Question

Solve $2\sqrt{\frac{x}{a}}+3\sqrt{\frac{a}{x}}=\frac{b}{a}+\frac{6a}{b}$

• A. $\frac{b}{2a}$ or $\frac{3a}{b}$
• B. $\frac{2a}{b}$ or $\frac{b}{3a}$
• C. $\frac{b^2}{4a}$ or $\frac{9ab}{b^2}$
• D. $\frac{b^2}{4a}$ or $\frac{9a^2}{b^2}$

Answer 1

The correct option is D $\frac{b^2}{4a}$ or $\frac{9a^2}{b^2}$

Let $\sqrt{\frac{x}{a}}=y$. The $\sqrt{\frac{a}{x}}=\frac{1}{y}$. Therefore,
$2y+\frac{3}{y}=\frac{b}{a}+\frac{6a}{b}$
or $2ab{y}^2-6a^{2}y-b^{2}y+3ab=0$ or $(2ay-b)(by-3a)=0$
i.e., $y=\frac{b}{2a}$ or $\frac{3a}{b}$
$\therefore \frac{x}{a}=\frac{b^2}{4a^2}$ or $\frac{9a^2}{b^2}$, i.e., $x=\frac{b^2}{4a}$ or $\frac{9a^3}{b^2}$