Question

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

• A. None
• B. 1
• C. 2
• D. 1 or 2

Hint:

When dealing with rational equations, eliminate denominators to convert the equation to a quadratic form.

Answer 1

The Correct Option is C

The given equation is: \[ \frac{A^2}{x} + \frac{B^2}{x-1} = 1. \] Multiplying both sides by \( x(x-1) \) to clear the fractions: \[ A^2(x-1) + B^2x = x(x-1). \] Simplifying: \[ A^2x - A^2 + B^2x = x^2 - x. \] Rearranging: \[ x^2 - (A^2 + B^2)x + A^2 = 0. \] This is a quadratic equation, which has two real roots. Hence, the number of real roots is 2.