For what value of #x# is #(3x^2)/(x+3)>(4x)/(x-2)#?

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Answer 1 · 2018-05-12T23:15:52

The answer
#x<[5-sqrt(61)]/(3)#
#x<[5+sqrt(61)]/(3)#
#x<0#

show below

#(3x^2)/(x+3)>(4x)/(x-2)#

#4x^2+12x>3x^3-6x^2#

#3x^3-10x^2-12x<0#

#x[3x^2-10x-12]<0#

#x<0#

now we will solve #[3x^2-10x-12]=0#

#x=ax^2+bx+c#

#a=3, b=-10 and c=-12#

#x=[-b+-sqrt(b^2-4ac)]/(2a)#

#x=[10+-sqrt(100+144)]/(6)#

#x=[5+-sqrt(61)]/(3)#

#x<[5-sqrt(61)]/(3)#

#x<[5+sqrt(61)]/(3)#