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