# How do you solve x^2+8x+1=0 by completing the square?

Jun 6, 2016

$x = - 4 - \sqrt{15}$ or

$x = - 4 + \sqrt{15}$

#### Explanation:

In ${x}^{2} + 8 x + 1 = 0$, we observe that coefficient of $x$ is $8$ and hence, we can complete the square by adding square of half of $8$ i.e. ${4}^{2} = 16$.

Hence ${x}^{2} + 8 x + 1 = 0$ can be written as ${x}^{2} + 8 x + 16 - 16 + 1 = 0$ or

${\left(x + 4\right)}^{2} - 15 = 0$ which can be written as

${\left(x + 4\right)}^{2} - {\left(\sqrt{15}\right)}^{2} = 0$

Now using identity ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$, this can be written as

$\left(x + 4 + \sqrt{15}\right) \left(x + 4 - \sqrt{15}\right) = 0$

Hence either $x + 4 + \sqrt{15} = 0$ i.e. $x = - 4 - \sqrt{15}$

or $x + 4 - \sqrt{15} = 0$ i.e. $x = - 4 + \sqrt{15}$