Using properties of proportion, solve for $x$ . Given that $x$ is positive :
$\frac{2 x + \sqrt{4 x^{2} - 1}}{2 x - \sqrt{4 x^{2} - 1}} = 4$

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Solution

Given expression is
$\frac{2 x + \sqrt{4 x^{2} - 1}}{2 x - \sqrt{4 x^{2} - 1}} = 4$

We use componendo and dividendo in above expression, we get
$\frac{2 x + \sqrt{4 x^{2} - 1} + 2 x - \sqrt{4 x^{2} - 1}}{2 x + \sqrt{4 x^{2} - 1} - (2 x - \sqrt{4 x^{2} - 1})} = 4$

$\frac{2 x}{\sqrt{4 x^{2} - 1}} = \frac{4 + 1}{4 - 1} = \frac{5}{3}$

Squaring and rearranging, we get:
$4 x^{2} \times 9 = 5 (4 x^{2} - 1)$
$\Rightarrow 36 x^{2} = 100 x^{2} - 25$
$\Rightarrow x = \pm \frac{5}{8}$
Choosing the positive value of $x$ we get $x = \frac{5}{8}$

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