The number of real solutions

**TheCoffeeMachine** · March 5th 2011, 03:48 AM

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 $n \geqslant 1$ .

**tonio** · March 5th 2011, 05:55 AM

Originally Posted by TheCoffeeMachine

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 $n$ .

.

**TheCoffeeMachine** · March 5th 2011, 06:11 AM

Originally Posted by tonio

.

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 $n\geqslant 2$ ?

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