$\begin{array}{l} L e t p, q, r b e r e a \ln n u m b e r s (p \neq q, r \neq 0) s u c h t h a t t h e r o o t s o f t h e e q u a t i o n \frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r} a r e e q u a l i n m a g n e t u d e b u t o p p o s i t e i n s i g n, \\ t h e n t h e s u m o f t h e s q u a r e s o f t h e r o o t s i s e q u a l t o \end{array}$

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

$2 (p^{2} + q^{2})$

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

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

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r} \\ \Rightarrow \frac{x + q + x + p}{(x + p) (x + q)} = \frac{1}{r} \\ \Rightarrow (2 x + p + q) r = x^{2} + (p + q) x + p q \\ \Rightarrow x^{2} + (p + q - 2 r) x + (p q - p r - q r) = 0 \\ α + β = - (p + q - 2 r), α β = p q - p r - q r \\ R o o t s a r e E q u a l i n m a g n i t u d e a n d \\ o p p o s i t e s i g n t h e n α + β = 0 \\ \Rightarrow - (p + q - 2 r) = 0 \\ \Rightarrow p + q = 2 r \\ N o w α^{2} + β^{2} = {(α + β)}^{2} - 2 α β \\ = 0 - 2 α β \\ = - 2 α β \\ = - 2 (p q - p r - q r) \\ = 2 r (p + q) - 2 p q \\ = (p + q) (p + q) - 2 p q \\ = {(p + q)}^{2} - 2 p q \\ = p^{2} + q^{2} \end{array}$