Minimal polynomial of $\frac{\sqrt{5}+1}{2}$ Minimal polynomial of $\frac{\sqrt{5}+1}{2}$
My try:$$x=\frac{\sqrt{5}+1}{2}$$
$$(2x-1)^2=5$$$$x^2-x-1=0$$ i.e of degree 2.Please check.
 A: One process of finding the minimal polynomial of $\alpha$ for a simple field extension $K(\alpha):K$ is to first find a polynomial with $\alpha$ as a root, then to prove (or at least argue) that it has the least degree among all such polynomials.
You have successfully found a polynomial, which is $x^2-x-1=0$.  Now, how to prove that this polynomial has the least possible degree, i.e. $2$?
When a quadratic polynomials is nominated as the minimal polynomial, we deploy the tactic mentioned by @HagenvonEitzen and attempt a proof by condradiction:
Assume that the minimal polynomial is of degree $1$, then it is of the form $x-\alpha$.  In our case, it's $x-\frac{\sqrt{5}+1}{2}$.  A minimal polynomial must have coefficients in $K$ ($\mathbb{Q}$ in our case), but $\frac{\sqrt{5}+1}{2}\not\in\mathbb{Q}$ because $\sqrt{5}\not\in\mathbb{Q}$.  Therefore, the minimal polynomial of $\frac{\sqrt{5}+1}{2}$ over $\mathbb{Q}$ cannot have degree $1$ and hence must have a degree of $2$ and above, which means that your polynomial
$$
  x^2-x-1=0
$$
is indeed the minimal polynomial of $x^2-x-1=0$ over $\mathbb{Q}$.
