# How do you solve using the completing the square method x^2+2x+4=0?

Apr 5, 2017

$x = - 1 \pm \sqrt{3} i$

#### Explanation:

Given:

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

Completing the square we get:

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

$\textcolor{w h i t e}{0} = {x}^{2} + 2 x + 1 + 3$

$\textcolor{w h i t e}{0} = {\left(x + 1\right)}^{2} + 3$

Note that for any Real value of $x$, we have:

${\left(x + 1\right)}^{2} \ge 0$

and so:

${\left(x + 1\right)}^{2} + 3 \ge 3$

To solve this quadratic we need to use Complex numbers.

The difference of squares identity can be written:

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

We use this with $a = \left(x + 1\right)$ and $b = \sqrt{3} i$ as follows:

$0 = {\left(x + 1\right)}^{2} + 3$

$\textcolor{w h i t e}{0} = {\left(x + 1\right)}^{2} - {\left(\sqrt{3} i\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(x + 1\right) - \sqrt{3} i\right) \left(\left(x + 1\right) + \sqrt{3} i\right)$

$\textcolor{w h i t e}{0} = \left(x + 1 - \sqrt{3} i\right) \left(x + 1 + \sqrt{3} i\right)$

Hence:

$x = - 1 \pm \sqrt{3} i$