(a) $x^2 - 4x + 1 = 0$
Sum of roots $= (2+\sqrt{3})+(2-\sqrt{3}) = 4$; Product of roots $= (2+\sqrt{3})(2-\sqrt{3}) = 4-3 = 1$.
Required equation: $x^2 - (\text{sum})x + (\text{product}) = 0 \Rightarrow x^2 - 4x + 1 = 0$.
For any quadratic with roots $\alpha$ and $\beta$, use $x^2 - (\alpha+\beta)x + \alpha\beta = 0$. The key trick here is recognising $(2+\sqrt3)(2-\sqrt3)$ as a difference of squares: $4-3=1$. Examiners award the mark for the correct option; showing the sum/product calculation ensures full credit even if asked as a short-answer variant.