How do you solve #sqrt(2x-7)>=5#? Precalculus Solving Rational Inequalities Polynomial Inequalities 1 Answer Babette Feb 4, 2017 #x>=16# Explanation: Square both sides #(sqrt(2x-7))^2>=5^2# #2x-7>=25# Add #7# to both sides #2x>=32# Divide both sides by #2# #x>=16# 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 1398 views around the world You can reuse this answer Creative Commons License