Invertibility of the $2\times 2$ symmetric matrix 
Let $a, b, c \in \mathbb{R}$ \ $ \{0\} $ be nonzero, real constants. For which value of $a, b, c$ does the matrix $$A = \begin{pmatrix}a & b\\ b & c\end{pmatrix} $$
have two non-zero eigenvalues?

I already found the fact that $ac \neq b^2$ but I am wondering if there is anything more.

Note the characteristic polynomial of $A$ is
\begin{align}
P\left (\lambda\right ) &= \left |\lambda I - A \right |\\
&= \left (\lambda - a\right )\left (\lambda - c \right ) - b^2\\
&= \lambda^2 - \left (a + c\right )\lambda + \left (ac - b^2\right )
\end{align}
In order for $A$ to have two distinct non-zero eigenvalues, we must have $ac - b^2 \neq 0$ and $\Delta_\lambda \neq 0$. Hence, we have
\begin{align}
\left (a+c\right )^2 - 4\left (ac - b^2\right ) &\neq 0\\
\left (a - c\right )^2 + 4b^2 &\neq 0
\end{align}
Note that this means we cannot have $a = c$ and $b = 0$ simultaneously (i.e. $A$ cannot be a multiple of the identity).
Therefore, we require $b^2 \neq ac$, and $A \neq kI$ for some real $k$.
We can easily check to confirm that these conditions are sufficient.
