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

Mar 16, 2016

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

Explanation:

1. First thing to do is transpose the number to the other side the equation;
${x}^{2} - 6 x - 5 = 0$
${x}^{2} - 6 x = 5$
2. Look for a possible number that can give a perfect square for the remaining terms and of course whatever number added or subtracted from the other side should be the same number and operation to the other side of the equation;
${x}^{2} - 6 x \textcolor{red}{+ 9} = 5 \textcolor{red}{+ 9}$
3. Then the find the factors; perfect square on the left side f the equation:
$\left(x - 3\right) \left(x - 3\right) = 14$
${\left(x - 3\right)}^{2} = 14$
4. Remove the square by putting $\sqrt{s} i g n$ on the left side and do the same to the right side:
$\sqrt{{\left(x - 3\right)}^{2}} = \pm \sqrt{14}$
5. Simplify the equation
$x - 3 = \pm \sqrt{14}$
6. Isolate $x$ by adding 3 both sides of the equation:
$x \cancel{- 3} \cancel{+ 3} = 3 \pm \sqrt{14}$
$x = 3 \pm \sqrt{14}$
7. The answer is $x = 3 + \sqrt{14}$ and $x = 3 - \sqrt{14}$