= Solution
Write $q=(x,X)^T$, $M=\begin{pmatrix}a&b\\b&c\end{pmatrix}$ and $K=\begin{pmatrix}\alpha&\beta\\\beta&\gamma\end{pmatrix}$. The linear equations are $M\ddot q+Kq=0$. A <normal mode> has all coordinates oscillating with one frequency and fixed relative amplitudes, $q=v\cos(\omega t+\varphi)$. Its squared frequency solves
$$
\det(K-\omega^2M)=(\alpha-a\omega^2)(\gamma-c\omega^2)-(\beta-b\omega^2)^2=0,
$$
and its amplitude vector is a nonzero null vector of $K-\omega^2M$. The assumed distinct positive roots give two normal frequencies, their positive square roots.
For the rods let $\theta_1,\theta_2$ be their small inclinations from the downward vertical. To first order, $\theta_1=x/a$ and $\theta_2=(X-x)/(2a)$. The first rod has kinetic <energy> $m\dot x^2/6$. The second has centre-of-mass horizontal velocity $(\dot x+\dot X)/2$ and angular velocity $(\dot X-\dot x)/(2a)$, giving
$$
T_2=\frac m4(\dot x+\dot X)^2+\frac m{12}(\dot X-\dot x)^2.
$$
Together these give $T=m(3\dot x^2+2\dot x\dot X+2\dot X^2)/6$. The centre-of-mass height changes give
$$
V=\frac{mga}{4}\theta_1^2+mga(\theta_1^2+\theta_2^2)
=\frac{mg}{4a}(6x^2-2xX+X^2),
$$
proving the displayed quadratic <Lagrangian>.
Here $M=(m/3)\begin{pmatrix}3&1\\1&2\end{pmatrix}$ and $K=(mg/(2a))\begin{pmatrix}6&-1\\-1&1\end{pmatrix}$. Set $\lambda=2a\omega^2/(3g)$. The <determinant> equation reduces to $5\lambda^2-17\lambda+5=0$, so
$$
\boxed{\omega_\pm^2=\frac{3g}{20a}(17\pm3\sqrt{21})}.
$$
The first row of the mode equation gives $X/x=(6-3\lambda)/(1+\lambda)$. The actual rod-angle ratio is therefore
$$
\frac{\theta_2}{\theta_1}=\frac{5-4\lambda}{2(1+\lambda)}.
$$
It is positive for the smaller root and negative for the larger root. Thus \b[the low-frequency mode has the rods in phase, and the high-frequency mode has them out of phase]. Using the rod-angle ratio is necessary because $X$ is the displacement of the end of the second rod, not that rod's inclination.
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