## College Algebra (10th Edition)

The solution is $\left\{-1, 3\right\}$.
Take the square root of both sides to obtain: $x-1 = \pm \sqrt{4} \\x-1= \pm \sqrt{2^2} \\x-1 = \pm 2$ Add 1 to both sides to obtain: $x = 1 \pm 2$ Split the solutions to obtain: $x_1 = 1+2 = 3 \\x_2 = 1-2 = -1$ Thus, the solution is $\left\{-1, 3\right\}$.