= Solution
The <Euler-Lagrange equation>[Euler--Lagrange equations] for the quadratic Lagrangian are
$$
m\ddot x+\mathsf Vx=0.
$$
For a <normal mode> $x(t)=a e^{i\omega t}$, they become the <generalized eigenvalue problem for small oscillations>
$$
\mathsf Va=m\omega^2a.
$$
Put $\lambda=m\omega^2/k$. With the stated spring constants,
$$
\frac{\mathsf V}{k}
=
\begin{pmatrix}
1+\varepsilon(1+\delta)&-\varepsilon\\
-\varepsilon&1+\varepsilon(1-\delta)
\end{pmatrix}
=(1+\varepsilon)I
+\varepsilon
\begin{pmatrix}
\delta&-1\\
-1&-\delta
\end{pmatrix}.
$$
The final matrix has <characteristic polynomial>
$$
\nu^2-(1+\delta^2),
$$
so its <eigenvalues>[eigenvalue] are $\nu_\pm=\pm\sqrt{1+\delta^2}$. Hence
$$
\lambda_\pm
=1+\varepsilon\pm\varepsilon\sqrt{1+\delta^2}
=1+\varepsilon\left(1\pm\sqrt{1+\delta^2}\right).
$$
The two <angular frequencies> therefore satisfy
$$
\boxed{
\omega_\pm^2=\lambda_\pm\frac{k}{m},
\qquad
\lambda_\pm=1+\varepsilon\left(1\pm\sqrt{1+\delta^2}\right)
}.
$$
This is the spectrum recorded by <normal modes of two equal masses between three springs>.
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