How do you solve #9x^2 - 16 = 0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Alan P. Apr 20, 2016 #(3x+4)(3x-4)# Explanation: In the general form #(a+b)(a-b)=a^2-b^2# Treating #9x^2# as #a^2# #rArr a=3x# and treating #16# as #b^2# #rArr b=4# #color(green)((3x+4)(3x-4))=9x^2-16# Answer link Related questions What are the different methods for solving quadratic equations? What would be the best method to solve #-3x^2+12x+1=0#? How do you solve #-4x^2+4x=9#? What are the two numbers if the product of two consecutive integers is 72? Which method do you use to solve the quadratic equation #81x^2+1=0#? How do you solve #-4x^2+4000x=0#? How do you solve for x in #x^2-6x+4=0#? How do you solve #x^2-6x-16=0# by factoring? How do you solve by factoring and using the principle of zero products #x^2 + 7x + 6 = 0#? How do you solve #x^2=2x#? See all questions in Comparing Methods for Solving Quadratics Impact of this question 11014 views around the world You can reuse this answer Creative Commons License