TITLE: Show $x^n+ax+b=0$ has most two solutions
QUESTION [5 upvotes]: For any real numbers $a$ and $b$ and even natural number $n$, Show $x^n+ax+b=0$ has most two solutions for all $x$ in $\mathbb{R}$.

let $f(x)=x^n+ax+b$. Set $f(x)$ and $a$ be $0$, then we have $x^n+b=0$. So $x=\pm b^{1/n}$. $f$ has two solutions. To yeld a contradiction. Now assume that $f$ has more than two solutions. As $f$ is a polynomial on $\mathbb{R}$, it is differentiable on $\mathbb{R}$; thus $f'(x)=nx^{n-1}+a$. Set $f'(x)=0$, then $0=nx^{n-1}+a\Rightarrow x=\left(\frac{-a}{n}\right)^{1/(n-1)}$. At $x=\left(\frac{-a}{n}\right)^{1/(n-1)}$, $f(\left(\frac{-a}{n}\right)^{1/(n-1)})$ is the only extreme value of $f$ which contradicts with $f$ has more than two solutions.

Can someone check this solution? I am not sure right or not. If not right, please give me a hint or suggestion. Thanks

REPLY [2 votes]: Observe that $p$ is concave up everywhere since $p''(x)=n(n-1)x^{n-2}\ $and since $n$ is even so is $n-2$. The result follows.