Chapter 1 - Section 1.2 - Quadratic Equations - 1.2 Assess Your Understanding - Page 101: 43

The solution set is {$2-\sqrt 2, 2+\sqrt 2$}.

$x^2-4x+2=0$ $a=1, b=-4, c=2$ $D=b^2-4ac=(-4)^2-4\cdot1\cdot2=16-8=8>0$. The equation has two real solutions. $x=\frac{-b\pm\sqrt {b^2-4ac}}{2a}=\frac{4\pm\sqrt 8}{2}$ Use quadratic formula. $x=\frac{4\pm2\sqrt 2}{2}$ Simplify $x=2\pm\sqrt 2$ $x=2+\sqrt 2$, $x=2-\sqrt 2$ The solution set is {$2-\sqrt 2, 2+\sqrt 2$}