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

2 Answers
Apr 30, 2018

Answer:

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

Answer:

#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.