How do you solve #x^2+4x+3=0# by completing the square?

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Answer 1 · 2016-03-28T00:05:02

#x = -3# or #x = -1#

Add #1# to both sides of the equation to get:

#x^2+4x+4 = 1#

The left hand side is now a perfect square:

#x^2+4x+4 = (x+2)^2#

So we have:

#(x+2)^2 = 1#

Hence:

#x+2 = +-sqrt(1) = +-1#

Subtract #2# from both ends to get:

#x = -2+-1#

That is:

#x = -3# or #x = -1#