How do you solve the quadratic #x-4/(3x)=-1/3# using any method? Precalculus Linear and Quadratic Functions Completing the Square 1 Answer Harish Chandra Rajpoot Jul 2, 2018 #x=-4/3, 1# Explanation: 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# Answer link Related questions What does completing the square mean? How do I complete the square? Does completing the square always work? Is completing the square always the best method? Do I need to complete the square if #f(x) = x^2 - 6x + 9#? How do I complete the square if #f(x) = x^2 + 4x - 9#? How do I complete the square if the coefficient of #x^2# is not 1? How do I complete the square if #f(x) = 3x^2 + 12x - 9#? If I know the quadratic formula, why must I also know how to complete the square? How do I use completing the square to describe the graph of #f(x)=30-12x-x^2#? See all questions in Completing the Square Impact of this question 2511 views around the world You can reuse this answer Creative Commons License