The number of real roots of the equation $$A^2/x + B^2/(x-1) = 1$$ , where A and B are real numbers not equal to zero simultaneously, is

The given equation can be written as : $$A^2 * (x-1) + B^2 * x = x^2 - x$$
=> $$x^2 + x(-1 - A^2 - B^2) + A^2 = 0$$
Discriminant of the equation = $$ (-1 - A^2 - B^2) ^2 - 4A^2$$
=$$ A^4 + B^4 + 1 - 2A^2 + 2B^2 + 2A^2B^2$$
= $$ A^4 + B^4 + 1 - 2A^2 - 2B^2 + 2A^2B^2 + 4B^2$$
= $$ (A^2 + B^2 - 1)^2 + 4B^2$$
>= 0, 0 when B =0 and A =1
Hence, the number of roots can be 1 or 2.
Option d) is the correct answer.