College Algebra, Chapter 1, 1.3, Section 1.3, Problem 26

Solve $\displaystyle 3x^2 - 6x - 1 = 0$ by completing the square.


$
\begin{equation}
\begin{aligned}

3x^2 - 6x - 1 =& 0
&& \text{Given}
\\
\\
3x^2 - 6x =& 1
&& \text{Add 1}
\\
\\
x^2 - 2x =& \frac{1}{3}
&& \text{Divide both sides by 3 to eliminate the coefficient of } x^2
\\
\\
x^2 - 2x + 1 =& \frac{1}{3} + 1
&& \text{Complete the square: add } \left( \frac{-2}{2} \right)^2 = 1
\\
\\
(x - 1)^2 =& \frac{4}{3}
&& \text{Perfect square}
\\
\\
x - 1 =& \pm \sqrt{\frac{4}{3}}
&& \text{Take square root}
\\
\\
x =& 1 \pm \sqrt{\frac{4}{3}}
&& \text{Add 1}
\\
\\
x =& 1 + \frac{2}{\sqrt{3}} \text{ and } x = 1 - \frac{2}{\sqrt{3}}
&& \text{Solve for } x
\\
\\
x =& \frac{3 + 2 \sqrt{3}}{3} \text{ and } x = \frac{3 - 2 \sqrt{3}}{3}
&& \text{Rationalize}



\end{aligned}
\end{equation}
$