All you did was right. Of course, since you've already deduced that $\det(A)\geq 0$, you already know it is not the case that $\det(A)$ can be any real value: it can't be negative! But it can be any nonnegative real value: given $r\gt 0$, take the diagonal matrix with diagonal entries $\sqrt{r}$ to get a matrix with that determinant.
So we are down to whether the matrix must have $bc=0$.
Say the characteristic polynomial is $t^2+2t+1 = (t+1)^2$. Then we can take $\left(\begin{array}{rr} 0 & -1\\ 1 & -2 \end{array}\right)$ and note that the characteristic polynomial is precisely $-t(-2-t)+1 = t^2 + 2t+1$, exactly what we want. However $bc=-1$.
(How did I come up with that? It's the "companion matrix" of $t^2+2t+1$; but you can come up with such a matrix for any quadratic: if you have $t^2+at+b$, write $t^2+at+b = t(t+a)+b = -t(-t-a)+b$, so put a $0$ and a $-a$ on the diagonal, and have the other two entries multiply to $-b$ and you are done; now just pick a polynomial with a double root).
Or you can obtain an example by starting with a matrix that is not diagonal and has repeated eigenvalues different from $0$, say $\left(\begin{array}{cc}1&1\\0&1\end{array}\right)$ and then conjugating by an appropriate invertible matrix, say
$\left(\begin{array}{cr} \frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & -\frac{1}{2} \end{array}\right) \left(\begin{array}{cc} \vphantom{\frac{1}{2}}1 & 1\\ \vphantom{\frac{1}{2}}0 & 1 \end{array}\right)\left(\begin{array}{cr} \vphantom{\frac{1}{2}}1 & 1\\ \vphantom{\frac{1}{2}}1 & -1 \end{array}\right) = \left(\begin{array}{cr} \frac{3}{2} & -\frac{1}{2}\\ \frac{1}{2} & \frac{1}{2}\end{array}\right)$ which again has $1$ as a repeated eigenvalue, but with $bc\neq 0$.