# How do you solve  x^2-2x-4=0 by completing the square?

Mar 3, 2018

color(purple)(x = sqrt5 + 1, -sqrt5 + 1 = 3.236, -1.236

#### Explanation:

${x}^{2} - 2 x - 4 = 0$

${x}^{2} - 2 x = 4$

${x}^{2} - 2 x + 1 = 4 + 1 = 5$

${\left(x - 1\right)}^{2} = {\left(\sqrt{5}\right)}^{2}$

$x - 1 = \pm \sqrt{5}$

color(purple)(x = sqrt5 + 1, -sqrt5 + 1 = 3.236, -1.236

Mar 3, 2018

$x = 1 \pm \sqrt{5}$

#### Explanation:

$\text{using the method of "color(blue)"completing the square}$

• " the coefficient of the "x^2" term must be 1 which it is"

• " add/subtract "(1/2"coefficient of x-term")^2" to"
${x}^{2} - 2 x$

$\Rightarrow {x}^{2} + 2 \left(- 1\right) x \textcolor{red}{+ 1} \textcolor{red}{- 1} - 4 = 0$

$\Rightarrow {\left(x - 1\right)}^{2} - 5 = 0 \leftarrow \textcolor{b l u e}{\text{add 5 to both sides}}$

$\Rightarrow {\left(x - 1\right)}^{2} = 5$

$\textcolor{b l u e}{\text{Take square root of both sides}}$

$\Rightarrow x - 1 = \pm \sqrt{5} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\text{add 1 to both sides}$

$\Rightarrow x = 1 \pm \sqrt{5} \leftarrow \textcolor{red}{\text{exact solutions}}$