Let $p,q$ and $r$ be real numbers $\left(p\ne q,r\ne 0\right)$, such that the roots of the equation $\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}$ are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to

Given quadratic equation is $\frac{1}{x+p }+\frac{1}{x+q}=\frac{1}{r}$.

Let $\alpha$ and $\beta$ be the roots of given equation.

$\Rightarrow \left(2x+p+q\right)r=\left(x+p\right)\left(x+q\right)$

$\Rightarrow x^2+\left(p+q-2r\right)x+pq-pr-qr=0$

Now, sum of roots $\left(\alpha +\beta \right)=\frac{-b}{a}=-\left(p+q-2r\right)$$$

$\Rightarrow -\left(p+q-2r\right)=0$    ($\because$Given that roots are equal in magnitude and opposite in sign)

$\Rightarrow p+q=2r ...\left(1\right)$

Product of roots $\left(\alpha \beta \right)=\frac{c}{a}=pq-pr-qr$

Now, ${\alpha }^2+{\beta }^2={\left(\alpha +\beta \right)}^2-2\alpha \beta$

$=0-2\left[pq-pr-qr\right]=-2pq+2r\left(p+q\right)$

$=-2pq+{\left(p+q\right)}^2=p^2+q^2$   $($$\because$ from $\left(1\right)$$)$

$=p^2+q^2$