Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.7 - Equations - Concept and Vocabulary Check - Page 105: 20

Answer

$2\pm\sqrt{2}$

Work Step by Step

$-(-4)=4,$ $(-4)^{2}=16$ $4(1)(2)=8$ $... x=\displaystyle \frac{4\pm\sqrt{16-8}}{2}$ $=\displaystyle \frac{4\pm\sqrt{8}}{2}$ $=\displaystyle \frac{4\pm\sqrt{4\cdot 2}}{2}$ $=\displaystyle \frac{4\pm 2\sqrt{2}}{2}$ $=\displaystyle \frac{2(2\pm\sqrt{2})}{2}$ = $2\pm\sqrt{2}$
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