This is Exercise 2.1.14 of ``Lecture on the geometry of Manifold."
Let $Z=\{(x,a,b,c) \in \mathbb{R}^{4}: a \neq 0, ax^2+bx+c=0 \}$. Show that $Z$ is a smooth submanifold of $\mathbb{R}^{4}.$
Clearly, if we just assume $Z' = \{(x,a,b,c) \in \mathbb{R}^{4}: ax^2+bx+c=0 \}$, then by letting $f:(x,a,b,c) \mapsto ax^{2}+bx+c$, which is clearly smooth and regular at $0$ (because $\frac{\partial f}{\partial c}=1$) we know that $Z'$ is an embedded smooth submanifold of codimension 1. However, in case of $Z$, if we construct similar map $F:(x,a,b,c) \mapsto (a,ax^{2}+bx+c)$, then we may not use the regular value theorem since it is not a level set of some value. Moreover, $U =\{(x,y)\in \mathbb{R}^{2}: x\neq 0, y=0 \}$ is not even a closed set nor open set. Now I'm stuck.
Could you give any hint for solving this exercise?
If you argued that $Z'$ is a smooth submanifold (which happens to be a closed subset of $\Bbb R^4$), then you're just observing that $Z$ is an open subset of $Z'$, which immediately makes it a smooth submanifold.