What are the roots of the equations `|x^(2) -x - 6| = |x + 2|` ?

Options

A. `-2.1.4`
B. 0.2.4
C. 0.1.4
D. `-2.2.4`

Answer

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

### Step 1: Identify the critical points
The expressions inside the absolute values change sign at certain points. We need to find the points where \( x^2 - x - 6 = 0 \) and \( x + 2 = 0 \).

1. **For \( x^2 - x - 6 = 0 \)**:
- Factor the quadratic: \( (x - 3)(x + 2) = 0 \)
- Roots are \( x = 3 \) and \( x = -2 \).

2. **For \( x + 2 = 0 \)**:
- The root is \( x = -2 \).

The critical points are \( x = -2 \) and \( x = 3 \).

### Step 2: Set up cases based on critical points
We will consider the following intervals based on the critical points:
1. \( x < -2 \)
2. \( -2 \leq x < 3 \)
3. \( x \geq 3 \)

### Step 3: Solve for each case

#### Case 1: \( x < -2 \)
In this interval, both expressions are negative:
- \( |x^2 - x - 6| = -(x^2 - x - 6) = -x^2 + x + 6 \)
- \( |x + 2| = -(x + 2) = -x - 2 \)

Setting them equal:
\[ -x^2 + x + 6 = -x - 2 \]
Rearranging gives:
\[ -x^2 + 2x + 8 = 0 \quad \text{or} \quad x^2 - 2x - 8 = 0 \]
Factoring:
\[ (x - 4)(x + 2) = 0 \]
Roots are \( x = 4 \) and \( x = -2 \). Since \( x < -2 \), there are no valid solutions in this case.

#### Case 2: \( -2 \leq x < 3 \)
In this interval, \( x + 2 \) is non-negative and \( x^2 - x - 6 \) is negative:
- \( |x^2 - x - 6| = -(x^2 - x - 6) = -x^2 + x + 6 \)
- \( |x + 2| = x + 2 \)

Setting them equal:
\[ -x^2 + x + 6 = x + 2 \]
Rearranging gives:
\[ -x^2 + 6 - 2 = 0 \quad \text{or} \quad -x^2 + 4 = 0 \]
This simplifies to:
\[ x^2 = 4 \]
Thus, \( x = 2 \) or \( x = -2 \). Both values are within the interval \( -2 \leq x < 3 \).

#### Case 3: \( x \geq 3 \)
In this interval, both expressions are non-negative:
- \( |x^2 - x - 6| = x^2 - x - 6 \)
- \( |x + 2| = x + 2 \)

Setting them equal:
\[ x^2 - x - 6 = x + 2 \]
Rearranging gives:
\[ x^2 - 2x - 8 = 0 \]
Factoring:
\[ (x - 4)(x + 2) = 0 \]
Roots are \( x = 4 \) and \( x = -2 \). Since \( x \geq 3 \), only \( x = 4 \) is valid.

### Step 4: Summary of solutions
The roots of the equation \( |x^2 - x - 6| = |x + 2| \) are:
- \( x = -2 \)
- \( x = 2 \)
- \( x = 4 \)

### Final Answer
The roots of the equation are \( x = -2, 2, 4 \).