# How do you solve x^2-6x-9=0 by completing the square?

Jan 19, 2018

#### Answer:

${\left(x - 3\right)}^{2} - 18$

#### Explanation:

${x}^{2} - 6 x - 9 = \left({x}^{2} - 6 x + 9\right) - 18$

${x}^{2} - 6 x + 9 = \left(x - 3\right) \left(x - 3\right) = {\left(x - 3\right)}^{2}$

${\left(x - 3\right)}^{2} - 18 = \left({x}^{2} - 6 x + 9\right) - \left(9 + 9\right) = {x}^{2} - 6 x + 9 - 9 - 9$

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

Jan 19, 2018

#### Answer:

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

#### Explanation:

The difference of squares identity can be written:

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

Use this with $A = \left(x - 3\right)$ and $B = 3 \sqrt{2}$ as follows:

$0 = {x}^{2} - 6 x - 9$

$\textcolor{w h i t e}{0} = {x}^{2} - 2 \left(x\right) \left(3\right) + {3}^{2} - 18$

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

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

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

Hence:

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