Let $p, q a n d r$ be real numbers $(p \neq q, r \neq 0)$ , such that 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 sum of squares of these roots is equal to :

p^{2} + q^{2}

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

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

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

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Solution

The correct option is A $p^{2} + q^{2}$
Converting the given equation in general form
we get
$(2 x + p + q) r = (x + p) (x + q) x^{2} + (p + q - 2 r) x + p q - p r - q r = 0 p + q = 2 r . . . . . . . . (1)$
$α = - β$
$(α + β)^{2} - 2 α β = α^{2} + β^{2}$
$[∵ (α + β) = 0]$
$α^{2} + β^{2} = - 2 α β$
$= - 2 (p q - p r - q r) = - 2 p q + 2 r (p + q) from (1) α^{2} + β^{2} = p^{2} + q^{2}$

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