# How do you solve x^2-2x-4=0 by factoring?

Oct 26, 2016

$x = 1 + \sqrt{5}$
OR
$x = 1 - \sqrt{5}$

#### Explanation:

The equation can be easily solved by factoring ${x}^{2} - 2 x - 4$ which is done by applying the quadratic formula.

$\textcolor{b l u e}{\delta = {b}^{2} - 4 a c} = {\left(- 2\right)}^{2} - 4 \left(1\right) \left(- 4\right) = 4 + 16 = 20$

$\textcolor{g r e e n}{{x}_{1} = \frac{- b + \sqrt{\delta}}{2 a}} = \frac{2 + \sqrt{20}}{2} = \frac{2 + \sqrt{4 \cdot 5}}{2} = \frac{2 + 2 \sqrt{5}}{2}$
${x}_{1} = \left(1 + \sqrt{5}\right)$

$\textcolor{b r o w n}{{x}_{2} = \frac{- b - \sqrt{\delta}}{2 a}} = \frac{2 - \sqrt{20}}{2} = \frac{2 - \sqrt{4 \cdot 5}}{2} = \frac{2 - 2 \sqrt{5}}{2}$
${x}_{2} = \left(1 - \sqrt{5}\right)$

${x}^{2} - 2 x - 4 = 0$
$\Rightarrow \left(x - \textcolor{g r e e n}{{x}_{1}}\right) \left(x - \textcolor{b r o w n}{{x}_{2}}\right) = 0$

$\Rightarrow \left(x - \left(1 + \sqrt{5}\right)\right) \left(x - \left(1 - \sqrt{5}\right)\right) = 0$

$\Rightarrow$
$x - \left(1 + \sqrt{5}\right) = 0 \Rightarrow x = 1 + \sqrt{5}$
OR
$x - \left(1 - \sqrt{5}\right) = 0 \Rightarrow x = 1 - \sqrt{5}$