## Algebra and Trigonometry 10th Edition

Solve the individual equaions obtained after breaking the expression into two separate equations. Case 1: $x^2-6 = -x$ $x^2+x-6=0$ (x+3)(x-2)=0 Plug in the Case 1 equation to see -3 is an extraneous solution. Hence x=2. Case 2: $x^2-6 = x$ $x^2-x-6=0$ (x-3)(x+2)=0 Plug in the Case 2 equation to see -2 is an extraneous solution. Hence x=3.