If the roots of the equation $\frac{1}{\mathrm{x}+\mathrm{p}}+\frac{1}{\mathrm{x}+\mathrm{q}}=\frac{1}{\mathrm{r}}$are equal in magnitude but opposite in sign, then the product of the roots will be

Given equation $\frac{1}{\mathrm{x}+\mathrm{p}}+\frac{1}{\mathrm{x}+\mathrm{q}}=\frac{1}{\mathrm{r}}$can be rewritten as
${\mathrm{x}}^2+\mathrm{x}(\mathrm{p}+\mathrm{q}-2\mathrm{r})+\mathrm{pq}-\mathrm{pr}-\mathrm{qr}=0$……………………….(i)
Let roots are $\alpha$ and - $\alpha$, then the product of roots
$-{\mathrm{\alpha }}^2=\mathrm{pq}-\mathrm{pr}-\mathrm{qr}=\mathrm{pq}-\mathrm{r}(\mathrm{p}+\mathrm{q})$………………………(ii)
and            sum of roots, $0=-(\mathrm{p}+\mathrm{q}-2\mathrm{r})$
$\Rightarrow \mathrm{r}=\frac{\mathrm{p}+\mathrm{q}}{2}$……………………….(iii)
On solving Eqs. (ii) and (iii), we get
$\begin{array}{r}-{\mathrm{\alpha }}^2=\mathrm{pq}-\frac{\mathrm{p}+\mathrm{q}}{2}(\mathrm{p}+\mathrm{q})\\ =-\frac{1}{2}\left\{(\mathrm{p}+\mathrm{q}{)}^2-2\mathrm{pq}\right\}\\ \Rightarrow {\mathrm{\alpha }}^2=-\frac{\left({\mathrm{p}}^2+{\mathrm{q}}^2\right)}{2}\end{array}$