How do you solve #4x^2 + 8x + 4 = 0#?

1 Answer
Mar 14, 2018

Answer:

The solution is #x=-1#.

Explanation:

First, divide both sides by #4#. Then, factor. Lastly, square root both sides and solve for #x#. Here's what that looks like:

#4x^2+8x+4=0#

#(4x^2+8x+4)/4=0/4#

#x^2+2x+1=0#

#x^2+x+x+1=0#

#color(red)x*x+color(red)x*1+color(blue)1*x+color(blue)1*1=0#

#color(red)x(x+1)+color(blue)1(x+1)=0#

#(color(red)x+color(blue)1)(x+1)=0#

#(x+1)^2=0#

#sqrt((x+1)^2)=sqrt0#

#x+1=0#

#x=-1#

This is the answer. Hope this helped!