# How do you solve by completing the square a^2+ 8a + 11 = 0?

Apr 30, 2018

$a = 4 + \sqrt{5} , 4 - \sqrt{5}$

#### Explanation:

${a}^{2} + 8 a + 11 = {\left(a + 4\right)}^{2} - 16 + 11 = {\left(a + 4\right)}^{2} - 5$

Checking:

$\left(a + 4\right) \left(a + 4\right) = {a}^{2} + 8 x + 16 - 5 = {a}^{2} + 8 x + 11$

Solve:

${\left(a - 4\right)}^{2} - 5 = 0$

${\left(a - 4\right)}^{2} = 5$

$a - 4 = \pm \sqrt{5}$

$a = 4 \pm \sqrt{5}$