Absolute value problem 1 by Dhaval Furia

Note that $|x^2 - x - 6| = |(x+2)(x-3)|$ .

For $x < - 2$ :

$\begin{aligned} |(x+2)(x-3)| & = x + 2 \\ (x+2)(x-3) & = x + 2 \\ x^2 - x - 6 & = x + 2 \\ x^2 - 2x - 8 & = 0 \\ (x+2)(x-4) & = 0 \end{aligned}$

There is no solution.

For $-2 \le x < 3$ :

$\begin{aligned} |(x+2)(x-3)| & = x + 2 \\ (x+2)(3-x) & = x + 2 \\ -x^2 + x + 6 & = x + 2 \\ x^2 & = 4 \\ \implies x & = \pm 2 \end{aligned}$

For $x \ge 3$ :

$\begin{aligned} |(x+2)(x-3)| & = x + 2 \\ (x+2)(x-3) & = x + 2 \\ x^2 - x - 6 & = x + 2 \\ x^2 - 2x - 8 & = 0 \\ (x+2)(x-4) & = 0 \\ \implies x & = 4 & \small \blue{\text{Since }x \ge 3} \end{aligned}$

Therefore the product of the roots is $-2\cdot 2 \cdot 4 = \boxed{-16}$ .

There are two possibilities :

(I) $x\geq 3$ or $x\leq -2\implies$ In this case, the equation is $(x+2)(x-3)=x+2\implies x=4,-2$

(II) $-2\leq x\leq 3\implies$ In this case the equation is $(x+2)(3-x)=x+2\implies x=2,-2$

Hence the product of the distinct values of $x$ is $4\times 2\times (-2)=\boxed {-16}$ .