# How do you solve x^2-6x-5=0 by completing the square?

May 31, 2018

$x = 3 + \sqrt{14}$ or $x = 3 - \sqrt{14}$

#### Explanation:

Writing
${x}^{2} - 2 \cdot 3 x + 9 - 5 - 9$
This is
${\left(x - 3\right)}^{2} = 14$
$| x - 3 | = \sqrt{14}$
so we get
$x = 3 + \sqrt{14}$
in the other case

$- x + 3 = \sqrt{14}$
$x = 3 - \sqrt{14}$

May 31, 2018

$x = \pm 14 + 3$

$x = + \sqrt{14} + 3 = 6.74 \mathmr{and} x = - 0.74$

#### Explanation:

${x}^{2} - 6 x - 5 = 0$
${x}^{2} - 6 x \text{ "=5" } \leftarrow$ move the constant to the right side.

Complete the square by adding $\textcolor{red}{{\left(- \frac{6}{2}\right)}^{2}}$ to both sides

${x}^{2} - 6 x \textcolor{red}{+ 9} = 5 \textcolor{red}{+ 9}$

$\text{ "(x-3)^2 = 14" } \leftarrow$ write as the square of a binomial

$\text{ "x-3 = +-sqrt14" } \leftarrow$ square root both sides.

$x = \pm 14 + 3 \text{ } \leftarrow$ isolate $x$

This gives two solutions for $x :$

$x = + \sqrt{14} + 3 = 6.74 \mathmr{and} x = - 0.74$