# How do you solve  1/(x^2 -3) = 8/x?

Apr 21, 2016

$x = \frac{1}{16} \pm \frac{\sqrt{769}}{16}$

#### Explanation:

Multiply both sides by $x \left({x}^{2} - 3\right)$ to get:

$x = 8 \left({x}^{2} - 3\right) = 8 {x}^{2} - 24$

Subtract $x$ from both sides and transpose to get:

$8 {x}^{2} - x - 24 = 0$

This is in the form $a {x}^{2} + b x + c = 0$ with $a = 8$, $b = - 1$ and $c = - 24$, which has roots given by the quadratic formula:

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

$= \frac{1 \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(8\right) \left(- 24\right)}}{2 \cdot 8}$

$= \frac{1}{16} \pm \frac{\sqrt{769}}{16}$