# How do you solve x^2+10x-1=0 by completing the square?

Apr 7, 2018

$x = \pm \sqrt{26} - 5$

#### Explanation:

Isolate all terms involving $x$ on one side and move the constant to the other side.

${x}^{2} + 10 x = 1$

Now, the original quadratic, ${x}^{2} + 10 x - 1 ,$ is in the form $a {x}^{2} + b x + c$ where $a = 1 , b = 10 , c = - 1$.

To complete the square, we first want to add ${\left(\frac{b}{2}\right)}^{2}$ to each side.

${\left(\frac{b}{2}\right)}^{2} = {\left(\frac{10}{2}\right)}^{2} = {5}^{2} = 25$, so we add $25$ to each side.

${x}^{2} + 10 x + 25 = 1 + 25$

The left side needs to be factored. Fortunately, since we added ${\left(\frac{b}{2}\right)}^{2}$ to each side, the factored form will simply be ${\left(x + \frac{b}{2}\right)}^{2} = {\left(x + 5\right)}^{2}$

${\left(x + 5\right)}^{2} = 26$

Now, take the root of both sides, accounting for positive and negative answers on the right.

$\sqrt{{\left(x + 5\right)}^{2}} = \pm \sqrt{26}$

$x + 5 = \pm \sqrt{26}$

Solving for $x$ yields

$x = \pm \sqrt{26} - 5$