How do you solve #x^4+x^2-6=0 #? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Noah G Aug 31, 2016 #x =+-sqrt(2)# Explanation: #(x^2 + 3)(x^2 - 2) = 0# #x^2 = -3 and x^2 = 2# #x = +-sqrt(-3) and x = +-sqrt(2)# #x = +-sqrt(2)# Hopefully this helps! 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 8383 views around the world You can reuse this answer Creative Commons License