TITLE: Finding the number of real roots of an unusual(!) equation
QUESTION [5 upvotes]: How many real roots does the below equation have?
\begin{equation*}
\frac{x^{2000}}{2001}+2\sqrt{3}x^2-2\sqrt{5}x+\sqrt{3}=0
\end{equation*}
A) 0  B) 11 C) 12 D) 1 E) None of these
I could not come up with anything.
(Turkish Math Olympiads 2001)

REPLY [8 votes]: Consider the discriminant of $f(X) = 2\sqrt{3}x^2-2\sqrt{5}x+\sqrt{3}=0$:
$$(-2\sqrt{5})^2 - 4(2\sqrt{3})\sqrt{3} = 20 - 24 < 0.$$
Therefore $f(x)$ has no real roots.
But $f(0) = \sqrt{3} > 0$, so $f(x) > 0$ everywhere.
Now combine this with $$\frac{x^{2000}}{2001} \geq 0.$$