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

x² - x - 6 = ( x + 2 ) ( x - 3 )
Case 1 : x² - x - 6 < 0
i.e. (x + 2)(x - 3) < 0
⇒ -2 < x < 3 and | x² - x - 6 | = - ( x² - x - 6 )
Therefore , | x² - x - 6 | = x + 2
= - (x + 2)(x - 3) = x + 2
⇒ (x - 3) = -1 ⇒ x = 2

Case 2 : : x² - x - 6 ≥ 0
i.e. (x + 2)(x-3) ≥ 0
⇒ x ≤ -2 or x ≥ 3
Now Checking for boundary conditions :
For x=-2 , | x² - x - 6 | = x + 2 , therefore , x=-2 is also the root. But for x=3 , | x² - x - 6 | ≠ x + 2.
Hence x=3 is NOT the root.
And for the interval x < -2 or x > 3 the expression | x² - x - 6 | = x² - x - 6
Therefore , | x² - x - 6 | = x + 2
= (x+2)(x-3) = x + 2
⇒ ( x - 3 ) = 1 ⇒ x = 4
∴ the root are -2, 2, and 4.
So the required product = (2)(-2)(4)=-16