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

Jun 18, 2018

Solution: $x = \left(- 4 + 3 \sqrt{2}\right) , x = \left(- 4 - 3 \sqrt{2}\right)$

#### Explanation:

${x}^{2} + 8 x - 2 = 0 \mathmr{and} {x}^{2} + 8 x = 2$ or

${x}^{2} + 8 x + 16 = 2 + 16$ or

${\left(x + 4\right)}^{2} = 18 \mathmr{and} \left(x + 4\right) = \pm \sqrt{18}$ or

$x + 4 = \pm 3 \sqrt{2} \mathmr{and} x = - 4 \pm 3 \sqrt{2}$

Solution: $x = \left(- 4 + 3 \sqrt{2}\right) , x = \left(- 4 - 3 \sqrt{2}\right)$[Ans]

Jun 18, 2018

Take the second coefficient, divide it by 2, and square it, to complete the square, and obtain the solutions:
$x = - 4 + \sqrt{18}$ and $x = - 4 - \sqrt{18}$

#### Explanation:

To complete the square, take the second coefficient (the one next to the $x$), divide it by 2, and square it. This will give you the number you need to complete the square.
In this case, that would be ${\left(8 \div 2\right)}^{2} = 16$

Add this number to both sides of the original equation:
$\left({x}^{2} + 8 x + 16\right) - 2 = 16$
${\left(x + 4\right)}^{2} - 2 = 16$
${\left(x + 4\right)}^{2} = 18$

This gives us 2 possible solutions:
$x + 4 = \pm \sqrt{18}$

$x = - 4 \pm \sqrt{18}$

Jun 18, 2018

$x = - 4 \pm 3 \sqrt{2}$

#### Explanation:

$\text{add 2 to both sides}$

${x}^{2} + 8 x = 2$

$\text{add "(1/2"coefficient of the x-term")^2" to both sides}$

${x}^{2} + 2 \left(4\right) x \textcolor{red}{+ 16} = 2 \textcolor{red}{+ 16}$

${\left(x + 4\right)}^{2} = 18$

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

$\sqrt{{\left(x + 4\right)}^{2}} = \pm \sqrt{18} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$x + 4 = \pm \sqrt{18} = \pm \sqrt{9 \times 2} = \pm 3 \sqrt{2}$

$\text{subtract 4 from both sides}$

$x = - 4 \pm 3 \sqrt{2} \leftarrow \textcolor{red}{\text{exact values}}$

$x \approx - 8.24 \text{ or "x~~0.24" to 2 dec. places}$