jee-main 2018 Q61
Finding roots or coefficients of a quadratic using Vieta's relations
Let $p , q$ and $r$ be real numbers ( $p \neq q , r \neq 0$ ), 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
(1) $p ^ { 2 } + q ^ { 2 }$
(2) $\frac { p ^ { 2 } + q ^ { 2 } } { 2 }$
(3) $2 \left( p ^ { 2 } + q ^ { 2 } \right)$
(4) $p ^ { 2 } + q ^ { 2 } + r ^ { 2 }$