# How do you solve x^2+6x-5=0 by completing the square?

##### 2 Answers
May 4, 2017

$x = 0.742 \mathmr{and} x = - 6.742$

#### Explanation:

For a quadratic $a {x}^{2} + b x + c = 0$, there are specific conditions required to be able to write it as the square of a binomial.

$a = 1 , \text{ } {\left(\frac{b}{2}\right)}^{2} = c$

${x}^{2} + 6 x - 5 = 0$

In this case, $a = 1$, but ${\left(\frac{6}{2}\right)}^{2} = 9 \ne - 5$

Move $- 5$ to the other side and add $9$ to both sides.

${x}^{2} + 6 x \textcolor{b l u e}{+ 9} = 5 \textcolor{b l u e}{+ 9}$

${\left(x + 3\right)}^{2} = 14 \text{ } \leftarrow$ find the square root of both sides.

$x + 3 = \pm \sqrt{14}$

$x = + \sqrt{14} - 3 \text{ or } x = - \sqrt{14} - 3$

$x = 0.742 \mathmr{and} x = - 6.742$

May 4, 2017

$x = - 3 \pm \sqrt{14}$

#### Explanation:

$\text{to solve using "color(blue)"completing the square}$

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

$\text{that is } {\left(\frac{6}{2}\right)}^{2} = 9$

$\Rightarrow \left({x}^{2} + 6 x \textcolor{red}{+ 9}\right) - 5 = 0 \textcolor{red}{+ 9}$

$\Rightarrow {\left(x + 3\right)}^{2} - 5 = 9$

$\text{add 5 to both sides}$

${\left(x + 3\right)}^{2} \cancel{- 5} \cancel{+ 5} = 9 + 5$

$\Rightarrow {\left(x + 3\right)}^{2} = 14$

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

$\sqrt{{\left(x + 3\right)}^{2}} = \textcolor{red}{\pm} \sqrt{14} \leftarrow \text{ note plus or minus}$

$\Rightarrow x + 3 = \pm \sqrt{14}$

$\text{subtract 3 from both sides}$

$x \cancel{+ 3} \cancel{- 3} = - 3 \pm \sqrt{14}$

$\Rightarrow x = - 3 \pm \sqrt{14}$