# Question #296de

Jan 8, 2018

$x \approx - 2.31 , - 0.19$

#### Explanation:

First we square both sides:
${\left(\frac{\sqrt{4}}{\sqrt{4 x + 9}}\right)}^{2} = {\left(\sqrt{8 x + 2}\right)}^{2}$

${\left(\sqrt{4}\right)}^{2} / {\left(\sqrt{4 x + 9}\right)}^{2} = {\left(\sqrt{8 x + 2}\right)}^{2}$

$\frac{4}{4 x + 9} = 8 x + 2$

Now we can rearrange and make it eqial to 0:
$4 = \left(8 x + 2\right) \left(4 x + 9\right)$
$\left(8 x + 2\right) \left(4 x + 9\right) - 4 = 0$

Now expand brackets:

$\left(8 x \cdot 4 x\right) + \left(2 \cdot 4 x\right) + \left(8 x \cdot 9\right) + \left(2 \cdot 9\right) - 4 = 0$

$32 {x}^{2} + 8 x + 72 x + 18 - 4 = 0$

$32 {x}^{2} + 80 x + 14 = 0$

We can remove out common factors:

$32 {x}^{2} + 80 x + 14 = 0$

$16 {x}^{2} + 40 x + 7 = 0$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- 40 \pm \sqrt{{40}^{2} - 4 \left(16 \cdot 7\right)}}{32}$

$= \frac{- 40 \pm \sqrt{1600 - 4 \left(112\right)}}{32}$

$= \frac{- 40 \pm \sqrt{1600 - 448}}{32}$

$= \frac{- 40 \pm \sqrt{1152}}{32}$

$= \frac{- 40 + \sqrt{1152}}{32} \mathmr{and} \frac{- 40 - \sqrt{1152}}{32}$

$\approx - 2.31 \mathmr{and} - 0.19$