How do you complete the square for #x^2 + 12x#?

1 Answer
May 17, 2015

The answer is #x=0#; #x=-12# .

Problem: Complete the square for #x^2+12x#.

Rewrite the equation as a trinomial.

#x^2+12x+0#

Move 0 to the right-hand side.

#x^2+12x=0#

Divide the coefficient of the #x#-term by 2, then square the result. Add the result to both sides.

#12/2=6#; #6^2=36#

#x^2+12x+36=36#

Factor the perfect square trinomial #x^2+12x+36# on the left-hand side.

#(x+6)^2=36#

Take the square root of both sides and solve for #x#.

#sqrt((x+6)^2)=+-sqrt36# =

#x+6=+-6#

#x=-6+6=0#

#x=-6-6=-12#

Check

If #x=0#:

#(0)^2+12(0)=0#

#0=0#

If #x=-12#:

#(-12)^2+12(-12)=0#

#144-144=0#

#0=0#