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

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

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


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

    Tonio




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