What is the number of real solutions of
`|x^(2) - x-6|=x+2`?

Options

A. 4
B. 3
C. 2
D. 1

Explanation

To solve the equation \( |x^2 - x - 6| = x + 2 \), we will consider two cases based on the definition of absolute value.

### Step 1: Identify the cases for the absolute value
The expression inside the absolute value, \( x^2 - x - 6 \), can be either positive or negative. We will consider two cases:
1. Case 1: \( x^2 - x - 6 \geq 0 \)
2. Case 2: \( x^2 - x - 6 < 0 \)

### Step 2: Solve Case 1 (\( x^2 - x - 6 \geq 0 \))
In this case, we can remove the absolute value:
\[ x^2 - x - 6 = x + 2 \]
Rearranging the equation gives:
\[ x^2 - x - x - 6 - 2 = 0 \implies x^2 - 2x - 8 = 0 \]

### Step 3: Factor the quadratic equation
Now we will factor the quadratic:
\[ x^2 - 2x - 8 = (x - 4)(x + 2) = 0 \]
Setting each factor to zero gives us:
\[ x - 4 = 0 \implies x = 4 \]
\[ x + 2 = 0 \implies x = -2 \]

### Step 4: Check which solutions are valid for Case 1
We need to check if these solutions satisfy the condition \( x^2 - x - 6 \geq 0 \):
- 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 5: Solve Case 2 (\( x^2 - x - 6 < 0 \))
In this case, we have:
\[ -(x^2 - x - 6) = x + 2 \]
This simplifies to:
\[ -x^2 + x + 6 = x + 2 \]
Rearranging gives:
\[ -x^2 + x - x + 6 - 2 = 0 \implies -x^2 + 4 = 0 \implies x^2 = 4 \]
Taking the square root gives:
\[ x = 2 \quad \text{or} \quad x = -2 \]

### Step 6: Check which solutions are valid for Case 2
We need to check if these solutions satisfy the condition \( x^2 - x - 6 < 0 \):
- 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 7: Summary of valid solutions
The valid solutions we found are:
- From Case 1: \( x = 4 \) and \( x = -2 \)
- From Case 2: \( x = 2 \)

Thus, the unique solutions are \( x = 4, -2, 2 \).

### Conclusion
The total number of real solutions is **3**.