How do you solve #y=x^2-12x+36#?

1 Answer
Jul 5, 2015

This is a perfect square trinomial:

#x^2-12x+36 = (x-6)^2#

So #y = 0# when #x=6#

Explanation:

Perfect square trinomials are of the form:

#a^2+-2ab+b^2 = (a+-b)^2#

In our case notice that the leading term #x^2# is square and the constant term #36 = 6^2# is square. So the question is: Is the middle term equal to #+-2*x*6 = +-12x# - Yes.