# Finding the number of real roots of an unusual(!) equation [closed]

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.

• I believe the 2nd derivative is always positive, so inflection never changes, which means you cannot have more than two zeros. Since two is not a choice, it must be zero or one. – Gregory Grant May 6 '15 at 16:53
• And it cannot be one because it is an even function. – John May 6 '15 at 23:54
• @John: Odd, I don't see any even function. – user21820 May 7 '15 at 0:47
• @user21820 Sorry, not even function, even degree. – John May 7 '15 at 15:15
• @John Ok but that's wrong too. $x \mapsto x^2$ has only one root. – user21820 May 9 '15 at 3:18

We have that $x^{2000} \geq 0$, because squares are nonnegative.

Further, we have $(x-\frac{1}{2}\sqrt{\frac{5}{3}})^2 \geq 0$. This gives $x^2-\sqrt{\frac{5}{3}}x+\frac{5}{12} \geq 0$

Therefore $2 \sqrt{3}x^2-2\sqrt{5}x+\frac{10}{12}\sqrt{3} \geq 0$

Therefore $2 \sqrt{3}x^2-2\sqrt{5}x+\sqrt{3} > 0$

Therefore $\frac{x^{2000}}{2001} + 2 \sqrt{3}x^2-2\sqrt{5}x+\sqrt{3} > 0$

Therefore there are no real roots.

• Nice Solution, Thanks! – Berkay May 6 '15 at 17:10

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 $$\frac{x^{2000}}{2001} \geq 0.$$

Hint: $2\sqrt{3} x^2 - 2\sqrt{5} x + \sqrt{3}$ is positive and has no roots.