How do you simplify #sqrt(2) / sqrt(8)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer sente May 2, 2016 #sqrt(2)/sqrt(8)=1/2# Explanation: Using the property that #sqrt(a)/sqrt(b) = sqrt(a/b)# for #b > 0# we have #sqrt(2)/sqrt(8) = sqrt(2/8) = sqrt(1/4) = sqrt(1)/sqrt(4) = 1/2# Answer link Related questions How do you simplify #\frac{2}{\sqrt{3}}#? How do you multiply and divide radicals? How do you rationalize the denominator? What is Multiplication and Division of Radicals? How do you simplify #7/(""^3sqrt(5)#? How do you multiply #(sqrt(a) +sqrt(b))(sqrt(a)-sqrt(b))#? How do you rationalize the denominator for #\frac{2x}{\sqrt{5}x}#? Do you always have to rationalize the denominator? How do you simplify #sqrt(5)sqrt(15)#? How do you simplify #(7sqrt(13) + 2sqrt(6))(2sqrt(3)+3sqrt(6))#? See all questions in Multiplication and Division of Radicals Impact of this question 1069 views around the world You can reuse this answer Creative Commons License