Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.2 Linear and Quadratic Functions - Exercises - Page 19: 59


The roots of both are reciprocals.

Work Step by Step

Applying the general law of solving a second degree polynomial, we get that the roots are $\frac{-b \pm \sqrt{b^2 - 4 a c}}{2a}$ and $\frac{-b \pm \sqrt{b^2 - 4 a c}}{2c}$ (quadratic equation) That is, the first roots multiplied with $\frac{1}{c}$ equal to the second roots multiplied with $\frac{1}{a}$. Hence, the roots of both are reciprocals.
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