How do you solve using completing the square method #-3x^2+30x-74#?

2 Answers
Aug 5, 2016

This cannot really be "solved" because it is not an equation and it is not "equal" to anything. It can only be written in a different form.

Aug 5, 2016

I got:

#y = -3(x - 5)^2 + 1#


You have:

#y = -3x^2 + 30x - 74#

What you can start with is to make the #x^2# term have a perfect-square coefficient (#1# is fine). That makes it easier to come up with a factor.

Then, halve the #x# term's coefficient and square it, adding it to both sides.

#y/3 = -x^2 + 10x - 74/3# (divide by 3)

#-y/3 = x^2 - 10x + 74/3# (switch signs)

Half of #-10# is #-5#, squared is #+25#. To make common denominators happen, #25 = 75/3#. Now add it to both sides.

#=> -y/3 + 75/3 = x^2 - 10x + 74/3 + 75/3#

#=> -y/3 = x^2 - 10x + 75/3 - 1/3#

#=> -y/3 = color(green)(x^2 - 10x + 25) - 1/3#

#=> -y/3 = color(green)((x - 5)^2) - 1/3#

Now you can return it back to #y# by reversing the two things we did in the first steps:

#=> y/3 = -(x - 5)^2 + 1/3#

#=> color(blue)(y = -3(x - 5)^2 + 1)#