Solvability of an equation

Question

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?

Answer 1

$x^n+ax+b=0\iff x^n=-b-ax\iff x=\sqrt[n]{-b-ax}$ 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.

Answer 2

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.