5
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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)

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  • 1
    $\begingroup$ 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. $\endgroup$ – Gregory Grant May 6 '15 at 16:53
  • $\begingroup$ And it cannot be one because it is an even function. $\endgroup$ – John May 6 '15 at 23:54
  • $\begingroup$ @John: Odd, I don't see any even function. $\endgroup$ – user21820 May 7 '15 at 0:47
  • $\begingroup$ @user21820 Sorry, not even function, even degree. $\endgroup$ – John May 7 '15 at 15:15
  • $\begingroup$ @John Ok but that's wrong too. $x \mapsto x^2$ has only one root. $\endgroup$ – user21820 May 9 '15 at 3:18
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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.

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  • $\begingroup$ Nice Solution, Thanks! $\endgroup$ – Berkay May 6 '15 at 17:10
8
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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 $$\frac{x^{2000}}{2001} \geq 0.$$

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7
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Hint: $2\sqrt{3} x^2 - 2\sqrt{5} x + \sqrt{3}$ is positive and has no roots.

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