Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 4 - Elementary Number Theory and Methods of Proof - Exercise Set 4.2 - Page 169: 30

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.
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