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Math problem II

  1. Nov 14, 2003 #1
    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)
     
  2. jcsd
  3. Nov 14, 2003 #2

    dduardo

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    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! + ....
     
  4. Nov 14, 2003 #3
    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?
     
  5. Nov 15, 2003 #4
    yup.
    thanks.
     
  6. Nov 15, 2003 #5
    but i use ( [squ] x - [squ] y ) to the power of two instead.
     
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