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Thread: solve

  1. #1
    Nov 2006


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  2. #2
    Super Member angel.white's Avatar
    Oct 2007
    Quote Originally Posted by perash View Post
    $\displaystyle x=\sqrt{x+1999\sqrt{x+1999\sqrt{x+1999\sqrt{x+1999 \sqrt{2000x}}}}}$

    1. Square both sides
    $\displaystyle x^{2}=x+1999\sqrt{x+1999\sqrt{x+1999\sqrt{x+1999\s qrt{2000x}}}}$

    2. Subtract x from both sides
    $\displaystyle x^{2}-x=1999\sqrt{x+1999\sqrt{x+1999\sqrt{x+1999\sqrt{20 00x}}}}$

    3. Divide by 1999
    $\displaystyle \frac{x^{2}-x}{1999}=\sqrt{x+1999\sqrt{x+1999\sqrt{x+1999\sqrt {2000x}}}}$

    4. Square both sides
    $\displaystyle (\frac{x^{2}-x}{1999})^{2}=x+1999\sqrt{x+1999\sqrt{x+1999\sqrt{ 2000x}}}$

    5. Simplify lefthand side
    $\displaystyle \frac{x^{4}-2x^{3}+x^{2}}{1999^{2}}=x+1999\sqrt{x+1999\sqrt{x+ 1999\sqrt{2000x}}}$

    6. Subtract x
    $\displaystyle \frac{x^{4}-2x^{3}+x^{2}}{1999^{2}}-x=1999\sqrt{x+1999\sqrt{x+1999\sqrt{2000x}}}$

    7. Common denominator and combine fractions
    $\displaystyle \frac{x^{4}-2x^{3}+x^{2}-1999^{2}x}{1999^{2}}=1999\sqrt{x+1999\sqrt{x+1999\ sqrt{2000x}}}$

    8. Divide by 1999
    $\displaystyle \frac{x^{4}-2x^{3}+x^{2}-1999^{2}x}{1999^{3}}=\sqrt{x+1999\sqrt{x+1999\sqrt {2000x}}}$

    9. Square both sides
    $\displaystyle (\frac{x^{4}-2x^{3}+x^{2}-1999^{2}x}{1999^{3}})^{2}=x+1999\sqrt{x+1999\sqrt{ 2000x}}$

    10. FOIL it out


    um, not going to finish, I just realized that this is annoying, and I am probably going to make a math error at some point, throwing the whole thing off. So I decided to look at the problem from a "is there an obvious answer" approach instead of a "how would I solve this" approach, and yes, the obvious answer is zero.
    Last edited by ThePerfectHacker; Nov 11th 2007 at 08:48 AM.
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  3. #3
    Global Moderator

    Nov 2005
    New York City
    Suppose that $\displaystyle x=\sqrt{2000x}$ then $\displaystyle \sqrt{2000x} = \sqrt{x+1999x } = \sqrt{x+1999\sqrt{2000x}}$. Then $\displaystyle \sqrt{x+1999\sqrt{2000x}}=\sqrt{x+1999\sqrt{x+1999 x}}$ substitute again $\displaystyle x=\sqrt{x+1999\sqrt{x+1999\sqrt{2000x}}}$. Keep on going until you get what you have. So we need to solve $\displaystyle x=\sqrt{2000x}$ which has a trivial solution (as angel-white says) $\displaystyle x=0$ and a non-trivial solution $\displaystyle x=2000$.
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