# How do you solve x^2 + 4x - 9 = 0 by completing the square?

Jan 11, 2017

$x = - 2 \pm \sqrt{13}$

#### Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Hence we find:

$0 = {x}^{2} + 4 x - 9$

$\textcolor{w h i t e}{0} = {x}^{2} + 4 x + 4 - 13$

$\textcolor{w h i t e}{0} = {\left(x + 2\right)}^{2} - {\left(\sqrt{13}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(x + 2\right) - \sqrt{13}\right) \left(\left(x + 2\right) + \sqrt{13}\right)$

$\textcolor{w h i t e}{0} = \left(x + 2 - \sqrt{13}\right) \left(x + 2 + \sqrt{13}\right)$

Hence:

$x = - 2 \pm \sqrt{13}$