Question

Solve for $x$ : $\frac{\sqrt{4x+1}+\sqrt{x+3}}{\sqrt{4x+1}-\sqrt{x+3}}=\frac{4}{1}$

Answer 1

We have,

$\frac{\sqrt{4x+1}+\sqrt{x+3}}{\sqrt{4x+1}-\sqrt{x+3}}=\frac{4}{1}$

Applying componendo and dividendo rule,

$\frac{\sqrt{4x+1}+\sqrt{x+3}+\sqrt{4x+1}-\sqrt{x+3}}{\sqrt{4x+1}+\sqrt{x+3}-\sqrt{4x+1}+\sqrt{x+3}}=\frac{4+1}{4-1}$

$\frac{2\sqrt{4x+1}}{2\sqrt{x+3}}=\frac{5}{3}$

$\frac{\sqrt{4x+1}}{\sqrt{x+3}}=\frac{5}{3}$

On squaring both sides, we get

$\frac{4x+1}{x+3}=\frac{5}{3}$

$12x+3=5x+15$

$12x-5x=15-3$

$7x=12$

$x=\frac{12}{7}$

Hence, this is the answer.