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

2 Answers
Apr 30, 2018

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

Apr 30, 2018

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.