# What is (x/4) - (2x/x+2) = 1 ?

Jun 27, 2018

$x = 5 + \sqrt{33}$ or $x = 5 - \sqrt{33}$

#### Explanation:

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

$\therefore \frac{x \left(x + 2\right) - 4 \left(2 x\right) = 4 \left(x + 2\right)}{4 \left(x + 2\right)}$

$\text{multiply both sides by}$$4 \left(x + 2\right)$

$\therefore x \left(x + 2\right) - 4 \left(2 x\right) = 4 \left(x + 2\right)$

$\therefore {x}^{2} + 2 x - 8 x = 4 x + 8$

$\therefore {x}^{2} + 2 x - 8 x - 4 x - 8 = 0$

$\therefore {x}^{2} - 10 x - 8 = 0$

$\text{quadratic formula:-}$

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

$\therefore a = 1 , b = - 10 , c = - 8$

$\therefore x = \frac{- \left(- 10\right) \pm \sqrt{{\left(- 10\right)}^{2} - 4 \left(1\right) \left(- 8\right)}}{2}$

$\therefore x = \frac{10 \pm \sqrt{132}}{2}$

$\therefore x = \frac{10 \pm \sqrt{33 \cdot 2 \cdot 2}}{2}$

$\therefore x = \frac{10 \pm 2 \sqrt{33}}{2}$

$\therefore x = 5 \pm \sqrt{33}$

$\therefore x = 5 + \sqrt{33} \mathmr{and} x = 5 - \sqrt{33}$