Let $\mathcal{F}$ be the family of nonzero polynomials of the form $p(x,y) = (ax^2 + bx + c)(dy^2 + ey + f)$, where $a,b,c,d,e,f \in \mathbb{Q}$. Let $E_p = \{(x,y) \in \mathbb{R}^2 \mid p(x,y) = 0\}$.
Question: Is $E = \bigcup_{p \in \mathcal{F}} E_p$ countable (that is, finite or countably-infinite)?
Here are my thoughts.
Fix $p(x,y) = (ax^2 + bx + c)(dy^2 + ey + f)$, where $a,b,c,d,e,f \in \mathbb{Q}$. Then $p(x,y) = 0$ if and only if either $ax^2 + bx + c = 0$ or $dy^2 + ey + f = 0$, which happens if and only if either $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ or $y = \dfrac{-e \pm \sqrt{e^2 - 4df}}{2d}$.
Now, the (complex) roots of $p(x,y)$ are all of the numbers in the collection $U_p = \bigcup_{\alpha, \beta \in \mathbb{C}} \{(\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \beta), (\alpha, \dfrac{-e \pm \sqrt{e^2 - 4df}}{2d})\}$.
Then, since $E_p \subseteq \mathbb{R}^2$, we have $E_p = U_p \cap \mathbb{R}^2$. So $E = \bigcup_{p \in \mathcal{F}} (U_p \cap \mathbb{R}^2)$.
But I am having some trouble proceeding. Why is $U_p \cap \mathbb{R}^2$ countable for each $p$?
I can see that there are countably-many polynomials $p \in \mathcal{F}$: each polynomial $p(x,y) = (ax^2 + bx + c)(dy^2 + ey + f)$ can be put in one-to-one correspondence with the tuple $(a,b,c,d,e,f) \in \mathbb{Q}^6$, which is a countable set. So if I can show that $E_p$ is countable for each $p$, then it will follow that $E = \bigcup_{p \in \mathcal{F}} E_p$ is the union of countably-many countable sets, hence also countable.
If there is a simpler proof that $\bigcup_{p \in \mathcal{F}} E_p$ is countable, then I'd appreciate any suggestions.
You're trying to prove a false statement: there are, in fact, uncountably many points in $E$. To see this, let $ax^2+bx+c$ have a solution $x=x_0$; then for any real $r$, the point $(x_0, r)$ is in $E$. So there are as many points in $E$ as there are real numbers - in particular, uncountably many.