Question: Solve: $\dfrac{\sqrt{x+4} + \sqrt{2} }{ \sqrt{x +4} - \sqrt{2}} = 2$

Solution:
Given,

$\dfrac{\sqrt{x+4} + \sqrt{2} }{ \sqrt{x +4} - \sqrt{2}} = 2$

$or, \sqrt{x+4} + \sqrt{2}  = 2(  \sqrt{x +4} - \sqrt{2})$

$or, \sqrt{x+4} + \sqrt{2} = 2 \sqrt{x +4} - 2\ sqrt{2})$

$or, 2\sqrt{2} + \sqrt{2} = 2\sqrt{x +4} - \sqrt{x +4}$

$or, 3\sqrt{2} = \sqrt{x +4}$

$or, \sqrt{x +4} = 3\sqrt{2}$

squaring both sides

$or, (\sqrt{x+4})² = (3\sqrt{2})²$

$or, x +4 = 9×2$

$or, x +4 = 18$

$or, x = 18-4$

$\therefore x = 14$
= Answer