= Solution
Substitution of the exponential ansatz gives the homogeneous <amplitude> system
$$
\begin{pmatrix}\omega_1^2-\omega^2&2i\Omega\omega\\-2i\Omega\omega&\omega_2^2-\omega^2\end{pmatrix}\begin{pmatrix}\bar x_1\\\bar x_2\end{pmatrix}=0.
$$
A nonzero mode requires zero <determinant>. Thus $\boxed{(\omega_1^2-\omega^2)(\omega_2^2-\omega^2)=4\Omega^2\omega^2}$ with the stated two effective stiffnesses. The off-diagonal product is positive, which determines the sign on the right.
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