We are tasked with solving the inequality:

$16 - x^2 \leq |x + 4|$

Step 1: Consider cases for the absolute value

The absolute value function $|x + 4|$ creates two cases that we need to solve separately:

1. Case 1: $x + 4 \geq 0$ (i.e., $x \geq -4$), where $|x + 4| = x + 4$.

   The inequality becomes: $16 - x^2 \leq x + 4$ Rearrange the inequality: $16 - x^2 - x - 4 \leq 0$ Simplifying: $-x^2 - x + 12 \leq 0$ Multiply through by $-1$ to get a standard quadratic form (note that this reverses the inequality): $x^2 + x - 12 \geq 0$ Factor the quadratic: $(x - 3)(x + 4) \geq 0$ Solve this using a sign analysis on the intervals determined by the roots $x = 3$ and $x = -4$:

   • For $x < -4$, both factors are negative, so the product is positive.
   • For $-4 < x < 3$, the product is negative.
   • For $x > 3$, both factors are positive, so the product is positive.

   Thus, the solution for this case is: $x \in (-\infty, -4] \cup [3, \infty)$ However, in this case, we must restrict $x \geq -4$ (since $x + 4 \geq 0$). This gives us: $x \in [-4, -4] \cup [3, \infty) = \{-4\} \cup [3, \infty)$

2. Case 2: $x + 4 < 0$ (i.e., $x < -4$), where $|x + 4| = -(x + 4) = -x - 4$.

   The inequality becomes: $16 - x^2 \leq -x - 4$ Rearrange the inequality: $16 - x^2 + x + 4 \leq 0$ Simplifying: $-x^2 + x + 20 \leq 0$ Multiply through by $-1$ to reverse the inequality: $x^2 - x - 20 \geq 0$ Factor the quadratic: $(x - 5)(x + 4) \geq 0$ Solve this using a sign analysis on the intervals determined by the roots $x = -4$ and $x = 5$:

   • For $x < -4$, both factors are negative, so the product is positive.
   • For $-4 < x < 5$, the product is negative.
   • For $x > 5$, both factors are positive, so the product is positive.

   Thus, the solution for this case is: $x \in (-\infty, -4] \cup [5, \infty)$ However, in this case, we must restrict $x < -4$, so the solution is: $x \in (-\infty, -4)$

Step 2: Combine both cases

From Case 1, we have $x \in \{-4\} \cup [3, \infty)$, and from Case 2, we have $x \in (-\infty, -4)$.

Thus, the combined solution is: $x \in (-\infty, -4) \cup \{-4\} \cup [3, \infty)$ Or more simply: $x \in (-\infty, -4] \cup [3, \infty)$

Final Answer:

The solution to the inequality is:

$x \in (-\infty, -4] \cup [3, \infty)$

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Would you like further details or have any questions about this solution? Here are five related questions to expand on this:

1. How do we solve inequalities involving absolute values in general?
2. Why do we need to break the inequality into different cases?
3. How does the sign analysis work in determining the intervals for quadratic inequalities?
4. How do we combine solutions from different cases?
5. Can we graph this solution on a number line?

Tip: When solving absolute value inequalities, always remember to break the inequality into cases based on the expression inside the absolute value.