Question: Solve: $\dfrac{\sqrt{x +2} - \sqrt{x -2}}{\sqrt{x +2} + \sqrt{x -2}} = \dfrac{1}{2}$

Solution:
Given,

$\dfrac{\sqrt{x +2} - \sqrt{x -2}}{\sqrt{x +2} +\sqrt{x -2}} = \dfrac{1}{2}$

$or, 2(\sqrt{x +2} - \sqrt{x -2}) = 1 ( \sqrt{x +2} +\sqrt{x -2})$

$or, 2\sqrt{x +2} - 2 \sqrt{x -2} = \sqrt{x +2} + \sqrt{x -2}$

$or, 2\sqrt{x +2} - \sqrt{x +2} = 2\sqrt{x -2} + \sqrt{x -2}$

$or, \sqrt{x +2} = 3 \sqrt{x -2}$

Squaring both sides

$or, (\sqrt{x+2})^2 = (3 \sqrt{x-2})^2$

$or, x +2 = 9(x -2)$

$or, x +2 = 9x -18$

$or, 18+2 = 9x -x$

$or, 20 = 8x$

$or, x = \dfrac{20}{8}$

$\therefore x = \dfrac{5}{2}$

= Answer