# Solve the following equation?

## $\frac{2 x}{2 {x}^{2} + 5 x + 2} > \frac{1}{x + 1}$

Jun 16, 2018

$- \frac{2}{3} < x < - \frac{1}{2}$ and $- 2 < x < - 1$ only

#### Explanation:

$\frac{2 x}{2 {x}^{2} + 5 x + 2} > \frac{1}{x + 1}$ where $x \ne - 1 , - \frac{1}{2} , - 2$

$\frac{2 x}{2 {x}^{2} + 5 x + 2} - \frac{1}{x + 1} > 0$

$\frac{2 x \left(x + 1\right) - \left(2 {x}^{2} + 5 x + 2\right)}{\left(x + 1\right) \left(2 {x}^{2} + 5 x + 2\right)} > 0$

$\frac{2 {x}^{2} + 2 x - 2 {x}^{2} - 5 x - 2}{\left(x + 1\right) \left(2 {x}^{2} + 5 x + 2\right)} > 0$

$\frac{- 3 x - 2}{\left(x + 1\right) \left(2 {x}^{2} + 5 x + 2\right)} > 0$

${\left(\left(x + 1\right) \left(2 {x}^{2} + 5 x + 2\right)\right)}^{2} \times \frac{- 3 x - 2}{\left(x + 1\right) \left(2 {x}^{2} + 5 x + 2\right)} > 0 \times {\left(\left(x + 1\right) \left(2 {x}^{2} + 5 x + 2\right)\right)}^{2}$

color (red) ("you need to multiply both sides by the denominator squared because you don't know whether your denominator is a positive or negative number. If it was negative, then your inequality sign would change."

$\left(- 3 x - 2\right) \left(x + 1\right) \left(2 {x}^{2} + 5 x + 2\right) > 0$

$\left(- 3 x - 2\right) \left(x + 1\right) \left(2 x + 1\right) \left(x + 2\right) > 0$

$- \left(3 x + 2\right) \left(x + 1\right) \left(2 x + 1\right) \left(x + 2\right) > 0$

graph{(-3x-2)(x+1)(2x+1)(x+2) [-10, 10, -5, 5]}

looking at the graph, we can see that
$- \frac{2}{3} < x < - \frac{1}{2}$ and $- 2 < x < - 1$ only