How do y simplify #sqrt648#? Algebra Radicals and Geometry Connections Simplification of Radical Expressions 1 Answer EZ as pi Jul 2, 2016 #18sqrt2# Explanation: Write 648 as the product of its prime factors first: #sqrt648 = sqrt(2xx2xx2xx3xx3xx3xx3xx3)# =#sqrt(2 xx 2^2 xx 3^4)" "# group even powers where possible. =#2xx 3^2 sqrt2# = #18sqrt2# 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 1496 views around the world You can reuse this answer Creative Commons License