Suppose $f(x) = x^{2k} + a_{2k-2} x^{2k-2} + \dots + a_2 x^2 + a_0 \in \mathbb R[x]$ with $k \in \mathbb N$. Let us assume $k \ge 2$. What can be said about the number of distinct real roots for $f(x)$?
By Fundamental Theorem of Algebra, it could be at most $2k$. But on the other hand, when I was playing with $x^4 + ax^2 + b$, it seems that if the roots are all real, it must be factored as $(x^2 - c^2)^2$ for some $c$ which would only give us two distinct real roots.
On
If $a_0\ne 0$ then the number of real roots is even.
If $a_0=0$, then the number of real roots is odd.
Observe that $f(x)$ can be written as $g(x^2)$ for a polynomial $g$ (with the same coefficients $a_{2k}$).
Every positive root $s$ of $g$ determines two real solutions of $f$: $\pm\sqrt s$.
On
What can be said easily is that the real roots come in pairs, as obviously with $x_0$ also $-x_0$ will be a root, and if $0$ is a root, it has an even degree $\ge 2$.
Also, $x^4-5x^2+4=(x^2-1)(x^2-4)=0$ has roots $\pm1, \pm2$.
Basically your equation has real roots $x_{2i-1}=-x_{2i}$ that correspond to the non-negative roots $y_i$ of $y^k+a_{2k-2}y^{k-1}+\ldots+a_2y+a_0=0$ via $x^2_{2i-1}=x^2_{2i}=y_i$.
Well, writing $g(x)=x^k+a_{2k-2}x^{k-1}+\dots+a_2x+a_0$, we have $f(x)=g(x^2)$. So, the roots of $f$ are just the square roots of the roots of $g$. This means that $f$ has two real roots for every positive root of $g$, and one more real root if $0$ is a root of $g$.
In particular, $f$ could have any number of distinct real roots from $0$ to $2k$. Indeed, to get any even number $2m$ of real roots ($0\leq m\leq k$), just let $g$ be a polynomial with exactly $m$ positive roots and for which $0$ is not a root. To get an odd number $2m+1$ of real roots ($0\leq m<k$), just let $g$ have $m$ positive roots and also have $0$ as a root.