Question

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

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

Answer 1

Given equation: $$\frac{2x + \sqrt{4x^2 - 1}}{2x - \sqrt{4x^2 - 1}} = \frac{4}{1}$$

Applying componendo and dividendo:

$$\frac{\left(2x + \sqrt{4x^2 - 1}\right) + \left(2x - \sqrt{4x^2 - 1}\right)}{\left(2x + \sqrt{4x^2 - 1}\right) - \left(2x - \sqrt{4x^2 - 1}\right)} = \frac{4 + 1}{4 - 1}$$

Simplifying both sides: $$\frac{4x}{2\sqrt{4x^2 - 1}} = \frac{5}{3}$$ $$\frac{2x}{\sqrt{4x^2 - 1}} = \frac{5}{3}$$

Squaring both sides: $$\frac{4x^2}{4x^2 - 1} = \frac{25}{9}$$

Cross-multiplying: $$36x^2 = 25\left(4x^2 - 1\right)$$

$$36x^2 = 100x^2 - 25$$

$$100x^2 - 36x^2 = 25$$

$$64x^2 = 25$$ 

$$x^2 = \frac{25}{64}$$

Taking the square root on both sides: $$x = \pm \frac{5}{8}$$

Since $$x$$ is given to be positive: $$x = \frac{5}{8}$$