Derive the Quadratic Equation

Here $$-\frac ba=\alpha+\beta=2+\sqrt3+2-\sqrt3=4$$ and
$$\frac ca=\alpha\beta=(2+\sqrt3)(2-\sqrt3)=2^2-3=1$$

So, the quadratic equation becomes $$x^2-4x+1=0$$

It seems to me that the sum and product relations for roots is much more specialized knowledge than is needed for this problem, although perhaps the problem was intended to be an application of these relations.

Just work backwards from "the solution":

$$x = 2 \pm \sqrt{3} \; \implies \; (x-2)=\pm\sqrt{3} \; \implies \; (x-2)^2 = \left(\pm \sqrt{3}\right)^2$$

$$\implies \; (x-2)^2 = 3,$$

and now you have a quadratic equation whose solution is $x = 2 \pm \sqrt{3}.$

$$ 0 = \left(x-\alpha\right)\left(x-\beta\right)=x^2 -\left(\alpha+\beta\right)x + \alpha \beta $$