Now that's a polynomial

Let $n=2014$ , then the given equation can be rewritten as:

$\sqrt{n}x^3 - (1+2n)x^2 + 2 \\ = \sqrt{n}x^3 - x^2 - 2nx^2 + 2 \\ = x^2(\sqrt{n}x - 1) - 2(nx^2 - 1) = 0$ .

Note that $nx^2-1=( \sqrt{n}x+1)( \sqrt{n}x-1 )$ . Now we an factor the last expression as $(\sqrt{n}x - 1)(x^2 - 2(\sqrt{n}x+1))=0$

This means that $\frac{1}{\sqrt{n}}$ is a root, and the other two roots are the roots of $x^2 - 2\sqrt{n}x - 2$ . Note that the constant term of the quadratic is negative, so one of the two roots is positive and the other is negative. In addition, by Vieta's Formulas, the roots sum to $2\sqrt{n}$ , so the positive root must be greater than $2\sqrt{n}$ in order to produce this sum when added to a negative value. Since $0 < \frac{1}{\sqrt{2014}} < 2\sqrt{2014}$ is clearly true, $x_2 = \frac{1}{\sqrt{2014}}$ and $x_1 + x_3 = 2\sqrt{2014}$ . Multiplying these values together, we find that $x_2(x_1+x_3) = \boxed{2}$ .

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We observe that $\sqrt{\frac{1}{2014}}$ is a root of given polynomial

Now sum of roots taken two at a time is zero because coefficient of $x$ is zero,i,e, $x_1x_2+x_2x_3+x_3x_1=0$ and also product of roots = $-2=x_1x_2x_3$ which is negative

This means there is exactly one negative root and two positive roots.

Since the sum of roots = $\frac{4029}{\sqrt{2014}}$

This means that the positive root other than $\frac{1}{\sqrt{2014}}$ is greater otherwise the sum of roots cannot be greater than $\frac{2}{\sqrt{2014}}$

So $x_2=\frac{1}{\sqrt{2014}}$

Now $x_1x_2+x_2x_3+x_3x_1=0$

$\implies x_2(x_1+x_3)=-x_1x_3=\frac{-x_1x_2x_3}{x_2}=\frac{\frac{2}{\sqrt{2014}}}{\frac{1}{\sqrt{2014}}}=\boxed{2}$