## Introductory Algebra for College Students (7th Edition)

$x=\left\{ 4-3\sqrt{2},4+3\sqrt{2} \right\}$
Taking the square root of both sides (the Square Root Property), the solutions to the given equation, $(z-4)^2=18 ,$ are \begin{array}{l}\require{cancel} z-4=\pm\sqrt{18} \\\\ z-4=\pm\sqrt{9\cdot2} \\\\ z-4=\pm\sqrt{(3)^2\cdot2} \\\\ z-4=\pm3\sqrt{2} \\\\ z=4\pm3\sqrt{2} .\end{array} Hence, $x=\left\{ 4-3\sqrt{2},4+3\sqrt{2} \right\} .$