How do you simplify #sqrt(30)/(sqrt(2)*sqrt(5)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 2 Answers James Jun 6, 2018 #[sqrt(30)/[(sqrt(2)*sqrt(5))]]=sqrt30/sqrt10=sqrt(30/10)=sqrt3# Explanation: Note that: #color(red)[sqrta/sqrtb=sqrt(a/b)]# #color(red)[sqrta*sqrtb=sqrt(a*b)]# #[sqrt(30)/[(sqrt(2)*sqrt(5))]]=sqrt30/sqrt10=sqrt(30/10)=sqrt3# Answer link Math Tutor Jun 6, 2018 #sqrt30/(sqrt2*sqrt5) = sqrt30/sqrt(2*5)=sqrt(30/10)=sqrt3# Explanation: #sqrt30/(sqrt2*sqrt5) = sqrt30/sqrt(2*5)=sqrt(30/10)=sqrt3# 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 2680 views around the world You can reuse this answer Creative Commons License