How do you solve #2(x - 1)^2 - 8(x - 1)^2= 0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Don't Memorise May 21, 2015 #(x-1)^2# is common to both the terms of the expression on the left hand side We can write the equation as #(x-1)^2*(2-8) = 0# #->-6(x-1)^2 = 0# Dividing both sides by #-6# we get: #->(cancel(-6)(x-1)^2)/cancel(-6) = 0/-6# #->(x-1)^2 = 0# #->x-1=0# #-> color(green)(x = 1# 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 1385 views around the world You can reuse this answer Creative Commons License