Option A: $k = \pm\sqrt{24}$
For equal roots, discriminant $= 0$: $k^2 - 4(\sqrt{3})(2\sqrt{3}) = 0 \Rightarrow k^2 - 24 = 0 \Rightarrow k = \pm\sqrt{24}$.
For equal roots, use $b^2 - 4ac = 0$. Here $a = \sqrt{3}$, $b = -k$, $c = 2\sqrt{3}$. So $k^2 = 4 \times \sqrt{3} \times 2\sqrt{3} = 24$, giving $k = \pm\sqrt{24}$. Remember both positive and negative values are valid.