## Elementary Algebra

{$-1, 3$}
Using Property 10.1, which states that for any non-negative real number $a$, $(x-b)^{2}=a$ can be written as $x-b=\pm\sqrt a$, we obtain: Step 1: $(x-1)^{2}=4$ Step 2: $x-1=\pm \sqrt 4$ Step 3: $x-1=\pm 2$ Step 4: $x-1=2$ or $x-1=-2$ Step 5: $x=2+1$ or $x=-2+1$ Step 6: $x=3$ or $x=-1$ The solution set is {$-1, 3$}.