Q.Solve the following for $x$, where $|x|$ is the modulus function, $[x]$ is the greatest integer function, $\{x\}$ is the fractional part function: $|x^2-x-6|=x+2$.

$|x^2-x-6|=x+2$ needs $x+2\ge0$, i.e. $x\ge-2$, since a modulus is never negative.

Case 1: $x^2-x-6=x+2 \implies x^2-2x-8=0 \implies (x-4)(x+2)=0 \implies x=4$ or $x=-2$.

Case 2: $-(x^2-x-6)=x+2 \implies -x^2+x+6=x+2 \implies -x^2+4=0 \implies x^2=4 \implies x=\pm2$. …

First require the RHS to be $\ge0$ (a necessary condition for any modulus equation), then split into the two sign cases of the …

Forgetting to check $x\ge-2$ upfront, or forgetting to verify each candidate root back in the original modulus equation (squa …