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of the last term be -, one root will be +1. For the proposed equation and the reciprocal have one root, the same in each, and 1 is the only quantity whose reciprocal is the same quantity; hence, since each of the other roots has the same sign as its reciprocal, the product of each root and its reciprocal must be positive; and, therefore, the last term of the equation, being the product of all the roots with their signs changed, must have a contrary sign to that of the root unity.

Hence a recurring equation of an odd degree may always be depressed to an equation of the next lower degree by dividing it by x+1, or x—1, according as the sign of the last term is + or -.

Corollary 5.-A recurring equation of an even degree may always be depressed to another of half the dimensions. For let the equation be

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dividing by r", and placing the first and last, the second and last but one, &c., in juxtaposition, we have

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the resulting equation is of the form

y+By+By+..... B-13+B=0;

and the original equation is reduced to an equation of half the dimensions.

EXAMPLES.

(1) Transform the equation 23-7x+7=0 into another whose roots shall be less than the reciprocals of those of the given equation by unity.

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.. 723+1422+7z+1=0 is the equation sought, where z+1:

Z

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or

(2) Find the roots of the recurring equation

x3—6x1+5x3+5x2-6x+1=0.

By Cor. 4, this equation has one root x=-1, and the depressed equation is x-7x3+12x2—7x+1=0.

Divide by x2, and arrange the terms as in Cor. 5; then

1

Put x+

x2+1/2 −7(x+1)+12=0. . . (A)
2+0-7

1

==z; then x2+=z2-2; hence, by substitution, (A) becomes

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22-2-7z+12=0;

22-7z+10=0;

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(3) Give the equation whose roots are the reciprocals of the roots of the equation

x-3x5-2x+3x3+12x2+10x-8=0.

(4) Find the roots of the recurring equation

5y5—4y1+3y3-3y2+4y—5—0.

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PROPOSITION III.

275. To transform an equation into another whose roots shall be any proposed multiple or submultiple of the roots of the given equation.

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Let x+A1x1+A 2x"¬2+... An-1+A=0 be any equation; then putting

y

y=mx, we have x= and by substituting this value of x in the given equa

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tion, and multiplying each term by m2, we have

y"+mà ̧y"—1+m2 à ̧y2+.... m2-1A-1y+m"A ̧=0;

an equation whose roots are m times those of the proposed equation. Hence we have simply to multiply the second term of the given equation by m, the third by m2, the fourth by m3, and so on, and the transformation is effected.

Corollary 1.-If the coefficient of the first term be m, then, suppressing m in the first term, making no change in the second, multiplying the third by m, the fourth by m2, and so on, the resulting equation will have its roots m times those of the given equation.

Corollary 2.-Hence, if an equation have fractional coefficients, it may be changed into another having integral coefficients, by transforming the given equation into another whose roots shall be those of the proposed equation multiplied by the product of the denominators of the fractions.

Corollary 3.-If the coefficients of the second, third, fourth, &c., terms of an equation be divisible by m, m2, m3, and so on, respectively, then m is a common measure of the roots of the equation.

EXAMPLES.

(1) Transform the equation 2x3-4x2+7x-3=0 into another whose roots shall be three times those of the proposed equation.

(2) Transform the equation 42—3x3—12x2+5x-1=0 into another whose roots shall be four times those of the given equation.

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(3) Transform the equation x+x2-7+2=0 into another whose roots shall be 12 times those of the given equation.

ANSWERS.

(1) 2x3-12x2+63x-81=0.

(2) x1—3x3—48x2+80x—64=0.
(3)x+4x2-36x+3456=0.

PROPOSITION IV.

276. To transform an equation into another whose roots shall be the squares of the roots of the proposed equation.

-2

Let +A+A 2x22+·······+An~1~+An=0 be any equation; then
-A11+Ax1¤2—......±An-1A,=0 is the equation whose roots are

the roots of the former, with contrary signs (Prop. VII., Art. 247).
Let a1, a2, A3, &c., be the roots of the former equation, and a1, —α,
&c., those of the latter; then we have

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(x+A+....)+(A12¬1+A3x-3+...)=(x—a1) (r—a2)(x-α3)....

(x+A+....)—(A1x-1+A3.23+...)=(x+a1)(x+a2)(x+a3).... Hence, by multiplying these two equations, we have

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Or

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-(A-2A, )2+(A ̧2—2A1A ̧+2A1)2— — ... &c., = (22—a2) (x2 — a22) (x2 — a32).... by actually squaring and arranging according to the powers of x. Now, for a write y, and we have

y"-(A-2A)y"1+(Ag2—2A1A3+2A1)}a ̈2—..... &c., =(y—a2)(y—af)

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...y"—(A,—2A)yTM¬1+(A ̧2—2A‚‚†2A,) yoo— ....=0 is an equation

whose roots are the squares of the roots of the given equation.

EXAMPLES.

(1) Transform the equation r3+3x2—6x—8=0 into another whose roots are the squares of those of the proposed equation.

or

Here x3-6x=−3x2+8 by transposition, and by squaring we have

x-12x1+36x2=9xa—48.x2+64

is the required equation.

.*. x6—21x+84x2—64=0,

y3-21y+84y-64-0

The roots of the given equation are −1, −4, 2; and those of the transformed equation are 1, 4, 16.

(2) 2+2+3x2+16x+15=0.

The transformed equation is

x+2x+33x2+23x2+166x-225=0,

which has (Art. 259) only one positive root, and therefore the proposed has only one real root.

(3) Transform the equation -x2—7x+15=0.

(4) Transform the equation 4—6x3+5x2+2x-10=0.
(5) Transform the equation *—4x3—8x+32=0.
(6) Transform the equation x-3x3-15x2+49x-12=0.

ANSWERS.

(3) y3-15y2+79y-225=0.

(4) y-26y+29y-104y+100=0.
(5) y-16y3-64y+1024=0.

(6) y1—39y3+495y2—2041y+144=0.

PROPOSITION V.

277. To transform an equation into another wanting any given term. By recurring to the transformed equation in Art. 251, note, in which the roots of the proposed are increased or diminished by a quantity represented by r, it will be seen that in order to know what value r must have to make the coefficient of any power of a disappear, it is only necessary to place the column of quantities by which that power is multiplied equal to zero, and the resulting equation, when resolved, will furnish the proper values of r. This equation will be of the 1° degree when it is required that the second term shall disappear, it will be of the 2o degree when the third is to disappear, and so on. The last term can be made to disappear only by means of an equation of the same degree as the proposed.

By removing the second term from a quadratic equation, we shall be immediately conducted to the well-known formula for its solution. Thus, the equation being

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and, that its second term may vanish, we must have

2r+A=0..r=—1A,

which condition reduces the transformed to

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which is the common formula for the solution of a quadratic equation.

PROPOSITION VI.

278. To transform an equation into one whose roots are the squares of the differences of the roots of the proposed equation.

If in the given equation, f(x)=0, we make x=a1+y, a1 being one of the roots, y will be the difference between a, and every other root. If we make x=a+y, y will be the difference between a, and every other root, and so on. But since a, a, &c., are roots of f(x)=0, they must satisfy it; hence

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If we eliminate a1 or a2, &c., between either of these equations (1) and the corresponding ones, ƒ(a1+y)=0, ƒ(a,+y)=0, &c., the final equation in y will be in each case the same, and is therefore the equation whose roots are the differences of the roots of the proposed equation. It is evidently the same thing to eliminate between f(r) and f(x+y).

The form of the equation f(x+y) is (Art. 251),

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The first term is identical with the proposed equation, and vanishes, and the whole is divisible by y; we thus deduce

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The equation (2) is of the m-1 degree, and by elimination with the proposed equation of the degree m will produce a final equation of the degree m(m—1), as will be hereafter shown. It is evident, indeed, that the roots being of the form a ̧—α, ɑ ̧—ɑ1, α1—A3, A3—A1, a2—a3, &c., will be equal in number to the permutations of m letters, two and two, which is m(m-1) (Art. 200). The factors m and m-1 will the one be even and the other odd, and the product m(m-1) must therefore necessarily be even; moreover, since if one root, a-a2, be represented by ẞ, another, a2-a1, will be represented by -ẞ, and the equation (2) will be composed of factors of the form (y-3) (y+ß)=y2—ß2; and hence will contain only even powers of y. It may therefore be written under the form

y2m+py2m2+qy2m-4+, &c., +12=0

and if we make y2=z, we have

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