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

Oct 20, 2015

$\textcolor{b l u e}{x = - 3}$ or $\textcolor{red}{x = 1}$

#### Explanation:

Firstly take LCM.

$\frac{{x}^{2} - 3}{x} = - 2$

${x}^{2} - 3 = - 2 x$

${x}^{2} + 2 x - 3 = 0$

Now mid term break.

${x}^{2} + \left(3 - 1\right) x - 3 = 0$

${x}^{2} + 3 x - x - 3 = 0$

$x \left(x + 3\right) - 1 \left(x + 3\right) = 0$

Taking common.

$\left(x + 3\right) \left(x - 1\right) = 0$

So either, $\textcolor{b l u e}{x + 3 = 0}$ or $\textcolor{b l u e}{x = - 3}$ OR $\textcolor{red}{x - 1 = 0}$ or $\textcolor{red}{x = 1}$.