Using componendo and dividendo, find value of x :

$\dfrac{\sqrt{3x + 4} + \sqrt{3x - 5}}{\sqrt{3x + 4} - \sqrt{3x - 5}} = 9.$

Applying componendo and dividendo we get,

$\Rightarrow \dfrac{\sqrt{3x + 4} + \sqrt{3x - 5} + \sqrt{3x + 4} - \sqrt{3x - 5}}{\sqrt{3x + 4} + \sqrt{3x - 5} - (\sqrt{3x + 4} - \sqrt{3x - 5})} = \dfrac{9 + 1}{9 - 1} \\[1em] \Rightarrow \dfrac{2\sqrt{3x + 4}}{2\sqrt{3x - 5}} = \dfrac{10}{8} \\[1em] \Rightarrow \dfrac{\sqrt{3x + 4}}{\sqrt{3x - 5}} = \dfrac{10}{8} \\[1em] \Rightarrow \dfrac{3x + 4}{3x - 5} = \dfrac{100}{64} \\[1em]$

Again applying componendo and dividendo we get,

$\Rightarrow \dfrac{3x + 4 + 3x - 5}{3x + 4 - (3x - 5)} = \dfrac{100 + 64}{100 - 64} \\[1em] \Rightarrow \dfrac{6x - 1}{9} = \dfrac{164}{36} \\[1em] \Rightarrow 36(6x - 1) = 1476 \\[1em] \Rightarrow 216x - 36 = 1476 \\[1em] \Rightarrow 216x = 1512 \\[1em] \Rightarrow x = \dfrac{1512}{216} \\[1em] \Rightarrow x = 7.$

Hence, x = 7.