The number of real roots of the equation $\displaystyle \frac{A^{2}}{x} +\frac{B^{2}}{x-1}=1$ where A and B are real numbers not equal to zero sim ultaneously, is :

Multiply the entire equation by x(x - 1) to clear the denominators, yielding the quadratic equation x^2 - x + (1 - A^2 - B^2)x - A^2 = 0, which rearranges to x^2 - (A^2 + B^2)x - A^2 = 0. The discriminant is D = (A^2 + B^2)^2 + 4A^2, which consists of squared terms and is always positive or zero. Since A and B are not zero simultaneously, D is strictly greater than zero, guaranteeing two distinct real roots for the quadratic equation. However, because x cannot be 0 or 1 based on the original denominators, if one of the found roots happens to be exactly 0 or 1, it must be discarded, meaning the original equation has either 1 or 2 real roots.

• Thanks
• Wrong explanation