jmc

The expression $x_2(x_1+x_3)$ is not symmetric in the roots $x_1,x_2,x_3,$ so Vieta's formulas can't be used directly to find its value. We hope that we can determine some of the values of the roots explicitly. Letting $a=\sqrt{2014},$ the equation becomes $$a{x}^3-(2a^2+1)x^2+2=0.$$We can rearrange this as $$(a{x}^3-x^2)-(2a^2{x}^2-2)=0,$$or $$x^2(ax-1)-2(ax-1)(ax+1)=0.$$Therefore, we have $$(ax-1)(x^2-2ax-2)=0.$$It follows that one of the roots of the equation is $x=\frac{1}{a},$ and the other two roots satisfy the quadratic $x^2-2ax-2=0.$ By Vieta's formulas, the product of the roots of the quadratic is $-2,$ which is negative, so one of the roots must be negative and the other must be positive. Furthermore, the sum of the roots is $2a,$ so the positive root must be greater than $2a.$ Since $2a>\frac{1}{a},$ it follows that $\frac{1}{a}$ is the middle root of the equation. That is, $x_2=\frac{1}{a}.$

Then $x_1$ and $x_3$ are the roots of $x^2-2ax-2=0,$ so by Vieta, $x_1+x_3=2a.$ Thus, $$x_2(x_1+x_3)=\frac{1}{a}\cdot 2a=\boxed{2}.$$