How do you solve #x+sqrt(-2x^2+9)=3# and identify any restrictions? Algebra Radicals and Geometry Connections Radical Equations 1 Answer G_Dub Apr 18, 2018 #x=0, 2, -(3sqrt2)/2<=x<=(3sqrt2)/2# Explanation: Restrictions: #-2x^2+9>=0# #=>-(3sqrt2)/2<=x<=(3sqrt2)/2# #x+sqrt(-2x^2+9)=3# #sqrt(-2x^2+9)=3-x# subtract #x# from both sides #-2x^2+9=9-6x+x^2# square both sides #3x^2-6x=0# #3x(x-2)=0# solve quadratic equation #x=0,x=2# 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 1699 views around the world You can reuse this answer Creative Commons License