Precalculus (6th Edition)

$\color{blue}{\left\{-29, 35\right\}}$
The numerator of the rational exponent is even so raising both sides to $\frac{5}{2}$ will not work. Raise both sides to the fifth power to obtain: $((x-3)^{2/5})^5=4^5 \\(x-3)^2=1024$ Take the square root of both sides to obtain: $\sqrt{x-3)^2}=\pm \sqrt{1024} \\x-3 = \pm \sqrt{32^2} \\x-3 = \pm 32$ Add $3$ to both sides of the equation to obtain: $\\x=3\pm 32 \\x_1=3-32=-29 \\x_2=3+32=35$ Upon checking both proposed solutions satisfy the original equation. Thus, the solutions set is $\color{blue}{\left\{-29, 35\right\}}$.