# How do you solve  1/(4 -x) = x/8?

May 30, 2018

Below

#### Explanation:

$\frac{1}{4 - x} = \frac{x}{8}$

$\frac{1}{4 - x} \times \frac{8}{8} = \frac{x}{8} \times \frac{4 - x}{4 - x}$

$\frac{8}{8 \left(4 - x\right)} = \frac{x \left(4 - x\right)}{8 \left(4 - x\right)}$

$8 = x \left(4 - x\right)$

$8 = 4 x - {x}^{2}$

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

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

$x = \frac{4 \pm \sqrt{16 - 4 \times 1 \times 8}}{2 \times 1}$

$x = \frac{4 \pm \sqrt{- 16}}{2}$

If you haven't learnt complex numbers yet, then you can ignore the following lines and just state that since the discriminant is less than 0, then there is no solution to the equation.

$x = \frac{4 \pm \sqrt{16 \times - 1}}{2}$

$x = \frac{4 \pm 4 i}{2}$

$x = 2 \pm 2 i$

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

$\left(x - \left(2 - 2 i\right)\right) \left(x - \left(2 + 2 i\right)\right) = 0$

$\left(x - 2 + 2 i\right) \left(x - 2 - 2 i\right) = 0$