How do you simplify #sqrt(63w^36)#? Algebra Radicals and Geometry Connections Simplification of Radical Expressions 1 Answer IDKwhatName Jul 15, 2017 #3sqrt(7)*w^18=3w^18sqrt(7)# Explanation: #sqrt(63w^36)=sqrt(63)*sqrt(w^36)# #sqrt(w^36)=(w^36)^(1/2)=w^(36/2)=w^18# #sqrt(63)=sqrt(a*b)#, where a nd b are factors of 63, and one is a perfect square number. Two factors of 63 are 9 and 7. #sqrt(63)=sqrt(9*7)=sqrt(9)*sqrt(7)=3sqrt(7)#. As #sqrt(63w^36)=sqrt(63)*sqrt(w^36)#, and we calculated thise vakues, it is just #3w^18sqrt(7)# 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 1468 views around the world You can reuse this answer Creative Commons License