Question

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

Answer 1

`x=-2+sqrt5,` `-2-sqrt5`

Explanation:

`x^2+4x-1=0`

Add `1` to both sides of the equation.

`x^2+4x=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.

`(4/2)^2=2^2=4`

`x^2+4x+4=1+4` =

`x^2+4x+4=5`

We now have a perfect square trinomial on the left side in the form `a^2+2ab+b^2=(a+b)^2`, where `a=x,` and `b=2`.

`(x+2)^2=5`

Take the square root of both sides.

`x+2=+-sqrt5`

Subtract `2` from both sides.

`x=-2+-sqrt5`

The two values for `x`.

`x=-2+sqrt5`

`x=-2-sqrt5`