= Solution
At coexistence a nonzero $Q_t$ must be stationary and have the same <free energy> as $Q=0$. Dividing $f(Q_t)=0$ by $6Q_t^2$ and $f'(Q_t)=0$ by $6Q_t$ gives
$$
a+bQ_t+3cQ_t^2=0,\qquad 2a+3bQ_t+12cQ_t^2=0.
$$
Eliminating $a$ yields $Q_t=-b/(6c)$ and then $b^2=12ac$. The allowed nonpositive root therefore gives
$$
\boxed{b_t=-\sqrt{12ac}=-2\sqrt{3ac},\qquad Q_t=\sqrt{\frac{a}{3c}}.}
$$
As $b$ is decreased through $b_t$, the equilibrium scalar order jumps \b[from $0$ for $b>b_t$ to $Q_t$ just on the $b<b_t$ side]. At $b=b_t$ both values coexist; the stable aligned branch below it is $Q_+$ from part (e). As a global-minimum check, at coexistence
$$
f(Q)=18cQ^2(Q-Q_t)^2\geq0.
$$
There are two equal minima and a barrier at $Q_t/2$, of height $a^2/(8c)$. Comparing these minima, rather than finding only the stationary-point discriminant, determines the transition.
This is the unrestricted Landau answer, compatible with the <physical range of a uniaxial nematic order parameter> when $a\leq3c$ so that $Q_t\leq1$. If one instead enforces $Q\in[-1/2,1]$ for arbitrary positive coefficients, negative $Q$ cannot win for $b\leq0$, and <bounded quartic nematic coexistence> minimizes $a/Q+3cQ$ over $0<Q\leq1$. For $a>3c$ its minimum is at the boundary, giving \b[$b_t=-(a+3c)$ and a jump $0\to1$]. No coefficients were supplied to choose this boundary case; the near-transition Landau approximation normally uses the interior result.
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