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Thread: 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
    $\displaystyle
    \sqrt{17-12{\sqrt2}}-\sqrt{17+12{\sqrt2}}=6$

    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 $\displaystyle \sqrt{9-12{\sqrt2}+8}-\sqrt{9+12{\sqrt2}+8}=6 $





    Why is this?



    Thank you!
    Last edited by Coach; Sep 26th 2007 at 03:59 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
    $\displaystyle
    \sqrt{17-12{\sqrt2}}-\sqrt{17+12{\sqrt2}}=6$

    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 $\displaystyle \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
    $\displaystyle (3 + \sqrt{8})^2 = 3^2 + 2 \cdot 3 \cdot \sqrt{8} + (\sqrt{8})^2 = 9 + 12\sqrt{2} + 8$

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

    -Dan
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  3. #3
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    Hello, Coach!

    Dan has the best approach . . .


    Prove that: .$\displaystyle \sqrt{17-12{\sqrt2}}-\sqrt{17+12{\sqrt2}}\:=\:6$
    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: .$\displaystyle (a + b\sqrt{2})^2\:=\:17 + 12\sqrt{2}$ . . . . where $\displaystyle a$ is positive and rational

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

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

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

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

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

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

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

    Using $\displaystyle a = 3$, we have: .$\displaystyle b = 2$


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

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

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

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