Question

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

Answer 1

Add one and then divide by 2

Explanation:

First add 1 to each side
`2x^2+4x+2=1`
Then divide each side by two:
`x^2+2x+1=1/2`
Then you see that `x^2+2x+1=(x+1)^2`
`(x+1)^2=1/2`
Take the square root of each side:
`x+1=1/sqrt(2)`
`x=1/sqrt(2)-1=(sqrt(2)-2)/2`

Answer 2

`x=-1-sqrt(2)/2`, and `x=-1+sqrt(2)/2`

Explanation:

First divide both sides of the equation by 2.

Equation 1
`x^2+2x+1/2=0`

Now add `1/2` to both sides of the equation. (see footnote)

`x^2+2x+1=1/2`

Now we can factor the left-hand side of the equation.

`(x+1)^2=1/2`

Now let's write the equation as the difference of two squares.

`(x+1)^2-1/2=0`

The square root of `1/2` is `sqrt(2)/2`, so we can factor the above expression as

`(x+1+sqrt(2)/2)(x+1-sqrt(2)/2)=0`

Therefore the solution to this equation is

`x=-1-sqrt(2)/2`, and `x=-1+sqrt(2)/2`

FOOTNOTE:

Why did I pick `1/2` to add to both sides of Equation 1? I looked at the coefficient for the `x` (linear) term in Equation 1 which is 2. I divided 2 by 2 and then squared the result. `(2/2)^2=1` is what I wanted the THIRD term on the left hand side of Equation 1 to be. This is why I added `1/2` to both sides. If, for example, the coefficient in for the `x` term had been 5 instead of 2, then I would have wanted the third term on the left-hand side of the equation to be `(5/2)^2=25/4` and I would have needed to add `23/4` to both sides of the equation.