# How do you solve x^2+4x-1=0 by completing the square?

##### 1 Answer
Jul 10, 2015

$x = - 2 + \sqrt{5} ,$ $- 2 - \sqrt{5}$

#### Explanation:

${x}^{2} + 4 x - 1 = 0$

Add $1$ to both sides of the equation.

${x}^{2} + 4 x = 1$

Force a perfect square trinomial on the left side by dividing the coefficient of the $x$ term and squaring the result. Add it to both sides of the equation.

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

${x}^{2} + 4 x + 4 = 1 + 4$ =

${x}^{2} + 4 x + 4 = 5$

We now have a perfect square trinomial on the left side in the form ${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$, where $a = x ,$ and $b = 2$.

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

Take the square root of both sides.

$x + 2 = \pm \sqrt{5}$

Subtract $2$ from both sides.

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

The two values for $x$.

$x = - 2 + \sqrt{5}$

$x = - 2 - \sqrt{5}$