Solve : `|x^(2)-x-6|=x+2 , x in R `

Answer

To solve the equation \( |x^2 - x - 6| = x + 2 \), we will break it down into cases based on the properties of absolute values.

### Step 1: Identify the expression inside the absolute value
The expression inside the absolute value is \( x^2 - x - 6 \). We can factor this quadratic expression:
\[ x^2 - x - 6 = (x - 3)(x + 2) \]
This means the roots are \( x = 3 \) and \( x = -2 \).

### Step 2: Determine the intervals for cases
The critical points divide the real line into three intervals:
1. \( (-\infty, -2) \)
2. \( [-2, 3] \)
3. \( (3, \infty) \)

### Step 3: Case 1: \( x \in (-\infty, -2) \) or \( x \in (3, \infty) \)
In this interval, \( x^2 - x - 6 \) is positive, so we can drop the absolute value:
\[ x^2 - x - 6 = x + 2 \]
Rearranging gives:
\[ x^2 - 2x - 8 = 0 \]
Factoring this quadratic:
\[ (x - 4)(x + 2) = 0 \]
Thus, the solutions are:
\[ x = 4 \quad \text{and} \quad x = -2 \]
- **Check**: \( x = 4 \) is in \( (3, \infty) \) (valid). \( x = -2 \) is not in this interval but is a boundary point.

### Step 4: Case 2: \( x \in [-2, 3] \)
In this interval, \( x^2 - x - 6 \) is negative, so we have:
\[ -(x^2 - x - 6) = x + 2 \]
This simplifies to:
\[ -x^2 + x + 6 = x + 2 \]
Rearranging gives:
\[ -x^2 + 4 = 0 \quad \Rightarrow \quad x^2 = 4 \]
Thus:
\[ x = 2 \quad \text{or} \quad x = -2 \]
- **Check**: Both \( x = 2 \) and \( x = -2 \) are in the interval \( [-2, 3] \) (valid).

### Step 5: Compile all valid solutions
The valid solutions from both cases are:
- From Case 1: \( x = 4 \)
- From Case 2: \( x = 2 \) and \( x = -2 \)

Thus, the complete solution set is:
\[ \boxed{4, 2, -2} \]