Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 11 - Quadratic Functions and Equations - 11.5 Equations Reducible to Quadratic - 11.5 Exercise Set - Page 732: 61



Work Step by Step

$ 3x^{4}+5x^{2}-1=0\qquad$...substitute $x^{2}$ for $u$ $ 3u^{2}+5u-1=0\qquad$... solve with the Quadratic formula. $u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ $u=\displaystyle \frac{-5\pm\sqrt{25+12}}{6}$ $u=\displaystyle \frac{-5\pm\sqrt{37}}{6}$ $u=\displaystyle \frac{-5+\sqrt{37}}{6} $or $u=\displaystyle \frac{-5-\sqrt{37}}{6}$ Bring back $x^{2}=u$. $x^{2}=\displaystyle \frac{-5+\sqrt{37}}{6}$ or $x^{2}=\displaystyle \frac{-5-\sqrt{37}}{6}$ $x=\pm\sqrt{\frac{-5\pm\sqrt{37}}{6}}$
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