determine the degree of $2+\sqrt3$ over $\Bbb Q$

Question

By using the binomial theorem for any $n\in\Bbb N$ $$(2+\sqrt3)^n=2^n+{n\choose1}2^{n-1}\sqrt3+{n\choose2}2^{n-2}3+{n\choose3}2^{n-3}3^{3/2}+...+{n\choose{n-1}}2\cdot3^{{n-1}\over2}+3^{n/2}\notin\Bbb Q$$

So the degree in infinite, is that correct?

Answer 1

$P(x)=(X-2)^2-3$ is the minimal polynomial so the degree is $2$.

Answer 2

$$(2+\sqrt{3})^2=7+4\sqrt{3}$$

$$(7+4\sqrt{3})-4(2+\sqrt{3})=-1\in\Bbb Q$$

Set $x=2+\sqrt{3}$, then $x^2-4x+1 =0$. Hence the degree of $x$ over $\Bbb Q$ is two.