Answer
See below.
Work Step by Step
1. Let $r, s$ be the solutions to the quadratic equation, we have:
$x^2+bx+c=(x-r)(x-s)$
2. Use the quadratic formula, we have $x=\frac{-b}{2}\pm\frac{\sqrt {b^2-4c}}{2}$ where $b,c$ are rationals.
3. Case1: assume $r=\frac{-b}{2}+\frac{\sqrt {b^2-4c}}{2}$ is a rational, we can conclude that $\sqrt {b^2-4c}=2r+b$ mus be a rational, thus $s=\frac{-b}{2}-\frac{\sqrt {b^2-4c}}{2}$ will also be a rational.
4. Case2: assume $r=\frac{-b}{2}-\frac{\sqrt {b^2-4c}}{2}$ is a rational, we can conclude that $\sqrt {b^2-4c}=-b-2r$ mus be a rational, thus $s=\frac{-b}{2}+\frac{\sqrt {b^2-4c}}{2}$ will also be a rational.
