# How do you solve x^2-10x-9=0?

Dec 13, 2016

$x = 5 \pm \sqrt{34}$

#### Explanation:

The difference of squares identity can be written:

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

Use this with $a = \left(x - 5\right)$ and $b = \sqrt{34}$ as follows:

$0 = {x}^{2} - 10 x - 9$

$\textcolor{w h i t e}{0} = {x}^{2} - 10 x + 25 - 34$

$\textcolor{w h i t e}{0} = {\left(x - 5\right)}^{2} - {\left(\sqrt{34}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(x - 5\right) - \sqrt{34}\right) \left(\left(x - 5\right) + \sqrt{34}\right)$

$\textcolor{w h i t e}{0} = \left(x - 5 - \sqrt{34}\right) \left(x - 5 + \sqrt{34}\right)$

Hence $x = 5 \pm \sqrt{34}$