# The number of real solutions

• Mar 5th 2011, 03:48 AM
TheCoffeeMachine
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 $\displaystyle n \geqslant 1$.
• Mar 5th 2011, 05:55 AM
tonio
Quote:

Originally Posted by TheCoffeeMachine
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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• Mar 5th 2011, 06:11 AM
TheCoffeeMachine
Quote:

Originally Posted by tonio
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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 $\displaystyle n\geqslant 2$?