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Math Help - The number of real solutions

  1. #1
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    The number of real solutions

    Prove the following, if true; otherwise, provide/derive a counterexample/contradiction:

    1. If  n\in\mathbb{N} and  x\in\mathbb{R}^{+}, then x^n+x^{\frac{1}{n}}+n = 0 has no real solutions for all n \geqslant 2.

    2. If  n\in\mathbb{N} and x \in\mathbb{R}^{+}, then x^n+x^{\frac{1}{n}}-n =  0 has a single real solution for all  n \geqslant 1.
    Last edited by TheCoffeeMachine; March 6th 2011 at 12:55 AM. Reason: Correction
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  2. #2
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    Quote Originally Posted by TheCoffeeMachine View Post
    Prove the following, if true; otherwise, provide/derive a counterexample/contradiction:

    1. If  n\in\mathbb{N}, then x^n+x^{\frac{1}{n}}+n = 0 has no real solutions for n \ge 1.


    With \displaystyle{n=1\,,\,\,x+x+1=0\Longrightarrow x=-\frac{1}{2}}

    Tonio




    2. If  n\in\mathbb{N}, then x^n+x^{\frac{1}{n}}-n =  0 has a single real solution for all  n.
    .
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  3. #3
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    Quote Originally Posted by tonio View Post
    .
    I'm sorry, it should have been n\geqslant 2 (and the second one for n\geqslant 1).
    I noticed and edited it while you were replying. Any ideas for n\geqslant 2?
    Last edited by TheCoffeeMachine; March 5th 2011 at 09:19 PM. Reason: Correction
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