Describe $\mathbb{R}[x]/(x^{2}+ax+b)$ where $a,b \in \mathbb{R}$ So obviously using quadractic formula, we have three cases for the roots which depends on the discriminant. I am not sure if I am right on this but,
Case 1: $a^{2}-4b>0$. 
Then we have two distinct real roots so this factors as
$$\mathbb{R}[x]/(x-\alpha) \oplus \mathbb{R}[x]/(x-\beta).$$
(My guess here is since $\alpha$ and $\beta$ are different, these two ideals are comaximal so I can apply the Chinese Remainder Theorem). But this is isomorphic to
$$\mathbb{R} \oplus \mathbb{R}.$$
Case 2: $a^{2}-4b < 0$.
In this case, we have a complex root so we have an extension of degree $2$. But obviously whatever root we have, this is contained in $\mathbb{R}[i]\cong \mathbb{C}$ so we must have this is $\mathbb{C}$ by the tower law since $\mathbb{R}[i]$ is also a degree $2$ extension of $\mathbb{R}$.
Case 3: $a^{2}-4b=0$.
Then we have a real root with multiplicity $2$. I am not sure what this is as if we have the root is $\alpha$, we get
$$\mathbb{R}[x]/(x-\alpha)^{2}$$
Since these roots are the same, $(x-\alpha)$ and $(x-\alpha)$ are not comaximal I believe so I don't know what this ring is.
 A: The first two cases are correct.
For the third case you can use that the map
$$
  \mathbb{R}[x] \to \mathbb{R}[x],
  \quad
  f(x) \mapsto f(x+\alpha)
$$
is a ring isomorphism which maps $x-\alpha$ to $x$, and thus induces an isomorphism
$$
  \mathbb{R}[x]/(x - \alpha)^2 \to \mathbb{R}[x]/(x^2),
  \quad
  \overline{f(x)} \mapsto \overline{f(x + \alpha)}.
$$
The resulting ring $\mathbb{R}[x]/(x^2)$ is the ring of dual numbers.
(If we rename the variable $x$ to $\varepsilon$ then the elements of $\mathbb{R}[\varepsilon]/(\varepsilon^2)$ are of the form $a + b \varepsilon$ with $a, b \in \mathbb{R}$ and $\varepsilon \neq 0$ (and $\varepsilon \notin \mathbb{R}$), but $\varepsilon^2 = 0$.)
It should probably also be pointed out that the three possible results
$$
  \mathbb{R} \times \mathbb{R},
  \quad
  \mathbb{C},
  \quad
  \mathbb{R}[x]/(x^2)
$$
are pairwise non-isomorphic:
The ring $\mathbb{R}[x]/(x^2)$ is the only one containing a nonzero nilpotent element (namely $x$) and $\mathbb{C}$ is the only one being a field (or even an integral domain).
