The sum of the roots of the equation `|x^(2)-x-6|=x+2` is

Options

A. 0
B. `-2`
C. 2
D. 4

Answer

To solve the equation \( |x^2 - x - 6| = x + 2 \), we need to consider two cases based on the properties of absolute values.

### Step 1: Identify Cases
The expression inside the absolute value can be either non-negative or negative. Therefore, we will split our solution into two cases:

- **Case 1:** \( x^2 - x - 6 \geq 0 \)
- **Case 2:** \( x^2 - x - 6 < 0 \)

### Step 2: Solve Case 1
In Case 1, since \( x^2 - x - 6 \geq 0 \), we can write:
\[ x^2 - x - 6 = x + 2 \]
Rearranging gives:
\[ x^2 - x - 6 - x - 2 = 0 \implies x^2 - 2x - 8 = 0 \]
Now, we can factor this quadratic equation:
\[ (x - 4)(x + 2) = 0 \]
Thus, the solutions for Case 1 are:
\[ x = 4 \quad \text{and} \quad x = -2 \]

### Step 3: Verify Case 1 Solutions
We need to check if these solutions satisfy \( x^2 - x - 6 \geq 0 \):
- For \( x = 4 \):
\[ 4^2 - 4 - 6 = 16 - 4 - 6 = 6 \geq 0 \quad \text{(valid)} \]
- For \( x = -2 \):
\[ (-2)^2 - (-2) - 6 = 4 + 2 - 6 = 0 \quad \text{(valid)} \]

### Step 4: Solve Case 2
In Case 2, since \( x^2 - x - 6 < 0 \), we can write:
\[ -(x^2 - x - 6) = x + 2 \]
This simplifies to:
\[ -x^2 + x + 6 = x + 2 \]
Rearranging gives:
\[ -x^2 + x - x + 6 - 2 = 0 \implies -x^2 + 4 = 0 \implies x^2 = 4 \]
Thus, the solutions for Case 2 are:
\[ x = 2 \quad \text{and} \quad x = -2 \]

### Step 5: Verify Case 2 Solutions
We need to check if these solutions satisfy \( x^2 - x - 6 < 0 \):
- For \( x = 2 \):
\[ 2^2 - 2 - 6 = 4 - 2 - 6 = -4 < 0 \quad \text{(valid)} \]
- For \( x = -2 \):
\[ (-2)^2 - (-2) - 6 = 4 + 2 - 6 = 0 \quad \text{(not valid)} \]

### Step 6: Collect Valid Solutions
The valid solutions from both cases are:
- From Case 1: \( x = 4, -2 \) (valid)
- From Case 2: \( x = 2 \) (valid)

### Step 7: Calculate the Sum of the Roots
The valid roots are \( 4, -2, 2 \). Now, we calculate the sum:
\[ \text{Sum} = 4 + (-2) + 2 = 4 \]

### Final Answer
The sum of the roots of the equation \( |x^2 - x - 6| = x + 2 \) is \( \boxed{4} \).