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

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

Vedclass · Accepted Answer

$p^2 + q^2$

Vedclass · Answer

(B) Given the equation: $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$Simplify the left side: $\frac{x + q + x + p}{(x + p)(x + q)} = \frac{1}{r}$Cross-multiply: $r(2x + p + q) = x^2 + (p + q)x + pq$Rearrange into standard quadratic form: $x^2 + (p + q - 2r)x + (pq - pr - qr) = 0$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 for a quadratic equation $ax^2 + bx + c = 0$,the sum of roots is $-b/a$. Thus,$-(p + q - 2r) = 0$,which means $p + q = 2r$.The sum of squares of the roots is $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.Since $\alpha + \beta = 0$,this simplifies to $\alpha^2 + \beta^2 = -2\alpha\beta$.From the product of roots formula,$\alpha\beta = pq - pr - qr$.Substituting $\alpha\beta$: $\alpha^2 + \beta^2 = -2(pq - pr - qr) = -2pq + 2pr + 2qr$.Since $2r = p + q$,substitute $2pr + 2qr = 2r(p + q) = (p + q)(p + q) = p^2 + 2pq + q^2$.Therefore,$\alpha^2 + \beta^2 = -2pq + (p^2 + 2pq + q^2) = p^2 + q^2$.

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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