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Math Help - A problem with roots

  1. #1
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    A problem with roots

    This is from one of the Finnish national Math competitions MAOL http://www.maol.fi/fileadmin/users/K...vastaukset.pdf




    prove that
     <br />
\sqrt{17-12{\sqrt2}}-\sqrt{17+12{\sqrt2}}=6

    I have a question

    1) The sample answer splits 17 into 9 and 8. Yet it moves the eight to the "ends" of the equation \sqrt{9-12{\sqrt2}+8}-\sqrt{9+12{\sqrt2}+8}=6





    Why is this?



    Thank you!
    Last edited by Coach; September 26th 2007 at 04:59 AM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Coach View Post
    This is from one of the Finnish national Math competitions MAOL http://www.maol.fi/fileadmin/users/K...vastaukset.pdf




    prove that
     <br />
\sqrt{17-12{\sqrt2}}-\sqrt{17+12{\sqrt2}}=6

    I have a question

    1) The sample answer splits 17 into 9 and 8. Yet it moves the eight to the "ends" of the equation \sqrt{9-12{\sqrt2}+8}-\sqrt{9+12{\sqrt2}+8}=6





    Why is this?



    Thank you!
    It's writing the problem in a suggestive form.

    Note that
    (3 + \sqrt{8})^2 = 3^2 + 2 \cdot 3 \cdot \sqrt{8} + (\sqrt{8})^2 = 9 + 12\sqrt{2} + 8

    and
    (3 - \sqrt{8})^2 = 3^2 - 2 \cdot 3 \cdot \sqrt{8} + (\sqrt{8})^2 = 9 - 12\sqrt{2} + 8

    -Dan
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  3. #3
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    Hello, Coach!

    Dan has the best approach . . .


    Prove that: . \sqrt{17-12{\sqrt2}}-\sqrt{17+12{\sqrt2}}\:=\:6
    In case you can't "eyeball" those radicals, here's a tip.


    Since the difference of those two radicals in an integer,
    . . we might suspect that the radicands are perfect squares.

    We conjecture: . (a + b\sqrt{2})^2\:=\:17 + 12\sqrt{2} . . . . where a is positive and rational

    The left side is: . a^2 + 2\sqrt{2}ab + 2b^2 \:=\:(a^2+b^2) + 2\sqrt{2}ab

    So we have: . (a^2+2b^2) + 2\sqrt{2}ab \:=\:17 + 12\sqrt{2}

    Equate rational and irrational components:
    . . a^2+2b^2\:=\:17 .[1]
    . . 2\sqrt{2}ab \:=\:12\sqrt{2}\quad\Rightarrow\quad ab = 6 .[2]
    and solve the system of equations.

    Solve [2] for b\!:\;\;b \:=\:\frac{6}{a}
    Substitute into [1]: . a^2 + 2\left(\frac{6}{a}\right)^2\:=\:17

    . . and we have: . a^4 - 17a^2 + 72 \:=\:0

    . . which factors: . (a^2-8)(a^2-9)\:=\:0

    . . and has roots: . a\;=\:\pm2\sqrt{2},\:\pm3

    Using a = 3, we have: . b = 2


    Therefore: . 17 + 12\sqrt{2} \:=\:(3 + 2\sqrt{2})^2 . . . . ta-DAA!

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  4. #4
    MHF Contributor red_dog's Avatar
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    We can use the following identities:
    \displaystyle\sqrt{A+\sqrt{B}}=\sqrt{\frac{A+\sqrt  {A^2-B}}{2}}+\sqrt{\frac{A-\sqrt{A^2-B}}{2}}

    \displaystyle\sqrt{A-\sqrt{B}}=\sqrt{\frac{A+\sqrt{A^2-B}}{2}}-\sqrt{\frac{A-\sqrt{A^2-B}}{2}}

    Now, \displaystyle\sqrt{17\pm 12\sqrt{2}}=\sqrt{17\pm \sqrt{288}} and plug A=17, \ B=288 in the formula above.
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