Question

If the roots of the equation $\frac{a}{x-a}+\frac{b}{x-b}=1$ are equal in magnitude but opposite in sign then $a+b$ is

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

The correct option is D $0$

Given equation is

$\frac{a}{x-a}+\frac{b}{x-b}=1$

$\frac{a(x-b)+b(x-a)}{(x-a)(x-b)}=1$

$⟹ax-ab+bx-ab=x^2-ax-bx+ab$

$⟹x^2-2x(a+b)+2ab=0$

let the roots of the equation be $\alpha ,-\alpha$

Then sum of roots $\alpha -\alpha =\frac{2(a+b)}{1}$

$⟹a+b=0$