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The product of the distinct roots of $|x^{2}-x-6|=x+2$ is

1. $-8$
2. $-24$
3. $-4$
4. $-16$

We have $x^2 – x – 6 = x+2 \qquad \to (1)$

and $-(x^2-x-6) = x+2 \qquad \to (2)$

From $(1),$ we get $x^2-2x -8 = 0 \implies (x-4)(x+2) = 0 \implies x = 4, -2$

From $(2),$ we get $x^2-4 = 0 \implies x^2 = 4 \implies x = \pm 2$

So, the roots of the equation are $4,-2,2$

Product of the roots $= -16$

Correct Option: D
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