How do you simplify #sqrt245#? Algebra Radicals and Geometry Connections Simplification of Radical Expressions 1 Answer EZ as pi Dec 28, 2016 #7sqrt5# Explanation: Write #245# as the product of its prime factors to find out what you are working with. #sqrt245 = sqrt(5xx7xx7)# Find square roots wherever possible. #sqrt(5 xx 7^2 )= 7sqrt5# Answer link Related questions How do you simplify radical expressions? How do you simplify radical expressions with fractions? How do you simplify radical expressions with variables? What are radical expressions? How do you simplify #root{3}{-125}#? How do you write # ""^4sqrt(zw)# as a rational exponent? How do you simplify # ""^5sqrt(96)# How do you write # ""^9sqrt(y^3)# as a rational exponent? How do you simplify #sqrt(75a^12b^3c^5)#? How do you simplify #sqrt(50)-sqrt(2)#? See all questions in Simplification of Radical Expressions Impact of this question 1818 views around the world You can reuse this answer Creative Commons License