If two sides of triangle are the roots of the equation$4x^2-\left(2\sqrt{6}\right)x+1=0$ and included angle is$60°$ , then the third side is

$\begin{array}{l}4x^2-2\sqrt{6}x+1=0\\ x=\frac{-(-2\sqrt{6})\pm \sqrt{24-16}}{2\left(4\right)}\end{array}$

$=\frac{2\sqrt{6}\pm \sqrt{8}}{8}$

$\begin{array}{l}=\frac{2\sqrt{6}+2\sqrt{2}}{8}=\frac{\sqrt{6}+\sqrt{2}}{4}\\ \therefore a=\frac{\sqrt{6}+\sqrt{2}}{4};b=\frac{\sqrt{6}-\sqrt{2}}{4}\end{array}$

$\cos C=\frac{b^2+a^2-c^2}{2ab}$

$\cos 60=\frac{{\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)}^2+{\left(\frac{\sqrt{6}+\sqrt{2}}{4}\right)}^2-c^2}{2.\frac{\sqrt{6}-\sqrt{2}}{4}.\frac{\sqrt{6}+\sqrt{2}}{4}}$

$\begin{array}{l}\frac{1}{2}=\frac{{(\sqrt{6}-\sqrt{2})}^2+{(\sqrt{6}+\sqrt{2})}^2-16c^2}{2(6-2)}\\ 4=2(6+2)-16c^2\end{array}$

$\begin{array}{l}4=16{(1-c)}^2\\ 1-c^2=\frac{1}{4}\end{array}$

$\begin{array}{l}{c}^2=\frac{3}{4}\\ c=\frac{\sqrt{3}}{2}\end{array}$