## Precalculus: Mathematics for Calculus, 7th Edition

$x = 3 \pm 2\sqrt2$
$Find$ $all$ $real$ $solutions$ $of$ $the$ $quadratic$ $equation:$ $x^2-6x+1 = 0$ Use the Quadratic Equation Formula: $x = \frac{-b \pm \sqrt {b^2-4ac}}{2a}$ $1x^2-6x+1$ $a = 1, b = -6, c = 1$ $x = \frac{-(-6) \pm \sqrt {(-6)^2- 4(1 \times 1)}}{2(1)}$ $x = \frac{6 \pm \sqrt {36- 4}}{2}$ $x = \frac{6 \pm \sqrt {32}}{2}$ [Note: $\sqrt {32} = \sqrt {16\times2} = \sqrt {16}\times\sqrt2$ = $4\sqrt2$] $x = \frac{6 \pm 4\sqrt {2}}{2}$ $x = \frac{6}{2} \pm \frac{4\sqrt2}{2}$ $x = 3 \pm 2\sqrt2$