Solvability of an equation Let $p\left(  x\right)  =x^{n}+ax+b$ and $a,b>0$. Is the equation $p\left(
x\right)  =0$ always solvable? Which are the solutions?
 A: $x^n+ax+b=0\iff x^n=-b-ax\iff x=\sqrt[n]{-b-ax}$ for odd n, and $x=\pm\sqrt[n]{-b-ax}$ for even n. 
$$n=2k+1:\qquad x=\sqrt[n]{-b-a\,\sqrt[n]{-b-a\,\sqrt[n]{-b-\ldots}}}$$
$$n=2k:\qquad\qquad x=\sqrt[n]{-b\pm a\,\sqrt[n]{-b\pm a\,\sqrt[n]{-b\pm\ldots}}}$$
where in the latter case the sign of a must be chosen so as to have a positive quantity under the radical sign.
A: I assume $n\ge2$.
If $n$ is odd, then $p(+\infty)=+\infty$ and $p(-\infty)=-\infty$, thus there is a (real) zero of $p(x)$. If $n$ is even, $p(\pm\infty)=+\infty$, so we look at the minima of $p$:
$$\frac{d}{dx}p(x)=nx^{n-1}+a$$
This has exactly one real zero at $x=\sqrt[n-1]{-\frac{a}{n}}$. Inserting into $p(x)$ we obtain:
$$p\left(\sqrt[n-1]{-\frac{a}{n}}\right)=\left(-\frac{a}{n}\right)^{\frac{n}{n-1}}+a\left(-\frac{a}{n}\right)^{\frac{1}{n-1}}+b$$
In order for $p(x)=0$ to have real solutions, this must be negative. The only term which is negative is the one in the middle, thus $n$ must satisfy:
$$\left(\frac{a}{n}\right)^{\frac{n}{n-1}}+b\le \left(\frac{a}{n}\right)^{\frac{1}{n-1}}$$
$$\Leftrightarrow a^{\frac{n}{n-1}}+n^{\frac{n}{n-1}}b\le a^{\frac{1}{n-1}}n^{\frac{n-1}{n-1}}=a^{\frac{1}{n-1}}n$$
I think that this last equation has to be solved numerically.
