How do you add #3sqrt2+4sqrt2#? Algebra Radicals and Geometry Connections Addition and Subtraction of Radicals 1 Answer GiĆ³ Mar 24, 2015 You consider #sqrt(2)# as a fixed quantity that you do not change and simply add the multiplier: #3sqrt(2)+4sqrt(2)=(3+4)sqrt(2)=7sqrt(2)# Answer link Related questions How do you add and subtract radicals? How is a radical considered a "like term"? How do you simplify #4\sqrt{3}+2\sqrt{12}#? How do you add #3""^3sqrt(2)+5""^3sqrt(16)#? How do you subtract #\sqrt{8x^3}-4x\sqrt{98x}#? How do you combine the radical #\sqrt{6}-\sqrt{27}+2\sqrt{54}+3\sqrt{48}#? How do you simplify #""^3sqrt{\frac{16x^5}{135y^4}}#? What is #sqrt(50)-sqrt(18)#? What is the square root of 50 + the square root of 8? What is #sqrt(8)+sqrt(18)-sqrt(32)#? See all questions in Addition and Subtraction of Radicals Impact of this question 9357 views around the world You can reuse this answer Creative Commons License