College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.4 - Radical Equations; Equations Quadratic in Form; Factorable Equations - 1.4 Assess Your Understanding - Page 118: 64

Answer

$x=\pm\sqrt{\dfrac{-5+\sqrt{41}}{2}}$

Work Step by Step

$\sqrt[4]{4-5x^{2}}=x...$ Raise both sides to the 4th power $4-5x^{2}=x^{4}$ $0=x^{4}+5x^{2}-4$ Substitute: $x^{2}=u$ $u^{2}+5u-4=0$ $u=\displaystyle \frac{-5\pm\sqrt{25+16}}{2}=\frac{-5\pm\sqrt{41}}{2}$ $u_{1}=\displaystyle \frac{-5-\sqrt{41}}{2}, u_{2}$=$\displaystyle \frac{-5+\sqrt{41}}{2}$ $x^{2}=\displaystyle \frac{-5-\sqrt{41}}{2}$ has no solutions $x^{2}=\displaystyle \frac{-5+\sqrt{41}}{2}$ $x=\pm\sqrt{\dfrac{-5+\sqrt{41}}{2}}$
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