Absolute Value Inequalities (Quadratics) I am currently struggling with the solutions of absolute value inequalities that involve quadratics. This is the example problem:
$$x|x + 5| \geq -6$$
I am able to find the solutions, but I struggle in interval notation. I considered graphing the two quadratic functions and find the shaded area as the solutions, but I still don't understand how the solution is $[-6,-3] \cup [-2, \infty]$. I understand $[-6,-3]$ but not the $\infty$ part. Yes, I can do this by plugging in values and checking if the solutions work but that is not efficient. What am I doing wrong? I appreciate anyone's help. 
https://www.desmos.com/calculator/7j8yamvbzv 
This is my graph I did to find the solutions.
 A: You should consider two separate cases: 
Case (a): $x+5 \geq 0$. In this case the inequality becomes $x^2+5x + 6 \geq 0$. The solutions of this quadratic inequality are $(-\infty,-3]\cup[-2, \infty)$. Taking in account that $x \geq -5$ gives $[-5,-3]\cup[-2,\infty)$ 
Case (b): Here $x+ 5 \leq 0$, which gives us $-x^2-5x + 6 \geq 0$. This inequality has solutions $[-6,1]$. Together with $x\leq -5$ this gives $[-6,-5]$.
Adding the solutions gives $[-6,-3]\cup[-2,\infty)$.
A: $$|x + 5| = \begin{cases}
x + 5 & \text{if $x \geq -5$}\\
-x - 5 & \text{if $x < -5$}
\end{cases}
$$
Case 1: $x \geq 5$ 
\begin{align*}
x|x + 5| & \geq -6\\
x(x + 5) & \geq -6\\
x^2 + 5x & \geq -6\\
x^2 + 5x + 6 & \geq 0\\
(x + 2)(x + 3) & \geq 0
\end{align*}
The expression on the left-hand side equals zero when $x = -2$ or $x = -3$.  It is positive when the factors are both positive or both negative.  Both factors are positive when $x > -2$.  Both factors are negative when $x < -3$.  Hence, if $x \geq -5$, the inequality is satisfied when $x \leq -3$ or $x \geq -2$.  
Since we require that $x \geq 5$ and $x \leq -3$ or $x \geq 5$ and $x \geq -2$, $-5 \leq x \leq -3$ or $x \geq -2$.  In interval notation, we write
\begin{align*}
\{x \in \mathbb{R} \mid -5 \leq x \leq -3\} & = [-5, -3]\\
\{x \in \mathbb{R} \mid x \geq -2\} & = [-2, \infty)\\
\end{align*}
Therefore, if $x \geq -5$ and satisfies the inequality $x|x + 5| \geq -6$, then
$$x \in [-5, -3] \cup [-2, \infty)$$ 
Case 2: $x < -5$
\begin{align*}
x|x + 5| & \geq -6\\
x(-x - 5) & \geq -6\\
-x^2 - 5x & \geq -6\\
x^2 + 5x & \leq 6\\
x^2 + 5x - 6 & \leq 0\\
(x + 6)(x - 1) & \leq 0
\end{align*}
The expression on the left-hand side equals zero when $x = -6$ or $x = 1$.  It is negative when the two factors have opposite signs, which occurs when $-6 <  x < 1$.  Hence, the inequality is satisfied if $x < -5$ and $-6 \leq x \leq 1$, so $-6 \leq x < -5$.  In interval notation, we write
$$\{x \in \mathbb{R} \mid -6 \leq x < -5\} = [-6, -5)$$
Therefore, if $x < -5$ and satisfies the inequality $x|x + 5| \geq -6$, then 
$$x \in [-6, 5)$$
Solution: The solution of the absolute value inequality $x|x + 5| \leq -6$ is the union of the solutions for the cases $x \geq 5$ and $x < 5$, so we obtain the solution set
$$S = [-5, -3] \cup [-2, \infty) \cup [-6, -5) = [-6, -3] \cup [-2, \infty)$$ 
