College Algebra, Chapter 1, 1.3, Section 1.3, Problem 36

Find all real solutions of $2x^2 - 8x + 4 = 0$.


$
\begin{equation}
\begin{aligned}

2x^2 - 8x + 4 =& 0
&& \text{Given}
\\
\\
2(x^2 - 4x + 2) =& 0
&& \text{Factor out 2}
\\
\\
x^2 - 4x =& -2
&& \text{Subtract 2}
\\
\\
x^2 - 4x + 4 =& -2 + 4
&& \text{Complete the the square: add } \left( \frac{-4}{2} \right)^2 = 4
\\
\\
(x - 2)^2 =& 2
&& \text{Perfect square}
\\
\\
x - 2 =& \pm \sqrt{2}
&& \text{Take the square root}
\\
\\
x =& 2 \pm \sqrt{2}
&& \text{Add 2}
\\
\\
x =& 2 + \sqrt{2} \text{ and } x = 2 - \sqrt{2}
&& \text{Solve for } x

\end{aligned}
\end{equation}
$