Explanation

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