## Precalculus (6th Edition) Blitzer

$\left\{1-\sqrt3, 1+\sqrt3\right\}$
RECALL: To solve $x^2 + bx=c$ by completing the square, add $\left(\frac{b}{2}\right)^2$ on both sides of the equation. Thus, to solve the given equation by completing the square, add $(\frac{-2}{2})^2=(-1)^2=1$ on both sides of the equation to obtain: $x^2-2x+1=2+1 \\x^2-2x+1=3 \\(x-1)^2=3$ Take the square root of both sides to obtain: $x-1=\pm \sqrt{3}$ Add 1 on both sides of the equation to obtain: $x = 1 \pm\sqrt3$ Two solutions: $x_1 = 1+\sqrt3 \\x_2 = 1-\sqrt3$ Therefore, the solution of the given equation is $\left\{1-\sqrt3, 1+\sqrt3\right\}$.