Suppose $L/K$ is an algebraic field extension. Take $\alpha\_1 \in L$. Then $\alpha\_1$ has minimal polynomial $f(x)$ over $K$. Let $\alpha\_2, ... \alpha\_k$ be the other roots of $f$ in $L$. The $\alpha\_i$ are known as algebraic conjugates.

Suppose I am given an $\alpha \in L$ and wish to find the minimal polynomial of $\alpha$ over $K$. It'd be helpful if I knew something about how the algebraic conjugates of $\alpha$ behaved. 

Let's take a concrete example. Suppose $L = \mathbb Q(\sqrt{3})$ and $K = \mathbb Q$, and I want to know the minimal polynomial of $2 + \sqrt{3}$. I could work it out by writing $ x = 2 + \sqrt{3}$, squaring, rearranging and squaring again. But this would quickly become cumbersome if we chose a more complicated extension and complicated $\alpha$. I know its a good idea to look at the product $ \prod (x - \alpha\_i) $, where the $\alpha\_i$ are algebraic conjugates of $\alpha$. What are the algebraic conjugates of $ 2 + \sqrt{3} $? I 'know' they are $ \pm 2 \pm \sqrt{3}$, but why is this? 

I suppose my question really comes down to:

i) Why are the conjugates of $2 + \sqrt{3}$ those mentioned above? Why do the symmetric sums and products involved in Vieta's formulas yield rationals in this case?

ii) Are there any other 'types' of $\alpha$ (and/or $L$) for which the form of its conjugates are known?

Thanks