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IN MEMORIAM
FLOR1AN CAJOR1
A SECOND COURSE
IN ALGEBRA
BY
WEBSTER WELLS, S.B.
PROFESSOR OF MATHEMATICS IN THE MASSACHUSETTS
INSTITUTE OF TECHNOLOGY
BOSTON, U.S.A.
D. C, HEATH & CO., PUBLISHERS
1909
Copyright, 1909, by Webster Wells
All rights reserved
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PREFACE
In the preparation of this text the author acknowledges
joint authorship with Robert L. Short, Technical High School,
Cleveland.
A knowledge of the more elementary parts of algebra is
presupposed. For this reason some definitions and rules for
operation are assumed as already known to the pupil.
Attention is called to the generalization and bringing together
of related topics. Chapter III is an example of this feature.
Here all forms of the exponent are treated. This gives oppor-
tunity to regard the logarithm as a decimal exponent and to
make the logarithmic operation laws intelligible. The intro-
duction of all linear equations and inequalities in Chapter II
shows their solution directly dependent upon the four funda-
mental operations. It is thought that the introduction of the
idea of functionality and of algebraic forms taken directly from
the calculus will be found helpful to those who expect to
pursue the study of mathematics further.
The treatment of factoring is thorough and so taken up that
Synthetic Division becomes the natural method for factoring
many higher forms and for solving equations of higher degree.
It is hoped that the treatment of variation as a proportion
will remove the reluctance with which most pupils approach
that subject in connection with their work in science.
In scope this text is sufficient preparation for most courses
in mathematics which require thorough knowledge of the
operations of algebra.
WEBSTER WELLS.
CONTENTS
CHAPTER PAGE
I. Fundamental Laws for Addition and Multiplication . 1
II. Addition, Subtraction, Multiplication, Division . . 4
Equivalent Equations 11
Equivalent Systems of Equations 16
Graphical Kepresentation 21
Inequalities 26
III. Exponents 32
Miscellaneous Examples 38
Logarithms 41
Properties of Logarithms .44
Use of Table 48
Applications 53
IV. Factors 57
Miscellaneous Examples 58
Factor Theorem ......... 60
Horner's Synthetic Division 63
Solutions by Factoring 65
Common Factors and Multiples 66
V. Fractions 73
a a
0'a 76
— , Ox 00, 00— oo 80
00
Ratio and Proportion 82
Variation . . .91
VI. Involution and Evolution . . . . . .97
Series, Binomial Theorem 108
Quadratic Surds - 117
VII. Imaginary Numbers . . . . . . . . 122
Graphs of Imaginaries . . . .... 125
VIII. Quadratic Equations . 128
Theory of Quadratic Equations ..... 136
v
VI
CONTENTS
IX.
X.
XI.
XII.
XIII.
XIV.
XV.
Graphs of Quadratic Equations
Discussion of Quadratic Equations
Problems in Physics
Factoring
Si mi ltaneous Quadratic Equations
Graphs .....
Series
Arithmetic Progressions .
Geometric Progressions
Infinite Series .
convergency and divergency
Summation ....
Differential Method
Interpolation
Undetermined Coefficients
Partial Fractions . . ,
Permutations and Combination
Determinants
Properties of Determinants
Evaluation ....
Solution of Equations
Theory of Equations .
Transformation of Equations
Descartes' Rule of Signs
Limits of Hoots .
Derivatives ....
Multiple Roots .
Sturm's Theorem
Solution of Higher Equations
Reciprocal or Recurring Equations
Binomial Equations
Cubic Equations .
Cardon's Method
Biquadratic Equations
EULBB'fl Method .
Horner's Method
PAET I
ALGEBRA
I. THE FUNDAMENTAL LAWS FOR ADDITION AND
MULTIPLICATION
1 . The Commutative Law for Addition.
If a man gains $8, then loses $3, then gains $6, and finally
loses $ 2, the effect on his property will be the same in what-
ever order the transactions occur.
Then, the result of adding +$8, -$3, + $6, and -$ 2,
will be the same in whatever order the transactions occur.
Then, omitting reference to the unit, the result of adding
-f 8, — 3, +6, and — 2 will be the same in whatever order the
numbers are taken.
This is the Commutative Law for Addition, which is :
The sum of any set of numbers will be the same in
whatever order they may be added.
2. The Associative Law for Addition.
The result of adding b + c to a is expressed a -f- (b + c),
which equals (b + c) + a by the Commutative Law for Addi-
tion (§ 1).
But (b -f- c) + a equals 6-f-c-f-a ; and b+ c-\-a equals a+b-\-c,
by the Commutative Law for Addition.
Whence, a + (b + c) = a + b + c.
Then, to add the sum of a set of numbers, we add the
numbers separately.
This is the Associative Law for Addition.
3. The Commutative Law for Multiplication.
The product of a set of numbers will be the same in
whatever order they may be multiplied.
1
2 ALGEBRA
The sign of the product of any number of terms is inde-
pendent of their order ; hence, it is sufficient to prove the
commutative law for arithmetical numbers.
Let there be, in the figure, a stars in each row, and a in a row.
b rows. *** •••
We may find the entire number of stars by multiply- *** •••
ing the number in each row, a, by the number of ***...
rows, b.
Thus, the entire number of stars is a x b. b rows.
We may also find the entire number of stars by multiplying the num-
ber in each vertical column, &, by the number of columns, a.
Thus, the entire number of stars is b x a.
Therefore, a x b = b x a,
which is the law for the product of two positive integers.
Again, let c, d, e, and / be any positive integers.
C P C X P '
Then, - x- - = ; for, to multiply two fractions, we
' d f dxf J
multiply the numerators together for the numerator of the
product, and the denominators together for its denominator.
O P P X c
Then, -x-== ; since the commutative law for multi-
d f fxd
plication holds for the product of two positive integers.
Hence, - x - = - x - ; which proves the commutative law
d f f d 1 *
for the product of two positive fractions.
4. The Associative Law for Multiplication.
To multiply by the product of a set of numbers, we
multiply by the numbers of the set separately.
The result of multiplying a by be is expressed a X (he),
which equals (be) x a, by the Commutative Law for Multi-
plication.
(be) x a equals bca, which equals abc by the Commutative
Law for Multiplication.
Whence, a x (6c) = abc
FUNDAMENTAL LAWS 3
This proves the law for the product of three numbers.
The Commutative and Associative Laws for Multiplication may be
proved for the product of any number of arithmetical numbers.
(See the author's Advanced Course in Algebra, §§ 18 and 19.)
5. The Distributive Law for Multiplication.
The law is expressed (a -f- b)c = ac + be.
We will now prove this result for all values of a, b, and c.
I. Let a and b have any values, and let c be a positive
integer.
Then, (a. -f b)c = (a + b) + (« + &) + ... to c terms
= (a + a+ ••• to c terms) -J- (p'+\b + ••• to c terms)
(by the Commutative and Associative Laws for Addition),
= ac + 6c.
II. Let a and 6 have any values, and let c = -, where e and
/are positive integers. *
Since the product of the quotient and divisor equals the dividend,
Then, (a + b) x - x / = (a + 6) x e = ae + 6e, by I.
Whence, (a + 6) x-x/=ax-x/+6x-x/.
Dividing each term by /, we have
(a+ &) x -= a x -+6 x.-.
^ / / /
Thus, the result is proved when c is a positive integer or a
positive fraction.
III. Let a and b have any values, and let c = — #, where #
is a positive integer or fraction.
(a + b)(-g) = - (a + b)g = - (ag + bg), by I and ft,
= - ag -bg = a(- g) + 6(- gr).
Thus, the distributive law is proved for all positive or nega-
tive, integral or fractional, values of a, 5, and c.
4 ALGEBRA
II. ADDITION, SUBTRACTION, MULTIPLICATION, DIVISION,
APPLICATIONS
6. Similar terms are those which do not differ at all or
differ only in their coefficients.
7. Any factor of a product may be considered the coefficient
of the product of the remaining factors.
8. To add two similar terms, write their coefficients with
the proper sign and affix the common literal part.
Ex. 1. Find the sum of ax and bx.
ax + bx = (a + b)x.
Ex. 2. Find the sum of 3 abcx and — 5 mcx.
3 abcx + (— 5 mcx) = (3ab — 5 m)cx.
This is equivalent to taking the common factor ex from the expression
3 abcx — 5 mcx.
9. To subtract two similar terms find what number added
to the subtrahend will produce the minuend. The number
added is called the difference. This is equivalent to chang-
ing the sign of the subtrahend and adding the result to the
minuend.
Ex. 3. Subtract 3 ax from 5 ax. 2 ax added to 3 ax is 5 ax.
Hence 2 ax is the difference.
Ex. 4. From 15 m take —8 m. Changing the sign (men-
tally) of —8 m, we have 15 m 4- 8 m = 23 m.
The written work should appear in this form :
15 m
— 8m
23 m
10. Three laws enter into multiplication of monomials :
TJie law of shjiis.
The. law of coefficients.
The late of ' exjumciitx.
FUNDAMENTAL PROCESSES 5
The product of two terms of like sign is positive ; the
product of two terms of unlike sign is negative.
To the product of the numerical coefficients annex the
letters ; giving to each an exponent equal to the sum of
its exponents in the factors.
The same three laws enter into division, except that quotient
takes the place of product and the exponent of the divisor is
subtracted from the exponent of the same letter in the divi-
dend. (Make a rule for division of monomials.) The reason
for such rule follows readily when division is defined as the
process of finding one of two numbers when their product and
one of the numbers are given.
11. An equation is a statement that two numbers are equal.
12. If an equation is true for all finite values of the un-
known numbers involved, it is an identical equation or identity.
13. If an equation is true only for a definite set of values
of the unknown numbers involved, it is an equation of condition.
14. An equation may not be true for any values of the un-
knowns involved. It is then said to have no roots.
15. If when a number is substituted for an unknown in an
equation, the equation becomes identical (§ 12) for that num-
ber, the equation is said to be satisfied.
The roots of an equation are the numbers which satisfy it.
A root of an equation is also called a solution of the equation.
16. Some principles used in the solution of equations are a
set of generally accepted truths called axioms. The axioms
most frequently in use are :
1. If the same number, or equal numbers, be added to
equal numbers, the resulting numbers will be equal.
2. If the same number, or equal numbers, be sub-
tracted from equal numbers, the resulting numbers
will be equal.
3. If equal numbers be multiplied by the same num-
ber, or equal numbers, the resulting numbers will be
equal.
6 ALGEBRA
4. If equal numbers be divided by the same number,
or equal numbers except 0, the resulting numbers will
be equal.
17. To solve an equation is to find its roots.
The following steps indicate the process :
$3-5 = 15. (1)
Add 5 to each member, (Ax. 1)
\x = 15 + 5 = 20. (2)
Multiply each member by 3, (Ax. 3)
2sc = 60. -( 3 )
Divide each member by 2, (Ax. 4)
x - 30. (4)
18. Two equations are equivalent when every solution of the
one is a solution of the other.
Thus equations (1), (2), (3), (4) are equivalent.
The axioms of algebra enable us to transform an equation
into an equivalent one which may be more easily solved than
the given one.
EXERCISE l
i. Add 3a- 2 b + 5c, b- 9a- 11 c, 3c + b- 2 a, b- c- a.
2. From the sum of 7 x — 8 ?/ -f- 4 z and -*2x + Bz+.f take
the sum of x — y — z and y + z — 9 x.
3. Add 3(?w + n)— 5s + £; — 8(m + n) + 4« — 11 s;
8 s - 9 (m + n) — 5 1 ; 6(m +w) — 4 g + 9 t.
4. From |p— fa-j-r take the sum of |j> + ^ a + f ?* and
\p-%q-\r.
5. Subtract a# 4- by + 02; from m 2 # — y + <2&
6. Subtract (c— d)»— (c+d)y from (2e+6d)aM-(4c — 3d)y,
7. Take mv -f a? from dkI — x 2 .
8. From 4aWc+5o6(c + d) — 9a*6c? take
(3 a + 5) &*c - aft (c H
FUNDAMENTAL PROCESSES 7
9. Simp] i f'y ( x* — 4 x 2 -f 5 x — 1) — (2 ar 3 + 5 a 2 — # — 7) +
(or 5 + 2 ar- 3 as + 2).
10. Simplify (x + 1) (x - 2) (a; - 3) - (x - 2) 2 + (ar 3 - 1).
11. Simplify (x + y) 4 - (x - y)\
12. Simplify [4 a; 2 -(2 x 4-5)] [2 x 2 - (x -3)].
13. Multiply 4 a; 2 + ^ — 2/ 2 by 3 a? 2 — 5 a;?/ -|- 4 1/ 2 .
14. Multiply a a; -|- by -f cz by bx — ay -\- cz.
15. Multiply 4 (m + ») s — 5 (wi 4- w) 4- 7
by (m 4- ^) 2 + 2(m + n)-f 1.
16. Multiply ar 2a+1 4- x a y h 4- y 2h by a; a — y h .
17. Expand (4 a 4- 3 &) 2 (4 a - 3 b) 2 .
18. Multiply 1 a 2 - \ ab + f b* by - f a 4- 1 6.
19. Multiply a 2 * 4- <&'&« 4- & 2e by a 2flr - a g b e 4- 7r e .
»20. Multiply x 2 — #!/ 4- 2/ 2 — xz — yz + z 2 by a; 4- y 4- 2.
21. Multiply x 2 + ax-\- bx 4- a& by x 4- c.
22. Divide 6x 6 ~ 19a; 5 + 12 a? 4 4- 5 ar 3 4-4 a; 2 - 6x -2
bv 2 a,* 2 3 a; 1
23. Divide a 12 + ft 12 by a 4 4- b\ J
24. Divide 32 m 5 — 243 w 3 by 2 m — 3 n.
25. Divide T i F a 3 + ^^ 3 byia4-|&.
26. Divide a 6n — 6 6w by a 2n 4- a*&" 4- b 2n .
27. Divide -i a^ 4- 3 7 g x l! + i !/ 3 by \ x 4- 1 y.
28. Divide 9 rV + 15 r 4 - 38 r 8 * - 8 s 4 - 26 r.s 3 by 5 r 8 4- 4 s 2 - ra
29. Divide 7 m 2 *+ 4 - 8 m'+V*- 1 - 12 n Ax ~ 2 by m x+2 - 2 n**~*
30. Divide a; 3 +(a + &) a; 2 - (6 a 2 - 5 ab) x + 6 a 2 b by a? 4- 3 a.
Solve the following equations and verify results :
31. (x + 2)(x-5) = x 2 -±x-4..
8 ALGEBRA
32. 6(o;-3)+5(4flj-7)+l =0.
34. 2 (3a: _2)_ i (3^-2) = i(3^-2)-17.
3S (2/ - 4) (2/ -f- 3) (2/ - 2) = (2/ - 1 ) 3 - 1.
Gt-\-T) 13 2* , t
6 15 21 5 3
37. ab + ax-\- 3b 2 — 2 a 2 = 4 be — bx-\-cx—c 2 —ac. Solve for #.
38. y — e = (x — d). Solve for x.
m
39. (a + b 4- c) (a? — 2 a) — (a — c) (a -f- &)
= ( a _6_ c )2_( a 2 + ^
ax— b , bx — c , ex — a A
40. 1 - 1 = 0.
a b e
19. It is sometimes convenient to indicate operations of
addition and subtraction. For this purpose parentheses are
used. The various forms of parentheses are :. parentheses (),
braces \ \, brackets [ ], and the vinculum .
A positive sign before parentheses indicates that the number
within is to be added. Hence, parentheses preceded by a -f
sign may be removed without changing the signs of the terms
within.
Ex. 2a + 3& + (3a-5& + c) = 2a + 3& + 3a-5& + c.
A negative sign before parentheses indicates that the num-
ber within the parentheses is to be subtracted. Hence, paren-
theses preceded by a — sign may be removed if the -f- signs of
the terms within be changed to — and the — signs to + (§ 9).
Ex. 1. 5a + 36 -(4 a + lb) « 5a +3 b - 4a — 7 b = a- 4 6.
Ex. 2. 5a+3 6-(-4a + 7&) = r>a+3&+4a-7 6 = 9a--l/>.
If the expression contains two or more parentheses, one
within the other, remove one at a time beginning with the
inner parentheses.
FUNDAMENTAL PROCESSES 9
Ex. 5 a + {3 a - (5 b + 2 a)} = ■
6 a + {3 a - 5 fr - 2 a} =
5 a + 3 a — 66 — 2 a =
6 a - 5 6.
EXERCISE 2
Simplify the following by removing the signs of aggregation,
and then uniting similar terms :
i. .9 m +(— 4 m + 6 n) — (3 m — n).
2. 2x-3y-[ox + y] + \-Sx-7y\. .
3. 4?/- 2 a 2 -[-4a; 2 - 7 xy + 5 y 2 ~] + (8 x 2 - 9 xy).
4. 3a 2 -5ab-l-±a 2 + 2ab-9b 2 l-7a 2 -6ab +.6 2 .
5. 5 a -(7 a -[9 a + 4]).
6. 7x-\-8y-10x-lly\.
7. 6 mn -f- 5 — ([ — 7 m/i — 3] — | — 5 mn — 11 \ ).
8. 2a-(-3 6+c-Ja-&j)-( 3a + 2c -[- 2& + 3c ]>
9 . 37_[4i_{i3_(56-28 + 7)}].
10. 9 m — (3 n -f J4 m — [n — 6 m] | — [m -f 7 nj).
11. In each of the above expressions find the value if
a == 1, b ±= — 2, c = — 3, m = 5, n = 2, a; = — 4, ?/ = — 1.
20. A number may be enclosed in parentheses preceded by
a + sign without changing the sign of its terms, but if a num-
ber is enclosed in parentheses preceded by a — sign, each plus
term placed in parentheses is changed to minus and each minus
term to plus.
EXERCISE 3
In each of the following expressions, enclose the last three
erms in parentheses preceded by a — sign :
1. a — b *- c -f d. 3. x -f- x 2 y — xy 2 — y 5 .
2. ?ft 3 + 2m 2 + 3m + 4. 4 . a 2 -4 6 2 + 12 6 - 9.
10 ALGEBRA
5. 4 x- — y 2 — 2 yz — z 2 . 7. x 2 — 2 xy + y 2 -f 3 x — 1 //.
6 . a 2 + // _ c 2 + c ^ 8. u 4 _5r* 3 -8n 2 + Gn + 7.
DEGREE OP A RATIONAL EXPRESSION
21 . A monomial is said to be rational and integral when it
is either a number expressed in Arabic numerals, or a single
letter with unity for its exponent, or the product of two or
more such numbers or letters.
Thus, 3 a 2 b s , being equivalent to 3 • a ■ a • b • b • ft, is rational and inte-
gral.
A polynomial is said to be rational and integral when each
3
term is rational and integral : as 2 ar ab -f- c 3 .
4
22. If a term has a literal portion which consists of a single
letter with unity for its exponent, the term is said to be of the
first degree.
Thus, 2 a is of the first degree.
The degree of any rational and integral monomial (§ 21) is
the number of terms of the first degree which are multiplied
together to form its literal portion.
Thus, 5 ab is of the second degree; 3 a 2 6 3 , being equivalent to
3 • a • a • b • b • ft, is of the fifth degree ; etc.
The degree of a rational and integral monomial equals the
sum of the exponents of the letters involved in it.
Thus, rt6 4 c 3 is of the eighth degree.
The degree of a rational and integral polynomial is the
degree of its term of highest degree.
Thus, 2 a 2 b — 3 c + d 2 is of the third degree.
23. If a rational and integral monomial (§ 21) involves a
certain letter, its degree with respect to it is denoted by its
exponent.
EQUIVALENT EQUATIONS 11
If it involves two letters, its degree with respect to them is
denoted by the sum of their exponents ; etc.
Thus, 2 ab 4 x 2 y s is of the second degree with respect to x and of the
fifth with respect to x and y.
24. An Integral Equation is one each of whose members is
a rational and integral expression (§ 21) ; as,
4a? — 5 = \y + 1.
A Numerical Equation is one in which all the known num-
bers are represented by Arabic numerals ; as,
2 x — 7 = x + 6.
25. If an integral equation (§ 24) contains one or more un-
known numbers, the degree of the equation is the degree of its
term of highest degree.
I Thus, if x and y represent unknown numbers,
ax — by — c is an equation gf the first degree ;
x 2 -f 4 x = — 2, an equation of the second degree ;
2 x 2 — 3 xy 2 = 5, an equation of the third degree ; etc.
A Linear, or Simple, Equation is an equation of the first
degree.
26. The equations of Exercise 1 were integral, first degree
in one unknown number, linear.
THEOREMS IN REGARD TO EQUIVALENT EQUATIONS
27. If the same expression be added to both members
of an equation, the resulting equation will be equivalent
to the first. ;
Let A=B (1)
be an equation involving one or more unknown numbers.
To prove the equation A + C = B + C, (2)
where C is any expression, equivalent to (1). -
Any solution of (1), when substituted for the unknown numbers,
makes A identically equal to B (§ 15).
It then makes A + C identically equal to B + C (§ 16, 1).
Then it is a solution of (2).
12 ALGEBRA
Again, any solution of (2), when substituted for the unknown num-
bers, makes A+C identically equal to B -f C.
It then makes A identically equal to B (§ 16, 2).
Then it is a solution of (1).
Therefore, (1) and (2) are equivalent.
The principle of § 16, 1, is a special case of the above.
28. The demonstration of § 27 also proves that
If the same expression be subtracted from both mem-
bers of an equation, the resulting equation will be equiva-
lent to the first.
The principle of § 16, 2, is a special case of this.
29. If the members of an equation be multiplied by
the same expression, which is not zero, and does not
involve the unknown numbers, the resulting equation
will be equivalent to the first.
Let A = B (1)
be an equation involving one or more unknown numbers.
To prove the equation A x <7 = B x (7, (2)
where C is not zero, and does not involve the unknown numbers, equiva-
lent to (1).
Any solution of (1), when substituted for the unknown numbers,
makes A identically equal to B.
It then makes A x C identically equal to B x C (§ 16, 3).
Then it is a solution of (2).
Again, any solution of (2), when substituted for the unknown num-
bers, makes A x C identically equal to B x C.
It then makes A identically equal to B (§ 16, 4).
Then it is a solution of (1).
Therefore, (1) and (2) are equivalent.
The reason why the above does not hold for the multiplier zero is,
that the principle of § 10, 4, does not hold when the divisor is zero.
The principle of § 10, 3, is a special case of the above.
30. If the members of an equation be multiplied by an ex-
pression which involves the unknown numbers, the resulting
equation is, in general, not equivalent to the first.
Consider, for example, the equation x + 2 = 3 x — 4. (1)
Now the equation
(x + 2) (x - 1) = (3 x - 4)<> - 1), (2)
EQUIVALENT EQUATIONS 13
which is obtained from (1) by multiplying both members by x — 1, is
satisfied by the value x = 1, which does not satisfy (1).
Then (1) and (2) are not equivalent.
Titus it is never allowable to multiply bath members of an
integral equation by an expression which involves the unknown
numbers; for in this way additional solutions are introduced.
31. If the members of an equation be divided by the
same expression, which is not zero, and does not involve
the unknown numbers, the resulting equation will be
equivalent to the first.
Let A=B (1)
be an equation involving one or more unknown numbers.
A B
To prove the equation — = — , (2)
where C is not zero, and does not involve the unknown numbers, equiva-
lent to (1).
Any solution of (1), when substituted for the unknown numbers,
makes A identically equal to B.
A B
It then makes — identically equal to — (§ 16, 4).
C
Then it is a solution of (2).
Again, any solution of (2), when substituted for the unknown num-
A B
bers, makes — identically equal to — •
It then makes A identically equal to B.
Then it is a solution of (1).
Therefore, (1) and (2) are equivalent.
The principle of § 16, 4, is a special case of the above.
32. If the members of an equation be divided by an expres-
sion which involves the unknown numbers, the resulting equa-
tion is, in general, not equivalent to the first.
Consider, for example, the equation
O + 2)(x - l) = (3x - 4) (a - 1). (1)
Also the equation x -}- 2 =: Sx — 4, -^ (2)
which is obtained from (1) by dividing both members by x — 1.
Now equation (1) is satisfied by the value x = 1, which does not sat-
isfy (2).
Then (1) and (2) are not equivalent.
I
14 ALGEBRA
It follows from this that it is never allowable to divide both
members of an in (earn I equation by an expression which in-
volves the unknown numbers ; for An this way solutions are lost.
33. If both members of a fractional equation be multi-
plied by the L.C.M. of the given denominators, the re-
sulting equation is in general equivalent to the first.
Let all the terms be transposed to the first member, and let
them be added, using for a common denominator the L. C. M.
of the given denominators.
The equation will then be In the form
- = 0. (1)
We will now prove the equation
-4 = 0, (2)
which is obtained by multiplying (1) by the L. C. M. of the given denomi-
nators, equivalent to (1), if A and B have no common factor.
Any solution of (1), when substituted for the unknown numbers,
A
makes — identically equal to 0.
Then, it must make A identically equal to 0.
Then, it is a solution of (2).
Again, any solution of (2), when substituted for the unknown num-
bers, makes A identically equal to 0.
Since A and B have no common factor, B cannot be when this solu-
tion is substituted for the unknown numbers.
Then, any solution of (2), when substituted for the unknown numbers,
A
makes — identically equal to 0, and is a solution of (1).
B
Therefore, (1) and (2) are equivalent, if A and B have no common
factor.
If A and B have a common factor, (1) and (2) are not equivalmt ;
consider, for example, the equations
£=JL -= 0, and x - 1 = 0.
a* - 1
The second equation is satisfied by the value x = 1, which does not
satisfy the first equation ; then, the equations are not equivalent.
EQUIVALENT EQUATIONS 15
34. A fractional equation may be cleared of fractions by
multiplying both members by any common multiple of the
denominators; but in this way additional solutions are often
introduced, and the resulting equation is not equivalent to the
first.
Consider, for example, the equation
% I % 2
x 2 — 1 x — 1
If we solve by multiplying both members by x 2 — 1, the L. C. M. of
x 2 — 1 and x — 1, we find x — — 2.
If, however, we multiply both members by (x 2 — l)(x — 1), we have
x s — x 2 + x 3 — x — 2 x 3 — 2 x 2 — 2 x + 2, or x 2 + x — 2 = 0.
The latter equation may be solved by using factors.
The factors of x 2 + x — 2 are x + 2 and x—1.
Solving the equation x -f 2 = 0, x = — 2.
Solving the equation x— 1=0, x = 1.
This gives the additional value x = 1 ; and it is evident that this does
not satisfy the given equation.
35. If both members of an equation be raised to the
same positive integral power (§ 66), the resulting equa-
tion will have all the solutions of the given equation,
and, in general, additional ones.
Consider, for example, the equation x = 3.
Squaring both members, we have
x 2 = 9, or x 2 - 9 = 0, or (x + 3) (x - 3) = 0.
The latter equation has the root 3, and, in addition, the root — 3.
We will now consider the general case.
Let A = B (1)
be an equation involving one or more unknown numbers.
Raising both members to the »th power, n being a positive integer, we
have A n = £*, or A* - B n = 0. (2)
Factoring the first number (§ 103, VII),
(A - B) (A n -* + A»~ 2 B + • • • + 5"- 1 ) = 0. (3)
Now, equation (3) is satisfied when A—B.
Whence, equation (2) has all the solutions of (1).
But (3) is also satisfied when
16 ALGEBRA
A n-l + A n-T B 4. ... 4. 7^-1 = ;
so that (2) lias also the solutions of this last equation, which, in general,
do not satisfy (1).
EQUIVALENT SYSTEMS OF EQUATIONS
36. Two systems of equations, involving two or more un-
known numbers, are said to be equivalent when every solution
of the first system is a solution of the second, and every solu-
tion of the second is a solution of the first.
are equations involving two or more unknown numbers,
the system of equations
A = 0,
mA + nB = 0,
where m and n are any numbers, and n not equal to
zero, is equivalent to the first system.
For any solution of the first system, when substituted for the un-
known numbers, makes A = and B — 0.
It then makes ^4 = and mA + nB — 0.
Then, it is a solution of the second system.
Again, any solution of the second system, when substituted for the
unknown numbers, makes A = and mA + nB = 0.
It therefore makes nB = 0, or B = Q.
Since it makes J. = and B = 0, it is a solution of the first system.
Hence, the systems are equivalent.
A similar result holds for a system of any number of equations.
Either m or n may be negative.
38. If either equation, in a system of two, be solved
for one of the unknown numbers, and the value found be
substituted for this unknown number in the other equa-
tion, the resulting system will be equivalent to the first.
Let ]— ** £
(2)
f A = B,
be equations Involving two unknown numbers, t and f.
Let W be the value at .< obtained by solving (1).
EQUIVALENT EQUATIONS • 17
Let F = Q be the equation obtained by substituting E for x in (2).
To prove the system of equations
(x=E, (3)
1 F = G ( 4 )
equivalent to the first system.
Any solution of the first system satisfies (3), for (3) is only a form of (1).
Also, the values of x and y which form the solution make x and E
equal ; and hence satisfy the equation obtained by putting E for x in (2).
Then, any solution of the first system satisfies (4).
Again, any solution of the second system satisfies (1), for (1) is only
a form of (3).
Also, the values of x and y which form the solution make x and E
equal ; and hence satisfy the equation obtained by putting x for E in (4).
Then, any solution of the second system satisfies (2).
Hence, the systems are equivalent.
A similar result holds for a system of any number of equations, in-
volving any number of unknown numbers.
39. The principles of §§ 27, 28, 29; 31, 33, 35, 36, and 37
hold for equations of any degree.
40. In the solution of an equation of Exercise 1, we replaced
each equation by an equivalent one more easily solved for the
unknown number.
41. Elimination is the process of deriving from a system of
two or more equations, a system containing one less unknown
number than the given system.
There are several methods of elimination, each method de-
pending on a process which w r ill form a second system equiva-
lent to the first.
42. A system of equations is called Simultaneous when each
contains two or more unknown numbers, and every equation
of the system is satisfied by the same set, or sets, of values
of the unknown numbers ; thus, each equation of the system
Jx + y = 6,
ix — y = 3,
is satisfied by the set of values x = 4, y = 1.
18 ALGEBRA
A Solution of a system of simultaneous equations is a set of
values of the unknown numbers which satisfies every equation
of the system ; to solve a system of simultaneous equations is
to find its solutions.
Ex. Solve (1) 2x + 6y = 9, ]
(2) x - y = 1, J
(1) 2x + 5y = %\
(8) 5x-5*/ = 5, J
- 0) 2* + 6|f = 9, ] 2s + 5y = 9,1
(4) (2a; + 5?/)+ 5s- by =9 + 6, j ' 7* = 14, J
System II is equivalent to system I, and system III is
equivalent to system II.
System III gives the required solution since (4) gives x = 2
and this value substituted in (1) gives y = 1.
Similarly it may be shown that elimination by substitution
and by comparison involve the deriving of equivalent systems
from the given system (§§ 37, 38).
Unless the equations of a given system are independent a
solution is not possible.
43. If two equations, containing two or more unknown num-
bers, are not equivalent, they are called Independent.
Consider the equations
(x + y = 5, (1)
1 x + y = 6. (2)
It is evidently impossible to find a set of values of x and y which shall
satisfy both (1) and (2).
Such equations are called Inconsistent.
EXERCISE 4
Solve the following equations, using Addition or Subtrac-
tion, Substitution or Comparison :
3x + 5y = 21. 2 — 7
1 '_ <=j>-3q.
I 8j> + g«=15.
[3^ + "^zl/ = 25.
3
|l5-2a+^=0.
5
fll*=W + l&
l2*-t* = 10.
5.
2#
+ 3.v
-6
# —
• ?/ -l
9
7
6 #
— 5 y
+ 10
5 a:
+ 3y
7 15'
' y — 3x = a.
I x + ^ 1/ = 9 a.
' a; + y + 2 3x — y ~ _x
~~4 17 ~~6'
9^5
\ X + V — 1 1/ N i
20 ALGEBRA
19.
2 a% — 4 by = a 2 — ab -f 2 6 2 .
gg-f-y.
fr**-*
-l = c + d. I a? v 1A o
20. J 4. — = 10f .
[ 2 # — y = 5 c — Id.
22. If 5 in. be added to the length and 3 in. to the breadth
of a certain rectangle, the area is increased by 120 sq. in., but
if 1 in. be subtracted from the length and 2 in. from the breadth,
the area is decreased by 70 sq. in. Find its dimensions.
23. 2 cu. ft. of water and 4 cu. ft. of ice together weigh
355 lb. The difference between the weights of 3 cu. ft. of
water and 2 cu. ft. of ice is 72 lb. 8 oz. Find the weights of a
cubic foot of each.
24. A masonry contractor held back $132.50 of the wages
due his men. His bricklayers earned $ 3 per day, and his hod
carriers $ 1.75 per day. Their combined wages for a day were
$ 256.25. He retained $ 1.50 from each bricklayer and $ 1 from
each hod carrier. How many carriers did he employ ?
25. A man rows a certain distance down stream at the rate
of 33 mi. an hour in 3^- hr. In returning it takes him 16 hr. to
reach a point 5 mi. below his starting point. Find the rate of
the current.
26. Two trains start toward each other, one from New York,
the other from Chicago. They meet in 10 hr., 40 min., the
distance between the two cities being 960 mi. If the first
train starts 3 hr. earlier than the second train, they will meet
9£ hr. after the second train starts. Find the rate of each
train.
27. A number lies between 300 and 100. If 18 is added to
the number, the last two digits change places with each other,
and if the number be divided by the number expressed by the
first two digits, the quotient is 10^ T . Find the number.
GRAPHICAL REPRESENTATION 21
28. Find two numbers whose difference is 93 and whose sum
divided by the smaller number gives a quotient of : 7 \
29. By the law of levers, the product of the weight W x by
the distance from W x to the fulcrum, F, is equal to the product
of the weight W 2 by the dis- ir, Jr .,
tance from W 2 to the fulcrum. '
A board resting across a pole balances when a 60-lb. boy is on
one end and a 100-1 b. boy on the other end. The board will
also balance if a 120-lb. boy sits 2 ft. from one end and a 60-lb.
boy sits 2 ft. from the other end. Find the length of the board.
30. If a regular hexagon is circumscribed about a given
circle, the difference between the areas of the hexagon and
circle is 32.24, and the sum of their areas is 660.56. Find the
radius of the circle.
GRAPHICAL, REPRESENTATION
44. A drawing or picture of given data or of an equation is
often of value.
45. Descartes (1596-1650) was the first mathematician to
apply measurement to equations.
It is impossible to locate absolutely a point in a plane. All
measurements are purely relative, and all positions in a plane
or in space are likewise relative. Since a plane is infinite in
length and breadth, it is necessary to have some fixed form
from which one can take measurements. For this form,
assumed fixed in a plane, Descartes chose two intersecting
lines as a coordinate system. Such a system of coordinates
has since his time been called Cartesian. It will best suit our
purpose to choose lines intersecting at right angles.
46. The Point. If we take any point M, its position is
determined by the length of the lines QM—x and PM?*p,
parallel to the intersecting lines OX and OF (Fig. £). The
values x= a and y = h will thus determine a point. The unit
of length can be arbitrarily chosen, but when once fixed remains
22
ALGEBRA
Y
Fig. 2.
the same throughout the problem under discussion. QM= x
and PM=y, we call the coordinates of the point M. x, measured
parallel to OX, is called the abscissa,
y, measured parallel to OF, is the
ordinate. OX and OFare the coordi-
nate axes. OX is the axis of x, also
called the axis of abscissas. OF is
the axis of y, also called the axis of
ordinates. 0, the point of intersec-
tion, is called the origin.
Two measurements are necessary
to locate a point in a plane.
For example, x = 2 holds for any point on the , t t y a
line AB (Fig. 2). But if in addition we demand
that y = 3, the point is fully determined by the in-
tersection of the lines AB and CD, any point on
CD satisfying the equation y = 3.
47. The Line. Consider the equation
x + y = 6.
In this equation, when values are assigned to x, we get a value
of y for every such value of x. When x =0, y=6; x=l, ?/ = 5;
x — 2, y = 4 ; x = 3, y= 3 ; x = 6, y = 1 ; etc., giving an infinite number
of values of x and 2/ which satisfy the equation.
Laying off these values on a pair of axes, as shown in
§ 46, we see that the points whose coordinates satisfy this
equation lie on the line AB (Fig. 3). It is readily seen that
there might be confusion as to the direction from the origin
in which the measurements should be taken. This is avoided
by a simple convention in signs. Negative values of x are
measured to the left of the ?/-axis, positive to the right. In
like manner, negative values of y are measured downward
from the #-axis, positive values upward. XO F, VOX'. X'OY',
Y'OX, are spoken of as the first, second, third, and fourth
quadrants respectively. (See Fig. 2.)
r>y plotting other equations of the first degree with fcwo un-
known quantities it will be seen that such an equation always
GRAPHICAL REPRESENTATION
23
represents a straight line. This line All | Fig, 3) is called the
graph of x + y=*6 and is the locus of all the points satisfying
that equation.
48. Now plot two simultaneous equations of the first degree
on the same axes, e.g. x + y = 6 and 2 x — 3 y = — 3 ( Fig. 1 ).
We see that the coordinates of the point of intersection have
the same values as the x and y of the algebraic solution of the
equations.
This is a geometric or graphical reason why there is but one
solution to a pair of simultaneous equations of the first degree
with two unknown numbers. A simple algebraic proof will
be given in the next article. Hereafter an equation of the
first degree in two variables will be called a linear equation.
A >
Y
X
o
l B
A
Y
s
\
L)
s
v*
y
\
y
/*
J
?
\
*
\
/
\
x
•*
o
\
B
Fig. 3. Fig. 4.
49. Algebraic Proof of the Principle of §48. Two simul-
taneous equations of the first degree cannot be satisfied by two
different sets of values for x and y. Given the equations
"•<•+ % =?$, (0
ex+ft = h. (2)
Eliminating ?/, («/— ^V = <"/ — hJl - (%)
Let Xi and x 2 be the roots of (3), different in value. Substituting
these roots, we have (of— eh)xi = cf — bh.
(of- eb)x 2 = cf- bh\
But X\ ^ .r 2 , .'. af-
; , whicli is impossible.
In. general, the plotting of two graphs on the same Efcses will
determine all the real solutions of the two equations, the
24
ALGEBRA
(1)
(2)
coordinates of each point of intersection of the graphs being
values of x and y which satisfy both equations.
50. It is well to introduce the subject of graphs by the use
of concrete problems which depend on two conditions and
which can be solved without mention of the word equation.
Professor F. E. Nipher, Washington University, St. Louis, proposes
the following :
t; A person wishing a number of copies of a letter made, went to a
typewriter and learned that the cost would be, for mimeograph work :
1 1.00 for 100 copies,
$2.00 for 200 copies,
$3.00 for 300 copies,
$4.00 for 400 copies, and so on.
" He then went to a printer and was made the following terms :
$2.50 for 100 copies,
$3.00 for 200 copies,
$3.50 for 300 copies,
$4.00 for 400 copies, and so on, a rise of 50 cents
for each hundred.
" Plotting the data of (1) and (2) on the same
axes, we have :
M The vertical axis being chosen for the price-units,
the horizontal axis for the number of copies.
"Any point online (1) will determine the price
for a certain number of mimeograph copies. Any
point on line (2) determines the price and cor-
responding number of copies of printer's work."'
Numerous lessons can be drawn from this problem. One is
that for less than 400 copies, it is less expensive to patronize
the mimeographer. For 400 copies, it does not matter which
party is patronized. For no copies from the mimeographer,
one pays nothing. How about the cost of no copies from the
printer? Why?
The graph offers an excellent method for the solution of
indeterminate equations in positive integers.
Ex. Solve 3x + 4ysm22 for positive integers. Plotting
the equation, we have
INEQUALITIES
25
Y
JL
4
1
*. *
O
Fig. 6.
We see that the line crosses the corner of a square only when x — 2
and x — 6. For all other integral values of x, y is fractional. The only
positive integral solutions are, therefore, x = 2, y = ±\ x = 6, |f=1-
This corresponds to the algebraic result.
51- In the equation y = — , y is dependent on x for its
value. That is, every change in x produces a change in y.
When two quantities are so related, the first is said to be a
Function of the second. Similarly y =f(x) 9 read y is a function
of x, means that y is equal to some expression in x. In place
of the equation represented by Fig. 6 one might have'
/(*) =
22-3x
f(x)=8-2x,
.F(a?)=±4-f-**
Make a graph of each of these two functions and find their
point of intersection.
EXERCISE 5
i. f(x) = lx-2±, find /(0), /(I), /(2), /(- 4), /(3f ).
2. 4> (*) = x 2 - 2 x + 1, find cf> (0), <£ (1), (2).
\
*The /(x) and F(x) mean simply different functions of fc m In these
same equations /(0) means the value of the function when is substi-
tuted for x in /(&) = 8 - 2 x, namely, /(0) = 8. Similarly /(l) = 8-2(1)
= 6,
26
ALGEBRA
Solve the following by means of graphs:
I 2 x — 5 // = — 16.
I 3 x •+• 7 7/ = 5. (
a?-5 2x-7/-l_2?/-2
hi
3
*(y)
+» — l
4
3 y -02
-44
F(»).w
LB
_ 7y-24 '
10
5a
-19
2-
3
-7a
INEQUALITIES
52. The Signs of Inequality, > and <, are read " is greater
than " and " is less than" respectively.
Thus, a > b is read " a is greater than b " ; a < b is read " a
is less than 6."
53. One number is said to be greater than another when
the remainder obtained by subtracting the second from the
first is a positive number.
One number is said to be less than another when the remain-
der obtained by subtracting the second from the first is a neg<*~
tive number.
Thus, if a — b is a positive number, a > b ; and if a — h is a
negative number, a < b.
54. An Inequality is a statement that one of two expressions
is greater or less than another.
The First Member of an inequality is the expression to the
left of the sign of inequality, and the Second Member is the
expression to the right of that sign.
Any term of either member of an inequality is called a term
of the inequality.
55* Two or more inequalities are said to subsist in the same
sense when the first member is the greater or the less in both.
Thus, a > b and c > d subsist in the same sense.
INEQUALITIES 27
PROPERTIES OF INEQUALITIES
56. An inequality will continue in the same sense
after the same number has been added to, or subtracted
from, both members.
For consider the inequality a>b.
By § 53, a — b is a positive number.
Hence, each of the numbers
{a + c) — (&+.c), and (a — c) — (b — c)
is positive, since each is equal to a — b.
Therefore, a + c > b -f c, and a — c > b — c. (§ 53)
57. It follows from § 56 that a term may be transposed
from one member of an inequality to the other by chang-
ing 1 its sign.
If the same term appears in both members of an inequality, affected
with the same sign, it may be removed.
58. If the signs of all the terms of an inequality be
changed, the sign of inequality must be reversed.
For consider the inequality a — b > c — d.
Transposing every term, d— c>b — a. (§57)
That is, b — a < d — c.
59. An inequality will continue in the same sense
after both nembers have been multiplied or divided by
the same positive number.
For consider the inequality a > b.
By § 53, a — b is a positive number.
Hence, if m is a positive number, each of the numbers
mCa — b) and a ~~ \ or ma— mb and — , is positive.
m m m
Therefore, ma>m?>, and — > — «
m m
60. It follows from §§ 58 and 59 that if both members of
an inequality be multiplied or divided by the same nega-
tive number, the sign, of inequality must be reversed-
61. If any number of inequalities, subsisting in the
same sense, be added member to member, the resulting
inequality will also subsist in the same sense.
28 ALGEBRA
For consider the inequalities a > ft, a' > ft', a" > ft", •••.
Each of the numbers, a — ft, a' — ft', a" — ft", •••, is positive.
Then, their sum a — ft + a' — ft' + a" — ft" + •••,
or a + a' + a" + ••• - (ft + ft' + ft" + •••),
is a positive number.
Whence, a + a' + a" + ••• > ft + ft' + b" + ....
If two inequalities, subsisting in the same sense, be subtracted mem-
ber from member, the resulting inequality does not necessarily subsist in
the same sense.
Thus, if a > ft and a 1 > ft', the numbers a — ft and a' — ft' are positive.
But (a — ft) — (a' — ft'), or its equal, (a — a') — (ft - ft'), may be posi-
tive, negative, or zero ; and hence a— a' may be greater than, less than,
or equal to ft — 6'.
62. If a > b and a' > ft', and each of the numbers a, a r , &, b\
is positive, then aa< ft', and a is positive,
aa'>ab' (§59). (1)
Again, since a>ft, and ft' is positive,
aft' > ftft'. (2)
From (1) and (2), aa' > ftft'.
63. If we have any number of inequalities subsisting in the
same sense, as a > b, a' > ft', a" > ft", •••, and each of the num-
bers a, a', a", •••,&, 6', &", •••, is positive, then
aa'a" ... > bb'b" ....
For by § 62,. aa'>bb'.
Also, a" > ft".
Then by § 62, aa'a" > ftft'ft".
Continuing the process with the remaining inequalities, we obtain
finall y aa'a"-- > ftft'ft".-.
64. Examples.
i. Find the limit of x in the inequality
9x
4x
+ 6y>lll.
+ 6y = ffi.
6x
6x
5x>
+ 9y =
45, a
74.
99.
[^EQUALITIES 29
Multiplying both members by 3 (§ 59), we have
21x-23<2x + 15.
Transposing (§ 57), and uniting terms,
19x<88.
Dividing both members by 19 (§ 59),
x<2.
( This means that, for any value of x < 2, 1 x — ^-* < — -f 5. ^
2. Find the limits of x and y in the following :
3x + 2p>37. (1)
2x-\-3y = 33. \2)
Multiply (1) by 3,
Multiply (2) by 2,
Subtracting (§ 56),
Multiply (1) by 2,
Multiply (2) by 3,
Subtracting, — 5 y > — 25
Divide both members by — 5, y < 5 (§ 60) .
(This means that any values of x and ?/ which satisfy (2), also satisfy
(1), provided x is > 9, and y < 5.)
3. Between what limiting valnes of x is x 2 — 4 # < 21 ?
Transposing 21, we have
x 2 -4x<21, if x 2 -4x-21 <0.
That is, if (x -f 3) (x — 7) is negative.
Now (x -f 3) (x — 7) is negative if x is between — 3 and 7 ; for if
x < — 3, both x -f 3 and x — 7 are negative, and their product positive ;
and if x > 7, both x + 3 and x — 7 are positive.
Hence, x 2 — 4 x < 21, if x > — 3, and < 7.
EXERCISE 6
Find the limits of x in the following :
1. (4* + 5) 2 -4<(8a; + 5)(2a;+3).
2 . (3 x + 2)(x + 3)- 4 x> (3 x -2)(x- 3) + 36.
3. (x + 4)(5 a - 2) + (2 * - 3) 2 > (3 x + 4) 2 - 78.
30 ALGEBRA
4 . (x-3)(x + 4)(x-5) < (x + l)(x-2)(x-3).
5. a\x - 1) < 2 6 2 (2 x - 1) - a&, if a - 2 & is positive.
Find the limits of x and y in the following :
ox-\- 6 y < 45. [ 7 * — 4 y > 11 .
3 x - 4 y = — 1 1 . J 3 . e + 7 // = : 15 .
8. Find the limits of x when
3 a;- 11 < 24- 11 a, and5s + 23<20a;+&
9. If 6 times a certain positive integer, plus 14, is greater
than 13 times the integer, minus 63, and 17 times the integer,
111 in 1 is 23, is greater than 8 times the integer, plus 31, what is
the integer?
10. If 7 times the number of houses in a certain village,
plus 33, is less than 12 times the number, minus 82, and 9
times the number, minus 43, is less than 5 times the number,
plus 61, how many houses are there?
11. A farmer has a number of cows such that 10 times
their number, plus 3, is less than 4 times the number, plus
79; and 14 times their number, minus 97, is greater than 6
times the number, minus 5. How many cows has he ?
12. Between what limiting values of x is x 2 -f 3 x < 4 ?
13. Between what limiting values of x is x 2 < 8 x — 1 5 '.'
14. Between what limiting values of x is 3 x 2 + 19 as < — 20 ?
65. If a and b are unequal numbers,
a*+V> 2ab.
For (a - ?0' 2 >° 5 6*i a ' 2 ~ 2 < fh + l)2 >°-
Transposing - 2 ab, . a 2 + V 2 >2 ab.
1. Prove that, if a does not equal 3,
(a-r-2)(tt-2) >0a- L8,
INEQUALITIES 31
By the above principle, if a does not equal 3,
a 2 + 9>6a.
Subtracting 13 from both members,
a 2 - 4 > (3 a - 13, or (a + 2) (a - 2) > a - 13.
2. Prove that, if a and & are unequal positive numbers,
a s + 6 3 > a 2 b + 6 2 a.
We have, a 2 + b 2 >2 ab, or a 2 - ab + & 2 > a&.
Multiplying both members by the positive number a + &,
a 3 + & 3 >a 2 & + & 2 a.
EXERCISE 7
i. Prove that for any value of x, except §,
3a;(3a;-10)>-25.
2. Prove that for any value of x, except $-,
* 4 o;(aj - 5) > 8 x - 49.
3. Prove that for any values of a and b, if 4 a does not
equal 3 6, (4a+ 3 6)(4 a - 3&)>66(4a - 36).
4. Prove that for any values of x and y, if o x does not
equal 4 y, ' 5x(ox-6y)>2 y(5 x-S y).
Prove that, if a and b are unequal positive numbers,
5. a 8 b + ab* > 2 a 2 b 2 .
6. a 3 +a 2 b+ab 2 +b s >2ab(a + b).
32 ALGEBRA
III. EXPONENTS
66. An Exponent is a number written at the right of and
above a number.
It is customary to speak of the number as raised to the
power indicated by the exponent.
67. The laws we shall develop are to hold for any exponent,
whether integral, fractional, positive, negative, or zero.
68. The number raised to the power is called the Base.
69. Meaning of a Positive Integral Exponent.
a 8 = a • a :• • a.
a 4 = a • a • a • a.
Similarly if m is a positive integer,
a m = a • a • a • • • • torn factors.
The following results have been proved to hold for any
positive integral values of m and n :
a m xa n = a m+n (F. C.) * (1)
(a-)" = a mn (F. C.) (2)
70. Meaning of a Fractional Exponent.
Let it be required to find the meaning of s .
If (1), § 69, is to hold for all values of m and n,
5 5 5 5,5,5 .
a 5 x a* x a* = a* * * = a 5 .
Then, the f&trc! power of a* equals a 5 . •
Hence, a* must be the cube root of a 5 , or a 3 = ^a 5 .
We will now consider the general case. p
Let it be required to find the meaning of a q , where j> and q
are any positive integers.
* F. C. refers to Wells's First Course in Algebra.
EXPONENTS 33
If (1), § 69, is to hold for all values of m and n,
P P P * + ? + ?+. "to* term* l Xq
a q xa q xa q x •• • to q factors = a q 9 q = a q = a*\
7'
Then, the gth power of a q equals oF.
p p
Hence, a q must be the qth. root of a p , or a q = ^/^p 9
Hence, in a fractional exponent, the numerator denotes
a r5ower, and the denominator a root.
For example, a* = -\/a?-, b^ = V& 5 ; x* =-%/x\ etc.
A Surd is the indicated root of a number, or expression,
which is not a perfect power of the degree denoted by the
index of the radical sign ; as V2, -\/5 9 or v^ -}- y. ,
The degree of a surd is denoted by its index ; thus, V«5 is a
surd of the third degree.
A quadratic surd is a surd of the second degree.
71. Meaning of a Zero Exponent.
If (1), § 69, is to hold for all values of m and n, we have
a m x a = a m+0 = a m .
Whence, a = — = 1.
a m
We must then define a as being equal to 1.
72. Meaning of a Negative Exponent.
Let it be required to find the meaning of a~ 3 .
If (1), § 69, is to hold for all values of m and n,
a~ 3 x a 3 = a" 3+3 = a° = 1 (§ 71).
Whence, a -3 = — •
a'
We will now consider the general case.
Let it be required to find the meaning of a~% where s repre-
sents a positive integer or a positive fraction.
34 ALGEBRA
If (1), § 69, is to hold for all values of m and n,
a~ 8 xa 8 = a~ s+8 =a () = 1 (§ 71).
Whence, a~ 8 = — •
a 8
We must then define a~ 8 as being equal to 1 divided by a 8 .
For example, a~ 2 = ~; a~i = — ; 3x~ 1 y~^ = — -; etc.,
a " a* xy*
73. It follows from § 72 that
Any factor of the numerator of a fraction may be
transferred to the denominator, or any factor of the
denominator to the numerator, if the sign of its ex-
ponent be changed.
Th a 2 b s = b s ^ aWc- 1 ^a 2 (T 4 te
US ' cd 4 a~ 2 cd 4 d* ~ b~ 3 c '
EXERCISE 8
Express with positive exponents :
i. a~ 2 b s . 5. 3xyz~ 2 . 9. 7 x 4 y~ 2 z.
2. xiy~ 2 z.' 6. 5c~^dk 10. 4 a~ G b~ 8 ck
3. 2m~ 4 w. 7- a- 2 xy~\ 11. 8wV.
4. a~ l b 4 c-\ 8. 3jr*?*. 12. rV~*f i
Transfer all literal factors from the denominators to the
numerators:
6x*
I3 - y
16.
1
2 a 2 6~ 3
19.
5 a*6~*
9 c" 3 d^
mn~ A
I4 ' 3^'
i7-
a 4
cd" 3 *
20.
4 in"W
** *£■
18.
3 (F. C). (1)
p p p p
(a~«b~«) q = (a*) q (b*y = a p &*. (2)
From (1) and (2), [(a&)*]« = (a*6«)«.
Taking the gth root of both members, we have
p p p
(ab) q = a*b«.
II. Let n = — s, where s is any positive integer or positive
fraction.
Then, (a&)-= -i- = -L(t M, J V) = o-*^.
(ab) s a 8 b s
EXERCISE 10
Find the values of the following :
1. (a s b^) 4 . 6. (25a 4 )~*.
1 \_f 7. (32al\/F^)i
2. '
x ' Zy J 8. (343 Vc-'SyK
3. (p'V'T 1 . f%xf^
\ 16 m~ 4
4. (a-V^c- 1 )-",
10.
5. (Vafy-*)*. \4afV 5
38 ALGEBRA
EXERCISE 11
Illustrative Examples.
Ex. 1. Reduce V| to its simplest form.
A surd is said to be in its simplest form when the expression
under the radical sign is rational and integral, is not a perfect
power of the degree denoted by any factor of the index of the
surd, and has no factor which is a perfect power of the same
degree as the surd.
§ = 2 8 . To be a perfect square the exponents of the factors of the
denominator must be even numbers. Hence multiplying both terms of
the fraction by 2, we have,
vTI
T'
Vs=Vts s
Ex. 2. Reduce V25 to its simplest form.
■v / 25=\ / V25= V5.
Ex. 3. Express 5 V7 entirely under the radical sign.
5V7= V5V7, or (5 2 ) 1(7)1-
By § 75, (5 2 )i(7)l = (5 2 . 7)1 = V175.
Ex. 4. Reduce (5)1, -\/3 to the same degree.
The L. C. M. of the indices of the roots is 12. Hence,
r^ = ^ / 3* = 3i\
The surds are now of the twelfth degree.
Ex. 5. Find the product of V45 and V72.
(45)1 (72)1 = (32 . 6)1 . ( 3 j . 2 3)i _ ( 2 2 . 3 4 . 5 . 2)*
= 2.3' 2 (5.2)2 = 18Vl0.
Reduce the following to their simplest form :
*■ -v^- 4- (72)1. 7 . 5(32 aV// >'.
2. (27 a 8 )*. 5 . ^128 a 4 ft*. 8. 7(80)1.
3. (45)1. 6. 3V25()a 2 .f. 9. (CSC,)'.
EXPONENTS 39
10. 4\Vl86\ I3 . (x-yXa&m 8 )*.
ii. V4a 2 -5arty. 14. (4 x 2 - 24 xb + 36 brf.
12. (a 3 -2a 2 6-fa6 2 )l 15. V(a?+3a?+2)(a?+6a?+8).
** (*)*• 18. (i)*. 20. 2V^.
17. (♦)*. 19. Ki)*- 2I /8a«\4
75;
22 . 1(1^)K 24. J^ZI.
mV25n 2 y *p-2q
, t \( 1 \ 4 ** ! A« 2 -& 2 \ }
Express entirely under the radical sign :
26. 2V5. , ' ■ / x ' \i
31. (a? + y) ) .
27. 3(2)*. V-y-J
28. a(bc 2 y. 32. Vm 2 -f mn — 2 n\
_____ m — n
29. 5 x V3 #y. ,
c + 4 / c 2 + 5c- 6 V
30. (a + 8ft)f— 1— V- 33 ' c-1^4-8c + 16y '
\a + o bj
Reduce the following to equivalent surds of the same degree :
34. V3, a/4. 37. 5Vx, 3 Vary.
35. (2)4, (i)i (5)i. 3»- («-»)*, («+*)*; &+*fy.
36. .(afy)*, («y)*, (®Y)*. 39- v^ + y 8 , Vz 4 -# 4 .
Simplify the following :
40. (18)4 + 3(50)4-2(72)4. 42. V^ + V'250 -2\Vl28.
41. 2(27)4-5(48)4+11(75)4. 43. f (12)*- f(*)* + (3)i
44. 8^80-2^/405 + 18^.
40 ALGEBRA
45. (24 a*x) * + 2 (54 ax) ' - 5 (6 a*x) • .
fatmFY famn*V . /aV 3 \*
46 ' [*r] "fawj + W '
47- a) i + 3( T V) i + ^V56.
48. ^^^~3(4a 2 x 2 )* + 5-v/8^V.
(a + 2/) 2 x 2 -y\ x J
Multiply the following:
50. V90 by V63. 56. Va 3 ^ by ^ax^f.
51. (35)* by (105)*. , 57. 2(5)* by 3(15)1
52. -^54 by -v^S. 58. 5^40 by 6(o)\
53. (3)* by (2)1 ftayi ,jtf\l / 2 y\1
54. a/2 by -^4. W A27&; \15aJ '
,55- (*)* by (I)* by (f)*. 60. VS^S • (2 a 2 )* . (6 r 5 )*.
61. 3 (2)* - 5 (3)4 by 4 (2)* + 3 (3)1
62. 5 V7 + 6 V2 by V7 - 4 V2.
63. 2 (8 a;)* - 9 (2 #)*' by (2 a) * - 3 (2 y)*.
64. 3 (a - 1)* + 4(2 a + 5)* by 2 (a- 1)* - 10(2 a + 5)*.
65. 5Vf-2V|by4Vf + 9V|.
Divide the following :
66. V72byV6. 7 o. (8 a*)'* by (16 a 4 )*.
67. 2Vl25by4V5. 7?- V722a? by V2j^
. 72. (4)* by (*)*.
68. (192)* by (12)*. • ,—1 T/ ,,_
V ; y ^ ; 73. 8 VaV by 6) l -*(?) l ^(*r + : >(^l
79. [(3a)ij* 84. (7Vn-5V3)(7Vn + 5V:V).
Express each of the following with a rational denominator :
85. 4- ' 89. !±2&
V2 v 2-V3
86. 1*. 9 o. 3(4)4^2^.
V5 2(3)*+ (10)*
^> 2 +»* 9 ( a + 6)*- («_*)*
88. a + ^ - 92. (^ + ^ + (^ 2 -.y 2 P
a - 6* (a 2 + y*)* - (x* - y 2 )*
LOGARITHMS
76. Any number may be expressed as a power of some num-
ber chosen as base.
For example, 4 = 2 2 , 8 = 2 3 , 64 = 2 6 , etc. Numbers between
4 and 8 would be expressed by 2 n where rt is 2 plus some frac-
tional number. In suclr a case the exponent is called the
Logarithm of the Number to the Base 2.
E.g. 2 is the logarithm of 4 to the base 2 ; 3 is the logarithm of 8 to
the base 2, etc.
77. The Common System of logarithms has 10 for its base.
Every positive arithmetical number may be expressed,
exactly or approximately, as a power of 10.
Thus, 100 = 10 2 ; 13 = 10 11139 -; etc.
42 ALGEBRA
When thus expressed, the corresponding exponent is called
its Logarithm to the Base 10.
Thus, 2 is the logarithm of 100 to the base 10 ; a relation
which is written log 10 100 = 2, or simply log 100 = 2. •
Logarithms of numbers to the base 10 are called Common
Logarithms, and, collectively, form the Common System.
They are the only ones used for numerical computations.
78. Any positive number, except unity, may be taken as the
base of a system of logarithms ; thus, if a x = m, where a and
m are positive numbers, then x = log ft m.
A negative number is nut considered as having a logarithm.
70. By §§ 71 and 72,
10° = 1, io- 1 ^ — = .1,
10
10* =10, io-« = A- = .oi,
10 2 = 100, 10- 3 = -i- - = .001, etc.
10 3 '
Whence, by the definition of § 76,
log 1 = 0, log .1 = - 1 = 9 - 10,
log 10 = 1, log .01 = - 2 = 8 - 10,
log 100 = 2, log .001 = - 3 = 7 - 10, etc.
The second form for log.l, log. 01, etc., is preferable in practice.
If no base is expressed, the base 10 is understood.
80. It is evident from § 79 that the common logarithm of
a number greater than 1 is positive, and the logarithm of a
number between and 1 negative.
81. If a number is not an exact power of 10, its common
logarithm can only be expressed approximately; the integral
part of the logarithm is called the characteristic, and the deci-
mal part the mantissa.
For example, log 13 = 1.1139.
Here ; the characteristic is 1, and the mantissa .1139.
-EXPONENTS 43
A negative logarithm is always expressed with a positive
mantissa, which is done by adding and subtracting 10.
Thus, the negative logarithm — 2.5863 is written 7.4137 — 10.
In this case, 7 — 10 is the characteristic.
The negative logarithm 7.4187 — 10 is sometimes written 3.4137 ; the
negative sign over the characteristic showing that it alone is negative, the
mantissa being always positive.
For reasons which will appear, only the mantissa of the
logarithm is given in a table of logarithms of number ; the
characteristic must be found by aid of the rules of §§ 82
and 83.
82. It is evident from § 79 that the logarithm of a number
between ± and 10 is equal to + a decimal ;
10 and 100 is equal to 1 + a decimal ;
100 and 1000 is equal to 2 -f- a decimal ; etc.
Therefore, the characteristic of the logarithm of a number
with one place to the left of the decimal point is ; with two
places to the left of the decimal point is 1 ; with three places
to the left of the decimal point is 2 ; etc.
Hence, the characteristic of the logarithm of a number
greater than 1 is 1 less than the number of places to the
left of the decimal point.
For example, the characteristic of log 906328.51 is 5.
83. In like manner, the logarithm of a number between
1 and .1 is equal to 9 + a decimal — 10 ;
.1 and ' .01 is equal to 8 -f a decimal — 10 ;
.01 and .001 is equal to 7 -j- a decimal — 10 ; etc.
Therefore, the characteristic of the logarithm of a decimal
with no ciphers between its decimal point and first significant
figure is 9, with —10 after the mantissa; of a decimal with
one cipher between its point and first significant figure is 8,
with —10 after the mantissa; of a decimal with two ciphers
between its point and first significant figure is 7, with — 10
after the mantissa; etc.
44 ALGEBRA
Hence, to find the characteristic of the logarithm of a
number less than 1, subtract the number of ciphers be-
tween the decimal point and first significant figure from
9, writing — 10 after the mantissa.
For example, the characteristic of log .007023 is 7, with — 10
written after the mantissa.
PROPERTIES OF LOGARITHMS
84. In ami system, the logarithm of 1 is 0.
For by § 71 , cfi** 1 ; whence, by § 78, log a l = 0.
85. In any system the logarithm of the base is 1.
For, a 1 = a ; whence, log a a = 1.
86. In any system whose base is greater than 1, the logarithm
of is — oo .*
For if a is greater than 1, a _Q0 = — = — — 0. (The discns-
«* . oo
sion of this form will be found in § 127.)
Whence, by § 78, log„ = — go.
No literal meaning can be attached to such a result as log a = — oo ; it
must he interpreted as follows :
If, in any system whose base is greater than unity, a number approaches
the limit 0, its logarithm is negative, and increases indefinitely in abso-
lute \alue.
87. In any system, the logarithm of a product is equal to the
sum of the logarithms of its factors.
Assume the equations
a x = m\ faj = log rt w,
[; whence, by § <8, {
a* = n ] [// = log a n.
Multiplying the assumed equations,
a x x a v = mn, or a x+y = mn.
Whence, log a mn = x -f- y = log a m -+- log,, ».
* oo stands for a number greater than any aarigned number. Sec § l *J<>.
EXPONENTS \~>
In like manner, the theorem may be proved for the product
of three or more factors.
By aid of § 87, the logarithm of a composite number may
be found when the logarithms of its factors are known.
Ex. Given log 2 = .3010, and log 3 = .4771 ; find log 72.
log 72 = log (2 x 2 x 2 x 3 x 3)
= tog 2 -f log2 4- log2 4- log3 4- log3
= 3 x log 2 4- 2 x log3 = .9030 4- .0542 = 1.8572.
EXERCISE 12
Given log 2 = .3010, log 3 = .4771, log 5 = .6990, log 7 = .8451,
find :
i. log 15. 4. log 125. 7. log 567. 10. -log 1875.
2. log 98. 5. log 315. 8. log 1225. 11. log 2646.
3. log 84. 6. log 392. 9. log 1372. 12. log 24696.
88. In any system, the logarithm of a fraction is equal
to the logarithm of the numerator minus the logarithm
of the denominator.
Assume the equations
a x = ml , [# = log a ra,
\ ; whence, \
a y = n J [y = \og a n.
Dividing the assumed equations,
— = — or a x y = — •
a y n' n
Whence, log a - = x — y = log a m — log a /i.
Ex. Given log 2 = .3010; find log 5.
log 5 = log i? = log 10 - log 2 = 1 - .3010 = .6990,
46 ALGEBRA
EXERCISE 13
Given log 2 = .3010, log 3 = .4771, log 7 = .8451, find :
i. log- 1 /. 4- log 245. 7. log |f. 10. log- 3 ™-
2. log- 2 f. 5. log85f. 8. log 375. 11. log 46f .
3. log 11$. 6. log 175. 9. log |f 12. log 2ft-
89. In any system, the logarithm of any power of a
number is equal to the logarithm of the number multi-
plied by the exponent of the power.
Assume the equation a x = m ; whence, x = log a m.
Raising both members of the assumed equation to the jith
power, aPX _ mP . w j ience? i g a m v — p X — p i g o m
90. In any system, the logarithm of any root of a num-
ber is equal to the logarithm of the number divided by
the index of the root.
For, log a Vm = log tt (m r ) = - log a m(§ 89).
91. Examples.
1. Given log 2 = .3010 ; find log 2*.
log I* =1* log 2 = - x .3010 = .5017.
o o
To multiply a logarithm by a fraction, multiply first by the numerator,
and divide the result by the denominator.
2. Given log 3 = .4771 ; find log V3.
log ^ = !°£» = i™ =.0506.
8 8
3. Given log 2 = .3010, log 3 = .4771, find log (2* x 3*).
By § 87, log (2* x 3*) a log 2* + log 3*
= - log 2 + - log 3 = .1003 + .5964 = .6067.
EXPONENTS 47
EXERCISE 14
Given log 2 = .3010, log 3 = .4771, Log 7 = .g4&, find;
I.
log 2 8 .
5-
log 42".
9-
log
oOl
13- log { s.
2.
log 5 7 .
6.
log 45*.
10.
log
(ft
14. log V54.
3-
log 3*.
7-
log 63*.
11.
log vs.
15. log \ 225.
4-
log 7 1
8.
log 98*.
12.
log
^7.
16. log v 1(32.
i7-
18.
log \/|.
log(f)*.
•
21. log
^2'
23-
loe w
19.
20.
log (3* x 100*).
log (5a/3).
22. log
2 i
o 6
24.
log £ .
92. In the common system, the mantissae of the log-
arithms of numbers having the same sequence of figures
are equal.
Suppose, for example, that log 3.053 = .4847.
Then, log 305.3 = log(100 x 3.053) = log 100 + log 3.053
= 2 +.4847 = 2.4847 ;
log .03053 = log (.01 x 3.053) = log .01 + log 3.053
= 8 - 10 + .4847 a 8.4847 - 10 ; etc.
It is evident from the above that, if a number be multiplied
or divided by any integral power of 10, producing another
number with the same sequence of figures, the mantissae of
their logarithms will be equal.
For this reason, only mantissae are given, in a table of Com-
mon Logarithms; for to find the logarithm of any number, we
have only to find the mantissae corresponding to its sequence
of figures, and then prefix the characteristic in accordance
with the rules of §§ 82 and 83.
This property of logarithms only holds for the •common
system, and constitutes its superiority over other systems for
numerical computation.
48 ALGEBRA .
93. Ex. Given log 2 =.3010, log 3 =.4771 ; find log .00432.
We have log 432 = log (2* x 8*) = 4 log 2 + 3 log 3 = 2.0353.
Then, by § 92, the mantissa of the result is .6858.
Whence, by § 83, log .00432 = 7.0353-10.
EXERCISE 15
Given log 2 = .3010, log 3 = .4771, log 7 = .8451, find :
1. log 2.7. 6. log .00000680. n. log 337.5.
2. log 14.7. 7. log .00125. 12. log 3.888.
3. log M. 8. log 5670. 13. log (4.5) 8 .
4. log .0162. 9. log .0000588. 14. log -y/SA.
5. log 22.5. 10. log .000864. 15. log (24.3)1
USE OF THE TABLE
94. The table (pages 50 and 51) gives the mantissas of
the logarithms of all integers from 100 to 1000, calculated to
four places of decimals.
95. To find the logarithm of a number of three figures.
Look in the column headed " No." for the first two signifi-
cant figures of the given number.
Then the required mantissa will be found in the correspond-
ing horizontal line, in the vertical column headed by the third
figure of the number.
Finally, prefix the characteristic in accordance with the
rules of §§ 82 and 83.
For example, log 108 = 2.2253 ;
log .344 - 9.5300 - 10 ; etc.
For a number consisting of one or two significant figures,
1 he column headed may be used.
Thus, let it be required to find log 88 and log 9.
\\y § ( ,)2, log 83 has the same mantissa as log 830j and log 9
the same mantissa as log 900.
Hence, log 83 = 1.9191, and log 9 =* 0.9542,
EXPONENTS 49
96. To find the logarithm of a number of more than three
figures.
i. Required the logarithm of 327.6.
We find from the table, log 327 = 2.5145,
. log 328 = 2.5150.
That is, an increase of one unit in the number produces an increase of
.0014 in the logarithm.
Then an increase of .0 of a unit in the number will increase the
logarithm by .6 x .0014, or .0008 to the nearest fourth decimal place.
Whence, log 327.6 = 2.5145 + .0008 = 2.£153.
In rinding the logarithm of a number, the difference between the next
less and next greater mantissae is called the tabular difference ; thus, in
Ex. 1, the tabular difference is .0014.
The subtraction may be performed mentally.
The following rule is derived from the above :
Find from the table the mantissa of the first three
significant figures, and the tabular difference.
Multiply the latter by the remaining figures of the
number, with a decimal point before them.
Add the result to the mantissa of the first three
figures, and prefix the proper characteristic.
In finding the correction to the nearest units' figure, the decimal por-
tion should be omitted, provided that if it is .5, or greater than .5, the
units' figure is increased by 1 ; thus, 13.26 would be taken as 13, 30.5 as
31, and 22.803 as 23.
2. Find the logarithm of .021508.
Mantissa 215 = .3324 Tab. diff. = 21
2 .08
.3326 Correction = 1.68 = 2, nearly.
The result is 8.3326 - 10.
EXERCISE 16
Find the logarithms of the following :
i. 64. 5. 1079. 9. .00005023. 13. 7.3165.
2. 3.7. 6. .6757. 10. .0002625. 14. .019608.
3. 982. 7. .09496. 11. 31.393. 15. 810.39.
4. .798. 8. 4.288. 12. 48387. 16. .0025446.
50
ALGKIJUA
No.
1
2
3
4
5
6
7
8
9
IO
0000
0043
0086
0128
0170
0212
0253
0294-
0334
o374
ii
0414
0453
0492
0531
0569
0607
0645
0682
0719
°755
12
0792
0828
0864
0899
0934
0969
1004
1038
1072
1 106
13
1 139
"73
1206
1239
1271
J 303
1335
1367
1399
1430
14
1461
1492
1523
J 553
1584
1614
1644
1673
'703
1732
15
1761
1790
1818
1847
1875
1903
1931
1959
1987
2014
16
2041
2068
2095
2122
2148
2175
2201
2227
2253
2279
17
2304
2330
2355
2380
2405
2430
2455
2480
2504
2529
18
2553
2577
2601
2625
2648
2672
2695
2718
2742
2765
19
2788
2810
2833
2856
2878
2900
2923
2945
2967
2989
20
3010
3032
3«54
3075
3096
3118
3139
3160
3181
3201
21
3222
3243
3263
3284
3304
3324
3345
3365
3385
3404
22
3424
3444
34^4
3483
3502
3522
354i
35 6 °
3579
3598
23
3617
3636
3655
3<>74
3692
3711
3729
3747
3766
3784
24
3802
3820
3^
3856
3874
3892
3909
3927
3945
3962
25
3979
3997
4014
4031
4048
4065
4082
4099
4116
4133
26
415°
4166
4183
4200
4216
4232
4249
4265
4281
4298
27
43H
433o
434<3
4362
4378
4393
4409
4425
4440
445 6
28
4472
4487
4502
4518
4533
4548
45 6 4
4579
4594
4609
29
4624
4639
4654
4669
4683
4698
47U
4728
4742
4757
30
477i
4786
4800
4814
4829
4843
4857
4871
4886
4900
31
49H
4928
4942
4955
4969
4983
4997
501 1
5024
5038
32
505 1
5°65
5079
5092
5^5
5119
5 J 32
5H5
5 J 59
5172
33
5185
5198
5211
5224
5237
5 2 5°
5263
5276
5289
5302
34
5315
5328
5340
5353
5366
5378
539i
5403
54i6
5428
35
5441
5453
5465
5478
549o
5502
55H
5527
5539
555i
36
5563
5575
5587
5599
5611
5 62 3
5635
5 6 47
5658
5670
37
5682
5 6 94
5705
57*7
5729
5740
5752
5763
5775
5786
38
5798
5809
5821
5832
5843
5855
5866
5877
5888
5899
39
591 1
5922
5933
5944
5955
5966
5977
5988
5999
6010
40
6021
6031
6042
6053
6064
6075
6085
6096
6107
6117
4i
6128
6138
6149
6160
6170
6180
6191
6201
6212
6222
42
6232
6243
6253
6263
6274
6284
6294
6304
631 |
6325
43
6335
6345
6 355
6365
6375
6385
6395
6405
6415
6425
44
6435
6444
6454
6464
6474
6484
M3
6503
6513
6522
45
6532
6542
6 55i
6561
6571
6580
6590
6599
6609
6618
46
6628
6637
6046
6656
6065
6675
6684
6693
6702
6712
47
6721
6730
6739
6749
6758
6767
6776
6785
6794
6803
48
6812
6821
6830
6839
6848
6857
6866
6875
6884
6S93
49
6902
691 1
6920
6928
(my
6946
6955
6964
6972
6981
5o
6990
6998
7007
7016
7024
7°33
7042
7050
7°59
7067
5i
7076
7084
7°93
7101
7110
71 1 S
7126
7 ! 35
7H3
7152
52
7160
7168
7'77
7185
7193
7202
7210
7 j 1 s
7226
7235
53
7243
7251
7259
72(7
7-^75
7284
7292
73°°
7308
7316
54
7324
7332
7340 7348
735 6
7.;<>4
7372
7380
7388
7396
No.
1
2 ! 3
4
5
6
• 7
8 9
EXPONENTS
51
No.
1
2
3 4
5
6
7 8
9
55
7404
7412
7419
7427
7435
7443
745 "
7459
7466
7474
56
7482
7490
7497
7505
7513
7520
7528
7536
7543
755 1
57
7559
7566
7574
7582
7589
7597
7604
7612
7619
7627
58
7634
7642
7649
7 6 57
7664
7072
7679
7686
7694
7701
59
7709
7716
7723
773i
7738
7745
7752
7760
7767
7774
60
7782
7789
7796
7803
7S10
7818
7825
7832
7839
7846
61
7853
7860
7868
7875
7882
7889
7896
7903
7910
7917
62
7924
793i
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
68
8325
8331
8338
8344
8351
8357
8363
8370
8376
8382
69
8388
8395
8401
8407
8414
8420
8426
8432
8439
8445
70
8451
8457
8463
8470
8476
8482
8488
8494
8500
8506
71
8513
8519
8525
8531
8537
8543
8549
8555
8561
8567
72
8573
8579
8585
8591
8597
8603
8609
8615
8621
8627
73
8633
8639
8645
8651
8657
8663
8669
8675
8681
8686
74
8692
8698
8704
8710
8716
8722
8727
8733
8739
8745
75
8751
8756
8762
8768
8774
8779
8785
8791
8797
8802
76
8808
8814
8820
8825
8831
8837
8842
8848
8854
8859
77
8865
8871
8876
88S2
8S87
8893
8899
8904
8910
8915
78
8921
8927
8932
8938
8943
8949
8954
8960
8965
8971
79
8976
8982
8987
8993
8998
9004
9009
9015
9020
9025
80
9031
9036
9042
9047
9053
9058
9063
9069
9074
9079
81
9085
9090
9096
9101
9106
9112
9117
9122
9128
9133
82
9138
9H3
9149
9154
9159
9165
9170
9175
9180
9186
83
9191
9196
9201
9206
9212
9217
9222
9227
9232
9238
84
9243
9248
9253
9258
9263
9269
9274
9279
9284
9289
85
9294
9299
9304
93°9
9315
9320
9325
9330
9335
9340
86
9345
935°
9355
9360
9365
937°
9375
9380
9385
939o
87
9395
9400
9405
9410
9415
9420
9425
9430
9435
9440
88
9445
945°
9455
9460
9465
9469
9474
9479
9484
9489
89
9494
9499
95°4
9509
95 J 3
95 l8
9523
9528
9533
9538
90
9542
9547
955 2
9557
9562
9566
9571
9576
958i
95 86
9 1
9590
9595
9600
9605
9609
9614
9619
9624
9628
9633
92
9638
9643
9647
9652
9657
9661
9666
9671
9675
9680
93
9685
9689
9694
9699
97°3
9708
97 J 3
9717
9722
9727
94
9731
9736
974i
9745
9750
9754
9759
9763
9768
9773
95
9777
9782
9786
9791
9795
9800
9805
9809
9814
9818
96
9823
9827
9832
9836
9841
9845
9850
9854
9859
9863
97
9868
9872
9877
9881
9886
9890
9894
9899
995
Next less mant. sa 5203 ; liunres corresponding, 880,
Tab. diff. 13)2. ()()(. IT, = .2, nearly.
1 3
70
EXPONENTS 53
By the above rule, there will be one otpber to be placed between the
decimal point and first significant figure ; the result is .03362.
The correction can usually be depended upon to only one decimal
place ; the division should be carried to two places to determine the last
figure accurately.
EXERCISE 17
Find the numbers corresponding to the following logarithms :
I.
0.S189.
6.
8.7954 - 10.
ii.
1.3019.
2.
7.6064 -
-10.
7-
6.5993 - 10.
12.
4.2527 -
-10.
3-
1.8767.
8.
9.9437 - 10.
*3-
2.0159.
4-
2.6760.
9-
0.7781.
14-
3.7264 -
10.
5-
3.9826.
10.
5.4571 - 10.
15-
4.4929.
APPLICATIONS
98. The approximate value of a number in which the opera-
tions indicated involve only multiplication, division, involu-
tion, or evolution may be conveniently found by logarithms.
The utility of the process consists in the fact that addition
takes the place of multiplication, subtraction of division,
multiplication of involution, and division of evolution.
i. Find the value of .0631 x 7.208 x .51272.
By § 87, log (.0631 x 7.208 x .51272)
s± log .0031 + log 7.208 + log .51272.
log .0631 = 8.8000 - 10
log 7.208= 0.8578
log. 51272 = 9.7099-10
Adding, log of result = 19.3677 - 20 = 9.3677 - 10. (See Note 1.)
Number corresponding to 9.3677 - 10 = .2332.
Note 1: If the sum is a negative logarithm, it should be written in
such a form that the negative portion of the characteristic may \& — 10.
Thus, 19.3677 - 20 is written 9.3677 - 10.
(In computations with four-place logarithms, the result cannot usually
be depended upon to more than four significant figures.)
54 ALGEBRA
2. Find the value of ■ ' *
7984
By § 88, log ^ = log 33(5.8 - log 7984.
log 336.8 = 12.5273 -10
log 7984 = 3.0022
Subtracting, log of result = 8i>251 - 10 (See Note 2.)
Number corresponding = .04218.
Note 2 : To subtract a greater logarithm from a less, or a negative
logarithm from a positive, increase the characteristic of the minuend by
10, writing — 10 after the mantissa to compensate.
Thus, to subtract 3.9022 from 2.5273, write the minuend in the form
12.5273 - 10 ; subtracting 3.9022 from this, the result is 8.6251 — 10.
3. Find the value of (.07396) 5 .
By § 89, log (.07396) 5 - 6 X, log .07396.
log .07396 = 8.8690 - 10
» 5^ '
44.3450 - 50
= 4.3450 - 10 = log .000002213.
4. Find the value of ^.035063.
By § 90, log ^035063 = 1 log .035063.
log .035063 = 8.5449 - 10
3 )28.5449 - 30 (See Note 3.)
9.5150 - 10 = log .3224.
Note 3 ; To divide a negative logarithm, write it in such a form that
the negative portion of the characteristic may be exactly divisible by the
divisor,* with — 10 as the quotient.
Tims, to divide 8.5449 — 10 by 3, we write the logarithm in the form
28.5449 - 30 ; dividing this by 3, the quotient is 9.5150 - 10.
EXERCISE 18
A negative number has no common logarithm (§ 78) ; if such numbers;
occur in computation, they may be treated ;is if they were positive, and
the sign of the result determined irrespective ol the logarithmic work.
Thus, in Ex. 3 of the following set, to tind the value of ( -96.86) X&S918
we find the value pi 96.86 x 3.8918, and put a — sign before the result.
BXPONEKT8 55
Find by logarithms the values of the following :
1.4.253x7.104. 4 54.029 X (-.0081487).
2. 6823.2 x .1634. 5 . .040764 x .12896.
3. (- 95.86) x 3.3918. 6. (-285.46) x (-.00070682).
„ 5978 -38.19
7 - 9J62* I0 - 10792' '* (8«.08)«.
8 . 21^58. fl . 670.48 . ^ C09 437)<.
45057 -5382.3
.06405 .000007913 ,o*okm
r» I2 . k. (3.625V.
.002037 .00082375
Arithmetical Complement
99. The Arithmetical Complement of the logarithm of a num-
ber, or, briefly, the Cologarithm of the number, is the logarithm
of the reciprocal of that number.
Thus, colog 409 = log ^ = log 1 - log 409.
log 1 = 10. - 10 (See Ex. 2, § 98.)
log 409= 2.6117
.-. colog 409= 7.3883-10.
Again, colog .067 = log-—- = log 1 — log .067
.067
log 1 = 10. - 10
log .067= 8.8261-10
•\ colog .067= 1.1739.
It follows from the above that the cologarithm, of a number
may be found by subtracting its logarithm from 10 — 10. _
The cologarithm may be found by subtracting the last significant figure
of the logarithm from 10 and each of the others from 0, — 10 being written
after the result in the case of a positive logarithm.
56 ALGEBRA
51384
Ex. Find the value of
8.708 x .0946
log - 51384 = log L 51384 x-Lxi)
b 8.708 x .01)40 V ^-< () « .0M6/
:log .51384 -flog— — + log- *
8.708 .0940
= log .51384 + colog 8.708 + colog .0946.
log .51384 = 9.7109 -10
colog 8.708 = 9.0001 -10
colog .0946 = 1.0241
9.7951 -10 = log .6239.
It is evident from the above example that, to find the loga-
rithm of a fraction whose terms are the products of factors, we
add together the logarithms of the factors of the numerator, and
the cologarithms of the factors of the denominator.
The value of the above fraction may be found without using cologa-
rithms, by the following formula :
log ^M = iog .51384 -log(8. 709 x .0946)
. 8. 709 x. 0946 8 5V J
e= log .51384 - (log 8.709 + log .0946).
The advantage in the use of cologarithms is that the written work of
computation is exhibited in a more compact form.
MISCELLANEOUS EXAMPLES
2^5
100. i. Find the value of
3*
log^? = log 2 + log i/6 + colog 3* (§ 99)
35
= log 2 + J log 5 + j colog 3.
log 2= .3010
IOg6= .6990; -3= .2330
colog 3 = 9.5229 - 10 ; x 9 = 9.(1024 - 10
.1864 =!(»-!.:
FACTORS 57
2. Find the value of J 1 / -- 03296 .
yi 7.962
^ •fiBW = llog« = 1 (log .03296 - log 7.902).
° * 7.902 3 7.962 3 V b y
log .03296 = 8.5180-10
log 7.962 = 0.9010
3)27.6170-30
9.2057- 10 = log. 1606.
The result is — .1606.
EXERCISE 19
Find by logarithms the values of the following:
207 8.5 x .05834 (- .076917) x 26.3
f' .3583 x 34o 3 ' .5478 x (- 3120.7) *
(J 6.08) x. 1304 a .8102 x (- 6.225) ,
'" (- 0721) x(- 17.976)'
15. V6xvl6x^2.
6 f- _^18_\*
1 ' V 8.7 x .0603J
•y/ .008546
-\/.0(>03867
8 (--14582)* ,
^/^00l' *?' 7^1000 -(.72346)*
17
IV. FACTORS
101. An irrational number is a numerical expression involv-
ing surds; as -^3, or 2 + V5 (§ 70).
102. A rational and integral expression is resolved into its
prime factors when further factoring would produce irrational
factors.
58 ALGEBRA
103. In the First Course we considered the following eight
types of factorable numbers:
TYPE FORMS
I. a 2 -b* =(a + b)(a-b).
II. a 2 + 2a6 + 6 2 ^(« + 6)(a+6),
a 1 -2ab + b 2 = (a - b) (a-b).
III. as 2 + ate + 6.
IV. aas' 2 + foe + c.
V. x 4 + ax 2 y* + y*.
VI. a 3 + & 3 =(a + &)(a 2 -a& + & 2 ),
a 3 -b*=(a- b)(a 2 + ab + b*).
VII. a n -b n ,
a n + 6 n .
VIII. aa? + a?/ -h as = a(» + y + «).
Of these types, IV is more readily factored by means of
VIII as follows:
Ex. Factor 6 x 2 - 7 x - 20.
Multiply — 20 by 6 (the coefficient of x 2 ). Factor — 120 so that the
sum of the factors is — 7 (the coefficient of x). These factors are —15, 8.
Then write a 9 0ft . , 1K , Q **%
6 x 2 — i x — 20 = 6 x 2 — 15 x + 8 x — 20.
Group by Type VIII, s= 3 jc(2 X - 5) + 4(2x - 5),
whence, 6 x 2 - 7 z - 20 = (2 x - 5) (3 x + 4) .
Type VI may be placed under Type VII.
EXERCISE 20
Factor :
i. 3a?-x-10. 5. a** + 4.
2. 4a 2 + 12a + 9. 6. 3 s + 8.
3. a 3 -?/ 3 . 7. a 2 + 96 2 -4c 2 + 6a&.
4. a 3 4- a 2 -2a -2. 8. a^+2^+2/ 2 +8(x+^)4-16.
FACTORS 59
9. b*wS-4>+1. 18. #+d+aM-«
10. 6a 2 -17a + 12. 19. ar 3 + 3# 2 + 3;r + 1.
11. 9a 4 -13;r 2 + 4. ; 20. 9ra 2 -36ran.
12. .^ + 7^-8. • 21. a»-a* + a*-l
13. ( a .-6) 2 -2(a-o)-35. 22. Ga 2 6 -4a 2 + 15afc - 10a.
14. m 2 + (a — Z>) m — a&. 23. a" 2 — 32.
15. m*-l. 24. 9 or 4 + 12 or 2 + 4.
16. (2a-36) 2 -(a-&) 2 . 25. 9a 2 - 30 ab + 25//- 4 c 2 .
17. p^-f^r 1 4- 12. 26. 9a 2 -25c 2 + 6 2 + 6aZ>.
27. 36 a 4 -61 a 2 + 25.
28. (3a-&) 2 -6a(3a-&) + 27a-96.
29. 2 a" 6 + 250. 32. a 6 + i/\
30. (7x + 2) + 3V7x + 2 + 2. 33. 16^ + 14^-15.
31. m(2a?-3)-4m 2 a? 2 + 9m 2 . 34. 25(?m + 3) 2 +10(m + 3) + l.
35. 8(2a-56)- 1 -12(2a-56)"M-4.
36. 143 A: 2 -103 A: + 14.
37 . Var + 4a-6 + 2a 2 -l + 4(2#-3).
38. g* + g 2 t 2 + t\ 42. *P+$.
39. x* + a 2 x — a 3 — az 2 . 43. a? 4 — 13 x 2 + 4.
40. c-3d-18-9d + 2c~*. 44- 9 ey- 16 e/ 3 .
41. r 4 -20r 2 + 99. 45. 304 v 2 + 25 v - 6.
46. (a> + l)' + 2(a> + l)* + l.
47. &(*+ff + H*+fff(*f fl-
ip. a 6 -64. 50- 25a 4 + a 2 ,+ l.
49. 6 or 4 — 41 x~hfr — 7 y. 51. 4+ a 3 — a 2 — 4 a.
52. m 4 — 1 + ra — m 3 .
53. 3(^ + l)+5(x- 2 -l) + (.r + l) 2 .
54. ear^ + lSar 1 -^. 57- ;> 2 -0-
55. a*+fti 58. 4a ? * +4 -4a* +2 + l.
56. **-#V. 59- a 2 x-9^ + 2a--18.
60 ALGEBRA
60. 4/>V + 20pr/-lG/A/-80^ / .
61. (x 2 - 2 x + 1) - (a + 1) 2 . 65. a 5 - 27 a 2 + 243 -9 a 3 .
62. 52m — 10m 2 — 10. 66. 2 2m -\- A x • 2 m — 21 x 2 .
63. s*»-j* 67. a + 2Va& + &.
64. a 8 -256. 68. (2 a - 3&) 2 - (3 a- 2 &) 2 .
69. a 2 + 2a&-f-c 2 -2&c-2ac+6 2 .
70. 27m 3 -54m 2 -f-36m-8. 72. 9 a 2 - 6a -4 6 2 - 45.
71. ar 5 + a? 4 + oj 8 4- # 2 4^ + l. 73. 1 + 2ab - a 4 - a 2 b 2 - 6 4 .
74. am 2 — m s 4 2 am 71 4- an 2 4- 2 m 2 n — mn 2 .
75- 2/ 2 + rz-2?/-z4-l.
FACTOR THEOREM
104. The Remainder Theorem.
Let it be required to divide px 2 4- qx 4- r by a; — a.
pas 2 + #x + r I x — a
px* - ap x \ px + ( ap + g)
(ap + g)a
(ap 4 q)x — p a 2 — ga
pa' 1 + qa + r, Remainder.
We observe that the final remainder,
pa 2 + qa-\-r,
is the same as the dividend with a substituted in place of x ;
this exemplifies the following law :
If any polynomial, involving x t be divided by x - a, the
remainder of the division equals the result obtained by
substituting a for x in the given polynomial.
This is called The Remainder Theorem.
To prove the theorem, let
px n 4- qx n ~ l 4- ••• +r#4s
be any polynomial involving x.
Let the division of the polynomial by # — a be (ferried on
until a remainder is obtained which does not contain x.
Let Q denote the quotient, and E the remainder.
FACTORS 61
Since the dividend equals the product of the quotient and
divisor, plus the remainder, we have
Q(x — a) + B'=:pa? + qx?- l 4- ... + rx + s.
Putting x equal to a, into the above equation, we have,
11 =2M n + qa n ~ l -j- ... +ra + s.
105. The Factor Theorem.
If any polynomial, involving x, becomes zero when as
is put equal to a, the polynomial has a? — a as a factor.
For, by § 104, if the polynomial is divided by x — a, the
remainder is zero.
106. Examples.
i. Find whether x — 2 is a factor of ar 3 — 5 ic 2 -f 8.
Substituting 2 for sc, the expression x s — 5 x 2 + 8 becomes
23 _ 5 . 22 + 8, or - 4.
Then, by § 104, if r 3 — 5 x 2 + 8 be divided by x — 2, the remainder is
— 4 ; and sc — 2 is not a factor.
2. Find whether m + wisa factor of
m 4 — 4 m*ra -f- 3 m 2 a 2 + 5 mn 3 — 2 ?* 4 . (1)
Putting m = — n, the expression becomes
7i 4 4- 4 n 4 -1- 2 ra 4 — 5 n* — 2 n 4 , or 0.
Then, by § 104, if the expression (1) be divided by ra 4- ?i, the re-
mainder is ; and m + n is a factor.
• 3. Prove that a is a factor of
(94-$,+ c)(afr + &c + ca) - (a + &)(& + c)(« + «■)•
Putting a ±= 0, the expression becomes
(b 4- c)frc - $(ft 4- c)c, or 0.
Then, by § 104, a — 0, 01 a, is a factor of the expression.
62 ALGEBRA
4. Factor x*-3x 2 - Ux- 8.
The positive and negative integral factors of 8 are 1, 2, 4, 8, — 1, — 2,
- 4, and - 8.
It is best to try the numbers in their order of absolute magnitude.
If x = 1, the expression becomes 1—3—14 — 8.
If x = — 1, the expression becomes — 1 — 3 + 14—8.
If x = 2, the expression becomes 8—12 — 28 — 8.
If x =— 2, the expression becomes —8—12+28 — 8, or 0.
This shows that x + 2 is a factor.
Dividing the expression by x + 2, the quotient is x 2 — 5 x — 4.
Then, x 3 - 3 x 2 - 14 a - 8 = (x + 2)0 2 - 5 x - 4).
EXERCISE 21
Factor the following :
1. a s + 8. 9. a n -6 n .
2. m 5 -f-?r\ 10. 2ar 3 -f5a; 2 — # — 6.
3. a; 6 - 729. 11. a 4 -ar* + 2a; 2 -4.
4. ar 3 + 5a; 2 -8a; + 2. - 12. 5a 3 -18a-4.
5. m 3 - 11 ?7i — 10. • 13. ar 3 + a,- 2 +-7 a; + 18.
6. a 4 — a 3 + 3 a — 14. 14. m 3 — 5 m 2 — 36.
7 . c 3_2c 2 -9. 15. A; 4 -5A; 2 + 3A;-2.
8. a; 4 -625.
Find without actual division :
16. Whether p — 1 is a factor of p 3 -+ 3p 2 — 4.
17. Whether a; -f- 2 is a factor of x A -f 3 x 3 — 4 a,\
18. Whether a? + 1 is a factor of 2 a? + 6 x 2 - 3 a? + 4.
19. Whether m — 3 is a factor of m 3 — 4 m — 15.
20. Whether a — 5 is a factor of a 3 — 3 a 2 — 5 a — 25.
21. Whether c - 2 is a factor of 3 c 3 — 9 c 2 + 5 c + 2.
22. Whether a is a factor of a(b — c) -f b (c — a) + c(a — b).
23. Whether c is a factor of a (b — c) + b (c — a) + c(a — b).
24. Whether x +- 1/ is a factor of a? (2 a? -f- 3 ?/) — y (3 a? +- 2 // ).
25. Whether b is a factor of a 2 (b - c) 2 + 6 2 (c - a) 2 + c 2 (a - b) 2 .
FACTORS 63
HORNER'S SYNTHETIC DIVISION
107. The method of synthetic division, or as it is sometimes
known, the method of detached coefficients, greatly abridges
the work of division, especially where binomial divisors are
concerned.
108. Divide ar 3 - 11 x 2 + 36 x - 36 by x - 3.
Writing dividend and divisor with coefficients only,
1-3
1-8 + 12 Quotient.
1-11 + 36-
-36
1- 3
- 8
- 8 + 24
+ 12-
-36
+ 12-
-36
Since the first term of each partial product is merely a repetition of
the term immediately above, it may be omitted.
We may also change the sign of the second term of the divisor if the
partial product is added instead of subtracted.
We then have
1 _ ii + 36 - 36 II +3
1+3 11-8 + 12
-8
-24
• +12
+ 36
Raise the numbers — 24, 36 now in the oblique column and the work
stands :
1- 11+36-361 + 3
4- 3-24 + 36
- +12
The quotient is x 2 — 8 x + 12.
If the last remainder is zero, x minus the divisor is a factor of the
expression.
64 • ALGEBRA
EXERCISE 22
Divide the following by synthetic division:
i . 2 x^ — 7 x 2 +- x + 10 by x — 2.
2. 3 a 4 — a 3 — 5 a 2 + 6 a + 7 by a -h 1.
3. a 4 -lla 3 + 29a 2 -9a+-14by a-7.
4. 4 m 3 — 17 m 2 ?i -f- 13 mn 2 -f- ft* by m — 3 fl.
5. 3a 5 + lla 4 _43^-4a; 2 -f lla-6 by » + 6.
6. 8^ 4 -35?; 3 + 7v 2 + 22v-8by v-4.
109. Divide ar 3 — 11 a; 2 + 36 x — 36 by x — 5, and by a — 7.
1 - 11 + 36 - 36 |6 1 - 11 + 36 - 36 [7_
■4. 5-30 4-30 -f 7-28 + 56
— 6+6—6 Remainder — 4 + 8 + 20 Remainder
(Quotient) > (Quotient)
A factor lies between x — 5 and x—7. It is found to be x — 6.
Then if in dividing by a binomial a remainder occurs, and if
the remainders arising from successive division by two binomi-
als are of opposite sign, a factor #— a lies between these two
binomials.
EXERCISE 23
i. Locate the root between 2 and 4 of x 3 — 17 x -f 24 = 0.
Locate roots of the following :
2. a 3 + 10a 2 + 17a-2S = 0.
3. a 4 + 3tt 3 -10a 2 + 3a + ir> = 0.
4. x 5 — 8x i -7x i + Mx 2 -5x + 4i) = 0.
5. 3ar 3 -26# 2 + 60<»-72 = 0.
6. m A - 2 m 3 - 19 m 2 + 12 m + 40 = 0.
FACTORS 65
SOLUTIONS
110. If the product of abc ••• to n factors = 0, at least one
of the factors must be zero.
Ex. 1. Let (x -2)(x- 3)(x + 4) = 0.
Then x — 2, x — 3, or x + 4 must equal zero.
The equation is satisfied by the root obtained by putting any one of
the factors equal to 0. Hence, x = 2, 3, or — 4 are the solutions of the
equation.
Ex. 2. Solve 5 2 *- 5* -12 = 0. . (1)
(5*-4)(5* + 3)=0. (2)
Whence, 5* - 4 = 0, 5* = 4, (3)
and 5* + 3 = 0, 5» = - 3. (4)
To solve (3) and (4), take the logarithms of each member of the
equations :
From (3) zlog5 = log 4 (§ 89) , (5)
and ^log4 = 1 0020 = 602 #
log5 .6990 699 v J
From (4) x log 5 = — log 3.
Ex. 3. Solve the equation .2* = 3.
Taking the logarithms of both members, xlog.2 — log 3.
Then x = ^^= A1U = -^2L = _ .6285+.
log. 2 9.3010-10 -.699
An equation of the form a x = b may be solved by inspection
if b can be expressed as an exact power of a.
Ex. 4. Solve the equation 16 x = 128.
We may write the equation (2 4 )* = 2 7 , or 2 4 * = 2 7 .
Then, by inspection, 4 x = 7 ; and x = J.
(If the equation were 16 x = — , we could write it (2 4 )* = — = 2" 7 ;
v 4 128 2 7
then 4x would equal — 7, and a; =— {.)
Xx-A
66 ALGEBRA
EXERCISE 24
Solve the following equations :
i. 13* = 8. 4 . .005038* = 816.3. 7. .2*+ 5 = .5*
2. .06* = .9. 5. 3 4 *- 1 = 4 2 *+ 3 . 8. 16* = 32.
3. 9.347* = .0625. 6. 7 3 *+ 2 = .8*. 9. 32* = ^ .
10. ( T V)^ = 8. 11. ay = Jj. 12. .04 2 *- 5 (.04)* -24 = 0.
13. 2 3 * + 7.2 2 *- 9.2* -63 = 0.
14. 3°"-5.3 2 '- 8.3* + 12 = 0.
15. II 4 * -5- II 2 * + 4 = 0.
16. 2 3 *+ 3 -6.2 2 *+ 2 + ll • 2* +1 -6 = 0.
17. .5 4 * - 2(.5) 3 * - 16 (.5) 2 * + 2 (.5)* + 15 = 0.
18. 2 3 * - 10- 2 2 *- 71 -2* -60 = 0.
19. a?-x 2 -9x + 9 = Q.
20. ar + (5c + 2c7)£ + 10ccZ = 0.
COMMON FACTORS AND MULTIPLES
111. A Common Factor of two or more expressions is a factor
of each of them.
112. The Highest Common Factor (H. C. F.) of two or more
expressions is their common factor of highest degree (§ 23).
113. A Common Multiple of two or more expressions is an
expression which is exactly divisible by each of them.
114. The Lowest Common Multiple (L. C. M.) of two or more
expressions is their common multiple of lowest degree.
Ex. 1. Find the H. C. F. of a 2 + 2 a - 3 and 1 - a 3 .
a' 2 +2a-3= (a-l)(a 4-3).
1-0,3 = (1_«)(1 + rt + a 2 ).
The factors of the first expression can be put in the form
- (1 - a )(8 + a).
Hence, the H. C. F. is 1 - a.
FACTORS 67
Ex. 2. Eequired the L. C. M. of
x 2 — 5x + 6, x 2 — 4x + 4, and X s — 9 x.
• * 2 -5x+6 = (£-3)(z-2).
x 2 -4x+4=: 0-2) 2 .
x* - 9 x = x(x + S)(x - 3).
It is evident by inspection that the L. C. M. of these expressions is
z(z-2) 2 (£ + 3)(jc-3).
115. "When the polynomials cannot be readily factored by
inspection, the H. C. F. and L. C. M. may be found by the fol-
lowing method.
The rule in Arithmetic for the H. C. F. of two numbers is :
Divide the greater number by the less.
If there be a remainder, divide the divisor by it; and
continue thus to make the remainder the divisor, and
the preceding divisor the dividend, until there is no
remainder.
The last divisor is the H. C. F. required.
Thus, let it be required to find the H. C. F. of 169 and 546.
169)546(3
507
39)169(4
156
13)39(3*
39
Then 13 is the H. C. F. required.
116. We will now prove that a rule similar to that of § 115
holds for the H. C. F. of two algebraic expressions.
Let A and B be two polynomials, arranged according to the
descending powers of some common letter.
Let the exponent of this letter in the first term of A be
equal to, or greater than, its exponent in the first term of B.
Suppose that B is contained in A p times, with a remainder
C; that C is contained in B q times, with a remainder D; and
that D is contained in C r times, with no remainder.
68 ALGEBRA
To prove that D is the H. C. F. of A and B.
The operation of division is shown as follows :
B)A(p
pB
C)B(q
qC_
D)C(r
rD
We will first prove that D is a common factor of A and B.
Since the minuend is equal to the subtrahend plus the remainder
( F - C "§ 4 °)' A^pB+C, (1)
B = qC+D, (2)
and C = rD.
Substituting the value of C in (2), we obtain
B = qrD + D= D(qr + 1). (8)
Substituting the values of B and C in (1), we have,
A =pD(qr + 1) +rD = D(pqr +p + r). (4)
From (3) and (4), D is a common factor of A and B.
We will next prove that every common factor of A and B
is a factor of D.
Let F be any common factor of A and B ; and let
A = mJP 7 , and B = nF.
From the operation of division, we have
C = A-pB, (5)
and D = B-qC. (6)
Substituting the values of ^4 and 7? in (5), we have
C = mF — pnF.
Substituting the values of B and C in (6) we have
D = nF— q(mF — pnF)= F(n —qm +pqn ).
Whence, F is a factor of D.
FACTORS ill)
Then, since every common factor of A and B is a factor of
]), and since D itself is a common factor of .1 and />, it follows
that D is the highest common factor of A and B.
We then have the following rule for the H. C. F. of two
polynomials, A and B, arranged according to the descending
powers of some common letter, the exponent of that letter in
the first term of A being equal to, or greater than, its exponent
in the first term of B:
Divide A by B.
If there be a remainder, divide the divisor by it;
and continue thus to make the remainder the divisor,
and the preceding divisor the dividend, until there is
no remainder.
The last divisor is the H. C. P. required.
It is important to keep the work throughout in descending powers of
some common letter ; and each division should be continued until the
exponent of this letter in the first term of the remainder is less than its
exponent in the first term of the divisor.
Note 1 : If the terms of one of the expressions have a common factor
which is not a common factor of the terms of the other, it may be re-
moved ; for it can evidently form no part of the highest common factor.
In like manner, we may divide any remainder by a factor which is not
a factor of the preceding divisor.
117. i. Find the H. C. F. of
Gx 2 - 25x + 14 and Ox*- 7 x 2 - 25 x + 18.
6x 2 -25x + 14)Gx*- 7a---25x+ 18(x + 3
6s* -25x 2 + Ux
18x 2 -39x
18s 2 -75s + 42
36x-24
In accordance with Note 1, we divide this remainder by 12, giving
x ~~ 2 ' 3x-2)()x 2 -25x + 14(2x-7
n.i- 2 - 4x
-21* - #
-'21 as + 14
Then, Sx — 2 is the H. C. F. required.
70 ALGEBRA
Note 2 : If the first term of the dividend, or of any remainder, is not
divisible by the first term of the divisor, it may be made so by multiply-
ing the dividend or remainder by any term which is not a factor of the
divisor.
2. Find the H. C. F. of
3 «8 + a 2 b _ 2 a tf an( i 4 a s b + 2 a 2 b 2 - ab s + ft 4 .
We remove the factor a from the first expression and the factor b from
the second (Note 1), and find the H. C. F. of
3 a 2 + ab - 2 V 2 and 4 a 3 + 2 a 2 ft - aft 2 + ft 3 .
Since 4 a 3 is not divisible by 3 a 2 , we multiply the second expression
by 3 (Note 2).
4 a 3 + 2 a 2 ft - aft 2 + ft 3
3^
3 a 2 + a & _ 2 ft 2 )l2 a 3 -f 6 a 2 ft - 3 ab 2 + 3 ft 3 (4 a
12q8 + 4a 2 ft-8aft 2
2 a 2 ft + 5 aft 2 + 3 ft 3
Since 2a 2 ft is not divisible by 3 a 2 , we multiply this remainder by
8 (Note 2).
2 a 2 b + 5 ab 2 + 3 ft 3
3
3 a 2 + aft - 2 ft 2 )6 a 2 ft + 15 aft 2 + 9ft 3 (2ft
6a 2 &+ 2 aft 2 - 4 ft 3
13 aft 2 + 13 ft 3
We divide this remainder by 13 b 2 (Note 1), giving a + b.
a + 6)3 a 2 + aft - 2 ft 2 (3 a - 2 6
3 a 2 + 3 aft
-2 aft
- 2 aft - 2 ft 2
Then, a + ft is the H. C. F. required.
Note 8 : If the first term of any remainder is negative, the sign 0!
each term of the remainder may be changed.
Note 4: If the given expressions have a common factor which can
be se£n by inspection, remove it, and find the H.C. P, of the resulting
expressions; the result, multiplied by the common factor, will be the
H. C. F. of the given expressions.
FACTORS 71
3. Find the H. C. F. of
2aJ 4 + 33 3 -6a 2 + 2ajand ffV^-W — 2flf — ft
Removing the common factor x (Note 4), we find the H. C. F. of
2x 3 + 8 as* - 8a; -f 2 and Gx 3 + 5x 2 - 2x - 1.
" 2x 3 + 3x 2 -()x + 2)6a* + 5*»- 2x-l(3
6x 3 + 0x 2 - 18a; + 6
-4x 2 + 16x- 7
The first term of this remainder being negative, we change the sign of
each of its terms (Note 3).
2x 3 + 3x 2 - 6x+ 2
2
4x 2 - 16x + 7)4x3 -f- 6x 2 - 12x + 4(x
4x 3 -16x 2 + 7x
22x 2 - 19x + 4
2
44 x 2 - 38x+ 8(11
44x 2 - 176x + 77
69 )138x-69
2x- 1
2x-l)4x 2 - 16x + 7(2x-7
4 x 2 — 2 x
- 14x
- 14x4- 7
The last divisor is 2x— 1 ; multiplying this by x, the H. C. F. of the
given expressions is x (2 x — 1).
(In the above solution, we multiply 2 x 3 -f 3x 2 — 6x -b 2 by 2 in order
to make its first term divisible by4x 2 ; and we multiply the remainder
22 x 2 — 19 x -h 4 by 2 to make its first term divisible by 4 x 2 .)
118. We will now show how to find the L. C. M. ot two ex-
pressions which cannot be readily factored by inspection.
Let A and B be any two expressions.
Let F be their H. C. F., and M their L. C. M.
Suppose that A = aF, and B = bF.
Then, AxB = abF 2 . (1>
Since .Pis the H. C. F. of A and #, a and b have no common factors ;
whence the L. C. M. of rti^and bF is abF.
That is, M=abF.
72 ALGEBRA
Multiplying each of these equals by F, we have
Fx M=ahF~. (2)
- From (1) and (2), A x B = F x M.
That is, the product of two expressions is equal to the product
of their H. C. F. and L C.;M.
Therefore, to find the L. C. M. of two expressions,
Divide their product by their highest common factor ; or,
Divide one of the expressions, by their highest common factor,
and multiply the quotient by the other expression.
Ex. Find the L. C. M. of
6x 2 -17x 2 + 12 and 12a 2 -4a-21.
6x 2 - 17 x + 12)12 x 2 - 4x-21(2
12 x 2 -34 x + 24
15 )30 x- 45
2x- 3)6x 2 -17x + 12(3«-4
6x 2 -9x
-Sx
-8x + 12
Then, the H.C.F. of the expressions is 2x — 3.
Dividing Ox* 2 — 17 x + 12 by 2 x — 3, the quotient is 3 x — 4.
Then, the L. C. M. is (3x- 4)(12x 2 - 4x- 21).
EXERCISE 25
Find the H. C. F. and L. C. M. of the following:
i. 2a 2 + a-6, 4 a 2 -8a +-3.
2. 6a; 2 -17a,' + 10, 9a? 2 -14^-8.
3 . x 2 -Cyx-27, x>-2x 2 -8x + 2l.
4. 6^ 2 -31^/ + 182/ 2 , 9 x 2 + 15 ^- 14 y\
5. 8a 2 + 6a-9, 6a 3 + 7a 2 -7a-6.
6. 4a 2 -lla-3, 8x A + 6x*-llx 2 -23x-5.
7. m 5 -4m 3 + ^ 2 -4, m 4 — 2 m 3 — m* + m + 2.
8. 12p 2 -19pg-21? 2 , Y2 l r + r, l ra 2 -{- 11a + 6.
15. (5a-3bf-(a + b)\ 72a 2 - 48 a& + 8& 2 .
16. a 3 + 6a 2 a + 12aa; 2 + 8ar 3 , 4 a * + 8 a 4 x - a V - 2 aV.
V. FRACTIONS
119. A Fraction is an indicated quotient written usually in
the form — , where a is the dividend, and is called the numera-
tor, and b the divisor, and called the denominator.
120. If the same factor be introduced into, or removed from,
both dividend and divisor, the quotient is not changed. Upon
this principle depends the reduction of fractions to either
higher or lower terms. The laws of sign for fractions are
those of ordinary division. The sign before the fraction de-
notes whether the quotient is to be added or subtracted.
REDUCTION OF FRACTIONS
121. Change of sign,
-+-«_ _ — cl _ _ + a _ y ♦— a
EXERCISE 26
Write each of the following in three other ways without
changing its value :
a
2'
2. S±». 3. 8 •
7 ° 2-x
2x-l &x-5
4 " x + 2 5- (x -3X^+4)
6 &2_ " 2
' 2b 2 -a?'
(ix-3y)(y-3x)^
(2y + x)(x-y)
74 ALGEBRA
122. Reduction to Lowest Terms. This is accomplished by
removing every factor common to both numerator and denomi-
nator. If numerator and denominator are not prime to each
other, it is possible generally to factor them by inspection.
When, however, the factors cannot be readily seen, the method
of § 117, known as the Euclidean method, may be used.
EXERCISE 27
Reduce the following to lowest terms :
27 a 8 + 8
9 x 2 + 12 x + 4
a 8 _ a *b _ ab A + h 5
a 4 -a?b-a 2 b 2 + ab 3
12z 2 + 16zy-3y 2
2 - Z2 "^ M , _,. • 5-
10 2* + 22/ -21 2/* 6 ' 14 as* + 14 as -281
2a* + 5ar°-2
x + 3
6a^-7^ 2 + 5
x — 2
x? — x 2 — 4 x
-6
a? + 7rf + 12x + 10
16 x 2 + 16 x -
32
Simplify the following:
7- ft'-g«% l^ 4
21.
1 _i_ 4 «ff + f ^ ^ + K 8
4 a; 2
x + 4: x — 1 a; + 2 a? 2 — a;- 16
# + 2 # — 3 #—5 a: 2 — 8 x + 1 5
1 a^4-2a;-ll\ 9
af'+5a;-14y * ^ + 343*
76 ALGEBRA
2 & 2 + 3a> + 2 4 , x 2 + l
2 4 .
SB + 1 a?" — 1 # 2 -|- 5 x + 6 # 12 a;
(2x 2 -2xy-2x)(x 2 -y 2 ) ^
X
x + y
cf + V a 4_ & 4 3a 2
■ a* + & 2 a 2 b + ab 2 a 4 - a?b + d 2 b 2 -ab 3 +b*
123. Under certain conditions a fraction may assume a form
the value of which is not readily seen. Such forms usually
occur in limiting values of fractions in which the unknown or
unknowns are considered variable.
124. A variable number, or simply a variable, is a number
which may assume, under the conditions imposed upon it, an
indefinitely great number of different values.
A constant is a number which remains unchanged throughout
the same discussion.
125. A limit of a variable is a constant number, the differ-
ence between which and the variable may be made less than
any assigned number, however small.
Suppose, for example, that a point moves from A towards B under the
condition that it shall move, during successive equal intervals of time,
first from A to O, halfway be-
tween i andi?; then to Z>, half- f L ? * f
way between C and B ; then to
E, halfway between D and B ; and so on indefinitely.
In this case, the distance between the moving point and B can be made
less than any assigned number, however small.
Hence, the distance from A to the moving point is a variable which
approaches the constant value A B as a limit.
Again, the distance from the moving point to B is a variable which
approaches the limit 0.
126. Interpretation of j •
Consider the series of fractions -, — , — '-, — — , •••.
3' .3' .03 ' .003'
FRACTIONS 77
Here each denominator after the iiist is one-tenth of the
preceding denominator.
It is evident that, by sufficiently continuing the series, the
denominator may be made less than any assigned number,
however small, and the value of the fraction greater than any
assigned number, however great.
In other words,
If the numerator of a fraction remains constant, while
the denominator approaches the limit 0, the value of
the fraction increases without limit.
It is customary to express this principle as follows :
a
The symbol go is called Infinity ; it simply stands for that which is
greater than any number, however great, and has no fixed value.
127. Interpretation of — •
00
Consider the series of fractions -, — , — —
3 ? 30' 300' 3000' *
Here each denominator after the first is ten times the pre-
ceding denominator.
It is evident that, by sufficiently continuing the series, the
denominator may be made greater than any assigned number,
however great, and the value of the fraction less than any
assigned number, however small.
In other words,
If the numerator of a fraction remains constant, while
the denominator increases without limit, the value of
the fraction approaches the limit 0.
It is customary to express this principle as follows T
«=0.
78 ALGEBRA
128. No literal meaning can be attached to such results as
a a A
- = oo , or — = ;
for there can be no such thing as division unless the divisor is
a finite number.
If such forms occur in mathematical investigations, they
must be interpreted as indicated in §§ 126 and 127. (Coin-
pare § 86.)
THE PROBLEM OF THE COURIERS
129. The following discussion will further illustrate the
form -, besides furnishing an interpretation of the form -•
The Problem of the Couriers.
Two couriers, A and B, are travelling along the same road in
the same direction, RB', at the rates of m and n miles an hour,
respectively. If at any time, say 12 o'clock, A is at P, and B
is a miles beyond him at Q, after how many hours, and how
many miles beyond P, are they together ?
B P Q Rl
I I I l
Let A and R meet x hours after 12 o'clock, and y miles beyond P.
They will then meet y — a miles beyond Q.
Since A travels mx miles, and B nx miles, in x hours, we have
f y — m#,
iy — a = nx.
Solving these equations, we obtain
We will now discuss these results under different hypotheses.
1. m>n.
In this case, the values of x and y are positive.
This means that the couriers meet at some time after 12, at some point
to the right of P.
FRACTIONS 79
This agrees with the hypothesis made ; for if m is greater than n. A is
travelling faster than B ; and he must overtake him at some point beyond
their positions at 12 o'clock.
2. m n or m < n.
In this case, x — and y = 0.
This means that the travellers are together at 12 o'clock, at the point P.
This agrees with the hypothesis made ; for if a = 0, and m and n are
unequal, the couriers are together at 12 o'clock, and are travelling at
unequal rates ; and they could not have been together before 12, and will
not be together afterwards.
4. m = n, and a not equal to 0.
In this case, the values of x and y take the forms - and —. re-
*• i
spectively.
If m — n approaches the limit 0, the values of x and y increase without
limit (§ 126) ; hence, if m = n, no fixed values can be assigned to x and y,
and the problem is impossible.
In this case, the result in the form - indicates that the given problem is
impossible.
This agrees with the hypothesis made ; for if m = n, and a is not zero,
the couriers are a miles apart at 12 o'clock, and are travelling at the same
rate ; and they never could have been, and never will be together.
5. m = rc, and a = 0.
In this case, the values of x and y take the form -•
If a = 0, and m = n, the couriers are together at 12 o'clock, and travel-
ling at the same rate.
Hence, they always have been, and always will be, together.
In this case, the number of solutions is indefinitely great ; for any
value of x whatever, together with the corresponding value of y, will
satisfy the given conditions.
In this case, the result in the form - indicates that the number of solu-
tions is indefinitely great.
Such form is called Indeterminate.
80 ALGEBRA
130. In § 129, we found that the form - indicated an ex-
pression which* could have any value ivhatever; but this is not
always the case.
Consider, for example, the fraction x ~~ a •
x 2 — ax
If x — a, the fraction takes the form -•
Now, x 2 -a? _ (x + a)(x-a) __ x+a .
x 2 — ax ic(x — a) x
which last expression is equal to the given fraction provided x does not
equal a.
The fraction x + a approaches the limit a "*" a , or 2, when x approaches
the limit a. x a
This limit we call the value of the given fraction ivhen x = a.
Then, the value of the given fraction when x = a is 2.
In any similar case, we cancel the factor which equals for the given
value of ac, and find the limit approached by the result when x approaches
the given value as a limit.
EXERCISE 28
Find the values of the following :
2 ax — 4 a 2 i o x 2 — 16 , A
2 or 3 -5 a; 2 . A 4a; 2 -4a; -3 ,
2. when a? =0. 4. — when.r=i|.
4:X 2 + 3x • 6af'-17a; + 12
s. — — 2— when x = — 2.
D a**-8a!*-f-16
6. — -~ J —- when x = 2.
ar* - 7 a; -f G
131. Other Indeterminate Forms.
Expressions taking the forms ||-, x 00 , or qo — 00, for oer-
tain values of the letters involved, are also iiulct erniiiiate.
FRACTIONS 81
i. Find the value of (ar 3 + 8) (l + -i— \ when x = - 2.
This expression takes the form x oo, when x =— 2 (§ 126).
Now, (x 3 + 8) ( 1 + -i-^ = x 3 + 8 + ^-ii?
V £ + 2/ x + 2
, = ic 3 + 8 + x 2 -2x + 4 = x 3 +x 2 — 2x+12.
The latter expression approaches the limit — 8+4 + 4 + 12, or 12,
when x approaches the limit — 2.
This limit we call the value of the expression when x =— 2 ; then, the
value of the expression when x = — 2, is 12.
In any similar case, we simplify as much as possible before finding the
limit.
1 2x
2. Find the value of -- when x = 1.
1 — x 1 — x 2
The expression takes the form oo — oo, when x = 1 (§ 126).
Now 1 2x := l+ x -2x = 1-s = 1 r
1 - x 1-x 2 1-x 2 1 - x 2 1 + x
The latter expression approaches the limit J when x approaches the
limit 1.
Then, the value of the expression when x = 1, is \.
132. Another example in which the result is indeterminate
is the following :
1 4- 2x
Ex. Find the limit approached by the fraction ^— when
x is indefinitely increased.
Both numerator and denominator increase indefinitely in absolute value
when x is indefinitely increased.
1 + 2
1 + 2x x
Dividing each term of the fraction by x, = =— = =
A — OX L
O
X
The latter expression approaches the limit r (§ 127), or — -, when
x is indefinitely increased.
In any similar case, we divide both numerator and denominator of the
fraction by the highest power of x.
82 ALGEBRA
EXERCISE 29
Find the limits approached by the following when x is in-
definitely increased :
' 4 + 5a?-3s 8 _ liii. a^-2a?-4
Find the values of the following :
1 12 * o
5. (2a; 2 -5#-3)f2 + -i-Vvhena = 3.
V X — 3J
RATIO AND PROPORTION
RAtlO
133. The Ratio of one number a to another number b is the
quotient of a divided by b.
Thus, the ratio of a to b is -; it is also expressed a : b.
b
The ratios here spoken of are but fractions under another
name, and have all the properties of fractions.
In the ratio a: b, a is called the first term, or antecedent, and
b the second term, or consequent.
If a and b are positive numbers, and a > b, - is called a
b
ratio of greater inequality ; if a < 5, it is called a ra/10 of less
inequality.
134. A ratio of greater inequality is decreased, and
one of less inequality is increased, by adding the same
positive number to each of its terms.
Let a and b be positive numbers, a being > b, and x a positive number.
Since a > &, ax > bx. (§ 50)
Adding ab to both members (§ 50),
ab 4- ax>ab + bx, or a(& -f x)>b(a +x).
RATIO AND PROPORTION 83
Dividing both members by b(b -f x), we have
j»j±|. _ (|M)
In like manner, if a < 6, - < 'L±*.
6 6 + x
PROPORTION
135. A Proportion is an equation whose members are equal
ratios.
Thus, if a : b and c : d are equal ratios,
a:b = c :d, or - =e - ,
is a proportion. The latter form is preferable.
136. In the proportion a: b = c : d, a is called the first term,
b the second, c the third, and d the fourth.
The first and third terms of a proportion are called the ante-
cedents, and the second and fourth terms the consequents.
The first and fourth terms are called the extremes, and the
second and third terms the means.
137. If the means of a proportion are equal, either mean is
called the Mean Proportional between the first and last terms,
and the last term is called the Third Proportional to the first
and second terms.
Thus, in the proportion a:b = b : c, b is the mean proportional
between a and c, and c is the third proportional to a and b.
The Fourth Proportional to three numbers is the fourth term
of a proportion whose first three terms are the three numbers
taken in their order.
Thus, in the proportion a:b = c:d, d is the fourth proportional to
a, b, and c.
138. A Continued Proportion is a series of equal ratios, in
which each consequent is the same as the next antecedent; as,
a : b = b \ c = c : d = d : e.
«_
6 =
~ d
ad :
= bc.
84 ALGEBRA
PROPERTIES OF PROPORTIONS
139. In any proportion, the product of the extremes is
equal to the product of the means.
Let the proportion be
Clearing of fractions,
140. From the equation ad = bc (§ 139), we obtain
be , ad ad -. 7 be
a = — , o = — , c = — , and a = — •
deb a
That is, in any proportion, either extreme equals the
product of the means divided by the other extreme ; and
either mean equals the product of the extremes divided
by the other mean.
141 . (Converse of § 139.) If the product of two numbers
be equal to the product of two others, one pair may be
made the extremes, and the other pair the means, of a
proportion.
Let ad = be.
Dividing by bd, ^ = H , or ■ ff = 5 .
° J bd bd* b d
In like manner, we may prove that
a_b
c d" 1
c - = f,etc.
d b
142. In any proportion, the terms are in proportion by
Alternation; that is, the means may be interchanged.
Let the proportion be
Then, by § 139,
Then, by § 141,
In like manner it may be proved that the extremes can be interchanged.
a _
b~
c
' d
ad =
■be.
« _
_b t
~ d
a _
b~
c
d 1
ad =
-.be.
b
d
RATIO AND PROPORTION 85
143. In any proportion, the terms are in proportion by
Inversion ; that is, the second term is to the first as the
fourth tef m is to the third.
Let the proportion be
Then, by § 139,
Whence, by § 141,
a c
It follows from § 143 that, in any proportion, the means can be written
as the extremes, and the extremes as the means.
144. The mean proportional between two numbers is
equal to the square root of their product.
Let the proportion be - = - •
b c
Then, by § 139, b 2 = ac, or b = Vac.
145. In any proportion, the terms are in proportion
by Composition ; that is, the sum of the first two terms
is to the first term as the sum of the last two terms is
to the third term.
Let the proportion be - = - •
b d
Then, ad = be.
Adding each member of the equation to ac,
ac + ad = ac + be, or a(c + d) — c(a + b).
By §141, a ± b = e_±d t
a c
ttt i a + b c -f d
We may also prove — r — = — ;—
b d
146. In like mariner we may also prove that the terms of
any proportion are in proportion by Division ; that is, the dif-
ference between the first two terms is to the first terin as the
difference between the last two terms is to the third term.
The proof is left to the student.
86 ALGEBRA
147. In any proportion, the terms are in proportion by
Composition and Division ; that is, the sum of the first
two terms is to their difference as the sum of the last
two terms is to their difference.
The proof is left to the student. Hint. — Divide the result of § 145 by
that of § 146.
148. In any proportion, if the first two terms be multi-
plied by any number, as also the last two, the resulting
numbers will be in proportion.
Let the proportion be - = - ; then, 2£ = *£.
b d mb nd
(Either m or n may be unity ; that is, the terms of either ratio may be
multiplied without multiplying the terms of the other.)
149. In any proportion, if the first and third terms be
multiplied by any number, as also the second and fourth
terms, the resulting numbers will be in proportion.
Let the proportion be - = - ; then, — = — •
b d nb nd
(Either m or n may be unity.)
1 50. In any number of proportions, the products of the
corresponding terms are in proportion.
Let the proportions be - — -, and - 3= .*!■"
b d f h
Multiplying, * x ? = ^ X # r OT 2« = ££.
b f d h bf dh
In like manner, the theorem may be proved for any number of
proportions.
151. In any proportion, like powers or like roots of the
terms are in proportion.
Let the proportion be - = - ; then, — = — •
b d b n d»
nt n,~
In like manner, = •
Vb Vd
RATIO AND PROPORTION 87
152. In a series of equal ratios, any antecedent is to its
consequent as the sum of all the antecedents is to the
sum of all the consequents.
Let
a: b = c.d = e:f.
Then, by § 139,
ad — 6c,
and
af = be.
Also,
ab = ba.
Adding,
a(b+d+f)=b(a + c + e).
Whence,
a :b = a + c + e \b + d + f.
(§ 141)
In like manner, the theorem may be proved for any number of equal
ratios.
153. If three numbers are in continued proportion,
the first is to the third as the square of the first is
to the square of the second.
Let the proportion be a : b = b : c ; or - = - .
b c
Then, ?x^? X «,or? = «!.
b c b b c b 2
154. If four numbers are in continued proportion,
the first is to the fourth as the cube of the first is
to the cube of the second.
Let the proportion be a :b = b :c = c :d ; or - =* - = -.
bed
Then, 2 x *x£=^xfx$ |OT ?»£
b e d b b b d 6 3
Similarly, it may be shown that if n numbers are in continued propor-
tion, the first antecedent is to the last consequent as the »th power of the
first antecedent is to the nth power of its consequent.
155. Examples.
i. If x : y = (x + zf : (y + z) 2 , prove z the mean proportional
between x and y.
From the given proportion, by § 139,
y(x + z) 2 = x(y + z)\
Or, x 2 y 4- 2 xyz + yz 2 = xy 2 4- 2 xyz + xz 2 .
Transposing, x 2 y — xy 2 = xz 2 — yz 2 .
Dividing by x — ?/, xy — z 2 .
Therefore, z is the mean proportional between x and y (§ 144).
88 ALGEBRA
The theorem of § 147 saves work in the solution of a certain
class of fractional equations.
2. Solve the equation 2 X ± | = ?A^.
^ 2z-3 26 + a
Regarding this as a proportion, we have by composition and division,
4 x _ 4b
6 ~ -'la
2x
26.
j
a
whence, x
86
a
3-
Pro
ve that if - :
b
c
then
a 2 -b 2 :
a 2 -
■3ab--
= c 2 -
-d 2 :
c 2 -
Scd.
Let
« _
6~
- = x, whence
d
i, a =
- bx ; then,
c 2
-1
a 2 - b 2
a 2 - Sab I
b 2 x/
W -
! -6 2
-3 6%
X 2
- 1
d 2
_ c 2
c 2 -
-tf 2
X 2 -
-3x
c 2
3c
- o Cd
d 2
d
Then,
a 2 -6 2
:a 2
-3a6
= c 2
-d 2
: c 2 —
3 6"d.
EXERCISE 30
i. Find the mean proportional between .0289 and 1.69.
2. Find the mean proportional between l T 7 ^ r and 12||.
3. Find the third proportional to if and 1|.
4. Find the fourth proportional to 9g\, 16^, and T 9 g.
5. Find the fourth proportional to m, n, and r.
6. Write in the form of a proportion : x 2 — 2 x -* 15 = a 2 .
Solve, using composition and division:
4#-f-5 _ a?-f5
4 # — 5 a? — 3
x — a b — c 2x — 3 5 # — 9
ai + a 3 (m + l) 2 -(w-lV
12. If - = - , show that a:c = b 2 : c 2 .
6 c
8.
RATIO AND PROPORTION . 89
• i 3 . fl^ry^cVtg-A^iaArt; »^^^y^»^^a^a»^-»r<
14. Find two numbers in the ratio of 2:.'{, sucli that the
sum of their squares shall be 208.
15. Find two numbers in the ratio of 3 : 1, such that the dif-
ference of their squares is 200.
16. Two numbers are in the ratio of 5 : 7. If 6 be added to
each, they will be in the ratio of 7 : 9. Find the numbers.
17. Two numbers are in the ratio of 2 : 5. If 4 be added to
each number, the resulting ratio will be twice the ratio had 4
been subtracted from each number. Find the numbers.
18. The difference between two numbers is 6, and the dif-
ference between their squares is 60. What is the ratio of their
sum to their difference ?
19. In similar figures in geometry, homologous sides are pro-
portional. If a pole 30 feet high casts a shadow 42 feet long,
how high must a pole be to cast a shadow 35 feet long ?
20. A ladder 40 feet long leans against the side of a build-
ing, with its foot 12 feet from the building. A second ladder,
40^ feet long, makes the same angle with the building as the
first ladder. How far is the foot of the second ladder from
the building ?
21. In the triangle ABC, MN is
drawn parallel to BC and divides the
other two sides proportionally. If
AM =12, 4M = 2 and 5(7=48, how
' AN 3'
long is AC? (M is the middle point of AB.) What is the
ratio of AN to MN?
22. The areas of any two similar figures are to each other
as the squares of their homologous .parts. If a regularTiexagon
has a side equal to 6 and an area of 54 V3, what is the area of
a regular hexagon whose side is 2 ?
90 ALGEBRA
23. The area of a circle is 6^ times that of another circle.
If the radius of the first circle is 5, what is the radius of the
second circle ?
24. If the altitude of a triangle is twice that of a similar
triangle, how do their areas compare ?
25. The volume of a rectangular solid is equal to the product
of its three dimensions, x, y, and z. If xyz — v and x : y : z
=5 a : b : c, filid to, y, and 2 in terms of a, b, c, and v.
26. Find three numbers in continued proportion whose sum
is 63, the second being 4 times the first.
27. Given the proportion - = - = -, where d = 81 and - = - .
_. , . , bed bo
Find a, 0, and c.
28. If 2 a — 36:4a-56 = 26— 3 c:46 — 5 c, prove 6 is the
mean proportional to a and c.
29. If3a + 56:4a-76 = 3c + 5d:4c-7d, prove ?==^
6 a
30. Find two numbers in the ratio of a to 6, such that if
3 increased by - they will be in the rat:
ix + 7 8x + ± 12x + l_Jx-l Solvefora; .
each be increased by - they will be in the ratio of e to /.
3 1
15 45 9(5 » + 2)
fl ± 6 + q-2ft a (2a^&) g + 3a6 i Solve for ^
a? ^ + a a 2 — a 2
33. A man borrows a certain sum, paying interest at the
rate of 5%. After repaying $180, his interest rate on the
balance is reduced to 4J%, and his annual interest is now less
by $ 10.80. Find the sum borrowed.
34. The digits of a certain number are three consecutive
numbers, of which the middle digit is the greatest, and the
first digit the least. If the number be divided by the sum of
its digits, the quotient is ^p. Find the number.
VARIATION 91
35. A certain number of apples were divided between three
boys. The first received one-half the entire Dumber, with one
apple additional, the second received one-third the remainder,
with one apple additional, and the third received the remain-
der, 7. How many apples were there ?
36. A freight train runs 6 miles an hour less than a pas-
senger train. It runs 80 miles in the same time that the
passenger train runs 112 miles. Find the rate of each train.
37. A and B each fire 40 times at a target ; A's hits are one-
half as numerous as B's misses, and A's misses exceed by 15
the number of B's hits. How many times does each hit the
target ?
t 38. A freight train travels from A to B at the rate of 12
miles an hour. After it has been gone 31 hours, an express
train leaves A for B, travelling at the rate of 45 miles an hour,
and reaches B 1 hour and 5 minutes ahead of the freight.
Find the distance from A to B, and the time taken by the
express train.
39. A tank has three taps. By the first it can be filled in
3 hours 10 minutes, by the second it can be filled in 4 hours
45 minutes, and by the third it can be emptied in 3 hours
48 minutes. How many hours will it take to fill it if all the
taps are open?
40. A man invested a certain sum at 3|%, and \\ this sum
at 4^% ; after paying an income tax of 5%, his net annual
income is $ 195.70. How much did he invest in each way ?
VARIATION
1 56. One variable number (§ 124) is said to vary directly as
another when the ratio of any two values of the first equals
the ratio of the corresponding values of the second.
It is usual to omit the word " directly •' and simply say that one nnin
ber varies as another.
92 ALGEBRA
Thus, if a workman received a fixed number of dollars per
diem, the number of dollars received in m days -will be to the
number received in n days as til is to n.
Then, the ratio of any two numbers of dollars received
equals the ratio of the corresponding numbers of days worked.
Hence, the number of dollars which the workman receives
varies as the number of days during which he works.
157. The symbol co is read "varies as" ; thus, ace b is read
" a varies as 6."
158. One variable number is said to vary inversely as
another when the first varies directly as the reciprocal of the
second.
Thus, the number of hours in which a railway train will
traverse a fixed route varies inversely as the speed ; if the
speed be doubled, the train will traverse its route in one-half
the number of hours.
159. One variable number is said to vary as two others
jointly when it varies directly as their product.
Thus, the number of dollars received by a workman in a
certain number of days varies jointly as the number which he
receives in one day, and the number of days during which he
works.
160. One variable number is said to vary directly as a sec-
ond and inversely as a third, when it varies jointly as the
second and the reciprocal of the third.
Thus, the attraction of a body varies directly as the amount
of matter, and inversely as the square of the distance.
161 . Ifxccy, then x equals y multiplied by a constant number.
Let x' and y' denote a fixed pair of corresponding values of x and y,
and x and y any other pair.
By the definition of § 156, - = - ; or, x - -y.
y y' y r
x'
Denoting the constant ratio — by m, we have
y'
x = my.
VARIATION 93
162. It follows from §§ 158, 159, 160, and 161 that :
1. If x varies inversely as y, x = — •
y
2. If x varies jointly as y and z, x = myz.
3. If x varies directly as y and inversely as z, a? = ^ •
z
1 63. Ifxccy, and y oc z, then xccz.
By § 161, if x cc y, x = my. (1)
And iiyazz, y — nz.
Substituting in (1), x — mnz.
Whence, by § 161, x*z.
164. Ifxccy when z is constant, and xccz when y is constant,
then xcc yz when both y and z vary.
Let y' and z' be the values of y and z, respectively, when x has the
value x'.
Let y be changed from y f to y", z remaining constantly equal to z\ "
and let x be changed in consequence from x' to X.
Then, by §156, ^=T/' W
Now, let z be changed from z' to z", y remaining constantly equal to
y", and let x be changed in consequence from Xto x .
Then, — = — . (2)
x" z" K J
Multiplying (1) by (2), ± = &L> (3)
x n y"z"
Now if both changes are made, that is, y from y' to y" and z from z' to
z", x is changed from x f to x", and yz is changed from y'z' to u"z".
Then by (3), the ratio of any two values of x equals the ratio of the
corresponding values of yz ; and, by § 156, xccyz.
The following is an illustration of the above theorem :
It is known, by Geometry, that the area of a triangle vaties as the
base when the altitude is constant, and as the altitude when the ba.se is
constant; hence, when both base and altitude vary, the area varies as
their product.
94 ALGEBRA
165. Problems.
Problems in variation are readily solved by converting the
variation into an equation by aid of §§ 161 or 162.
i. If x varies inversely as y, and equals 9 when y = S, find
the value of x when y = 18.
If x varies inversely as y, x =— (§ 162).
y
Putting x = 9 and y = 8, 9 = — , or m = 72.
8
Then, x = — ; and, if y = 18, x = — = 4.
V 18
Since variation is simply another way of stating a proportion, the prob-
lems in variation may be solved readily by means of proportion.
E.g. In the above problem ^
xcc -,
y
x.= ™.
y
This equation is true for any assigned values of the variables.
Then, ■ x x =-, (1)
2/i
x 2 =™- (2)
, 2/2
Dividing (1) by (2) Q = V* (3)
x 2 2/i
which is in the form of inverse proportion. Substituting the given values
of x and y in (3) , we have q .. 8
x 2 8 '
9 . 8
whence x 2 = ■ = 4.
18
2. Given that the area of a triangle varies jointly as its
base and altitude, what will be the base of a triangle whose
altitude is 12, equivalent to the sum of two triangles whose
bases are 10 and 6, and altitudes 3 and 9, respectively ?
Let J5, //, and A denote the base, altitude, and area, respectively, of
any triangle, and B' the base of the required triangle.
Since A varies jointly as /> and //, A = mlUI (§ 162).
Therefore, the area of the first triangle is m x 10 x 3, or 30 ???,, and the
area of the second is m x 6 x 9, or 54 m.
VARIATroX 95
Then, the area of the required triangle is 30 m -f 54 m, or 84 m.
But, the area of the required triangle is also m x B' x 12.
Therefore, 12 mB' = 84 m, or B' = 7.
Or using proportion and letting A\ — area of first triangle, A 2 = area of
second, A s = area of third.
A 3 = Ai + A 2
Ai = mB 1 H 1 . (l)
A 2 = mB 2 H 2 . (2)
A s = mB 3 H 3 . (3)
Adding (1) and (2)
A x + A 2 = m^BiHi + B 2 H 2 ). (4)
Dividing (4) by (3)
Ai+A 2 = m^BxHi + B 2 II 2 )
A s m(B s H 3 )
or, 1= B l H l + B*H 3 m (5)
Substituting the given values of B and if in (5) we have
1 = 10 . 3 + 6 ■ 9
12 £ 3 '
whence, B 3 = 7.
EXERCISE 31
i. If xvzy, and # = 3 when ?/ = 12, what is the value of x
when y = 2S?
2. If y&x 2 , and 2/ = 4 when x = l, what is the value of y
in terms of x 2 ?
3. If 2/ varies inversely as x, and y = 4 when # == — 3, w r hat
is the value of y when x = 2 ?
4. If & varies directly as y and inversely as 2, and x = £
when # = -§ and 3 = f , what is the value of x when y = £ and
5. If a; varies jointly as y and z and # = — 20 when y = 2
and 3 = 8, what is the value of # when ?/ = — £ and z<== 16 ?
6. If (3 x + 4) oc (2 ?/ — 5) when # = — 1 and y = 4, w T hat is
the value of x when ?/ = 19 ?
96 ALGEBRA
7. If x 2 varies inversely as y 8 , when x = 4 and ?/ = 2 f what
is the value of 2/ when a? = ? / f ?
8. If x equals the sum of two numbers, one of which varies
directly as y and the other inversely as z 2 , and x = 47 when
y = — 16 and 3 = 2, and a; = 2 when y = — 2 and z = 1, find the
value of x when y = 3 and z = \*
9. The area of a triangle varies jointly as its base and
altitude. If the area of a triangle whose base is 6 and whose
altitude is 9 is 27, what is the base of a triangle whose area is
44 and whose altitude is 11 ?
10. The distance through which a body falls from rest
varies as the square of the time during which it falls. If a
body falls 900 feet in 7.5 seconds, how many feet will it fall
in 16 seconds ?
11. The illumination from a source of light varies inversely
as the square of the distance from the source. How far must
an object 20 feet from the light be moved in order that it may
receive twice as much light ?
12. A circular plate of lead, 17 inches in diameter, is melted
and formed into three circular plates of the same thickness.
If the diameters of two of the plates are 8 and 9 inches
respectively, find the diameter of the other; it being given
that the area of a circle varies as the square of its diameter.
13. A cow tied to a stake by a rope 24 yards long will graze
over the area within her reach in three days. She breaks her
rope and, in repairing it, it is shortened 1^ feet. In how many
days will she graze over the new area ?
14. A pump supplying the water for a building has a 10-ineh
stroke and a cylinder 4 inches in diameter. It is not possible
to increase the number of strokes of the pump, nor to increase
the length of the cylinder. By how much must the diameter
be increased if 50% is added to the capacity of the pump'/
(The volumes of cylinders vary as the product of the base and
altitude.)
INVOLUTION AND EVOLUTION 97
VI. INVOLUTION AND EVOLUTION
166. We have already given (Chapter III) the involution
and evolution of monomials. We will now consider involution
and evolution of polynomials.
167. Square of a polynomial. By actual multiplication
(a + b + c) 2 = a 2 + b 2 + c 2 + 2 ab + 2 ac + 2 be.
In like manner
(a + b + c + df
= a 2 + &Vf- c 2 + d 2 + 2 a& + 2 ac + 2 ad + 2 6c + 2 6cZ + 2 cd,
and so on for the square of any polynomial.
The law observed may be stated as follows :
The square of a polynomial is equal to the sum of
the squares of its terms, together with twice the prod-
uct of each term by each of the following terms.
Ex. Expand (2 x 2 - 3 x - 5) 2 .
The squares of the terms are 4 sc 4 , 9 x 2 , and 25.
Twice the product of the first term by each of the following terms gives
the results — 12 x :3 and - 20 x 2 .
Twice the product of the second term by the following term gives the
result 30 x.
Then, (2 x 2 - 3 x - 5) 2 = 4 x 4 + 9 x 2 + 25 - 12 x* - 20 x 2 + 30 x
= 4 x 4 - 12 cc 3 - 11 x 2 + 30 x + 25.
168. Cube of a binomial. By actual multiplication
(a -f bf = d 3 + 3 a 2 b + 3 a& 2 + Jft
That is, the cube of the sum of two numbers is equal to
the cube of the first, plus three times the square of the
first times the second, plus three times the first times
the square of the second, plus the cube of the second.
In like manner, the cube of the difference of two
numbers is equal to the cube of the first, minus three
times the square of the first times the second, plus three
times the first times the square of the second, minus
the cube of the second.
The cube of a trinomial may be found by the above method,
if two of its terms be enclosed in parenthesis, and regarded as
a single term.
98 ALGEBRA
169. Square Root of any Polynomial Perfect Square.
By § 167, (a + b + cf = a 2 + 2 ab + V 1 + 2 ac + 2 be + c 2
= a 2 + (2a + &)6 + (2a + 2& + c)e. (1)
Then, if the square of a trinomial be arranged in order of
powers of some letter :
I. The square root of the first term gives the first term of
the root, a.
II. If from (1) we subtract a 2 , we have
(2a + 6)H(2« + 2Hc)c # (2)
The first term of this, when expanded, is 2 ab ; if this be
divided by twice the first term of the root, 2 a, we have the
next term of the root, b.
III. If from (2) we subtract (2a + b)b, we have
(2a + 2& + c)c. (3)
The first term of this, when expanded, is 2 ac; if this be
divided by twice the first term of the root, 2 a, we have the
last term of the root, c.
IV. If from (3) we subtract (2a + 2& + c)c, there is no
remainder.
Similar considerations hold with respect to the square of a
polynomial of any number of terms.
170. The principles of § 169 may be used to find the square
root of a polynomial perfect square of any number of terms.
Let it be required to find the square root of
4 x 4 + 12 x 3 - 7 x 2 - 24 x + 16.
4 x 4 + 12 x s - 7 x 2 - 24 x + 16 1 2 x 2 + 3 x - 4
a 2 = 4 a*
2 a + & = 4 x 2 + 3 .x
a a
12 x 8 - 7 a 2 - 24 X + 16, 1st Rem.
2a+2& + c = 4.r 2 + 6x — 4
-4
- 16 x 2 - 24 x 4- 16, 2d Rem.
_ 16x* — 24x + 16
2 x 2 + 3 x — 4 is called the square root and 2 a the first trial divisor.
2 a + b is the first complete divisor.
INVOLUTION AND EVOLUTION 99
We then have the following rule for extracting the square
root of a polynomial perfect square :
Arrange the expression according to the powers of
some letter.
Extract the square root of the first term, write the
result as the first term of the root, and subtract its
square from the given expression, arranging the re-
mainder in the same order of powers as the given ex-
pression.
Divide the first term of the remainder by twice the
first term of the root, and add the quotient to the part
of the root already found, and also to the trial divisor.
Multiply the complete divisor by the term of the root
last obtained, and subtract the product from the re-
mainder.
If other terms remain, proceed as before, doubling the
part of the root already found for the next trial divisor.
171. Cube Root of any Polynomial Perfect Cube.
By §168, ( a +& + c) 3 =[(a + 6)+c] 3
= (a -f bf + 3(a + b) 2 c + 3(« + b)c 2 -f.c 3
= a 3 + 3 a 2 b + 3 ab 2 + V + 3(a + b) 2 c + 3(a + by + c 3
= a 3 + (3 a 2 + 3 ab + b 2 )lj + [3(« + bf+ 3(a + b)c + c^c. (1)
Then, if the cube of a trinomial be arranged in order of
powers of some letter :
I. The cube root of the first term gives the first term of the
cube root, a.
II. If from (1) we subtract a 3 , we have
(3a 2 + 3ab + b 2 )b + [3(a + b) 2 + 3(a + b)c + c 2 ]c. (2)
The first term of this, when expanded, is 3a 2 6; if this be
divided by three times the square of the first term of the root,
3a 2 , we have the next term of the root, b.
III. If from (2) we subtract (3 a 2 + 3 ab 4- &*)&, we have
I3(a + by + 3(a + b)c + c 2 ]c. (3)
100 ALGEBRA
The first term of this, when expanded, is 3a 2 c; if this be
divided by three times the square of the first term of the root,
3 ,
there is no remainder.
Similar considerations hold with respect to the cube of poly-
nomials of any number of terms.
172. The principles of § 171 may be used to find the cnbe
root of a polynomial perfect cube of any number of terms.
Let it be required to find the cnbe root of
aj6 + 6a 5 + 3iB 4 -28aj 8 -9a? 2 + 54aj-27.
x 6 H-6^+3x 4 -28x 3 -9x 2 +o4x-27
= x 6
3 « 2 +3 ab + b 2 = 3 x*+6 x 3 +4 x 2
2x
6x 5 + 3x*-28x 3 -9x 2 +54x-27
0x r > + 12x*4-'8x 3
S(a + b) 2 = Sx^ + 12x ii +12x 2
3(a + 6)c + c' 2 =_ - 9x 2 -18x+9
8s* + 12a;*+ 3x 2 -18x+9
- 9;e4_;30x 3 -9x 2 + 54x-27
- 9x 4 -36\x 3 -9x 2 +54x-27
The first term of the root is the cube root 5f x 6 , or x 2 .
Subtracting the cube" of x 2 , or x 6 , from the given expression, the first
remainder is 6 x 5 + 3 x 4 — 28 x 3 — 9 x 2 + 54 x — 27.
Dividing the first term of this by three times the square of the first
term of the root, 3x 4 , we have the next term of the root, 2x (§ 171, II).
Now, 3 ab + b 2 equals 3 x x 2 x 2 x + (2 x) 2 , or 6 x 3 + 4 x 2 .
Adding this to 3x 4 , multiplying the result by 2x, and subtracting the
product, 6x 5 + 12 x 4 + 8x 3 , from the first remainder, gives the second
remainder, - 9x 4 - 30 x 3 - 9x 2 -f 54 x- 27 (§ 171, III).
Dividing the first term of this by three times the square of the first
term of the root, 3x 2 , we have the last term of the root, — 3.
Now, 8(a + b) 2 equals 3(x 2 + 2 x) 2 , or 3 x 4 + 12x 3 + 12 x 2 ; 3(« + b)c
equals 3(x 2 + 2x)( — 3), or — 9x 2 — 18 x ; and tfi = 9.
Adding these results, we have 3x 4 -f 12 x 3 + 3x 2 — 18 x + 9.
Subtracting from the second remainder the product of this by — 8, »./' t and :> > .'"* -f 12 X s + 12 x 2 are called trial divisor*,
and the expressions 3 x 4 + x 3 -f 4 x 2 and 3 x 4 + 12 x 8 + 3 x 2 — 18 x -f 9
complete divisors.
INVOLUTION AM) EVOLUTION 101
We then have the following rule for finding the cube root of
a polynomial perfect cube :
Arrange the expression according to the powers of
some letter.
Extract the cube root of the first term, write the result
as the first term of the root, and subtract its cube from
the given expression ; arranging the remainder in the
same order of powers as the given expression.
Divide the first term of the remainder by three times
the square of the first term of the root, and write the
result as the next term of the root.
Add to the trial divisor three times the product of the
term of the root last obtained by the part of the root
previously found, and the square of the term of the root
last obtained.
Multiply the complete divisor by the term of the
root last obtained, and subtract the product from the
remainder.
If other terms remain, proceed as before, taking three
times the square of the part of the root already found for
the next trial divisor.
EXERCISE 32
Find the square roots of the following :
i. 4 a 4 + 12 a*b - 7 a 2 b 2 - 24 db* + 16 b\
2. 49m 4 -5m 2 -42m 3 + l + 6m.
3. 9a 2 -24a&-36«c + 166 2 + 48&c + 36c 2 .
5. x 6 + 5x i + Ux s -6x 5 + l-4x-2x 2 .
6. 4m4-25m i -12m* + 16-24,y,l.
7 . 64 c 2 -80 c -23 + 9c" 2 + 30 c" 1 .
8. 4 x + 9 y~ A + 4 xhj~ 2 + 24 $#"* - 1 % V 1 -
9 . 6 yz~ 2 + ±x- 2 + y 2 -4 x-hj - 12 x~ l z~ 2 + 9 z~\
102 ALGEBRA
10. 4 a 3 + 29 a - 4 J + 21 a 2 - 20 a * + 4 - 18 a*
Find the cube roots of the following :
ii. 343 %* - 441 x>y 4 189 xy 2 - 27 y\
12. a 6 -9 0^4 21 a 4 4 9 ar* - 42 a? 2 - 36 ar- 8.
13.. 18 a 4 - 13 a 3 + 1 + 8 a G 4 9 a 2 - 3 a - 12 a 5 .
14. 54 m 5 4 44 ra 3 + 1 + 27 m fi 4 63 m 4 4 6 m 4 21 m 2 .
5 3 3 27
16. 64 ah- 6 - 240 a/>" 4 c + 300 ah~ 2 c 2 - 125 c 3 .
17. 8 .s 3 4 36 s 2 4 18 « - 81 - 27 s" 1 4 81 s~ 2 - 27 s~ 3 :
1 8. 21 a* - 54 a* + 27 ai + 63 a - 44 a* 4 1 - 6 a*.
19. a?" 3 - 3 x~hf 4 3 ar 1 ?/ - z 3 - 3 ar 2 z -.y* 4- 6 arty*« - 3#s
+ 3 x- V _ 3 y ^
20. a 4 6 aV 1 4 12 ah~ 2 4- 8 6~ 3 4 3 a V 2 4 12 aVV" 2
+ 12 6" 2 c- 2 4 3 a V 4 4- 6 Ir^r* 4 c" 6 .
Find the fourth roots of the following :
21. 81 a 10 - 36 a* k x~* 4 6 a 5 *- 1 * - $ a*ar 5 4 gV *"*•
22. a? 8 - 12 x- 7 4 50a; 6 -72 x 5 -21 a* 4 4- 72 a 8 + 50 a 2 4 12 a + 1.
Find the sixth roots of the following :
23. 64 m 12 - 192 m 10 4- 240 m 8 - 160 m« 4- 60 m 4 - 1 2 m s 4 1.
24. a 3 - 3 a*6* 4 V <** - f a *(t 4 1 j «?; 6 - A «^ ¥ + * l * & 9 -
173. Square Root of any Integral Perfect Square.
The square root of an integral perfect square may be found
in the same way as the square root of a polynomial.
INVOLUTION AND EVOLUTION 103
We have the following rule for finding the square root of an
integral perfect square :
Separate the number into periods of two digits each,
beginning with the units' place.
Find the greatest square in the left-hand period, and
write its square root as the first digit of the root ; sub-
tract the square of the first root digit from the left-hand
period, and to the result annex the next period.
Divide this remainder, omitting the last digit, by twice
the part of the root already found, and annex the quo-
tient to the root, and also to the trial divisor.
Multiply tne complete divisor by the root digit last
obtained, and subtract the product from the remainder.
If other periods remain, proceed as before, doubling
the part of the root already found for the next trial
divisor.
Note 1 : It sometimes happens that, on multiplying a complete divisor
by the digit of the root last obtained, the product is greater than the
remainder.
In such a case, the digit of the root last obtained is too great, and one
less must be substituted for it.
Note 2 : If any root digit is 0, annex to the trial divisor, and annex
to the remainder the next period.
Ex. Required the square root of 15376.
1
a 2 = 1
2 a + b = 200 + 20
b = 20
'53'76
00 00
53 76
44 00
100 + 20 + 4
= a + b + c
= (2a + b)b
2a + 26 + c = 200+40 + 4
4
9 76
9 76
= (2a + 2b + c)c
Pointing the number in accordance with the rule of § 173, we find that
there are three digits in its square root.
Let a represent the hundreds' digit of the root, with two ciphers
annexed ; b the tens' digit, with one cipher annexed ; and c the units'
digit.
Then, a must be the greatest multiple of 100 whose square is less than
15376 ; this we find to be 100.
Subtracting a 2 , or 10000, from the given number, the result is 5376.
104 ALGEBRA
Dividing the remainder by 2 a, or 200, we have the quotient 26+ ; which
suggests that b equals 20.
Adding this to 2 a, or 200, and multiplying the result by b, or 20, we
have 4400 ; which, subtracted from 5376, leaves 97(1.
Since this remainder equals (2 a + 2 b + c)c, we can get c approxi-
mately by dividing it by 2 a + 2 &, or 200 + 40.
Dividing 976 by 240, we have the quotient 4+ ; which suggests that
c equals 4.
Adding this to 240, multiplying the result by 4, and subtracting the
product, 976, there is no remainder.
Then 124 is the square root.
Omitting the ciphers for the sake of brevity, and condensing the opera-
tion, we may arrange the work of the example as follows:
1'53'76[124
1
22
53
44
2441976
976
CUBE ROOT OP AN ARITHMETICAL NUMBER
174. The cube root of 1000 is 10 ; of 1000000 is 100, etc.
Hence, the cube root of a number between 1 and 1000 is be-
tween 1 and 10 ; the cube root of a number between 1000 and
1000000 is between 10 and 100 ; etc.
That is, the integral part of the cube root of an integer of
one, two, or three digits contains one digit; of an integer of
four, five, or six digits contains two digits ; and so on.
Hence, if a point be placed over every third digit of an
integer, beginning at the units' place, the number of
points shows the number of digits in the integral part
of its cube root.
175. Cube Root of any Integral Perfect Cube.
The cube root of an integral perfect cube may be found in
the same way as the cube root of a polynomial.
INVOLUTION AND EVOLUTION 105
Required the cube root of 124871G8.
12487108 1 200 + 30 + 2
q 8 = 8000000 1 = a + b + c
3 a 2 = 120000
Sab = 18000
b 1 = 900
138900
30
4487168
4167000
3(a + b) 2 = 158700
3(a + b)c = 1380
c 2 = 4
160084
2
320168
320168
Pointing the number in accordance with the rule of § 174, we find
that there are three digits in the cube root.
Let a represent the hundreds' digit of the root, with two ciphers
annexed ; b the tens' digit, with one cipher annexed ; and c the units'
digit.
Then, a must be the greatest multiple of 100 whose cube is less than
12487168 ; this we find to be 200.
Subtracting a 3 , or 8000000, from the given number, the result is
4487168.
Dividing this by 3 a 2 , or 120000, we have the quotient 37+ ; which sug-
gests that b equals 30.
Adding to the divisor 120000, 3 ab, or 18000, and 6 2 , or 900, we have
138900.
Multiplying this by &, or 30, and subtracting the product 4167000 from
4487168, we have 320168.
Since this remainder equals [3 (a + &) 2 + 3(a + b)c + c 2 ]c (§ 171, III),
we can get c approximately by dividing it by 3(a + ft) 2 , or 168700.
Dividing 320168 by 158700, the quotient is 2+ ; which suggests that c
equals 2.
Adding to the divisor 158700, 3 (a + 6)c, or 1380, and c 2 , or 4, we have
160084 ; multiplying this by 2, and subtracting the product, 320168, there
is no remainder.
Then, 200 + 30 + 2, or 232, is the required cube root.
176. Omitting the ciphers for the sake of brevity^ and con-
densing the process, the work of the example of § 175 will
stand as follows ;
106 ALGEBRA
1200
180
9
1389
4487
4167
158700
1380
4
160084
320168
320168
The numbers 120000 and 158700 are called trial divisors, and the num-
bers 138900 and 160084 are called complete divisors.
We then have the following rule for finding the cube root of
an integral perfect cube :
Separate the number into periods by pointing every
third digit, beginning with the units' place.
Find the greatest cube in the left-hand period, and
write its cube root as the first digit of the root ; subtract
the cube of the first root digit from the left-hand period,
and to the result annex the next period.
Divide this remainder by three times the square of
the part of the root already found, with two ciphers
annexed, and write the quotient as the next digit of the
root.
Add to the trial divisor three times the product of the
last root digit by the part of the root previously found,
with one cipher annexed, and the square of the last root
digit.
Multiply the complete divisor by the digit of the root
last obtained, and subtract the product from the re-
mainder. -*
If other periods remain, proceed as before, taking
three times the square of the part of the root already
found, with two ciphers annexed, for the next trial
divisor.
Note 1 : Note 1, § 173, applies with equal force to the above rule.
Note 2 : If any root-figure is 0, annex two ciphers to the trial divisor,
and annex to the remainder the next period.
INVOLUTION AND EVOLUTION 107
177. In the example of § 175, the first complete divisor is
3a 2 +3ab + b 2 . (1)
The next trial divisor is 3 (a -f b) 2 , or 3 a 2 -f 6 ab + 3 b 2 .
This may be obtained from (1) by adding to* it its second
term, and double its third term.
That is, if the first number and the double of the sec-
ond number required to complete any trial divisor be
added to the complete divisor, the result, with two
ciphers annexed, will give the next trial divisor.
This rule saves much labor in forming the trial divisors.
Ex. Find the cube root of 157464.
157464 |_54_
125
7500
600
16
8116
32464
32464
EXERCISE 33
Find the square roots of the following :
i. 5294601. 3. .00098596. 5. .0037319881.
2. 68.7241. 4. 567762.25.
Find the cube roots of the following :
6. 658503. 9. .000070444997.
7. 1953125. 10. .000001601613.
8. 748.613312.
Find the first four figures of the square roots of :
11. 3. 12. f 13. if- 14. f 15- iM
Find the first four figures of the cube roots of :
16. 5. 17. 16. 18. J. 19- -~- 20 - *V
108 ALGEBRA
OTHER POWERS
178. A Series is a succession of terms.
A Finite Series is one having a limited number of terms.
An Infinite. Series is one having an unlimited number of
terms.
179. In §§ 103 and 168 we gave rules for finding the square
or cube of any binomial.
The Binomial Theorem is a formula by means of which any
power of a binomial may be expanded into a series.
180. Proof of the Binomial Theorem for a Positive Integral
Exponent.
The following are obtained by actual multiplication :
(a + x) 2 = a 2 + 2 ax + x 2 ;
(a + x) s = a?+3a 2 x + 3ax 2 + x i ', .
(a + x) 4 = a 4 + 4 a 3 x + 6 cPx 2 -f 4 ax 3 + x 4 ; etc.
In these results, we observe the following laws :
1. The number of terms is greater by 1 than the exponent
of the binomial.
2. The exponent of a in the first term is the same as the
exponent of the binomial, and decreases by 1 in each succeed-
ing term.
3. The exponent of x in the second term is 1, and increases
by 1 in each succeeding term.
4. The coefficient of the first term is 1, and the coefficient of
the second term is the exponent of the binomial.
5. If the coefficient of any term be multiplied by the expo-
nent of a in that term, and the result divided by the exponent
of x in the term increased by 1, the quotient will be the
coefficient of the next following term.
181. If the laws of § 180 be assumed to hold for the expan-
sion of (a-f- x) n , where n is any positive integer, the exponent
of a in the first term is n, in the second term n — 1, in the
third term n — 2, in the fourth term n — 3, etc.
INVOLUTION AND EVOLUTION 109
The exponent of x in the second term is 1, in the third term
2, in the fourth term 3, etc.
The coefficient of the first term is 1*; of the second term n.
Multiplying the coefficient of the second term, n, by n — 1,
the exponent of a in that term, and dividing the result by
the exponent of x in the term increased by 1, or 2, we have
Vl \ u ~ ) as the coefficient of the third term ; and so on.
1.2 '
Then, (a + x) n = a n + na n ~ l x + n ^"^ 1 ) ^- 2 x 2
n(n-l)(n-2) _^, ■ (1)
1.2-3 V '
Multiplying both members of (1) by a + a, we have
(« + x)^ 1 = «»+! + na n x -4- w(n ~" X) a n ~W + W C W ~ 1 )C»- 2 ) g >-2 g 8 + ...
1 • 2i 1 • '- • o
+ a n x+ na n ~ 1 x 2 + n ( n ~ ^ a n ~ 2 x* + — .
1 • ii
Collecting the terms which contain like powers of a and x, we have
(a + z) n+1 = a n+1 + (n + l)a n x + pO*- 1 ) + nla*- 1 * 2
r n(n-l)(n-2) . n(n-l) "] ^^ ■ ...
L 1-2.3 1-2 J
= a"* 1 + (n + l)a n x + np-=-^ + lla"" 1 * 2
1-2 L 3 J
Then, (a 4- x) n+1 = ^ n+1 + O 4- l)a n x + n r^-Ha"" 1 * 2
= a** 1 + (n + l)a"x + ( n + *)* o"-^ 2
(w+l)ii(n-l) 2z 3 + .... (2)
1.2-3 K
110 ALGEBRA
It will be observed that this result is in accordance with
the laws of § 180 ; which proves that, if the laws hold for any
power of a -f x whose exponent is a positive integer, they also
hold for a power whose exponent is greater by 1.
But the laws have been shown to hold for (a -f- x) 4 , and
hence they also hold for (a + x) 5 ; and since they hold for
(a -f- x) 5 , they also hold for (a + #) 6 ; and so on.
Therefore, the laws hold when the exponent is any positive
integer, and equation (1) is proved for every positive integral
value of n.
Equation (1) is called the Binomial Theorem.
In place of the denominators 1 • 2 ; 1-2-3, etc., it is usual to write
[2, [3, etc.
The symbol |_n, read " factorial w," signifies the product of the natural
numbers from 1 to n, inclusive.
The method of proof exemplified in § 181 is known as Mathematical
Induction.
182. Putting a — 1 in equation (1), § 181, we have
(i+ a? )» = i+n^+ ^r 1 > g »+ n ( n -y n -- 2 ) ^+...-
\2_ [3
183. In expanding expressions by the Binomial Theorem,
it is convenient to obtain the exponents and coefficients of the
terms by aid of the laws of § 180.
i . Expand (a + x) 5 .
The exponent of a in the first term is 6, and decreases by 1 in each
succeeding term.
The exponent of x in the second term is 1, and increases by 1 in each
succeeding term.
The coefficient of the first term is 1 ; of the second, 5.
Multiplying 5, the coefficient of the second term, by 4, the exponent of
a in that term, and dividing the result by the exponent of x increased by
1, or 2, we have 10 as the coefficient of the third term ; avid so on.
Then, (a + a) 5 = a 5 + 5 a 4 x + 10 a' 6 x +10 aV + 6 ax v + x 6 .
INVOLUTION AND EVOLUTION 111
It will be observed that the coefficients of terms equally distant from
the ends of the expansion are equal ; this law will be proved in § 185.
Thus the coefficients of the latter half of an expansion may be written
out from the first half.
If the second term of the binomial is negative, it should be
enclosed, negative sign and all, in parentheses before applying
the laws.
2. Expand (1 — xf.
= 16 4-6.1&. (-x) + 15.1*. (-a-) 2 + 20- 1". (-*)•
+ 15 • I 2 . ( - X)4 + 6 . 1 • (- X) 5 + (- *)•
= 1 - 6 x + 15 x 2 -20 x 3 + 15 x 4 - 6 x 5 + x 6 .
If the first term of the binomial is an arithmetical number, it is con-
venient to write the exponents at first without reduction ; the result
should afterwards be reduced to its simplest form.
If either term of the binomial has a coefficient or exponent
other than unity, it should be enclosed in parentheses before
applying the laws.
3. Expand (3 m 2 — Vn) 4 .
(3 m 2 - Vny = [(3 m 2 ) + (- ?ii)] 4
= (3 ra 2 )* + 4 (3 m 2 ) 3 ( - ni) + 6(3 m 2 ) 2 (- ni) 2
+ 4 (3 m 2 ) ( - niy + (- nty
= 81 m 8 - 108 m*fȣ + 54 m*n$ - 12 m 2 n + nl
A trinomial may be raised to any power by the Binomial
Theorem, if two of its terms be enclosed in parentheses, and
regarded as a single term ; but for second powers, the method
of § 167 is shorter.
4. Expand (x 2 -2x- 2) 4 .
( X 2 _ 2 x - 2)4 = [(x 2 _ 2x) + (- 2)]*
= (x 2 - 2 x) 4 + 4 (x 2 - 2 x) 3 (- 2) + 6 (x 2 - 2 x) 2 *;- 2)*
+ 4(x 2 -2x)(-2) 3 + (-2) 4
= x 8 - 8 x? + 24 x« - 32 x 6 + 16x*
_ 8 ( x 6 - 6 x 5 + 1 2 x 4 - 8 X s )
+ 24(x 4 -4x 3 + 4x 2 ) -32(x 2 -2x)+ 16
= x 8 - 8 x 1 + 16 xe + 16 x 5 - 56x 4 -32x 3 + 64 x 2 + 64x + 16.
112 ALGEBRA
EXERCISE 34
Expand the following :
2. (x-yf. \ 3n J
3 . (i-*p. ««• (2c-*-icr^.
W ' ' 16. (1 - a 2 )' 2 .
5. (a 2 - ft)\ \ _ ' , _
18. (a + 6) 16 .
7. (3 m — 4 n) 4 . .. ~
8. (^-2g) 6 . I9 ' (2 a " t+ i~=fJ
9. (ar 2 + ^) 5 20 . (a^-7/V) 11 .
10. (2a~* + ^) 7 . 21. (a + ^-c) 4 .
/^ \6 22. (x 2 - 2 a - 3) 4 .
12.
23. (m 2 - 2 m + I) 4 -
a 2 & 2 \ 8 24. (a 2 + x + l) 5 .
& ay 25. (1 + c + c 2 )
,2\6
184. To find the rth or general term in the expansion of
(a + x) n .
The following laws hold for any term in the expansion of
(a + x) n , in equation (1), § 181 :
1. The exponent of x is less by 1 than the number of the
term.
2. The exponent of a is n minus the exponent of x %
3. The last factor of the numerator is greater by 1 than the
exponent of a.
4. The last factor of the denominator is the same as the
exponent of x.
Therefore in the rth term, the exponent of x will be r — 1.
The exponent of a will be n — (r — 1), or n — r + 1.
The last factor of the nu 111 crater will be n — r + 2.
The last factor of the denominator will be r — 1.
INVOLUTION AND EVOLUTION 113
Hence, the rth term
n(n - 1) O - 2) . • . (n - r + 2) n _ r+1 r-1 w i
In finding any term of an expansion, it is convenient to obtain
the coefficient and exponents of the terms by the above laws.
Ex. Find the 8th term of (3 a* - &- 1 ) 11 .
We have, (3 ah - 6" 1 ) 11 = [(3 ah) + (- ft" 1 )] 11 . ■ ■
In this case, n = 11, r = 8.
The exponent of (— b' 1 ) is 8 — 1, or 7.
The exponent of (3 a%) is 11 — 7, or 4.
The first factor of the numerator is 11, and the last factor 4 + 1, or 5.
The last factor of the denominator is 7.
Then, the 8th term = U • 10 ■ 9-8- 7 -6- 5 J w _ & _n 7
1.2.3.4.5.6.7^ ;i ;
= 330(81 a 2 ) (- 6" 7 ) = - 26730 a 2 &" 7 .
If the second term of the binomial is negative, it should be enclosed,
sign and all, in parentheses before applying the laws.
If either term of the binomial has a coefficient or exponent other than
unity, it should be enclosed in parentheses before applying the laws.
Find the : exercise 35
x. 5th term of (a + b) K ^ q{ _ JV*
2. Tth term of (x-y) 10 . 3 xj '
3. 6th term of (1 - a) 11 . 8 . 6th term of f^ + -^
lr
4. 4th term of (a 2 — & 3 ) 8 .
5. 8th term of (d - 2 #J» g . 5th term of ( /« _Ji)\
6. 10th term of (m' s + r> ) * ,. ,1 * p / /- « , N7
v 2/ io. 4th term of (xVy — -f y-i) 7 .
185. Multiplying both terms of the coefficient, in (1), § 184,
by the product of the natural numbers from 1 to n — r -f- 1, in-
clusive, the coefficient of the rth term becomes
w(n-l)--(n — r + 2)-(tt-r + l)-..2-l [n
\r — 1 x 1 • 2 ... (n - r + 1) " [r — 1 \ n — r + l '
114 ALGEBRA
Since the number of terms in the expansion is n -j- 1, the rth
term from the end is the (n — r -f 2)th from the beginning.
Then, to find the coefficient of the rth term from the end, we
put in the above formula n — r -f 2 for r.
Then, the coefficient of the rth term from the end is
or
\n — r + 2 — l \n — (n — r + 2)+l ' \n — r + 1 \r — 1
Hence, in the expansion of (a + a?) w , the coefficients of
terms equidistant from the ends of the expansion are
equal.
186. It was proved in § 181 that, if n is a positive integer,
(a + x) n = a n + na n ~ l x + n ( n — 1 ) a n - 2 z 2
v J 1-2
, n (n — l)(n — 2) _- o .
+ v la 2 q + '"'
If n is a negative integer, or a positive or negative fraction,
the series in the second member is infinite (§ 178) ; for no one
of the expressions n — 1, n — 2, etc., can equal zero ; in this
case, the series gives the value of (a + x) n , provided it is
convergent.
As a rigorous proof of the Binomial Theorem for Fractional and Nega-
tive Exponents is too difficult for pupils at this stage of their progress, the
author has thought best to omit it ; any one desiring a rigorous algebraic
proof of the theorem, will find it in the author's Advanced Course in
Algebra, § 575.
187. Examples.
In expanding expressions by the Binomial Theorem when
the exponent is fractional or negative, the exponents and
coefficients of the terms may be found by the laws of § 180,
which hold for all values of the exponent.
I. Expand (a + xy to five terms.
The exponent of a in the first term is §, and decreases by 1 in each
succeeding term.
INVOLUTION AND EVOLUTION 115
The exponent of x in the second term is 1, and increases by 1 in each
succeeding term.
The coefficient of the first term is 1 ; of the second term, J.
Multiplying §, the coefficient of the second term, by — |, the exponent
of a in that term, and dividing the product by the exponent of x increased
by 1, or 2, we have — I as the coefficient of the third term ; and so on.
Then, (a + «)*- a% + f ef*as - \a~^x L + ,^a~*x 8 - ^gT^o 4 + •••.
2. Expand (1 + 2 x~^)~ 2 to five terms.
Enclosing 2 z* in parentheses, we have
(1+2 x~V 2 = [1 + (2 aT*)]" 2
= l" 2 - 2 • 1-3 . (2 x~i) + 3 • I" 4 • (2 x~i) 2
- 4 • I" 5 • (2 x"i) 3 + 5 • I" 6 • (2 x~i)* - •••
= l-4x"U 12z- 1 -32x"£ + 80x- 2 +
By writing the exponents of 1, in expanding [1 -f (2x~^)] -2 , we can
make use of the fifth law of § 180.
3. Expand to four terms.
^V 1 - 3 x*
Enclosing a -1 and — 3 x* in parentheses, we have
-_JL_ = J = [(«-i) + (-3xi)]-*
V a-i-3x* (a" 1 -3 a*)*
= (a" 1 ) - * - 1 (a- 1 )"^- 3 xi) + { 0-T^(- 3 xi) 2
-H(a-i)- : ¥(-3a*)«+...
= as + afxi + 2 afet + - 1 /- a'^x + ♦•••
EXERCISE 36
Expand each, of the following to five terms :
1. (a + x)*. 4- Va-b. 7. (a* + 2 &)*
1
2. (1 + *)-*. 5 * (a + a;) 5 ' 8 - (rf-4^)"*.
3. (1 — x) *. VI — x x~* + 3 y
116 ALGEBRA
I2 . !___. 14. (m*-3rr*)-3.
10. (m~ 3 + *
n. V[(a~ 2 — 66-c) 7 ]. \2/ «/ xSva 41 J
188. The formula for the rth term of (a + x) n (§ 184) holds
for fractional or negative values of n, since it was derived from
an expansion which holds for all values of the exponent.
Ex. Find the 7th term of (a - 3 x~^)~K
Enclosing — 3 aft in parentheses, we have
(a - 3 x"i)"i = [a + (- 3 jc~f)]"i.
The exponent of (— 3 #~t) is 7 — 1, or 6.
The exponent of a is — ± — 6, or — &
The first factor of the numerator is — J, and the last factor — ^ + 1,
or - ¥•
The last factor of the denominator is 6.
Hence, the 7th term
1.23.4.5.6
3» v y 9
_. , _ EXERCISE 37
Find the :
a-V(-3ari)«
1. 6th term of (a -f *)*. 6. 11th term of V(m 4- n) 5 -
K4 , . - , 7N -i 7. 7th term of (a~ 2 -2 &V 2 .
2.- 5th term of (a — 0) ». ' v }
3. 7th term of (1 + x)~\ S ' 8th term of ,_ * r '
* v ^ ; (rf + y 2 ) 4
4. 8th term of (1 - &)*. 9. 10th term of (ar 5 + y *)'*
5. 9th term of (a - x)~ 3 . 10. 6th term of (a* - 2 6" 4 )"^
11. 5th term of (m+3n- 8 )*
1
12. 9th term of
^[(a 3 + 3&-fy]
INVOLUTION AND EVOLUTION 117
189. Extraction of Roots.
The Binomial Theorem may sometimes be used to find the
approximate root of a number which is not a perfect power of
the same degree as the index of the root.
Ex. Find V25 approximately to five places of decimals.
The nearest perfect cube to 25 is 27.
We have v"25 = ^27-2 = [(3 8 ) + ( - 2)]*
= (3 3 ) } + i(3 8 )"*(- 2)- i(3 3 )~*(- 2)2
= 3--
+ A(3 3 ) T (-2)«-.
2 4 40
3 . 3 2 9 • 35 81 • 3 8
Expressing each fraction approximately to the nearest fifth decimal
place, we have
^/25 = 3 - .07407 - .00183 - .00008 = 2.92402.
We then have the following rule :
Separate the given number into two parts, the first of
which is the nearest perfect power of the same degree as
the required root, and expand the result by the Binomial
Theorem.
If the ratio of the second term of the binomial to the first is a small
proper fraction, the terms of the expansion diminish rapidly ; but if this
ratio is but little less than 1, it requires a great many terms to insure any
degree of accuracy.
EXERCISE 38
Find the approximate values of the following to five places
of decimals :
i. Vl7. 2. V51. 3. ^60. 4. ^14. 5. \/84. 6. a/35.
PROPERTIES OF QUADRATIC SURDS
190. A quadratic surd (§ 70) cannot equal the sum of
a rational expression and a quadratic surd.
For, if possible, let (a)? = b -f (c)*,
1 1
where 6 is a rational expression, and (a) 2 and (c) 1 quadratic surds.
118 ALGEBRA
Squaring both members, a = 6 2 + 2 6(c)' 2 -f c,
or, 2 6(c)* = a- 6 2 - c.
Whence, (c) z = a ~ ft2 ~ c .
26
That is, a quadratic surd equal to a rational expression.
But this is impossible ; whence, (a) 2 cannot equal 6 +(c) 2 .
191. If a +(&)* = c + (cf )*, where a and c are rational ex-
pressions, and (5)* and (V21.
5. 7 + 4(3)* 13. 113-12(85)1
6. 17-12V2. 14. 366 + 24V2I0.
7. 2 + (3)4. 15. 540 -14 VII.
8. l + i V3.
195. Solution of Equations having the Unknown Numbers under
Radical Signs.
1. Solve the equation V# 2 — 5 — x = — 1.
Transposing — x, Vx 2 — 5 = x — 1.
Squaring both members, cc 2 — 5 = # 2 — 2j+1.
Transposing, 2 x = G ; whence, g = 3.
(Substituting 3 for x in the given first member, and taking the positive
value of the square root, the first member becomes
V9"^5" - 3 = 2 - 3 = - 1 ;
which shows that the solution x = 3 is correct.)
We then have the following rule :
Transpose the terms of the equation so that a surd term
may stand alone in one member; then raise both mem-
bers to a power of the same degree as the surd.
If surd terms still remain, repeat the operation.
The equation should be simplified as much as possible before perform
ing the involution.
INVOLUTION AND EVOLUTION 121
2. Solve the equation V2 x — 1 -+■ V2 x + 6 = 7.
Transposing V2 x — 1, a/2 x + 6 = 7 — V2 x — 1.
Squaring, 2x4-0 = 49 - 14 V2 x - 1 + 2 x — 1.
Transposing, 14 v 2 x — 1 = 42, or V2 x — 1 = 3.
Squaring, 2 x — 1 = 9 ; whence, x = 5.
3. Solve the equation Vcf — 2 — Va7=
V# — 2
Clearing of fractions, x — 2 — Vx 2 — 2 x = 1.
Transposing, — Vx' 2 — 2 x = 3 — x.
Squaring, x 2 — 2 x = 9 — 6 x + x 2 .
Transposing, 4 x = 9, and x = - • .
4
(If we put x = - , the given equation becomes
If we take the positive value of each square root, the above is not a
true equation.
Authorities differ as to whether it is allowable in such instance to
choose the negative value for one of the square roots. It seems more
consistent to adhere to the signs expressed in the given equation. If this
rule is followed, the above equation has no solution.
EXERCISE 40
Solve the following equations ; verify each root :
1. Vx + 5 + 2 = 5. 2. Va + 7 — Vx= 1.
3. Vx 2 + ±x _3-Va 2 + a + 6:=0.
4. -^ + 11 + 4 = 7. 8 . V^-WS+8 = - 12
Vtf-f-8
5. Var* — 11 -f 1 = #. /- , 1 1
V a? 4- 1 1 _ V if + 19
6. Vx - 28 = 14 - Vx. VS-3 Vs%2
/- , /T k 12 _ V4a; + 5 + Va; + 3
7. Vif+VlO— if = — . 10. — n ' ! — — 5.
VlO — x Vi a + — V» + 3
122 ALGEBRA
ii 3y2 a ? + 4 = 3V2a? + 2 _ a Va + # + Va — # _ 1
V2 a; V2 # -|- 1 Va + # — Va — x *
13- Va? + m+ Va? — w = V2 m — ri + 3 #.
14. Va — y-\- V& — y= Va~^
15. V2s + 3-V3s + 3=-Vs-10.
16. A/r+Wf^"= 1 + x. 19- Va 2 -5a;-8 = V.r-4.
17. x\U-l- V# = a. 20 Vb 2 + x + Vc 2 + x = b^
■y/b 2 + x — Vc- -\-x c
l8 V6-f a? + V^ = &
V& -\-x — Va
VII. IMAGINARY NUMBERS
196. If a number involves an indicated even root of a nega-
tive number, it is called imaginary. Such numbers depend
upon a new unit, V— 1 or (— 1)* ; as V— 2, V — 3.
197. An imaginary number of the form V— a is called a
pure imaginary number, and the sum of a real and an imagi-
nary is called a complex number ; as a + b V— 1.
198. Meaning of a Pure Imaginary Number.
If Va is real, we define Va as an expression such that,
when raised to the second power, the result is a.
To find what meaning to attach to a pure imaginary number,
we assume the above principle to hold when Va is imaginary*
Thus, V— 2 means an expression such that, when rais ed to
the second power, the result is — 2 ; that is, (^
(_2^) 2 =-2.
In like manner, ( V^l) 2 = (- 1*) 2 = - 1 ; etc.
IMAGINARY NUMBERS 123
OPERATIONS WITH IMAGINARY NUMBERS
199. By § 198, (V^) 2 = (-5*) 2 =--5. (1)
Also, (V5V^l) 2 =(V5) 2 (V^l) 2 =5(-l) = -5, (2)
or (V=5) 2 =(5*) s •(-l^) 2 =5(-l) = -5.
From (1) and (2), (V^5) 2 = ( V5 V^) 2 .
Whence, V 11 ^ = -y/5V-l, or 5*(- 1)*.
Then, every imaginary square root can be expressed as the
product of a real number by V— 1. It is advisable to reduce
every imaginary to this form before performing the indicated
operations.
V— 1 is called the imaginary unit ; it is often represented by i.
In all operations with imaginary numbers, it is advisable to
reduce the number to the form a + bi where a and b are real.
Ex. Add V^4 and V-36.
V^4 = 2*\ V- 36 = 6 i.
2t + 6 i = 8 *\ or 8v^l, or 8 (- l)i
The Powers of » : V — 1 = i 1 ;
(V"=t:) 3 --v^i=^;
Note that the even powers of i are real, the odd powers
imaginary, the fifth power like the first power, the sixth
power like the second, etc.
Ex. 1. Multiply V-^2 by V^S.
v r ^2 = iv / 2, v^3 = iV3.
(iV2) (iVS) = <*V5=- \/S, or -(€)*.
124 ALGEBRA
Ex. 2. Divide (-40)2 by (-5)'.
(- 40)4 = i (40)2, (- §)♦ = i (5)*
-^— % = (8)* = 2 (2)*, or 2 v2.
EXERCISE 41
Simplify the following :
i. V^l^+V^l.
2. 2V := ^9 + 4V^25-3V^36.
3. 2V ::r 3-3V^27 + 5V : = : 12.
4. 7 V^a 2 - 3 V- 49 a 2 - 2 V^4a 2 .
5. iV^8 + iV^32-lV^162.
6. Add2 + V : ^to3-2v r=: 27.
7. From 8-6 V- 121 subtract 5 + 2 V-169.
8. From a - V26-6 2 -l take 6 - V2 a - a 2 - 1.
Multiply the following :
9. V^byV^T. 11. -V^Wby V^~6. .
10. V^by V-144. 12. _V-432by-V := ~75.
13. V — a 2 , V — 5 2 , an d — V — c 2 .
14. 2+V-3by 3-4V :r 3.
15. 5-2V-Tby 4-3i.
16. 4t"VaJ — SiVy by 9i'V# + V — 2/.
Expand the following :
17. (2-V^3) 2 . 18. (3V^2 + 2V" : ^3) 2 .
19. (2V^4-3V^)(2V^-3V'^7).
20. (a — V— 6) 8 .
IMAGINARY NUMBERS 125
Divide the following:
21. V^^byV^. 23. -Vl^by-V^
22. V— 54 by — V— 3. 24. — V— 96a6 by V— Sab.
25. 6^V6~VS84 by -V^H
GRAPHICAL REPRESENTATION OF IMAGINARY NUMBERS
200. Let be any point in the straight line XX 1 .
We may suppose any positive real
number, + a, to be represented by /^*\
the distance from to A, a units to 2 A' -& 0+^ A x
the right of O in OX.
Then any negative real number, — a, may be represented
by the distance from to A', a units to the left of in OX'.
201. Since — a is the same as (-f-a)x(- 1), it follows
from § 200 that the product of + a by — 1 is represented by
turning the line OA, which represents the number + a,
through two right angles, in a direction opposite to the motion
of the hands of a clock.
Then, in the product of any real number by — 1, we may
regard — 1 as an operator which turns $he line which repre-
sents the first factor through two right angles, in a direction
opposite to the motion of the hands of a clock.
202. Graphical Representation of the Imaginary Unit i (§ 196).
By the definition of § 198, — 1 = 1 x ii
Then, since one multiplication by i, fol-
lowed by another multiplication by i, turns
the line which represents the first factor
X- 1 — ^ 1 2 through two right angles, in a direction
opposite to the hands of a clock, we may
regard multiplication by 1 as turning the
line through one right angle, in the same
direction.
Thus, let XX and YY' be straight lines intersecting at right angles
at O.
Y
•B
c-
+ai
+i
S .
nr
0-f-a A
-1
c'.
-ai
-B'
f
126
ALGEBRA
Then, if + a be represented by the line OA, where A is a units to the
right of in OX, + ai may be represented by OB, and — ai by OB',
where B is a units above, and B' a units below, O, in 77'.
Also, + i may be represented by OC, and — i by OC, where C is one
unit above, and C one unit below, 0, in 77'.
203. Graphical Representation of Complex Numbers.
We will now show how to represent the complex number
a + bi.
Y
■ B
rO
,V
*>A
I
*>
c^.
/X
C
v>'
B
r'
EXERCISE 42
Represent the following graphically :
i. 3 I 2. — 61 3. 4 + i.
5. 2-5i. 6. -5-:W.
4 . _l + 2 t\
7. -7 + 4t.
IMAGINARY NUMBERS
127
204. Graphical Representation of Addition.
We will now show how to represent the result of adding b to
=— 7, c = 3 ; substituting in (1),
• = 3 or -.
22 4 2
EXERCISE 44
Solve by the first method : (Verify each result.)
i. ar- 12^4- 32 = 0. 6. *2 + *_30 = 0.
z 2 + 7 z- 30 = 0. 7. 6 z 2 4-4 = -11 z.
4^-Tf — 8. 8 4* 2 -3* = 7.
16x 2 -8x-35=0.
9 . ?_ 2 _^_M = o.
3 m 2 -26 = 9 m 2 -80. 3 2 6
3a; 2 2a?-6 = 1 2_
x 2 — 7#4-6 x — 6
Solve by second method : (Verify each result.)
ii. (3fc4-2)(2&H-3) = (fc-3)(2fc-4).
12. 30 - 3°
07 iC-f 1
13. Vm -f 2 4- V3 m + 4 = 8.
14. (y-3y-(y + 2)* = -65.
15. V5 4- x 4- V5 — x = -
16.
d-2
iVote i : In solving equations involving fractions or radicals reject any
root which does not satisfy the given equation.
132 ALGEBRA
Solve by means of the formula in §213, (1): (Verify each
result.)
17. 3x 2 -2x = A0. 5 13 1
18. 9x 2 + 18x=-8. . 6 * 9z " 1S
1 1 a 2 - 17
20.
x + 3 x-5 x 2 -2x-15
21 . y~ c y + c^ if-Bc 2 ^
y + c y — c y 2 — +7a?-26 = 58-7a.
28. 5 (a + 2)* + (a? + 2) = 36.
29. (a 2 + 4a4-2) 2 = 31+2(a 2 + 4a + 2).
30. a 4 -8ar J + 10# 2 + 24#-315 = 0.
31. What number is that to which if you add its square the
sum will be 42 ?
32. A rectangular field is 40 rods longer than it is wide. By
doubling its length and decreasing its width by 15 rods, the area
is unchanged. Find dimensions of the field.
32- The difference between two numbers is 7, and the dif-
ference between their cubes is 1267. Find the numbers.
34. The denominator of a fraction is 3 more than its numera-
tor and by adding the fraction to its reciprocal the sum is
2^. What is the fraction ?
QUADRATIC EQUATIONS 135
35. There is a number consisting of two digits whose sum
is 11. If from the number 3 times the product of the digits
is subtracted, the remainder will equal the sum of the digits.
Find the number.
36. A man sells goods for $120, gaining a per cent equal
to I the cost of the gooo^s. What was the cost of the goods ?
37. A picture 13 inches by 8 inches is surrounded by a
frame of uniform width whose area is 162 square inches.
Find the width of the frame.
38. A man put $ 2400 in a savings bank which paid inter-
est semiannually. At the end of a year he found that he had
to his credit $ 2496.96. What interest did the bank pay ?
39. A number of people plan an excursion which is to cost
them $ 30. It is found later that 3 of the party cannot go,
which increases the cost 50 cents to each member. How many
are there in the party and what did each one pay ?
40. A and B start together for a 6-mile walk. A's rate per hour
is \ mile more than B's, and he finds he can reach his destina-
tion in 24 minutes less time than B. What is the rate of each ?
41. An open rectangular box is 8 inches high. Its length
is 4 more than its width. Its volume is 768 cubic inches.
Find its inside dimensions.
42. In a given circle APB, a perpen-
dicular DP, dropped from a point P in
the circumference to the diameter AB,
is a mean proportional between the seg-
ments, AD and DB, of the diameter. If
the radius of the circle is 12 and DP is
2V5, how far is D from B?
43. An open rectangular box 5 inches deep (inside measure)
is made of 1-inch lumber. Its length is 1 inch less than twice
its width. The difference between the volumes when inside and
outside measurements are taken is 271 cubic inches. How much
sheet metal will be needed for lining the sides and bottom of
the box ?
136 ALGEBRA
44. Two lines AB and CD intersect at in such a manner
that AO-OB=CO- OD. If CD = U,AO = 15, and AB = 18,
find CO.
45. A has a lease on a square room. He sublet to B a part
10 feet wide along one entire side of the room, at a rental of
$ 160 per month. The part of the room retained by A contained
704 square feet. How much rental per square
foot did B pay ? Explain your negative roots.
46. A tangent, PT, to a circle is a mean pro-
portional between the whole secant PD and the
external segment PE. If PT is 12, the radius
5, and PD passes through the center, find PE.
47. The upper base and the altitude of a trapezoid are equal,
the lower base is 20 and the area is 112. Find the upper base.
48. The length of a rectangle is V2 more than the side of a
given square, and its breadth is V2 less than a side of the
same square. The area of the rectangle is 1. Find the di-
mensions of the rectangle correct to three decimal places.
THEORY OF QUADRATIC EQUATIONS
214. Number of Roots.
A quadratic equation cannot have more than two dif-
ferent roots.
Every quadratic equation can be reduced to the form
ax 2 + bx + c = 0. *
If possible, let this have three different roots, n, ft, and r$.
Then, an 2 + br x + c = 0, (1)
ar 2 2 + br 2 + c = 0, (2)
and ar s 2 + br 3 + c = 0. (3)
Subtracting (2) from (1), a{r x 2 - r 2 2 ) + &(r* - r 2 ) = 0.
Then, a(n + r%) (n. — r 2 ) 4- &Oi — r 2 ) = 0,
or, (n — r 2 ) (ari + ar 2 + 6) — 0.
Then, by § 110, either ri — r 2 =* 0, or ar\ -f oft + ?> = 0.
But ri — r 2 cannot equal 0, for, by hypothesis, r*i and r 2 are different.
QUADRATIC EQUATIONS
137
Whence,
ari + ar-2, + 6 = 0.
(4)
Iii like manner, by subtracting (3) from (1), we have
ari + an + 6 = 0. (5)
Subtracting (5) from (4), ar 2 — an = 0, or r 2 — n = 0.
But this is impossible, for, by hypothesis, r 2 and n are different ;
hence, a quadratic equation cannot have more than two different roots.
215. The graphs of quadratic equations can be readily con-
structed by the method used in §§ 44-48.
Construct the graph of
x-
■6 = 0.
Placing the first member of the equation equal to y, we have
x 2 — x — 6 = y.
a)
(2)
Assigning values to x, we obtain corresponding values of y. For
example, Substituting x = in (2), we have y = - 6,
Substituting x = 2 in (2), we have y = — 4, etc.
x 2 — x — 6 = ?y
2/
-6
-6i
(-1)
-6
- 51 (-B)
-4
(O)
- 2J (D)
(E)
6
-51
(G)
-4
(H)
W
6
Y
\
\
\
\
X'
K )
O
/ E
>■
Yc
H
c
/P
A
Y'
X 2 _ x _ e s- 0,
(£-3)(x + 2) =0,
x = 3 or
Solving
or
we have, x — 3 or — 2.
These values, x = 3, x = — 2, are the abscissas of the points where the
curve crosses the x-axis, the curve showing in a graphical way why a
quadratic equation has two roots.
138 ALGEBRA
The graph of every equation of the form x 2 + px — q = or
ax 2 -f bx -f- c = is a curve of the above form and is called a
parabola.
216. Sum of Roots and Product of Roots.
Let r x and r 2 denote the roots of ax 2 -f- bx + c = 0.
By § 213, (1), n = -ft + V5'-4qc and ^ = - b - Vb 2 - 4 flC .
2 a 2 a
Adding these values, n -f ra == — — - = — -•
2 a a
Multiplying them together,
nn = ^-{b^-iac) ( 103 j = 4_ae = c .
4 a 2 4 a 2 a
Hence, if a quadratic equation is in the form ax 2 -f bx
+ c = 0, the sum of the roots equals minus the coefficient
of x divided by the coefficient of a? 2 , and the product of
the roots equals the independent term divided by the
coefficient of a? 2 .
217. Formation of Quadratic Equations.
By aid of the principles of § 216, a quadratic equation may
be formed which shall have any required roots.
For, let ri and r 2 denote the roots of the equation
ax 2 + bx + c = 0, or a 2 + — + - = 0. (1)
a a
Then by § 216, 5 = _ n - r 2 , and G - = nr 2 .
a a
Substituting these values in (1), we have
x 2 — r\X — r 2 x + 7"ir 2 = 0.
Or, (x — r x ) (x — r 2 ) = 0.
Therefore, to form a quadratic equation which shall have
any required roots,
Subtract each of the roots from 05, and place the prod-
uct of the resulting expressions equal to zero.
Ex. Form the quadratic whose roots shall be 4 and — -J.
By the rule, (« - 4) (x + J) = 0.
Multiplying by 4, {x - 4) (4 x + 7) = ; or, 4 x 2 - 9 x - 28 = 0.
QUADRATIC EQUATIONS 139
DISCUSSION OF GENERAL EQUATION
218. The roots of a quadratic equation may take several
forms :
1. The roots may be rational, unequal, of the same sign.
2. The roots may be rational, unequal, of opposite sign.
3. The roots may be rational, equal.
4. The roots may be irrational, unequal.
5. The roots may be irrational, equal.
6. The roots may be irrational and the number under the
radical sign negative.
These forms and the causes for their existence are at once
seen when one considers the formula in § 213.
By § 213, the roots of ax 2 + bx -f- c == are
— b 4- V& 2 — 1 ac -, — b — -\/b 2 — 4 ac
r, = :£ — ■ and r 2 =
1 2a 2a
We will now discuss these results for all possible real values
of a, b, and c.
I. b 2 — 4 ac positive.
In this case, i\ and r 2 are real and unequal.
II. &*-4ac = 0.
In this case, r x and r 2 are real and equal.
III. c = 0.
In this case, the equation takes the form
ax 2 + bx = : whence x = or
a
Hence, the roots are both real, one being zero.
IV. b = 0, and c = 0.
In this case, the equation takes the form ax 2 = 0.
Hence, both roots equal zero.
V. b 2 — 4 ac negative.
• In this case, r x and r 2 are imaginary (§ 196).
VI. 6 = 0.
In this case, the equation takes the form
ax 2 -f c = ; whence, a? = ± \/
140 ALGEBRA
If a and c are of unlike sign, the roots are real, equal in
absolute value, and unlike in sign.
If a and c are of like sign, both roots are imaginary.
The roots are both rational, or both irrational, according as
b 2 — 4 ac is, or is not, a perfect square.
219. It is evident that irrational roots, whether real or
imaginary, must occur in conjugate pairs.
That is, in an equation of the form of ax* -f bx + c = 0, where
a, b, c are real, if one root is of the form k -f y7t the other
must be k — ^/h where k and h are real.
EXERCISE 46
Find by inspection the sum and product of the roots of the
following :
i. x 2 — 2 x — 35 = 0. 5- x 2 -f ax — bx = ab.
2. x 2 + 15^ + 36 = 0. 6. cdx 2 + d 2 x = c 2 x + cd.
3. 2r J + 7a;-4 = 0. 7- # 2 -2V2a-2 = 0.
4- 5a 2 -13a = -6.
8. One root of 8 x 2 — 2 a? — 15 = is — 1^; find the other.
9. One root of 6X 2 + 11 x— 2 = is ^; find the other.
10. One root of 2ar 5 -8a; 2 +2a4-12=0 i s 2 ; find the others.
11. One root of m 3 — 7 m + 6 = is — 3 ; find the others.
12. If r Y and r 2 are the roots of x 2 -f x -f- 1 = 0, what does
r 2 -+- r 2 2 equal ? r? -f- ^ 3 ?
Form the equations whose roots are :
13- 2, 3. 18. a, 6 a.
14. — 1, 4. 19. a 4- V&, a — V&.
15. i-3. 20. 2+V^3, 2-V^3.
16. -|, — £ 21. 3c-d, — 2c+5f7.
17. 0, -£.
QUADRATIC EQUATIONS
141
22.
23-
V2k-5Vg y^2k±jyVg t
2 ' 2
6,-1,0.
Determine by inspection the nature of the roots of the
following :
24. x 2 + 7 x + 12 = 0. 30.
25. x 2 + 8x = -16. 31.
26. £ 2 + 2a-l = 0. 32.
27. ^ + 2a + 3 = 0. 33-
28. 2 a; 2 + 7 a=3. 34-
29. 4 a 2 -16 = 0. 35-
2a 2 = 15z + 18.
x 2 - x = 12.
10a 2 -z = 2.
23 a? - 6 = 7 a 2 .
16 x 2 + 24 x + 9 = 0.
5x 2 + 3x = -2.
GRAPHS
220. The nature of the roots discussed in § 218 is illustrated
by the use of graphs :
Y
Fig 9
Y
1
I
Fig "■
1
y
-8
-9
-8
-5
7
-5
7
1
X
1
2
8
4
5
-2
-3
y
1
1
4
9
16
9
16
\
X
\
\
/
\
/
\
/
\
/
i
1
x'
O
X
2
/
I
3
\
/
\
/
4
\
/
\
/
5
\
/
\
/
-1
\
I
\
1
-?
\
1
\
3
\
\
/
\
1
X'
\
/
f
X
V
/
y'
xl
Fig. 1. #2_ 2^-8 =
6 2 -4ac>0
Fig. 2. #2-2a? + l=0
ft 2 - 4 ac = O
142
ALGEBRA
Fig
3
X
y
3
1
2
2
3
3
6
4
11
- 1
6
-2
11
Y
Snip
1111
xj o x
y'
-3| 18
3. x*-23C + 3 = Q
b 2 - 4 ac < O
In Fig. 1, the curve crosses the a>axis at points whose abscis-
sas are 4, — 2 ; the abscissas of these points being the values of
x found in solving the equation. In Fig. 2, the intersection
points coincide and we have two values of x each equal to 1.
In Fig. 3, the curve and the a>axis do not coincide.
EXERCISE 47
Plot the curves :
i. f(x) = x 2 + 6 x + 8. (§§ 47, 220.)
2. f(x) = x 2 -6x + 8. 4 . f(x) = x 2 -6x + 9.
3. f(x) =x 2 -9. 5. f(x) = x 2 + 2 x + 4.
221. Many problems in Physics are dependent on the laws
of proportion and variation. The solution of such problems
is often obtained more readily by graphical means than by
algebraic solution.
QUADRATIC EQUATIONS
143
Ex. 1. Graphical representation of a direct proportion.
When a man is running at a constant speed, the distance
which he travels in a given time is directly proportional to
d s
his speed. The algebraic expression of this relation is — = — ,
ord = ms. (See §161.) d ' 2 *
Now, if we plot successive values of the distance, d, which correspond
to various speeds, a, in precisely the same
manner in which we plotted successive , *
values of x and y in §§ 44-48, we obtain
as the graphical picture of the relation
between s and d a straight line passing
through the origin. (See Fig. 1.)
This is the graph of any direct
proportion.
Fig. l.
Ex,
P
.2.
r \
• i x.
j J i L— L-Ulpa
Graphical representation of an inverse proportion.
The volume which a given body
of gas occupies when the pressure
to which it is subjected varies has
been found to be inversely propor-
tional to the pressure under which
the gas stands ; we have seen that
the algebraic statement of this re-
V P>
lation is — - = tt*
If we plot successive values of V and
P in the manner indicated in §§ 44-48, we
obtain a graph of the form shown in
Fig. 2.
This is the graphical representa-
' tion of any inverse proportion ; the
curve is called an equilateral hyper-
• bola.
Ex. 3. The path traversed by a
' falling body projected horizontally.
i i » 4
5 6
7 8 r
Fig.
2.
V 2 P x '
or V =
m
P'
V=l
P = m.
V=5,
p_m
5
F=2,
P — Wl
2 '
r=6,
P -VI.
(3
V=3
3
V-=l,
p_m
7
F=4
p_m
4 '
F=8,
8
144
ALGEBRA
When a body is thrown horizontally from the top of a tower,
if it were not for gravity, it would move on in a horizontal
direction indefinitely, traversing exactly the same distance in
each succeeding second.
Hence, if V represents the velocity of projection, the horizontal dis-
tance, If, which it would traverse in any number of seconds, t, would
be given by the equation H— Vt.
On account of gravity, however, the body is pulled downward, and
traverses in this direction in any number of seconds a distance which is
given by the equation 8 = \ gt 2 .
To find the actual path taken by the body, we have only to plot
successive values of II and S, in the manner in which we plotted the
successive values of x and ?/, in §§ 44-48.
Thus, at the end of 1 second the vertical distance Si is given by
#i = i 9 x l 2 = \ g ; at the end of 2 seconds we have, #2 = J g x 2 2 = \g ;
at the end of 3 seconds, #3 = \ g x 3' 2 = § g ; at the end of 4 seconds,
ft = .
On the other hand, at the end of 1 second
we have H\ = V \ at the end of 2 seconds,
H 2 — 2 V ; at the end of 3 seconds,
H 3 =S V; at the end of 4 seconds, H 4 =4 V.
If, now, we plot these successive values
of H and S, we obtain the graph shown in
Fig. 3.
This is the path of the body ; it is a
parabola. (§ 226, Ex. 2.)
Ex. 4. GrapJi of relation between
the temperature and pressure existing
within an air-tight boiler containing
only water and water vapor.
One use of graphs in physics is to express a relation which
is found by experiment to exist between two quantities, which
cannot be represented by any simple algebraic equation.
For example, when the temperature of an air-tight boiler
which contains only water and water vapor is raised, the pres-
sure within the boiler increases also; thus we find by direct
experiment that when the temperature of the boiler. is 0° centi-
grade, the pressure which the vapor exerts will support a
column of mercury 4.6 millimeters high.
Fig. 3,
QUADRATIC EQUATIONS
145
700
000
When the temperature is raised to 10°, the mercury column rises to
9.1 millimeters ; at 30° the column is 31.5 millimeters long, etc.
To obtain a simple and compact picture of the relation between tem-
perature and pressure, we plot a succession of temperatures, e.g. 0°, 10°,
20°, 80°, 40°, 50°, 60°, 70°, 80°, 90°,
100°, in the manner in which we
plotted successive values of x in §§ 44-
48, and then plot the corresponding
values of pressure obtained by experi-
ment in the manner in which we
plotted the ?/'s in §§ 44-48 j we obtain
the graph shown in Fig. 4.
From this graph we can find at 200"
once the pressure which will exist
within the boiler at any temperature.
For example, if we wish to know
the pressure at 75° centigrade, we
observe where the vertical line which
passes through 75° cuts the curve and then run a horizontal line from this
point to the point of intersection with the line OP.
This point is found to be at 288 ; hence the pressure within the boiler
at 75° centigrade is 288 millimeters.
O 10 20 SO 40 50 00 TU
Fig. 4.
EXERCISE 48
PROBLEMS IN PHYSICS
i. When the force which stretches a spring, a straight wire,
or any elastic body is varied, it is found that the displacement
produced in the body is always directly proportional to the
force which acts upon it ; i.e. if e^ and d 2 represent any two
displacements, and f x and f 2 respectively the forces which pro-
duce them, then the algebraic statement of the above law is
If a force of 2 pounds stretches a given wire .01 inch, how
much will a force of 20 pounds stretch the same wire ?
2. If the same force is applied to two wires of the same
length and material, but of different diameters, D Y and D 2 , then
(i)
146 ALGEBRA
the displacements +l) = 0..
SIMULTANEOUS QUADRATIC EQUATIONS
223. In solving simultaneous quadratic equations involving
two unknown numbers it is necessary to eliminate one of the
unknowns as was done in simultaneous linear equations.
The elimination of an unknown number from two equations
of the second degree will often produce an equation of the
fourth degree with one unknown number which cannot be
solved by the ordinary methods. The following general direc-
tions will lead to the solution of many types.
224. Case I. When each equation is in the form
ax 2 + by 2 = c.
In this case, either x 2 or y 2 can be eliminated by addition
or subtraction (§ 42, II, III).
Case II. When each equation is of the second degree, and
homogeneous; that is, when each term involving the unknown
numbers is of the second degree with respect to them (§ 23).
150 algp:bra
The equations may then be solved as follows :
Ex. Solve the equations J ' ^'
I x 2 + 2/ 2 = 29. (2)
Dividing (1) by (2), x2 ~^ x V = A r 29 z 2 -58 xy = 5 a? + 5 i/ 2 .
a? + ?/ 2 29
Then, 5 y 2 + 58 a;y - 24 z 2 = 0, or (5 y - 2 a:) (y + 12 *) = 0.
2 a*
Placing 5 y — 2 x = 0, ^ = ^ ; substituting in (1),
5
z 2 _ ȣL _ 5? or 34 = 25.
5
Then, a; = ± 5, and y = — = ± 2.
5
Case III. When the given equations are symmetrical with
respect to x and y ; that is, when x and y can be interchanged
without changing the equation.
Equations of this kind may be solved by combining them in
such a way as to obtain the values of x + y and x — y.
\x + y = 2. (1)
i. Solve the equations {
\ xy=-15. (2)
Squaring (1), x 2 + 2 xy + y 2 = 4.
Multiplying (2) by 4, \xy =— 60.
Subtracting, x 2 — 2 xy -f y 2 == 64.
Extracting square roots, x — y = ± 8. (3)
Adding (1) and (3), 2 as = 2 ± 8 = 10 or - 6.
Whence, x = 5 or — 3.
Subtracting (3) from (1), 2 ?/ = 2 T 8 = - 6 or 10.
Whence, y = — 3 or 5.
The solution is x = 5, y = — 3 ; or, x = — 3, y = 5.
The above method offers the most desirable form of solution
and should be employed when possible.
If one equation is of the second degree, the other of the
first degree, and they are not symmetrical, Case IV should be
used.
QUADRATIC EQUATIONS 151
Case IV. When one equation is of the second degree and the
other of the first.
Equations of this kind may be solved by finding one of the
unknown numbers in terms of the other from the first degree
equation, and substituting this value in the other equation.
r, a i ^ \2x 2 -xy = 6y. (1)
Ex. Solve the equations \ w
J x + 2y = l. (2)
From (2), 2 y 5= 7 - x, or y = I— £. (*)
Substituting in (1), 2x 2 -x( 7 -^-\ = qH-^JSV
Clearing of fractions, 4 x 2 — 7 x -f x 2 = 42— 6 x, or 5 x 2 — x=42.
Solving, x = 3 or - — .
5
Substituting in (3) , y = 1=1? or 1±J£ = 2 or — .
..... 2 2 10
The solution is x = 3, y = 2 ; or x = — ^, y = fg.
Certain examples where one equation is of the third degree and the
other of the first may be solved by the method of Case IV.
225. Special Methods for the Solution of Simultaneous Equa-
tions of Higher Degree.
No general rules can be given for examples which do not
come under the cases just considered; various artifices are
employed, familiarity with which can only be gained by
experience.
v-y=i9. (i)
i. Solve the equations
\x 2 y-xy 2 =6. (2)
Multiply (2) by 3, 3 x 2 y - 3 xy 2 = 18. (3)
Subtract (3) from (1), x 3 - Sx 2 y + Sxy 2 - y 3 = 1.
Extracting cube roots, x — y = l. (4)
Dividing (2) by (4), xy = 6. ~ (5)
Solving equations (4) and (5) by the method of § 224, Case III, we
find % ss 3, y = 2 j or, x = — 2, y =— 3.
152 ALGEBRA
. ( x 8 -\- y 8 = 9 xy.
2. Solve the equations 1
^ I a; + = 6.
Putting x = u -f v and y = u — v,
(u + v) 3 + (w-v) 3 = 9(w4-v)(w-'y), or, 2 w 8 + 6 uv 2 = 9(u 2 - v 2 ); (1)
and (u -f a) + (m — «?) = 6, 2 w = 6, or w = 3.
Putting w = 3 in (1), 54 + 18 v 2 = 9 (9 - t> 2 ).
Whence, v 2 = 1, or v = ± 1.
Therefore, x = w + v =3 ±1=4 or 2 ;
and y = u — v = 3=Fl=2or4.
The solution is x = 4, ?/ = 2 ; or, x = 2, y = 4.
The artifice of substituting u + v and tt — v for £ and ?/ is advantageous
in any case where the given equations are symmetrical (§ 224, Case III)
with respect to x and y. See also Ex. 4.
3. Solve the equations
x* + if + 2x + 2y = 23. (1)
xy = 6. (2)
Multiplying (2) by 2, 2xy= 12. (3)
Add (1) and (3), x 2 + 2xy + y 2 + 2x + 2y = 35.
Or, (x + ?/) 2 + 2(x + ?/)=35.
Completing the square, (x + y) 2 + 2 (x + y) +1 = 36.
Then, (x + y) + 1 = ± 6 ; and x + y = 5 or — 7. (4)
Squaring (4), x 2 + 2 x^ + y 2 = 25 or 49.
Multiplying (2) by 4, 4 xy = 24.
Subtracting, x 2 — 2 xy + ?/ 2 = 1 or 25.
Whence, x — y = ± 1 or ± 5. (5)
Adding (4) and (5), 2x = 5±l, or-7±6.
Whence, x = 3, 2, - 1, or — 6.
Subtracting (5) from (4), 2 y = 5 T 1 ,. or - 7 =F 5.
Whence, y = 2, 3, - 0, or - 1.
The solution isx = 3, ?/ = 2 ; x = 2, y = 3 ; x =— 1, ?/ = — (J ; or x = — 6,
QUADRATIC EQUATIONS
153
f x A + y* = 97.
4. Solve the equations -j
4 H la? +2/ =-1.
Putting x = u + v and y = u — v,
(u + t>) 4 + (fS - *)* = 97, or 2 «* + 12 m¥ + 2 «* = 97, (1)
and (w + a) + (w — a) = — 1, 2 w = — 1, or w = — J.
Substituting value of U in (1), J + 3 v 2 + 2 v 4 = 97.
Solving this,
25 _ 31
or ; and v =± - or ±
V-31
2
Then,x = n+^-l±g, or -1^^-^1=2,-3, or ~ 1±V - :il :
2 2' 2 2 ' 2
and y = ^-^-l T g r _l V-31^_ 3 r -l^V33T
2 2 2 2 2
The solution is x = 2, ?/ = — 3; x=-3, ?/ = 2 ; x- — l + — ~,
_1_V-31 _ -l-V-31 B -i+V-31
w = ; or x = , y = !
y 2 . 2 2
MISCELLANEOUS EXAMPLES
EXERCISE 50
Solve the following equations and verify each result :
3-
(2xy + x = — 36.
1 xy — 3 y = — 5.
6.
[a 2 + 2/ 2 + z-2/ = 32.
1 xy = 6.
1-1=12.
( x 2 — 3 xy — 4: y 2 = 0.
I3x-5y =46.
ja2__2?/ 2 + 3a: = -8.
a;
7- ■
8. i
^ ?/__io;
rtfi 3
U2_2 r °-42/ = -2.
154
10.
i5-
i8.
19.
22.
23-
k ,2 3 b
x a
(x 4 + y 4 = 17.
{ x — y =0.
(±d + k-3dk = -6.
\d-5k + 2dk = 10.
(x 2 + ±xy = 13.
[2xy + 9y 2 =87. .
17.
ALGEBRA
r 1
, 1
1
+ —
+
# 2
spy
.T
12.
ji
, 1
-
-h- =
s.
[x
2/
13.
14.
16.
:49.
a? + y *=» 35.
Va? 4- S/y = 5.
11 a; 2 — ay — y 2 = 45.
7a; 2 + 3a^-2?/ 2 = 20.
r ^ 2 - 24 xy + 95 = 0.
13.
3a?+2y 3a?-2 y
3#— 2 y
[8y 2 + 3a 2 = 29,
- = 6a 2 .
a#
a; + y = 5 a#y.
'- + ± = -19 0.3.
x 6 y 6
- + - = — a.
3 2/
e 2 + 9Z 2 + 4e = 9.
e j 4. 2 £ = - 2.
3 x 2 - 5 ay = 2 a 2 + 1 3 a b - 7 b 2 ,
x + y = 3(a — 6).
41
3a,-+2y 20
. a; 2 ?/ + a;?/ 2
2a 8 + 24a.
= 2 a 3 - 8 a.
of 4- yf
jr*y + i
V2a 2 -9 = 3y + 6.
Va? 4 — 17 y 2 = x 2 — 5.
24.
25.
26.
27.
28.
29.
5x-2y = l.
4 a; + 3 z = — 5.
a? 2 ?/ -}- #?/ 2 = 56.
» + y = - 1.
x[ , y 2 = 19
y a 6 *
a; y 6
f3a; 2 + 3?/ 2 = 10a*/.
[x y 3
# 2 # -f y 2 x = 42.
[ X y b
J 5 f + ga - 3 s 2 = 27.
l4^-4gs + 3s 2 = 72.
QUADRATIC EQUATIONS 155
+ 4xy-3y = 42. . ( x A + tftf + y A = 481.
' 2 y 2 — xy + 5 y = — 10. [ a? 2 — #?/ -f y 2 = 37.
■16 aY- 104 xy=-105. J J 9 aj 2 -13«y-3aj= -123.
| a; _ 2/ = _2. 33 ' 1b?/4-42/ 2 + 22/ = 125.
* Divide the first equation by the second.
EXERCISE 51
i. Find two numbers whose product is 112 and whose dif-
ference is 6.
2. A rectangular field has a perimeter of 104 rods and an
area of 4 acres. Find its dimensions.
3. The square of the sum of two numbers minus four times
their product equals 49, and the difference of their squares
equals 175. What are the numbers ?
4. The sum of the cubes of two numbers is 855 ; and if the
sum of the numbers be multiplied by their product, the result
will be 840. What are the numbers ?
5. There is a number consisting of two digits, the sum of
whose squares is 80 ; and if the sum of the digits be multiplied
by 4, the number will be expressed with its digits reversed.
What is the number ?
6. A man loaned a sum of money at 6 % for a given time
and received $240 interest; if he had loaned the same sum for
two years longer at the rate represented by the first number
of years, he would have received $40 more than at first. Find
the time and the amount loaned.
7. If 5 be added to the denominator and subtracted from
the numerator of a certain fraction, it will be expressed by its
reciprocal ; and the difference of the squares of numerator and
denominator equals 65. What is the fraction ?
8. A number consists of three digits, the second of which
is twice the first. The sum of the squares of the digits equals
156 ALGEBRA
89, and if 99 be subtracted from the number, the digits will be
reversed. What is the number ?
g. A man buys two pieces of cloth, each containing as many-
yards as its price per yard in cents, and he pays $ 41 for the
whole amount. If the prices for the two pieces of cloth had
been interchanged, his bill would have been $1 less. How
many yards of each did he buy and what was the price per
yard ?
io. Two squares have together an area of 613 square rods.
If the side of the first square were decreased by 6, and that of
the second increased by 1, their perimeters would be in the
ratio of 2 to 3. Find the side of each square.
ii. There are two numbers whose sum decreased by the
square root of their product is 13 ; and the sum of their squares
increased by their product is 481. Find the numbers.
12. Two boys count their pennies. They find that the
product of the numbers representing them is 84, and that the
square of their sum decreased by twice their difference is 351.
How many did each have ?
13. There are two numbers whose difference is 819, and
the difference of their cube roots is 3. What are the
numbers ?
14. There is a difference of one hour's time in two trains
which go from A to B, the rate of the first train being 5 miles
an hour more than that of the second train. If the speed of
each train were increased 2 miles per hour, the difference
in time from A to B would be decreased 7 minutes 80
seconds. Find the distance from A to B and the rate of
each train.
15. The difference of the perimeters of a square and a circle
is 5.752 feet and the circle contains 81.86 square feet more
than the square. Find the radius of the circle and the side
of the square.
QUADRATIC EQUATIONS
157
16. In an isosceles triangle the product of the base and one
leg is 108, and the difference between the squares of the base
and leg is 52. Find the altitude of the triangle.
17. The perimeter of a rectangle is 46 inches. If its length
be increased 3 inches, its area will be 153 square inches. Find
its dimensions. Is there more than one such rectangle ? Ex-
plain.
18. If the sum of the denominator and numerator of a cer-
tain fraction be divided by their difference, the quotient is 9.
But if the product of the numerator and denominator be di-
vided by their sum, the quotient is 2 with a remainder of 2.
Find the fraction. What principle of proportion is illustrated
in this problem ? If this principle is applied, are simultaneous
equations necessary ?
226. It was noted in §§ 224, 225, that two second degree
equations had four solutions, or pairs of values for x and y, that
a second degree and a first degree equation had two solutions,
that if imaginary roots entered they were always in pairs.
The geometric explanation for this is readily seen if the
equations are plotted.
Ex. 1. Consider the equation x 2 -f- y 2 = 25.
This means that, for any point on the
graph, the square of the abscissa, plus the
square of the ordinate, equals 25.
But the square of the abscissa of any
point, plus the square of the ordinate, equals
the square of the distance of the point from
the origin ; for the distance is the hypotenuse
of a right triangle, whose other two sides are
the abscissa and ordinate.
Then the square of the distance from of
any point on the graph is 25 ; or, the distance
from of any point on the graph is 5.
Thus, the graph is a circle of radius 5, having its center at 0.
(The graph of any equation of the form x 2 -f y 2 = a is a circle.)
Y
B
s
H
/
\
r
/
•
\
X
/
/
/
1
/
X
A
I
j
\
/
\
/
V
*•
^
y
a
158
ALGEBRA
Ex. 2. Consider the equation y 2 = 4 x + 4
Ha;-*), */ 2 = 4, or y =±2.. (^, i?)
If x = 1, y 2 = 8, or y = ± 2 V2. (C, Z>)
If x=- 1, y = 0, Etc. (E)
The graph extends ^definitely to the right of
77'.
If x is negative and < — 1, y 2 is negative, and
therefore y imaginary; then, no part of the
graph lies to the left of E.
(The graph of Ex. 2 is a parabola ; as also is
the graph of any equation of the form y' 2 = ax or y 2 = ax + b.)
Ex. 3. Consider the equation x 2 -+- 4 y 2 = 4.
In this case it is convenient to first
locate the points where the graph inter-
sects the axes.
If y = 0, x 2 = 4, or x = ± 2. (.4, 4')
If x = 0, 4 y 2 = 4, or y = ± 1. (J3, i?')
Putting x — ± 1 , 4 ?/ 2 = 3, ?/ 2 = J, or
V3
(C, A C, 2>')
|
Y
d!
B
X
C
X
A'
?
v
Wf
b'
D
,
y'
i
If x has any value > 2, or < — 2, y 2 is
negative, and y imaginary ; then, no part of the graph lies to the right of
A, or left of A'.
If y has any value > 1, or < — 1, x 2 is negative, and x imaginary ; then,
no part of the graph lies above B, or below B 1 -
(The graph of Ex. 3 is an ellipse ; as also is the graph of any equation
of the form ax 2 + by 2 — c. )
Ex. 4. Consider the equation x 2 — 2 y 2 -
1.
Y
3
B
A
1
X
A
V
s
1
s
1
/
s
p
V
**
s
V
Y,
Here x 2 - 1 = 2 y 2 , or y 2 = -
1
If ac = ± 1, 2/ 2 = 0, ory =0. (A'. A)
If x has any value between 1 and
— 1, y 2 is negative, and // imaginary.
Then, no part of the graph l&Ofl bt-
tween ^4 and A'.
If »: =±2, **9t|t or y=±y/l.
(2?, C, 5', C)
QUADRATIC EQUATIONS
159
The graph has two branches BAC and B'A'C, each of which extends
to an indefinitely great distance from O.
(The graph of Ex. 4 is a hyperbola; as also is the graph of any
equation of the form ax 2 — by 2 = c, or xy = a.)
Ex. 1. Consider the equations
227. Graphical Representation of Solutions of Simultaneous
Quadratic Equations.
\y 2 = ±x,
[3x-y = 5.
The grapjh of y 2 = 4 x is the parabola AOB.
The graph of 3 x — y = 5 is the straight line AB,
intersecting the parabola at the points A and B,
respectively.
To find the coordinates of A and J5, we proceed
as in § 48 ; that is, we solve the given equations.
The solution is x = 1, y = — 2 ; or, x = -^ 5 ,
y = ^ (§224, IV).
It may be verified in the figure that these are
the coordinates of A and 2?, respectively.
Hence, if any two graphs intersect, the coordinates of any point
of intersection form a solution of the set of equations represented
by the graphs.
— Y- j r*C-
S i±j ~3.
III IstlUl^IZ"
:z: = Ei^s;::E:::
— -Y'-t ^s:-
Ex. 2. Consider the equations
x 2 + y 2 = 17,
xy = 4:.
The graph of x 2 -f y 2 = 17 is the circle
AD, whose centre is at O, and radius Vl7.
The graph of xy = 4 is a hyperbola,
having its branches in the angles XOY
and J'07', respectively, and intersecting
the circle at the points A and B in angle
XOY, and at the points C and D in
angle X'OY'.
The solution of the given equation is
X = 4, ?/ = 1 ; se = 1,^ = 4;
x — — 1, y =— 4 ; and # =— 4, y = — 1.
It may be verified in the figure that these are the coordinates of A, 2?,
C, and D, respectively.
V
\! 1
* B
,J
t t
r
*N
/
\
\
'
L
S k
1
\
J
1
I
\
\
«*.
X
I
X
c
L
1
J
y
f
•
/
'
X
'
- C1P
;
1 1
v
!
160
ALGEBRA
2.
x 2 + 4 y 2 = 4.
a — y = l.
4-
x 2 -4y = -7.
2x + 3y = ±.
5-
9x 2 + y 2 = U$.
xy = -8.
6.
Y
'"" R
<
^
X'
5B """
ziss
Y'N
EXERCISE 52
Find the graphs of the following sets of equations, and in
each case verify the principle of § 227 :
'x 2 + y 2 = 29.
xy = 10.
2x 2 + 5y 2 = 53.
3 x 2 - 4 f = - 24.
x 2 + y 2 = 13.
4:X — 9y = 6.
228. i£c. 1. Consider the equations
/a 2 + 4 ^ = 4, (1)
>2x + 3y = -5. (2)
The graph of x 2 -f 4 1/' 2 = 4 is the ellipse
The graph of 2x + 3?/=— 5 is the
straight line CD.
If 2/ or x is eliminated between these
two equations, we find that the resulting
equation containing one unknown num-
ber is such that if all the terms are transposed to one member, that mem-
ber is a trinomial perfect square. Hence, the equation has equal roots
and the line and curve are tangent at A ( 218, II).
If in Ex. 1, § 228, the second equation had been 2x + 3y = — 10(2),
the roots would have been imaginary and the line would not have met the
ellipse.
REVIEW EXAMPLES
EXERCISE 53
X 2
Scale : \ inch.
Vl-« 2 +
i. Rednce
Vl-ar 2
■ to the form
1-x 2
2. Reduce
i+ j
i +
(x - a) 2
x-\- a
x—a
QUADRATIC EQUATIONS 161
3. Reduce 0* + O 2 - (e x -e~ x ) 2 tQ the form / _2_
(e x + e _x ) 2 \e x + T* to unity.
a; + V# 2 -\-y 2 x + V# 2 + y 2
ax
( a 2 __ /p2\f j[
8. Reduce v y to the form — >
a If a V 1 Va 2 -^ 2
9. 5 = Vr+X 2 . If JT= V2a - ?/ ~^ find s = JS
2/ v y
2
10. Reduce — — to the form — ,
x 3 1
11. Reduce _ — - to the form _
X s — x 4 — 6 25
1-
4ar»
1 - 2 x*\ 2
12. Reduce V2 a# — x 2 to the form
a (x — a) 2
162 ALGEBRA
13. 2 tan x + (tan xf — 3 = ; find tan x.
14. 2 cos x H = 3 ; find cos x.
cos a;
15. — h 2 cot x = - VI + cot 2 # ; find cot x.
cot a? 2
16. Eeduce ( a + + («+0-
Therefore, 2 8 = n(a + T), and S m £ (a + I). (II)
m
232. Substituting in (II) the value of I from (I), we have
£**?[2a+-<»-- l) TK~> —T-> "^ t0 19 terms '
1 T P
I0 - 7Vf> £R> o£> "''i t0 47 terms '
2o 50 25
233. The j#rs£ term, common difference, number of terms, last
term, and sum of the terms are called the elements of the
progression.
If any three of the five elements of an arithmetic progres-
sion are given, the other two may be found by substituting the
known values in the fundamental formulae (I) and (II), and
solving the resulting equations.
i . Given a = — f, w = 20, S = — f ; find d and I.
Substituting the given values in (II),
- | = 10 (- | + I) or - i =- | +1 ; then, I = J - J = |.
Substituting the values of a, w, and I in (I), f =— J + 19 , # = — 136 6 ; find n and d.
9. Given a = .4, I = 34.6, n = 20 ; find cZ and 5.
10. Given S = 18.15, d = . 02, a = .23; find Z and n.
1 1 . Given S = ^— — , d = — -^ , w = 15 ; find a and I.
12. Given n = 26, d = , Z = ~ ; find # and a.
8 8 '
ARITHMETIC PROGRESSION 167
234. From (I) and (II), general formulae for the solution of
examples like the above may be readily derived.
Ex. Given a, d, and S\ derive the formula for n.
By §232, 2^=«[2a + (»-l)d], or dn 2 + (2a-d)n = 2S.
This is a quadratic in w, and may be solved by the method of § 213;
multiplying by 4 d, and adding (2 a — d) 2 to both members,
4 -&* '" to 7 terms.
5-
h h h * • • to 6 terms.
6.
- h -tV> — Hi ••' t0 8 terms.
241. If any three of the five elements of a geometric pro-
gression are given, the other two may be found by substituting
the given values in the fundamental formulae (I) and (II), and
solving the resulting equations.
But in certain cases the operation involves the solution of
an equation of a degree higher than the second ; and in others
the unknown number appears as an exponent, the solution of
which form of equation can usually only be effected by the aid
of logarithms (§ 110).
i. Given a = — 2, n = 5, I = — 32 ; find r and S.
Substituting the given values in (I), we have
— 32 = — 2 r 4 ; whence, r 4 = 16, or r = ± 2.
Substituting in (II),
If r = 2, S = 2(-32)-(-2) = _ 64 2 _ _ 62
2-1
If r= _o ig= (-2)(-32)-(-2) = 64 + 2 = ^ 22
-2-1 -3
The solution is r = 2, S = - 62 ; or, r = - 2, S =- 22.
The interpretation of the two answers is as follows :
If r = 2, the progression is — 2, — 4, — 8, — 16, — 32, whose sum is
-62.
If r =— 2, the progression is — 2, 4, — 8, 16, — 32, whose sum is — 22.
GEOMETRIC PROGRESSION 171
2. Given a = 3, r = - £, # = VsV" ; find w and Z.
Substituting in (II), ^ = -**~ 8 = L±i.
° l ; ' 729 -}-l 4
Whence, 1 + 9 = tfiff - t or, I = V# - 9 =- 7 fc.
Substituting the values of a, r, and I in (I),
- th = H- i)"" 1 ; or, (- J)"- 1 =- nVy.
Whence, by inspection, n — 1=7, orn = 8.
From (I) and (II) general formulae may be derived for the solution of
cases like the above.
If the given elements are n, Z, and S, equations for a and r may be
found, but there are no definite formulas for their values.
The same is the case when the given elements are a, n, and S.
The general formulae for n involve logarithms ; these cases are discussed
in § 110.
EXERCISE 59
i. Given r = 2, n = 12, S = 4095 ; find a and I
2. Given a = 2, r = — 3, Z = 1458 ; find n and S.
3. Given Z = — -g^, a = — if , n = 10 ; find r and S.
4. Given a » f, J a 3584, # = **£** ; find r and rc.
5. Given r = i, w = 5, Z = T ^ ; find a and S.
6. Given ^ = -A||-i^ a = -64, r=i; find w and I
7. Given a, Z, and # ; derive the formula for r.
8. Given r, Z, and n ; derive the formulae for a and S.
9. Given a, n, and Z ; derive the formulae for r and S.
10. Given #, n, and r ; derive the formulae for a and L
242. Sum of a Geometric Progression to Infinity.
The limit (§ 125) to which the sum of the terms of a decreas-
ing geometric progression approaches, when the number of
172 ALGEBRA
terms is indefinitely increased, is called the sum of the series to
infinity.
Formula (II), § 239, may be written
a— rl
1 — r
It is evident that, by sufficiently continuing a decreasing
geometric progression, the absolute value of the last term may
be made less than any assigned number, however small.
Hence, when the number of terms is indefinitely increased,
Z, and therefore rl, approaches the limit 0.
Then, the fraction a ~~ r approaches the limit — ^— •
Therefore, the sum of a decreasing geometric progression to
infinity is given by the formula
Sfatr^z* (HI)
1 — r
Ex. Find the sum of the series 4, — f, - 1 /, ... to infinity.
o
Here a = 4, r =
3
Substituting in (III), S= 4 = — .
V J 1+f 5
To find the value of a repeating decimal.
This is a case of finding the sum of a decreasing geometric
series to infinity, and may be solved by formula (III).
Ex. Findthe value of .85151-...
We have, .85151 ••• = .8 + .051 -f .00051 + ....
The terms after the first constitute a decreasing geometric progression
in which a = .051, and r — .01.
Substituting in (III) , S = ,051 = '— = — = — •
1-.01 .99 990 330
Then the value of the given decimal is r 8 ^ + ^, or |f J.
GEOMETRIC PROGRESSION 173
EXERCISE 60
Find the sum to infinity of the following :
T 9 2 2 . . . 4 _ 8 _ 1 R _ 3 2 .
2. 1, -i,i,-. 5- -.3, .12, -.048, -..
3- fj I? TV* '"' *i ~~ ** & ***•
Find the values of the following:
7- .4777.-.. 8. .8181.-.. 9- .5243243-... io. .207575-...
243. Geometric Means.
We define inserting m geometric means between two numbers,
a and b, as finding a geometric progression of m + 2 terms,
whose first and last terms are a and b.
Ex. Insert 5 geometric means between 2 and ^ff.
We find a geometric progression of 7 terms, in which a = 2, and
I = iff ; substituting n = 7, a = 2, and Z = }|| in (I),
i|8 a* ^ r* ; whence r 6 = 7 %%, and r = ± f
The result is 2, ± f, |, ± if, |f, ± ^, fff.
244. Let x denote the geometric between a and b.
Then. - = -, or x 2 = ab.
a x
Whence, x = ^Jab.
That is, the geometric mean between two 7i'umbers is equal to
the square root of their product.
245. Problems.
i. The sixth term of an arithmetic progression is f, and the
fifteenth term is -L 6 -. Find the first term.
Ry § 230, the sixth term is a + 5 d, and the fifteenth term a -f 14 d .
, , «'". f a+ 5d = J. . (1)
Then, by the conditions, \ ■
la + Ud=\*-. r (2)
Subtracting (1) from (2), 9 d = }j whence, d = \.
Substituting in (1), a + § = |j whence, a = -f.
174 ALGEBRA
2. Find four numbers in arithmetic progression such that
the product of the first and fourth shall be 45, and the product
of the second and third 77.
Let the numbers be x — 3 y, x — y, x + y, and x -f 3 y.
x 2 -9y 2 = 45.
Then by the conditions, I x< * 9 y2 ~ 45,
I x 2 - w 2 = 77.
Solving these equations, x = 9, y — ± 2 ; or, x = — 9, y = ± 2 (§ 224)
Then the numbers are 3, 7, 11, 16 ; or, —3, — 7, — 11, — 15. •
In problems like the above, it is convenient to represent the unknown
numbers by symmetrical expressions.
Thus, if five numbers had been required, we should have represented
them by x — 2 ?/, x — y, x, x + y, and x + 2 y.
3. Find 3 numbers in geometric progression such that their
sum shall be 14, and the sum of their squares 84.
Let the numbers be represented by a, ar, and af 2 .
a + qr + ar 2 = 14. (1)
a 2 + a 2 r 2 +a 2 r i = te. (2)
Divide (2) by (1), a - ar + ar 2 = 6. , (3)
Then, by the conditions,
2
ar
= 8, or r
_4
a
(4)
a
a
= 14
, or a 2 -
-10 a
+ 16
= 0.
a ■■
= 8 or 2.
r
_4
8
or 4 - =
2
lor
2
2.
Subtract (3) from (1),
Substituting in (1),
Solving this equation,
Substituting in (4) ,
Then, the members are 2, 4, and 8.
EXERCISE 61
1. The seventh term of an A. P. is ■££, the twenty-first term
is - 3 ^ $ . Find the fifteenth term.
2. Show that the sum of the odd integers from 1 to 999 is
the square of their number.
3. The first term of an A. P. is 1, the sum of the third and
ninth terms is 32. Find the sum of the first thirteen terms.
GEOMETRIC PROGRESSION 175
4. The sum of the first ten terms of an A. P. is to the sum
of the first seven terms as 29 to 14. Find the ratio of the
common difference to the first term.
5. There are four numbers, such that the first three form a
G. P., the last thr^ee form an A. P. The sum of the first three
is 73, of the last three 192. The difference between the second
and fourth is 112. Find the numbers.
6. How many arithmetic means are inserted between — f
and f , when their sum is 2 ^- ?
7. Find four numbers in A. P., such that the sum of the
first and second shall be — 1, and the product of the second
and fourth 24.
8. A traveller sets out from a certain place, and goes 7
miles the first hour, 1\ the second hour, 8 the third hour, and
so on. After he has been gone 5 hours, another sets out and
travels 16 J miles an hour. How many hours after the first
starts are the travellers together ?
9. If a person saves $ 120 each year, and puts the sum at
simple interest at 3\°/o at the end of each year, to how much
will his property amount at the end of 18 years ?
10. A ball is dropped from a window 32 feet above the
pavement. Assuming the ball to be perfectly elastic and that
on each rebound it rises to within \ of its former height, how
far does it travel before coming to rest ?
11. Two men travel from P to Q, leaving P at the same
time. The distance from P to Q is 63 miles. The first
travels 1 mile the first hour, 2 miles the second hour, 4 miles
the third hour, and so on. The second travels 11 miles the
first hour, 10| miles the second hour, 10 1 miles the third
hour, and so on. Which is first to arrive at Q ?
12. Find the geometric mean between .0729 and .0529.
13. Find the geometric mean between — and
2.2o o76.
176 ALGEBRA
14. Find the geometric mean between — TjW an d ?L.
xy-y 2 xy
15. Find the geometric mean between a 2 — 4 a + 4 and
4 a 2 + 4 a + 1.
16. The product of the first five terms of a G. P. is 243.
Find the third term.
17. The digits of a number of three figures are in geometric
progression. If units' and tens' digits are interchanged, the
number formed exceeds the original number by 36. The sum
of the digits is 14. Find the number.
18. A man travels 445*- miles. He travels 10 miles the first
day, and increases his speed one-half mile in each succeeding
day. How many days does the journey require ?
19. An A. P. has 19 terms such that the sum of the three
middle terms is 3, and the sum of the first term and the last
two terms is — 13. Find the series.
20. Find the number of arithmetic means between 1 and 69,
such that the ratio of the last mean to the first mean is 13.
21. Find an A. P. of 17 terms such that the sum of the first
three terms is to the last term as 3 to 13, the first term being
unity.
22. The sum of three successive terms of a geometric pro-
gression is 39 and the sum of their squares is 819. Find the
series.
23. The sum of how many terms of the series 1, 3, 9 •••, is
3280 ?
24. Show that in any G. P., if each term is subtracted from
the succeeding term, the differences form a G. P.
25. Find three numbers in A. P., such that the square of the
first added to the product of the other two gives 16, and the
square of the second added to the product of the other two
gives 14.
PAET II
X. INFINITE SERIES
246. Infinite Series (§ 178) may be developed by Division,
or by Evolution.
Let it be required, for example, to divide 1 by 1 — x.
1 — se)l(l + x + x 2 + ■
1-x
X
x-x 2
. = l+x + x 2 + x? + -•.
Then,
Again, let it be required to find the square root of 1 + #•
(1)
1+x
1 + ?_£l + ...
2 8
1
2 + ^
2
X
X -
r 2
2 + x-
x*\
8|
X 2
4
Then, vTTx = l +£-!-+••
(2)
It should be observed that the series, in (1) and (2), do not give the
values of the first members for every value of x ; thus, if x is a very
large number, they evidently do not do so.
EXERCISE 62
Expand each of the following to four terms :
T 5-f-4a l+3#
2.
l + 3aj
1 + 3 a + x 2 *
3-
4.
5x
5 — X — 3X 2
179
180 ALGEBRA
5. Vl + 3a. 8. Va 3 + b s .
6. VI -5x 2 . 9- V^ + l.
7. Va 2 + 6 2 . io. V9a 2 -16 6 2 .
CONVERGENOY AND DIVERGENCY OP SERIES
247. An infinite series is said to be Convergent when the
sum of the first n terms approaches a fixed finite number as a
limit (§ 125), when n is indefinitely increased.
An infinite series is said to be Divergent when the sum of
the first n terms can be made numerically greater than any
assigned number, however great, by taking n sufficiently great.
248. Consider, for example, the infinite series
1 -{- x + x 2 -{- X s + •••.
I. Suppose x — a?!, where x x is numerically < 1.
The sum of the first n terms is now
1 + Xl + Xl >+ ... + Xl »-i = l^£L n (§ 103, VII).
1 — Xi
If n be indefinitely increased, x" decreases indefinitely in
absolute value, and approaches the limit 0.
Then the fraction approaches the limit .
1 — x x 1 — x 1
That is, the sum of the first n terms approaches a fixed finite
number as a limit, when n is indefinitely increased.
Hence, the series is convergent when x is numerically < 1.
II. Suppose 05 = 1.
In this case, each term of the series is equal to 1, and the
sum of the first n terms is equal to n ; and this sum can be
made to exceed any assigned number, however great, by taking
n sufficiently great.
Hence, the series is divergent when x=l.
INFINITE SERIES 181
III. Suppose x m — 1.
In this case, the series takes the form 1 — 1 + 1 — 1 + ...,
and the sum of the first n terms is either 1 or according as n
is odd or even.
Hence, the series is neither convergent nor divergent when
08-1.
An infinite series which is neither convergent nor divergent
is called an Oscillating Series.
IV. Suppose x = x ly where x is numerically > 1.
The sum of the first n terms is now
1 + Xl + Xl 2 + ... + Xi n-l = ^LJZI (§ 103 , VII).
x x — 1
x n — 1
By taking n sufficiently great, — can be made to numer-
al — 1
ically exceed any assigned number, however great.
Hence, the series is divergent when x is numerically > 1.
249. Consider the infinite series
l+x + x 2 + x s -\ ,
developed by the fraction ■ (§ 246).
1 — x
Let x= .1, in which case the series is convergent (§ 248).
The series now takes the form 1 + .1 +.01 4- .001 -f- ••• , while
the value of the fraction is — or — .
.9 9
In this case, however great the number of terms taken, their
sum will never exactly equal y*-.
But the sum approaches this value as a limit; for the series
is a decreasing geometric progression, whose first term is 1, and
ratio .1 ; and, by § 242, its sum to infinity is , or — .
Thus, if an infinite series is convergent, the greater the num-
ber of terms taken, the more nearly does their sum approach
182 ALGEBRA
the value of the expression from which the series was
developed.
Again, let x =.10, in which case the series is divergent.
The series now takes the form' 1 +10 + 100 + 1000 H ,
while the value of the fraction is — , or — -.
In this case the greater the number of terms taken, the
more does their sum diverge from the value — J.
Thus, if an infinite series is divergent, the greater the num-
ber of terms taken, the more does their sum diverge from the
value of the expression from which the series was developed.
It follows from the above that an infinite series cannot be
used for the purposes of demonstration if it is divergent.
SUMMATION OF SERIES
250. The Summation of an infinite literal series is the pro-
cess of finding an expression from which the series may be
developed.
RECURRING SERIES
251. Consider the infinite series
l + 2oj + 3a 2 + 4ar 3 + 5a 4 +....
Here (3 x 2 ) - 2 x(2 x) + af (1) = 0,
(4 X s ) - 2 x(S x 2 ) + x\2 x) = 0, etc.
That is, any three consecutive terms', as, for example, 2 x, 3 x 2 ,
and 4: X s , are so related that the third, minus 2x times the
second, plus x 2 times the first, equals 0.
252. A Recurring Series is an infinite series of the form
Oo + a^ + o^H — ,
where any r + 1 consecutive terms, as for example
a n x , a u _iX , a n _ 2 x , •••, a n _ r x ,
INFINITE SERIES 188
are so related that
a n x n + px (a^x"-' 1 ) + qx 2 (a n _^x n - 2 ) + ... + sx r (a n _ r x n ~ r ) = ;
p, q } • ••, s being constants.
The above recurring series is said to be of the rth order, and
the expression
1 + px -{-px 2 -f • • • + sx r
is called its scale of relation.
The recurring series of § 251 is of the second order, and its
scale of relation is 1 — 2 x + x 2 .
An infinite geometric series is a recurring series of the first order.
Thus, in the infinite geometric series
1 + x+x 2 +x* + .--,
any two consecutive terms, as for example x % and ic 2 , are so related that
(x 3 ) — x(x' 2 ) = ; and the scale of relation is 1 — x.
253. To find the scale of relation of a recurring series.
If the series is cf the first order, the scale of relation may be
found by dividing any term by the preceding term, and sub-
tracting the result from 1.
If it is of the second order, a , a 19 a 2 , a 3 , •••, its consecutive
coefficients, and 1+px + qx 2 its scale of relation, we shall have
'« 3 + qa 2 + ra^ = 0,
a 5 +pa 4 + qa 3 + ra 2 = ;
from which p, q, and r may be determined.
To ascertain the order of a series, we may first make trial of
a scale of relation of three terms.
184 ALGEBRA
If the result does not agree with the series, try a scale of
four terms, five terms, and so on until the correct scale of rela-
tion is found.
If the series is assumed to be of too high an order, the equa-
tions corresponding to the assumed scale will not be inde-
pendent (§ 43).
254. To find the sum (§ 250) of a recurring series when its
scale of relation is known.
Let 1 + px-\-qx 2 be the scale of relation of the series
a + a x x + a 2 x 2 -f- •••.
Denoting the sum of the first n terms by & ni we have
S n = a + ctix 4 a 2 x 2 + \- a n -ix n -\
Then, pxS n = pa^x + pa\x 2 + • • • + pa n -2X n ~ l + pa n -\X n ,
and qx 2 S n = qa x 2 + ••• -h qan-sx 11 - 1 -f qa n -2X n + qa n -\X n+l .
Adding these equations, and remembering that, by virtue of the scale
of relation,
«2 +pai 4- qa = 0, • •, a n -i +pa n - 2 + qa n -z = 0,
the coefficients of x 2 , x s , • ••, a: n_1 become 0, and we have
8 n 0. +px + qx 2 ) = a + («i +pa )x 4- (pa n -i 4- qa n -2)x n 4- ^n-i« n+1 .
Whence,
• g _ q 4-(«i 4- P«o)a +(pqn-i + ga n -2)x n + ga n -iz w+1 . ^n
1 + px 4- ## 2
which is a formula for the sum of the first n terms of a recurring series of
the second order.
If x is so taken that the given series is convergent, x n and x n + l approach
the limit 0, when n is indefinitely increased, and the fraction (1) ap-
proaches the limit
1 + px 4* qx 2
If this fraction be expanded into an infinite series by division, we
obtain the given series ; but it is only when the series is convergent that
it expresses the value of the fraction.
INFINITE SERIES 185
Then, the sum of the given series (§ 250) is given by the formula
# - ftp +Oi 4- p«n)a; i / 2 )
1 + px -f qx l
If q = 0, the series is of the first order, and «i +p«o = ; then
1 + px
which is a formula for the sum of a recurring series of the first order.
(Compare § 242.)
In like manner, we shall find the formula
8 _ tto 4- (fli 4- pao)x 4- (ff2 + ff«i 4- gao)x 2 ^ 4 n
1 +px 4- gx 2 4- rx 3
for the sum of a recurring series of the third order.
255. A recurring series is formed by the expansion in an
infinite series of a fraction, called the generating fraction.
The operation of summation reproduces the fraction, the
process being the reverse of that of § 26S.
256. Ex. Find the sum of the series
2 + x + 5x 2 + 7a? + 17x 4 + ....
To determine the scale of relation, we first assume the series to be of
the second order (§ 253).
Substituting a - 2, a\ — 1, a 2 = 5, a s — 7, in (1), § 253,
J 5 + p+2q=0,
1 7 + r op 4- q = Q.
Solving these equations, p— — 1, tf — -2.
To ascertain if 1 — x — 2 x 2 is the correct scale of relation, consider the
fifth term.
Since 17 x 4 + (- x) (7 x 3 ) + (- 2 x 2 ) (5 x 2 ) is equal to 0, it follows that
1 — x — 2 x 2 is the correct scale.
Substituting the values of a , «i, i>, and q in (2),
j y_ 24(l-2)g _ 2-x
l-x-2x 2 l-x-2x 2 '
The result may be verified by expansion.
186 ALGEBRA
EXERCISE 63
Find the sum of the following:
l 4-r-# + 7# 2 — 5ar 3 +19^H .
2 . l-13x-23x 2 -85x i -239x 4 + ....
3 . l+5x + 21x 2 + 85x 3 + Mlx i + ....
4 . 5-13a + 35a 2 -97ar 3 -h275a 4 + ....
5 . 3 + 10# + 36^ + 136ar J + 528a 4 + ....
6. 3 + x + 33x 2 + 109x* + 657x i + ....
7 . 14. 2 #- 3 a; 2 + 6^r 3 -7 a; 4 + 10 a; 5 -11 a 6 + ....
8. l-2x-a 2 -7ar 3 -18# 4 -59;r 5 -181a 6 + ....
THE DIFFERENTIAL METHOD
257. If the first term of a series be subtracted from the
second, the second from the third, and so on, the series formed
is called the first order of differences of the given series.
The first order of differences of this new series is called
the second order of differences of the given series ; and so on.
Thus, in the series
1, 8, 27, 64, 125, 216, ..,
the successive orders of differences are as follows :
1st order, 7, 19, 37, 61, . 91, ....
2d order, 12, 18, 24, 30, ....
3d order, 6, 6, 6, ....
4th order, 0, 0, ....
The Differential Method is a method for finding any term, or
the sum of any number of terms of a series, by means of its
successive orders of differences.
258. To find any term of the series
a U a 2) a 3) a 4) '") a »l a n+l> '"•
The successive orders of differences are as follows:
1st order, a 2 — «i, 03 — «2, «4 — «8i •••» «n+i — «n, ••••
2d order, a 3 — 2a» + <*h a * — 2 a * + a ^ '"-
3d order, a 4 — 3 a 3 + 3 a 2 — «i> ••• J etc.
INFINITE SERIES 187
Denoting the first terms of the 1st, 2d, 3d, •••, orders of dif-
ferences by d ly d 2 , d 3 , •••, respectively, we have
di = a 2 — a\ ; whence, a 2 = «i + d\.
d 2 = as — 2 a 2 + «i ; whence,
as = — «i + 2 ^ 2 + d 2 = — a\ + 2 « x + 2 c?i + d 2 = cti + 2 dfi + (^.
e? 3 = « 4 — 3 a 3 + 3 a 2 — «i ; whence,
«4 = «i — 3 a 2 + 3 a 3 + (? 3 = «i -f- 3 di + 3 d 2 + d 3 ; etc.
It will be observed, in the values of a 2 , a 3 , and a 4 , that the
coefficients of the terms are the same as the coefficients of the
terms in the expansion by the Binomial Theorem of a -J- a; to
the first, second, and third powers, respectively.
We will now prove by Mathematical Induction that this law
holds for any term of the given series.
Assume the law to hold for the ?ith term, a n ; then the coef-
ficients of the terms will be the same as the coefficients of the
terms in the expansion by the Binomial Theorem of a + x to
the (n — l)th power ; that is,
a n =a 1 +(n-l)d 1 + ^^^^ld 2
+ (n-l)(» -2)^-3) ^^ (1)
lit
If the law holds for the nth term of any series, it must also
hold for the nth term of the first order of differences ; or,
a n+1 -a n = d l + (n-l)d 2 + ( n - 1 ^ n - 2 ^ d 3 + .... (2)
Adding (1) and (2), we have
a n+1 = a 1 + [(n-l)+iyi 1 + 1 ^±[(n-2) + 2]d 2
+ »- 1 ^ 2 ) [(n-3) + 3]d < +...
= fll + ndl + !l^ ^ (3)
If. l£
This result is in accordance with the above law.
188 ALGEBRA
Hence, if the law holds for the nth. term of the given series,
it holds for the (n + l)th term ; but we know that it holds for
the fourth term, and hence it holds for the fifth term; and
so on.
Therefore, (1) holds for any term of the given series.
If the differences finally become zero, the value of a n can be obtained
exactly.
259. To find the sum of the first n terms of the series
Ctiy 0,9, &3, & 4 , & 5 , ••*. {±J
Let S denote the sum of the first n terms.
Then S is the (n + l)th term of the series
0, ft, ai + «2, cii + a 2 + as, •••• (2)
The first order of differences of (2) is the same as (1) ; whence, the
2d order of differences of (2) is the same as the 1st order of differences
of (I), the 3d order of (2) is the same as the 2d order of (1), and so on.
Then, if di, d 2 , •••, represent the first terms of the 1st, 2d, • •-, orders
of differences of (1), cti, di, d 2 , ..., will be the first terms of the 1st, 2d,
3d, •••, orders of differences of (2).
Putting a\ — 0, d\ = «i, d 2 — c?i, etc., in (3), § 258,
< y = nfl 1 + < w t " 1 ^ i + < w " 1 ^ w " 2 > d a + -. (3)
\1 . 12.
260. Ex. Find the twelfth term, and the sum of the first
twelve terms, of the series 1, 8, 27, 64, 125, •••.
Here, n = 12, a\ xs 1.
Also, di = 7, d 2 = 12, -0 = C084.
1.2 1.2-3 1.2-3.4
INFINITE SERIES 189
261. Piles of Shot.
Ex. If shot be piled in the shape of a pyramid with a tri-
angular base, each side of which exhibits 9 shot, find the num-
ber in the pile.
The number of shot in the first five courses are 1, 3, 6, 10, and 15,
respectively ; we have then to find the sum of the first nine terms of the
series 1, 3, 6, 10, 15, • •• ; the successive orders of differences are as
follows :
1st order, . 2, 3, 4, 5, ....
2d order, . 1, 1, 1, ....
3d order, . 0, 0, ....
Putting n = 9, d! = 1, di = 2, d 2 = 1 in (3), § 259,
S = 9 + — • 2 + 9 ' 8 ' 7 • 1 = 165 .
1-2 1-2-3
. EXERCISE 64
i. Find the first term of the sixth order of differences of
the series 3, 5, 11, 27, 67, 159, 375, • ••.
2. Find the 15th term, and the sum of the first 15 terms, of
the series 1, 9, 21, 37, 57, •••.
3. Find the 14th term, and the sum of the first 14 terms, of
the series 5, 14, 15, 8, —7, •••.
4. Find the sum of the first n multiples of 3.
5. If shot be piled in the shape of a pyramid with a square
base, each side of which exhibits 25 shot, find the number in
the pile. <
6. Find the 13th term, and the sum of the first 13 terms, of
the series 1, 3, 9, 25, 57, 111, ....
7. Find the 10th term, and the sum of the first 10 terms, of
the series 4, - 2, 10, 4, - 56, - 206, ....
8. Find the sum of the squares of the first n multiples of 2.
9. Find the 71th term, and the sum of the first n terms, of
the series 1, — 3, — 13, — 17, —3, 41, ....
190 ALGEBRA
10. Find the number of shot in a pile of 9 courses, with a
rectangular base, if the number of shot in the longest side of
the base is 24.
ii. Find the number of shot in a truncated pile of 8
courses, with a rectangular base, if the number of shot in
the length and breadth of the base are 20 and 14, respec-
tively.
12. Find the 12th term, and the sum of the first 12 terms, of
the series 1, 13, 49, 139, 333, 701, 1333, ....
13. Find the 9th term, and £he sum of the first 9 terms, of
the series 20, 4, -36, -132, -356, -820, -1676, ....
14. Find the sum of the fourth powers of the first n natural
numbers.
15. Find the number of shot in a pile with a rectangular
base, if the number of shot in the length and breadth of the
base are m and n, respectively.
16. How many shot are contained in a truncated pile of n
courses, whose bases are triangles, if the number of shot in
each side of the upper base is m ?
INTERPOLATION
262. Interpolation is the process of introducing between the
terms of a series other terms conforming to the law of the
series.
Its usual application is in finding intermediate numbers be-
tween those given in Mathematical Tables.
The operation is effected by giving fractional values to n in
(1), § 258.
The method of Interpolation rests on the assumption that a
formula which has been proved for an integral value of n,
holds also when n is fractional.
INFINITE SERIES 191
263. Ex. Given V5 = 2.2361, V0 = 2.4495, V7 = 2.6458,
V8 = 2.8284, • • • ; find Vo\3.
In this case the successive orders of differences are :
.2134, .1963, .1826, ...
-.0171, -.0137, ....
.0034, ....
Whence, d x = .2134, d 2 =-.0171, d 3 = .0034, ....
Now, the required term is distant 1.3 intervals from V5.
Substituting n = 2.3 in (1), § 258, we have, approximately,
V6\3 = 2.2361 + 1.3 x .2134 + 1 ' 3 x f ( - .0171)
1.3 x .3x-.7 >0034
1x2x3
= 2.2361 + .2774 - .0033 - .0002 = 2.5100.
EXERCISE 65
i. Given log 26 = 1.4150, log 27 = 1.4314, log 28 = 1.4472,
log 29 = 1.4624, • . . ; find log 26.7.
2. Given r#$[ = 4.49794, ^92 = 4.51436, ^93 = 4.53066,
^94 = 4.54684, • -.; find y/WS.
3. The reciprocal of 35 is .02857; of 36, .02778; of
37, .02703; of 38, .02632; etc. Find the reciprocal of
36.28,
4. Given log 124 = 2.09342, log 125 = 2.09691, log 126
= 2.10037, log 127 = 2.10380, • • . ; find log 125.36.
5. Given 21 3 = 9261, 22 8 = 10648, 23 3 = 12167, 24 3 = 13824,
and 25 3 = 15625 ; find the cube of 21f
6. Given log 61 = 1.78533, log 62 = 1.79239, log 63 = 1.79934,
log 64 = 1.80618, ... ; find log 63,527,
192 ALGEBRA
XI. UNDETERMINED COEFFICIENTS
THE THEOREM OF UNDETERMINED COEFFICIENTS
264. An important method for expanding expressions into
series is based on tLe following theorem :
If the series A + Bx + Cx 2 + Dx* + ••• is always equal
to the series A' + B'x + Cx 2 + D'x 3 + •••, when x has any
value which makes both series convergent, the coef-
ficients of like powers of x in the series will be equal ;
that is, A = A', B = B<, C = C.
265. Before giving the proof of the Theorem of Undeter-
mined Coefficients, we will prove two theorems in regard to
infinite series.
First, if the infinite series
a + bx + cx 2 + dx 2 -f •••
is convergent for some finite value of x, it is finite for this
value of x (§ 247), and therefore finite when x = 0.
Hence, the series is convergent when x = 0.
Second, if the infinite series
ax + bx 2 + ex 5 + •••
is convergent for some finite value of x, it equals when x = 0.
For, ax + bx 2 -f ex 3 + ••• is finite for this value of x, and
hence a + bx + ex 2 + • • • is finite for this value of x.
Then, a + bx + cx 2 -\- ..« is finite when # = 0; and therefore
»(a + bx + cx 2 + •••), or ax + bx 2 + cx 3 + •••, equals when
x = 0.
266. Proof of the Theorem of Undetermined Coefficients.
The equation
A + Bx+Cx 2 +Dx*+ ... =A' + B'x+C'x°- + D'x»+ ... (1)
is satisfied when x has any value which makes both members
convergent; and since both members are convergent when
x = (§ 265), the equation is satisfied when x = 0.
UNDETERMINED COEFFICIENTS 193
Putting x = 0, we have, by § 265,
Bx+Cx* + Dx* + .-• =0, and B'x + Ox 2 + DW + ••• =0.
Whence, A = A'.
Subtracting A from the first member of (1), and its equal A'
from the second member, we have
Bx+Cx* + Dx* + ...=B'x'+C'x 2 + D'x' + ....
Dividing each term by x,
B+Cx + Dx 2 + ... =B' + Cx + D'x*+ .... (2)
The members of this equation are finite for the same values
of x as the given series (§ 265).
Then, they are convergent, and therefore equal, for the same
values of x as the given series.
Then the equation (2) is satisfied when x = 0.
Putting x = 0, we have B = B'.
Proceeding in this way, we may prove C= (7, etc.
267. The theorem of § 264 holds when either or both of the
given series are finite.
EXPANSION OP FRACTIONS
2 3 x 2 — x 3
268. i. Expand in ascending powers of x.
1 — 2 x -f 3 x 1
Assume 2 ~ 8 ^ ~ ^ = A + Bx + Cx 2 + B& + Ex* + — , (1)
1 — 2 x + 3 x 2
where ^4, J5, 0, Z>, E, •••, are numbers independent of x.
Clearing of fractions, and collecting the terms in the second member
involving like powers of x, we have
* 4 +-. (2)
2-Sx 2 -x 3 = A+ B\x+ C
-2A\ -2B
*: 2 + D
-2C
+ 3B
x 3 + E
-2D
4-3(7
A vertical line, called a bar, is often used in place of parentheses.
Thus, " + B I x is equivalent to (B — 2 A)x.
-2,4
194 ALGEBRA
The second member of (1) must express the value of the fraction for
every value of x which makes the series convergent (§ 249) ; and there-
fore equation (2) is satisfied when x has any value which makes the
second member convergent.
Then, by § 267, the coefficients of like powers of x in (2) must be
equal ; that is,
A= 2.
B-2A = 0;0r, £ = 2.4 = 4.
C-2B + 3A = -S; or, C = 2 B- 3 A - 3 = .- 1.
Z>_2 C+3.B=- 1; or, Z) = 2 C- 3 2? - 1 = - 15.
E-2D+3C= 0;or, E=2D-3C =-27; etc.
Substituting these values in (1), we have
2 _ 3 x 2 - x
1 — 2 x + 3 x 2
= 2+4x-x 2 -\5xS-27x* .
The result may be verified by division.
The series expresses the value of the fraction only for such values of x
as make it convergent (§ 249).
If the numerator and denominator contain only even powers
of x, the operation may be abridged by assuming a series con-
taining only the even powers of x.
2 -I- 4 a? 2 x 4
Thus, if the fraction were — — — ■ , we should assume
it equal to A + B x 2 + C x A + D x G + E ar s -f — .
In like manner, if the numerator contains only odd powers
of 05, and the denominator only even powers, we should assume
a series containing only the odd powers of x.
If every term of the numerator contains x, we may assume a
series commencing with the lowest power of x in the numerator.
If every term of the denominator contains x, we determine
by actual division what power of x will occur in the first term
of the expansion, and then assume the fraction equal to a series
commencing with this power of x, the exponents of x in the
succeeding terms increasing by unity as before.
2. Expand — in ascending powers of x.
3 x -— xr
undeterminp:d coefficients 195
Dividing 1 by 3 ic 2 , the quotient is — - ; we then assume,
o
= Ax-' 2 + Bx~* + C + Dx + Ex 2 + — . (3)
3 x 2 - x 3
Clearing of fractions.
l=3^ + 3#|x + 3C|x 2 + 3.D|z 3 + 3.E'
_ A - i? - C - i>
z 4 + «
Equating coefficients of like powers of x,
SA = \,3B-A = 1 3C-B = 0,3D-C = 0,3E-D = 0', etc.
Whence, A = \, B=l, = 1, D = ±, **^.~.
Substituting in (8), _£_ -^f| + i + Jf^*-
In Ex. 1, E = 2 D — 3 C ; that is, the coefficient of se 4 equals twice the
coefficient of the preceding term, minus three times the coefficient of the
next but one preceding.
It is evident that this law holds for the succeeding terms ; thus, the
coefficient of x 5 is 2 x (- 27) - 3 x (- 15), or - 9.
After the law of coefficients has been found in any expansion, the terms
may be found more easily than by long division ; and for this reason the
method of § 268 is to be preferred when a large number of terms is
required.
The law for Ex. 2 is that each coefficient is one-third the preceding.
EXERCISE 66
Expand each of the following to five terms in ascending
powers of x :
4 + 2a 3 + x + x 2 1 + 2 a; + 4 .r 2
' l-Vx + 3-x*'
5 4- .6 x 2
2-3x 2 + x 3 '
1 _ 6 tf + 4 g 8
z+Jf^2&
1-23
l-8aj
l + 5x
2 + x 2
1-2 x 2
5 x
5+2^-7^
4 + X-3X 2
1-2Z 2
x _ 5 x * 4. 8 x 4
2 _ 3 .T + z 2
4 x 2 - 3 .t 3
10.
7. ~ — : » 11.
4 8 12
1-r-Gar 9 ' ' 6-4z-5ar 3 ' ' 2r J + 4ar 5 + a 6
196
ALGEBRA
EXPANSION OF SURDS
269. Ex. Expand Vl — x in ascending powers of x.
Assume Vl - x = A + Bx + Cx' 2 -f Dx* + E& + • ••.
Squaring both members, we have, by § 167,
(1)
l-x = A*
+ 2AB
x+ B 2 \x 2
+ 2AC\ + 2AD
x*+ C 2
+ 2AE
+ 2BD
\ + 2BC
Equating coefficients of like powers of x,
A 2 = 1; or, A=l.
2AB = -1; or, B=- _I=-i.
2A 2
x* + -
0; or, C=-
& + 2AC
2AD+2BC
C 2 + 2AE + 2BD = ; or, E
2A
0; or, B=-^ =
A
1
"8*
t
16*
C 2 + 2BD_
2A
128'
etc.
Substituting these values in (1), we have
x _ x? _ x*_ _ Sac 4
VI - x = 1 - 2 g 16 128 "
The result may be verified by Evolution.
The series expresses the value of Vl — x only for such values of x as
make it convergent.
EXERCISE 67
Expand each of the following to five terms in ascending
powers of x:
i. Vl + 2 x.
2. Vl - 3 X.
3- Vl-4a + a; 2 . 5. Vl-f6z.
4. Vl -f x — x 2 . 6. Vl — x — 2 a:-.
PARTIAL, FRACTIONS
270. If the denominator of a fraction can be resolved into
factors, each of the first degree in x } and the numerator is of a
UNDETERMINED COEFFICIENTS 197
lower degree than the denominator, the Theorem of Undeter-
mined Coefficients enables us to express the given fraction as
the sum of two or more partial fractions, whose denominators
are factors of the given denominator, and whose numerators
are independent of x.
271. Case I. No factors of the denominator equal.
19 x 4- 1
i. Separate — — — into partial fractions.
1 (3x-l)(5x + 2) '
Assume 19 * +1 = A + B , (1)
where A and B are numbers independent of as.
Clearing of fractions, 19 x + 1 = A(JS x + 2) -f B(?> x — 1).
Or, 19x+l = (5A+SB)x + 2A-B. (2)
The second member of (1) must express the value of the given fraction
for every value of x.
Hence, equation (2) is satisfied by every value of x ; and by § 267, the
coefficients of like powers of x in the two members are equal.
That is, 5i + 35 = 19,
and 2A-B = 1.
Solving these equations, we obtain A = 2 and B = 3.
Substituting in (1), 19 x + 1 = ? l ?
(8 x - 1) (6 x + St; 3 x - 1 5 x + 2
The result may be verified by finding the sum of the partial
fractions.
2. Separate -^ into partial fractions.
2x — x~ — x 3
The factors of 2 a - x 2 - %* are sc, 1 - x, and 2 + x (§ 103, III, VIII).
Assume then, x+A — _ A + _JL_ _( *?_i.
2 x — x 2 — x 3 x 1 — x 2 + x
Clearing of fractions, we have
x + 4 = ^4(1 -x)(2 + x) + Bx(2 + x) + Cx(l - x).
This equation, being satisfied by every value of x, is satisfied when x = 0*
198 ALGEBRA
Putting x = 0, we have 4 = 2 A, or A — 2.
Again, the equation is satisfied when x = 1.
5
Putting x = 1, we have 5 = 3 J5, or i? = --
3
The equation is also satisfied when x.= — 2.
Putting x = — 2, we have 2^—6 C, or C = — J.
The „, * + 4 = ? + _L + ^i_ = ? + 6 1
' 2x — x 2 — x s x 1 — x 2 + x x S(l — x) 3(2 + x)
To find the value of A, in Ex. 2, we give to x stick a value as will
make the coefficients of B and C equal to zero ; and we proceed In a
similar manner. to find the values of B and C.
This method of finding A, B, and C is usually shorter than that used
in Ex. 1.
Case II. All the factors of the denominator equal,
x * - jla? + 26 .
Let it be required to separate — — -~ — into partial
fractions.
Substituting y -f 3 for a;, the fraction becomes
(y + 3) 2 - ll(y +3)+26 = y2-5y + 2 = l 5 | 2
2/ 3 2/ 3 y y' 2 y*
Replacing y by x — 3, the result takes the form
1 5 2
as _ 3 (a _ 3)2 (x - 3) 8
This shows that the given fraction can be expressed as the sum of
three partial fractions, whose numerators are independent of .r, and
whose denominators are the powers of x — 3 beginning with the first and
ending with the third.
Similar considerations hold with respect to any exam pic 1
under Case II ; the number of partial fractions in any case
being the same as the number of equal factors in the denomi-
nator of the given fraction.
6 x 4- 5
Ex. Separate — — into partial fractions.
(3x + 5) 2 l
In accordance with the above principle, we assume the given fraction
UNDETERMINED COEFFICIENTS 199
equal to the sum of two partial fractions, whose denominators are the
powers of 8 x + 5 beginning with the first and ending with the second.
That is, 6 * + 5 ,= A +- B
Ex. Separate — - — ^~ 2 ■ into partial fractions.
(3x + 5) 2 3x + 5 (3x + 5) 2
Clearing of fractions, 6 x + 5 = 4.(3 a; + 5) + B.
= SAx + 5A + B.
Equating coefficients of like powers of x,
3 A = 6,
and 5 A + 5 = 5.
Solving these equations, J. = 2 and i? = — 5.
Whence _6x±6_ = _2 5
(3 x + 5) 2 3 x + 5 (3 x + 5) 2
Case III. Some of the factors of the denominator equal.
a ? - 4 a? -f 3 ..
a?(a? + l) 2
The method in Case III is a combination of the methods of Cases I and
II ; we assume,
x 2 - 4 x + 3 __ A B C
x(x + l) 2 x x + 1 (x + l) 2 '
Clearing of fractions,
x 2 — 4x + 3 = 4(x + l) 2 + J3x(x + jj + 0a .
= (A + B)x 2 + (2A + B+C)x + A.
Equating coefficients of like powers of x,
A + B = l,
2A+ J5+ C = -4,
and A = 3.
Solving these equations, A = 3, J5 = — 2, and (7 = — 8.
Whence, *-** + « = ? _ _^_ _ 8 .
x(x + l) 2 X x+1 (x + 1) 2
The following general rule for Case III will be found convenient :
X
A fraction of the form should be assumed
equal to {x + a)(.x+b)-(.x + my
* .+ * +... + ^ + _JL_ + ...+ _^_+..
x + a x + b x + m (x + m) 2 (x + m) r
200 ALGEBRA
single factors like x + a and x + b having single partial fractions cor-
responding, arranged as in Case I ; and repeated factors like (x + m)
having r partial fractions corresponding, arranged as in Case II.
272. If the degree of the numerator is equal to, or greater
than, that of the denominator, the preceding methods are
inapplicable.
In such a case, we divide the numerator by the denominator
until a remainder is obtained which is of a lower degree than
the denominator.
X s _ g x 2 _ j
Ex. Separate into an integral expression and
partial fractions.
Dividing x 8 — 3 x 2 — 1 by x 2 — x, the quotient is x — 2, and the re-
mainder — 2 x — 1 ; we then have
a*-3a*-l =a; 2 , -2S-1 , (1)
X 2 — X X 2 — X
O y i
We can now separate into partial fractions by the method
x 2 — x
1 3
of Case I ; the result is
x x— 1
Substituting in (1), ^ i = x _2 + --- e
x x — 1
Another way to solve the above example is to combine the methods of
§§ 268 and 271, and assume the given fraction equal to
Ax + B + C + _D^.
x x— 1
273. If the denominator of a fraction can be resolved into
factors partly of the first and partly of the second, or all of the
second degree, in x, and the numerator is of a lower degree
than the denominator, the Theorem of Undetermined Coeffi-
cients enables us to express the given fraction -as the sum of
two or more partial fractions, whose denominators are factors
of the given denominator, and whose numerators are inde-
pendent of x in the case of fractions corresponding to factors
UNDETERMINED COEFFICIENTS 201
of the first degree, and of the form Ax + B in the case of
fractions corresponding to factors of the second degree.
The only exceptions occur when the factors of the denominator are of
the second degree and all equal.
Ex. Separate into partial fractions.
x 3 + l
The factors of the denominator are x + 1 and x 2 — x + 1.
Assume then -1— = -A_ + fx +V . (1)
x* + l x+l x 2 -x+l
Clearing of fractions, 1 = A(x 2 — x + 1) + (Bx + C) (x + 1).
Or, % ^ (4 + B)x 2 + (- A + B + C)x + A + C.
Equating coefficients of like powers of x,
A + B = 0,
-A + B+C = 0,
and .4 + C = 1.
Solving these equations, A = J, I? = — J, and (7 = f.
1 1 x-2
Substituting in (1),
x 3 + l 8(x + l) 3(£ 2 -x+l)
EXERCISE 68
Separate into partial fractions :
% -1 ■ 6 a 8 + 4a 2 + 2a; + 3
a? _ 9 3 + 20 (a; 2 + 1) (a; 2 + x + 1)
15x-2T „ 3a;-7
7-
10x 2 + ^-21 ^-2^-8
2a? + 3 8 6x?-12
a-*-a&-12 a^-5a; 2 -f-4
12 a; + 18 « 2 ~15a? + 3
^ + 3a 2 -18x* ar 9 -3x-28'
43a-31 5a 2 + 16a;-2
30 a 2 -12 a; -306 ^ + 4^-3^-18
202
ALGEBRA
12.
13.
14.
18.
Sx i -^16x z -10x 2 -^2Sx-\-ll
"* 2x* + x-3
59x-5S
12 a; 2 -25 a + 12*
2^-11^4- 19
x*-§x* + 12x-%
2x s -3x-8
(x> + x-2) 2 ' .
X 3 — X
15.
16.
17.
2x* + x 2 + 5x
(x i + 2x + l)(x 2 -x + l)
x* + 2a?-9x 2 + 7x
x 4 — 4:X s + 6x 2 — 4:X + l
12ft 4 + 19ar*-7a:
4a; 4 + 1
2ar>-2
5L-? 19. **-*. 20
x (x 2 + x j r \)+2(x 2 +x+l) # 4 +a; 2 +-l
2x-Z
4tx*—x
REVERSION OP SERIES
274. To revert a given series y = a + bx m +- cx n +- ••• is to
express # as a series proceeding in ascending powers of y.
Ex. Revert the series y = 2x — 3x 2 -\-4:X s — 5x 4 + •••.
Assume x = Ay + By 2 + Cy z + Zty 4 + •••.
Substituting in this the given value of y,
x = A(2 x - 3 x 2 + 4x 3 - 5 z 4 + •••)
+ 5(4 x 2 + 9 £ 4 - 12 x 3 + 16 x* + — )
-f (7(8 x 3 - 36 x 4 + .-•) + 2>(16 x 4 + ■
(1)
)+"
That is, x = 2 ^4x - 3 .4 1 x 2 + 4 .4
+ 45! - 12 B
■ + 8(7
x 3 — 5 ^4
+ 25 5
-36(7
+ 16 2)
sc 4 +
Equating coefficients of like powers of x,
2.4 = 1;
4 4-125 + 8 (7=0;
- 5 .4 + 25 B - 36 (7 + 16 D = ; etc.
Solving, A s J, 5 = f , (7 = Ai *> = f¥s, etc.
Substituting in (1) , x = J y + f y 2 + ^ y 3 + T 3 2 \ y 4 + • • -.
If the even powers of x are wanting in the given series, the
operation may be abridged by assuming x equal to a series
containing only the odd powers of y.
pp:rmutations and combinations 203
EXERCISE 69
Revert each of the following to four terms :
i. 2/ = ^ + 3^ + 5x 3 + 7^+ .... 3< y = x + f + f + tj r ....
2. y = x-2x 2 + 3x 3 -4:X 4 + .... 2 3 4
4 . y = 2x + 5x 2 + 8x 3 + llx i + ....
/y» /V»2 /y»3 /y»4 /y, /yvJ ,y»3 /y»4
5 ' 2/ = 2"4 + 6-8 + "" 6 - 2/ = |I + [3 + [4 + |5 + --
7. 2/=2o;-4^+6a) 5 -8^+.., 8. y = | + J + f + f + ""
XII. PERMUTATIONS AND COMBINATIONS
275. The different orders in which things can be arranged
are called their Permutations.
Thus, the permutations of the letters a, b, c, taken two at a
time, are ab, ac, ba, be, ca, cb\ and their permutations, taken
three at a time, are abc, acb, bac, bca, cab, cba.
276. The Combinations of things are the different collections
which can be formed from them without regard to the order
in which they are placed.
Thus, the combinations of the letters a, b, c, taken two at a
time, are ab, be, ca ; for though ab and ba are different permu-
tations, they form the same combination.
277. To find the number of permutations of n different things
taken tivo at a time.
Consider the n letters, a, b, c, •••.
In making any particular permutation of two letters, the
first letter may be any one of the n ; that is, the first place can
be filled in n different ways.
After the first place has been filled, the second place can be
filled with any one of the remaining n — 1 letters.
Then, the whole number of permutations of the letfers taken
two at a time is n(n — 1).
We will now consider the general case.
204 ALGEBRA
278. To find the number of permutations of n different things
taken r at a time.
Consider the n letters a, b, c, •••.
In making any particular permutation of r letters, the first
letter may be any one of the n.
After the first place has been filled, the second place can be
filled with any one of the remaining n — 1 letters.
After the second place has been filled, the third place can be
filled in n — 2 different ways.
Continuing in this way, the rth place can be filled in
n — (r — 1), or n — r + 1 different ways.
Then, the whole number of permutations of the letters taken
r at a time is given by the formula
n P r = n(n-l)(n-2)-..(n-r + l). (1)
The number of permutations of n different things taken r at a time is
usually denoted J)y the symbol n P r .
279. If all the letters are taken, r = n, and (1) becomes
n P n = n(n-l)(n-2):-3-2-l = \n 1 _ (2)
Hence, the number of permutations of n different things
taken n at a time equals the product of the natural num-
bers from 1 to n inclusive. (See note, § 181.)
280. To find the number of combinations of n different things
taken r at a time.
The number of permutations of n different things taken r at
a time is n(n-l)(n~2) ... (n-r-f 1) (§ 278).
But, by § 279, each combination of r different things may
have \r permutations.
• Hence, the number of combinations of n different things taken
rata time equals the number of permutations divided by [r.
That is, n C r = n(n-l)(n-2)..-(n-r + l) ; Q])
\r
The number of combinations of n different things taken r at a time is
usually denoted by the symbol „O r .
PERMUTATIONS AND COMBINATIONS 205
281 . Multiplying both terms of the fraction (3) by the prod-
uct of the natural numbers from 1 to n — r inclusive, we have
C = n(n-l) ...( n -r + l)-(w-r) -2-l _ b ■
[r X 1 • 2 • • • (n — r) \r_ \ n — r
which is another form of the result.
282. The number of combinations of n different things
taken r at a time equals the number of combinations
taken n — r at a time.
For, for every selection of r things out of n, we leave a selec-
tion of n — r things.
The theorem may also be proved by substituting n — r f or r, in the
result of § 281.
283. Examples.
i. How many changes can be rung with 10 bells, taking 7 at
a time ?
Putting n = 10, r = 7, in (1), § 278,
10P7 = 10- 9. 8- 7- 6. 5. 4 = 604800.
2. How many different combinations can be formed with 16
letters, taking 12 at a time ?
By § 282, the number of combinations of 16 different things, taken 12
at a time, equals the number of combinations of 16 different things, taken
4 at a time.
Putting n = 16, r = 4, in (3), § 280,
16.15.14.13 = 182(X
16 1.2.3-4
3. How many different words, each consisting of 4 consonants
and 2 vowels, can be formed from 8 consonants and 4 vowels ?
The number of combinations of the 8 consonants, taken 4 at^a time, is
8 - 7 '°- 5 ,or70.
1- 2-3-4
206 ALGEBRA
The number of combinations of the 4 vowels, taken 2 at a time, is
i^ 3 ,or6.
1-2'
Any one of the 70 sets of consonants may be associated with any one
of the 6 sets of vowels ; hence, there are in all 70 x 6, or 420 sets, each
containing 4 consonants and 2 vowels.
But each set of 6 letters may have ]_6, or 720 different permutations
(§ 279).
Therefore, the whole number of different words is
420 x 720, or 302400.
EXERCISE 70
i. How many different permutations can be formed with
14 letters, taken 6 at a time ?
2. In how many different orders can the letters in the word
triangle be written, taken all together ?
3. How many combinations can be formed with 15 things,
taken 5 at a time ?
4. A certain play has 5 parts, to be taken by a company of
12 persons. In how many different ways can they be
assigned ?
5. How many combinations can be formed with 17 things,
taken 11 at a time ?
6. How many different numbers, of 6 different figures each,
can be formed from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, if each
number begins with 1, and ends with 9 ?
7. How many even numbers, of 5 different figures each, can
be formed from the digits 4, 5, 6, 7, 8 ?
8. How many different words, of 8 different letters each,
can be formed from the letters in the word ploughed, if the
third letter is o, the fourth w, and the seventh e ?
PERMUTATIONS AND COMBINATIONS 207
9. How many different committees, of 8 persons each, can
be formed from a corporation of 14 persons ? In how many
will any particular individual be found?
10. There are 11 points in a plane, no 3 in the same straight
line. How many different quadrilaterals can be formed, having
4 of the points for vertices ?
11. From a pack of 52 cards, how many different hands of
6 cards each can be dealt ?
12. A and B are in a company of 48 men. If the company
is divided into equal squads of 6, in how many of them will A
and B be in the same squad ?
13. How many different words, each having 5 consonants
and 1 vowel, can be formed from 13 consonants and 4 vowels ?
14. Out of 10 soldiers and 15 sailors, how many different
parties can be formed, each consisting of 3 soldiers and 3
sailors ?
15. A man has 22 friends, of whom 14 are males. In how
many ways can he invite 16 guests from them, so that 10 may
be males ?
16. From 3 sergeants, 8 corporals, and 16 privates, how many
different parties can be formed, each consisting of 1 sergeant,
2 corporals, and 5 privates ?
17. Out of 3 capitals, 6 consonants, and 4 vowels, how many
different words of 6 letters each can be formed, each beginning
with a capital, and having 3 consonants and 2 vowels ?
18. How many different words of 8 letters each can be
formed from 8 letters, if 4 of the letters cannot be separated ?
How many if these 4 can only be in one order ?
19. How many different numbers, of 7 figures each, can be
formed from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, if the first, fourth,
and last digits are odd numbers ?
208 ALGEBRA
284. To find the number of permutations of n things which are
not all different, taken all together.
Let there be n letters, of which p are a's, q are &'s, and r are
c's, the rest being all different.
Let N denote the number of permutations of these letters
taken all together.
Suppose that, in any particular permutation of the n letters,
the p a's were replaced by p new letters, differing from each
other and also from the remaining letters.
Then, by simply altering the order of these p letters among
themselves, without changing the positions of any of the other
letters, we could from the original permutation form [p differ-
ent permutations (§ 279).
If this were done in the case of each of the JV original per-
mutations, the whole number of permutations would be N x \p.
Again, if in any one of the latter the q 6's were replaced by
q new letters, differing from each other and from the remain-
ing letters, then by altering the order of these q letters among
themselves, we could from the original permutation form \q
different permutations ; and if this were done in the case of
each of the N x \p permutations, the whole number of permu-
tations would be N x ]/) x | q.
In like manner, if in each of the latter the r c's were re-
placed by r new letters, differing from each other and from
the remaining letters, and these r letters were permuted
among themselves, the whole number of permutations would be
Nx [px )_7 X ]r.
We now have the' original n letters replaced by n different
letters.
But the number of permutations of n different things taken
n at a time is I n (§ 279).
Therefore, N x \p X I q X I r = I n : or, N= — =~- •
L L_ L 1_ \p\q\r
Any other case can be treated in a similar maimer.
PERMUTATIONS AND COMBINATIONS 209
Ex. How many permutations can be formed from the let-
ters in the word Tennessee, taken all together?
Here there are 4 e's, 2 n's, 2 s's, and 1 1.
Putting in the above formula n = 9, p = 4, q = 2, r = 2, we have
l& _5.6.7.8.9 = 378a
L4(_2|2 2 ' 2
EXERCISE 71
i. In how many different orders can the letters of the word
denomination be written ?•
2. There are 4 white billiard balls exactly alike, and 3 red
balls, also alike; in how many different orders can they be
arranged ?
3. In how many different orders can the letters of the word
independence be written ?
4. How many different signals can be made with 7 flags, of
which 2 are blue, 3 red, and 2 white, if all are hoisted for each
signal ?
5. How many different numbers of 8 digits can be formed
from the digits 4, 4, 3, 3, 3, 2, 2, 1 ?
6. In how many different ways can 2 dimes, 3 quarters, 4
halves, and 5 dollars be distributed among 14 persons, so that
each may receive a coin?
285. To find for what value of r the number of combinations
of n different things taken r at a time is greatest.
By § 280, the mimber of combinations of n different things,
taken r at a time, is
r _ n(tt-l)-(w-r+2)(n-r + l) m
* ' L2.3...(r-l)r ' W
210 \i.(,i:i;i; \
Also, the number of combinations of n different things, taken
r — 1 at I t line, is
»(n \) ■■■ \, t (r hill n(n I) .,
The expression (1) Is obtained by multiplying the expres-
Bion (2) by ■ — , or — i.
The latter expression decreases as r increases.
If. then, we Bud the values of (I) corresponding to the Val-
ues L| 2j • ». ■••• Of /*. the results will OOntinnally increase so
i « /• ! 1 . ' -,
Long as is > l.
/'
I. Suppose a even; and let n »2m, where mn ia a positive
integer.
Then, ™ — iX- beoomes — ■ - —
r r
If, w > 2w '- r + 1 becomes '" j 1 ,and is 1.
2 m — r +1 . i ,", i
If r-m + 1, - ; . becomes r and is - I.
Then, ,0! will bave its greatest value when r»HM
II. Suppose N Oddj and let n ■ % 2m I 1. where m is a posi-
tive integer,
1 hen. becomes " '
r r
if r ». - - - j " beoomes " -j and is >l,
r m
[fr«m+l, - - beoomes - 4~ti and equals 1,
r »< 4- 1
2 m r I 4> tw
If r in I 2, - - beoomes ^, and is <1.
r
DETERMINANTS
211
Then, n C r will have its greatest value when r equals in or
1 n
— or —
+ 1.
m -f- 1 ; that is,
Then, n (7 r will have its greatest value when r equals
n 4- 1
or — ~— ; the results being the same in these two cases.
Li
XIII. DETERMINANTS
286. The solution of the equations
| a Y x + b^y = c lf
\ a& -f- bgj = c 2 ,
i s x — b & — b & y = °2 a l ~ C i a 2 .
afii — ajbi a Y b 2 — a 2 &i
The common denominator may be written in the form
a l9 h
-1
(1)
This is understood as signifying the product of the upper
left-hand and lower right-hand numbers, minus the product of
the lower left-hand and upper right-hand.
The expression (1) is called a Determinant of the Second Order.
The numerators of the above fractions can also be expressed as deter-
minants ; thus,
b-zCi — b\C 2 =
Cl,
bi
, and c 2 a\ -
- Ci& + aj) s i\ — dib^s + ajb x c 2 — ajbfr
with results of similar form for y and z.
212
ALGEBRA
The denominator of (1) may be written in the form
a 2 , b 2 , c 2 . (2)
This is understood as signifying the sum of the products of the
numbers connected by lines
parallel to a line joining the
upper left-hand corner to the
lower right-hand, in the fol-
lowing diagram, minus the
sum of the products of the
numbers connected by lines
parallel to a line joining
the lower left-hand corner
to the upper right-hand.
The expression (2) is called a Determinant of the Third Order.
The numerator of (1) can also be expressed as a determinant, as follows :
ds, b 3 , c 8
as may be verified by expanding it by the above rule.
EXERCISE 72
Evaluate the following:
2.
14 15
9 12
2 x — ?/ 2 x -f- y
2x + y
2x-y
4 3
2 1
8 7
4-
3 4 5
9 12
6 7 8
•
5-
15 12
8 4
6 5
2
4
6.
12 9
2
5 4
DETERMINANTS
213
Show that
6 8 10
7 5 10
11 6 10
8 5 4
12 9 6
.
14 16 7
Show that
7-4 8
3. 6 -9
8-5 1
3
= 6
10
10
10
10
+ 11
10
10
-4 8
7 8
7
-4
3
-5 13
+ 6
8 13
+ 9
8
-5
It is found in geometry that if the vertices of a triangle are
at the points x = 2, ?/ = 3 ; a? = 4, y = 5; x =— 1, y = 4, the
area of the triangle is found to be
2 3 1
, the abscissas (§ 46) forming the first
Area :
4 5 1
-14 1
column of the determinant, the ordinates the second column,
and the third column being l's.
Find the areas of the triangles whose vertices are at the
following points:
io.* x=— 2, y=l\ x=£, y=l; x=2, y=6. Make diagram.
ii. x = — 4, 2/ = 3; a = 4, ?/ = 3 ; x = 2,y = —7. Make
diagram.
12. x = 2, 2/ = 4; x = 8, y = A-, x=— 1, y = — 9. Make
diagram.
13. x = — 5, y = S- y x = 5, y = 3-, x = 0, y=—3. Make
diagram.
14. a=-2, y = 8; a = -2, y=-2; # = 5, 2/=~ 4 -
Make diagram.
* If your area is negative, its absolute value is the area sought. A
change in the order of selecting the vertices will change the sign of the
area, but not the absolute value of the area.
214
ALGEBRA
288. The numbers in the first, second, etc., horizontal lines
of a determinant are said to be in the first, second, etc., rows,
respectively ; and the numbers in the first, second, etc., vertical
columns, in the first, second, etc., columns.
The numbers constituting the determinant are called its
elements, and the products in the expanded form its terms.
Thus, in the determinant (2), of § 287, the elements are a \, 2) a \, 3
a 2, 1? ®>2, 2? ^2, 3
%, 1) ^3, 2? #3, 3
The products of the elements taken three at a time, subject to the
restriction that each product shall contain one and only one element
from each row, and one and only one from each column, the first
suffixes being written in the order 1, 2, 3, are
0\, 1 #2, 2 #3, 3? #1, 1 #2, 3 «3, 2, (L\, 2 0>2, 1 #3, 3? #1, 2 #2, 3 #3, 1? #1, 3 #2, 1 #3, 2>
and «i,3 a 2 , 2 #3, i-
In the first of these there are no inversions in the second suffixes ; in
the second there is one, 3 before 2 ; in the third there is one ; in the
fourth, two ; in the fifth, two ; in the sixth, three.
Then by the rule of § 290, the first, fourth, and fifth products are
positive, and the second, third, and sixth are negative ; and the ex-
panded form is
0\, 1 0,2, 2 053, 3 — &1, 1 0,2, 3 #3, 2 — #1, 2 #2, 1 0%, 3 + Q>\, 2 &2, 3
*2, 1>
^2, 2?
H % n
-% n
x n, 1)
and
a i, 2?
tt », 1
««,2
l 2, nj
Since the second suffixes of the first determinant arc tlic
same as the first suffixes of the second, if the first determinant
be expanded by the rule of § 290, and the second by the rule
of § 293, the results will be the same.
Therefore the determinants are equal.
DETERMINANTS
217
295. A determinant is changed in sign if any two con-
secutive rows, or any two consecutive columns, are
interchanged.
Consider the determinants
a b c
b a
c
d e f
and
e d
f
g h Jc
h g
k
Evaluating each determinant as in Exercise 72, we have
aek + dhc + bfg — gee — bdk — fha
and bdk -f egc + afh — hdc — aek — fgb,
each term of the second determinant being the negative of the correspond-
ing term in the first. The second determinant has therefore the same
absolute value as the first, but of opposite sign. In a manner similar to
that used in § 294, it may be shown that this property holds for determin-
ants of the nth order.
It follows, from §§ 294 and 295, that if two consecutive
rows are interchanged, the sign of the determinant is changed.
296. A determinant is changed in sign if any two
rows, or any two columns, are interchanged.
Consider the m letters a, b, c, •••, e, f, g.
By interchanging a with &, then a with c, and so on in succession with
each of the m — 1 letters to the right of a, a may be brought to the right
of g.
Then, by interchanging g with /, then g with e, and so on in succession
with each of the m — 2 letters to the left of g, g may be brought to the
left of b.
That is, a and g may be interchanged by (m — 1) + (ni — 2), or
2 m — 3, interchanges of consecutive letters ; that is, by an odd number
of interchanges of consecutive letters.
It follows from the above that any two rows, or any two columns, of
a determinant may be interchanged by an odd number of interchanges
of consecutive rows or columns.
But every interchange of two consecutive rows or columns changes
the sign of the determinant (§ 295). **
Therefore the sign of the determinant is changed if any two
rows, or any two columns, are interchanged.
218
ALGEBRA
297. Cyclical interchange of rows or columns.
By n — 1 successive interchanges of two consecutive rows, the
first row of a determinant of the nth order may be made the last.
Thus, by § 295, the determinant
a 2 , b<
is equal to (— l) n
a n ,
The above is called a cyclical interchange of rows.
In like manner, by n — 1 successive interchanges of two con-
secutive columns, the first column of a determinant of the nth
order may be made the last.
298. If two rows, or two columns, of a determinant
are identical, the value of the determinant is zero.
Let D be the value of a determinant having two rows, or two columns,
identical.
If these rows, or columns, are interchanged, the value of the resulting
determinant is - D (§ 296).
But since the rows, or columns, which are interchanged are identical,
the two determinants are of equal value.
Hence, D — — D ; and this equation cannot be satisfied by any value
of D except 0.
Ex. Evaluate the following determinant :
a a d
= abk 4- bed 4- cea — cbd — abk — ace.
= 0.
299. If each element in one column, or in one row, is
the sum of m terms, the determinant can be expressed
as the sum of m determinants.
Consider the determinant
a 2, 1>
a 2, r)
(1)
;
"1, r>
a 2, rl
8*»rj "" a n,
Multiplying each element in the rth column by m, we have
«i,i, •••, mai,n *-i «i,
«2,1,
ma2,r,
«2,
(i)
(2)
Let «i iP ••• a g , r ••• a n , s be the absolute value of one of the terms of (1).
Replacing a q , r by ma q> ,., the absolute value of the corresponding term
of (2) is mai, p ••• a q>r ••• «»,«•
It is evident from this that the determinant (2) equals m times the
determinant (1).
Ex. Consider the values of
9
h .
k
• Evaluating, m(aek -f- bfg + chd — ceg — dbk — afh) and
maek + mbfg + mchd — mceg — dmbk —mafli, which are identical.
a
d g
ma
d
b
e h
and
mb
e
c
f *
mc
f
220
ALGEBRA
301 . If all the elements in any column, or row, be mul-
tiplied by the same number, and either added to, or sub-
tracted from, the corresponding elements in another
column, or row, the value of the determinant is not
changed.
Let the elements in the rth column of the following deter-
minant be multiplied by m, and added to the corresponding
elements in the qth column.
«1> •'•■> ®q,
'"t
Ctr,
t &n
h, •>•$ b q , •••,
b r ,
, &n
(1
#i? •••? 1c q <) •••?
fori
, fC n
We then obtain the determinant
0>U '"•) ®q + ma r,
-•»
Or,
'**!
&n
6i, ••, bq + mhr,
'"5
&n
-'
bn
(2
ki, • •, k q + mkr,
-..,
fori
***?
K
which by §§ 299 and 300, is equal to
Q>ll *"% Qqi "'$ Qri '"i Q>n
«1,
'" i
«r,
, a n ••
', «n
b\, —, b qj •••, & PJ •••, b n
+ m
hi
b r ,
rCr,
5r, ••
*» tin
Kit '*•$ Kq-j
.. ?
AV, "**} "'W
But the coefficient of m is zero (§ 298).
Whence, the determinant (2) is equal to (1).
302. Minors.
If the elements in any m rows and any m columns of a
determinant of the r*th order be erased, the remaining elements
form a determinant of the (n — m)th order.
This determinant is called an mth Minor of the given deter-
minant. h „ n „
***j ^2, n
(1)
By § 290, the absolute values of the terms which involve
a h x are obtained by forming all possible products of the ele-
ments taken n at a time, subject to the restrictions that the
first elements shall be a h x and that each product shall contain
one and only one element from each row except the first, and
one and only one from each column except the first.
It is evident from this that the coefficient of a h j in (1) may
be obtained by forming all possible products of the following
elements taken n — 1 at a time,
^2,2? ^2,3? '"> ®2,n
a 3, 2? a 3, 3> ***? a 3, n
subject to the restriction that each product shall contain one
and only one element from each row, and one and only one from
each column, writing the first suffixes in the order 2, 3, • •, n,
and making each product + or — according as the number of
inversions in the second suffixes is even or odd.
Then by § 290, the coefficient of a lt x is
Ojo o j ^2 3? * * *? ^2 n
tt 3, 2) a 3, 3> "'J a 3,n
that is, the minor obtained by erasing the first row and the
first column of the given determinant.
222
ALGEBRA
304. By aid of § 303, a determinant of any order may be
expressed as a determinant of any higher order.
1, 0, 0, 0,
Thus,
1,
o,
o,
a% %
hi
Cl
() 7
Cll,
hi,
Cl
«2,
&2,
c 2
=
0,
«2?
&2,
c 2
0*,
6» ?
c 3
0,
az,
&•,
c 3
0, 1, 0, 0,
0, 0, GL\, b\, C\
0, 0, a 2 , b 2 , c 2
0, 0, a z , b 3 , c 3
etc.
305. Coefficient of any Element of a Determinant.
To find the coefficient of b 3 , in the determinant
"■■i,
K
bo,
CO
(-1)3
%> ^3? °3
By two interchanges of consecutive rows, the last row may be made
the first ; thus, by § 295, the determinant equals
a 3 , &3, c 3
(-1)2 ai , h, ft • (2)
a*L, b 2 , C 2
By interchanging the first two columns, the determinant (2) equals
b 3 , az, c s
(- 1) 3 6i, oi, 4 • (3)
b 2 , a 2 , c 2
By § 303, the coefficient of b 3 , in (3), is
a\, ci
a 2 , c 2
That is, the coefficient of the element in the third row and second
column equals (— 1)*+ 2 multiplied by that minor of (1) which is obtained
by erasing the third row and second column.
We will now consider the general case ; to find the coefficient of a*, r in
the determinant
a\,i, •••, <*i,r» •••» «i,
ak, i,
a n ,u
aic,r
', a n ,ri
«*,*
(4)
DETERMINANTS
223
By k — 1 interchanges of consecutive rows, and r — 1 interchanges of
consecutive columns, the element a&, r may be brought to the upper left-
hand corner.
Thus, by § 295, the determinant equals
Then, by § 303, the coefficient of a A , r is
(_ 1)*+r-2
$n, 1» ***? ^n, n
But
(_ l)*+r-2 - (_ !)t+r +(_ 1)« = (_ 1)
fc+r
Hence, the coefficient of the element in the ftth row
and rth column equals (— l)* +r ? multiplied by the minor of
(4) which is obtained by erasing the Zcth row and rth
column.
306. By aid of § 305, a determinant of any order may
be expressed in terms of determinants of any lower
order.
Thus, since every term of a determinant contains one and only one
element from the first row, we have,
a u &i, ci, di
#2? &2> Cli d'l
«3, 63, c 3 , dj
#4, 64, C4, C?4
|&2»
c 2 ,
$2
«1 63,
C3,
<*3
-h
|&4i
C4,
^4
«2» C2, ^3
«4, C4, (?4
«2>
&2»
<^2
«3,
fc.
C*3
rr*
«4,
64,
1
1
1
7.
x,
■4,
y>
-T,
6,
6. 3,
6,
Plot this line (§51).
y = 5, and as = 6, y
4,
2/,
12,
= 0.
= 0.
Are the points x — — 4, ?/ = — 7,
and as = 8, ?/ = 6, on this line?
8.
15, 12, 11,
14, -4, -8,
16, 10, -2,
10, 12, 3,
6
25
5
6
9-
io.
Are the points x = 3,
= — 12, on this line?
0, 0, 3
7,
9,
o,
8,
o,
o,
10
o,
o,
-6,
-8
o,
3,
5,
9
5,
o,
7,
8-
4,
9,
o,
16
8,
16,
4,
226
ALGEBRA
4, 0,
12, 10
8,
t
0, 0,
0, 3, 0,
0, 6, 15,
18
II.
10, 0, 6, 4
24, 0, 3, 12
12.
0, 12, 98,
0, 2, 5,
104
6
5, -4,
0,
15
-4, 5,
15,
13. Show that
12, 0,
17, -9,
5,
-9,
4
3
=
0, 12,
-9, 17,
4, 5
3, -9
6, 8,
8,
-2
8, 6,
-2, 8
14, 3,
-13
1
-13, 21,-
-17, -11
14. Show that
8,-5,
21, 24,
21
-17
2
1
=
14, 8,
3, -5,
21, -6
24, -9
-6, -9,
-11
p 2
1, 2,
1, 2
5, 6,
12,
16
5, 2,
3, 16
15. Show that
7, 9, -
18, 27,
-16,
36;
-15
9 :
= 108
7, 3,
2, 1,
-4, -15
1, 1
2, 9, -
-20,
-11
2,3,
-5, -11
16, 8,
7,
-4
16. Evaluate
8, 6,
7, 5,
3,
-11,
2
3
-4, 7,
8,
in, y, n,
X
17. Evaluate
x, y, n,
«i »» y,
m
m
m, n,
y,
X
It is shown in geometry that if a straight line passes through
the points x = x l9 y = y ly x — x 2 , y = y 2 , where x x , x 2 , y ly y 2 are
definite values of x and 2/, the equation of the line is found
from the determinant
1
1 =0.
1
X
y
x x
V\
x 2
2/2
DETERMINANTS
227
Find the equations of the lines passing through the following
points :
18. as = 1, y = 3; #=— 2, y s4
19. x = 0, y = 8 ; x = 8, y = 0.
20. # = — 1, 2/ = 2 ; a? = 3, y = — 1.
308. Let A r , B n • ••, iT r , denote the coefficients of the ele-
ments a,., b r , •••, k r , respectively, in the determinant.
hi ^1?
Kq
K,
"n
(1)
Then, since every term of the determinant contains one and
only one element from the first column, the value of the
determinant is
A 1 a 1 + A 2 a 2 -\- ••• + A n a n .
In like manner, the value of the determinant also equals
, BA + B 2 b 2 + . .. + B H b n , • • •, KJh + KM 2 + • • . 4- &Jc*
309. If m lf m 2 , • ••, m n are the elements in any column of the
determinant (1), of § 308, except the first,
A^in^ + A 2 m 2 -+- • • • -f A n m n
is the value of a determinant, which differs from (1) only in
having m 1? m 2 , •••, m n instead of a lt a 2 , •••, a n as the elements in
the first column.
Then A 1 m 1 + A 2 m 2 -f ••• + A n m n = ;
for it is the value of a determinant which has two columns
identical.
In like manner, if m 1? ra 2 , •••, m n are the elements in any
column of the determinant (1), except the second, ^
Bim^ + B 2 m 2 + • • • + 5 n m n = ;
and so on.
228
ALGEBRA
SOLUTION OP EQUATIONS
310. Let it be required to solve the following system of n
linear, simultaneous equations, involving n unknown numbers :
0&1 +
+ q&r +
+ hx H = p lm
la„a 1 + ••• + q,fl r + ■•■• 4-&A=2>„.
(1)
(2)
(3)
Let Q u Q 2 , • ••, Q n denote the coefficients of the elements #i, q 2 , •••, q n
respectively, in the determinant
D--
Q2,
»j k 2
•j kn
Multiplying equations (1), (2), ..., (3) by § b Q 2 , • •, Q m respectively,
and adding, we have
xiiQiai + Q 2 a 2 + ••• + Q n a n ) + •••
+ Xr{Qiqi + §2^2 + ••• + Q n q n ) + •"
+ x n (Qiki + Q 2 k 2 + -. + Q n k n ) = QiPi + Q 2 p 2 +
+ §nPn.
By § 309 the coefficient of each unknown number, except x r , is zero.
By § 308, the coefficient of x r is D ; also, the second member is the
value of a determinant which differs from D only in having pi, p 2 , •••» Pn
instead of q u q 2 , •••, q n as the elements in the rth column ; denoting the
latter by D n we have
x r D = D r , and x r =
I)
311. i£#. Find the value of y from the equations
(3x-5y + 7z = 28.
i2a + 6x n ~ 2 H \-q n _ x x+q n = 0. (2)
Equation (2) must also have at least one root.
Let b be this root ; then (2) may be written
(x-b) (x n ~ 2 + r 3 x n ~ s + • • • + *Li? + r n ) = 0,
and the equation may be solved by placing
x - b = 0,
and x n ~ 2 -f r s x n ~ 3 -\ h r n x x + r n = 0.
After Ti — 1 binomial factors have been divided out, we shall
arrive finally at an equation of the first degree,
x — 7c = ; whence, x = 7c.
Therefore, the given equation has the n roots a, 6, • • •, 7c.
The roots are not necessarily unequal.
m
Ex. x 3 - 3 x 2 + 3 x - 1 = 0.
Whence, (x - l)(x - l)(x - 1) =
and x = l, or 1, or 1.
232 ALGEBRA
317. Depression of Equations.
It follows from § 316 that, if m roots of an equation of the
nth degree are known, the equation may be depressed to an-
other equation of the (n — m)th degree, which shall contain the
other n — m roots.
Thus, if all but two of the roots of an equation are known,
these two may be obtained from the depressed equation by the
rules for quadratics.
Ex. Two roots of the equation 9 x 4 — 37 x 2 — 8 x + 20 = are
2 and — f ; find the others.
By § 314, the first member of the given equation is divisible by
(a- 2) (3« + 5), or Sx 2 - x - 10.
Dividing 9 x 4 — 37 x 2 — 8 x + 20 by 3 x 2 — x — 10, the quotient is
3x 2 + x-2.
Then the depressed equation is
3 x 2 + x - 2 = 0.
Solving by the rules for quadratics, x = } or — 1.
EXERCISE 75
Find whether :
i. One root of ar 3 — x 2 — 32 x + 60 = is 5; if so, find the
others.
2. One root of 2 a? - 6 x 2 - 2 x + 6 = is 3 ; if so, find the
others.
3. Two roots of 4 x 4 - 12 x 3 - 13 x 2 + 45 x - 18 = are - 2
and 3 respectively ; if so, find the others.
4. One root of 3 X s — 5 x 2 + 2 x — 4 = is 2 ; if so, find the
others.
5. Two roots of 14 a 4 + 65 ar* - 222 x 2 + 65 x + 14 = are 2
and — 7 ; if so, find the others.
6. Two roots of x 4 + 4 x s - 6 x 2 + 24 x - 72 = are - 6 and
2; if so, find the others.
7. Two roots of x 4 — 4 X s + 3 x 2 + 4 a; — 4 = are 2 and — 1 ;
if so, find the others.
THEORY OF EQUATIONS 233
8. Three roots of 2 x* + 29 x 4 + 1*8 or 3 + 379 a; 2 + 394 x + 120
= are — 2, — 3, — £ j if so, find the others.
9. Two roots of a* 4 + 12 a 8 + 34 x 2 - 12 x - 35 = are 1 and
— 7 ; if so, find the others.
10. Two roots of 5 x* - 18 a) 3 +72z - 120 = are 5 and - 4 ;
if so, find the others.
318. Formation of Equations.
It follows from § 316, that if the roots of
x n +p x x n ~ l -\ t- Pn _ lX + Pn =zO
are a, b, •••, k, the equation may be written in the form
(x — a)(x — b) • • • (x — k) = 0.
Hence, to form an equation which shall have any required
roots,
Subtract each root from a?, and place the product of
the resulting expressions equal to zero.
Ex. Form an equation having the roots 1, i, and — f .
By the rule (x- l)(x -DO + }) = 0.
Multiplying the terms of the second factor by 2, and of the third by 3,
(x-l)(2z-l)(3£ + 5) =0.
Expanding, 6x 3 + % 2 — 12 x + 5 = 0.
EXERCISE 76
Form equations having the roots :
I- 1,-4,6. 7 . _ m) m±v^.
2. 3,-1,-21 4
3-2,3,5,0. 8. 3±V2,-3 ± V2.
4. -1,-2,7,-8. 9- «, -~>- & >-J-
5- 2, 9, - J, |. ^ 4±2V3 -2±V3
6. 4,4, -I, -£. 3 ' 3
234 ALGEBRA
319. Composition of Coefficients.
By § 318, the equation of the nth degree whose roots are a,
b, c, d, • ••, ft, Z, m, is
(x — a)(x — b)(x — c)(x — d) ••• (x — m) = 0. (1)
By actual multiplication, we obtain
(x — a) (x — b) — x 2 — (a + b)x + ab ;
(x — a) (x — b) (x — c)
= x* — (« + & + c)x 2 + (ab -f be -f ca)x — a&c ;
(# — a) (x — 6) (x — c) (x — d)
= x i —(a + b + c + d)x* -f (a& + ac+ ad + bc + bd + cd)x 2
— (abc + abd + aceZ + 6cd)x + abed = ; etc.
When all the factors of the first member of (1) have been multiplied
together, the result will be in the form
x n -\-piX n ~ l + p 2 x n - 2 + p 3 x n - s -f ••• +p n ;
where pi = - (a + b -f c + ••• + k + I -f m) ;
p 2 = ab + ac + 6c + • • • -f Zm ;
p 3 = — (abc + «6c? + acd + ••• + Mm);
p n = ± abed ••• klm, according as n is even or odd.
Hence, in an equation of the nth degree in the general form,
The coefficient of the second term is equal to minus
the sum of all the roots.
The coefficient of the third term is equal to the sum of
the products of the roots, taken two at a time,
The coefficient of the fourth term is equal to minus the
sum of the products of the roots, taken three at a time ;
etc.
The last term is equal to plus or minus the product of
all the roots, according as n is even or odd.
320. It follows from § 319 that, if an equation of the nth
degree is in the general form,
If the second term is wanting, the sum of the roots is 0.
If the last term is wanting, at least one root is 0.
THEORY OF EQUATIONS 235
If the last term is an integer, it is divisible by every integral
root.
EXERCISE 77
In each of the following, find the sum of the roots, and the
product of the roots :
i. x z -Sx 2 + l§x-12 = b. 2. ar 3 - 31 x- 30 = 0.
3. 4^-12^ + 3^ + 13^-6=0.
321. If all but one of the roots of an equation of the ?ith
degree in the general form are known, the remaining root may
be found by changing the sign of the coefficient of the second
term of the given equation, and subtracting the sum of the
known roots from the result ; or, by dividing the last term of
the given equation if n is even, or its negative if n is odd, by
the product of the known roots.
If all but two are known, the coefficient of the second term
of the depressed equation may be found by adding the sum of
the known roots to the coefficient of the second term of the
given equation ; and the last term of the depressed equation
may be found by dividing the last term of the given equation
by plus or minus the product of the known roots according as
n is even or odd.
Ex. Two roots of the equation 9 x A — 37 x 2 - 8 a; + 20 =
are 2 and — f ; what are the others ?
We first put the equation in the general form by dividing each term
by 9.
It then becomes sc* - - 3 / x 2 — § x + ^ = 0.
Since there is no sc 3 term, the coefficient of the second term is 0.
Then the coefficient of the second term of the depressed equation is
+ 2 - § or 1
The coefficient of the last term of the depressed equation is
Solving, x = § or — 1.
236 ALGEBRA
EXERCISE 78
i. One root of x 3 + 7 x 2 — 5x — 75 = is — 5 ; find the others.
2. Three roots of 4 a; 4 — 55 x 2 — 45 # + 36 = are 4, —3, \\
find the other.
3. Four roots of 20 x 5 - 108 x 4 + 225 or 3 - 224 x 2 + 105 a - 18
= are 1, 1, -§, f ; find the other.
4. Three roots of or 5 - a 4 - 15 X s + 25 ar> + 14 x - 24 = are 2,
1, — 4 ; find the others.
5. Two roots of x* - ax* + (2 a - 7 a 2 - l)rf + (a 3 - 2 a 2 + a)*
+ 6 a 4 — 12 a 3 + 6 a 2 = are a — 1 and 3 a; find the others.
322. Fractional Roots.
An equation in the general form with integral coeffi-
cients cannot have as a root a rational fraction in its
lowest terms.
Let the equation be
x n +p 1 af t ~ 1 +p 2 x n 2 H VPn-l* +Pn = 0,
where p l9 p 2 , '" } p n are integral.
If possible, let -, a rational fraction in its lowest terms, be a root of
b
the equation ; then,
Multiplying each term by 6 n_1 , and transposing,
b
By hypothesis, a and b have no common divisor ; hence, a n and b have
no common divisor.
We then have a rational fraction in its lowest terms equal to an integral
expression, which is impossible.
Therefore, the equation cannot have as a root a rational fraction in its
lowest terms.
THEORY OF EQUATIONS 237
323. Imaginary Roots.
If the imaginary number a + hi is a root of an equation
in the general form, with real coefficients, its conjugate
(a - bi) is also a root.
Let the equation be
X n + p lX n-l + . . . +p n _ lX +p n = 0, (1)
where p 1? •••, p n are real numbers.
Since a + bi is a root of (1), we have
(a + bi)» + pi(a + bi)"- 1 + •- +Pn-i{a+bi) + p n = 0.
Expanding by the Binomial Theorem, we have, by § 180,
a n + na n- m _ riin-J^ %2 _ n(n - l)(n - 2) _
[2 >(S
+ pja n ~ l + (n- \)a-m - ^ - *) 0* - 2) aW _ 3&2 1
+ ... +j£»-i(a + 50 +p w = 0. (2)
Collecting the real and imaginarj^ terms, we have a result of the form
P+Qi = 0. (3)
Here, P stands for the sum of all the terms containing a alone, together
with all the terms containing even powers of i ; and Qi for all terms
containing odd powers of i.
In order that equation (3) may hold, we must have
P = 0, and Q = 0.
Now substituting a — bi for x in the first member of equation (1),
it becomes
(a - 60" +Pi(a - bi)*- 1 + - +p n -i(a - bi) fp n . (4)
Expanding by the Binomial Theorem, we shall have a result which
differs from the first member of (2) only in having the second, fourth,
sixth, etc., terms of each expansion, or those involving i as a factor,
changed in sign.
Then, collecting the real and imaginary terms, the expression (3) is
equal to p _ q^
where P and Q have the same meanings as before.
But since P = and Q *= 0, we have P— Qi =s 0.
Whence, a — bi is a root of (1). *•
The above proof holds without change when a equals zero ; thus the
theorem holds for any pure imaginary number, of the form bi.
238 ALGEBRA
324. The product of the factors of the first member of
equation (1), § 323, corresponding to the conjugate imaginary
roots a+bi and a — bi is
[x _ (a + bi)J_x - (a-bi)-] (§ 318)
= (x — a — bi)(x — a + bi)
= (x- a) 2 - (bi) 2 = (x - a) 2 + b 2 ;
and is therefore positive for every real value of x.
325. It follows from §§ 316 and 323 that every equation of
odd degree has at least one real root ; for an equation cannot
have an odd number of imaginary roots.
TRANSFORMATION OF EQUATIONS
326. To transform an equation into another which shall have
the same roots with contrary signs.
Let the equation be
x n +Pif l +p 2 x n ~ 2 + ••• + Pn-iX +p n = 0. (1)
Substituting — y for as, we have
(- y) n +pi(- y) 71 - 1 +m- y) n ~ 2 + - +p»-i(- y) +p n = o.
Dividing each term by (— 1) M , we have
Or, yn_ piy n-l + p2 yn-2 ± Pn-l2/ T Pn = J (2)
the upper or lower signs being taken according as n is odd or even.
It follows from (1) and (2) that the desired transformation
may be effected by simply changing the signs of the alternate
terms commencing with the second.
If the equation is incomplete., any missing term must be supplied with
the coefficient zero before applying the rule.
THEORY OF EQUATIONS 239
327. Ex. Transform the equation a; 3 — 10 x + 4 = into
another which shall have the same roots with contrary
signs.
The equation may be written x s + • x 2 — 10 x + 4 = 0.
Then, by the rule, the transformed equation is
X 3_ .x 2 - 10x-4 = 0, or x 3 -10x-4=0.
EXERCISE 79
Transform each of the following into an equation which
shall have the same roots with contrary signs :
i. x s -.6x 2 + 12x-8 = 0. 2. x*-6x 3 + ±x 2 -9x + 16 = 0.
3. x 7 + 5x 5 — 3x i + x — 4: = 0.
328. To transform an equation into another whose roots shall
be respectively m times those of the first.
Let the equation be
X n +^> 1 X n ~ 1 -\-p 2 X n ~ 2 + • • • +P n -iX +p n = 0.
Putting mx = y, that is, 2- for x, we have
m
+ Pn = 0.
Multiplying each term by m n ,
y n -\-pi i my n - 1 -\-p 2 m 2 y n - 2 + ... + p n -\m n - l y + p n m n = 0.
Hence, to effect the desired transformation, multiply the
second term by m, the third term by m 2 , and so on.
' Ex. Transform the equation X s -\-7 x 2 — 6 = into another
whose roots shall be respectively 4 times those of the first.
Supplying the missing term with the coefficient zero, and applying the
rule, we have
x 3 + 4 . 7 x 1 + 4* • Ox - 4« • 6 = 0, or x 3 + 28 x 2 - 384 = 0.
240 ALGEBRA
329. To transform an equation with fractional coefficients into
another whose coefficients shall be integral, that of the first term
being unity.
The transformation may be effected by transforming the
equation into another whose roots shall be respectively m
times those of the first (§ 328) ; we then give to m such a
value as will make every coefficient integral.
By giving to m the least value which will make every coeffi-
cient integral, the result will be obtained in its simplest form.
Ex. Transform the equation X s — = into an-
H 3 36 108
other whose coefficients shall be integral, that of the -first term
being unity.
By § 328, the equation
^."V-^ + ^O
3 36 108
has its roots respectively m times those of the given equation.
It is evident, by inspection, that the least value of m which will make
every coefficient integral, is 6.
Putting m = 6, we have
x s _ 2 x 2 - x + 2 = 0,
whose roots are 6 times those of the given equation.
330. To transform an equation into another whose roots shall
be respectively those of the first increased by m.
Let the equation be
x n + Pl x n ~ l + . • • +p n _ x x +p H = 0. (1)
Putting x -f m = y, that is, y — m for x, we have
(y — m) n +pi(y — m) n - l + \-Pn-i(y — m)+p n = 0. (2)
Expanding the powers of y — m by the Binomial Theorem, and collect-
ing the terms involving like powers of y, we shall have a result of the
yn + qiy n-i _|_ .. . _+_ q n _ iy + Qn = o, (3)
whose roots are respectively those of the given equation increased by m.
THEORY OF EQUATIONS 241
Ex. Transform the equation or 3 — 7 # -f 6 = into another
whose roots shall be respectively those of the first increased by 2.
Substituting y — 2 for cc,
(j/-2)3-7G,-2) + 6 = 0.
Expanding, and collecting the terms involving like powers of y, we
have y8-6y* + 6y + 12 = 0.
331. If m and the coefficients of the given equation are
integral, the coefficients of the transformed equation may be
conveniently found by the following method.
Putting x + m for y in (3), we obtain
(x + m) n + qi (x + m)» * + ••• + g„_i(x + m) + q n = 0, (4)
which must, of course, take the same form as (1) on expanding the
powers of x + m, and collecting the terms involving like powers of x.
Dividing the first member of (4) by x + w, we have
(x + m)"- 1 + q x (x + m)»- a -f .-. +q*-% C* + w) + g„_i (5)
as a quotient, with a remainder q n .
Dividing (5) by x + m, we have the remainder q n -i; etc.
Hence, to obtain the coefficients of the transformed equation :
Divide the first member of the given equation by
oo + m ; the remainder will be the last term of the required
equation.
Divide the quotient just found by oc+ in ; the remainder
will be the coefficient of the next to the last term of the
transformed equation ; and so on.
Ex. Transform the equation sc 3 — 7#4-6 = into another
whose roots shall be respectively those of the first increased by 2.
Dividing x s — 7 x + 6 by x + 2, we have the quotient x 2 — 2 x — 3, and
the remainder 12 (§ 108).
Dividing x 2 — 2 x — 3 by x + 2, we have the quotient x — 4, and the
remainder 5. N
Dividing x — 4 by x + 2, we have the remainder — 6.
Then, the transformed equation is
X 3_6x 2 + 5x + 12 = 0.
Compare Ex., § 330.
242 ALGEBRA
332. To transform an equation into another whose roots
shall be those of the first diminished by ra, we change y — m to
y -\-m in the method of § 330, and x -f- m to x — m in the rule
of § 331.
EXERCISE 80
i. Transform x 2 — x — 12 = into an equation whose roots
shall be, respectively, 5 times the first. Verify your results.
2. Transform ar 3 -f- x 2 — 14 x — 24 = into an equation whose
roots shall be, respectively, twice those of the first. Verify
results.
3. Transform x^-^Sx 2 — 23 x — 210 = into an equation
whose roots shall be, respectively, \ times the first.
Transform each of the following into an equation with in-
tegral coefficients, that of the first term being unity :
4. 6 or 3 -11 x 2 -14 a; + 24 = 0. Verify result.
5. 8x* + Ux 2 -5x-2 = Q.
6. 2a 4 -13ar 3 -91ar J + 390a + 216 = 0.
7. 90^-flll^ + 25^ 2 -12cc-4 = 0.
8. x A + — - — -— = 0.
7 14 196
9. Transform a^-f-lO x 2 + 7 x— 18 = into an equation whose
roots shall be, respectively, those of the first diminished by 4.
10. Transform x* - 3 X s — 19 x 2 + 27 x + 90 = into an equa-
tion whose roots shall be, respectively, those of the first in-
creased by 3.
333. To transform the equation
x n +p 1 x n ~ 1 -\ f- p n _iX -f p n =
where p l is not zero, into another whose second term nkatt be
wanting.
THEORY OF EQUATIONS 243
Expanding the powers of y — m in the first member of (2),
§ 330, and collecting the terms involving like powers of y, we
have
y n + (l\ — mn)y n ~ l -\ = 0.
If m be so taken that p Y — mn = 0, whence m = — , the coeffi-
n
cient of y n ~^ will be zero.
Hence, the desired transformation may be effected by sub-
stituting in the given equation y — -in place of x.
Ex. Transform x 5 — 6x 2 + 9 x — 6 = into an equation whose
second term shall be wanting.
_ ft
Substituting y or y + 2, in place of x, we have
o
(?/ + 2)» _ 6 (y + 2)2 + 9 (y + 2) - 6 = 0.
Then, y s + 6 y* + 12 y + 8 - 6 ?/ 2 - 24 y - 24 + 9 y + 18 - 6 = 0,
or y 3 — 3 y — 4 = ;
whose roots are those of the given equation diminished by 2.
EXERCISE 81
Transform each of the following into, an equation whose
second term shall be wanting :
i. ^-6a 2 + 4a-l=0. 3. x 4 + 12 ^+2^-3 = 0.
2. x 3 + 5x 2 + 8 = 0. 4. x 5 -x i + 7x-l = 0.
DESCARTES' RULE OF SIGNS
334. If an equation of the nth degree is in the general form
(§ 312), a Permanence of sign occurs when two successive terms
have the same sign, and a Variation of sign occurs when two
successive terms have opposite signs.
Thus, in the equation x G — 3 x 4 — X s -f 5 x + 1 = 0, there are
permanences and two variations.
,
244 ALGEBRA
335. Descartes' Rule of Signs.
No equation, whether complete or incomplete, can
have a greater number of positive roots than it has
variations of sign ; and no complete equation can have
a greater number of negative roots than it has perma-
nences of sign.
Let an equation in the general form have the following signs :
+ + - + 00 ,
the missing terms being supplied with zero coefficients.
If we introduce a new positive root a, we multiply this by x — a (§ 318) ;
writing only the signs which occur in the process, we have
123456789
+ +0-+00-- (1)
+ -
+
+
—
+
—
—
—
—
+
-
+
+
+
m
—
—
+
-
—
m
+
1
2
3
4
5
6 7
8
9
10
(2)
9 10
Here m signifies a term which may be +, 0, or — .
Now, in (1), let a dot be placed over the first minus sign, then over
the next plus sign, then over the next minus sign, and so on.
The number of dots shows the number of variations ; thus in (1) there
are three variations.
In the above result, we observe the following laws :
I. Directly under each dotted term of (1) is a term of (2)
having the same sign.
Thus, the terms numbered 4, 5, and 8, in (1) and (2), have
the same sign.
II. The last term of (2) is of opposite sign to the term
directly under the last dotted term of (1).
The above laws are easily seen to hold universally.
By the first law, however the term marked m is taken, there
are at least as many variations in the first eight terms of (2)
as in (1) ; and by the second law, there is at least one varia-
tion in the remaining terms of (2).
Hence, the introduction of a new positive root increases the
number of variations in the equation by at least one.
THEORY OF EQUATIONS 245
If, then, we form the product of all the factors correspond-
ing to the negative and imaginary roots of an equation, multi-
plying the result by the factor corresponding to each positive
root introduces at least one variation.
Hence, the equation cannot have a greater number of posi-
tive roots than it has variations of sign.
To prove the second part of Descartes' Rule, let — y be sub-
stituted for x in any complete equation.
Then since the signs of the alternate terms commencing
with the second are changed (§ 326), the original permanences
of sign become variations.
But the transformed equation cannot have a greater number
of positive roots than it has variations.
Hence, the original equation cannot have a greater number
of negative roots than it has permanences.
In all applications of Descartes' Rule, the equation must contain a
term independent of x ; that is, no root must equal zero ; for a zero root
cannot be regarded as either positive or negative.
336. It follows from the last part of § 335, and from § 326,
that in any equation, whether complete or incomplete, the
number of negative roots cannot exceed the number of varia-
tions in the equation which is formed from the given equation
by changing the signs of the alternate terms commencing with
the second.
337. In any complete equation, the sum of the number of
permanences and variations is equal to the number of terms
less one, or to the degree of the equation.
That is, the sum of the number of permanences and varia-
tions is equal to the number of roots (§ 316).
Hence, if the roots of a complete equation are all real, the
number of positive roots equals the number of variations, and the
number of negative roots equals the number of permanences.
An equation whose terms are all positive can have »o posi-
tive root ; and a complete equation whose terms are alternately
positive and negative can have no negative root.
246 ALGEBRA
338. Ex. Determine the nature of the roots of
x 3 + 2 x + 5 = 0.
There is no variation, and consequently no positive root.
Changing the signs of the alternate terms commencing with the second,
we have x s + 2 x - 5 = 0. . (See Note, § 326. )
Here there is one variation ; and therefore the given equation cannot
have more than one negative root (§ 330).
Then since the equation has three roots (§ 316), one of them must be
negative and the other two imaginary.
If two or more successive terms of an equation are wanting, it follows
by Descartes' Rule that the equation must have imaginary roots.
EXERCISE 82
If the roots of the following are all real, determine their
SlgnS: i. ^ + 10^ + 7^-18 = 0.
2. a 4 -3ar 3 -19# 2 + 27a + 90=0.
3. 36x?-mx s + '27x 2 + 7x-3 = 0.
4. ar 5 -4a 4 -5a 3 + 20a; 2 + 4a-16=0.
5. 2a 4 -13ar J -91a 2 + 390a + 2iG = 0.
Determine the nature of the roots of the following :
6. 2^ + 0^+2^-12 = 0.
7. # 4 + 3ar ? + 7ar J + 6a; + 4 = 0.
8. a 4 -2^-9 = 0.
9. x>-2x 4 + 4x s -8x 2 + 16x-16=0.
io. x 7 + 3 x 4 + 5x 2 + 2 = 0.
LIMITS TO THE ROOTS
339. To find a superior limit to the positive roots oj an
equation.
The following examples illustrate the method of finding a
superior limit to the positive roots of an equation.
•
THEORY OF EQUATIONS 247
1. Find a superior limit to the positive roots of
x* - 3 x 2 + 2 x - 5 = 0.
Grouping the positive and negative terms, we can write the first mem-
ber in the form
a*(a-3)+2(a:- J). (1)
It is evident that if x equals or exceeds 8, the expression (1) is positive
Hence, no root of the given equation equals or exceeds 3, and 8 is a
superior limit to the positive roots.
2. Find a superior limit to the positive roots of
a 4 -15 a 2 - 10 a+ 24 = 0.
2 x^ x^
We separate the first term into the parts - r - and — , and write the first
member in the form
/?*!_L5 xA +^_M)aA + 24, or ^(2 x 2 - 45) + -(a* - 30) + 24.
It is evident from this that no root can be so great as 5 ; hence, 5 is a
superior limit to the positive roots.
If we had written the first member in the form
^_ 15^2 \ + ht_ io x \ +2 4, or £- 2 (z 2 - 30) + -<> 3 - 20) + 24,'
we should have found 6 as a superior limit to the positive roots.
2 x 4 sc 4 sc 4 sc 4
Thus, separating x 4 into — and — , instead of — and — , gives a smaller
So A A
limit.
340. To find an inferior limit to the negative roots of an
equation.
First transform the equation into another which shall have
the same roots with contrary signs (§ 326).
The superior limit to the positive roots of the -transformed
equation, obtained as in § 339, with its sign changed, will
be an inferior limit to the negative roots of the given equa-
ion.
248 ALGEBRA
Ex. Find an inferior limit to the negative roots of
tf + 2 x* + 5 x 2 - 7 = 0.
Transforming the equation into another which shall have the same
roots with contrary signs (§ 326), we have
x b + 2z 3 -5z 2 -f 7 = 0. (lj
We can write the first member in the form
x 2 (x 3 -5) + 2z 3 + 7.
It is evident from this that no root of (1) can be so great as 2 ; hence,
— 2 is an inferior limit to the negative roots of the given equation.
By grouping the x b and x 2 terms in (1), we obtain a smaller limit than
if we group the x 3 and x 2 terms.
EXERCISE 83
In each of the following, find a superior limit to the positive
roots, and an inferior limit to the negative :
, i. ar J + 3a 2 + a~-4 = 0.
2 . x A + 5x*-15x-9 = 0.
3. x 4 + 3x* -5 x -8 = 0.
4. 3x*-5x 2 -8x-7 = 0.
5. ^-4a 4 + 6ar 3 + 32a 2 __ 15^ + 3 = 0.
6. 2x 5 + 5x i + 6x i -13x 2 -25x + 4; = 0.
7. In the equation X s — 2 x 2 — 3a? + l = 0, prove 3 a supe-
rior limit to the positive roots, and — 2 an inferior limit to the
negative.
8. In the equation 2 a 8 + 5 aj 2 — 7 x — 3 = 0, prove — 4 an
inferior limit to the negative roots, and find a superior limit to
the positive.
9. In the equation x 4 -f 3 x* — 9 x 2 + 12 x — 10 = 0, prove 3
a superior limit to the positive roots, and — G an inferior limit
to the negative.
THEORY OF EQUATIONS 249
DERIVATIVES
341. If we take the polynomial
ax n + bx"- 1 + cx n ~ 2 + •••,
multiply each term by the exponent of x in that term, and
then diminish the exponent by 1, the result
nax n ~ l -f (n — l)bx n ~ 2 + (n — 2)cx n ~ z + ...
is called the first derivative of the given polynomial.
The first derivative of the first derivative is called the sec-
ond derivative of the given polynomial ; the first derivative
of the second derivative is called the third derivative ; and
so on.
Ex. Find the successive derivatives of 3 X s — 9 x 2 — 12 x -f 2.
The first is 9 x 2 - 18 x - 12.
The second is 18 x— 18.
The third is 18.
The fourth is 0.
It will be understood hereafter that when we speak of the derivative of
an expression, we mean the first derivative.
EXERCISE 84
Find the successive derivatives of :
i. 5x 2 + $x-7. 4 8^ -3 a 2 + 2.
2. 3^-7^ + 2. 5 . 6x 6 -5x 5 + 4:X s -3x 2 + 27.
3. 9z 3 -7a 2 + 15a;-l. 6. x 5 - a? 4- 10 X s + 5 x 2 - 7 x.
MULTIPLE ROOTS
342. If an equation has two or more roots equal to a, a is
said to be a Multiple Root of the equation.
In the above case, a is called a double root, a tripte root, a
quadruple root, etc., according as the equation has two roots,
three roots, four roots, etc., equal to a.
250 ALGEBRA
343. Let the roots of the equation
&+P&-* +&#*>+ -. +p n = (1)
be a, b, c, d, •••.
Then, by §> 318, we have
x n + piic"- 1 + P2X n ~ 2 + ••• = (£-«)(£- &)(£ — c) •••.
Putting x + h in place of x, we obtain
(x + ft)" +pi(x + ft)*" 1 +p 2 (x + ft) n ~ 2 + •••
= (ft + x - a) (ft + x - b)(h + x - c) •••. (2)
Expanding the powers of x + ft by the Binomial Theorem, the coeffi-
cient of h in the first member of (2) is
nx n ~ l +pi(n — l)x n ~ 2 +p 2 (n - 2)x n ~ s + ••• ; (3)
which, we observe, is the first derivative of the first member of (1).
Again, it is evident from § 319 that the coefficient of h in the second
member of (2) is
(x — b) (x — c) (x — d) ••• to n — 1 factors
+ (x — a)(x — c)(x — d) ••• to n — 1 factors
+ (x — a) (x — b) {x — d) • • • to n — 1 factors + • •• . (4)
Since equation (2) is true for every value of ft, by § 264 these coeffi-
cients of ft in the two members are equal.
Now if b = a, that is, if equation (1) has two roots equal to «, every
term of (4) will be divisible by x — a, and therefore the expression (3)
will be divisible by x — a.
Hence, the equation formed by equating (3) to zero will have one root
equal to a (§ 315).
In like manner, if c = b = a, that is, if (1) has three roots
equal to a, the equation formed by equating (3) to zero will
have two roots equal to a ; and so on.
Hence, if any equation of the form (1) has m roots
equal to a, the equation formed by equating to zero the
derivative of its first member -will have m — 1 roots equal
to a.
344. It follows from § 343 that, to determine the existence
of multiple roots in an equation of the form
PoX n +PlX n - 1 + ••' +Pn-\X+Pn = 0,
we proceed as follows :
THEORY OF EQUATIONS 251
Find the II. C. F. of the first member and its derivative.
If there is no H. C. F., there can be no multiple roots.
If there is a II C. F., by equating it to zero and solving the
residting equation, the required roots may be obtained.
It is to be observed that the number of times that each root
occurs in the given equation exceeds by one the number of
times that it occurs in the equation formed by equating the
H. C. F. to zero.
Ex. Find all the roots of
a 5 +0* - 9 a? - 5a? + 16s + 12 = 0. (1)
The derivative of the first member is
5z 4 + 4x 3 -27x 2 -10x4 16.
The H. C. F. of this and the first member of (1) is x 2 — x — 2.
Solving the equation x 2 — x — 2 = 0, x = 2 or — 1.
Then, the multiple roots of (1) are 2, 2, — 1, and — 1.
Subtracting the sum of 2, 2, — 1, and — 1 from — 1, the remaining
root is - 3 (§ 321).
EXERCISE 85
Find all the roots of the following :
2. x* + 6x 3 -llx 2 -60x + 100 = 0.
3. 9a? + 105 a* + 343 x +343 =0,
4. 4x i + 32x s + 63x 2 -8x-16 = 0.
5. af + brt-ll ar 3 - 49 a 2 + 160 a; -100 = 0.
6. x 4 + 3a? 3 + 4a; 2 + 3aj + l = 0.
345. The equation x n — a = can have no multiple roots ;
for the derivative of x n — a is nx n ~ l , and x n — a and nx"' 1 have
no common factor except unity.
Hence, the n roots of x n = a are all different-
It follows from this that every expression has two'different
square roots, three different cube roots, and, in general, n dif-
ferent nth. roots.
252 ALGEBRA
LOCATION OF ROOTS
346. If two real numbers, a and b, not roots of the
equation
^+^ tt !+ .,.. +P nl v+p n = 0, (1)
when substituted for a? in the first member, give results
of opposite sign, an odd number of roots of the equation
lie between a and b.
Let a be algebraically greater than b.
Let d, • ••, g be the real roots of (1) lying between a and b, and h, • ••, &,
the remaining real roots.
Let x n -f piX*~ l + ••• +i?n-]X +p n be denoted by X.
Then, by § 318,
Xr=(x-d) ... (x-g) -(x-h) .-. (*-*) ■ Y; (2)
where F denotes the product of the factors corresponding to the imagi-
nary roots, if any, of (1).
Substituting a, and then &, for x in (2), the second member becomes
(a-d) ... (a-g)-(a-h) ••• ((*-*)• F, (3)
and (p- d )... (b-g).(b-h) - {b-k).Y"' y (4)
where Y f and y" denote the values of Y when sc is put equal to a and 6,
respectively.
Since a is greater than b, each of the numbers d, •••, g is less than a
and greater than b.
Whence, each of the factors a — d, •••, a — g is +, and each of the
factors b — d, • ■ • , b — g is — .
Again, since none of the numbers h, •••, k lie between a and 5, the
expression (a — K) • • • (a — k) has the same sign as the expression
(b-h) ... (b-k).
Also, y and Y" are positive ; for the product of the factors corre-
sponding to a pair of conjugate imaginary roots of (1) is positive for every
real value of x (§ 324).
But by hypothesis, the expressions (3) and (4) are of opposite sign.
Hence, the number of factors b — d, •••, b — g -must be odd;
that is, an odd number of roots lie between a and b.
If the numbers substituted differ by unity, it is evident that the inte-
gral part of at least one root is known.
THEORY OF EQUATIONS 253
Ex. Locate the roots of or 3 -j- x 2 — 6 x — 7 = 0.
By Descartes' Rule (§ 835), the equation cannot have more than one
positive, nor more than two negative roots.
The values of the first member for the values 0, 1, 2, 3, — 1, — 2, and
— 3 of x are as follows :
J5=0; -7. x = 2;— 7. a; = - 1 ; — 1. x = - 3 ; - 7.
x = l; -11. x = 3; 11. sc = - 2 ; 1.
Since the sign of the first member is — when x = 2, and -f when x = 3,
one root lies between 2 and 3.
The others lie between — 1 and — 2, and — 2 and — 3, respectively.
In locating roots by the above method, first make trial of the numbers
0, 1, 2, etc., continuing the process until the number of positive roots de-
termined is the same as has been previously indicated by Descartes' Rule.
Thus, in the above example, the equation cannot have more than one
positive root ; and when one has been found to lie between 2 and 3, there
is no need of trying 4, or any greater positive number.
The work may sometimes be abridged by finding a superior limit to
the positive roots, and an inferior limit to the negative roots of the given
equation (§§ 339, 340), for no number need be tried which does not fall
between these limits.
EXERCISE 86
Locate the roots of the following :
i. x? + 4:X 2 -6 = 0. 5. x 4 + 3x*-±x-l = 0.
2 . a?-7 x * + 6x + 5 = 0. 6. x 4 + ar*-19 x 2 - 17 x + 1 = 0.
3. ^ + 3^-7^ + 2 = 0. 7- x 4 -4x 3 + 6 z-2 = 0.
4. f + 4a; 2 + «-3 = 0. 8. # 4 - 7 ^ + # + 4 = 0.
9. Prove that the equation x 4 — 5x? — 7 x — 2 = has one
root between 2 and 3, and at least one between and —1.
10. Prove that the equation x 4 — 3 xP + x 2 — 3 x — 4 = has
one root between and — 1, and at least one between 3
and 4.
11. Prove that the equation ar ? - r -5a;-f-4 = has one root
between and — 1.
254
alc;ebra
347. The method of § 346 is not sufficient to deal with
every problem in location of roots.
Let it be required, for example, to locate the roots of
By § 325, the equation has at least one real root.
By Descartes' Rule, it has no positive root.
Putting x equal to 0, — 1, — 2, — 3, the corresponding values
of the first member are 1, 1, 1, and — 5, respectively.
Then, the equation has either one root or three roots between
— 2 and — 3 ; but the methods already given are not sufficient
to determine which.
Sturm's Theorem (§ 350) affords a method for determining
completely the number and situation of the real roots of an
equation.
It is more difficult of application than the method of § 346,
and should be used only in cases which the latter cannot
resolve.
348. Graphical Representation.
The graph of an expression of higher degree than the sec-
ond, with one unknown number, may be found as in § 51.
Ex. Find the graph of
[f( x) . , x>- 2 x 2 -2 z + 3.
Put/O) = z 3 - 2 x 2 - 2 x + 3.
If 3=0, /(a) =3.
If x = l,/(x) = 0.
Ifx = 2,/(x) = -l.
If x = -2,/(x) = -9.
Ifx=-l,/(x)=2.
If x = 3,/(x) =6.
etc.
The graph is the curve ABC, which extends in either direction to an
indefinitely great distance from XX.
THEORY OF EQUATIONS 255
349. Graphical Location of Roots.
The principle of § 220 holds for the graph of the first mem-
ber of an equation of higher degree than the second, with one
unknown number.
Thus, the graph of § 348 intersects XX' at x = 1, between
x = 2 and x = 3, and between x = — 1 and x = — 2.
And the equation X s — 2 x 2 — 2^+3=0 has one root equal
to 1, one between 2 and 3, and one between — 1 and — 2.
This may be verified by solving the equation ; the factors of the first
member are x — 1 and x 2 — x — 3.
This method of locating roots is simply a graphical repre-
sentation of the process of § 346, and is subject to the limita-
tions stated in § 347.
If the graph is tangent to XX', the equation has two or
more equal roots (compare § 220, Fig. 2) ; if it does not inter-
sect XX', the equation has no real root.
The note to the example of § 346 applies with equal force to the
graphical method of locating roots.
EXERCISE 87
Locate the roots of the following graphically :
i. a^ -3* — 1 = 0. 4- x 3 - 8 ^ + 19 a -12 = 0.
2. a 4 + 2z 2 + 3 = 0. 5 . aj 8 + 7aj 2 4-14a? + 8 = 0.
3. ^-7ar 9 + 12a-5 = 0. 6. x'-Sx 2 - 2 x + 5 = 0.
350. Sturm's Theorem.
Let a^+jPi^-f ••• -fiV-i#+P» = (1)
be an equation from which the multiple roots have been re-
moved (§ 343).
Let x n -{- p^ 91 " 1 + ••• +p n ~ l x+p n be denoted by f(x), and let
f(x) denote the first derivative of f(x) (§ 341).
Dividing f(x) by f(x), we shall obtain a quotient #i> with
a remainder of a degree lower than that of fi(x).
Denote this remainder, ivith the sign of each of its terms
256 alokbka
changed, by f 2 (x) 9 and divide f(x) by f(x) 9 and so on ; the
operation being precisely the same as that of finding the
H. C. F. of f(x) and fi(x) 9 except that the signs of the terms
of each remainder are to be changed, while no other changes
of sign are permissible. '
Since, by hypothesis, fix) = has no multiple roots, f(x)
and fi(x) have no common divisor except 1 (§ 343) ; and we
shall finally obtain a remainder/,^) independent of x.
The expressions f(x),f 1 (x) 9 f 2 (x), •••,/„(»), are called Sturm's
Functions.
The successive operations are represented as follows :
f(x) = Q l f(x)-f 2 (x) 9 (2)
fx(p) = Q 2 f 2 (x) -f(x) 9 (3)
Mx) = Qsf*(x)-fi(x), ( 4 )
fn-2(x) = Q n _ } f n . 1 (x)-f n (x).
We may now enunciate Sturm's Theorem :
Let two real numbers, a and b, be substituted in place
of x in Sturm's Functions, and the signs noted.
The difference between the number of variations of
sign (§ 334) in the first case and that in the second is
equal to the number of real roots of /(as) =0 lying be-
tween a and b.
The proof of the theorem depends upon the following
principles :
I. Tivo consecutive functions cannot both become for the
same value of x.
For if, for any value of x 9 f(x)—0 and f 2 (x) = 0, then by
(3), ./<$(#) = ; and since f 2 (x) = and f(x) = 0, by (4) f 4 (x) = ;
continuing in this way, we shall finally have/„(#) = 0.
But by hypothesis, f n (x) is independent of x 9 and conse-
quently cannot become for any value of x.
Hence, no two consecutive functions can become for the
same value of x.
THEORY OF EQUATIONS 257
II. If any function, except f(x) and f n (x), becomes for any
value of x, the adjacent functions have opposite signs for this
value of x.
For if, for any value of x, f 2 (x) = 0, then, by (3), we must
have f(x) = —f 3 (x) for this value of x.
Therefore,/^) and f(x) must have opposite signs for this
value of x ; for, by I, neither of them can equal zero.
III. Let c be a root of the equation f(x) = 0, where f(x)
is any function except f(x) aj\o\f n (x).
By Ilyf^x) &ndf r+1 (x) have opposite signs when x= c.
Let h be a positive number, so taken that no root of f r -i(x)
= 0, or f r+ i(x) = lies between c — h and c + h.
Then, as x changes from c—h to c + k, no change of sign
takes place in f r _i(x), or f r+ i(x) ; while f r (x) reduces to zero,
and changes or retains its sign according as the root c occurs
an odd or even number of times in f(x) = 0.
Therefore, for values of x between c — h and c, and also
for values of x between c and c + h, the three functions
f-i( x )> fX x )> an( i fr+i( x ) present one permanence and one
variation. *
Hence, as x increases from c—h to c + h, no change occurs •
in the number of variations in the functions f_i(x), f r (x), and
f r+ i(x) ; that is, no change occurs in the number of variations
as x increases through a root of f r (x) = 0.
IV. Let c be a root of the equation f(x) = ; and let h be a
positive number so taken that no root of f(x) = lies between
c — h and c + h.
Then as x increases from c — h to c -f h, no change of sign
takes place in f(x), while f(x) reduces to zero, and changes
sign.
Now if we put x = c — h in (1), the first member becomes
Expanding the powers of c — h by the Binomial Theorem,
258 ALGEBRA
and collecting the terms involving like powers of h, we
have .
c 1i +p l c n * + ••• +p n -ic+p n
-hlnc^ + in-l^c" 2 + ••• +p n -i]
+ terms involving h 2 , h 3 , • ••, h n . (5)
But since c is a root of f(x) = 0, we have by (1),
Also, it is evident that the coefficient of — h is the value of
f x (x) when c is substituted in place of x; let this be denoted by
A ; then (5) reduces to
— JiA -f terms involving 7i 2 , 7i 3 , • • •, h n . (6)
In like manner, the value of f(x) when x is put equal to
c + h, is
+ 7^yl + terms involving h 2 , 7i s , • • • h n . (7)
Now, if 7i be taken sufficiently small, the signs of the ex-
pressions (6) and (7) will .be the same as the signs of their
first terms, — h A and -f hA, respectively.
Hence, if h be taken sufficiently small, the sign of (6) will
be contrary to the sign of A, and the sign of (7) will be the
same as the sign of A.
Therefore, for values of x between c — h and c, the functions
/ (x) and fi(x) present a variation, and for values of x between
c and c + h they present a permanence.
Hence, a variation is lost as x increases through a root of
the equation f(x) = 0.
We may now prove Sturm's Theorem ; for as x increases
from b to a, supposing a algebraically greater than b, a varia-
tion is lost each time that x passes through a root of f(x) = 0,
and only then; for when x passes through a root of f r (x) =0,
where f r (x) is any function except f(x) and /»(#), no change
occurs in the num ber of variations.
Hence, the number of variations lost as x increases from b
to a is equal to the number of real roots of A r =0 included
between a and b.
THEORY OF EQUATIONS 259
351. It is customary, in applications of Sturm's Theorem,
to speak of the substitution of an indefinitely great positive
number for x, in an expression, as substituting -f- co for x ; and
the substitution of a negative number of indefinitely great
absolute value as substituting — oo for x.
The substitution of + oo and — go for x in Sturm's Func-
tions determines the number of real roots of f(x) — 0.
The substitution of + oo and for x determines the number
of positive real roots, and the substitution of — oo and the
number of negative real roots.
Since Sturm's Theorem determines the number of real roots
of an equation, the number of imaginary roots also becomes
known (§ 316).
352. If a sufficiently great number be substituted in place
of x in the expression
f(x)=p d x n +p l x n ~ 1 + .-. + Pn _ lX +p n ,
the sign of the result will be the same as the sign of its first
term, p x n .
It follows from the above that :
If + oo be substituted in place of oc in /(a?), the sign of
the result will be the same as the sign of its first term.
If -oo be substituted in place of as in /(as), the sign of
the result will be the same as, or contrary to, the sign of
the first term, according as the degree of / (as) is even or
odd.
353. Examples.
i. Determine the number and situation of the real roots of
aj 3_2ar-x4-l=0.
Here, f(x) = x* -2 oc? - x + \, and f x (x) = 3 x 2 - 4 x - 1.
In the process of finding / 2 (x), / 3 (x), etc., any positive numerical
factors may be omitted or introduced at pleasure, for the sign of the
result is not affected thereby ; in this way fractions may be avoided.
260 ALGEBRA
In the present case, we multiply /(x) by 3, to make its first term
divisible by 3 x 2 .
3x 3 -4x 2 - x
-2x 2 - 2x + 3
3
-6a; 2 - 6x + 9(-2
— 6x 2 + 8x + 2
7 )-14x+7
- 2 x + 1 Then, / 2 (x) = 2 x - 1.
3 X 2_ 4 x _i
2
2x- 1)6 x 2 - 8x-2(3x
6x 2 - 3x
- 6x-2
2
_10x-4(-6
— 10 x + 5
~^9 Then,/ 3 (x)=9.
Substituting — oo for x in /(x), /i(x), /2(x), and /s(x), the signs are
— , + , — , and + , respectively (§ 352); substituting for x, the signs
are +, — , — , +, respectively ; and substituting + oo for x, the signs are
all + .
Hence, the roots of the equation are all real, and two of them are posi-
tive and the other negative (§ 351).
We now substitute various numbers to determine the situation of the
roots :
/w
AW
Mx)
/•(*)
x— —
GO,
-
+
—
+ *
3 variations.
x= —
1,
—
+
-
+
3 variations.
x = 0,
+
-
—
+
2 variations.
X = 1,
-
—
+
+
1 variation.
35 = 2,
—
+
+
+
1 variation.
x = 3,
+
+
+
+
no variation.
X = 00
»
+
+
+
+
no variation.
We then know that the equation has one root between and — 1, one
between and 1, and one between 2 and 3.
2. Determine the number and situation of the real roots of
4 x 3 — 6 x — 5 = 0.
THEORY OF EQUATIONS 261
' Here, /(x) = 4x 3 — 6x— 5; and f x (x) = 12 x 2 — 0, or 2 x 2 - 1, omitting
the factor 6.
• 2x 2 -l)4x 3 -6x-5(2x
4 x 3 - 2 x
-4x-5
2 x 2 - 1
2
Then,/ 2 (x) = 4x + 5.
4x + 5)4x 2 - 2(x
4 x 2 + 5 x
— 5x— 2
4
-20x- 8(-
- 20 x - 25
17
-5
Then, / 3 (x) = - 17.
The last step in the division may be omitted; for we only need to
know the sign of /s(x), and it is evident by inspection, when the re-
mainder — 5 x — 2 is obtained, that the sign of / 3 (x) will be — .
/(*)
/iW
h(x)
Mx)
X = — QO ,
—
+
—
—
2 variations.
x = 0,
—
-
+
—
2 variations.
x = l,
—
+
+
-
2 variations.
SB = 2,
+
+
+
-
1 variation.
X = GO ,
+
+
+
—
1 variation.
Therefore, the equation has a real root between 1 and 2, and two
imaginary roots.
In substituting the numbers, it is best to work from in either direc-
tion, stopping when the number of variations is the same as has been
previously found for + oo or — co , as the case may be.
EXERCISE 88
Determine the number and situation of the real roots of :
i. x s + 2x 2 -x-l = 0. 5- x*-4:X 2 + x+3 = 0.
2. x s + 3x-5 = 0. 6. a 4 -8a 2 -8# + l = 0.
3. x 3 -5x 2 +2x + 6 = 0. 7- a 4 + 2 ar*- 5ar-10a-3=0.
4 . x s + x 2 -15x-28 = 0. 8. a 4 -r-3ar*-3z + l = 0.
262 ALGEBRA
XV. SOLUTION OF HIGHER EQUATIONS
354. Synthetic Division (§ 107) not only abbreviates the
process of division, but its application is of importance in the
solution of many forms of higher equations containing either
commensurable or incommensurable roots.
COMMENSURABLE ROOTS
355. We use the term commensurable root, in Chapter XV,
to signify a rational root expressed in Arabic numerals.
356. By § 322, an equation of the nth degree in the general
form (§ 312), with integral numerical coefficients, cannot have
as a root a rational fraction in its lowest terms.
Therefore, to find all the commensurable roots of such an
equation, we have only to find all its integral roots.
Again, by § 320, the last term of an equation of the above
form is divisible by every integral root.
Hence, to find all the commensurable roots, we have only to
ascertain by trial which integral divisors of the last term are roots
of the equation.
The trial may be made in two ways :
I. By substitution of the supposed root.
II. By dividing the first member of the equation by the
unknown number minus the supposed root (§ 315).
In this case, the operation may be conveniently performed
by Synthetic Division (§ 107).
In the case of small numbers, such as ±1, the first method
may be the most convenient.
The second has the advantage that, when a root has been
found, the process gives at once the depressed equation (§ 317)
for obtaining the remaining roots.
Work may sometimes be saved by finding a superior limit to the
SOLUTION OF HIGHER EQUATIONS 263
positive, and an inferior limit to the negative, roots (§§ 339, 340);
for no number need be tried which does not fall between these
limits.
Descartes' Rule of Signs (§ 335) may also be advantageously employed
to shorten the process.
Any multiple root should be removed (§ 343) before applying either
method.
Ex. Find all the roots of x 4 - 15 x 2 + 10 x + 24 = 0.
By Descartes' Rule, the equation cannot have more than two positive
roots.
Changing the signs of the alternate terms commencing with the second,
we have z 4 - 15 x 2 — 10 x + 24 = 0.
Then, the given equation cannot have more than two negative roots
(§ 336).
The integral divisors of 24 are ±1, ±2, ±3, ±4, ±6, ±8, ±12, and
±24.
By substitution, we find that 1 is not, and that — 1 is, a root of the
equation.
Dividing the first member by x — 2, x — 3, etc., by the method ex-
plained in § 108, we have
1 + o - 15 + 10 + 24 j_2 1 + - 15 + 10 + 24 [3
2 4 -22-24 3 9 -18-24
2-11-12, Rem. 3 - 6 - 8, Rem.
The work shows that 2 and 3 are roots of the given equation ; and
since the equation cannot have more than two positive roots, these are
the only positive roots.
The remaining root may be found by dividing 24 by the product of
! — 1, 2, and 3 (§ 321), or by the same process as above.
Dividing the first member by x + 2, x + 3, etc., we have
1 + 0-15+10+24 [ -2 1 + 0-15+10+24 [^-3
-2 4 22 - 64 -3 9 18-84
-2-11 32-40 -3- 6 28-60
1 +o_ 15 +10 +241-4
-4 16 - 4-24
-4 16
The work shows that the remaining root is — 4.
264 ALGEBRA
357. By § 829, an equation of the nth degree in the general
form, with fractional coefficients, may be transformed into
another whose coefficients are integral, that of the first term
being unity.
The commensurable roots of the transformed equation may
then be found as in § 356.
* Ex. Find all the roots of 4 a? - 12 x 2 + 27 x - 19 = 0.
Dividing through by the coefficient of x 8 , we have
4 4
Proceeding as in § 329, it is evident by inspection that the multiplier 2
will remove the fractional coefficients ; the transformed equation is
X 3_ 2 . 3 x 2 + 22 .21* -2 3 • - = 0,
4 4
or, X s - 6 x 2 + 27 x - 38 = ; (1)
whose roots are those of the given equation multiplied by 2.
By Descartes' Rule, equation (1) has no negative root.
The positive integral divisors of 38 are 1, 2, 19, and 38.
Dividing the first member by x — 1, x — 2, etc., we have
1 _ 6 + 27 - 38 |_1_ 1 - 6 + 27 - 38 |_2
1 -5 22 2-8 38
- 5 22 - 16 - 4 19
The work shows that 2 is a root of (1).
The remaining roots may now be found by depressing the equation;
it is evident from the right-hand operation above that the depressed equa-
tion is x 2 — 4 x + 19 = 0.
Solving by the rules for quadratics, x = 2 ± V— 15.
Then, the three roots of (1) are 2 and 2 ± V— 15.
Dividing by 2, the roots of the given equation are 1 and 1 ± V— 15.
EXERCISE 89
Find all the commensurable roots of the following, and the
remaining roots when possible by methods already given.
i. ,r s -9x 2 + 23x-15 = Q. 3. aj 8 + 12a? 2 + 44a? + 48 = 0.
2 . aj3_8x 2 -h 5a> + 14 = 0. 4. x* + Ax 2 - 9»-36 = 0.
SOLUTION OF HIGHER EQUATIONS ' 265
5. 3ar 3 H-4ar>-13a;-f 6 = 0.
6. 4a 8 +16a 2 -7a>-39=0.
7. a 4 +10ar 3 + 3ox 2 + o0z + 24 = 0.
8. a 4 - 5 &» + 20 a?- 16 = 0.
9. a*-15a* + 65a?-105x + 5± = 0.
10. aj 4 + 8a^4-ll^ 2 -32a!-60 = 0.
11. x i -2x*-l7x 2 + 18x + 72 = 0.
12. 4a 4 -12a 8 -9ar 9 + 47x-30 = 0.
13. 6a* 4 -7a 8 -37a 2 + 8a;4-12 = 0.
14. ar 5 + 8a; 4 -7ar 3 -103a 2 + 69x + 18 = 0.
15. 3a 4 -f-2ar 5 -18x 2 + 8 = 0.
16. x 4 + # - 6 x 2 + 16 a? - 32 = 0.
♦ RECIPROCAL OR RECURRING EQUATIONS
358. A Reciprocal Equation is one such that if any number
is a root of the equation, its reciprocal is also a root.
It follows from the above that, if - be substituted for x in
x
a reciprocal equation, the transformed equation will have the
same roots as the given equation.
359. Let
x n +p l x n ~ 1 +p^ n - 2 + .- +p H -jx?+p n -iX+p H = (1)
be a reciprocal equation.
Putting - for x, the equation becomes
x
x 11 x n ~ x x"- 1 X 1 X
Clearing of fractions, and reversing the order of the terms,
PnX n +Pn-\X n ~ 1 +i>n-2# n ~ 2 4" '— "f P»X 2 + Pl% + 1=0.
266 • ALGEBRA
Dividing through by p n ,
x n + Pj}=_\ X n-l +P_rL=2 x n-2 + ... + P?^ + Si X + -I = 0. (2)
Pn Pn Pn Pn Pn
By § 358, this equation has the same roots as (1) ; and hence the fol-
lowing relations must hold between the coefficients of (1) and (2),
p 1 =^,p 2 =^,..-, P n-2=^,Pn-i= P -±,p n = ±-. (3 )
Pn Pn Pn Pn Pn
From the last equation, p n % = 1 ; whence, p n = ± 1.
Substituting the value of p n , the equations (3) become
Pi - ± Pn-l, P2 = ± Pn-2, "• 5
all the upper signs, or all the lower signs, being taken together.
We then have four varieties of reciprocal equations :
1. Degree odd, and coefficients of terms equally distant
from the extremes of the first member equal in absolute value
and of like sign ; as, x 3 — 2 x 2 — 2 x + 1 = 0.
2. Degree odd, and coefficients of terms equally distant
from the extremes of the first member equal in absolute value
and of opposite sign ; as, 3 x 5 -f 2 x 4 — X s + x 2 — 2 x — 3 = 0.
3. Degree even, and coefficients of terms equally distant
from the extremes of the first member equal in absolute value
and of like sign ; as, x* — 5 X s + 6 x 2 — 5 x -f- 1 = 0.
4. Degree even, and coefficients of terms equally distant
from the extremes of the first member equal in absolute value
and of opposite sign, and middle term wanting ; as,
2^ 6 + 3aj 5 -7^ 4 + 7^ 2 -3a?-2 = 0.
On account of the properties stated above, reciprocal equa-
tions are also called Recurving Equations,
360. Every reciprocal equation of the first variety may be
written in the form
SOLUTION OF HIGHER EQUATIONS 267
or, i ) (^ + l)+i>i(^- 1 +^)-hp 2 (^- 2 + ^)+ ••• =0; (1)
or, p (x n + 1) +p&(&~* + !) +P^(x n ~ 4 + 1) + ... = ;
the number of terms being even.
Since n is odd, each of the expressions x n + 1, x n ~ 2 + 1, etc., is divisible
by x+ 1.
Therefore, — 1 is a root of the equation (§ 315).
Dividing the first member of (1) by x + 1, the depressed equation is
PoO w_1 — x n ~ 2 + x n ~ s — ••• + x 2 — X + 1)
+ Pl(X n ~ 2 — X n " 3 + £ n ~ 4 — ••• +X S - x 2 + x)
+ p 2 (x n ~ s - x n ~* + x n ~ 5 - ••• + x 4 — X s + z 2 ) + ... = 0.
Or, jpo^ n_1 + (Pi - Po)£ n ~ 2 + (P2-.P1 +Po)x n ~* + ...
+ (JP2 - Pi + Po)x 2 + (Pi-po)x+p = 0;
which is a reciprocal equation of the third variety.
361. Every reciprocal equation of the second variety may
be written in the form
PoX n -f pjX** 1 + p 2 x n ~ 2 + • • • — peps 2 — p l x—p = 0,
or, p (x n - 1) -f p^x 71 - 1 - x) + p 2 (aj n " 2 - # 2 ) + • • • = 0, (1)
or, Po(x n — 1) + p&ix 71 - 2 — 1) + p 2 x 2 (x n ~ 4 — 1) + • • • =0.
Since each of the expressions x n — 1, x n ~ 2 — 1, etc., is divisible by
x — 1, + 1 is a root of the equation.
Dividing the first member of (1) by x — 1, the depressed equation is
Po(x n ~ l + x n ~ 2 H- x n ~ 3 + ... + x 2 + x + 1)
+ PiOc n ~ 2 + x n ~ 3 +x"~ 4 -f ..." + x s + a* 2 + a)
+ P20 n-3 + £"~ 4 + x n - 5 + ... +x 4 4-a^ + x 2 ) + ••• =0,
or, poo 71-1 + (Pi + Po)^ n ~ 2 + (pa -f Pi +Po)x n ~ s + •••
+ (Pi + Pi + Po)x 2 + (p -f p ).* + Po = ;
which is a reciprocal equation of the third variety.
268 ALGEBRA
362. Every reciprocal equation of the fourth variety may
be written in the form
p (x n -1)+ PiO"" 1 - *) 4-i> 2 (^" 2 - a 8 ) 4- • • • = 0, (1)
or, p (x n - 1) +PxX{ar-* - 1) +p&\x n ~* - 1) 4- • • • = ;
the number of terms being even (§ 359).
Since each of the expressions x n — 1, x n ~ 2 — 1, etc., is divisible by
x 2 — 1, both 1 and — 1 are roots of the equation.
Dividing the first member of (1) by x 2 — 1, the depressed equation is
p (x n ~ 2 + x n ~ 4 + x n ~ & + • •• 4 z 4 4 x 2 4- 1)
+ lh(x w - 8 4 £ TC-5 4 x n " 7 + h x 5 4 « 3 + £)
4 P2(x n ~ 4 4 £ w_6 4 x n ~ 8 4- ... 4 x 6 4 x 4 4 x 2 ) + •• • = 0,
or, i> ^ n - 2 + piz*- 3 4 (i>2 4 Po)x TO " 4 4 —
+ (p 2 4p )^ 2 +W» + Po = ° ;
which is a reciprocal of the third variety.
363. Every reciprocal equation of the third variety
may be reduced to an equation of half its degree.
Let the equation be
i>o^ m 4-Pi^ m-1 4- r-p m _2^ m+2 +i>m-i^ w+1 +i>m^
+P m -ix m - 1 4-i> m _2^ w " 2 + ••• +PiX +p Q = 0.
Dividing each term by x m , the equation may be written
Po[x m + ±-)+ Pl (x m - l + -^—)+ ...
V x™) \ x m -^j
+p m - 2 (x 2 4 i] +*-i(* +£) +P- = 0. (1)
Put x+-=y.
x
Then, a; 2 4 ~ = fa +-V- 2 = y 2 - 2 ;
x 2 \ x]
= y(y 2 -2)-y = y3-3y;
SOLUTION or HIGHER EQUATIONS 269
= y(y s - 3 y)- (y 2 - 2) = y* - 4 y* + 2 ; etc.
In general,
x'- V &/\ z' -1 / \ x r - 2 J
an expression of the rth degree with respect to y.
Substituting these values in (1), the equation takes the form
w m + + 9a 2 + 12 a? -144 = 0.
3. ar* + 72o; + 152 = 0. 8. tf + x 2 - 3a + 36 = 0.
4. ar l -12a 2 + 21a-10 = 0. 9. a?-2a?-15x + 36 = 0.
5. .^-3a; 2 + 48aj + 52 = 0. 10. ^-4^ + 8 x- 8 = 0.
11. Find one root of x 3 + x — 2 = 0.
a 3 fo 2
370. If a is negative, and — numerically greater than — , the
IF 2 ¥ . .
expression \-r~^~o~ 1S ima S inar y»
In such a case, Cardan's method is of no practical value ;
for there is no method in Algebra for finding the cube root of
a binomial surd.
In this case, which is called the Irreducible Case, Cardan's
method is said to fail.
It is possible, in cases where Cardan's method fails, to find
the roots by a method involving Trigonometry.
But practically it is easier to find them by the method of
§ 356, or by Horner's method (§ 374), according as the equa-
tion has or has not a commensurable root.
SOLUTION OF HIGHER EQUATIONS 273
BIQUADRATIC EQUATIONS
371. A Biquadratic Equation is an equation of the fourth
degree, containing but one unknown number.
372. Euler's Method for the Solution of Biquadratics.
l>y § 333, every biquadratic can be reduced to the form
x 4 + ax 2 + bx + c = 0. (1)
Let x = u + y + z.
Then, x 2 = u 2 + y 2 + z 2 + 2 uy + 2 yz + 2 zu,
or, x 2 — (u 2 + y 2 + z 2 ) = %(uy + yz + zu).
Squaring both members, we have
x 4 - 2 x 2 (u 2 + y 2 + 2 2 ) + (u 2 + 2/ 2 + 2 2 ) 2
-. 4(w 2 y 2 + y 2 z 2 + z 2 w 2 + 2 wy 2 ^ + 2 w 2 */z + 2 i/2 2 tt)
= 4(tt 2 y 2 + y 2 z 2 + 2 2 w 2 ) + 8 uyz(u + y + z).
Substituting x for u + y + z and transposing,
x 4 - 2 x 2 (w 2 + y 2 + a») - 8 wysx
+ O 2 + y 2 + z 2 ) 2 - 4(«V + V 2 z 2 + z 2 u 2 ) = 0.
This equation will be identical with (1) provided
a=-2(u 2 + y 2 + z 2 ), (2)
b = — 8 uyz, or w^2 = , (3)
8
and c as (?< 2 + y 2 + 2 2 ) 2 - 4 O 2 ?/ 2 + y 2 z 2 + z 2 w 2 ) • (4)
By (2), u 2 + 2/ 2 + z 2 = - £ ; and, by (3), u 2 y 2 z 2 =s£.
2 o4
Also, by (4) , u 2 y 2 + y 2 ^ 2 + z 2 u 2 = (^ 2 + ^ 2 + s 2 ) 2 - c .
4
Then, hV + */ 2 z 2 + z 2 u 2 = -i^— = ^-^ •
By § 319, the cubic equation whose roots are u 2 , y 2 , and 2 2 is
£3 _ ( tt 2 + ?/2 + 2,2)^2 + (^2 + j^S + ^2)$ _ yi^A _ 0.
Putting for it 2 + y 2 -f z 2 , w 2 y 2 + y 2 z' 2 -f z 2 ?< 2 , and u 2 y 2 z 2 , the values
given above, this becomes
t> + °fi + <£=A°t-£=o. (5)
2 16 04 v '
^74 ALGEBRA
If Z, m, and n represent the roots of this equation, we have u 2 — Z,
y 2 = m, and z 2 = n ; or, w = ± VZ, ?/ = ± Vm, 2 = ± Vtt.
Now x = u + y + z; and since each of the numbers u, y, and z has two
values, apparently x has etyfa values.
But by (3), the product of the three terms whose sum is a value of x
must be
8
Hence, the only values of x are, when b is positive,
— Vl — Vm — Vw, — y/l + Vm + Viz,
Vz~— Vm -f Vn, and Vz"-f- Vm — Vn ;
and when 5 is negative,
Vz + Vm -f Vw, Vz~— Vm — V»,
— Vz + Vm — Vra, and — y/l — Vm + Vn.
Equation (5) is called the auxiliary cubic of (1).
i2x. Solve the equation
a 4 -46ar-24a + 21 = 0.
Here, a = - 46, b - - 24, c = 21.
Whence, fl 2 - 4 c _ m and 6* = Q
16 ' 64
Then the auxiliary cubic is
£ _ 23 £ 2 + 127 * - 9 = 0.
By the method of § 356, one value of t is 9.
Dividing the first member by t - 9, the depressed equation is
4 t 2 - 14 t + 1 = 0.
Solving, t = 7 ± V49 - 1 = 7 ± 4 V3.
Proceeding as in § 193, we have
V(7 ± 4 V3) = V(4 ± 2 Vl2 + 3) = 2 ± V3.
Then since b is negative, the four values of x are
3 + 2 + V3 + 2 - V3, 3 - 2 - V3 - 2 + V3,
- 3 + 2 + V3 - 2 + V3, and - 3 - 2 - V3 + 2 - V3.
That is, 7,-1, -3 + 2 V3, and - 3 - 2 V3.
SOLUTION OF HIGHER EQUATIONS 275
EXERCISE 93
Solve the following :
i. x 4 -60x 2 + 80a; + 384 = 0.
2 . x 4 -Ux 2 + l6x + 192 = 0.
3. a 4 -40a 2 + 64a + 128 = 0.
4. x 4 - 54.x 2 -216a-243 = 0.
INCOMMENSURABLE ROOTS
373. We will now show how to find the approximate numeri-
cal values of those roots of an equation which are not com-
mensurable (§ 355).
374. Horner's Method of Approximation.
Let it be required to find the approximate value of the root
between 3 and 4 of the equation
x 3 -3x 2 -2x + 5 = 0.
We first transform the equation into another whose roots
shall be respectively those of the first diminished by 3, by the
second method explained in § 332.
The operation is conveniently performed by Synthetic Division (§ 108).
1 _3 _2 +5 [8
3 0-6
1st quotient, 1 — 2, — 1 1st Rem.
3 9
2d quotient, 1 3, 7 2d Rem.
3
6, 3d Rem.
The transformed equation is y z + 6y 2 -\- 7 y — 1=0. (1)
We know that equation (1) has a root between and 1. ^
If, then, we neglect the terms involving y* and y 2 , we may obtain an
approximate value of y by solving the equation 7 y — 1 = ; thus, approxi-
mately, y = .1 and x = 8.1.
276 ALGEBRA
Transforming (1)
into an equation
whose roots shall be
respectively
3se of (1) diminished by .1, we have
1+6
+ 7
-1
ill
.1
6.1
.61
7.61
.761
- .239
.1
6.2
.1
6.3
.62
8.23
The transformed equation is
z* + 6.3 2 2 + 8.23 2 - .239 = 0. (2)
Neglecting the z* and z 2 terms, we have, approximately,
8.23
Thus, the value of x to two places of decimals is 3.12.
The work is usually arranged in the following form, the coefficients of
the successive transformed equations being denoted by (1), (2), (3), etc.
a)
(2)
-3
(1)
-2
-2
9
7
.61
7.61
(1)
(2)
(3)
+ 5 |3.128
3
3
3
3
6
-6
-1
.761
- .239
.167128
- .071872
•1
6.1
.1
(2)
.62
8.23
.1264
6.2
8.3564
.1
.1268
6.3
(3)
8.4832
.02
6.32
.02
6.34
.02
6.36
(3)
Dividing .071872 by 8.4832, we have .008 + , and the value of x to three
places of decimals is 3. 128.
SOLUTION OF HIGHER EQUATIONS 277
The process can be continued until the root has been found
to any desired degree of precision.
We derive from the above the following rule for finding the
approximate value of a positive incommensurable root :
Find by § 346, or by Sturm's Theorem (§ 350), the
integral part of the root. (Compare § 347.)
Transform the given equation into another whose
roots shall be respectively those of the first diminished
by this integral part.
Divide the absolute value of the last term of the trans-
formed equation by the absolute value of the coefficient
of the first power of the unknown number, and write the
approximate value of the result as the next figure of the
root.
Transform the last equation into another whose roots
shall be respectively those of the first diminished by the
figure of the root last obtained, and divide as before for
the next figure of the root ; and so on.
In practice, the work may be contracted by dropping such decimal
figures from the right of each column as are not needed for the required
degree of accuracy.
In determining the integral part of the root, it will be found convenient
to construct the graph of the first member of the given equation.
375. To find an approximate value of a negative incom-
mensurable root, change the signs of the alternate terms of the
equation commencing with the second (§ 326), and find the
corresponding positive incommensurable root of the trans-
formed equation.
The result with its sign changed will be the required nega-
tive root.
376. In finding any particular root-figure by the method of
§ 374, we are liable, especially in the first part of the" process,
to get too great a result ; the same thing occasionally happens
when extracting square or cube roots of numbers.
278 ALGEBRA
Such an error may be discovered by observing the signs of
the last two terms of the next transformed equation ; for since
each root-figure obtained as in § 374 must be positive, the last
two terms of the transformed equation must be of opposite sign.
If this is not the case, the last root-figure must be diminished
until a result is obtained which satisfies this condition.
Let it be required, for example, to find the root between and — 1 of
the equation x s + 4 x 2 — 9 x — 5 = 0.
Changing the signs of the alternate terms commencing with the second
(§ 326), we have to find the root between and 1 of the equation
x s _ 4 X 2 _ 9 x + 5 = o.
Dividing 5 by 9, we have . 5 suggested as the first root-figure ; but it
will be found that in this case the last two terms of the first transformed
equation are — 12.25 and — .375.
This shows that .5 is too great ; we then try .4, and find that the last
two terms of the first transformed equation are of opposite sign.
The work of finding the first three- root-figures is shown below.
1-4 - 9 +5 1^469
.4 - 1.44 - 4.176
-3.6 -10.44 (1) .824
A - 1.28 - .713064
-3.2 (1) -11.72 (2) .110936
.4 - .1644
(1) _ 2.8 ' - 11.8844
.06 - .1608
- 2.74 (2) - 12.0452
.06
-2.68
.06
(2) - 2.62
The required root is — .469, to three places of decimals.
377. Sometimes too small a number is suggested for the
first root-figure.
Let it be required, for example, to find the root between and 1 of the
equation
s 8 - 2 x 2 + 3 x - 1 = 0.
SOLUTION OF HIGHER EQUATIONS 279
Dividing 1 by 3, we have .3 suggested as the first root-figure.
-2
+ 3
-1 ^3
.3
- .51
.747
- 1.7
2.49
- .253
.3
-1.4
- .42
2.07
.3
■
-1.1
The number suggested by the next division is greater than .1; showing
that too small a root-figure has been taken.
378. If the coefficient of the first power of the unknown
number in any transformed equation is zero, the next figure of
the root may be obtained by dividing the absolute value of the
last term by the absolute value of the coefficient of the square of
the unknown number, and then taking the square root of the result.
For if the transformed equation is if + ay 2 -f- b = 0, it is evi-
dent that, approximately, ay 2 + b = 0, or y =1
• a
We proceed in a similar manner if any number of consecu-
tive terms immediately preceding the last term are zero.
Horner's method may be used to find any root of a number approxi-
mately ; for to find the nth root of a is the same thing as to solve the
equation x 11 — a = 0.
379. If an equation has two or more roots which have the
same integral part, the first decimal root-figure of each must be
obtained by the method of § 346, or by Sturm's Theorem.
If two or more roots have the same integral part, and also
the same first decimal root-figure, the second decimal root-figure
of each must be obtained by the method of § 346, or by Sturm's
Theorem ; and so on.
Horner's method may be used to determine successive figures in the
integral, as well as in the decimal, portion of the root.
If all but one of the roots of an equation are known, the^. remaining
root may be found by changing the sign of the coefficient of the second
term of the given equation, and subtracting the sum of the known roots
from the result (§ 321).
280 ALGEBRA
EXERCISE 94
Find the root between :
i. 1 and 2, of a 8 — 9 a 2 + 23 x -16 = 0.
2. 4 and 5, of x 3 — 4 # 2 — 4 x -J- 12 = 0.
3. Oand -1, of ar* + 8 a 2 -9 a- 12 = 0.
4. -2and -3, of a^-3 a 2 - 9 # + 4 = 0.
5. 3and4, of X s - 6 x 2 + 15 x -19 = 0.
6. Oandl, of x * + x> + 2 x 2 - x-l = 0.
7. 2 and 3, of # 4 -3 ar* + 4 a- 5 = 0.
8. -land -2, of a 4 - 2^-3 a 2 + a- 2 = 0.
Find all the real roots of the following :
9. x? + 2x 2 -x-l = 0. 12. a 4 + 2ar*-5 = 0.
10. x*-2x*-7x-l = 0. 13. a? - x 2 + 2x- 1 = 0.
11. ar s -5a 2 + 2a + 6 = 0. 14. x 4 -6 x 2 + 11 x + '21 = 0.
Find the approximate values of the following:
15. ^3. 16. \$t 17. ^7.
380. We may now give general directions for finding the
real roots of any equation of the form
x n +PlX n-i + ... + Pn _ lX +p n = 0,
with integral numerical coefficients :
1. Determine by Descartes' Rule (§ 335) limits to the num-
ber of positive and negative roots.
2. Find all the commensurable roots, if any, as explained
in § 356.
3. If possible, locate the incommensurable roots by the
method of § 346.
4. If the incommensurable roots are not all located in this
way, apply Sturm's Theorem (§ 350), observing that, if the first
member and its first derivative have a common factor, the
given equation has multiple roots (§ 343).
5. Approximate to the decimal portions of the incommen-
surable roots by Horner's method (§ 374).
INDEX
Abscissa, 22.
Addition, commutative law, 1.
Affected quadratic, 128.
Aggregation, signs of, 6.
Alternation, 84.
Arithmetical complement, 55.
Arithmetic mean, 167.
Arithmetic progression, 163.
Associative law, addition, 1 ; multipli-
cation, 2.
Axioms, 5.
Binomial, cube of, 97; equations, 270;
surds, 118; theorem, 108; theorem,
rth term, 112.
Biquadratic equations, 273.
Cardan's method, 271.
Characteristic, 42.
Circle, 157.
Coefficients, 4; composition of, 234;
of determinant, 222 ; undetermined,
192.
Combinations, 203.
Commensurable roots, 262.
Common factor, 66.
Common logarithms, 42.
Common multiple, 66.
Commutative law, 1.
Complement, arithmetical, 55.
Completing the square, 128.
Complex number, 122, 126.
Composition, 85.
Composition and division, 86.
Composition of coefficients, 234.
Condition, equation of, 5.
Conjugates, 140.
Constant, 76.
Continued proportion, 83,
Convergent series, 180,
Coordinates, 22.
Cube of binomial, 97; root of num-
bers, 104; root of polynomial, 99.
Cubic equations, 270.
Degree, 11, 33.
Derivatives, 249.
Descartes' rule for signs, 243.
Determinants, 211; definition of, 214;
evaluation of, 224; minors, 220;
properties of, 216.
Difference, 4.
Differential method, 186.
Direct proportion, graph, 143.
Discussion of quadratics, 139.
Distributive law, multiplication, 3.
Divergent series, 180.
Division, synthetic, 63, 85.
Elimination, 17.
Ellipse, 158.
Equations, binomial, 270; biquadratic,
273; cubic, 270; definition, 5; equiv-
alent, 6, 11, 16; formation, 233;
higher, 262; inconsistent, 18; inde-
pendent, 18; identical, 5; integral,
10; linear, 11, 23; numerical, 10;
quadratic, 128 ; quadratic form, 133 ;
radical, 120; reciprocal, 265; simple,
11; simultaneous, 17; simultaneous
quadratic, 149; solution of, 5, 18;
theory of, 230; transformation of,
238.
Equivalent equations, 6, 11, 16.
Euler's method, 273.
Evolution, 98.
Evolution of determinants, 224.
Expansion of surds, 196.
Exponents, 32.
Expression, degree of, 10, 11 ; rational,
10.
Factors, 57, 66, 147.
Factors, type forms, 58.
281
282
INDEX
Factor theorem, 60.
Finite series, 108.
Formation of equations, 233.
Formula, quadratic, 130.
Fourth proportional, 83.
Fractional exponent, 32.
Fractional roots, 236.
Fractions, 73; generating, 185;
tial, 196; reduction of, 74.
par-
Generating fraction, 185.
Geometric, means, 171; progression,
166.
Graphs, 21, 254; direct proportion,
143; imaginaries, 125; inverse pro-
portion, 143; quadratic equations,
137, 141; simultaneous quadratics,
157.
Higher equations, 262.
Highest common factor, 66.
Hindoo method, 130.
Horner's method, 275.
Horner's synthetic division, 63.
Hyperbola, 158.
Identity, 5.
Imaginaries, 122; graphs, 125; roots,
140, 237.
Incommensurable roots, 275.
Inconsistent equations, 18.
Independent equations, 18.
Indeterminant forms, 76, 80.
Induction, mathematical, 110, 187.
Inequalities, 26.
Inferior limit, 247.
Infinite series, 108, 179.
Integral equation, 10.
Integral exponent, 32.
Interpolation, 190.
Inverse proportion, graph, 143.
inversion, 85.
Involution, 97.
Irrational number, 57.
Irrational roots, 139.
Limit, 76.
Limit to roots, 246.
Line, 22.
Linear equation, 11.
Location of roots, 252, 255.
Logarithms, 41.
Logarithm table, 50.
Lowest common multiple, 66.
Mantissa, 42.
Mathematical induction, 110, 187.
Mean proportional, 83.
Minors, 220.
Multiple, common, 66.
Multiple roots, 249.
Multiplication, commutative law, 1;
distributative law, 3,
Negative exponent, 33.
Negative sign, 8.
Number, irrational, 57.
Order of difference, 186.
Ordinate, 22.
Origin, 22.
Oscillating series, 181.
Parabola, 158.
Parentheses, 8.
Partial fractions, 196.
Permutations, 203.
Physics problems, 145.
Piles of shot, 189.
Point, 21.
Polynomial, cube root of, 99; square
of, 97 ; square root of, 98.
Positive sign, 8.
Powers of i, 123.
Progressions, 163.
Properties of determinants, 216; of
inequalities, 27 ; of logarithms, 44.
Proportion, 83.
Pure imaginary, 122.
Pure quadratic, 128.
Quadratic equations, 128; discussion
of, 139; graph of, 137, 141; theory
of, 136.
Quadratic, factoring, 147; formula,
130; surds, 33, 117.
Radical equations, 120.
Ratio, 82.
Rational expression, 10.
Reciprocal equation, 2(>5.
INDEX
283
Recurring equations, 265.
Reduction of fractions, 74.
Remainder theorem, 60.
Reversion of series, 202.
Roots, 5; commensurable, 262; extrac-
tion of, 97, 98, 102, 117 ; fractional,
236; imaginary, 140, 237; incom-
mensurable, 275; limits of, 246;
location of, 252, 255; multiple, 249.
rth term, 112.
Scale of relation, 185.
Series, 108, 163, 183 ; convergent, 180 ;
recurring, 182; reversion of, 202;
summation of, 182.
Shot, piles of, 189.
Similar terms, 4.
Simple equations, 11.
Simultaneous equations, 17, 149;
quadratic equations, 149.
Solution, 5, 18; by determinants, 228.
Square, completion of, 128.
Square of numbers, 102; of polyno-
mial, 97 ; root of polynomial, 98.
Straight line, 23.
Sturm's theorem, 255.
Summation of series, 182.
Superior limit, 246.
Surds, 33, 117, 118 ; expansion of, 196.
Symmetrical forms, 150.
Synthetic division, 63.
Systems of equations, 16.
Tables, Logarithm, 50.
Term, rth, 112.
Terms, similar, 4.
Theorem, binomial, 108; Sturm's, 255.
Theory of equations, 230.
Theory of quadratic equations, 136.
Third order of determinants, 212.
Third proportional, 83.
Transformation of equations, 238.
Type forms, factors, 58.
Undetermined coefficients, 192.
Variable, 76.
Variation, 91.
Zero exponent, 33.
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