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If 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 product of the roots will be

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a

\frac{p^{2} + q^{2}}{2}

b

- \frac{(p^{2} + q^{2})}{2}

c

\frac{p^{2} - q^{2}}{2}

d

- \frac{(p^{2} - q^{2})}{2}

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detailed solution

Correct option is B

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

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