Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.4 Quadratic Equations - 1.4 Exercises - Page 122: 51


$\color{blue}{\left\{\dfrac{1-\sqrt5}{2}, \dfrac{1+\sqrt5}{2}\right\}}$.

Work Step by Step

RECALL: The quadratic equation $ax^2+bx+c=0$ can be solved using the quadratic formula: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ The given equation has: $a=1, b=-1, c=-1$ Substitute these values into the quadratic formula to obtain: $x=\dfrac{-(-1) \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)} \\x=\dfrac{1\pm\sqrt{1+4}}{2} \\x=\dfrac{1\pm\sqrt{5}}{2}$ Thus, the solution set is $\color{blue}{\left\{\dfrac{1-\sqrt5}{2}, \dfrac{1+\sqrt5}{2}\right\}}$.
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