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

Apr 30, 2018

Add one and then divide by 2

#### Explanation:

First add 1 to each side
$2 {x}^{2} + 4 x + 2 = 1$
Then divide each side by two:
${x}^{2} + 2 x + 1 = \frac{1}{2}$
Then you see that ${x}^{2} + 2 x + 1 = {\left(x + 1\right)}^{2}$
${\left(x + 1\right)}^{2} = \frac{1}{2}$
Take the square root of each side:
$x + 1 = \frac{1}{\sqrt{2}}$
$x = \frac{1}{\sqrt{2}} - 1 = \frac{\sqrt{2} - 2}{2}$

Apr 30, 2018

$x = - 1 - \frac{\sqrt{2}}{2}$, and $x = - 1 + \frac{\sqrt{2}}{2}$

#### Explanation:

First divide both sides of the equation by 2.

Equation 1
${x}^{2} + 2 x + \frac{1}{2} = 0$

Now add $\frac{1}{2}$ to both sides of the equation. (see footnote)

${x}^{2} + 2 x + 1 = \frac{1}{2}$

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

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

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

${\left(x + 1\right)}^{2} - \frac{1}{2} = 0$

The square root of $\frac{1}{2}$ is $\frac{\sqrt{2}}{2}$, so we can factor the above expression as

$\left(x + 1 + \frac{\sqrt{2}}{2}\right) \left(x + 1 - \frac{\sqrt{2}}{2}\right) = 0$

Therefore the solution to this equation is

$x = - 1 - \frac{\sqrt{2}}{2}$, and $x = - 1 + \frac{\sqrt{2}}{2}$

FOOTNOTE:

Why did I pick $\frac{1}{2}$ 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. ${\left(\frac{2}{2}\right)}^{2} = 1$ is what I wanted the THIRD term on the left hand side of Equation 1 to be. This is why I added $\frac{1}{2}$ 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 ${\left(\frac{5}{2}\right)}^{2} = \frac{25}{4}$ and I would have needed to add $\frac{23}{4}$ to both sides of the equation.