Question: Solve: $\dfrac{\sqrt{x} + \sqrt{7}}{\sqrt{x} - \sqrt{7}} = 3$


Solution:
Given,

$\dfrac{\sqrt{x} + \sqrt{7}}{\sqrt{x} - \sqrt{7}} = 3$

$or, \sqrt{x} - \sqrt{7} = 3(\sqrt{x} - \sqrt{7})$

$or, \sqrt{x} - \sqrt{7}= 3\sqrt{x} - 3\sqrt{7}$

$or, 3\sqrt{7} + \sqrt{7} = 3\sqrt{x} - \sqrt{x}$

$or, 4\sqrt{7} = 2\sqrt{x}$

$or, 2\sqrt{x} = 4\sqrt{7}$

$or, \sqrt{x} = \dfrac{4\sqrt{7}}{2}$

$or, \sqrt{x} = 2\sqrt{7}$

[ Squaring both sides ]

$or, (\sqrt{x})^2 = (2\sqrt{7})^2$

$or, x = 4 * 7$

$\therefore x = 28$
= Answer