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

$x=\pm4$
Since $x^2=a$ implies $x=\pm\sqrt{a}$ (the Square Root Property), then the solutions of the given equation, $x^2=16 ,$ are \begin{array}{l}\require{cancel} x=\pm\sqrt{16} \\\\ x=\pm\sqrt{(4)^2} \\\\ x=\pm4 .\end{array}