# Math problem II

denian
i tried this. but fail to get the answer

find the square root of 11 - 6*(square root of 2) and write the solution in surd notation.

answer provided by the book :
3 - (square root of 2)

## Answers and Replies

Staff Emeritus
My calculator gives the answer to be 3-sqrt(2).

Why don't you try using binomial expansion to solve the problem.

sqrt(11-6*sqrt(2)) = sqrt(11)*(1+((-6/11)*sqrt(2))^(1/2)

x = (-6/11)*sqrt(2)
n = 1/2

1 + n*x + (n*(n-1)*x^2)/2! + (n*(n-1)*(n-2)*x^3)/3! + ....

gnome
You want 11 - 6&radic;2 to be a perfect square of some binomial. Since you have that -6&radic;2 term, look for a binomial such that
(a - b&radic;2)2 = 11 - 6&radic;2

What happens when you multiply out
(a - b&radic;2)2 ?

Can you solve for a and b?

Does that help?

denian
yup.
thanks.

denian
but i use ( [squ] x - [squ] y ) to the power of two instead.