Page images
PDF
EPUB

Form of any Equation.

by any power of x[1], and used to denote the remaining

[blocks in formation]

Now the given equation is, by hypothesis, satisfied by the

[merged small][merged small][ocr errors][merged small][merged small][merged small][merged small]

The terms not multiplied by 2[1], or a power of z[1], must, therefore, cancel each other; and the first member of the given equation becomes

Px[1],

which is divisible by [1], or its equal x x'.

162. Corollary. If the equation

xn + axn-1 + &c. = 0

is divided by x -x', the first term of the quotient is n-1; and if the coefficients of x2-2, 2-3, &c. in the quotient

are denoted by a', b', &c., the quotient is

xn-1 + α' χη-2 + b xn-3 + &c.;

and the equation is

(x - x') (xn-1 + a' xn-2 + bxn-3 + &c.) = 0;

which is satisfied either by the value of x,

x = x',

or by the roots of the equation

xn-1 + a' xn-2 + bxn-3 + &c. = 0.

If now x" is one of the roots of this last equation, we may have in the same way

xn-1+a'xn-2+&c. =(x-x") (xn-2+a" xn-3+&c.)=0,

Number of the Roots of an Equation.

and the given equation becomes

(x-x') (x - x) (xn-2+a" xn-3+&c.) = 0;

which is satisfied by the value of x",

x = x";

so that z" is a root of the given equation.

By proceeding in this way to find the roots x'', x", &c., the given equation may be reduced to the form

(х - х') (х-х") (x - x) (x-x) &c. = 0, in which the number of factors х-х, х x", &c. is the same with the degree n of the given equation; and, therefore, the number of roots of an equation is denoted by the degree of the equation; that is, an equation of the third degree has three roots, one of the fourth degree has four roots, &c.

163. Scholium. Some of the roots x', x", x', &c. are often equal to each other, and in this case the number of unequal roots is less than the degree of the equation. Thus the number of unequal roots of the equation of the 9th degree,

(x - 7) (x + 4)3 (x - 1)5 = 0,

is but three, namely, 7, 4, and 1, and yet it is to be regarded as having 9 roots, one equal to 7, three equal to - 4, and five equal to 1.

[blocks in formation]

would appear to have but one root, that is,

n

Imaginary Roots.

but it must, by the preceding reasoning, have n roots, or rather, the nth root of a must have n different values.

[merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small]

must be divisible by x - 1, and we have

x3

1 = (x - 1)(x2 + x + 1) = 0.

Now the roots of the equation

are

x2 + x + 1 = 0

x=(-1+√ - 3), and = (-1--3).

Hence the required roots are

x = 1, = + (−1+√-3), and = (-1--3).

3. Find the four roots of the equation

[blocks in formation]
[blocks in formation]

must be divisible by x - 1, and we have

x5 - 1 = (x - 1) (x4 + x3 + x2 + x + 1) = 0.

Now the roots of the equation

x4 + x3 + x2 + x + 1 = 0

can be found by the following peculiar process.

Divide by x2, and we have

x2+x+1++

1

1

0.

[ocr errors]

x2

[merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

which, being substituted in the preceding equation, gives

y2 + y-1 = 0;

the roots of which are

y=(-1+√5), and = (-1-5).

But the values of x deduced from the equation

[blocks in formation]

are

Solution of Equations of a peculiar Form.

x= [y+(y2-4)], and = [y- (y2 -4)],

in which y, being substituted, gives

x = [-1-√5±√(−10 + 2√5)],

and = [-1-√5±√(-10-25)].

6. Find the six roots of the equation

x6 = 1.

Ans. x = 1, = − 1, = 1 (−1±√-3),
and = (1 + √-3).

We might proceed in the same way to higher equations, such as the 8th, 9th, 12th, &c.; but, since much more simple solutions are given by the aid of trigonometry, this subject will be postponed to a more advanced part of the

course.

165. Corollary. Before proceeding farther, we may remark that the method of solution used in the last example of the preceding article may be applied to any equation of an even degree, in which the successive coefficients of the different powers of x are the same, whether the equation is arranged according to the ascending or according to the descending powers of x, as is the case in the following equation.

Ax2 + B x2n-1 + C x 2 n −2+ &c.
+ Cx2 + B x + A = 0.

EXAMPLES.

1. Solve the equation

Ax + Bx3 + Cx2 + Bx + A = 0.

Solution. Divide by x2, and we have

« PreviousContinue »