MathematicsMedium129×since 2002Q5129Let p, q and r be real numbers (p ≠\ne= q, r ≠\ne= 0), such that the roots of the equation 1x+p+1x+q=1r{1 \over {x + p}} + {1 \over {x + q}} = {1 \over r}x+p1+x+q1=r1 are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to :Ap2+q22{{{p^2} + {q^2}} \over 2}2p2+q2Bp² + q²C2(p² + q²)Dp² + q² + r²Check answerSkip