# How do you solve using the completing the square method x^2 - x - 1 = 0?

May 25, 2017

$x = \frac{1 \pm \sqrt{5}}{2}$

#### Explanation:

Completing the square is a bit tricky, the first step is to remove any multiplier on the highest term. Since our multiplier is 1, we are good to go.

Now we must get the constant to the other side. So we add 1 to each side.

${x}^{2} - x - 1 \textcolor{red}{+ 1} = 0 \textcolor{red}{+ 1}$
${x}^{2} - x = 1$

Now we take ${\left(\frac{b}{2}\right)}^{2}$ where $b$ is the coefficient in front of the second term, or the term that is not the highest order. In this case it is -1.

${\left(- \frac{1}{2}\right)}^{2} = \frac{1}{4}$

Add this number to both sides.

${x}^{2} - x \textcolor{red}{+ \frac{1}{4}} = 1 \textcolor{red}{+ \frac{1}{4}}$

Now we should be able to correctly solve the left side so it is one term squared.

${x}^{2} - x + \frac{1}{4} = \frac{5}{4}$
${\left(x - \frac{1}{2}\right)}^{2} = \frac{5}{4}$

From here we solve for $x$. Square root both sides, and add the 1/2 to get your final answer.

$x - \frac{1}{2} = \pm \frac{\sqrt{5}}{2}$
$x = \frac{1 \pm \sqrt{5}}{2}$