Let p, q and r be real numbers (p ≠ q, r ≠ 0) , such that the roots...

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

Let $\alpha and$\beta$be the roots of given equation.</p><p>$⇒ \left(2x + p + q\right)r = \left(x + p\right) \left(x + q\right)

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

Now, sum of $roots \left(\alpha + \beta\right) = \dfrac{- b}{a} =- \left(p + q - 2r\right)</p><p>$⇒ - \left(p + q - 2r\right)$= 0$ ($∵Given that roots are equal in magnitude and opposite in sign)$

$⇒ p + q = 2r . . . \left(1\right)$

Product of roots \left(\alpha\beta\right)=\dfrac{c}{a} = pq - pr - qr$</p><p>Now,${\alpha}^{2} + {\beta}^{2} = {\left(\alpha + \beta\right)}^{2} - 2\alpha\beta$</p><p>$= 0 - 2 \left[pq - pr - qr\right] = - 2pq + 2r \left(p + q\right)$</p><p>$= - 2pq + {\left(p + q\right)}^{2} = {p}^{2} + {q}^{2} (∵$from \left(1\right))$

$= {p}^{2} + {q}^{2}$