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

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Answer 1 · 2016-08-03T19:15:49

#x=6#

Subtract #12x# from both sides giving:

#x^2-12x+36=0#

Note that #(-6)xx(-6)=+36" and "-6-6=-12#

Factorizing

#(x-6)(x-6) =0-> (x-6)^2=0 #

#=>x=6#
Tony B

Answer 2 · 2016-08-03T19:17:32

Double root x = 6

#x^2 - 12x + 36 = (x - 6)^2#
There is a double-root, x = 6.