# How do you solve x^2 - 4x + 7 =0 by completing the square?

Aug 10, 2017

$x = 2 \pm \sqrt{3} i$

#### Explanation:

Given:

${x}^{2} - 4 x + 7 = 0$

While completing the square we will find that this takes the form of the sum of a square and a positive number. As a result it has no solution in real numbers, but we can solve it using complex numbers.

The imaginary unit $i$ satisfies ${i}^{2} = - 1$

The difference of squares identity can be written:

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

We can use this with $a = \left(x - 2\right)$ and $b = \sqrt{3} i$ as follows:

$0 = {x}^{2} - 4 x + 7$

$\textcolor{w h i t e}{0} = {x}^{2} - 4 x + 4 + 3$

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

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

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

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

Hence the two roots are:

$x = 2 + \sqrt{3} i \text{ }$ and $\text{ } x = 2 - \sqrt{3} i$