To solve the inequation \( \sqrt{x+2} > \sqrt{8-x^2} \), we will follow these steps:
### Step 1: Square both sides
Since both sides of the inequality involve square roots, we can square both sides to eliminate the square roots. However, we must ensure that both sides are non-negative.
\[
\sqrt{x+2} > \sqrt{8-x^2} \implies x + 2 > 8 - x^2
\]
### Step 2: Rearrange the inequality
Now, we rearrange the inequality to bring all terms to one side:
\[
x + 2 - 8 + x^2 > 0 \implies x^2 + x - 6 > 0
\]
### Step 3: Factor the quadratic expression
Next, we factor the quadratic expression \( x^2 + x - 6 \):
\[
x^2 + x - 6 = (x - 2)(x + 3) > 0
\]
### Step 4: Determine the critical points
The critical points from the factors are \( x = 2 \) and \( x = -3 \). We will use these points to test intervals.
### Step 5: Test intervals
We will test the intervals determined by the critical points: \( (-\infty, -3) \), \( (-3, 2) \), and \( (2, \infty) \).
1. **Interval \( (-\infty, -3) \)**: Choose \( x = -4 \)
\[
(-4 - 2)(-4 + 3) = (-6)(-1) = 6 > 0 \quad \text{(True)}
\]
2. **Interval \( (-3, 2) \)**: Choose \( x = 0 \)
\[
(0 - 2)(0 + 3) = (-2)(3) = -6 < 0 \quad \text{(False)}
\]
3. **Interval \( (2, \infty) \)**: Choose \( x = 3 \)
\[
(3 - 2)(3 + 3) = (1)(6) = 6 > 0 \quad \text{(True)}
\]
### Step 6: Combine the results
From our tests, the solution to the inequality \( (x - 2)(x + 3) > 0 \) is:
\[
x \in (-\infty, -3) \cup (2, \infty)
\]
### Step 7: Ensure the original conditions are met
Next, we need to ensure that both \( \sqrt{x+2} \) and \( \sqrt{8-x^2} \) are defined and non-negative:
1. **For \( \sqrt{x+2} \)**:
\[
x + 2 \geq 0 \implies x \geq -2
\]
2. **For \( \sqrt{8-x^2} \)**:
\[
8 - x^2 \geq 0 \implies x^2 \leq 8 \implies -2\sqrt{2} \leq x \leq 2\sqrt{2}
\]
### Step 8: Find the intersection of the intervals
Now we need to find the intersection of the intervals \( (-\infty, -3) \cup (2, \infty) \) with \( [-2, 2\sqrt{2}] \):
- From \( (-\infty, -3) \): The intersection is \( [-2, -3) \) (but since -3 is not included, we only consider \( -2 \)).
- From \( (2, \infty) \): The intersection is \( (2, 2\sqrt{2}] \).
### Final Solution
Thus, the solution set is:
\[
x \in [-2, -3) \cup (2, 2\sqrt{2}]
\]
