Given the Cartan Matrix $$\begin{pmatrix}2&-1 \\ -3& 2\end{pmatrix}$$ one can find all the positive roots via the following algorithm.
Let $\beta$ be the long root and $\alpha$ the short root. Let further $r,q\in\mathbb{Z}_{\ge 0}$ such that $\beta-r\alpha\in \Phi$ and $\beta+q\alpha\in \Phi$. We then have $$\langle \beta,\alpha \rangle = r-q.$$
$\alpha$-string through $\beta$: Since both roots are positive $\beta-\alpha$ is not a root, therefore $r=0$. We further can calculate $\langle\beta,\alpha \rangle = -3$ so we get four roots: $$\beta, \beta+\alpha, \beta+2\alpha,\beta+3\alpha.$$
$\beta$-string through $\beta+3\alpha$: Since for $r=1$ we get $3\alpha$ which is not a root, we have $r=0$. Further $\langle \beta+3\alpha,\beta\rangle =-1$, so we get: $$\beta + 3\alpha, 2\beta+3\alpha.$$ Only $2\beta+3\alpha$ is a new root.
Since $\alpha$ is also a root we have now in total $6$ roots, are all positive and there are in total 12 roots (from the root diagram), so we are done, with the solution set: $$\{\alpha, \beta, \beta+\alpha, \beta+2\alpha,\beta+3\alpha, 2\beta+3\alpha\}.$$
In Georgis book he presents a way of obtaining this result much quicker (see page 118, eq. (8.59)), via the diagram:
In his description $\alpha_1$ is the short root and $\alpha_2$ is the long root. The $k$ is supposed to represent some kind of level and hes mentioned $SU(2)$ a couple of times in his attempted explanation, but I can hardly make any sense of it..
Can somebody explain to me how this diagram is constructed and how I'm supposed to read off the positive roots from it?