Using properties of proportion, solve for x. Given that x is positive :

$\dfrac{2x + \sqrt{4x^{2} - 1}}{2x - \sqrt{4x^{2} - 1}} = 4$

Given,

$\dfrac{2x + \sqrt{4x^{2} - 1}}{2x - \sqrt{4x^{2} - 1}} = 4$

Applying Componendo and Dividendo, we get :

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

Squaring both sides, we get :

$\Rightarrow \Big(\dfrac{2x}{\sqrt{4x^2 - 1}}\Big)^2 = \Big(\dfrac{5}{3}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{4x^2}{4x^2 - 1}\Big) = \Big(\dfrac{25}{9}\Big) \\[1em] \Rightarrow 9(4x^2) = 25(4x^2 - 1) \\[1em] \Rightarrow 36x^2 = 100x^2 - 25 \\[1em] \Rightarrow 100x^2 - 36x^2 = 25 \\[1em] \Rightarrow 64x^2 = 25 \\[1em] \Rightarrow x^2 = \dfrac{25}{64} \\[1em] \Rightarrow x = \sqrt{\dfrac{25}{64}} \\[1em] \Rightarrow x = \dfrac{5}{8}$

Hence, x = $\dfrac{5}{8}$.