Can we find $x$ such that $\det[x^2 A + x B + C] = 0$?

Question

If $\mathbf{A}, \mathbf{B}, \mathbf{C} \in \mathbb{C}^{4 \times 4}$ are invertible matrices, is there a way to find $x$ such that the following determinant vanishes?

$$\det \left( x^2 \mathbf{A} + x \mathbf{B} + \mathbf{C}\right) = 0$$

Answer 1

If we allow complex $x$ then yes. If not then in 1-dimensional case it is not that difficult to find numbers $A,B,C$ such that $Ax^2 + Bx + C$ has no real roots. For complex numbers - determinant can be expressed as a sum of products of entries over permutations. So, for some numbers $M,N,K$ the question is if $Mx^2 + Nx + K$ has a root. Which is of course true. Though there could be a degenerate case when $M$ and $N$ are zero and $C$ is not. This happens in 1 dimensional case when $A=0, B=0$, and $C=1$. In words the full answer is if $Ax^2 + Bx + C$ is not a constant matrix then yes. Otherwise, only if $C=0$.