# How do you solve x^2 + 8x + 2 = 0 by completing the square?

Jun 28, 2017

$x = - 4 \pm \sqrt{14}$

#### Explanation:

$\text{express as } {x}^{2} + 8 x = - 2$

$\text{to "color(blue)"complete the square}$

add (1/2"coefficient of x-term")^2" to both sides"

$\text{that is add " (8/2)^2=16" to both sides}$

$\Rightarrow {x}^{2} + 8 x \textcolor{red}{+ 16} = - 2 \textcolor{red}{+ 16}$

$\Rightarrow {\left(x + 4\right)}^{2} = 14$

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

$\sqrt{{\left(x + 4\right)}^{2}} = \pm \sqrt{14} \leftarrow \text{ note plus or minus}$

$\Rightarrow x + 4 = \pm \sqrt{14}$

$\text{subtract 4 from both sides}$

$x \cancel{+ 4} \cancel{- 4} = \pm \sqrt{14} - 4$

$\Rightarrow x = - 4 \pm \sqrt{14}$

Jun 28, 2017

Move +2 to the right side of the equation.
${x}^{2} + 8 x = - 2$

Then halve the coefficient of x.
${x}^{2} + \left(\frac{8}{2}\right) x = - 2$

Then square that same coefficient.
${x}^{2} + {\left(\frac{8}{2}\right)}^{2} x = - 2$

Since ${\left(\frac{8}{2}\right)}^{2}$ = $16$ we can put that number into the equation.
So, ${x}^{2} + 8 x + 16 = - 2$

When you find the number to complete the square you must add it to both sides of the equation.
So, ${x}^{2} + 8 x + 16 = - 2 + 16$
= ${x}^{2} + 8 x + 16 = 14$

Then factorise ${x}^{2} + 8 x + 16$ = $\left(x + 4\right) \left(x + 4\right)$

Therefore, the answer is: $\left(x + 4\right) \left(x + 4\right) = 14$