About number of algebra of dimension 2 over a field

Question

I got stuck in this exercise. I would be so grateful if anyone can give me a hint for it. Thanks a lot

Let k be a field, A a k-algebra of dimension 2. Choose $t \in A, t \notin k \subset A$. Then $A$ is generated by $1$ and $t$ and so there is an isomorphism of algebras $k[x]/(x^2+ax+b) \cong A$ for some $a, b \in k$. Which of the following are true:

1. if $k = \mathbb{Q}$, there are infinitely many non-isomorphic possibility for $A$
2. if $k = \mathbb{F_p}$, there are two possibilities for $A$ up to isomorphism
3. if $k = \mathbb{F_p}$, there are three possibilities for $A$, up to isomorphism
4. if $k$ is algebraically closed, there are two possibilities for $A$, up to isomorphism

I know that number 1) is true, because we just need to choose $a$ and $b$ such that the roots of $(x^2+ax+b)$ is in $k$. Also, because $A$ is vector space with dimension $2$ over $k$, so there must be a unique $a$ and $b$ such that $(t^2+at+b) = 0$. In that case $k[x]/(x^2+ax+b) \cong A$. So in my opinion, there is only one possibility for $a$ and $b$, but it is not true. Can anyone give me some explanation for it. Thanks a lot.