Let p, q, and r be real numbers (p≠q, r≠0), such that the roots of the equation 1/(x+p) + 1/(x+q) = 1/r are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to.

1x+p+1x+q=1rx+p+x+q(x+p)(x+q)=1r(2x+p+q)r=x2+px+qx+pqx2+(p+q𕒶r)x+pq−pr−qr=0Letαandβbe the roots.⟹α+β=−(p+q𕒶r)...[1]⟹αβ=pq−pr−qr...[2]Roots are equal in magnitude and opposite in sign⟹α+β=0.⟹−(p+q𕒶r)=0...[3]α2+β2=(α+β)2𕒶αβ=(−(p+q𕒶r))2𕒶(pq−pr−qr).. (from [1] and [2])=p2+q2+4r2+2pq𕒸pr𕒸qr𕒶pq+2pr+2qr=p2+q2+4r2𕒶pr𕒶qr=p2+q2+2r(2r−p−q).. (from [3])=p2+q2+0=p2+q2Hence, answer is option (B)