= Solution
A real symmetric traceless $2$ by $2$ matrix has eigenvalues $\lambda$ and $-\lambda$ and can be written
$$
Q_{ij}=2\lambda\left(n_in_j-\frac12\delta_{ij}\right)
$$
for a unit <nematic director> $\mathbf n\sim-\mathbf n$. Since $Q_{ij}Q_{ji}=2\lambda^2$, the uniform <Landau free energy> is
$$
\mathbb F_{\rm bulk}=2a\lambda^2+4b\lambda^4.
$$
For $a<0<b$, its nonzero minima satisfy
$$
4a\lambda+16b\lambda^3=0,
$$
and hence
$$
\boxed{\lambda_0=\sqrt{-\frac a{4b}}.}
$$
The signs $\pm\lambda_0$ amount to exchanging the two orthogonal eigenvectors, so the chosen convention takes $\lambda_0\geq0$.
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