How do you complete the square for #: x^2 + 6x#?

1 Answer
May 16, 2015

#(x+3)^2 = x^2+6x+9#

So #x^2 +6x = (x+3)^2 - 9#

If you were actually trying to solve #x^2+6x=0#, then completing the square would be a round about way of doing it, but would look something like this:

#0 = x^2+6x = x^2+6x+9-9 = (x+3)^2-9#

Adding 9 to both sides you get #(x+3)^2 = 9#

So #x+3 = +-sqrt(9) = +-3#

Then subtracting 3 from both sides, you get

#x = -3+-3#

In other words #x = 0# or #x = -6#.