If $\dfrac{\sqrt{x + 2} + \sqrt{x - 3}}{\sqrt{x + 2} - \sqrt{x - 3}} = 5$, prove that x = 7.

Given,

$\dfrac{\sqrt{x + 2} + \sqrt{x - 3}}{\sqrt{x + 2} - \sqrt{x - 3}} = 5$

Applying Componendo and Dividendo, we get :

$\Rightarrow \dfrac{(\sqrt{x + 2} + \sqrt{x - 3}) + (\sqrt{x + 2} - \sqrt{x - 3})}{(\sqrt{x + 2} + \sqrt{x - 3}) - (\sqrt{x + 2} - \sqrt{x - 3})} = \dfrac{5 + 1}{5 - 1}\\[1em] \Rightarrow \dfrac{2\sqrt{x + 2}}{2\sqrt{x - 3}} = \dfrac{6}{4}\\[1em] \Rightarrow \dfrac{\sqrt{x + 2}}{\sqrt{x - 3}} = \dfrac{3}{2}\\[1em] \Rightarrow \Big(\dfrac{\sqrt{x + 2}}{\sqrt{x - 3}}\Big)^2 = \Big(\dfrac{3}{2}\Big)^2$

Squaring both sides, we get :

$\Rightarrow \dfrac{x + 2}{x - 3} = \Big(\dfrac{9}{4}\Big) \\[1em] \Rightarrow 4(x + 2) = 9(x - 3) \\[1em] \Rightarrow 4x + 8 = 9x - 27 \\[1em] \Rightarrow 9x - 4x = 27 + 8 \\[1em] \Rightarrow 5x = 35 \\[1em] \Rightarrow x = \dfrac{35}{5} = 7.$

Hence, proved that x = 7.