Question

The product of the distinct roots of |x² - x - 6 | = x + 2 is

• A. -8
• B. -24
• C. -4
• D. -16

Answer 1

The Correct Option is D

Solve for all real values of \(x\) satisfying the equation:

\[ |x^2 - x - 6| = x + 2 \]

Step 1: Factor the Quadratic

We start by factoring: \[ x^2 - x - 6 = (x + 2)(x - 3) \]

Step 2: Analyze Cases Based on Absolute Value

Case 1: When \(x^2 - x - 6 < 0\)

This means: \[ (x + 2)(x - 3) < 0 \] The solution to this inequality is: \[ -2 < x < 3 \] Within this interval: \[ |x^2 - x - 6| = -(x^2 - x - 6) = -[(x + 2)(x - 3)] \] Now equating: \[ -(x + 2)(x - 3) = x + 2 \Rightarrow - (x - 3) = 1 \Rightarrow x = 2 \]

Case 2: When \(x^2 - x - 6 \geq 0\)

\[ (x + 2)(x - 3) \geq 0 \Rightarrow x \leq -2 \quad \text{or} \quad x \geq 3 \]

Boundary Check:

• At \(x = -2\): \(|x^2 - x - 6| = |-2^2 + 2 - 6| = |-4| = 4\), and \(x + 2 = 0\). So this does NOT satisfy.
• But actually \(|x^2 - x - 6| = 4\) and \(x + 2 = 0\) → contradiction. Let's recheck:

\[ x = -2 \Rightarrow |4 + 2 - 6| = |0| = 0, \quad x + 2 = 0 \Rightarrow \text{Match! So } x = -2 \text{ is valid.} \]

\[ x = 3 \Rightarrow |9 - 3 - 6| = |0| = 0, \quad x + 2 = 5 \Rightarrow \text{Not equal. So } x = 3 \text{ is not a root.} \]

Check Beyond Boundaries:

For \(x > 3\), we consider: \[ |x^2 - x - 6| = x^2 - x - 6 \] So the equation becomes: \[ x^2 - x - 6 = x + 2 \Rightarrow x^2 - 2x - 8 = 0 \Rightarrow (x - 4)(x + 2) = 0 \Rightarrow x = 4 \text{ or } x = -2 \]

Since \(x = -2\) is already found valid, and \(x = 4 > 3\), it's also valid.

Final Roots:

• \(x = -2\)
• \(x = 2\)
• \(x = 4\)

Step 3: Product of Roots

\[ (-2) \cdot 2 \cdot 4 = \boxed{-16} \]