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

Options

A. two positive roots
B. two real roots
C. three real roots
D. four real roots

Answer

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

### Step 1: Set up the cases
The absolute value equation \(|A| = B\) can be rewritten as two separate equations:
1. \(A = B\)
2. \(A = -B\)

In our case, we have:
1. \(x^2 - x - 6 = x + 2\)
2. \(x^2 - x - 6 = -(x + 2)\)

### Step 2: Solve Case 1
**Case 1:** \(x^2 - x - 6 = x + 2\)

Rearranging the equation:
\[ x^2 - x - 6 - x - 2 = 0 \]
\[ x^2 - 2x - 8 = 0 \]

Now, we can factor this quadratic:
\[ (x - 4)(x + 2) = 0 \]

Setting each factor to zero gives us the roots:
\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \]
\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \]

### Step 3: Solve Case 2
**Case 2:** \(x^2 - x - 6 = -(x + 2)\)

Rearranging the equation:
\[ x^2 - x - 6 + x + 2 = 0 \]
\[ x^2 - 4 = 0 \]

Factoring this gives us:
\[ (x - 2)(x + 2) = 0 \]

Setting each factor to zero gives us the roots:
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \]

### Step 4: Compile the roots
From Case 1, we found the roots \(x = 4\) and \(x = -2\).
From Case 2, we found the roots \(x = 2\) and \(x = -2\).

### Final Roots
The unique roots of the equation \(|x^2 - x - 6| = x + 2\) are:
\[ x = -2, \quad x = 2, \quad x = 4 \]