Question

Using properties of proportion, solve for x:

$\left(iii\right)\frac{3x+\sqrt{9x^2-5}}{3x-\sqrt{9x^2-5}}=5$

Answer 1

Step: Solve for x

If $\frac{a}{b}=\frac{c}{d}then\frac{a+b}{a-b}=\frac{c+d}{c-d}$

Given, $\frac{3x+\sqrt{9x^2-5}}{3x-\sqrt{9x^2-5}}=5$

$\Rightarrow \frac{3x+\sqrt{9x^2-5}+3x-\sqrt{9x^2-5}}{3x+\sqrt{9x^2-5}-3x+\sqrt{9x^2-5}}=\frac{5+1}{5-1}$

$\Rightarrow \frac{6x}{2\sqrt{9x^2-5}}=\frac{6}{4}$

$\Rightarrow \frac{x}{\sqrt{9x^2-5}}=\frac{1}{2}$

Squaring both sides, we get

$\Rightarrow \frac{x^2}{9x^2-5}=\frac{1}{4}$

$\Rightarrow 9x^2-5=4x^2$

[cross multiplication]

$\Rightarrow 9x^2-4x^2=5$

$\Rightarrow 5x^2=5$

$\Rightarrow x^2=1$

$\Rightarrow x=1or-1$

Hence, the value of $xis1or-1$.