If the roots of the equation frac{1}{x + p} + | vedclass

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(B) The given equation is $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$.Multiplying by $r(x + p)(x + q)$,we get $r(x + q) + r(x + p) = (x + p)(x +

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

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(B) The given equation is $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$.Multiplying by $r(x + p)(x + q)$,we get $r(x + q) + r(x + p) = (x + p)(x + q)$.$rx + rq + rx + rp = x^2 + px + qx + pq$.$x^2 + x(p + q - 2r) + (pq - rp - rq) = 0$.Let the roots be $\alpha$ and $-\alpha$. Since the sum of the roots is zero,the coefficient of $x$ must be zero:$p + q - 2r = 0 \implies r = \frac{p + q}{2}$.The product of the roots is $\alpha(-\alpha) = -\alpha^2 = pq - r(p + q)$.Substituting $r = \frac{p + q}{2}$ into the product expression:$-\alpha^2 = pq - \frac{p + q}{2}(p + q) = pq - \frac{(p + q)^2}{2}$.$-\alpha^2 = \frac{2pq - (p^2 + q^2 + 2pq)}{2} = -\frac{p^2 + q^2}{2}$.

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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