## Algebra: A Combined Approach (4th Edition)

$x = (-2 + 3\sqrt 2, -2 - 3\sqrt 2)$
$(x + 2)^2 = 18$ $x + 2 = ±\sqrt 18$ (Use Radical Property) $x + 2 = ±3\sqrt 2$ (Simplify the radical) $x = -2 ± 3\sqrt 2$ (Subtract 2 from both sides) Check 1: Let $x = -2+3\sqrt 2$ $(x+2)^2 = 18$ $((-2+3\sqrt 2)+2)^2$ could be $18$ $(-2+3\sqrt 2+2)^2$ could be $18$ $(3\sqrt 2)^2$ could be $18$ $9\times2$ could be $18$ $18=18$ True. Check 2: Let $x = -2+3\sqrt 2$ $(x+2)^2 = 18$ $((-2-3\sqrt 2)+2)^2$ could be $18$ $(-2-3\sqrt 2+2)^2$ could be $18$ $(-3\sqrt 2)^2$ could be $18$ $9\times2$ could be $18$ $18=18$ True. Therefore the solution set is $(-2 + 3\sqrt 2, -2 - 3\sqrt 2)$