## Precalculus: Mathematics for Calculus, 7th Edition

$x=2\pm\sqrt{2}$
$x^{2}-4x+2=0$ Let's take the independent term to the right side of the equation: $x^{2}-4x=-2$ Let's remember, that in order to complete the square, we need to add $(\dfrac{b}{2})^{2}$ to both sides of the equation. $b$ is the coefficient of the first degree term. In this case $b=-4$ $x^{2}-4x+(\dfrac{-4}{2})^{2}=-2+(\dfrac{-4}{2})^{2}$ $x^{2}-4x+4=-2+4$ $x^{2}-4x+4=2$ We have a perfect square trinomial on the left side of the equation. We factor it and the equation becomes: $(x-2)^{2}=2$ Take the square root of both sides of the equation: $\sqrt{(x-2)^{2}}=\sqrt{2}$ $x-2=\pm\sqrt{2}$ Solve for $x$: $x=2\pm\sqrt{2}$