We are asked to solve the inequality:

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

This involves two cases due to the absolute value. We will break it down into two separate inequalities:

Case 1: $x^2 + 2x - 12 \leq x + 2$

1. Subtract $x + 2$ from both sides: $x^2 + 2x - 12 - (x + 2) \leq 0$ Simplify: $x^2 + x - 14 \leq 0$

2. Now solve the quadratic inequality: Factor the quadratic expression: $x^2 + x - 14 = (x - 3)(x + 4)$ So the inequality becomes: $(x - 3)(x + 4) \leq 0$

3. Solve this using a sign chart for the intervals determined by the roots $x = 3$ and $x = -4$:

   • The critical points are $x = -4$ and $x = 3$.
   • Test the intervals $(-\infty, -4)$, $(-4, 3)$, and $(3, \infty)$:
     • For $x < -4$, both factors are negative, so the product is positive.
     • For $-4 < x < 3$, one factor is negative and the other positive, so the product is negative.
     • For $x > 3$, both factors are positive, so the product is positive.

   Therefore, the solution to this inequality is: $-4 \leq x \leq 3$

Case 2: $-(x^2 + 2x - 12) \leq x + 2$

1. Multiply through by $-1$ (and reverse the inequality): $x^2 + 2x - 12 \geq -(x + 2)$ Simplify: $x^2 + 2x - 12 + x + 2 \geq 0$ $x^2 + 3x - 10 \geq 0$

2. Solve the quadratic inequality: Factor the quadratic expression: $x^2 + 3x - 10 = (x - 2)(x + 5)$ So the inequality becomes: $(x - 2)(x + 5) \geq 0$

3. Solve this using a sign chart for the intervals determined by the roots $x = 2$ and $x = -5$:

   • The critical points are $x = -5$ and $x = 2$.
   • Test the intervals $(-\infty, -5)$, $(-5, 2)$, and $(2, \infty)$:
     • For $x < -5$, both factors are negative, so the product is positive.
     • For $-5 < x < 2$, one factor is negative and the other positive, so the product is negative.
     • For $x > 2$, both factors are positive, so the product is positive.

   Therefore, the solution to this inequality is: $x \leq -5 \quad \text{or} \quad x \geq 2$

Combining the solutions:

• From Case 1, we have $-4 \leq x \leq 3$.
• From Case 2, we have $x \leq -5$ or $x \geq 2$.

The final solution is the intersection of these two cases: $-4 \leq x \leq -5 \quad \text{or} \quad 2 \leq x \leq 3$

Thus, the solution is: [ x \in [-4, -5] \cup [2, 3] ]

Would you like further clarification on any step?

Here are 5 questions to explore this concept further:

1. How do we graph absolute value inequalities?
2. What are the conditions for using the sign chart method?
3. Can absolute value inequalities always be split into two cases?
4. How do quadratic inequalities differ from linear ones?
5. Why is it necessary to reverse the inequality when multiplying by a negative number?

Tip: Always check the boundaries when solving inequalities with absolute values, as they can change the direction of the inequality.