if the roots of the equation `1/ (x+p) + 1/ (x+q) = 1/r` are equal in magnitude but opposite in sign, show that p+q = 2r & that the product of roots is equal to `(-1/2)(p^2+q^2)`.

If the roots of the equation 1/(x+p) + 1/(x+q) = 1/r are equal in magnitude but opposite in sign and its product is alpha

If the roots of the equation 1/(x+p) + 1/(x+q) = 1/r are equal in magnitude but opposite in sign and its product is alpha

IF the roots of the equation (1)/(x+p) +(1)/(x+q) =(1)/(r ) are equal in magnitude and opposite in sign , then p+q +r=

If the roots of the equation (1)/(x+p)+(1)/(x+q)=(1)/(r) are equal in magnitude and opposite in sign, then (p+q+r)/(r)

Let p, q and r be real numbers (p ne q,r ne 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.

If the roots of the equation 1/(x+p)+1/(x+q)=1/r are equal in magnitude and opposite in sign, then (A) p+q=r (B)p+q=2r (C) product of roots=- 1/2(p^2+q^2) (D) sum of roots=1