How do you simplify #sqrt5(2sqrt5 - sqrt4)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer VNVDVI Apr 1, 2018 #10-2sqrt5# Explanation: Distribute #sqrt(5):# #sqrt(5)(2sqrt5-sqrt4)=2sqrt(5)sqrt(5)-sqrt(4)sqrt(5)# Recall that #sqrtasqrta=a# So, we have #2(5)-sqrt(4)sqrt(5)# #sqrt(4)=2,# so we end up with #10-2sqrt5# 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 1092 views around the world You can reuse this answer Creative Commons License