## College Algebra (10th Edition)

The solution set is $\left\{-3, 3\right\}$.
Raise both sides of the equation to the 4th power to obtain: $x^2+16 = (\sqrt{5})^4 \\x^2+16=25$ Subtract 16 on both sides of the equation to obtain: $x^2=25-16 \\x^2=9$ Take the square root of both sides to obtain: $x = \pm\sqrt{9} \\x = \pm \sqrt{3^2} \\x = \pm 3$ Thus, the solution set is $\left\{-3, 3\right\}$.