Multiple choice

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

  1. None
  2. $1$
  3. $2$
  4. $1$ or $2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Equation: A^2(x-1) + B^2x = x(x-1) => A^2x - A^2 + B^2x = x^2 - x => x^2 - (1+A^2+B^2)x + A^2 = 0. This is a quadratic equation. The discriminant D = (1+A^2+B^2)^2 - 4A^2. Since A, B are real and not both zero, this can be positive or zero, leading to 1 or 2 real roots.

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