How do you solve #4x^2=9x#?

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Answer 1 · 2016-05-28T18:06:18

#x = 0, 9/4 #

#4x^2 = 9x#
#4x^2 - 9x = 0 #
#x(4x-9)=0#
#x=0#
#4x-9=0#
#4x=9#
#x= 9/4#

Answer 2 · 2018-08-10T02:40:32

#x=0# and #x=9/4#

We have a second degree term, so this is beginning to look like a quadratic. Let's subtract #9x# from both sides to get

#4x^2-9x=0#

Both terms have an #x# in them, so we can factor that out to get

#x(4x-9)=0#

Setting both factors equal to zero, we get

#x=0# and #x=9/4#

Hope this helps!