How do you simplify # sqrt (a^5 / 2)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer James May 20, 2018 #sqrt (a^5 / 2)=sqrt(a^5)/sqrt2=sqrt(a^4*a)/sqrt2# #=[sqrt(a^4)*sqrta]/sqrt2=[a^2*sqrta]/sqrt2# Explanation: Note that #color(red)[sqrt(a/b)=sqrta/sqrtb]# #sqrt (a^5 / 2)=sqrt(a^5)/sqrt2=sqrt(a^4*a)/sqrt2# #=[sqrt(a^4)*sqrta]/sqrt2=[a^2*sqrta]/sqrt2# 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 1371 views around the world You can reuse this answer Creative Commons License