College Algebra, Chapter 1, 1.1, Section 1.1, Problem 12

Check whether a.) $\displaystyle x = \frac{b}{2}$ or b.) $\displaystyle x = \frac{1}{b}$ is a solution of the equation $\displaystyle x^2 - bx + \frac{1}{4}b^2 = 0$

a.) $\displaystyle x = \frac{b}{2}$

$
\begin{equation}
\begin{aligned}
\left( \frac{b}{2} \right)^2 - b \left( \frac{b}{2} \right) + \frac{1}{4} b^2 &= 0 && \text{Substitute } x = \frac{b}{2}\\
\\
\frac{b^2}{4} - \frac{b^2}{2} + \frac{b^2}{4} &= 0 && \text{Get the LCD}\\
\\
\frac{b^2 - 2b^2 + b^2}{4} &= 0 && \text{Combine like terms}\\
\\
\frac{0}{4} &= 0 && \text{Simplify}\\
\\
0 &= 0
\end{aligned}
\end{equation}
$

So $\displaystyle x = \frac{b}{2}$ is the solution to the equation.

b.) $\displaystyle x = \frac{1}{b}$

$
\begin{equation}
\begin{aligned}
\left( \frac{1}{b} \right)^2 - \cancel{b} \left( \frac{1}{\cancel{b}} \right) + \frac{1}{4} b^2 &= 0 && \text{Substitute } x = \frac{1}{b}\\
\\
\frac{1}{b^2} - \frac{1}{1} + \frac{b^2}{4} &= 0 && \text{Get the LCD}\\
\\
\frac{4-4b^2 + b^4}{4b^2} &\neq 0
\end{aligned}
\end{equation}
$

So $\displaystyle x = \frac{1}{b}$ is not the solution to the equation.