College Algebra, Chapter 1, 1.3, Section 1.3, Problem 28

Solve $\displaystyle x^2 = \frac{3}{4} x - \frac{1}{8}$ by completing the square.


$
\begin{equation}
\begin{aligned}

x^2 =& \frac{3}{4} x - \frac{1}{8}
&& \text{Given}
\\
\\
x^2 - \frac{3}{4} x =& \frac{-1}{8}
&& \text{Subtract } \frac{3}{4} x
\\
\\
x^2 - \frac{3}{4} x + \frac{9}{64} =& \frac{-1}{8} + \frac{9}{64}
&& \text{Complete the square: add } \left( \frac{\displaystyle \frac{-3}{4}}{2} \right)^2 = \frac{9}{64}
\\
\\
\left(x - \frac{3}{8} \right)^2 =& \frac{1}{64}
&& \text{Perfect square}
\\
\\
x - \frac{3}{8} =& \pm \sqrt{\frac{1}{64}}
&& \text{Take square root}
\\
\\
x =& \frac{3}{8} \pm \frac{1}{8}
&& \text{Add } \frac{3}{8}
\\
\\
x =& \frac{1}{2} \text{ and } x = \frac{1}{4}
&& \text{Solve for } x

\end{aligned}
\end{equation}
$