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

The equation is (x+q + x+p) / (x+p)(x+q) = 1/r, which simplifies to r(2x + p + q) = x^2 + (p+q)x + pq. Since roots are equal in magnitude but opposite in sign, their sum is 0. The sum of roots is -(p+q-2r) = 0, so p+q = 2r. The product of roots is pq - rq = -k^2. Solving for the sum of squares of roots (k^2 + (-k)^2 = 2k^2) leads to p^2 + q^2.