Question

Given the quadratic equation, x-bx-36=0, if the sum of the reciprocals of the roots is 4/9, what is the sum of the roots?

Answer 1

First we need to look for a polynomial where the recipricals of the roots of your equation are the roots. Let r1 and r2 be the roots of your equation. Then if:\[f(x)=x^2-bx-36\]it should be clear that:\[f(\frac{1}{x})\]should be the function that has the reciprocals as roots. This yields:\[f\left(\frac{1}{x}\right)=\left(\frac{1}{x}\right)^2-b\left(\frac{1}{x}\right)-36=0\]\[\iff 1-bx-36x^2 =0\]The sum of the roots of this equation is:\[-\frac{-b}{-36}=-\frac{b}{36}=\frac{4}{9}\Longrightarrow b = -16\]So our original polynomial was f(x) = x^2+16x-36, in which the sum of the roots is -16.

Answer 2

You can see this works out since the roots of the polynomial x^2+16x-36 are -18 and 2, and \[-\frac{1}{18}+\frac{1}{2}=\frac{4}{9}\]as stated in the problem.

Answer 3

Thank You!!!