How do you simplify #sqrt(50/3)+sqrt(3/6)#? Algebra Radicals and Geometry Connections Addition and Subtraction of Radicals 1 Answer Daphne_456456-Minec May 22, 2018 (Someone check this please.) #sqrt(3/6)=sqrt(1/2)=1/sqrt(2)=sqrt2/2# #sqrt(50/3)=(5sqrt2)/sqrt3=(5sqrt6)/3# #sqrt2/2+(5sqrt6)/3=(3sqrt2)/6+(10sqrt6)/6=(3sqrt2+10sqrt6)/6# 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)#? How do you add #3sqrt2+4sqrt2#? What is the square root of 50 + the square root of 8? See all questions in Addition and Subtraction of Radicals Impact of this question 1666 views around the world You can reuse this answer Creative Commons License