Solve `|x^2-x-2|+|x+6|=|x^2-2x-8|`

Explanation

To solve the equation \( |x^2 - x - 2| + |x + 6| = |x^2 - 2x - 8| \), we will analyze the expressions inside the absolute values and determine the intervals where each expression is positive or negative.

### Step 1: Identify the critical points
First, we need to find the points where each expression inside the absolute values equals zero.

1. For \( x^2 - x - 2 = 0 \):
\[ (x - 2)(x + 1) = 0 \implies x = 2, -1 \]

2. For \( x + 6 = 0 \):
\[ x = -6 \]

3. For \( x^2 - 2x - 8 = 0 \):
\[ (x - 4)(x + 2) = 0 \implies x = 4, -2 \]

The critical points are \( -6, -2, -1, 2, 4 \).

### Step 2: Determine the intervals
The critical points divide the number line into the following intervals:
1. \( (-\infty, -6) \)
2. \( (-6, -2) \)
3. \( (-2, -1) \)
4. \( (-1, 2) \)
5. \( (2, 4) \)
6. \( (4, \infty) \)

### Step 3: Analyze each interval
We will evaluate the signs of the expressions in each interval.

1. **Interval \( (-\infty, -6) \)**:
- \( x^2 - x - 2 > 0 \)
- \( x + 6 < 0 \)
- \( x^2 - 2x - 8 > 0 \)
\[ |x^2 - x - 2| + |x + 6| = (x^2 - x - 2) - (x + 6) = x^2 - 2x - 8 \]
This holds true.

2. **Interval \( (-6, -2) \)**:
- \( x^2 - x - 2 > 0 \)
- \( x + 6 > 0 \)
- \( x^2 - 2x - 8 > 0 \)
\[ |x^2 - x - 2| + |x + 6| = (x^2 - x - 2) + (x + 6) = x^2 + 4 \]
This does not hold.

3. **Interval \( (-2, -1) \)**:
- \( x^2 - x - 2 > 0 \)
- \( x + 6 > 0 \)
- \( x^2 - 2x - 8 < 0 \)
\[ |x^2 - x - 2| + |x + 6| = (x^2 - x - 2) + (x + 6) = x^2 + 4 \]
This does not hold.

4. **Interval \( (-1, 2) \)**:
- \( x^2 - x - 2 < 0 \)
- \( x + 6 > 0 \)
- \( x^2 - 2x - 8 < 0 \)
\[ |x^2 - x - 2| + |x + 6| = -(x^2 - x - 2) + (x + 6) = -x^2 + 2x + 8 \]
This holds true.

5. **Interval \( (2, 4) \)**:
- \( x^2 - x - 2 > 0 \)
- \( x + 6 > 0 \)
- \( x^2 - 2x - 8 < 0 \)
\[ |x^2 - x - 2| + |x + 6| = (x^2 - x - 2) + (x + 6) = x^2 + 4 \]
This does not hold.

6. **Interval \( (4, \infty) \)**:
- \( x^2 - x - 2 > 0 \)
- \( x + 6 > 0 \)
- \( x^2 - 2x - 8 > 0 \)
\[ |x^2 - x - 2| + |x + 6| = (x^2 - x - 2) + (x + 6) = x^2 + 4 \]
This holds true.

### Step 4: Combine the intervals
The solution to the equation is the union of the intervals where the equation holds true:
\[ x \in (-\infty, -6) \cup (-1, 2) \cup (4, \infty) \]

### Final Solution
Thus, the final solution is:
\[ x \in (-\infty, -6) \cup (-1, 2) \cup (4, \infty) \]