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 $r(x+q) + r(x+p) = (x+p)(x+q)$.$rx + r

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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 $r(x+q) + r(x+p) = (x+p)(x+q)$.$rx + rq + rx + rp = x^2 + px + qx + pq$.Rearranging the terms to form a quadratic equation: $x^2 + x(p + q - 2r) + (pq - pr - qr) = 0$.Let the roots be $\alpha$ and $-\alpha$ as they are equal in magnitude but opposite in sign.The sum of the roots is $\alpha + (-\alpha) = 0$.From the quadratic equation,the sum of roots is $-(p + q - 2r) = 0$,which implies $p + q - 2r = 0$,or $r = \frac{p + q}{2}$.The product of the roots is $\alpha \cdot (-\alpha) = -\alpha^2$.From the quadratic equation,the product of roots is $pq - r(p + q)$.Substituting $r = \frac{p + q}{2}$ into the product expression:Product $= pq - \frac{p + q}{2}(p + q) = pq - \frac{(p + q)^2}{2} = \frac{2pq - (p^2 + 2pq + q^2)}{2} = \frac{2pq - p^2 - 2pq - q^2}{2} = -\frac{p^2 + q^2}{2}$.

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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