College Algebra, Chapter 1, 1.3, Section 1.3, Problem 2

Explain how you would use each method to solve the equation $x^2 - 4x - 5 = 0$

a.) By factoring


$
\begin{equation}
\begin{aligned}

x^2 - 4x - 5 =& 0
&& \text{Given}
\\
\\
(x - 5)(x + 1) =& 0
&& \text{Factor}
\\
\\
x - 5 =& 0 \text{ or } x + 1 = 0
&& \text{Zero Product Property}
\\
\\
x =& 5 \text{ or } x = -1
&& \text{Solve}

\end{aligned}
\end{equation}
$


b.) By completing the square


$
\begin{equation}
\begin{aligned}

x^2 - 4x - 5 =& 0
&& \text{Given}
\\
\\
x^2 - 4x =& 5
&& \text{Add 5}
\\
\\
x^2 - 4x + 4 =& 5 + 4
&& \text{Complete the square: add } \left( \frac{-4}{2} \right)^2 = 4
\\
\\
(x - 2)^2 =& 9
&& \text{Perfect Square}
\\
\\
x - 2 =& \pm \sqrt{9}
&& \text{Take square root}
\\
\\
x =& 2 \pm 3
&& \text{Add two}
\\
\\
x =& 5 \text{ and } x = -1
&& \text{Solve}

\end{aligned}
\end{equation}
$


c.) By using the Quadratic Formula


$
\begin{equation}
\begin{aligned}

x^2 - 4x - 5 =& 0
&& \text{Given}
\\
\\
x =& \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-5)}}{2(1)}
&& \text{Apply the formula } x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\\
\\
x =& \frac{4 \pm \sqrt{16 + 20}}{2}
&& \text{Simplify}
\\
\\
x =& \frac{4 \pm 6}{2}
&&
\\
\\
x =& 2 \pm 3
&& \text{Solve for } x
\\
\\
x =& 5 \text{ and } x = -1
&&


\end{aligned}
\end{equation}
$