# If $\frac{x^2-bx}{ax-c} = \frac{k-1}{k+1}$ has roots, whose magnitude is equal but signs are opposite.

If $$\frac{x^2-bx}{ax-c} = \frac{k-1}{k+1}$$ has roots, whose magnitude is equal but signs are opposite.

Answer is $$\frac{a-b}{a+b}$$

I used cross multiplication and since the roots are opposite in sign, on adding the roots, the total must be zero. But this is a long method. Please tell me any shorter method to solve this problem.

Assuming we need the value of $$k,$$
$$(k+1)x^2-x[b(k+1)-a(k-1)]+c(k-1)=0$$
So, if $$\alpha$$ is a root, $$-\alpha$$ will be the other
$$\implies \alpha+(-\alpha)=\dfrac{b(k+1)-a(k-1)}{k+1}$$
$$\implies\dfrac{k-1}{k+1}= \dfrac ab$$
• I am getting $\frac{k+1}{k-1} x^2 -x(b\frac{k+1}{k-1} -a) -c =0$ – sawan kumawat Apr 26 at 8:42