## Algebra and Trigonometry 10th Edition

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