How do you solve #x^3>=9x^2#? Precalculus Solving Rational Inequalities Polynomial Inequalities 1 Answer sente Dec 27, 2016 #x in [9, oo)uu{0}# Explanation: #x^3>=9x^2# #=> x^3-9x^2>=0# #=> x^2(x-9)>=0# #x^2(x-9)# has two roots: #x=0# and #x=9#. We will check what happens in the intervals on each side of them. Case 1: #x in (-oo, 0)# #=> x^2 > 0 and x-9 < 0# #=> x^2(x-9) < 0# Case 2: #x in (0, 9)# #=> x^2>0 and x-9 < 0# #=> x^2(x-9) < 0# Case 3: #x in (9, oo)# #=> x^2 > 0 and x-9 > 0# #=> x^2(x-9) > 0# Thus we have #x^2(x-9)>=0# if #x=0# or #x in [9, oo)# Answer link Related questions What are common mistakes students make when solving polynomial inequalities? How do I solve a polynomial inequality? How do I solve the polynomial inequality #-2(m-3)<5(m+1)-12#? How do I solve the polynomial inequality #-6<=2(x-5)<7#? How do I solve the polynomial inequality #1<2x+3<11#? How do I solve the polynomial inequality #-12<-2(x+1)<=18#? How do you solve the inequality #6x^2-5x>6#? How do you solve #x^2 - 4x - 21<=0# A) [-3, 7] B) (-∞, -3] C) (-∞, -3] [7, ∞) D) [7, ∞)? How do you solve quadratic inequality, graph, and write in interval notation #x^2 - 8x + 15 >0#? How do you solve #-x^2 - x + 6 < 0#? See all questions in Polynomial Inequalities Impact of this question 1885 views around the world You can reuse this answer Creative Commons License