# How do you solve the equation x^2-6x+9=8 by completing the square?

##### 1 Answer
Nov 25, 2016

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

#### Explanation:

${x}^{2} - 6 x + 9 = \textcolor{w h i t e}{a} 8$
$\textcolor{w h i t e}{a a a {a}^{2} a a} - 9 \textcolor{w h i t e}{a} - 9 \textcolor{w h i t e}{a a a}$Subtract 9 from both sides

${x}^{2} - \textcolor{red}{6} x \textcolor{w h i t e}{a a a} = - 1$

Divide the coefficient of the $x$ term $\textcolor{red}{6}$ by 2 and square the result.

$\frac{\textcolor{red}{6}}{2} = \textcolor{\lim e g r e e n}{3} \implies {\textcolor{\lim e g r e e n}{3}}^{2} = \textcolor{b l u e}{9}$

Add $\textcolor{b l u e}{9}$ to both sides.

${x}^{2} - 6 x + \textcolor{b l u e}{9} = - 1 + \textcolor{b l u e}{9}$

${x}^{2} - 6 x + 9 = 8$

Note that you got the exact equation you started with! The next step is to factor the left side into the square of a binomial. The left side of this equation started out in factorable form, but I demonstrated all the steps anyway for future reference.

Factor the left side

$\left(x - 3\right) \left(x - 3\right) = 8 \textcolor{w h i t e}{a a a}$

#Rewrite as the square of a binomial.

${\left(x - \textcolor{\lim e g r e e n}{3}\right)}^{2} = 8 \textcolor{w h i t e}{a a a}$Note that $\textcolor{\lim e g r e e n}{3}$ is the number you got when you divided the coefficient of the x term by 2.

Square root both side.

$\sqrt{{\left(x - 3\right)}^{2}} = \sqrt{8}$

$x - 3 = \sqrt{4 \cdot 2}$

$x - 3 = \pm 2 \sqrt{2}$
$\textcolor{w h i t e}{a} + 3 \textcolor{w h i t e}{a a a a a a a a} + 3$

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