= Solution
Use the column-coroot convention $C_{ij}=2(\alpha_i,\alpha_j)/(\alpha_j,\alpha_j)$, consistent with the printed <matrix> and the stated shorter first root. It gives
$$
\frac{2(\alpha,\beta)}{\|\beta\|^2}=-1,\qquad\frac{2(\beta,\alpha)}{\|\alpha\|^2}=-3,\qquad\|\beta\|^2=3\|\alpha\|^2.
$$
The <root string> through $\beta$ in the $\alpha$ direction contains $\beta,\beta+\alpha,\beta+2\alpha,\beta+3\alpha$. Reflecting $3\alpha+\beta$ in $\beta$ gives $3\alpha+2\beta$. Together with $\alpha$ and their negatives these give
$$
\boxed{\Phi=\pm\{\alpha,\ \beta,\ \alpha+\beta,\ 2\alpha+\beta,\ 3\alpha+\beta,\ 3\alpha+2\beta\}.}
$$
There are no further roots. Indeed the simple reflections act on coefficient pairs by $(a,b)\mapsto(-a+3b,b)$ and $(a,b)\mapsto(a,a-b)$; the displayed twelve-element set is closed under both. Every positive nonsimple root can be reflected to a <positive root> of smaller height: choose a <simple root> with positive <inner product> as in (a), and its positive integral Cartan pairing decreases height. Repetition reaches a <simple root>, so every root lies in an orbit already contained in the list. This is the <G2 root system>.
With $\alpha=e_1$, the <inner product> and length give the two reflected possibilities
$$
\boxed{\beta=-\frac32e_1+\frac{\sqrt3}{2}e_2\quad\text{or}\quad\beta=-\frac32e_1-\frac{\sqrt3}{2}e_2.}
$$
\Image[/past-exam-of-the-lie-algebras-and-representation-theory-of-symmetric-groups-course-of-the-university-of-oxford/2009-g2-roots.png]
{title=The twelve roots of G2 with the shorter simple root horizontal}
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