How do you solve #\sqrt { 6x - 4} = \sqrt { 3}#? Algebra Radicals and Geometry Connections Radical Equations 1 Answer James · Stefan V. Apr 12, 2018 #x = 7/6# Explanation: #sqrt(6x-4)=sqrt(3)# #6x-4 = 3 " "# Square both sides to cancel the radicals #6x = 7" "# Add #4# to both sides #x = 7/6" "# Divide both sides by #6# Answer link Related questions How do you solve radical equations? What are Radical Equations? How do you solve radical equations with cube roots? How do you find extraneous solutions when solving radical equations? How do you solve #\sqrt{x+15}=\sqrt{3x-3}#? How do you solve and find the extraneous solutions for #2\sqrt{4-3x}+3=0#? How do you solve #\sqrt{x^2-5x}-6=0#? How do you solve #\sqrt{x}=x-6#? How do you solve for x in #""^3sqrt(-2-5x)+3=0#? How do you solve for x in #sqrt(42-x)+x =13#? See all questions in Radical Equations Impact of this question 1556 views around the world You can reuse this answer Creative Commons License
