How do you simplify #(sqrt10 - 9) ^2#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Konstantinos Michailidis May 21, 2016 We are going to use the identity #(A-B)^2=A^2-2*A*B+B^2# Hence we have that #(sqrt10-9)^2=(sqrt10)^2-2*sqrt10*9+9^2= 10-18*sqrt10+81=91-18*sqrt10# 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 827 views around the world You can reuse this answer Creative Commons License