How do you solve #x^2 - 4x = 0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Faisal Mar 28, 2018 #(x+2sqrtx)(x-2sqrtx)# Explanation: = #x^2-4(sqrtx)^2# = #x^2-(2sqrtx)^2# *Use* #a^2-b^2=(a+b)(a-b)# = #(x+2sqrtx)(x-2sqrtx)# 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 1979 views around the world You can reuse this answer Creative Commons License