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.

This is my graph I did to find the solutions.

• Welcome to MathSE. Please type your question rather than posting an image since images cannot be searched. This MathJax tutorial explains how to typeset mathematics on this site. Sep 16, 2018 at 20:47
• While graphing does provide answers (possibly approximate due to the nature of finding points on a picture), of a more rigorous nature would be breaking the problem down into cases. Your inequality asks when the product $x|x+5|$ is greater than or equal to $-6$. One might think of $x|x+5|$ as a continuous function, so points where $x|x+5| = -6$ exactly play a role as boundaries of regions (intervals) where the inequality is satisfied. Sep 16, 2018 at 21:20

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 $[-6,-3]\cup[-2,\infty)$.

$$|x + 5| = \begin{cases} x + 5 & \text{if x \geq -5}\\ -x - 5 & \text{if x < -5} \end{cases}$$

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 $$S = [-5, -3] \cup [-2, \infty) \cup [-6, -5) = [-6, -3] \cup [-2, \infty)$$