If the roots of the equation  $\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}$                         

are equal in magnitude but opposite in sign, then their product is

The equation (1) can be written as

$c(x+b+x+a)=(x+a)(x+b)$

or $x^2+(a+b-2c)x+ab-ac-bc=0$                    (2)

Let $\alpha$ and $-\alpha$ be the roots of (2) then

$0=\alpha +(-\alpha )=a+b-2c\Rightarrow c=\frac{1}{2}(a+b)$

Also, $2\alpha (-\alpha )=2ab-2(a+b)c=2ab-(a+b{)}^2$

$\begin{array}{l}=-\left(a^2+b^2\right)\\ \Rightarrow \alpha (-\alpha )=-\frac{1}{2}\left(a^2+b^2\right)\end{array}$