How do you solve #(3x - 1) ^ { 2} = 5#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Juan A. · Stefan V. Mar 17, 2017 #x =(+- sqrt(5) +1)/ 3# Explanation: #(3x -1)^2 = 5# #3x -1 = +- sqrt(5)# #3x = +- sqrt(5) + 1# #x =(+- sqrt(5) + 1)/ 3# 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 1293 views around the world You can reuse this answer Creative Commons License
