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Q.
For the equation $\frac{1}{x+a}-\frac{1}{x+b}=\frac{1}{x+c}$, if the product of roots is zero, then sum of roots is
Complex Numbers and Quadratic Equations
Solution:
$\frac{1}{x+a}-\frac{1}{x+b}=\frac{1}{x+c}$
${\frac{b-a}{x^{2}+(b+a) x+a b}=\frac{1}{x+c}}$
or $x^{2}+(a+b) x+a b=(b-a) x+(b-a) c$
or $x^{2}+2 a x+a b+c a-b c=0$
Since product of the roots $=0$
$a b+c a-b c=0 $
$\Rightarrow a=\frac{b c}{b+c}$
Thus, sum of roots $=-2 a=\frac{-2 b c}{b+c}$