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

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 will break it down into cases based on the definition of absolute value.

### Step 1: Factor the quadratic expression
First, we need to factor the expression inside the absolute value:
\[ x^2 - x - 6 = (x - 3)(x + 2) \]
The roots of the equation \( x^2 - x - 6 = 0 \) are \( x = 3 \) and \( x = -2 \).

### Step 2: Set up cases based on the roots
We will consider three cases based on the critical points \( x = -2 \) and \( x = 3 \).

#### Case 1: \( x < -2 \)
In this case, both \( x - 3 \) and \( x + 2 \) are negative, so:
\[ |x^2 - x - 6| = -(x^2 - x - 6) = - (x - 3)(x + 2) = -x^2 + x + 6 \]
Thus, the equation becomes:
\[ -x^2 + x + 6 = x + 2 \]
Simplifying gives:
\[ -x^2 + 6 = 2 \implies -x^2 = -4 \implies x^2 = 4 \implies x = -2 \text{ or } x = 2 \]
Since \( x < -2 \), there are no valid solutions from this case.

#### Case 2: \( -2 \leq x < 3 \)
In this case, \( x + 2 \) is non-negative and \( x - 3 \) is negative, so:
\[ |x^2 - x - 6| = -(x^2 - x - 6) = - (x - 3)(x + 2) = -x^2 + x + 6 \]
Thus, the equation becomes:
\[ -x^2 + x + 6 = x + 2 \]
Simplifying gives:
\[ -x^2 + 6 = 2 \implies -x^2 = -4 \implies x^2 = 4 \implies x = -2 \text{ or } x = 2 \]
Both values \( x = -2 \) and \( x = 2 \) are valid in this case.

#### Case 3: \( x \geq 3 \)
In this case, both \( x - 3 \) and \( x + 2 \) are non-negative, so:
\[ |x^2 - x - 6| = x^2 - x - 6 \]
Thus, the equation becomes:
\[ x^2 - x - 6 = x + 2 \]
Simplifying gives:
\[ x^2 - 2x - 8 = 0 \]
Factoring gives:
\[ (x - 4)(x + 2) = 0 \]
Thus, \( x = 4 \) or \( x = -2 \). Since \( x \geq 3 \), the valid solution from this case is \( x = 4 \).

### Step 3: Collect all valid solutions
From all cases, the valid solutions are:
- From Case 2: \( x = -2, 2 \)
- From Case 3: \( x = 4 \)

Thus, the roots of the equation \( |x^2 - x - 6| = x + 2 \) are:
\[ \boxed{-2, 2, 4} \]