEATICS 199
88, x-1
34.
30.
X — 2/ = 6.
31.
xy = c,
X -\- y = a.
32.
x-^ + y-^ = 2.
33.
3,
35.
36.
37.
PROBLEMS
2/-1
/ + y+l _13
x^-x+1 43'
4a;^-13a;y+9V=9,
^¥ — y^ = ^'
x^-\-2xy=16,
2>x'-4.xy-\-2f= 6.
a-« = ^ + 37,
x^y z^xy^ -\- 12.
(Reject all results which do not satisfy the conditions of the problems.)
1. Find two numbers whose difference is 6 and the differ-
ence of whose squares is 120.
2. The sum of two numbers is 20 and the sum of their
squares is 202. Find the numbers.
3. Find two numbers whose sum plus their product is 132
and whose quotient is 3.
4. It takes k)^ rods of fence to inclose a rectangular lot
whose area is one acre. Find the dimensions of the lot.
5. The area of a right triangle is 180 square feet and its
hypotenuse is 41 feet. Find the legs.
6. The area of a pasture containing 15 acres is doubled by
increasing its length and its breadth by 20 rods. What were
the dimensions at first ?
7. The difference of the areas of two squares is 495 square
feet and the difference of their perimeters is 60 feet. Find
a side of each square.
8. The area of a rectangular field is 43^ acres and one
diagonal is 120 rods. Find the perimeter of the field.
200 SECOND COUESE IN ALGEBRA
9. The value of a certain fraction is |-. If the fraction is
squared and 44 is subtracted from both the numerator and the
denominator of this result, the value of the fraction thus
formed is -^^. Find the original fraction.
10. Two men together can do a piece of work in 4|- days.
One man requires 4 days less than the other to do the work
alone. Find the number of days each requires alone.
11. The perimeter of a rectangle is 250 feet and its area
is 214 square yards. Find the length and the width.
12. The base of a triangle is 8 inches longer than its alti-
tude and the area is 1^ square feet. Find the base and altitude
of the triangle.
13. The volumes of two cubes differ by 316 cubic inches
and their edges differ by 4 inches. Find the edge of each.
14. The perimeter of a right triangle is 80 feet and its area
is 240 square feet. Find the legs and the hypotenuse.
15. The perimeter of a rectangle is 7 a and its area is a^.
Find its dimensions.
16. A man travels from A to B, 30 miles, by boat and from
B to C, 120 miles, by rail. The trip required 6 hours. He
returned from C to B by a train running 10 miles per hour
faster, and from B to A by the same boat. The return trip
took 5 hours. Find the rate of the boat and of each train.
17. There were 1400 fewer reserved seats at a certain sale
than of unreserved seats, and the price of the latter was
15 cents less than the price of the former. The total proceeds
were |490, of which |250 came from the reserved seats. Find
the number of each kind of seats and the price of each.
18. If a two-digit number be multiplied by the sum of its
digits, the product is 324; and if three times the sum of
its digits be added to the number, the result is expressed by
the digits in reverse order. Find the number.
SYSTEMS SOLVABLE BY QUADEATICS 201
19. The sum of the radii of two circles is 31 inches and the
difference of their areas is 155 tt square inches. Find the radii.
20. The yearly interest on a certain sum of money is $42.
If the sum were |200 more and the interest 1% less, the
annual income would be |6 more. Find the principal and
the rate.
21. A vheelman leaves A and travels north. At the same
time a second wheelman leaves a point 3 miles east of A and
travels east. One and one-third hours after starting, the short-
est distance between them is 17 miles, and 3|- hours later the
distance is 53 miles. Find the rate of each.
22. A starts out from P to Q at the same time B leaves Q
for P. When they meet, A has gone 40 miles more than B.
A then finishes the journey to Q in 2 hours and B the journey
to P in 8 hours. Find the rates of A and B and the distance
from P to Q.
23. A leaves P going to Q at the same time that B leaves
Q on his way to P. From the time the two meet, it requires
6| hours for A to reach Q, and 15 hours for B to reach P.
Find the rate of each, if the distance from P to Q is 300 miles.
GEOMETRICAL PROBLEMS
1. The sides of a triangle are 6, 8, and 10. Find the
altitude on the side 10.
Hint. From the accompanying fig-
ure we easily obtain the system
(x^ + y^ = S6,
\ (10 - x)2 + ?/2 = 64.
2. The sides of a triangle are
8, 15, and 17. Find the altitude
of the triangle on the side 17 and the area of the triangle.
3. The sides of a triangle are 11, 13, and 20. Find the
altitude on the side 20 and the area of the triangle.
UE
202
SECOND COURSE IK ALGEBRA
4. The sides of a triangle are 13, 14, and 15. Find the
altitude on the side 14 and the area of the triangle.
5. The sides of a triangle are 12, 17, and 25. Find the
altitude on the side 12 and the area of the triangle.
6. Find correct to two decimals the altitude on the side 16
of a triangle whose sides are 16, 20, and 24 respectively.
7. The parallel sides of a trapezoid are 14 and 26 respec-
tively, and the two nonparallel sides are 10 each. Find the
altitude of the trapezoid.
Hint. Let ABCD be a trapezoid. Draw CE parallel to DA and CF
perpendicular to AB.
Then
EC = 10,
and
AE = 14,
and
EB = 26- 14, or 12.
If we let EF = x, FB must equal 12 — x ; then we can obtain the
system of equations
rx2 + ?/ :::: IQO,
\ (12 - X)2 + 2/2 = 100.
8. The two nonparallel sides of a trapezoid are 10 and 17
respectively, and the two bases are 9 and 30 respectively. Find
the altitude of the trapezoid.
9. The bases of a trapezoid are 15 and 20 respectively,
and the two nonparallel sides are 29 and 30. Find the alti-
tude of the trapezoid and the area.
10. The sides of a trapezoid are 7, 9, 20, and 24. The sides
24 and 9 are the bases. Find the altitude and the area.
11. The sides of a trapezoid are 21, 27, 40, and 30. The
sides 21 and 40 are parallel. Find the altitude and the area
of the trapezoid.
SYSTEMS SOLVABLE BY QUADEATICS 203
12. The sides of a trapezoid are 23, 85, 100, and x. The
sides 23 and 100 are the bases, and each is perpendicular to
the side x. Find x
and the area of the
trapezoid.
13. The area of a
triangle is 1 square
foot. The altitude
on the first side is
16 inches. The sec-
ond side is 14 inches
longer than the third.
Find the three sides.
L B
14. A rectangular
tank is 8 feet 6 inches long .and 6 feet 8 inches wide. A board
10 inches wide is laid diagonally on the floor. What two equa-
tions must be solved to determine the length of the longest
board that can be thus laid ?
HiKTS. Let BR
the triangle AKL.
X and DE = y. The triangle DKE is similar to
CHAPTER XVI
PROGRESSIONS
111. A sequence of numbers. In all fields of mathe-
matics we frequently encounter groups of three or more
numbers, selected according to some law and arranged in
a definite order, whose relations to each other and to other
numbers we wish to study.
There is an unlimited variety of such groups, or suc-
cessions, of numbers. Only two simple types will be
considered here.
112. Arithmetical progression. An arithmetical progression
is a succession of terms in which each term after the first is
formed by adding the same number to the preceding one.
Thus, if a denotes the first term and d the common number added,
any arithmetical progression is represented by
a, a -h d, a -{• 2 d, a + ^ d, a + 4: d, ' • .
This common number d is called the common difference
and may be any number, positive or 7ieijative, It may be
found for any given arithmetical progression by subtract-
ing any term from the term which follows it.
The numbers 3, 7, 11, 15, • • • form an arithmetical progression,
Muce any term, after the first, minus the preceding one gives 4. Simi-
larly, 12, 6, 0, — 6, — 12, • • • is an arithmetical progression, since
any term, after the first, minus the preceding one gives the com-
mon difference — 6. In like manner 5, 5, 6^, • • • is an arithmetical
progression whose common difference is 1^.
204
PKOGEESSIONS 205
ORAL EXERCISES
State the first four terms of the arithmetical progression if :
1. a = 2,d = 3. 6. a = 100, d = -10.
2. a = 5, d = 4.. 7. a = 20x, d = — 2x.
3. a = 10, d= 6. S. a =x, d = 2x.
^. a = 20, d = 5. 9. a = 17 m, d = — 2 m.
5. a = 18, ^--3. 10. a=V5,d = l-\-V5.
From the following select the arithmetical progressions, and
in each arithmetical progression find the common difference :
11.1,10,19,.... 16. 8,9f,10|,....
12. 4, 12,36,.... 17. 15,3, -12,.-..
13. 19, 11, 3, . . .. 18. 6a, 10a, 14«, • • ..
14. 9, 12, 1, 16, . . ..' 19. 18 a, 14 a, 12 a, • . ..
15. 3, ^, -V-, .... 20. 5 a, 5 a + 3, 5 a + 6, . . ..
21. Sx, X — 2, —X — 4:, • . ..
22. 2x-l, X, 1, 2-cc, ...."^ ^'
23. — 5 Va, — 2 Va, Va, 4 Va, ....
24. 5V^-1, 4V^-2, 3V^-3, ....
113. The last or Trth term of an arithmetical progression.
In the arithmetical progression
a, a + d, a + 2 d, a + 3 (f, a + 4 ff, • . .
one observes that the coefficient of d in each term is 1
less than the number of the term. Hence the ?ith, or
general, term is a -\- (n — 1) d. li I denotes the nth term,
we have
l=a+(n-i)d.
206 SECOND COURSE IN ALGEBRA
EXERCISES
Eind the required terms of the following arithmetical
progressions :
1. The fifteenth term of 2, 7, 12, • • •.
Solution. Here a = 2, c? = 5, and n = 15.
Hence I = a -{■ (n —lyd
becomes Z = 2 + 14 • 5 = 72.
2. The eighth term of 1, 4, 7, • • •.
3. The eleventh term of 15, 9, 3, - 3, • • ..
4. The twenty -first term of a, 4 a, 7 <7, • • •.
5. The sixteenth term of Qcc, jr, — 7ic, • • •.
6. The sixth term of ^^-, f , |, • • ••
7. The eleventh term of — y, |-, 1, • • • .
8. The sixth term of V3, 4 V3, 7 Vs, • • •.
9. The eighth term of 3 Vs, V5, - V5, - 3 Vs, ....
10. The eighth term of 5 + a, 7+ 3 a, 9 + 5 «, . . -.
11. The twelfth and the twentieth terms of 0, — 5x + 2,
4 -10a;, ....
12. The tenth term of V^ + 3, 5 Va +7, 9 Va + 11, . . ..
13. The tenth and the twentieth terms of 9 Va — 3, 5 Va,
Va 4- 3, ....
■yj^ 5 "v £C 9 "v £C
14. The fourteenth term of -—> — - — > — r— 5 ' ' '•
z z z
15. The ninth and the twelfth terms of V3, — 7=? 3 Vs, . . ..'
V3
V iC 3 V cc
16. The eighth and the sixteenth terms of -jr- 4- 1, —75 h 2,
5 Vcc
+ 3,
17. The fifteenth term of ^^ - 12, - 5, ^^^ ^ ""^
PROGRESSIONS 207
18. The twenty-ninth term of a, a -\- d, a -\- 2 d, - - -.
19. The mth term of a, a -^ d, a -\- 2 d, • - • .
20. The (m — l)th term ot a, a -\- d, a -{- 2d, • • •.
21. The (n - 2)th term oi a, a -\- 5, a -{-10, ".
o _
22. The (n - 5)th term of —^, 3 V3, 5 Vs, • • -.
23. rindthe(7i-3)thtermofV5-l,2V5-2,3(V5-l), ....
24. Find the nth term of -> ? 2 — -? • • ••
a a a
25. The first and third terms of an arithmetical progression
are 2 and 22. Find the seventh term ; the nth term.
26. The first and second terms of an arithmetical progres-
sion are r and s respectively. Find the third term and the
nth. term.
27. The edges of a box are consecutive even integers with
n the least. Express in terms of n, (a) the sum of the edges ;
(b) the area of the faces ; (c) the volume.
28. A body falls 16 feet the first second, 48 feet the next,
80 feet the next, and so on. How far does it fall during the
twelfth second ? during the nth second ?
29. The digits of a certain three-digit number are in arith-
metical progression. If their sum is 24 and the sum of their
squares is 194, find the number.
114. Arithmetical means. The arithmetical means between
two numbers are numbers which form, with the two
given ones as the first and the last term, an arithmetical
progression.
The insertion of one or more arithmetical means between
two numbers is performed as in the following examples:
208 SECOND COURSE IN ALGEBEA
EXABIPLES
1. Insert four arithmetical means between 7 and 72.
Solution. 1= a + (n — l)d.
There will be six terms in all.
Therefore 72 = 7 + (6 ~ 1) d.
Solving, ^ = 13.
The required arithmetical progression is 7, 20, 33, 46, 59, 72.
2. Insert one arithmetical mean between h and k.
Solution. I = a -\- (n — 1) d.
There will be three terms in all.
Therefore k = h + 2d.
Solving, d = —^'
Therefore the progression is h, h -\ — , k, or 7i, — -— , k.
It follows from the above that the arithmetical mean between two
numbers is one half of their sum.
EXERCISES
Insert one arithmetical mean between :
1. 3 and 15. 2. 3 and 27. 3. h and 5 h. 4. m and n.
Insert two arithmetical means between :
5. 2 and 8. 7. 49 and 217. 9. x and 3 //.
6. 5 and 29. 8. x and x -{- 6 y. 10. h and k.
Insert three arithmetical means between :
11. 10 and 34. 13. - 12a; and Ux.
12.-5 and 31. 14. h and k.
15. Insert seven arithmetical means between —13 and 131.
16. Insert five arithmetical means between — | and ^^.
PROGRESSIONS 209
17. Insert four arithmetical means between — 2 Vs and
18 V5.
18. Insert five arithmetical means between 7 a — Sb and
13 a + 9&.
/- 27
19. Insert six arithmetical means between VS and p=-
2V3
3
20. Insert two arithmetical means between 7= and
4V3-6. 1-^
21. In going a distance of 1 mile an engine increased its
speed uniformly from 15 miles per hour to 25 miles per hour.
What was the average velocity in miles per hour during that
time ? How long did it require to run the mile ?
22. At the beginning of the third second the velocity of a
falling body is 64 feet per second, and at the end of the third
second it is 96 feet per second. What is its average velocity
in feet per second during the third second ? How many feet
does it fall during the third second ?
23. The velocity of a body falling from rest increases uni-
formly and is 32 feet per second at the end of the first second.
What is the average velocity in feet per second during the
first second ? How many feet does the body fall during the
first second ? the second second ?
24. Find the average length of twenty lines whose lengths
in inches are the first twenty odd numbers.
25. Find the average length of fifteen lines whose lengths in
inches are given by the consecutive even numbers beginning
with 58.
26. With the conditions of Problem 23 determine the average
velocity per second of a body which has fallen for 12 seconds.
27. A certain distance is separated into ten intervals, the
lengths of which are in arithmetical progression. If the shortest
interval is 1 inch and the longest 37 inches, find the others.
210 SECOND COUKSE IN ALGEBEA
115. Sum of a series. The indicated sum of several
terms of an arithmetical progression is called an arith-
metical series. The formula for the sum of n terms of
an arithmetical series may be obtained as follows:
S = a+{a + d^+(a + ^d)+ . . . J^(l-2d)-^(l-d)+l. (1)
Reversing the order of the terms in the second member
of (1),
S = l+(l-d)+(l-2d^+ . . .+(a + 2c?)4-(a + (^)-h«. (2)
Adding (1) and (2),
2^ = (a + Z) + (a + + (^ + 0+---+(^ + + (« +
+ (a + Z) = n(6)^ + Z).
Therefore S= -(« + /).
EXAMPLE
Required the sum of the integers from 7 to 92, inclusive.
Solution, n = 86, a = 7, Z = 92.
Substituting in 5 = - (a + Z),
Therefore the sum of the integers from 7 to 92 is 4257.
' ORAL EXERCISES
Using the formula S = -(a-\-l)j find the sum of the following :
1. The six-term series in which a = 2 and I =17.
2. The ten-term series in which a =1 and I = 46.
3. The twelve-term series in which a = — 12 and I = 21.
4. The seven-term series in which a = S and I = 63.
5. The twenty-term series in which a = 5 and I = 86.
6. The twelve-term series in which a = — 175 and I =125.
progeessio:n^s 211
Yt
The formula S = -(a + l) enables one to find the sum of
a series if the first term, the last term, and the number of
the terms are given. If, however, the last term is not given,
but instead the common difference is given or evident, we
can use a formula obtained by substituting m. S = -(a -\- V)
the value of I from the formula l = a-\-{n — 1) c?, on page 205.
VI
Then s = j.\a -^ a -^ (n -V)d\
or S = -[2a + (n-l)tf].
JL
EXAMPLE
Required the sum of the first fifty-nine terms of the progression
2,9, 16,-.. .
Solution, n = 59, a = 2, and d = l.
Substituting in 5 = ^ [2 a + (/i - 1) rf],
5- = ^ (4 + 58 • 7) = ^ (410) = 12,095.
EXERCISES
Find the sum of the following :
1. The first eight terms of the series in which a = 2 and c? = 4.
2. The first ten terms of the series in which a. = 8 and d = 5.
3. The first nine terms of the series in which a = 1 and
d = ll.
4. The first -fifteen terms of the series in which a = 17
and d = S.
5. The first twenty terms of the series in which a = 100 and
d=-5.
212 SECOND COURSE IN ALGEBRA
6. The first eight terms of the series 2 + 4 4- 6 + • • • .
7. The first ten terms of the series 1 + 9 -h 17 + • • •.
8. The first ten terms of the series — 8 + (— 4) + + • • • .
9. The first eighteen terms of the series 1 + 5 + 9 + • • •.
10. The first twenty terms of the series 10 + 8 + 6 -f • • -.
11. The first twelve terms of 1 + | + 2, • • ..
12. The first twelve terms of 15 + 12i + 10 H .
13. The first one hundred integers.
14. The first one hundred even numbers.
15. The first one hundred odd numbers.
16. Show that the sum of the first n even numbers is n{n + 1).
17. Show that the sum of the first n odd numbers is n^.
18. Find the sum of the even numbers between 247 and 539.
19. How many of the positive integers beginning with 1
are required to make their sum 861 ?
Hint. Substitute in the formula *S = - [2 a + (n — 1) d] and solve for n.
20. How many terms must constitute the series 7 + 10+13
+ ... in order that its sum may be 242 ?
21. Beginning with 90 in the progression 78, 80, 82, • • .,
how many terms are required to give a sum of 372 ?
22. The second term of an arithmetical progression is — 2
and the eighth term is 22. Find the eleventh term.
23. Find the sum of the first t terras of -> > -^ • • ••
a a a
24. If Z = 25, a = 1, and <^ = 4, find n and s.
25. If a = - 20, 6^ = 11, and s = 216, find n and I.
26. It d = — 9, n = 15, and s = 0, find a and I.
PEOGKESSIOiS^S 213
27. The first and second terms of an arithmetical progres-
sion are h and k respectively. Find the sum of n terms.
28. If 5 = 9 A, a = 12 - 10 h, and ti = 9, find I and d.
29. If s = 66 V3, a = - 4 V3, and ^Z = 2 Vs, find n and I.
30. A clock strikes the hours but not the half hours. How
many times does it strike in a day ?
31. A car running 15 miles an hour is started up an incline,
which decreases its velocity ^ of a foot per second. («) In how
many seconds will it stop ? (J)) How far will it go up the
incline ?
32. A car starts down a grade and moves 3 inches the first
second, 11 inches the second second, 19 inches the third second,
and so on. (a) How fast does it move in feet per second at the
end of the thirtieth second ? (h) How far has it moved in the
thirty seconds ?
33. An elastic ball falls from a height of 24 inches. On
each rebound it comes to a point \ inch below the height
reached the time before. How often will it drop before coming
to rest ? Find the total distance through which it has moved.
34. The digits of a three-digit nmnber are in arithmetical
progression. The first digit is 3 and the number is 20^ times
the sum of its digits. Find the number.
35. A clerk received $60 a month for the first year and a
yearly increase of |75 for the next nine years. Find his salary
for the tenth year and the total amount received.
36. If a man saves |200 a year and at the end of each
year places this sum at simple interest at 6^, what will be
the amount of his savings at the time of the sixth annual
deposit ?
37. Assuming that a ball is not retarded by the air, deter-
mine the number of seconds it will take to reach the ground if
214 SECOND COURSE IN ALGEBEA
dropped from the top of the Washington Monument, which is
555 feet high. With what velocity will it strike the ground ?
Hint. See Exercise 28, p. 207.
38. A ball thrown vertically upward rose to a height of
256 feet. In how many seconds did it begin to fall ? With
what velocity was it thrown ?
39. By Exercise 28, p. 207, it is seen that a falling body obeys
the law of an arithmetical progression. Show from the data of
that exercise that the general formula S = -(2 a -\- (n — 1) d)
at''
becomes the special one S = ^y which is used in physics for
all such problems.
40. A ball thrown vertically upward returned to the ground
6 seconds later. How high did it rise ? With what velocity
was it thrown ?
41. A and B start from the same place at the same time and
travel in the same direction. A travels 20 miles per hour. B
goes 30 miles the first hour, 26 miles the second, 22 miles the
third, and so on. When are they together ?
Note. In the earliest mathematical work known a problem is
found which involves the idea of an arithmetical progression. In
the papyrus of the Egyptian priest Ahmes, who lived nearly two
thousand years before Christ, we read in essence, " Divide 40 loaves
among 5 persons so that the numbers of loaves that they receive shall
form an arithmetical progression, and so that the two who receive
the least bread shall together have one seventh as much as the
others." From that time to this, the subject has been a favorite one
with mathematical writers, and has been extended so widely that it
would require many volumes to record all of the discoveries regard-
ing the various kinds of series.
116. Geometrical progression. A geometrical progression is
a succession of terms in which each term after the first
is formed by multiplying the preceding one always by the
same number.
PEOGRESSIONS 215
Thus, if a denotes the first term and r the common multiplier, then
any geometrical progression is represented by a, ar, ar^, ar^, ar\ • • • .
The common multiplier is called the ratio. It is evident
from the above that the ratio r in a geometrical progression
is found by dividing any term by the preceding one.
The numbers 2, 10, 50, 250, • • • form a geometrical progression,
since any term, after the first, divided by the preceding one gives the
same number 5. Similarly, the numbers 3,-3 V2, 6,-6 V2, • • •
form a geometrical progression, since any term, after the first,
divided by the preceding one gives the common ratio — V2.
ORAL EXERCISES
Determine which of the following are geometrical progres-
sions and state the ratio in each case :
1. 1, 3, 9, 27, .... 7. i Ve, VO, V54, ....
2. 2,4,16,.... 8. Vf, -V|,2,....
3 2, 6, 18,.... 9. 7a,S5a, 175a,....
4. 5, 1, i, •••• 10. 8V5, -2V5, V5, ....
5. 18, - 3, i . . .. U. 5x% lOa^^ 20a:^ • • ..
6. 2, 1 V2, i V2, . . .. 12. 3y\ 12y, ^Sy\ . . ..
13. Find the condition under which a, b, and c form a
geometrical progression.
State in order the first four terms of the geometrical
progression in which the first term is
14. 1 and the ratio is 4. 17. 64 and the ratio is J.
15. 3 and the ratio is 10. 18. — 243 and the ratio is ^.
16. — 3 and the ratio is 2. 19. 2 and the ratio is V3.
216 SECOND COURSE IN ALGEBRA
117. The nth term of a geometrical progression. Since
by definition any geometrical progression is represented by
a, ar, af", ar\ - - -,
it is evident that the exponent of r in any term is 1 Iqss
than the number of the term. Therefore, if t„ denotes the
nth, or general, term of any geometrical progression,
EXERCISES
1. Find the fifth term of 4, 12, 36, • • -.
Solution. Here a = 4, r = 3, n — 1 = 4.
Substituting these values in the formula ty^ = ar^^-'^,
<^ = 4.3* = 324.
2. Find the eighth term of 1, 2, 4, • • ..
3. Find the tenth term of 3, 6, 12, • • -.
4. Find the eighth term of 3, 2, |-, • • •.
5. Find the twelfth term of 7, - 14, 28, • • -.
6. Find ^^ of 15, -5, +|, ....
27 a^
7. Find t^ of 12 a", 9 a^ -^, ••-.
— 2 c* — 5
8. Find t^ of — g— , - 1, Y^» . . ••
9. Find fg of 4 V2, 4, 2 V2, . • -.
1 1 V2
10. Find t„ of
:> t:'
^' 2V2 2 -2
11. Write the twentieth term of $4.12(1.01), $4.12(1.01)^,
H.12(1.01)«, ....
PROGRESSIONS 217
12. Since the nth. term of a geometrical progression is ar'^~^,
what is the (n - l)th term ? the (n - 2)th ? the (n - 3)th ? the
(n + l)th ? the (n + 2)th ?
13. The first and second terms of a geometrical progression
are m and ti respectively. Find the next two terms.
118. Geometrical means. Geometrical means between two
numbers are numbers which form, with the two given ones
as the first and the last term, a geometrical progression.
Thus ar and ar^ are the geometrical means in a, ar, ar^, a?-^.
EXAMPLES
1. Insert two real geometrical means between 9 and 72.
Solution. There are four terms in the geometrical progression,
and a = 9, n = 4, and tn = t^= 72.
Substituting these values in t^ = ar*^-^,
72 = 9 r3.
Whence r = 2, and — 1 ± V— 3.
The required geometrical progression is 9, 18, 36, 72.
2. Insert one geometrical mean between h and k.
Solution. There are three terms in the progression, and a = //,
w = 3, and t„ = k.
Substituting these values in tn = ar"-i, we have
h — Ji-r^.
Solving, r=±A/--
Hence the progression is ^, ± ^-y/-, ^, or A, ± VA^, k.
It follows from the above that the geometrical mean between
two numbers is the square root of their product.
K£
218 SECOND COUESE IN ALGEBKA
ORAL EXERCISES
Insert one geometrical mean between :
1. 1 and a^. 3. 1 and 4ic^ 5. a and a*.
2. 1 and x^. 4.3 and 75. 6. a^ and d^.
Insert two geometrical means between :
7. 1 and icl 10. 1 and 125. 13. x"- and x^.
8. 1 and 2«. * 11. a and a\ 14. 2 and 16.
9. 1 and 27. 12. x and x\ 15. 5 and 40.
EXERCISES
Obtain progressions involving real terms only :
1. Insert two geometrical means between 21 and 168.
2. Insert two geometrical means between 15 and 405.
3. Insert three geometrical means between 3 and 243.
4. Insert one geometrical mean between 9 and 81.
5. Insert one geometrical mean between aP' and a*.
6. Insert three geometrical means between — 9 and — 144.
7. The fourth term of a geometrical progression is 16, the
eighth term is 256. Find the tenth term.
8. The second term of a geometrical progression is 4 Vs,
the fifth term is ^. Find the first term and the ratio.
9. Show that the geometrical means between a and«c are
± Vac.
10. The first and fourth terms of a geometrical progression
are a and c respectively. Find the second and thii-d terms.
11. Insert three geometrical means between h and k.
12. The sum of the first and fourth terms of a geometrical
progression is 56. The second term is 6. Find the four terms.
PROGRESSIONS
219
B D
10, find BD and DC.
13. In the accompanying figure, ABC is a right triangle
and AD i?, perpendicular to tl^e hypotenuse BC. Under these
conditions the length oi AD is always a geometrical mean
between the lengths of BD and DC.
{a) If BD = 4: and DC = 9,
find AD.
(b) If BD = S and BC = 21,
find AD.
(c) If BC = 25 and AD
14. In the accompanying fig-
ure, AB touches and AD cuts the
circle. Under such conditions
the length of AB is always a
geometrical mean between the
lengths of AC and AD.
(a) If AD = 36 and ^C = 4,
-find AB.
(b) If DC = 90 and AB = 24, find AC and AD.
119. Geometrical series. The indicated sum of n terms
of a geometrical progression is called a geometrical series. The
process of obtaining in its simplest form the. expression for
this sum is often called finding the sum of the series.
The expression for the sum is derived as follows:
Let Sji denote the indicated sum.
S = a-\- ar :{- ar^ -h •
ar'
-\-ar"-'-\-ar"
= a(l-\-r + r^ +
1
_^^«-3_^^r^-2^^n-l^^^
S„ = a
d
(1)
(2)
1-r
since the polynomial in (1) is equal to the fraction in the
parenthesis of (2) (see Exercise 18, p. 39).
a — ar"
Hence
s„ =
1-r
220 SECOND COURSE IN ALGEBRA
EXERCISES
1. Find the suni of the first nine terms of 3, — 6, 12, • • ••
„ , , . ,^ a — ar^
Solution. Sn =
1 — r
By the conditions, a = 3, r =— 2, and n = 9.
Substituting,
__3-_H-^
l-(-2)
2. Find the sum of 1, 5, 25, • • • to nine terms.
3. Find S^for 2,-4, 8, • • •.
4. Find S^ for 40, 20, 10, ••..
5. Find ^g for - 180, 90, - 45, . • •.
6. Find 5^ for I, 1,1-,....
7. Find S^^ for a% a\ a\.".
8. Fina S^ for 2 Vs, 6, 6 Vs, • • -.
9. Find >S^ for 4, 12, 36, • • ..
10. Find 5„ for 125, - 25 Vs, 25, ....
11. Find >s;,_i for 3, 12, 48, . . •.
12. Find S,, for 3, - 15, 75, ....
13. Find »S„,_2 for x, 4:x\ 16x',- - -.
14. Show that for a geometrical progression S,i = — ;-•
Hint. Substitute I for the factor rtr"-i in the last term of the
luimerator in the formula of Exercise 1.
15. A rubber ball falls from a height of 60 inches and on
each rebound rises 60% of the previous height. How far does
it fall on its sixth descent? Through what distance has it
moved at the end of the sixth descent ?
120. Infinite geometrical series. If the number of terms
of a geometrical series is unlimited, it is called an infinite
geometrical series.
PEOGEESSIONS 221
In the progression 2, 4, 8, • • • the ratio is positive and
greater than 1, and each term is greater than the term
preceding it. Such a progression is said to be increasing.
Obviously the sum of an unlimited number of terms in
such an increasing geometrical progression is unlimited.
If r >1 (read "r is greater than 1"), the sum can be made
as large as we please by taking enough terms.
In the progression 2, 1, i, i, i, Jg, • • . the ratio is
positive and less than 1, and each term is less than the term
preceding it. Such a progression is said to be decreasing.
Though the number of terms of such a geometrical pro-
gression be unlimited, the sum of as many terms as we
choose to take is always less than some definite number.
The sum of the first three terms of the series 2 + 1+^+:^+^
+ • • • is 3^ ; of four terms, is 3| ; of five terms, is 3| ; of six terms,
is 3^f ; of seven terms, is 3|^, etc. Here for any number of terms
the sum is always less than 4.
The limit of the sum of the series 2 + l4-i+i+i+t's"+*"
can be seen by reference to the following diagram :
D E^FGX
i— -r-i
Here the terms of the above series are represented by intervals on the
line AXj which is four units in length; that is, AB = 2, BC — 1,
CD = I, DE= 1, EF= I, FG = ^V etc.
A study of the several intervals reveals the fact that as we pass
from A toward X each new interval is just one half of the then unused
portion of AX. It follows, then, that the distance laid off gradually
approaches, though it never equals, AX, for at no time is an interval
adjoined which is more than one half of that which then remained.
Therefore there will always be an unused interval between the
extremity of the last interval used and the point X.
Since the sum of the intervals laid off approaches but never quite
equals the interval AX, we have in the above diagram an illustration
of the fact that the sum of the series 2+l+i + ;|^ + ^ + TV+ •••
approaches, yet always remains less than, 4.
222 SECOND COURSE IN ALGEBRA
a — ar^
The formula for the sum of a geometric series S^ —
may be written as the difference of two fractions,
1 — r 1 — r
For the series 2 + 1 + J + | + • • • it follows that
Now (1)2=1, (1)8=1, (-1-)".= ^^, .(J/ = 3:V Conse-
quently (|)^ becomes very small if n is taken very great.
Therefore 2(|)^, the numerator of the last fraction in (3),
decreases and approaches zero as n increases without limit.
Since the denominator of the fraction remains ^ while
the numerator approaches zero, the value of the fraction
decreases and approaches zero as n increases. Then if
>S^oo denotes S^ where n has increased without limit, we
may write 9
Soo approaches » or 4.
2
In general, if r < !('' if r is numerically less than 1"), the
numerical value of fraction approaches zero as n
1 — r
increases without limit. Under such conditions the formula
S,, = -z becomes aSoo = -z •
1— rl— r \ — r
This means that for r numerically less than 1, S^
approaches •> but for any definite value of n it is
always numerically less than this number.
Hence whenever we speak of the sum of such a series
we mean the limit which the sum approaches as n increases
indefinitely.
PROGRESSIONS
223
EXERCISES
Find the sum of the following series :
8. 5 + V54-I4-
1. 4 + 1 + 1 + . .
Solution. So
Substituting, So
i-i
9- l + ; + 3 +
(x>l)
2. l + ^ + i + ....
3. 3 + (-l) + i + ---.
4. 5-|-(-2)+A + ....
5. 2 + V2 + H .
_ 5a 5a
6. 5a + — + — +....
7. l + x + a;2+ .... (x<l)
10. .121212-.
Hint. .121212 =
11. .666...
12. .151515....
13. .3232....
14. 25.2727...
15. .71515....
16. .3108108....
17. A flywheel whose circumference is 5 feet makes 80
revolutions per second. If it makes 99% as many revolu-
tions each second thereafter as it did the preceding second,
how far will a point on its rim have moved by the time it is
about to stop ?
Note. In the study of geometrical progressions we have seen
that the sum of the infinite series 1 + a; + a:^ + a;^ + . . . is a definite
number when x has any value less than 1. But it has no finite value
when X is equal to or greater than 1 ; that is, we have an expression
which we cannot use arithmetically unless x has a properly chosen
value. If we were studying some problem which involved such a
series, it would be a matter of the most vital importance to know
whether the values of x under discussion were such as to make the
series meaningless.
This question of distinguishing between expressions which con-
verge (that is, the sum of whose terms approaches a limit) and those
which do not has an interesting history. Newton and his followers in
the seventeenth century dealt with infinite series and always assumed
224 SECOND COURSE IN ALGEBRA
that they converged, as, in fact, most of them did. But as more
complicated series came into use it became more difficult to tell from
inspection whether they meant anything or not for a given value
of the variable.
It was not until the beginning of the nineteenth century that
Gauss, Abel, and Cauchy, in Germany, Norway, and France respec-
tively, began to study this subject effectively and to devise far-
reaching tests to determine the values of x for which certain series
converge to a finite limit. It is said that on hearing a discussion
by Cauchy in regard to series which do not always converge, the
astronomer La Place became greatly alarmed lest he had made use
of some such series in his great work, ** Celestial Mechanics." He
hurried home and denied himself to all distractions until he had
examined every series in his book. To his intense satisfaction they
all converged. In fact it has often been observed that a genius can
safely take chances in the use of delicate processes, which seem
very foolish and unsafe to a man of ordinary ability.
MISCELLANEOUS EXERCJSES
1. The digits of a certain three-digit number are in geo-
metrical progression. The sum of the digits is 13. The number
divided by the sum of its digits gives a quotient of 10 and a
remainder of 9. Find the number.
2. What per cent of the eleventh term of the progression
1, 2, 4, 8, . . • is the eleventh term of 1, 101, 201, • • • ?
3. Compare the sum of the first ten terms of tfce first pro-
gression in Exercise 2 with the sum of the first ten terms of
the second progression.
4. What' meaning may be attached to (a) $600.(1.06)'?
ih) $500 . (1.05)^? (c) $500 . (1.03)«?
5. What will $100 amount to in three years, with interest
at 4%, compounded annually? in five years ?
6. What will $100 amount to in two years, with interest at
6%, compounded semiannually ? in three years ?
PROGRESSIONS 225
7. For eacli of ten successive years a man saves $200 from
his salary and places it in a savings bank where it draws 4<^
interest, compounded annually. Find the total amount of his
savings at the time of his fifth annual deposit. Indicate (with-
out multiplication) the amount of his deposit at the beginning
of the eleventh year.
8. A loan of S dollars is to be repaid in four equal annual
payments ot p dollars each. Find j9 if money is worth r%.
Solution. The sum due at beginning of second year is
Sa + r)-p, (1)
The sum due at beginning of third year is
[^(l + r)-p](l+r)-p. (2)
The sum due at beginning of fourth year is
{[S(l + r) - ;>] (1 + r) - ;. } (1 + r) - p. (3)
The sum due at beginning of fifth year is
[{['^(1 + -i^] (1 + ^) -i>} a + -i>] a + r) -p. (4)
By the conditions of the problem, (4) = 0, for all the debt has
then been paid. Setting (4) equal to zero and simpUfying,
S(l + ry-p{l + ry-p(l-{-ry-p{l+r)-p = 0. (5)
Solving (5) ior p,
S(l + ry
(1 + ry + (1 + ry + (1 + r) + 1
r
But the denominator in (6) is a geometrical series whose sum
by the formula *S„, = is -'^ ^ ■-
1— r . r
Substituting this last value for the denominator of (6),
^ (1 + ry-l ^^
In the general case, if we have n annual payments, the exponent
Sr(l + r^'^
4 in (7) would be replaced by n, and then p = -——^ ^ —
p= .. ,„.s,.rv:;^ , - w
226
SECOND COUESE IN ALGEBRA
9. A loan of flOOO is to be repaid in three equal annual
payments, interest at 6%. Find the payment required.
10. A loan of |2000 bearing interest at 5% is to be repaid
in five equal annual payments. Find the payment required.
11. The machinery of a certain mill cost |10,000. The owner
figures that the machinery depreciates 10% in value each succes-
sive year. What was the estimate on the value of the machinery
at the end of the sixth year ?
12. A vessel containing wine was emptied of one third of
its contents and then filled with water. This was done four
times. What portion of the original contents was then in the
vessel ? (J
13. In the adjacent figure the tri-
angle DEF is formed by joining the
mid-points of AB, BC, and CA respec-
tively. Triangles GHI and KLAI are
formed in like manner. If ABC is
equilateral, prove that the successive
perimeters of the triangles form a
geometrical progression, li AB = 5, find the sum of the
perimeters of all the triangles which may be so formed.
14. Each square in the adjacent fig-
ure except the first is formed by join-
ing the mid-points of the ftquare next
larger. It AB = 4:, show that the perim-
eters of the squares form a decreasing
geometrical progression. Find the sum
of the perimeters of all the squares
which may be so drawn?
CHAPTER XVII
THE BINOMIAL THEOREM
121. Powers of binomials. The following identities are
easily obtained by actual multiplication :
(^ j^hy = d^-\-2ah + h\ (1)
(a Jrhy = a^ + Z a% + Zal^ + lA (2)
(a 4- 6)4 = ^4 + 4 a% + 6 a.262 + 4 aS^ _^ h\ (3)
(a + 6)5 = ^5 _^ 5 ^46 +10 ^352 + 10 ^263 _^ 5 a64 + 65. (4)
If a -f- 6 is replaced by a — h, the even-numbered terms
in each of the preceding expressions will then be negative
and the odd-numbered terms will be positive.
122. The expansion of (a+by. The form of the expan-
sion for the general case will now be mdicated :
The first term is a" and the last is 6".
The second term is -h na"~^b.
The exponents of a decrease hy 1 in each term after the
first.
The exponents of b increase hy 1 in each term after the
second.
The product of the coefficient in any term and the 'exponent
of a in that term., divided hy the exponent of b increased hy i,
gives the coefficient of the next term.
The sign of each term- is -\- if a and b are positive ; the
sign of each even-numhered term is — if b alone is negative,
227
228 SECOND COURSE IN ALGEBRA
According to the above statement we have
(a + b)" = a" + 7 a"-^b+ "^f ~ ^ a"-^bi'
1 • 1*2
n(n-l)(n-2)
■^ 1.2.3 ''""^ + . . . + &^.
This expresses in symbols the law known as the binomial
theorem. The theorem holds for all positive values of n and
also, with certain limitations, for negative values.
Note. The coefficients of the various terms in the binomial
expansion are displayed in a most elegant form as follows :
1
1 1
12 1
13 3 1
14 6 4 1
In this arrangement each row may be derived from the one above it
by observing that each number is equal to the sum of the two num-
bers, one to the right and the other to the left of it, in the line above.
Thus 4=1 + 3, 6 = 3 + 3, etc. The next line is 1 5 10 10 5 1.
The successive lines of this table give the coefficients for the expan-
sions of (a + &)"■ for the various values of n. The numbers in the last
line of the triangle are seen to be the coefficients when n = 4 ; the
next line would give those for n = 5. This array is known as Pascal's
triangle, and was published in 1665. Tt was probably known to
Tartaglia nearly a hundred years before its discovery by Pascal.
ORAL EXERCISES
What is the second term in the expansions of Exercises 1-4 ?
1. (a + by\ 2. (a-b)"". 3. (a-\-by\ 4. (a - bfK
Assuming that the terms in Exercises 5-10 occm- in an ex-
pansion of the binomial a -\- b, find (a) the exponents, (b) the
coefficient in the next following term of the expansion.
5. 10a%\ 7. 15a*b\ 9. 252 a^b^.
6. Sab\ 8. 11 . 5 . 9a'b\ 10. 20 a^b\
THE BINOMIAL THEOKEM 229
EXERCISES
Expand by the rule :
1. (a + b)\ 3. (a + 1)'. 5. {a + 3)«.
2. {a - If. 4. (a + 2)^ 6. (2 - af.
Obtain the first four terms of the following :
7. (a + by\ 8. (a + b)"". 9. (a + 1)^. 10. (a - 2)2^.
Expand :
11. (a^ + 2by.
Hints. To avoid confusion of exponents first write
(a2)& + (a2)4 (2 6)1 + (^2)3 (2 6)2 + (a2)2 (2 6)3 + (a2)i (2 6)* + (2 6)5.
Then in the spaces left for them put in the coefficients according to
the rule of this section.
Finally, expand and simplify each term.
12. («.^ + 2)«. 13. (a^-24)'. 14- (^ J- 15. (^)'.
Obtain in simplest form the first four terms of the following ;
16. (a^ + 3 b)"". /a^ 2^Y
17. (a'^-Sby. ' ^^' ^*/
19. (t + -^) • 23. [^ + ^J.
20. (^-^)- 24.(1 + -J
■■&-sf' -('-;:
25. Write the first six terms of the expansion ol (a + by\
and evaluate it for n = 1, n = 2, n = S, n = 4:. How does the
number of terms compare with n ? What is the value of each
coefficient after the (t^ -h l)th ? Why does not the expansion
extend to more than five terms when n = 4^?
230 SECOND COURSE IN ALGEBRA
Compute the following to two decimal places. (In each case
carry the computation far enough to be certain that the terms
neglected do not affect the second decimal place.)
26. (l.l)^*'. 29. (.98)".
Hint. (1.1)1° z= (1 + .l)io etc. Hint. (.98)" = (1 - .02)" etc.
27. (5.2)«. 30. (4.9)^
28. (1.06)«. 31. (2.9)1
123. Extraction of roots by use of the binomial expansion.
The expansion in section 122 may be verified for any par-
ticular integral value of n without difficulty by direct mul-
tiplication, as in section 121. But if n has a negative or
fractional value, a laborious proof is required to show that
the expansion is still valid when a is numerically greater
than h. Since none of the factors of the coefficients, as n,
n — 1, n — 2, vanish for fractional or negative values of n^
it appears that for such exponents the expansion is an
infinite series.
For example,
(a + h)^ = a^ + \arh -\'a~h'^ + -^^ . a" V+ • • ..
By giving n the values |- or J, one can compute the
square root or the cube root of a number to any required
degree of accuracy.
In such computations it is desirable to let the number
which corresponds to a in the binomial exceed the one
corresponding to h by as much as possible and at the same
time to have a** an integer.
Note. 'The process of extracting the square root and even the
cube root by means of the binomial expansion was familiar to the
Hindus more than a thousand years ago. The German Stifel (1486-
1567) stated the binomial theorem for all powers up to the seven-
teenth, and also extracted roots of numbers by this method.
THE BINOMIAL THEOEEM 231
EXERCISES
Eind the first four terms of the following :
1. (l + x)\ 3. (S-x)i. 5. (2 + x)^.
2. (2-\-x)K 4. (l + x)k 6. (S-x)^.
Eind to at least three decimals by the binomial theorem :
7. (27)i
Solution. (27)^ = (25 + 2)^
= 25^+ i.25-2. 2-^-25-t.22
+ V^.25-^.23
= 5 + .2 - .004 + .00016 . • • = 5.196 + .
It is proved in more advanced books that when the terms of
an infinite series are alternately plus and minus, and each term is
numerically less than the preceding one, the value of the entire sum
from a given term on cannot exceed that term. This fact renders
these so-called " alternating series " especially convenient for com-
putation, since a definite limit of error is known at each stage of
the computation. In this example the error cannot exceed .00016.
8. (lT)k 9. (28)1 10. (38)i 11. (78)^ 12. (125)i
13. (61)i
Solution. (61)^ = (64 - 3)i
= 64i - J- • 64- 1 . 3 - ^ . 64- ^ . 3^
-/r-64-t.33-f,-.
= ^ ~ T^?r — tttV? — t-^At^w + . . .
= 4 - .0625 - .00097 - .00000259 • • • = 3.93653 - .
Here three terms give the result to five figures.
14. (79)i
Hint. (79)^ =: (81 - 2)i = 81^- -| • 81"^ • 2+ . • ..
Here (81 — 2)^ yields more accurate results with fewer terms than
does (64 + 15)i.
15. (28)i 16. (66)i 17. (30)i 18. (700)i
232 SECOND COUESE IN ALGEBRA
124. The factorial notation. The notation 5 ! or [5 sig-
nifies 1 . 2 . 3 . 4 . 5, or 120. Also 4! =1 . 2 . 3 • 4 = 24.
In general, t^ ! = 1 • 2 • 3 • 4 . . • (w — 2) (/i — 1) n.
The symbol nl or [w is read "factorial ti."
In the factorial notation the denominators of the fourth and
fifth terms of the expansion of (a + by become 3 1 and 4 I respec-
tively (see formula, p. 228).
EXERCISES
Evaluate :
1. 6!. 3. 5!. 2!. 5. 4! -3! • 2!.
2. 4!. 3!. 4. 6!--2!. 6. ti! ^(/i -1)!-
^ , ^ n(n-l)(n-2) -.- (n-r-^2) ^
Evaluate -^ ^ — , \^, — ^' when:
(r-l)l
7. n=7,r = 5. 10. n = 20, r =15.
8. 71=15, r = S. 11. n=lS,r=17.
9. 71 = 21, r = 12. 12. n = 10, r = 11.
125. The rth term of (a + &)". According to the bmomial
theorem the fifth term of the expansion on page 228 is
n(n-l)(n- 2) (n - S)a''-^^
. "^ 4!
•
If we note carefully this term and the directions on
page 227, we can write down, from the considerations
that follow, any required term without writing other
terms of the expansion.
The denominator of the coefficient of the fifth term is 4 I
From the law of formation the denominator of the sixth
term would be 5 !, of the seventh term 6 1, etc. Conse-
quently in the rth term the denominator of the coefficient
would be (r —1)1.
THE BINOMIAL THEOEEM 233
The numerator of the coefficient of the fifth term contains
the product of the four factors ?^(7^— l)(/i— 2)(?i— 3).
The numerator of the sixth term would contain these four
and the additional factor 7i — 4. Similarly, the last factor
in the numerator of the seventh term would be n — 5, etc.
Hence the last factor in the rth term would be 7i — (r — 2),
and the numerator of the coefficient of the rth term is
7i(n-l)(^-2)(^-3) ... (w-r + 2).
The exponent of a in the fifth term is n — 4, and in the
sixth term it would be n — 5, etc. Therefore in the rth
term the exponent of a is n — (r — 1), or n — r + 1.
The exponent of b in the fifth term is 4, in the sixth
term is 5, etc. Therefore in the rth term the exponent of
h is r — 1.
The sign of any term of the expansion (if n is a posi-
tive integer) is plus if the binomial is a + b. If the
binomial is a — b, the terms containmg the odd powers of
b will be negative. In other words, the sign in such cases
depends upon whether the exponent r — 1 is odd or even.
Hence the rth term (r not equal to 1) of (« + b^^^ equals
n(n-l)(/2-2)(n-3). • .(n-r+2)
(r-1)!
.a"-'+^lf-K (1)
If we wanted the twelfth term, we would in using (1)
substitute 12 for r.
EXERCISES
Write the indicated terms :
1. Fifth term of (a + bf^. 2. Sixth term of (a + bf.
Solution. Substituting 10 for n
,-. • .^ f 1 /ix • 3. Fourth term of (a + ^)^.
and o for r m the formula (1) gives ^ ' ^
lQ-9-8-7 ^ 10-9 -8 -7 4. Seventh term of (a - bf
4! 4-3.2
= 210 a«&*. 5. Eighth term of (a — bf^
234 SECOKB COURSE IN ALGEBRA
6. Fourth term of (a + -) • 8. Sixth term of ( ] •
7. Fifth term of (a^ - hf". 9. Middle term of {x" - xf.
10. Seventh term of (-; ) •
(v-.-^r
11. Fifth term of
Find the coefficient of :
12. x^ in (1 + xy. 14. x^^ in (x^ -f- 1)'^
13. x^ in (1 + xY' 15- x^' in (^^ - »"')"
Note. The binomial theorem occupies a remarkable place in the
history of mathematics. By means of it Napier was led to the dis-
covery of logarithms, amd its use was of the greatest assistance to
Newton in making his most wonderful mathematical discoveries.
But to-day the results of Newton and of Napier are explained with-
out even so much as a mention of the binomial theorem, for simpler
methods of obtaining these results have been discovered.
It was Newton who first recognized the truth of the theorem, not
only for the case where n is a positive integer, which had long been
familiar, but for fractional and negative values as well. He did not
give a demonstration of the general validity of the binomial develop-
ment, and none even passably satisfactory was given until that of
Euler (1707-1783). The first entirely satisfactory proof of this
difl&cult theorem was given by the brilliant young Norwegian Abel
(1802-1829).
CHAPTER XVIII
RATIO, PROPORTION, AND VARIATION
126. Ratio. The ratio of one number a to o, second
number b is the quotient obtained by dividing the first
by the second, or -. The ratio of ^ to 5 is also written
J. ^
a: 0.
It follows from the above that all ratios of two numbers
are fractions and all fractions may be regarded as ratios.
Thus -' — ' — ' and — — : are ratios as well as fractions.
5 ^d c-d V3
Since ratios like the above are fractions, operations
which may be performed on fractions may be performed on
these ratios. Hence the value of a ratio is not changed
by multiplying or dividing both numerator (antecedent)
and denominator (consequent) by the same number.
a
Thus - = and - = t-
h b' X ^2.
y
EXERCISES
Simplify the following ratios by considering them as frac-
tions and reducing the fractions to lowest terms :
,2
-i^-^^^-i^*^^^
2. 1 kilometer : 1 mile. (1 kilometer = .62 miles.)
3. 1 liter : 1 quart. (1 liter = .001 cubic meter ; 1 quart =
^^ cubic inches ; and 1 meter = 39.4 inches.)
235
236 SECOND COURSE IN ALGEBRA
4. A city lot 100 x 160 feet : 1 acre. (1 acre = 43,560 square
feet.)
5. Area of printed portion of this page : total area of
the page.
6. If |24,000 is divided between two men so that the
portions received are to each other as 5:7, how much does
each receive?
Hint. Let 5 x and 7x be the required parts.
7. Separate 690 into four parts which are to each other
as 2 : 5 : 7 : 9.
X X -\- S
8. Show that -< ;: if x is positive.
X -\- 2 X -^ 5
Hint. Reduce the given fractions to respectively equivalent fractions
having a common denominator, then compare the numerators of the
fractions so obtained.
9. Arrange the ratios 3 : 4 and 7 : 9 in decreasing order of
magnitude.
127. Proportion. A proportion is a statement of equality
between two ratios. Four numbers, a, h, <?, and d, are in
proportion if the ratio of the first pair equals the ratio of
the second pair.
This proportion is written
a: b = c: a or r = -: •
b a
Though both forms are equations, the second is the
more familiar one and for this reason is preferable.
' Note. By the earlier mathematicians ratios were not treated as
if they were numbers, and the equality of two ratios which we
know as a proportion was not denoted by the same symbol as other
kinds of equality. The usual sign of equality for ratios was : : , a
EATIO, PROPOETION, AND VARIATION 237
notation which was introduced by the Englishman Oughbred in
1631 and was brought into common use by John Wallis about 1686.
The sign = was used in this connection by Leibnitz (1646-1716)
in Germany, and by the Continental writers generally, while the
English clung to Oughtred's notation.
ci c
In the proportion t = -^ the first and fourth terms (a, cZ)
ci
are called the extremes, and the second and third terms
(^, (?) are called the means.
128. Mean proportional. A mean proportional between two
numbers a and h is the number m if — = — • It follows
that Tn^ = ab^ or m = ± Va6.
129. Third proportional. A third proportional to two num-
bers a and h is the number t \i - = --
t
130. Fourth proportional. A fourth proportional to three
ci c
numbers a, 5, and c is the number / if - = — .
131. Test of a proportion. Since a proportion is an
equality between two ratios (fractions), it is therefore
an equation. Hence any operation which may he performed
on an equation may he performed on a proportion, (See
Axioms, p. 16.)
Thus, in the proportion — = - both members may be multiplied
h d
by hd, giving ad = he. Here the first member is the product of the
extremes of the proportion, and the second member is the product
of the means.
Therefore in any proportion the product of the extremes equals the
product of the means.
132. Proportions from equal products. The numbers
which occur in a pair of equal products may be used in
various ways as the terms of a proportion.
238 SECOND COURSE IN ALGEBRA
Thus, if ad = be,
we may write either
a c ^ _ ^
Proof, li a- d = h- c is divided by hd, we obtain
ad he a e /^\
— = — , or - = — (1)
hd hd h d
li a ' d = b ' c is divided by ed, we obtain
c d'
(2)
If the means in (1) are interchanged, (2) is obtained.
This process of obtaining (2) from (1) is called alternation.
li a- d = b- c is divided by ac, we obtain
- = -■ (3)
a e
If the fractions in (1) are inverted, (3) is obtained.
This process of obtaining (3) from (1) is called inversion.
EXERCISES
1. Find a mean proportional between: (a) 3 and 27;
(b) y and -; (c) — and -; (d) ^ and xy.
2. Find a third proportional to: (a) 9 and 6; (b) 180
and 60; (c) 216 and 36.
3. Find the fourth proportional to : (a) 14, 10, and 7 ;
(b) 27, 3, and 36; (c) 96, 12, and 8.
4. Form three proportions from each of the following
equations : (a) 5 x = ^ y ; (b) (a 4- 3) • 2 = (a + 1) • 3 ;
(c) dip' — y^ — T^ — s\
2 10 2 20
5. Write by alternation: (a) - = — 5 (b) - = —
6. Write by m version: («)- = -—? (o) - = t^tt-
•^ ^^6 12^^ic30
RATIO, PROPOETION, AND VAEIATION 239
a c
If - = -, prove the following and state the corresponding
theorems in words
7.
a
h
G
d
h
d
8.
a
c
9.
10.
11.
b'' d''
-\a _ Vc
a -\- b c -\- d
b d
Hint. Add 1 to each member
.a c
of - = —
b d
12.
a -i- b c -\-
a
by inversion
Hint. Write -
b d
and apply hint of Exercise 11.
13.
14.
15.
a — b _c — d
b ^ d
a — b G — d
a
a-\-b
G
c-\-d
a — b c — d
The proportions given in Exercises 11, 13, and 15 are said
to be derived from a :b = g: d by addition, subtraction, and
addition and subtraction respectively.
If -[=-,'> show that the following equalities are true :
di
^^ 5a 5c
,« 2a 2g
7b Id
18.
19.
P
ac
bd
5a -\-b _5g -\- d
5a — b 5g — d
20.
21.
22.
23.
b^
d^
b'
a''-7P
d^
c'-ld^
2 -(^2
2ab 2Gd
7a2_3^2 7c2_3^2
5ab
5Gd
24.
a'-^-ab-Jrb^ _a'- ah -f- b''
g' + cd -\-d''~ c'-Gd-\-d''
240 SECOND COUKSE IN ALGEBRA
25. In the proof which follows give the reason for each
step and state the result as a theorem :
o_c_e a-\-c-\-e _a _c _e
l^d^f ^^' b + d+f~b~d~f
Proof. Setting each of the given ratios above equal to r,
- = r, - = r, and - = r. (1)
Then from (1), a = hr, c = d?^ e =fr. (2)
Adding in (2), a + c + e = br -]- dr +fr. (3)
Factoring in (3), a + c + e = (h + d +fy. (4)
Therefore 1^7^ = '' ^^^
Hence by (1) and (5),
a-\-c + e_a_c_e
bTJTf ~b~d~f
26. If T = 3 = - 5 show that = - or -•
Solve, using theorems of proportion :
27. 2a^:(ic + 8) = 10:3. 29. 5 : 4 = (.t - 3):(£r - 4).
28. 25:x = x:169. 30. (15 + x):(15 - «) = 13 :17.
31. (10 + £c):(20 + 3cc) = (10-ic):-3ir.
32. V^ + 2 ^^+1
■y/x-2 x-7
33. Show that the mean proportional between two numbers
is the geometric mean between these numbers.
PROBLEMS
1. The surface of a sphere is 4 7r72^ If S represents the
surface of a sphere, R its radius, and D its diameter, show
2. Find the ratio of the surfaces of two spheres whose
radii are in the ratio 1:10,
KATIO, PEOPORTION, AND VAKIATION 241
3. If the diameter of the earth is 7920 miles and that of
Mars is 4230 miles, find the ratio of their surfaces.
4. If the diameter of the moon is 2160 miles, find the ratio
of its surface to that of the earth.
5. The volume of a sphere is — - — • If V represents the
o
volume of a sphere, R its radius, and D its diameter, show that
for any two spheres, y j^s j)s
6. The diameter of the sun is approximately one hundred
and nine times the diameter of the earth. Find the ratio of
their volumes.
7. From Exercises 3 and 5 find the ratio of the volumes of
Mars and the earth.
8. Find the ratio of the volumes of the earth and the moon.
9. The areas of two similar triangles are to each other
as the squares of any two corresponding lines. If the corre-
sponding sides of two similar triangles are 12 and 20 and
the area of the first is 90 square inches, find the area of the
second.
10. The areas of two similar triangles are 147 and 300
respectively. If the base of the first is 10, find the correspond-
ing base of the second.
11. If ^jBC is any triahgle and KR is a line parallel to BC,
meeting AB at ^-and AC at R,
then
area ABC _ AB^ _ AC^ _ BC^
area AKR ~ AK^ ~ AR^ ~ KR^'
If in the accompanying figure
area ABC = 225 square inches,
area AKR = 81 square inches, and AB = W inches, find AK,
242
SECOND COUESE IN ALGEBRA
12. In the figure of Exercise 11, if ABC = 845 square inches,
BC = 13 inc*hes, and KR = 7 inches, find the area AKR.
13. If in the figure of Exercise 11 triangle AKR equals ^j
of the trapezoid KBCR and AC = 15 inches, find AR and RC.
14. In Exercise 13 substitute ^ for 2^ and solve for AR to
two decimals.
15. If in the figure of Exercise 11 the triangle is equivalent
to the trapezoid and AK = 10, find KB to two decimals.
16. A certain flagpole casts a shadow 45 feet long at the
same time that a near-by post 8 feet high casts a shadow
4: J feet long. Find the height of the pole.
17. The bisector of an angle of a triangle divides the oppo-
site side into segments which are proportional to the adjacent
sides. In triangle ABC ii AB = 15, BC = 24:, and CA = 25,
find the segments of i^C made by the bisector of angle A.
\S. If a plane be passed par-
allel to the base of a pyramid
(or cone), as in the accompany-
ing figure, cutting it in KRL,
then pyramid D — ABC : pyra-
mid D - KRL = DH^ : DS^, etc.
If in the adjacent figure the
volumes of the pyramids are 8
and 27 cubic inches respectively,
and the altitude DH equals 15
inches, find DS.
19. If DH in the accompany-
ing figure is 18 inches and the volume of one pyramid is one
third the volume of the other, find DS to two decimals.
20. In the accompanying figure the frustum is seven eighths
of the whole pyramid, (a) If DH equals 16, find DS-, (b) if
DH = X, find DS.
RATIO, PROPORTION, AND VARIATION 243
21. If a plane parallel to the base divides the whole pyra-
mid into two parts having equal volumes and DH = 75, find
to two decimals the parts into which the plane divides DH.
22. The volumes of two similar figures are to each other as
the cubes of any two corresponding edges. Compare the vol-
umes of two similar solids, the edge of one of which is 60%
greater than the corresponding edge of the other.
23. Compare the radii of two spheres whose volumes are
to each other as 125 : 27.
133. Variation. The word quantity denotes anything
which is measurable, such as distance, rate, time, and area.
Many operations and problems in mathematics deal with numeri-
cal measures of quantities, some of which are fixed and others con-
stantly changing. Such problems as deal with the relation of the
numerical measures of at least two changing quantities are called
problems in variation.
The theory of variation is really involved in proportion.
This fact will become evident after a study of the illustra-
tions of the different kinds of variation here given.
The equation x='^y may refer to no physical quantities
whatever, yet it is possible to imagine y as taking on in
succession every possible numerical value, and the value
of X as changing with every change of ?/, and consequently
always being three times as great as the corresponding
v^lue of y. In this sense, which is strictly mathematical,
X and y are variables.
The symbol for variation is oc, and x cc y is read " x
varies directly as ?/ " or '' x varies as ?/."
134. Direct variation. One hundred feet of copper wire
of a certain size weighs 32 pounds. Obviously a piece of
the same kind 200 feet long would weigh 64 pounds, a
piece 300 feet long would weigh 96 pounds, and so on.
244 SECOND COURSE IN ALGEBRA
Here we have two variables, W^ (weight) and L (length),
so related that the value of ^depends on the value of X,
and in such a way that W increases proportionately as L
increases. That is, W is directly proportional, or merely pro-
portional, to L, Hence, if W^ and W^ are any two weights
corresponding to the lengths L^ and L<^ respectively,
W,: W, = L,:L,. (1)
The fact expressed by (1) can be stated in the form of
a variation, thus: WccL.
In general, if x cc ^, and x and «/ denote ani/ two corre-
sponding values of the variables, and x-^ and ?/j a particular
pair of corresponding values of these variables,
then £ = 1. (2)
From (2), x = (^\^. (3)
X
But -^ is a constant, being the quotient of two definite
numbers.
Call this constant K, and (3) may be written
x = Ky.
That is, if one variable varies as a second, the first always
equals the second multiplied hy some co7istant.
Thus for the copper wire just mentioned, W= j'VV ^» ^^ ^ ^•
Here, though W varies as L varies, W is always equal to L multi-
plied by the constant 5%.
EXERCISES
1.. If X QC y, and x = S when ^ = 8, find x when y =12.
Solution. The variation is direct. ■"
Therefore - = —
X 12
Solving, a; = 4^.
RATIO, PEOPORTION, AJ^D VARIATION 245
2. li x^y, and ic = 8 when y = 15, find y when x = 10.
3. If. xcc 2/, and x = h when y = k, find y when x = r.
4. If a? oc 2/, and a? = 2 when y = 5, find iT.
135. Inverse variation. If a tank full of water is emptied
in 24 minutes through a smooth outlet in which the area
of the openmg A is, 1 square inch, an outlet in which A
is 2 square inches would empty the tank in one half the
time, or in 12 minutes; and an outlet in which ^ is 3 square
inches would empty the tank in 8 minutes.
Suppose it possible to increase or decrease A at will.
When A is doubled t is halved ; when A is trebled t is
divided by 3 ; and so on. We then have in t (the time
required to empty the tank) and in A (the area of the
opening) two related variables such that if A increases, t
will decrease proportionally, while if A decreases, t will
increase proportionally.
Now let t^ and ^2 be any two times corresponding to the
areas A^ and A^ respectively; then
t^:% = A^:A^. (1)
The letters and the subscripts in (1) say : The first time
is to the second time as the second area is to the first area.
The proportion (1) may be written ^^ : ^2 = — - —
A^ A^
where the subscripts on the ^'s and those on the ^'s come
in the same order.
First, from (1), t^. A^ = t^. A^. • (2)
Dividing (2) by ^1^2^ ^ = ^' (3)
Whence '<:];) = <i;)- w
246 SECOND COURSE IN ALGEBRA
Therefore f ^ : ^2 "= — • — (^)
In the form of a variation (5) becomes ^ oc —
In general, x varies inversely as y when x varies as the
reciprocal of ?/ ; that is,
.«i. (6)
And if X and 7/ denote any two corresponding values
of the variable, and x^ and 1;/^ a particular pair of corre-
sponding values,
xix-. = — : — (7)
Whence — = ^, or ^«/ = x-^y^. (8)
Vi y
But x^y^ is a constant, being the product of two definite
numbers. Call this constant K.
Then (8) becomes xy = K.
That is, if one variable varies inversely as another, the
product of the two is a constant.
EXERCISES
1. If £c varies inversely as y, and x = S when y = 5, find x
when y = 15.
Solution. The variation is inverse.
Hence 8 : a; = - : — r •
5 lo
Solving, a: = 2§.
2. If a; QC -5 and x = 1 wIkd // - '-^'). fnid x when ?/ = 10.
RATIO, PROPORTIOK, AND VARIATION 247
3. If y QC -J and y = h wlien z = k, find y when z = r. .
4. If m QC -J and m = 2 when ??, = — ? find m when n = 12.
5. If f oc -J and i^ = 2 when tz, = 8, find n when ?^ = 8.
n
6. If ?/J QC -? and w = 100 when d = 4000, find w when
c^ = 5000.
7. If ^ QC -J and z^ = 4 when r = 25, find /C
r
8. If z^ oc -, and w = 200 when d = 4000, find ii:.
cc
136. Joint variation. If the base of a triangle remains
constant while the altitude varies, the area will vary as
the altitude. Similarly, if the base varies while the alti-
tude remains constant, the area w411 vary as the base. If
both base and altitude vary, the area varies as the product
of the two ; that is, the area of the triangle varies jointly
as the base and altitude. Further, if at any time A denotes
the area of a variable triangle, and h^ and 5^ the corre-
sponding altitude and base, then
A = ^- (1)
If A^ denotes the area at ani/ other time, and h^ and b^
the corresponding altitude and base, then
A = ^- (2)
Now (1)^(2) gives A^: A^ = hj)^: h^h^.
In the form of a variation this last proportion becomes
Acchb.
248 SECOND COURSE IN ALGEBEA
In general, any variable x varies jointly as two others,
y and 2, if x<^yz', (1)
that is, if X varies as the product of the two.
If X varies jointly as y and 2, and if x^ y^ and z denote
any corresponding values of the variables, while x-^, y^^ and
z^ denote a particular set of such values, then
L=yL. (2)
From (2), x = i^^yz. (3)
But in (3) the fraction — ^ is a constant, since o^j, y^^
and 2j are particular values of the variables x^ «/, and z.
Calling this constant JT, we may write xocyz as the
equation x=Kyz.
One variable may vary directly as one variable (or several varia-
bles) and inversely as another (or several others). Also one variable
may vary as the square, or the cube, or the square root, or the recip-
rocal, or as any algebraic expression whatever involving the other
variable (or variables).
EXERCISES
1. If £c varies jointly as y and z, and a? = 24 when 2/ = 6 and
^ = 8, find X when y = l^ and ^ = 4.
Solution. The variation is joint.
24 6-8
Therefore — = -— — - •
X 18-4
Solving, X = 36.
2. If a? varies as yz, and £c = 10 when y = 15 and z = 6, find
X when y — ^ and z = %.
3. If yl oc hh, and A = 30 when A = 5 and ^^ = 12, find A
when h — 1 and 5 = 10.
RATIO, PROPORTION, AND VARIATION 249
4. It* A oc hb, and A = 48 when li = 8 and b = 12, find /v.
5. li X varies directly as y and inversely as z, and ic = 10
when y = 5 and z = 27, find x when 2/ = 12 and s = 36.
6. If F varies directly as T and inversely as P, and V = 80
when P = 30 and T = 300, find P when T = 400 and V = 40.
PROBLEMS
1. The weight of any object below the surface of the earth
varies directly as its distance from the center. An object
weighs 172 pounds at the surface of the earth. What would
be its weight (a) 1000 miles below the surface? (b) 3000
miles below the surface ? (c) at the center of the earth ?
(Radius of the earth = 4000 miles.)
2. The distance which sound travels varies directly as the
time. A soldier measures with a stop watch the time elapsing
between the sight of the flash of an enemy's gun and the
sound of its report. If sound travels 1100 feet per second, how
far off was the enemy Avhen the observed time was 5| seconds ?
3. When the volume of air in a bicycle pump is 24 cubic
inches, the pressure on the handle is 30 pounds. Later when
the volume of air is 20 cubic inches, the pressure is 36 pounds.
Assume that a proportion exists here, determine whether it is
direct or inverse, and find the volume of the air when the
pressure is 42 pounds.
4. The distance (in feet) through which a body falls from
rest varies as the square of the time in seconds. If a body falls
16 feet in 1 second, how far will it fall in (a) 3 seconds ?
(b) 10 seconds?
5. The intensity (brightness) of light varies inversely as
the square of the distance from the source of the light.
A reader holds his book 3 feet from a lamp and later 6 feet
distant. At which distance does the page appear the brighter ?
How many times as bright ?
KK
250 SECOND COURSE IN ALGEBRA
6. A lamp shines on the page of a book 5. feet distant.
Where must the book be held so that the page will receive
twice as much light ? four times as much light ?
7. The area of a circle varies as the square of its radius
The area of a certain circle is 154 square inches and its radius
is 7 inches. Find the radius of a circle whose area is 616 square
inches.
8. The area illuminated on a screen by a spot light varies
directly as the square of the distance from the source of
the light to the screen. If the lighted area at a distance of
40: feet is a circle of diameter 10 feet, find the diameter of the
illuminated circle at a distance of 15 feet.
9. The weight of an object above the surface of the earth
varies inversely as the square of its distance from the center
of the earth. An object weighs 172 pounds at the surface of
the earth. What would it weigh (ci) 1000 miles above the
surface ? (h) 3000 miles above the surface ? (<•) 5000 miles
above the surface ?
10. How far above the surface of the earth would a 150-pound
object have to be placed so that its weight would be reduced
one third ?
11. The weight of a sphere of given material varies directly
as the cube of its radius. Two spheres of the same material
have radii 3 inches and 5 inches respectively. The first weighs
8 pounds. Find the weight of the second.
12. The time required by a pendulum to make one vibra-
tion varies directly as the square root of its length. If a
pendulum 100 centimeters long vibrates once in 1 second,
find the time of one vibration of a pendulum 81 centimeters
long.
13. Find the length of a pendulum which vibrates once in
2 seconds : once in 7 seconds.
RATIO, PROPORTION, AND VARIATION 251
14. The pressure of wind on a plane surface varies jointly
as the area of the surface and the square of the wind's velocity.
The pressure on 1 square foot is .9 pound when the rate of
the wind is 15 miles per hour. Find the velocity of the wind
when the pressure on 1 square yard is 72.9 pounds.
15. The pressure of water on the bottom of a containing
vessel varies jointly with the area of the bottom and the depth
of the water. When the water is 1 foot deep the pressure on
1 square foot of the bottom is 62.5 pounds. Find the pressure
on the bottom of a circular tank of 14 feet diameter in whicdi
the water is 10 feet deep.
16. The cost of ties for a railroad varies directly as the
length of the road and inversely as the distance between the
ties. The cost of ties for a certain piece of road, the ties being
2 feet apart, was $1320. Find the cost of ties for a piece twenty
times as long as the first if tlie ties are 2^ feet apart.
CHAPTER XIX
LOGARITHMS ,
137. Introduction. Logarithms were invented to shorten
the work of extended numerical computations which involve
one or more of the operations of multiplication, division,
involution, and evolution. Their use has decreased the
labor of computing to such an extent that many calcula-
tions which would require hours without the use of loga-
rithms can be performed with their aid in one tenth of
the time or less.
138. Definition of logarithm and base. If we write the
equation ^^^y^ ^^^
we express therein the essential relation between a number,
n, and its logarithm, /, for a given base, h. In the notation
of logarithms this is written
logt« = ?, (2)
and it is read ''the logarithm oi'n to the base h equals /."
We can define verbally in one statement both logarithm
and base as follows:
The logarithm of a given mimher is the exponent in the
power to which ariother number^ called the base^ 7nu8t be
raised in order to equal the given number.
It is important to realize that equations (1) and (2) are
merely two different ways of expressing precisely the same
relations, one the exponential way, the otlier the logarithmic^
262
LOGARITHMS 253
Above all it is necessary to keep in mind the fact that a
logarithm is an exponent.
Thus in 32 = 2^ the given number is 32, the base is 2, and tlie
logarithm is 5 ; that is, log2 32 = 5.
139. Systems of logarithms. The base of the common^ or
Brifigs, system of logarithms is 10. Hence a table of common
logarithms is really a table of exponents of the number 10.
Since the greater portion of these exponents are approxi-
mate values of irrational numbers, it follows that compu-
tations by means of logarithms give only approximate
results. Tables exist, however, in which each logarithm is
given to twenty or more decimals ; hence practically any
desired degree of accuracy can be obtained by using the
proper table. This system is used in numerical work almost
exclusively. The table on pages 266-267 is a table of
common logarithms carried to four decimal places.
The only other system of logarithms used in computa-
tions is called the natural system. It has for its base the
irrational number 2.7182 -f, which is usually denoted by
the letter e and is used mainly for theoretical purposes.
- It can be proved that the laws given on pages 91-92,
governmg the use of rational exponents, hold for irrational
exponents. In the work on logarithms this fact will be
assumed.
ORAL EXERCISES '
1. If 8 = 2\ x = ? log^S = ?
2. If 1000 = 10^« = ?logj^ 1000 = ?
3. legale == ? log3 81 = ? log^625 = ?
4. If b^=U,h = ? If h^ = 243, b = ?
5. 10^ = ? logj^lO = ? 8. 3« = ? loggl = ?
6. lO'^ = ? log.olOO = ? 9. 5« = ? loggl = ?
7. W = ? log^^l = ? 10. n'=? log;,l= ?
25^
SECOND COUKSE IN ALGEBRA
Number
Base
Logarithm
Number
Base
Logarithm
11.
8
2
?
24.
?
4
3
12.
32
2
?
25.
V
6
2
13.
64
4'
o
26.
9
V
J
14.
125
5
V
27.
8
•/
3
2
15.
1000
10
?
28.
16
?
3
16.
16
'
?
2
29.
2
4
?
17.
81
?
2
30.
3
9
?
18.
81
?
4
31.
2
8
?
19.
40
f
2
32.
3
81
?
20.
216
f
3
33.
T^
49
<>
21.
V
10
3
34.
8
■1.
•/
22.
?
10
»>
35.
4
8
V
23.
?
1(
3
1
36.
.1
10
•>
liead in tlic notation of lojjiiritlinis :
37. 300 =10-^*^ 42. 1730 = 103-*^.
38. 65:::=10^«^ 43. 173 = W--^«^
39. 4=:10"". 44. 1.73 = 10-'«.
40. 1 = 10^ 45. .173 = 10-1 + •-'^\
41. .10 = 10-^ • 46. .0173 = 10- '- + •-««.
Read Exercises 47-49 and 54-56 as powers of 10 :
47. log 3 = .48. 48. log 20 = 1.301. 49. log 4.9 = .69.
50. log^^lOO + log^^lOOO 4- log,olO,000 = ?
51. log^,10 + log^^.Ol - logj^l = ? 54. log 490 = 2.69.
52. log^S + log327 4- log,l = ? 55. log .0049 = - 3 + .69.
. 53. log39 + log^64 = ? 56. log 381 = 2.58.
LOGARITHMS 255
Biographical Note. John Napier. Although many scientists have
been honored with titles on account of their discoveries, very few of the
titled aristocracy have become distinguished for their mathematical
achievements. A notable exception to this rule is found in John Napier,
Lord of Merchiston (1550-1617), who devoted most of his life to the
problem of simplifying arithmetical operations. Napier was a man of
wide intellectual interests and great activity. In connection with the
management of his estate he applied himself most seriously to the study
of agriculture, and experimented with various kinds of fertilizers in a
somewhat scientific manner, in order to find the most effective means of
reclaiming soil. He spent several years in theological writing. When the
danger of an invasion by Philip of Spain was imminent he invented
several devices of war. Among these were powerful burning mirrors,
and a sort of round musket-proof chariot, the motion of which was con-
trolled by those within, and from which guns could be discharged through
little portholes. "
But by far. the most serious activity of Napier's life was the effort to
shorten the more tedious arithmetical processes. He invented the first
approximation to a computing machine, and also devised a set of rods,
often called Napier's bones, which were of assistance in multiplication.
His crowning achievement, however, was the invention of logarithms, to
which he devoted fully twenty years of his life.
140. Steps preceding computation. Before computation
by means of the table can be taken up, two processes
requiring considerable explanation and practice must be
mastered.
/. To find from the table the logarithm of .a given iiumher.
II. To find from the table the number corresponding to a
given logarithm,
141. Characteristic and mantissa. Unless a number is
an exact power of 10, its logarithm consists of an integer
and a decimal.
This fact is illustrated in Exercises 37-46, p. 254.
The integral part of a logarithm is called its characteristic.
The decimal part of a logarithm is called its mantissa.
Log 200 = 2.301. Here 2 is the characteristic and .301
is the mantissa.
256 SECOND COURSE IN ALGEBRA
The characteristic of any number is obtained not from
a table of logarithms but by an inspection of the number
itself, according to rules which will now be derived.
104 = 10,000; that is, the log 10,000 = 4.
103 = 1000;
that is, the log 1000
= 8.
102 = 100;
that is, tlie log 100
= 2.
101=10;
that is, the log 10
= 1.
100 = 1;
that is, the log 1
= 0.
10-i=.l;
that is, the log .1
= -1.
10-2 = .01;
that is, the log .01
= —2.
10-3 =.001;
that is, the log .001
= -3.
The preceding table indicates between what two integers
the logarithm of a number less than 10,000 lies. This
determines the characteristic.
Since 542 lies between 100 and 1000 (that is, between
102 and 103), log 542 must lie between 2 and 3 and must
equal 2 (characteristic) plus a decimal (mantissa).
And since .0045 lies between .001 and .01 (that is,
between 10"^ and 10"'^), log .0045 = — 3 plus a positive
decimal or — 2 plus a negative decimal.
For the determination of the characteristic of a positive
number we have the following rules:
/. The characteristic of a number greater than 1 is one less
than the number of digits to ike left of the decimal point,
II. The characteristic of a number less than 1 is negative
and numerically one greater than the number of zeros between
the decimal point and the first significant jigwre.
Accordingly the characteristic of 2536 is 3; of 6 is 0;
of .4 is -1; of .032 is - 2 ; of .00086 is -4.
JOHN NAPIER
LOGARITHMS
25
ORAL
EXERCISES
What is
the characteristic
of the following :
1. 347.
5. 35.
9. 97.2.
13.
.00972.
2. 9864.
6. 972.
10. 9.72.
14.
30.467.
3. 95.
7. 9720.
11. .972.
15.
.5000.
4. 7.
8. 97200.
12. .0972.
16.
.000375.
The table on pages 2G6-267 gives the mantissas of
numbers from 10 to 999. Before each mantissa a decimal
point is understood.
The numbers 5420, 542, 5.42, .0542, and .000542 are
spoken of as composed of the same significant digits in the
same order. They differ only m the position of the deci-
mal pohit, and consequently their logarithms to the base 10
will have different characteristics, but they Avill have the
same mantissa.
The last two points are easily illustrated b}^ any two
numbers Avhich have the same significant digits in the
same order.
log 5.42 = .734, or 5.42 = 10-'34;
5.42 . 102 _ 542 = 10-1 . 102 _ 102.734.
Therefore log 542 = 2. 734.
The property just explamed does not belong to a system
of logarithms in which the base is any number other than 10.
Thus, if the base is 100, the most convenient number after
10, the logarithms of 5420, 542, 54.2, and 5.42 are respec-
tively 1.8670, 1.3670, .8670, and .3670. While a certain
regularity in characteristic and mantissa can be seen here,
it is obvious that the rules for obtaining them would not
be so simple as they are for the base' 10. Moreover, it can
be seen that tables of a given accuracy are far shorter when
the base is 10 than they would be with any other base.
258 SECOND COURSE IN ALGEBRA
142. Use of the table. To obtain the logaritlim of a
imiiiber of three or fewer significant figures from tlie
table, we have the
Rule. Determine the characteristic hy inspection.
Find in colmnn N the first two sif/nificant figures of the
given 7iimiber. In the roiv with these and iii the column headed
hy the third figure of the number, find the required mantissa.
ORAL EXERCISES
Find tlie logarithm of the following :
1. 263. 4. 56. 7. 3.7. 10. 7.
2. 375. 5. 560. 8. 3700. 11. 932.
3. 729. 6. 37. 9. 5. 12. .932.
Solution. The characteristic of .932 is — 1 and the mantissa is
.9694. Hence log .932 = — 1 + .9694. This is usually written in the
abhreviated form, 1.9694. The mantissa is always kept positive in
order to avoid the addition and subtraction of both positive and
negative decimals, which in ordinary practice contain from three to
five figures. Negative characteristics, being integers, are compara-
tively easy to take care of. (The student should note that log .932
is really negative, being —1-1- .9694, or — .0306.)
13. .563. 15. .00376. 17. .0202. 19. 3.86.
14. .0637. 16. .00468. 18. 725000. 20. .987.
143. Interpolation. The process of finding the logarithm
of a number not found in the table, from the logarithms
of two numbers which are found there, or the reverse of
this process, is called interpolation.
If we desire the logarithm of a number not in the table,
say 7635, we know that it lies between the logarithms of
7630 and 7640, which are given in the table. Since 7635
is halfway between 7630 and 7640, we assume, though it
is not strictly true, that the required logarithm is halfway
LOGAllITHMS 259
between their logarithms, 3.8825 and 3.8831. In order to
find log 7635 we first look up log 7630 and log 7640 and
then take half (or .5) their difference (this difference
may usually be taken from the column headed D) and
add it to log 7630. This gives
log 7635 = 3.8825 +.5 x.0006 = 3.8828. .
Were, we finding log 7638, we should take .8 of the
difference between log 7630 and log 7640 and add it to
log 7630 as follows :
log 7638 = 3.8825 +.8 X .0006
= 3.8825 +.00048
= 3.8825 +.0005
= 3.8830.
Observe that in using four-place tables one should not
carry results to five figures. If the fifth figure is 5, 6, 7,
8, or 9, omit it and increase the fourth figure by 1 ; that
is, obtain results to the nearest figure in the fourth place.
For finding the logarithm of a number we have the
Rule. Prefix the proper characteristic to the mantissa of the
first three significant figures.
Then multiply the difference hettveen this mantissa and the
next greater mantissa in the table (^called the tabular differ-
ence') by the remaining figures of the number preceded by a
decimal point.
Add the product to the logarithm of the first three figures^
taking the nearest decimal in the fourth place.
In this method of interpolation we have assumed that
the increase in the logarithm is directly proportional to
the increase in the number. As has been said, this is
not strictly true, yet the results here obtained are nearly
always correct to the fourth decimal place.
260 SECOND COURSE IN ALGEBRA
EXERCISES
Find the logaritliin of the following :
1. 3625. 5. 646.8. 9. 705.50. 13. 30.07.
' 2. 464.7. 6. 82.543. 10. 3.0075. • 14. 3.1416.
3. 52.73. 7. 10.101. 11. .00286. 15. 2.71828.
4. 42.75. 8. 500.35. 12. .0007777. 16. .0023456.
144. Antilogarithms. An antilogarithm is the number cor-
responding to a given logarithm. Thus antilog 2 equals 100.
If we desire the antilogarithm of a given logarithm,
say 4.7308, we proceed as follows: The mantissa .7808 is
found in the row which has 53 in column N and in the
column which has 8 at the top. Hence the first three sig-
nificant figures of the antilogarithm are 538. Since the
characteristic is 4, the number must have five digits to the
left of the decimal pohit. Thus antilog 4.7308 = 53,800.
Therefore, if the mantissa of a given logarithm is found
in tlie table, its antilogarithm is obtained by tlie
Rule. Find the roiv and the colmnn in which the given
mantissa lies. In the row found take the two figures which
are in column N for the first two significant figures of the
antilogarithm and for the third figure the number at the top
of the column in tvhich the mantissa stands.
Place the decimal point as ind/'cafcd li/ ihc rltaractenstic.
ORAL EXERCISES
Find the antilogarithm of the following :
1. 3.8768. 6. 7.5866-10. 10. 4.1335.
2. 1.8035. Hint. 7. 58«0 - 10 = 8.58(5(5. 11. 6.9154.
3. .5763. 7.9.2455-10. 12.8.3464.
4. 1.3747. 8. 4.1335 - 10. 13. 0.8882.
5. 2.7649. 9. 5.7875 - (>. 14. 5.9689.
LOGARITHMS 261
If the mantissa of a given logarithm, as 1.5271, is not
in the table, the antilogarithm is obtained by interpolation
as follows:
The mantissa 5271 lies between
.5263, the mantissa of the sequence 336,
and .5276, the mantissa of the sequence 337.
Therefore the antilogarithm of 1.5271 lies between 33.6
and 33.7. Since the tabular difference is 13 and the dif-
ference between .5263 and .5271 is 8, the mantissa .5271
lies -^^ of the way from .5263 to .5276. Therefore the re-
quired antilogarithm lies ^^ of the way from 33.6 to 33.7.
Then antilog 1.5271 = 33.6 + ^83 x .1, •
and 33.6 +.061= 33.66.
Therefore when the mantissa is not found in the table
we have the
Rule. Write the number of three figures corresponding to the
lesser of two mantissas hetiveen which the given mantissa lies.
Subtract the less mantissa, from the given one and divide
the remainder by the tabular difference to two decimal places.
If the second digit is 5 or more., increase the first digit by 1;
if less than 5., omit it.
Annex the resulting digit to the three already found and
place the decimal point where indicated by the characteristic.
EXERCISES
Find the antilogarithms of the following :
1. 1.5523. 5. 1.2566. 9. 9.2664 - 10.
2. 2.3821. 6. 7.3572 - 10. 10. .7729.
3. 0.6790. 7. 9.8327 - 10. 11. 7.1060 - 10.
4. 2.5720. 8. 5.9613 - 8. 12. 6.2318 -:- 10.
262
SECOND COURSE IN ALGEBRA
145. Multiplication. Multiplication by logarithms de-
pends on the
Theorem. The logarithm of the product of tivo numbers
is the sum of the logarithms of the numhers.
. That is, for the numbers a and x
\og,,{a . X) = log^a -h log^a-.
Proof. Let log^a = l^, (1)
and log/^x = h. (2)
From (i), a = bh. (3)
From (2), x = hk. (4)
(3) X (4), ax = ¥1 + 12. (5)
Therefore lt)g^«.r = li-^ U
= logf,a + \ogf,x.
Solution.
EXERCISES
Perform the indicated operation by logarithms
1. 18 X 25.
log 18 =1.2553
log25 = 1.3979
log (18 X 25) = 2.G532 (adding)
antilog 2.6532 = 450.
6. 386 X 27.
7. 432 X 263.
8. 589 X 375.
9. 4326 X 497.
2. 37 X 28.
3. 29 X 9.
4. 9.8 X 6.
5. 42 X 3.3.
10. 2870 X 3754.
11. 286.7 X 2.391.
12. 3.412 X 2.526.
13. 432 X .574.
Solution.
log 432 = 2.6355 = 2.6355
log .574 = 1.7589 = 9.7589
10
log (432 X .574) = 2.3944 = 12.3944 - 10 (adding)
autilog 2.3944 = 247.9.
LOGARITHMS 263
Since the mantissa is always positive, any number carried over from
the tenths' cohimn to the units' column is positive. This occurs in
the preceding solution, where .6 + -7 = 1.3, giving + 1 to be added
to the sum of the characteristics + 2 and — 1, in the units' column.
Mistakes in such cases will be few if the logarithms with negative
characteristics be written as in the 9—10 notation on the right.
In the preceding example and in others which follow, two methods
are given for writing the logarithms which have negative character-
istics. This is done to illustrate those cases in which the second of
th<5^two ways is preferable. It should be understood that in practice
one, but not necessarily both, of these methods is to be used.
14. 385 X .647. 19. .6381 x -.01897.
15. 571 X 073 Hint. Determine by inspection the
sign of the product. Then operate as
16. 37.6 X .00865. if all signs were positive.
17. .0476 X 673. 20. 675 x - .0286.
18. .07325 X 6.354. 21. -.437 x -.0046.
146. Division. Division by logarithms depends on the
Theorem. The logarithm of the quotient of two numbers
is the logarithm of the dividend minus the logarithm of the
divisor.
That is, for the numbers a and x
logfc
a
' X
= logf^a - log^.r.
Proof. Let
i
log^a = /j,
(1)
(2)
From (1),
a = Uu
(•^)
From (2),
X = hh.
(4)
(3) -^(4),
X
Therefore
= logta - logftA-.
264 SECOND COURSE IN ALGEBRA
EXERCISES
Divide, iising logarithms :
1. 891^27.
Solution. log 891 = 2.9499
log 27 =1.4314
log (891 -^ 27) = 1.5185 (subtracting)
antilog 1.5185 = 33.
2. 96-5-32. 5. 439-- 27.1. 8. 9896 -- 52.78.
3. 888 -T- 47. 6. 3860 -^ 4.32. 9. 6732 -f- 7.81.
4. 976^361. 7. 4627^281. 10. 3.26 -- .0482.
Solution, log 3.26 = 0.5132 = 10.5132 - 10
log. 0482 = 2.6830=8^6830-10
log (3.26 -i- .0482) = 1.8302 = 1.8302 -
antilog 1.8302 = 67.64.
11. 2.35-^.0683. _ 347 x (- 625)
12. 4.86 --.751. * 346
13. .0635-^.277. 473.2 x 4.78
14. .2674 --3.66. * -68.3
15. .07882 H- 68.72. 9.63 x -0892
16. 356 X 392-- 128. ^^' .00635
147. Involution. Involution by logarithms depends on the
Theorem. The logarithm of the rth power of a number is
r times the logarithm of the number.
That is, for the numbers r and x, log^a^ = r log^a;.
Proof. Let logftx = I. (1)
Then x = ¥. (2)
Raising both members of (2) to the rth power,
xr = bri.
Therefore logta:^ = rl
= r log&a:.
LOGARITHMS 265
EXERCISES
Compute, using logarithms :
1. (2.73)«.
Solution. log 2.73 = .4362.
log (2.73)3 = 1.3086 (multiplying by 3).
antilog 1.3086 = 20.33.
2. (6.32)*. 3. (34.26)1 4. (6.715)1
5. (.425)1
Solution. log .425 = 1.0284 = 9.0284 - 10.
log (.425)3 = 2.8852 = 28.8852 - 30.
antilog 2.8852 = .07677.
6. (.362)1 ' 9. (486.2)2 . (3.85)3.
7. (.0972)1 10. (.375)^ • (62.5)1
8. (.003597)^ 11. (^.25)^ - (1.232)3.
148. Evolution. Evolution by means of logarithms de-
pends on the
Theorem. The logarithm of the real rth root of a number
is the logarithm of the number divided hy r.
That is, for the real numbers r and w, logf^^n= — logjn.
Proof. Let log,/* = f- (1)
Then n = bK (2)
Extracting the rth root of both members of (2),
1 11
(ny = (h^Y = hr, (3)
1 7 log, n
Therefore log. (nV = - = -^^^^ • (4)
^ r r
RE
266
SECOND COURSE IN ALGEBRA
N
1
2
3
4
5
6
7
8
9
D
10
0000
0043
0086
0128
0170
0212
0253
0294
0334
0374
42
11
0414
0453
0492
0531
0569
0607
0645
0682
0719
0755
38
12
0792
0828
0864
0899
0934
0969
1004
1038
1072
1106
35
13
1139
1173
1206
1239
1271
1303
1335
1367
1399
1430
32
14
1461
1492
1523
1553
1584
1614
1644
1673
1703
1732
30
15
1761
1790
1818
1847
1875
1903
1931
1959
1987
2014
28
16
2041
2068
2095
2122
2148
2175
2201
2227
2253
2279
20
17
2304
2330
2355
2380
2405
2430
2455
2480
2504
2529
25
18
2553
2577
2601
2625
2648
2672
2695
2718
2742
2765
24
19
2788
2810
2833
2856
2878
2900
2923
2945
2967
2989
22
20
3010
3032
3054
3075
3096
3118
3139
3160
3181
3201
21
21
3222
3243
3263
3284
3304
3324
3345
3365
3385
3404
20
22
3424
3444
3464
3483
3502
3522
3641
3560
3579
3598
19
23
3617
3636
3655
3674
3692
3711
3729
3747
3766
3784
18
24
3802
3820
3838
3856
3874
3892
3909
3927
3945
3962
18
25
3979
3997
4014
4031
4048
4065
4082
4099
4116
4133
17
26
4150
4166
4183
4200
4216
4232
4249
4266
4281
4298
16
27
4314
4330
4346
4362
4378
4393
4409
4'425
4440
4456
16
28
4472
4487
4502
4518
4533
4548
4564
4579
4594
4609
16
29
4624
4639
4654
4669
4683
4698
4713
4728
4742
4757
15
30
4771
4786
4800
4814
4829
4843
4857
4871
4886
4900
14
31
4914
4928
4942
4955
4969
4983
4997
5011
6024
5038
14
32
5051
5065
6079
6092
5105
5119
5132
5145
6159
6172
13
33
5185
5198
5211
6224
5237
5250
5263
5276
6289
6302
13
34
5315
5328
6340
5353
6366
6378
6391
5403
5416
5428
13
35
5441
5453
6465
6478
6490
5502
6514
5527
6639
5551
12
36
5563
5575
5587
6599
5611
5623
5635
6647
5658
5670
12
37
5682
5694
5705
6717
6729
5740
5752
5763
6775
5786
12
38
5798
5809
5821
6832
6843
6855
6866
6877
6888
5899
11
39
5911
5922
5933
6944
6955
6966
6977
6988
6999
6010
11
40
6021
6031
6042
6053
6064
6075
6085
6096
6107
6117
11
41
6128
6138
6149
6160
6170
6180
6191
6201
6212
6222
10
42
6232
6243
6253
6263
6274
6284
6294
6304
6314
6325
10
43
6335
6345
6355
6365
6375
6385
6395
6405
6415
6425
10
44
6435
6444
6454
6464
6474
6484
6493
6503
6513
6522
10
45
6532
6542
6551
6561
6571
6680
6590
6599
6609
6618
10
46
6(J28
6637
6646
6666
666.5
6676
6684
6693
6702
6712
9
47
6721
6730
6739
6749
6758
6767
6776
6785
6794
6803
9
48
6812
6821
6830
6839
6848
6857
6866
6875
6884
6893
9
49
6902
6911
6920
6928
6937
6946
6965
6964
6972
6981
9
60
6990
6998
7007
7016
7024
7033
7042
7050
7059
7067
9
51
7076
7084
7093
7101
7110
7118
7126
7135
7143
7152
8
52
7160
7168
7177
7185
7193
7202
7210
7218
7226
7235
8
53
7243
7251
7259
7267
7275
7284
7292
7300
7308
7316
8
64
7324
7332
7340
7348
7356
7364
7372
7380
7388
7396
8
LOGARITHMS
267
N
1
2
3
4
5
6
7
8
9
D
55
7404
7412
7419
7427
7435
7443
7451
7459
7466
7474
8
56
7482
7490
7497
7505
7513
7520
7528
7536
7543
7551
8
57
7559
7566
7574
7582
7589
7597
7604
7612
7619
7627
8
58
7634
7642
7649
7657
7664
7672
7679
7686
7694
7701
59
7709
7716
7723
7731
7738
7745
7752
7760
7767
7774
60
7782
7789
7796
7803
7810
7818
7825
7832
7839
7846
61
7853
7860
7868
7875
7882
7889
7896
7903
7910
7917
62
7924
7931
7938
7945
7952
7959
7966
7973
7980
7987
63
7993
8000
8007
8014
8021
8028
8035
8041
8048
8055
64
8062
8069
8075
8082
8089
8096
8102
8109
8116
8122
65
8129
8136
8142
8149
8156
8162
8169
8176
8182
8189
66
8195
8202
8209
8215
8222
8228
8235
8241
8248
8254
67
8261
8267
8274
8280
8287
8293
8299
8306
8312
8319
6
68
8325
8331
8338
8344
8351
8357
8363
8370
8376
8382
6
69
8388
8395
8401
8407
8414
8420
8426
8432
8439
8445
6
70
8451
8457
8463
8470
8476
8482
8488
8494
8500
8506
6
71
8513
8519
8525
8531
8537
8543
8549
8555
8561
8567
6
72
8573
8579
8585
8591
8597
8603
8609
8615
8621
8627
6
73
8633
8639
8645
8651
8657
8663
8669
8675
8681
8686
6
74
8692
8698
8704
8710
8716
8722
8727
8733
8739
8745
6
75
8751
8756
8762
8768
8774
8779
8785
8791
8797
8802
6
76
8808
8814
8820
8825
8831
8837
8842
8848
8854
8859
6
77
8865
8871
8876
8882
8887
8893
8899
8904
8910
8915
6
78
8921
8927
8932
8938
8943
8949
8954
8960
8965
8971
6
79
8976
8982
8987
8993
8998
9004
9009
9015
9020
9025
5
80
9031
9036
9042
9047
9053
9058
9063
9069
9074
9079
5
81
9085
9090
9096
9101
9106
9112
9117
9122
9128
9133
5
82
9138
9143
9149
9154
9159
9165
9170
9175
9180
9186
5
83
9191
9196
9201
9206
9212
9217
9222
9227
9232
9238
5
84
9243
9248
9253
9258
9263
9269
9274
9279
9284
9289
5
85
9294
9299
9304
9309
9315
9320
9325
9330
9335
9340
5
86
9345
9350
9355
9360
9365
9370
9375
9380
9385
9390
5
87
9395
9400
9405
9410
9415
9420
9425
9430
9435
9440
5
88
9445
9450
9455
9460
9465
9469
9474
9479
9484
9489
5
89
9494
9499
9504
9509
9513
9518
9523
9528
9533
9538
5
90
9542
9547
9552
9557
9562
9566
9571
9576
9581
9586
5
91
9590
9595
9600
9605
9609
9614
9619
9624
9628
9633
5
92
9638
9643
9647
9652
9657
9661
9666
9671
9675
9680
5
93
9685
9689
9694
9699
9703
9708
9713
9717
9722
9727
5
94
9731
9736
9741
9745
9750
9754
9759
9763
9768
9773
5
95
9777
9782
9786
9791
9795
9800
9805
9809
9814
9818
5
96
9823
9827
9832
9836
9841
9845
9850
9854
9859
9863
5
97
9868
9872
9877
9881
9886
9890
9894
9899
9903
9908
4
98
9912
9917
9921
9926
9930
9934
9939
9943
9948
9952
4
99
9956
9961
9965
9969
9974
9978
9983
9987
9991
9996
4
268 SECOND COURSE IN ALGEBKA
EXERCISES
Compute, using
logarithms :
1. -y/sje.
Solution.
log 376 = 2.5752.
log V 376 = .8584 (dividing by 3).
Then
antilog .8584 = 7.218.
2. ^783
3. S/1435. 4. ^3421
5. A/.000639.
Solution.
log .000639 = 4.8055.
If one divided 4.8055 as it stands by 3, he would be likely to
confuse the negative characteristic and the positive mantissa. This
and other difficulties may easily be avoided by adding to the char-
acteristic and subtracting from the resulting logarithm any integral
multiple of the index of the root which will make the characteristic
positive.
Thus log .000639 = 2.8055 - 6.
iog\/.000639 = .9:^52-2 (dividing by 3).
Then antilog 2.9352 = .08614.
6. VX)756. 11. (- 6.387)^. 14. a/269.
7.^.0007624 ,^^3-^^ 15. v/i74^.
t: 12. -
8. V.005679. M (3.423)«
^- ^^^-^^ • 1 (43.56)^.7.984
10. (4.925)'. N (7.623)« ' 17. ^^V? ^-07241.
18. Determine the logarithms of 5732, 573.2, 57.32, and
5.732 to the base 10 and to the base 100. Compare the results.
What fact about logarithms do these results emphasize ?
LOGARITHMS 269
Note. The preceding four-place table will usually give results
correct to one half of one per cent. Five-place tables give the man-
tissa to five decimal places of the numbers from 1 to 9999 and, by
interpolation, the mantissa of numbers from 1 to 99,999. Such tables
give results correct to one twentieth of one j)er cent, a degree of
accuracy which is sufficient for most engineering work.
Six-place tables give the mantissa to six decimals for the same
range of numbers as a five-place table, but the labor of using a six-
place table is much greater than that of using a five-place one.
Seven-place tables contain the mantissas of the numbers from
1 to 99,999. Such tables are needed in certain kinds of engineering
work and are of constant use in astronomy.
In place of a table of logarithms engineers often use an instru-
ment called a slide rule. This is really a mechanical table of
logarithms arranged ingeniously for rapid practical use. Results can
be obtained with such an instrument far more quickly than with an
ordinary table of logarithms, and that without recording or even
thinking of a single logarithuL. A slide rule ten inches long usually
gives results correct to three figures. In work requiring greater
accuracy a larger and more elaborate instrument which gives a five-
figure accuracy is used.
149. Exponential equations. An exponential equation is an
equation in which the unknown occurs in an exponent.
Many exponential equations are readily solved by means
of logarithms, since log a^ =x log a.
Thus let ci^=c. Then xloga = logc. Whence a: = logc-f- loga.
EXAMPLE
Solve for x 8"^ = 324.
Solution. log 8-^ = log 324,
or x log 8 = log 324.
Whence . = l2£^ = 1^ = 2.78+.
logs .9031
The student must overcome his hesitation actually to
divide one logarithm by another if, as here, it is necessary.
270 SECOND COURSE IN ALGEBRA
MISCELLANEOUS EXERCISES
1. Can you find the logarithm of a negative number to a
positive base ? Explain.
Find, without reference to the table, the numerical values of;
2. log^O. 6. 51og^9.
3. log^8. 7. log^8H-31og3 4.
4. log32. 8. 21og^81-41og3,27.
5. 4 log^27. 9. 3 log^l25 + 2 log,25 - 2 log^o.
10. 4 log,(i) - 5 log/Jy) + 2 log^T^-
Simplify :
11. log I + log if 13. log -V- + log U - log i-
12. log ^V - log f |. 14. 2 log 3 + 3 log 2.
15. 31og4 + 41og3-21og6.
Show that :
16. log la — ^J= log (a + x) + log (a — .r) — log a.
17. log V^'-x^ = J [log(« + x)-\- log(«. - x)].
18. log Vr(.s' - a) = I [log s + log (.s -a)y
> s — a
lllogs + log(.s - h) + log(.v - c) - log{s - u)^
Solve, using logarithms (obtain results to four figures) :
20. The circumference of a circle is 2 wR. (tt = 3.1416,
li = radius.)
(a) Eind the circumference of a tdrcle whose radius is 85 .
inches.
(b) Eind the radius of a circle whose circumference is 3281
centimeters.
LOGARITHMS 271
21. The area of a circle is ttR^.
(a) Find the area of a circle whose radius is 5.672 feet.
(b) Find the radius of a circle whose area is 67.37 square
feet.
22. The area of the surface of a sphere is 4 irR^.
(a) The radius of the earth is 3958.79 miles. Find its
surface.
(b) Find the length of the equator.
23. The volume of a sphere is —
o
(a) Find the radius of a sphere whose volume is 86 cubic feet.
(/>») Find the diameter of a sphere whose volume is 47 cubic
inches.
24. If the hypotenuse and one leg of a right triangle are
given, the other leg can always be computed by logarithms.
In the adjacent figure let a and c be given and x required.
Then
X = Vc^ - a' = V{r-\-a)(c-a).
Whence
logic = i log(c + «) 4- 1^ logO; - a).
(a) The hypotenuse of a right triangle is 377 and one leg
is 288. Find the other leg.
(b) The hypotenuse of a right triangle is 1285 and one leg
is 924. Find the other leg.
25. The area of an equilateral triangle whose side is s is
— V3. Find in square feet the area of an equilateral triangle
whose side is 34.23 inches.
Solve for x :
26. 3^ = 25. 30. 3 = (1.04)-^. 34. 10*^ "^ = 3.
27. 64^ = 4. 31. 2^ = 64. 35. 8-^ + '^ = 6.
28. 16-^ = 1024. 32. 42*^+1 = 84. 36. (.3)"^ = 5.
29. (- 2y = 64. 33. 3^ + ^ = 6561. 37. (.07)-^ = 9.
272 SECOND COURSE IN ALGEBRA
Find the number of digits in :
38. (a) 3^2; (b) 2^-, (c) 2« • 3« . 5'.
39. In how many years will $1 double itself at 3% interest
compounded annually ? *
Solution. At the end of one year the amount of $1 at 3% is $1.03 ;
at the end of two years it is |(1.03) (1.03), or $(1.03)2 ; at the end of
three years it is $(1.03)^, and at the end of x years it is $(1.03)^.
If X is the number of years required, (1.03)^ = 2.
Taking the logarithms of both members of the equation,
X log 1.03 = log 2.
o 1 • log 2 .3010 ^^^ . t^
Solvmg, x = - — ^ = = 23.5 + . *&
^ log 1.03 .0128 V^
40. In how many years will $1 double itself at 6% interest
compounded annually?
41. In how many years will any sum of money treble itself
at 4^ interest compounded annually ?
42. In how many years will $450 double itself at 3-|-%
interest compounded annually ?
43. In how many years will $4000 amount to $7360.80 at
5% interest compounded annually ?
44. About 300 years ago the Dutch paid $24 for the island
of Manhattan. At 4*^ compound interest, what would this
payment amount to at the present time ?
45. In how many years will $12 double itself at 3%
interest compounded semiannually ?
46. Show that the amount of P dollars in t years at ?•%
interest compounded annually is P (1 + ry ; compounded semi-
annually is P(H--) ; compounded quarterly is pfl + jj ;
and compounded monthly is P f 1 + —
* In making computations of this nature by the aid of logarithms, care
must be exercised not to retain more significant figures in the jesult than
are given with accuracy by the process.
LOGARITHMS 273
47. Find the amount of |5000 at the end of four years,
interest at 4% compounded {(i) annually; (b) semiannually;
(c) quarterly.
48. Find the amount of |4.12 at the end of five years,
interest at 4^^, compounded quarterly.
49. Set up and solve the equation used to determine the
amount which should be paid for a |5 certificate to be paid
in five years, interest at 4'^, compounded quarterly.
Note. It is not a little remarkable that just at the time when
Galileo and Kepler were turning their attention to the laborious
computation of the orbits of planets, Napier should be devising a
method which simpUties these processes. It was said a hundred
years ago, before astronomical computations became so complex as
they now are, that the invention of logarithms, by shortening the
labors, doubled the effective life of the astronomer. To-day the
remark is well inside the truth.
In the presentation of the subject in modern textbooks a loga-
rithm is defined as an exponent. But it was not from this point of
view that they were first considered by Napier. In fact it was not
till long after his time that the theory of exponents was understood
clearly enough to admit of such application. This relation was
noticed by the mathematician Euler, about one hundred and fifty
years after logarithms were invented.
It was by a comparison of the terms of certain arithmetical and
geometrical progressions that Napier derived his logarithms. They
were not exactly like those used commonly to-day, for the base
which Napier used was not 10. Soon after the publication (1614)
of Napier's work, Henry Briggs, an English professor, was so much
impressed with its importance that he journeyed to Scotland to con-
fer with Napier about the discovery. It is probable that they both
saw the necessity of constructing a table for the base 1 0, and to this
enormous task Briggs applied himself. With the exception of one
gap, which was filled in by another computer, Briggs's tables form
the basis for all the common logarithms which have appeared from
that day to this.
The square roots and the cube roots on the following page are
corrected to the nearest digit in the third decimal place.
274
SECOND COURSE IN ALGEBRA
No.
Squares
Cubes
Square
Roots
Cube
Roots
No.
Squares
Cubes
Square
Roots
Cube
Roots
1
1
1
1.000
1.000
51
2,601
132,651
7.141
3.708
2
4
8
1.414
1.260
52
2,704
140,608
7.211
3.733
3
9
27
1.7.32
1.442
53
2,809
148,877
7.280
3.756
4
16
64
2.000
1.587
54
2,916
157,464
7.348
3.780
5
25
125
2.236
1.710
55
3,025
166,375
7.416
3.803
6
36
216
2.449
1.817
56
3,136
175,616
7.483
3.826
7
49
343
2.646
1.913
57
3,249
185,193
7.550
3.849
8
64
512
2.828
2.000
58
3,364
195,112
7.616
3.871
9
81
729
3.000
2.080
59
3,481
205,379
7.681
3.893
10
100
1,000
3.162
2.154
60
3,600
216,000
7.746
3.915
11
121
1,331
3.317
2.224
61
3,721
226,981
7.810
3.936
12
144
1,728
3.464
2.289
62
3,844
238,328
7.874
3.958
13
169
2,197
3.606
2.351
63
3,969
250,047
7.937
3.979
14
196
2,744
3.742
2.410
64
4,096
262,144
8.000
4.000
15
225
3,375
3.873
2.466
65
4,225
274,625
8.062
4.021
16
256
4,096
4.000
2.520
66
4,356
287,496
8.124
4.041
17
289
4,913
4.123
2.571
67
4,489
300,763
8.185
4.062
18
324
5,832
4.243
2.621
68
4,624
314,4.32
8.246
4.082
19
361
6,859
4.359
2.668
69
4,761
328,509
8.307
4.102
20
400
8,000
4.472
2.714
70
4,900
^43,000
8.367
4.121
21
441
9,261
4.583
2.759
71
5,041
357,911
8.426
4.141.
22
484
10,648
4.690
2.802
72
5,184
373,248
8.485
4.160
23
529
12,167
4.796
2.844
73
5,329
389,017
8.544
4.179
24
576
13,824
4.899
2.884
74
5,476
405,224
8.602
4.198
25
625
15,625
6.000
2.924
75
5,625
421,875
8.660
4.217
26
676
17,576
5.099
2.962
76
5,776
438,976
8.718
4.236
27
729
19,683
5.196
3.000
77
5,929
. 456,533
8.775
4.254
28
784
21,952
5.292
3.037
78
6,084
474,552
8.832
4.273
29
841
24,389
5.385
3.072
79
6,241
493,039
8.888
4.291
30
900
27,000
5.477
3.107
80
6,400
512,000
8.944
4.309
31
961
29,791
5.568
3.141
81
6,561
531,441
9.000
4.327
32
1,024
32,768
5.657
3.175
8'.i
6,724
551,1368
9.055
4.344
33
1,089
35,937
5.745
3.208
83
6,889
571,787
9.110
4.362
34
1,156
39,304
5.831
3.240
84
7,056
592,704
9.165
4.380
35
1,225
42,875
5.916
3.271
85
7,225
614,125
9.220
4.397
36
1,296
46,656
6.000
3.W2
86
7,396
636,a56
9.274
4.414
37
1,369
50,653
6.083
3.3;52
87
7,569
6.58,503
9.327
4.431
38
1,444
54,872
6.164
3.:362
88
7,744
681,472
9.^381
4.448
39
1,521
59,319
6.245
3.391
89
7,921
704,969
9.4M
4.465
40
1,600
64,000
6.325
3.420
90
8,100
729,000
9.487
4.481
41
1,681
68,921
6.403
3.448
91
8,281
753,571
9.539
4.498
42
1,764
74,088
6.481
3.476
92
8,464
778,688
9.592
4.514
43
1,849
79,.507
6.557
3.503
93
8,649
804,357
9.644
4.531
44
1,936
&5,184
6.6;^3
3.530
M
8,836
830,584
9.695
4.547
45
2,025
91,12,5
6.708
3.557
95
9,025
857,375
9.747
4.563
46
2,116
97,336
6.782
3.583
96
9,216
884,7.%
9.798
4.579
47
2,209
103,823
6.856
3.609
97
9,409
912,673
9.849
4.595
48
2,304
110,592
6.928
3.6;m
98
9,604
941,192
9.899
4.610
49
2,401
117,649
7.000
3.659
99
9,801
970,299
9.950
4.626
50
2,500
125,000
7.071
3.684
100
10,000
1,000,000
10.000
4.642
INDEX
Abel, 224, 234
Addition, algebraic, 2 ; commuta-
tive law of, 2 ; of fractions, 53
Ahmes, 214
Alternation, 238
Antecedent, 235
Antilogarithm, 260
Apothem, 125
Axes, X- and y-, 69
Axiom, 16
Axioms, 16
Base, 252, 257
Binomial theorem, 227 ; extraction
of roots by, 230 ; rth term of
(a + 5)", 232-233
Binomials, difference of two
squares, 28; powers of, 227;
sum or difference of two cubes,
34
Briggs, 253, 273
Brouncker, William, 60
Cardan, 36, 165
Cauchy, 224
Characteristic, 255
Checking, rule for, 19
Circle, 179
Complex number, 159
Conjugate iniaginaries, 162
Conjugate real radicals, 120
Consequent, 235
Constant, 244, 246, 248
Coordinates of a point, 70
Cubes of numbers, 274
Decimals, repeating, 110
Diophantos of Alexandria, 81
Discriminant of a quadratic equa-
tion, 172
Distance, x~ and y-. 69
Division, by logarithms, 263 ; rule
for, 7
Elimination, 73
Ellipse, 180
Equation, definition of an, 15 ; of
condition, 15 ; derived, 77 ; inde-
pendent, 77; root of an, 16 ; satis-
fying an, 16
Equations, definition and typical
solution of irrational, 153 ; equiv-
alent, 17 ; exponential, 269 ; for-
mation of, with given roots, 168 ;
homogeneous, 190 ; with imagi-
nary roots, 136, 164 ; indetermi-
nate, 80 ; involving fractions,
61 ; linear, solution by addition
and subtraction, 75 ; in one un-
known, rule for solving, 19 ;
rule for the solution of irrational,
154 ; in several unknowns, 80 ;
solution of, by factoring, 42 ;
special cases, 76 ; use of division
in, 196
Euclid, 45, 123
Euler, 234, 273
Evolution, by logarithms, 265 ; law
of, 92
Exponent, logarithm as an, 253 ;
meaning of a fractional, 109 ;
meaning of a negative, 93 ; mean-
ing of a zero, 93
Exponents, fundamental laws of,
91
Expressions, integral, 25; reducible
to difference of two squares, 33
Extremes, 237
Factor, highest common, 44
Factor Theorem, 36
Factorial notation, 232
Factoring, definition of the process,
25 ; general directions for, 39 ;
solution of equations by, 42
Factors, prime, 25
Fraction, changes of sign in a, 49
275
276
SECOND COUESE IN ALGEBRA
Fractions, addition and subtrac-
tion of, 53 ; complex, 59 ; division
of, 57 ; equivalent, 52 ; multi-
plication of, 56 ; operations on,
47
Function, definition of a, 129;
graph of a cubic, 132 ; graph of a
linear, 130 ; graph of a quadratic,
131 ; notations for a, 130
Functions, 129 ; names of, 129
Galileo, 273
Gauss, 134, 159, 165, 224
Graph, of a cubic function, 132 ;
of a linear function, 130 ; of a
quadratic equation in two vari-
ables, 177
Graphical representation of nu-
merical data, 184
Graphical solution of an equation
in one unknown, 134; the process
of, 135
Graphical solution of a linear sys-
tem, 69, 70
Graphical solution of a quadratic
system in two variables, 181
Hindu mathematicians, 123
Hyperbola, 178
Identity, 15
Imaginaries, addition and subtrac-
tion of, 159 ; conjugate, 162 ; defi-
nitions, 158 ; division of, 162 ;
equations with imaginary roots,
164 ; factors involving, 166 ; mul-
tiplication of, 160 ; note on the
use of, 166
Imaginary, 110
Imaginary roots, 136
Index, 109
Infinite geometrical series, 220
Integral expressions, 25
Interpolation, 258
Inversion, 238
Involution, by logarithms, 264 ;
law of, 92
Irrational numbers, 109
Kepler, 180, 273
Klein, Professor Felix, 134
La Place, 224
Leibnitz, Gottfried Wilhelm, 237
Line, straight, 130
Logarithms, definition of, 252 ;
antilogarithm, 260 ; interpola-
tion, 258 ; systems of, 253 ; table
of, 266-267
Mantissa, 255, 263
Means, 237; arithmetical, 207; geo-
metrical, 217
Multiple, lowest common, 50
Multiplication, law of, 5 ; rule for,
by logarithms, 262
Napier, John (Lord of Merchiston),
234, 255, 273
Newton, 137, 223, 234
Numbers, classification of, 110;
complex, 159; imaginary, 110;
irrational, 109; orthotomic, 158;
pure imaginary, 158 ; rational,
109; real, 110
Operations, order of fundamental, 1
Origin, 70
Oughtred, 237
Parabola, 177
Parentheses, removal of, 8
Pascal's triangle, 228
Polynomials, with a common
monomial factor, 26 ; factored
by grouping tenns and taking
out a conmion binomial factor,
27
Powers, sum or difference of two
like, 37
Probability curve, 187
Problems, solution of, 21
Products, important special, 10
Progressions, connnon difference
of arithmetical, 204 ; definition
of arithmetical, 204; definition of
geometrical, 214 ; nth term of
arithmetical, 205 ; nth term
of geometrical, 216 ; ratio of
geometrical, 215
Proportion, 236 ; test of a, 237
]*roportional, fourth, 237; mean,
237; third, 237
INDEX
277
Proportions, derived by addition,
by subtraction, and by addition
and subtraction, 239 ; from equal
products, 237
Quadratic equation, character of
the roots of, 172 ; comparison of
the various methods of solution
of the, 146 ; formation of, with
given roots, 168 ; number of roots
of, 1 74 ; relation between the roots
and the coefficients of, 169 ; solu-
tion of, by completing the square,
139 ; solution of, by formula, 144
Quadratic expressions, factors of,
175
Quadratic trinomial, 30 ; rule for
solving, 30 ; the general, 31; rule
for solving the general, 82
Quantity, 243
lladicals, 109 ; addition and sub-
traction of, 116 ; algebraic sign
of. 111; conjugate, 120 ; division
of, 119 ; factors involving, 124 ;
multiplication of real, 117; simi-
lar, 116; simplification of , 112
Radicand, 109
Ratio, 235
Rational expressions, 25
Rational numbers, 109
Rationalizing factor, 119
Remainder Theorem, 35
Root, cube, of arithmetical numbers,
274 ; cube, of monomials, 101 ; defi-
nition of cube, 101 ; definition of
square, 101; principal, 101; rule
for extracting square, 106; square,
of arithmetical numbers, 106,
274 ; square, of a monomial, 101 ;
square, of polynomials, 103 ;
square, of surd expressions, 122
Roots, imaginary, 136 ; imaginary,
for a cubic equation, 137; table
of square and cube, 274
Series, arithmetical, 210 ; geomet-
rical, 219 ; infinite geometrical,
220 ; sum of, 210
Slide rule, 269 •
Squares of numbers, 274
Stifel, 230
Subtraction, of fractions, 53 ; of
polynomials, 3
Surd, 110
Systems, determinate, in three
variables, 82 ; equivalent, 196 ;
incompatible or inconsistent, 76 :
simultaneous, 76 ; solution by
addition and subtraction, 75 ; so-
lution by substitution, 74; solu-
tion of, when one equation is
linear and the other quadratic,
188; solution of, when both are
quadratic, 191 ; solution of a lin-
ear, in two variables, by graphs,
69 ; special methods for solution
of, 194 ; symmetric, 193
Table, of cubes and squares, 274 ;
of logarithms, 266-267
Tartaglia, 36, 228
Terms, similar, 2
Transposition, 17
Trinomials, perfect squares, 28
Variable, 130, 244
Variation, 243; direct, 243; inverse,
245 ; joint, 247
Wallis, John, 60, 98, 237
Zero as an exponent, meaning of,
93
VB 35926
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