How do you solve #x^2/(x^2+x)>=0#? Precalculus Solving Rational Inequalities Solving Rational Inequalities on a Graphing Calculator 1 Answer Cesareo R. Sep 11, 2016 #-1 < x le 0# Explanation: Calling #y = x^2+x>0# we have #x^2+x-y > 0# or #(-1-sqrt(1+4y))/2 < x < (-1+sqrt(1+4y))/2# The smallest interval is for #y = 0# so #-1 < x le 0# Answer link Related questions How do I solve the rational inequality #(x-4)/(x+5)<4# using a TI-84? How do I solve the rational inequality #(x+10)/(3x-2)<=3# using a TI-84? How do I solve the rational inequality #(x+2)/(2x+1)>5# using a TI-84? How do I solve the rational inequality #(3x-2)/(x+2)<=1/3# using a TI-83? How do I solve the rational inequality #(x^2-1)/(x+1)<2# using a TI-83? How do I solve the rational inequality #(x^2-x-6)/(x+2)<=-3# using a TI-83? How do you solve rational inequalities? How do you solve the inequality #(x^2-2x-24)/(x^2-8x-20)>=0#? How do you solve the inequality #x^3-x^2-6x>0#? How do you solve #(16-x^2)/(x^2-9)>=0#? See all questions in Solving Rational Inequalities on a Graphing Calculator Impact of this question 1852 views around the world You can reuse this answer Creative Commons License