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Math Help - Math Contest Problem

  1. #1
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    Math Contest Problem

    Could anyone explain this problem to me? If x is a positive real number and x^2 + 1/(x^2) = 3, then x^3 + 1/(x^3) = ? Thanks in advance!
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  2. #2
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    Quote Originally Posted by eric1299171 View Post
    Could anyone explain this problem to me? If x is a positive real number and x^2 + 1/(x^2) = 3, then x^3 + 1/(x^3) = ? Thanks in advance!
    A contest you are doing now or an old problem?
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  3. #3
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    It's a problem from the 2007 Luzerne County Math Contest.
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  4. #4
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    Quote Originally Posted by eric1299171 View Post
    Could anyone explain this problem to me? If x is a positive real number and x^2 + 1/(x^2) = 3, then x^3 + 1/(x^3) = ? Thanks in advance!
    \displaystyle x^2\cdot\left[x^2+x^{-2}-3\right]=0\Rightarrow x^4-3x^2+1=0

    w^2=x^4\Rightarrow w=x^2

    w^2-3w+1=0

    \displaystyle w=\frac{3\pm\sqrt{5}}{2}=\frac{3\pm \sqrt{5}}{2}\Rightarrow x=\pm\sqrt{w}=\pm\sqrt{\frac{3\pm \sqrt{5}}{2}}
    Last edited by dwsmith; February 5th 2011 at 03:13 PM. Reason: forgot to square the -3
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  5. #5
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    If you use the quadratic formula, wouldn't you get a 5 under the square root instead of a -7?
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  6. #6
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    Quote Originally Posted by eric1299171 View Post
    If you use the quadratic formula, wouldn't you get a 5 under the square root instead of a -7?
    Correct, I forgot to square.
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  7. #7
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    Thanks. =) Then after you solve for x how do you figure out what x^3 + 1/(x^3) is? I tried plugging back in but it turned really ugly.
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  8. #8
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    You can do it without explicitly solving for x. We have
    \left(x+\dfrac{1}{x}\right)^2=x^2+\dfrac{1}{x^2}+2  =5. Since x is positive, x+1/x=\sqrt{5}.

    Now expand \left(x+\dfrac{1}{x}\right)^3.
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  9. #9
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    Thank you so much! My math teacher and I were working on this for hours. =)
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