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

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(B) Given the equation $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$.Multiplying by $r(x+p)(x+q)$,we get $r(x+q) + r(x+p) = (x+p)(x+q)$.$rx + rq +

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

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(B) Given the equation $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$.Multiplying by $r(x+p)(x+q)$,we get $r(x+q) + r(x+p) = (x+p)(x+q)$.$rx + rq + rx + rp = x^2 + px + qx + pq$.$2rx + r(p+q) = x^2 + (p+q)x + pq$.Rearranging into standard quadratic form $ax^2 + bx + c = 0$:$x^2 + (p+q-2r)x + (pq - rp - rq) = 0$.Let the roots be $\alpha$ and $-\alpha$ since 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 $p+q-2r = 0$,so $2r = p+q$.The product of the roots is $\alpha 	imes (-\alpha) = -\alpha^2$.From the equation,the product of roots is $pq - r(p+q)$.Substituting $r = \frac{p+q}{2}$ into the product:Product $= pq - (\frac{p+q}{2})(p+q) = pq - \frac{(p+q)^2}{2}$.Product $= \frac{2pq - (p^2 + 2pq + q^2)}{2} = \frac{2pq - p^2 - 2pq - q^2}{2} = -\frac{p^2 + q^2}{2}$.

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,what is their product?

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