Proposition: Allowed scalar parameters

The set

is empty, a singleton, or all of $\mathbb{R}$. More precisely:

1. if $\beta =0$, it is all of $\mathbb{R}$ when $\alpha =0$ and empty otherwise;

2. if $\beta \ne 0$, choose $z_0$ with $\beta (z_0)\ne 0$. The only possible parameter is

   $$c_0=-\frac{\alpha (z_0)}{\beta (z_0)}.$$

   It is allowed exactly when $c_0\in \mathbb{R}$ and $\alpha +c_0\beta =0$ on every relation test.

Proof. If two distinct real parameters solve (eq:affine), their difference times $\beta$ is zero, hence $\beta =0$; then $\alpha =0$. Otherwise there is at most one solution. The displayed alternatives follow directly.