How do you simplify #sqrt468 #? Prealgebra Exponents, Radicals and Scientific Notation Square Root 1 Answer Shwetank Mauria Jun 20, 2017 #sqrt468=6sqrt13# Explanation: #sqrt468# = #sqrt(2xx2xx3xx3xx13)# = #sqrt(ul(2xx2)xxul(3xx3)xx13)# = #2xx3xxsqrt13# = #6sqrt13# Answer link Related questions How do you simplify #(2sqrt2 + 2sqrt24) * sqrt3#? How do you simplify #sqrt735/sqrt5#? How do you rationalize the denominator and simplify #1/sqrt11#? How do you multiply #sqrt[27b] * sqrt[3b^2L]#? How do you simplify #7sqrt3 + 8sqrt3 - 2sqrt2#? How do you simplify #sqrt(48x^3) / sqrt(3xy^2)#? How do you simplify # sqrt ((4a^3 )/( 27b^3))#? How do you simplify #sqrt140#? How do you simplify #sqrt216#? How do you simplify #sqrt540#? See all questions in Square Root Impact of this question 4306 views around the world You can reuse this answer Creative Commons License