= Solution
Let $x_1,x_2$ be <displacements> from equilibrium, with positive directions chosen alike. <Hooke's law> and <Newton's second law> give
$$
m_1\ddot x_1=-(k_1+k_2)x_1+k_2x_2,\qquad m_2\ddot x_2=k_2x_1-(k_2+k_3)x_2.
$$
For a <normal mode> $x_j=a_j\cos\Omega t$, the <determinant> condition is
$$
\bigl(k_1+k_2-m_1\Omega^2\bigr)\bigl(k_2+k_3-m_2\Omega^2\bigr)-k_2^2=0.
$$
Solving the quadratic for $\Omega^2$ gives
$$
\boxed{\Omega_\pm^2=\frac12\left[\frac{k_1+k_2}{m_1}+\frac{k_2+k_3}{m_2}\pm\sqrt{\left(\frac{k_1+k_2}{m_1}-\frac{k_2+k_3}{m_2}\right)^2+\frac{4k_2^2}{m_1m_2}}\right]}.
$$
The <angular frequencies> are the positive square roots. Positive masses and <spring constants> make the stiffness <quadratic form> $k_1x_1^2+k_2(x_2-x_1)^2+k_3x_2^2$ positive definite, ensuring both squared <frequencies> are positive. Distinct <spring constants> do not require equal masses or any symmetry approximation.
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