# How do you solve 0 = x^2 + 10x + 5?

May 12, 2016

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

#### Explanation:

Complete the square and use the difference of squares identity:

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

with $a = \left(x + 5\right)$ and $b = 2 \sqrt{5}$ as follows:

$0 = {x}^{2} + 10 x + 5$

$= {\left(x + 5\right)}^{2} - 25 + 5$

$= {\left(x + 5\right)}^{2} - 20$

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

$= \left(\left(x + 5\right) - 2 \sqrt{5}\right) \left(\left(x + 5\right) + 2 \sqrt{5}\right)$

$= \left(x + 5 - 2 \sqrt{5}\right) \left(x + 5 + 2 \sqrt{5}\right)$

Hence:

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

May 12, 2016

$x = - 5 + 2 \sqrt{5} \textcolor{w h i t e}{\text{XXX")orcolor(white)("XXX}} x = - 5 - 2 \sqrt{5}$

#### Explanation:

The easiest way for this particular example is to use the quadratic formula:

For an equation of the form $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$
the solutions are given by $x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - 4 \textcolor{red}{a} \textcolor{g r e e n}{c}}}{2 \textcolor{red}{a}}$

For the given example

$\textcolor{red}{a} = 1$
$\textcolor{b l u e}{b} = 10$ and
$\textcolor{g r e e n}{c} = 5$

So
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- 10 \pm \sqrt{{10}^{2} - 4 \left(1\right) \left(5\right)}}{2 \left(1\right)}$

$\textcolor{w h i t e}{\text{XXXX}} = \frac{- 10 \pm 4 \sqrt{5}}{2}$

$\textcolor{w h i t e}{\text{XXXX}} = - 5 \pm 2 \sqrt{5}$