Two different roots for $P(x) = x^4+ax^2+bx+c$ 
Let $a, b, c \in \mathbb{R}$ and $a > 0$. Also let $P: \mathbb{R} \to \mathbb{R}$, $P(x) = x^4+ax^2+bx+c$. Show that the function has at most two different roots.

My assumption was to use Bolzano's theorem, but I couldn't figure out how to use it here. Also I'm curious if we could use something like Vieta's formula here? Any help would be appreciated.
 A: We have
$$P'(x)=4x^3+2ax+b$$
and $$P''(x)=12x^2+2a>0$$
this means $P$ is a convex function over $\mathbb{R}$, and has at most two roots.
A: Let me try to explain this in detail by simplifying the question as follows: 
Let $a=1>0, |b|=1, |c|=1$. Then we will see 4 different cases and use Descarte's Rule of Signs (*) (http://web.cs.iastate.edu/~cs577/handouts/polyroots.pdf)
$\textbf{Case 1}$: $b=c=1$. Then, $P(x)=x^4+x^2+x+1$ and by (*) it doesn't have any positive roots. However, $P(-x)=x^4+x^2-x+1$ this has 2 sign changes that implies that the function has 2 negative roots. Hence, in total, $2$ real roots.
$\textbf{Case 2}$: $b=1,c=-1$. Then, $P(x)=x^4+x^2+x-1$ and by (*) it has one sign change $\rightarrow$ one positive root. Also, $P(-x)=x^4+x^2-x-1$ this has 1 sign change that implies that the function has 1 negative root1. Hence, in total, $2$ real roots.
$\textbf{Case 3}$: $b=-1,c=1$. Then, $P(x)=x^4+x^2-x+1$ and by (*) it has one sign change $\rightarrow$ one positive root. However, $P(-x)=x^4+x^2+x+1$ this doesn't have any sign changes $\rightarrow$ no negative roots. Hence, in total, $1$ real roots.
$\textbf{Case 4}$: $b=-1,c=-1$. Then, $P(x)=x^4+x^2-x-1$ and by (*) it has one sign change $\rightarrow$ one positive root. Also, $P(-x)=x^4+x^2+x-1$ this has one sign change $\rightarrow$ 1 negative root. Hence, in total, $2$ real roots.
