Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 1 - 1.4 - Quadratic Equations and Applications - 1.4 Exercises - Page 111: 73

Answer

$$x=−4±2\sqrt5$$

Work Step by Step

Using the quadratic formula, we obtain: $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ $$x=\frac{-8\pm \sqrt{(8)^2-4(-4)(1)}}{2(1)}$$ $$x=−4±2\sqrt5$$
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