College Algebra, Chapter 1, 1.3, Section 1.3, Problem 44

Find all real solutions of $\displaystyle x^2 - 4x + 1 = 0$.


$
\begin{equation}
\begin{aligned}

x^2 - 4x + 1 =& 0
&& \text{Given}
\\
\\
x^2 - 4x =& -1
&& \text{Subtract 1}
\\
\\
x^2 - 4x + 4 =& -1 + 4
&& \text{Complete the square: add } \left( \frac{-4}{2} \right)^2 = 4
\\
\\
(x - 2)^2 =& 3
&& \text{Perfect square}
\\
\\
x - 2 =& \pm \sqrt{3}
&& \text{Take the square root}
\\
\\
x =& 2 \pm \sqrt{3}
&& \text{Add 2}
\\
\\
x =& 2 + \sqrt{3} \text{ and } x = 2 - \sqrt{3}
&& \text{Solve for } x
\\
\\
y =& \frac{1 + \sqrt{5}}{4} \text{ and } y = \frac{1 - \sqrt{5}}{4}
&& \text{Simplify}

\end{aligned}
\end{equation}
$