# How do you solve x^2 + 5x - 2 = 0?

Jan 5, 2017

$x = - \frac{5}{2} \pm \frac{\sqrt{33}}{2}$

#### Explanation:

The difference of squares identity can be written:

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

We can solve the given quadratic by completing the square and using the difference of squares identity with $a = \left(2 x + 5\right)$ and $b = \sqrt{33}$.

I will multiply it by $4$ to start with, to cut down on the need for arithmetic involving fractions:

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

$\textcolor{w h i t e}{0} = 4 {x}^{2} + 20 x - 8$

$\textcolor{w h i t e}{0} = {\left(2 x\right)}^{2} + 2 \left(2 x\right) \left(5\right) + 25 - 33$

$\textcolor{w h i t e}{0} = {\left(2 x + 5\right)}^{2} - {\left(\sqrt{33}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(2 x + 5\right) - \sqrt{33}\right) \left(\left(2 x + 5\right) + \sqrt{33}\right)$

$\textcolor{w h i t e}{0} = \left(2 x + 5 - \sqrt{33}\right) \left(2 x + 5 + \sqrt{33}\right)$

$\textcolor{w h i t e}{0} = 4 \left(x + \frac{5}{2} - \frac{\sqrt{33}}{2}\right) \left(x + \frac{5}{2} + \frac{\sqrt{33}}{2}\right)$

Hence:

$x = - \frac{5}{2} \pm \frac{\sqrt{33}}{2}$