The given inequality is:

$|x + 12| \leq |x^2 - 8|$

Solution Steps:

Step 1: Analyze Absolute Values

The absolute value inequality can be written in terms of cases:

1. $x^2 - 8 \geq 0$ (i.e., $x^2 \geq 8$)
2. $x^2 - 8 < 0$ (i.e., $x^2 < 8$)

We'll handle each case separately.

---

Step 2: Case 1 — When $x^2 \geq 8$

In this case, the absolute values simplify to: $|x + 12| \leq x^2 - 8$ Breaking this into two conditions: $x + 12 \leq x^2 - 8 \quad \text{and} \quad -(x + 12) \leq x^2 - 8$

1. From $x + 12 \leq x^2 - 8$: $x^2 - x - 20 \geq 0$ Factoring: $(x - 5)(x + 4) \geq 0$ The solution is: $x \in (-\infty, -4] \cup [5, \infty)$

2. From $-(x + 12) \leq x^2 - 8$: $x^2 + x - 20 \geq 0$ Factoring: $(x - 4)(x + 5) \geq 0$ The solution is: $x \in (-\infty, -5] \cup [4, \infty)$

Combining the two, we find for this case: $x \in (-\infty, -5] \cup [-4, -4] \cup [5, \infty)$

---

Step 3: Case 2 — When $x^2 < 8$

In this case, $|x^2 - 8| = -(x^2 - 8) = 8 - x^2$. Thus: $|x + 12| \leq 8 - x^2$ Breaking this into two conditions: $x + 12 \leq 8 - x^2 \quad \text{and} \quad -(x + 12) \leq 8 - x^2$

1. From $x + 12 \leq 8 - x^2$: $x^2 + x + 4 \leq 0$ This quadratic has no real solutions, as its discriminant is negative ($1^2 - 4(1)(4) < 0$). Hence, no solutions arise from this condition.

2. From $-(x + 12) \leq 8 - x^2$: $x^2 - x - 20 \geq 0$ We solved this earlier and found: $x \in (-\infty, -4] \cup [5, \infty)$

However, since $x^2 < 8$, the valid interval is: $x \in [-\sqrt{8}, \sqrt{8}]$

---

Step 4: Combine the Results

From the two cases:

1. Case 1 gave $x \in (-\infty, -5] \cup [-4, -4] \cup [5, \infty)$,
2. Case 2 gave $x \in [-\sqrt{8}, \sqrt{8}] \cap \text{Case 1 regions}$.

The final solution is: $x \in [-4, -4]$

Would you like a detailed explanation of any step?

5 Related Questions:

1. How do you solve inequalities involving absolute values in general?
2. What is the significance of the discriminant when solving quadratic equations?
3. How do you find intersections of intervals on a number line?
4. What are the main properties of absolute values?
5. How can graphing help solve inequalities like this one?

Tip:

Always test your solution by substituting values from each interval into the original inequality to verify!