How do you add #sqrt32+9sqrt2-sqrt18#? Prealgebra Exponents, Radicals and Scientific Notation Square Root 1 Answer Shwetank Mauria Apr 29, 2016 #sqrt32+9sqrt2-sqrt18=10sqrt2# Explanation: #sqrt32+9sqrt2-sqrt18# = #sqrt(ul(2xx2xx2xx2)xx2)+9sqrt2-sqrt(ul(3xx3)xx2)# = #2xx2xxsqrt2+9sqrt2-3sqrt2# = #4sqrt2+9sqrt2-3sqrt2# = #(4+9-3)sqrt2# = #10sqrt2# 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 #sqrt468 #? 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#? See all questions in Square Root Impact of this question 2116 views around the world You can reuse this answer Creative Commons License