# How do you solve x^2+5x+4<=0?

Jun 26, 2016

$x \in \left[- 4 , - 1\right]$

#### Explanation:

${x}^{2} + 5 x + 4 = \left(x + 4\right) \left(x + 1\right)$

This is zero when $x = - 4$ or $x = - 1$

The quadratic has a positive leading coefficient, so is is an upright parabola, crossing the $x$ axis at $\left(- 4 , 0\right)$ and $\left(- 1 , 0\right)$.

Hence any $x \in \left[- 4 , - 1\right]$ satisfies the required inequality and any value of $x$ outside that closed interval does not.

graph{x^2+5x+4 [-10, 10, -5, 5]}