Eigenvalues changing under a change in the potential

If I have computed the eigenvalues of the Hamiltonian (i.e. the allowed energies) for the potential $V(x) = \dfrac{1}{2}m\omega^2x^2$ to be $E_n = \hbar\omega(n \dfrac{1}{2})$ then if the potential is changed to $V(x) = \dfrac{1}{2}m\omega^2x^2 Kx$ for some real K then am I correct in thinking that the allow energies won't change because we can complete the square and obtain a potential of the same form (translated) plus some constant. But constants in the energy are arbitrary (?) and so are translations so they won't change the energies?

Thanks

this isn't quite right, the energies themselves will be different but what will not change is the overall behaiviour of the systen.

i haven't worked out the schrodinger equation fully for this examole but following as per your example you will have a potential of the form (ax b)^2 V_0 via completing the square, where V_0 is a constant offset.

it can be shown that adding a constant offset to the potential doesnt change the eigenvalues (the only difference is that your wavefunction picks up an extra exponential phase factor). ignoring the phase offset, now you will be solving the schrodinger equation for a slightly modified potential (ax b)^2. the procedure is the same but i don't think the energy eigenvalues will be exactly the same anymore due to the linear scaling (admittedly i havent tried this myself so could have gone wrong somewhere)

I'm confused as to why, ignoring the phase, the solutions won't be the same. If they're defined over all of the real life and the potential is effectively just translated by a finite distance the eigenfunctions won't just have the same translation but the same energies as before.