How do you solve #9x^2 = 27x#?

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Answer 1 · 2016-07-16T04:35:21

#x = 3# and #x = 0#

We can rewrite #9x^2 = 27x# by subtracting #27x# from both sides and make it equal to #0#.

#9x^2 = 27x#

#9x^2 - 27x = 0#

Factoring out a #9# and #x# we now get

#9x(x-3) = 0#

We now have two separate factors, namely #9x = 0# and #x-3 = 0#, so we can simply solve for #x#.

#9x = 0 -> x = 0#

#x-3=0 -> x =3#

So our solutions are #x = 3# and #x = 0#.

To check our answer, we can also graph #9x^2-27x#.

graph{9x^2-27x [-2.82, 4.973, -1.955, 1.942]}

Answer 2 · 2016-07-16T04:37:34

#x=0# or #x=3#

#9x^2=27x# #hArr#

#9x^2-27x=0# or

#9x(x-3)=0# or dividing by #9#

#x(x-3)=0# i.e. either #x=0# or #x-3=0#

Hence #x=0# or #x=3#