Confused over how the eigenvalues were obtained in a question involving a second order polynomial

Question

The characteristic equation is as follows:
$E^2-hbar omega E -hbar^2omega^2alpha^2 = 0 $
The eigenvalues I get are $frac{hbaromega(1pmsqrt{1+4alpha^2})}{2}$, however, I am being told the solution is $E_{0}(alpha) = hbaromega(1-sqrt{1+alpha^2})$ and $E_{1}(alpha)=hbaromega(1+sqrt{1+alpha^2)}$
I can't see how they are getting rid of both the 4 in from of $alpha^2$ and the denominator 2. At this point I am wondering the the provided solution is incorrect, which I hope is not the case, and there is some trick I am missing here.

Michael Seifert · Answer

The given solution is wrong.  You can see this by taking the $alpha = 0$ case of the provided Hamiltonian:
$$
H = hbar omega begin{bmatrix} 0 & 0 \ 0 & 1 end{bmatrix}
$$
The eigenvalues of this matrix are obviously $0$ and $hbar omega$ (since the matrix is diagonal, we can just read off the eigenvalues.)  Your answer reduces to these values in the $alpha = 0$ case, while the official answer reduces to $0$ and $2 hbar omega$.
This tactic, by the way, is generally quite useful in physics:  when faced with a discrepancy between two results, look for a simpler case where the answer is obvious and which will allow you to distinguish between the two.

Confused over how the eigenvalues were obtained in a question involving a second order polynomial

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