How do you simplify #sqrt 80 / sqrt 5#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 2 Answers Konstantinos Michailidis Sep 17, 2015 It is #sqrt80/sqrt5=(sqrt((5*16)/5))=sqrt(16)=4# Answer link Alippiun Sep 17, 2015 #4# Explanation: #sqrt80/sqrt5# can also be written as #sqrt((80/5)# or #(80/5)^(1/2)# Simplify #80/5# or in other terms, divide #80# by #5#; #=sqrt((cancel(80)16)/(cancel(5)1))# #=sqrt16# #=4# 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 5398 views around the world You can reuse this answer Creative Commons License