If the length of the two sides of a triangle are the roots of the equation $\;$ $x^2 - 2 \sqrt{3}x + 2 = 0$ $\;$ and if the included angle between them has measure $\dfrac{\pi}{3}$, then find the perimeter of the triangle.
Let the given triangle be $ABC$ (as shown in the figure) with sides $b$ and $c$ given by the roots of the quadratic equation $x^2 - 2 \sqrt{3}x + 2 = 0$.
The included angle between sides $b$ and $c$ is $\angle A = \dfrac{\pi}{3}$ (given).
Roots of the quadratic equation $\;$ $x^2 - 2 \sqrt{3}x + 2 = 0$ $\;$ are
$\begin{aligned}
x & = \dfrac{2 \sqrt{3} \pm \sqrt{\left(2 \sqrt{3}\right)^2 - 4 \times 1 \times 2}}{2 \times 1} \\\\
& = \dfrac{2 \sqrt{3} \pm \sqrt{12 - 8}}{2} \\\\
& = \dfrac{2 \sqrt{3} \pm 2}{2} \\\\
& = \sqrt{3} \pm 1
\end{aligned}$
In $\triangle ABC$, let $b = \sqrt{3} + 1$ and $c = \sqrt{3} - 1$
By cosine rule, $\;$ $\cos A = \dfrac{b^2 + c^2 - a^2}{2 b c}$
$\begin{aligned}
\text{i.e. } a^2 & = b^2 + c^2 - 2 b c \cos A \\\\
& = \left(\sqrt{3} + 1\right)^2 + \left(\sqrt{3} - 1\right)^2 - 2 \times \left(\sqrt{3} + 1\right) \left(\sqrt{3} - 1\right) \cos \left(\dfrac{\pi}{3}\right) \\\\
& = 3 + 1 + 2 \sqrt{3} + 3 + 1 - 2 \sqrt{3} - 2 \times 2 \times \dfrac{1}{2} \\\\
& = 6
\end{aligned}$
$\therefore \;$ $a = \sqrt{6}$
$\therefore \;$ Perimeter of $\triangle ABC$ $= a + b + c = \sqrt{6} + \sqrt{3} + 1 + \sqrt{3} -1 = 2 \sqrt{3} + \sqrt{6}$