How do you simplify #sqrt18/(sqrt8-3) #? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Tony B May 21, 2016 #-12-9sqrt(2)" "# Using 'primary root' only #-12+-9sqrt(2)" "#All solutions. Explanation: Using #a^2-b^2=(a+b)(a-b)# Multiply by 1 but in the form of #1=(sqrt(8)+3)/(sqrt(8)+3)# #(sqrt(18)(sqrt(8)+3))/((sqrt(8)-3)(sqrt(8)+3))# #(sqrt(18)(sqrt(8)+3))/(8-3^2)# But #sqrt(8)=2sqrt(2)" and "sqrt(18)=3sqrt(2)# #(3sqrt(2)(2sqrt(2)+3))/(8-3^2)# #color(brown)("Note that from above,"-3^2" is different to "(-3)^2)# #(12+9sqrt(2))/(-1)# #-12-9sqrt(2)" "# Using 'primary root' only #-12+-9sqrt(2)" "#All solutions. Answer link Related questions How do you simplify #\frac{2}{\sqrt{3}}#? How do you multiply and divide radicals? How do you rationalize the denominator? What is Multiplication and Division of Radicals? How do you simplify #7/(""^3sqrt(5)#? How do you multiply #(sqrt(a) +sqrt(b))(sqrt(a)-sqrt(b))#? How do you rationalize the denominator for #\frac{2x}{\sqrt{5}x}#? Do you always have to rationalize the denominator? How do you simplify #sqrt(5)sqrt(15)#? How do you simplify #(7sqrt(13) + 2sqrt(6))(2sqrt(3)+3sqrt(6))#? See all questions in Multiplication and Division of Radicals Impact of this question 950 views around the world You can reuse this answer Creative Commons License