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Q.
If the roots of the equation $\frac{a}{x-a}+\frac{b}{x-b}=1$ are equal in magnitude and opposite in sign, then
Complex Numbers and Quadratic Equations
Solution:
Given equation is $\frac{a}{x-a}+\frac{b}{x-b}=1$
$\Rightarrow a(x-b)+b(x-a)=(x-a)(x-b)$
$\Rightarrow x^{2}-x(a+b)+a b=a x-a b+b x-ab$
$\Rightarrow x^{2}-2 x(a+b)+3 a b=0$
So, sum of roots $=\alpha+(-\alpha)=2(a+b)$
or $a+b=0$