= Solution
At onset the <determinant> has a double root, so
$$
k_c^2=\frac{u_0\kappa_2-s\kappa_1}{2\kappa_1\kappa_2}=\frac{u_0d^2-s}{2d\sqrt{\kappa_1\kappa_2}}.
$$
Squaring the neutral-boundary equation and canceling the positive factors gives $u_0=s/[d(2\sqrt{2s}-d)]$. Substitution yields
$$
\boxed{k_c^2=\frac{1}{\sqrt{\kappa_1\kappa_2}}\frac{s(d-\sqrt{2s})}{d(2\sqrt{2s}-d)}.}
$$
For a positive branch with $u_0>1/2$, the actual neutral curve lies strictly between $\sqrt{2s}$ and $2\sqrt{2s}$. If $d<\sqrt{2s}$, then $d<d_c(u_0)$ for every such branch: a homogeneously stable state cannot have this <Turing instability>. If $d>2\sqrt{2s}$, then $d>d_c(u_0)$ for every finite positive $u_0$, so every homogeneously stable positive state is spatially unstable. The negative value yielded by the displayed formula outside its boundary domain is not a physical critical <wavenumber> and does not imply renewed stability. One must also keep $u_0<s$, which can restrict the attainable neutral curve further.
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