Question

The number of real roots of the equation \[ \frac{\sqrt{x}}{\sqrt{1 - x}} + \frac{\sqrt{1 - x}}{\sqrt{x}} = \frac{13}{6} \] is:

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

Convert expressions using symmetric substitutions for radicals.

Answer 1

The Correct Option is B

Let $\sqrt{x} = a$, then $\sqrt{1 - x} = b$, and $a^2 + b^2 = 1$ Given: \[ \frac{a}{b} + \frac{b}{a} = \frac{13}{6} \Rightarrow \frac{a^2 + b^2}{ab} = \frac{13}{6} \Rightarrow \frac{1}{ab} = \frac{13}{6} \Rightarrow ab = \frac{6}{13} \] So: \[ (a^2 + b^2)^2 = a^4 + b^4 + 2a^2b^2 = 1 \Rightarrow \text{Use identity to find values satisfying this} \] Thus 2 roots satisfy the original domain condition $(0, 1)$.