How do you solve using the completing the square method x^2 + 6x + 7 = 0?

Mar 31, 2017

${x}_{1} \approx - 1.6 , {x}_{2} \approx - 4.4 \text{ to 1 dec. place}$

Explanation:

To use the method of $\textcolor{b l u e}{\text{completing the square}}$

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

$\Rightarrow {x}^{2} + 6 x {\textcolor{red}{+ 3}}^{2} + 7 = 0 {\textcolor{red}{+ 3}}^{2}$

$\Rightarrow {x}^{2} + 6 x + 9 = 9 - 7$

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

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

$\sqrt{{\left(x + 3\right)}^{2}} = \pm \sqrt{2}$

$\Rightarrow x + 3 = \pm \sqrt{2}$

$\Rightarrow {x}_{1} = \sqrt{2} - 3 \approx - 1.6 \text{ to 1 dec. place}$

$\Rightarrow {x}_{2} = - \sqrt{2} - 3 \approx - 4.4 \text{ to 1 dec. place}$