Page:Elementary algebra (1896).djvu/474

456 Rh it is clear that we shall always obtain an equation having the same coefficients; this equation is therefore the original equation with some one of the roots a, b, c, \ldots k substituted for x.

Let us take for example the equation

P+ pe + px + ps=0;

and let a, b, c be the roots; then

a+b+c=— py ab + ac + be =+ py abe = — pz.

Multiply these equations by a^2, —l-a, 1 respectively and add; thus a= — pe’ — pt — Py that is, + pe + pa + ps; = 0,.

which is the original equation with a in the place of x. The above process of elimination is quite general, and is applicable to equations of any degree.

573. If two or more of the roots of an equation are connected by an assigned relation, the properties proved in Art. 570 will sometimes enable us to obtain the complete solution.

Ex. 1. Solve the equation 4x^3 — 24° + 23z + 18 = 0, having given that the roots are in arithmetical progression.

Denote the roots by a-b, a, a+b; then the sum of the roots is 3a; the sum of the products of the roots two at a time is 3a2-b2; and the product of the roots is a(a2-b2); hence we have the equations

8a=6, 3a? Y= 243, a(a? — b*) 33

from the first equation we find a = 2, and from the second b = 3, and since these values satisfy the third, the three equations are consistent. Thus the roots are -{1}{3}, 2, {9}{3}.

Ex. 2. Solve the equation 242% — 142? -—63z+445=0, one root being double another. Denote the roots by a, 2a, b; then we have

3a+b= 7, 20?+4+ 3ab =— 4, 2b =— }Y.

From the first two equations, we obtain

8a?-—2a—3=0;


 * . a=} or —}, and b = — 3 or #§.