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Math problem II

  • Thread starter denian
  • Start date
  • #1
160
0
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

  • #2
dduardo
Staff Emeritus
1,890
3
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! + ....
 
  • #3
1,036
1
You want 11 - 6√2 to be a perfect square of some binomial. Since you have that -6√2 term, look for a binomial such that
(a - b√2)2 = 11 - 6√2

What happens when you multiply out
(a - b√2)2 ?

Can you solve for a and b?

Does that help?
 
  • #4
160
0
yup.
thanks.
 
  • #5
160
0
but i use ( [squ] x - [squ] y ) to the power of two instead.
 

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