Answer
Since \(R\) is an infinite set that is at most countable, it follows that \(R\) is **countably infinite**.
Work Step by Step
**Solution Sketch**
Consider all equations of the form
\[
x^2 + b\,x + c = 0
\quad\text{where}\quad b,c \in \mathbb{Z}.
\]
Let \(R\) be the set of **all** solutions (real or complex) to these equations. We want to show \(R\) is countable.
---
### 1. Enumerate All Possible \((b,c)\) Pairs
Since \(b\) and \(c\) range over \(\mathbb{Z}\), the set of all pairs \((b,c)\) is \(\mathbb{Z} \times \mathbb{Z}\). We know \(\mathbb{Z}\times \mathbb{Z}\) is **countably infinite**.
---
### 2. Each Equation Has at Most 2 Solutions
For a fixed pair \((b,c)\), the quadratic equation \(x^2 + bx + c = 0\) has at most two solutions (in the complex plane, or possibly fewer if we only consider real solutions and the discriminant is negative).
Hence the set of solutions for each \((b,c)\) is **finite** (size at most 2).
---
### 3. A Countable Union of Finite Sets Is Countable
Form the union of solution sets over all \((b,c) \in \mathbb{Z}\times \mathbb{Z}\). Symbolically:
\[
R
\;=\;
\bigcup_{(b,c)\in \mathbb{Z}\times \mathbb{Z}}
\{\text{solutions of }x^2+bx+c=0\}.
\]
- We have **countably many** pairs \((b,c)\).
- For each pair, the set of solutions is **finite** (size at most 2).
A **countable union** of finite sets is **countable**. Thus \(R\) is at most countable.
---
### 4. \(R\) Is Infinite
Clearly, \(R\) is not finite, because for instance the equation \(x^2 - nx + 0=0\) has the solutions \(x=0\) or \(x=n\), and \(n\) can be any integer. Hence there are infinitely many distinct solutions in \(R\).
