# Solve the equation x^2+5x+4 ge 0 ?

Nov 30, 2017

$x \le - 4$ or $x \ge - 1$

#### Explanation:

We have:

${x}^{2} + 5 x + 4 \ge 0$

We first solve the equation:

${x}^{2} + 5 x + 4 = 0 \implies \left(x + 4\right) \left(x + 1\right) = 0$
$\implies x = - 1 , - 4$

Ther using appropriate software or a graphical calculator we examine the graph:
graph{ y=x^2+5x+4 [-10, 10, -5, 5]}

And if we add shading we seek the range in which the curve lies above the $x$-axis.
graph{ (y-x^2-5x-4)(sqrt(y))^2 <= 0 [-10, 10, -5, 5]}

and we see from the graph that:

$\left\{\begin{matrix}x \le - 4 \\ - 4 < x < - 1 \\ x \ge - 1\end{matrix}\right. \implies \left\{\begin{matrix}{x}^{2} + 5 x + 4 \ge 0 \\ {x}^{2} + 5 x + 4 < 0 \\ {x}^{2} + 5 x + 4 \ge 0\end{matrix}\right.$

Thus we have:

$x \le - 4$ or $x \ge - 1$