College Algebra, Chapter 1, 1.3, Section 1.3, Problem 76

Solve the equation $\displaystyle bx^2 + 2x + \frac{1}{b} = 0 (b \neq 0)$ for $x$.


$
\begin{equation}
\begin{aligned}

bx^2 + 2x + \frac{1}{b} =& 0
&& \text{Given}
\\
\\
bx^2 + 2x =& \frac{-1}{b}
&& \text{Subtract } \frac{1}{b}
\\
\\
x^2 + \frac{2}{b} x =& \frac{-1}{b^2}
&& \text{Divide both sides by } b
\\
\\
x^2 + \frac{2}{b} x + \frac{1}{b^2} =& \frac{-1}{b^2} + \frac{1}{b^2}
&& \text{Complete the square: add } \left( \frac{\displaystyle \frac{2}{b}}{2} \right)^2 = \frac{1}{b^2}
\\
\\
\left( x + \frac{1}{b} \right)^2 =& 0
&& \text{Perfect square}
\\
\\
x + \frac{1}{b} =& \pm \sqrt{0}
&& \text{Take the square root}
\\
\\
x =& \frac{-1}{b}
&& \text{Solve for } x



\end{aligned}
\end{equation}
$