= Solution
Complete the square in the <Hermitian matrix> polynomial:
$$
H=\left(A+\frac\alpha2I\right)^2+\left(\beta-\frac{\alpha^2}4\right)I.
$$
For every nonzero <vector> $v$,
$$
v^\dagger Hv=\left\|\left(A+\frac\alpha2I\right)v\right\|^2+\left(\beta-\frac{\alpha^2}4\right)\|v\|^2>0.
$$
Thus $H$ is a <positive-definite matrix>, so $\boxed{0\text{ is not an eigenvalue of }H}$. In <finite-dimensional spectral theorem> terms, each <eigenvalue> is
$$
h_j=\lambda_j^2+\alpha\lambda_j+\beta=\left(\lambda_j+\frac\alpha2\right)^2+\beta-\frac{\alpha^2}4>0.
$$
If $\alpha^2\geq4\beta$, the scalar <polynomial> has real <roots of a polynomial>
$$
r_\pm=\frac{-\alpha\pm\sqrt{\alpha^2-4\beta}}2.
$$
An allowed nonzero root can be an <eigenvalue> of $A$, a <Hermitian matrix> which is also an <invertible matrix>, in which case $H$ has a zero <eigenvalue>. For example, $\alpha=2$, $\beta=1$, $A=-I$ gives $H=0$. The weaker inequality does not force failure for every $A$; it removes the general positive lower bound. There is also the harmless corner $\alpha=\beta=0$, for which $H=A^2$ stays positive definite because $A$ is invertible, even though the strict discriminant hypothesis is absent.
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