Question

Let p, q and r be real numbers $(p \neq q, r \neq 0)$, such that the roots of the equation $\frac{1}{x + \rho} + \frac{1}{x + q} = \frac{1}{r}$ are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to :

• A. $\frac{p^2 + q^2}{2}$
• B. $p^2 + q^2$
• C. $2(p^2 + q^2)$
• D. $p^2 + q^2 + r^2$

Answer 1

Answer: (B)

$(2x + p + q) r = (x + p) (x + q) $
$x_2+ (p + q - 2r) x + pq - pr - qr = 0$
$p + q = 2r$ ..............(i)
$\alpha_2 + \beta_2= (\alpha + \beta)_2 - 2 \alpha; \beta$
$= 0 - 2 [pq - pr - qr] = - 2pq + 2r (p + q) = - 2pq + (p + q)_2 = p_2 + q_2$