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

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(A) Given the equation: $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$ Multiply both sides by $r(x + p)(x + q)$ to clear the denominators: $r(x + q

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

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(A) Given the equation: $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$ Multiply both sides by $r(x + p)(x + q)$ to clear the denominators: $r(x + q) + r(x + p) = (x + p)(x + q)$ $rx + rq + rx + rp = x^2 + px + qx + pq$ $2rx + r(p + q) = x^2 + (p + q)x + pq$ Rearranging into standard quadratic form $ax^2 + bx + c = 0$: $x^2 + (p + q - 2r)x + (pq - r(p + q)) = 0$ Let the roots be $\alpha$ and $-\alpha$. For a quadratic equation $x^2 + Bx + C = 0$,the sum of roots is $-B$. Here,$\alpha + (-\alpha) = 0$,so the coefficient of $x$ must be $0$. $p + q - 2r = 0$ $2r = p + q$ $r = \frac{p + q}{2}$

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 $r = ......$

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