If $\dfrac{\sqrt{x + 5} + \sqrt{x - 16}}{\sqrt{x + 5} - \sqrt{x - 16}} = \dfrac{7}{3}$, prove that x = 20.

Given,

$\dfrac{\sqrt{x + 5} + \sqrt{x - 16}}{\sqrt{x + 5} - \sqrt{x - 16}} = \dfrac{7}{3}$

Applying Componendo and Dividendo, we get :

$\Rightarrow \dfrac{(\sqrt{x + 5} + \sqrt{x - 16}) + (\sqrt{x + 5} - \sqrt{x - 16})}{(\sqrt{x + 5} + \sqrt{x - 16}) - (\sqrt{x + 5} - \sqrt{x - 16})} = \dfrac{7 + 3}{7 - 3} \\[1em] \Rightarrow \dfrac{(\sqrt{x + 5} + \sqrt{x - 16} + \sqrt{x + 5} - \sqrt{x - 16})}{(\sqrt{x + 5} + \sqrt{x - 16} - \sqrt{x + 5} + \sqrt{x - 16})} = \dfrac{10}{4} \\[1em] \Rightarrow \dfrac{2\sqrt{x + 5}}{2\sqrt{x - 16}} = \dfrac{5}{2} \\[1em] \Rightarrow \dfrac{\sqrt{x + 5}}{\sqrt{x - 16}} = \dfrac{5}{2} \\[1em]$

Squaring both sides, we get :

$\Rightarrow \Big(\dfrac{\sqrt{x + 5}}{\sqrt{x - 16}}\Big)^2 = \Big(\dfrac{5}{2}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{x + 5}{x - 16}\Big) = \Big(\dfrac{25}{4}\Big) \\[1em] \Rightarrow 4(x + 5) = 25(x - 16) \\[1em] \Rightarrow 4x + 20 = 25x - 400 \\[1em] \Rightarrow 25x - 4x = 400 + 20 \\[1em] \Rightarrow 21x = 420 \\[1em] \Rightarrow x = \dfrac{420}{21} \\[1em] \Rightarrow x = 20.$

Hence, proved that x = 20.