How do you simplify #sqrt5 times sqrt5#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer sente Apr 18, 2016 #sqrt(5)xxsqrt(5)=5# Explanation: By definition, #sqrt(x)# is a value whose square is #x#, that is, #sqrt(x)xxsqrt(x)=x# for any #x#. If #x>=0#, then #sqrt(x)# denotes the unique non negative value with that property. 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 2028 views around the world You can reuse this answer Creative Commons License