# How do you solve x^2-6x+11=0 by completing the square?

Jun 25, 2018

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

#### Explanation:

Let's first subtract $11$ from both sides to get

${x}^{2} - 6 x = - 11$

When we complete the square of an equation of the form

$a {x}^{2} + b x = - c$

We take half of our $b$ value, square it, and add it to both sides.

Our $b$ value is $- 6$, half of that is $- 3$, and that value squared is $9$. Adding it to the left and right gives us

$\textcolor{s t e e l b l u e}{{x}^{2} - 6 x + 9} = - 11 + 9$

What I have in blue can be factored as ${\left(x - 3\right)}^{2}$. This gives us

${\left(x - 3\right)}^{2} = - 2$

Taking the square root of both sides gives us

$x - 3 = \sqrt{- 2}$

Which can be rewritten as

$x - 3 = \sqrt{- 1} \cdot \sqrt{2}$

Note that $i = \sqrt{- 1}$. With this definition in mind, we now have

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

Adding $3$ to both sides gives us

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

Hope this helps!