Let p, q and r be real numbers (p ne q, r ne | vedclass

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(B) Given the equation: $\frac{1}{x+p} + \frac{1}{x+q} = \frac{1}{r}$Simplifying the left side: $\frac{x+q+x+p}{(x+p)(x+q)} = \frac{1}{r}$$\frac{2x+p+

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$p^2 + q^2$

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(B) Given the equation: $\frac{1}{x+p} + \frac{1}{x+q} = \frac{1}{r}$Simplifying the left side: $\frac{x+q+x+p}{(x+p)(x+q)} = \frac{1}{r}$$\frac{2x+p+q}{x^2+(p+q)x+pq} = \frac{1}{r}$Cross-multiplying: $r(2x+p+q) = x^2+(p+q)x+pq$Rearranging into standard quadratic form $ax^2+bx+c=0$:$x^2 + (p+q-2r)x + (pq-pr-qr) = 0$Let the roots be $\alpha$ and $\beta$. Since the roots are equal in magnitude but opposite in sign,$\alpha = -\beta$,which implies $\alpha + \beta = 0$.From the sum of roots formula,$\alpha + \beta = -(p+q-2r)$.Setting this to zero: $p+q-2r = 0$,so $p+q = 2r$.We need to find the sum of squares of the roots,$\alpha^2 + \beta^2$.Since $\alpha = -\beta$,$\alpha^2 + \beta^2 = \alpha^2 + (-\alpha)^2 = 2\alpha^2$.Also,$\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$.Since $\alpha+\beta = 0$,the sum of squares is $-2\alpha\beta$.From the product of roots formula,$\alpha\beta = pq-pr-qr$.Thus,$\alpha^2 + \beta^2 = -2(pq - pr - qr) = -2pq + 2pr + 2qr$.Since $2r = p+q$,substitute $2pr+2qr = 2r(p+q) = (p+q)(p+q) = (p+q)^2$.So,$\alpha^2 + \beta^2 = -2pq + (p+q)^2 = -2pq + p^2 + q^2 + 2pq = p^2 + q^2$.

Let $p, q$ and $r$ be real numbers $(p \ne q, r \ne 0),$ 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:

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