The product of the distinct roots of ∣x2 − x − 6∣ = x + 2 is
Explanation:
∣x2 − x − 6∣ = x + 2
∴ ∣(x − 3)(x + 2)∣ = x + 2
Case 1: x < −2. i.e. (x − 3)(x + 2) > 0
(x - 3)(x + 2) = x + 2
∴ x = 4. (which is rejected since 4 is not less than −2)
Case 2: x = −2.
This is a real root of this equation.
Case 3: −2 < x < 3. i.e. (x − 3)(x + 2) < 0
- (x - 3)(x + 2) = x + 2
∴ x = 2.
Case 4: x = 3.
This does not satisfy the equation so x = 3 is not a root of this equation.
Case 5: x > 3. (x − 3)(x + 2) > 0
(x − 3)(x + 2) = x + 2
∴ x = 4.
Required product = (−2) × 2 × 4 = −16.
Hence, option (a).
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