How do you solve the quadratic #x-4/(3x)=-1/3# using any method?

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Answer 1 · 2018-07-02T15:20:45

#x=-4/3, 1#

Given that

#x-4/{3x}=-1/3#

#x-4/{3x}+1/3=0#

#\frac{3x^2-4+x}{3x}=0#

#\frac{3x^2+x-4}{3x}=0#

#\frac{3x^2+4x-3x-4}{3x}=0#

#\frac{x(3x+4)-(3x+4)}{3x}=0#

#\frac{(3x+4)(x-1)}{3x}=0#

#(3x+4)(x-1)=0\quad (\forall \ x\ne 0)#

#x=-4/3, 1#