= Solution
Write $q=(x,X)^{\mathsf T}$, $M=\begin{pmatrix}a&b\\b&c\end{pmatrix}$ and $K=\begin{pmatrix}\alpha&\beta\\\beta&\gamma\end{pmatrix}$. The <Euler-Lagrange equations> give $M\ddot q+Kq=0$. A <normal mode> is a motion $q=v\cos(\omega t+\varphi)$ with fixed nonzero <vector> $v$; its <normal frequency> is $\omega>0$. Substitution gives $(K-\omega^2M)v=0$. Thus the squared <normal frequencies> are the assumed positive distinct roots of
$$
(ac-b^2)\lambda^2-(\alpha c+\gamma a-2\beta b)\lambda+(\alpha\gamma-\beta^2)=0.
$$
For either root, solve the associated two linear equations for the mode ratio.
For the rods, the small angles are $\theta=x/a$ and $\phi=(X-x)/(2a)$. The first rod has <kinetic energy> $m\dot x^2/6$. The second rod's center has horizontal <velocity> $(\dot x+\dot X)/2$ and its <angular velocity> is $(\dot X-\dot x)/(2a)$. Its center-of-mass <moment of inertia> is $2ma^2/3$, giving
$$
T_2=\frac m4(\dot x+\dot X)^2+\frac m{12}(\dot X-\dot x)^2.
$$
Adding these <kinetic energies> gives $T=m(3\dot x^2+2\dot x\dot X+2\dot X^2)/6$. The first center rises by $a\theta^2/4$, and the second by $a\theta^2/2+a\phi^2/2$. Multiplying by the respective masses and $g$ gives
$$
V=\frac{mg}{4a}\bigl[5x^2+(X-x)^2\bigr]
=\frac{mg}{4a}(6x^2-2xX+X^2).
$$
Therefore the stated <Lagrangian> is $T-V$.
Set $\mu=2a\omega^2/(3g)$. The mode equation is
$$
\det\left[\begin{pmatrix}6&-1\\-1&1\end{pmatrix}
-\mu\begin{pmatrix}3&1\\1&2\end{pmatrix}\right]
=5\mu^2-17\mu+5=0.
$$
Hence
$$
\boxed{\omega_\pm^2=\frac{3g}{20a}(17\pm3\sqrt{21}).}
$$
The mode ratio is $r=X/x=(6-3\mu)/(1+\mu)$, so the ratio of the rods' angles is $\phi/\theta=(r-1)/2=(5-4\mu)/[2(1+\mu)]$. It is positive for $\mu_-=(17-3\sqrt{21})/10$ and negative for $\mu_+=(17+3\sqrt{21})/10$. Thus the lower-frequency <normal mode> has the rods \b[in phase], and the higher-frequency <normal mode> has them \b[out of phase].
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