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 731: 15


$x=\displaystyle \pm\frac{\sqrt{5}}{2}$ or $x=\pm 1$

Work Step by Step

$ 4x^{4}-9x^{2}+5=0\qquad$...substitute $x^{2}$ for $u$ $ 4u^{2}-9u+5=0\qquad$... solve with the Quadractic formula. $a=4,\ b=-9,\ c=5$ $ u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $-9,\ a$ for $4$ and $c$ for $5$. $ u=\displaystyle \frac{-(-9)\pm\sqrt{(-9)^{2}-4\cdot(5)\cdot 4}}{2\cdot 4}\qquad$... simplify. $u=\displaystyle \frac{9\pm\sqrt{81-80}}{8}$ $u=\displaystyle \frac{9\pm\sqrt{1}}{8}$ $ u=\displaystyle \frac{9\pm 1}{8}\qquad$... the symbol $\pm$ indicates two solutions. $u=\displaystyle \frac{9+1}{8}=\frac{10}{8}=\frac{5}{4}$ or $u=\displaystyle \frac{9-1}{8}=\frac{8}{8}=1$ Bring back $x$. $x^{2}=\displaystyle \frac{5}{4}$ or $x^{2}=1$ $x=\displaystyle \pm\frac{\sqrt{5}}{2}$ or $x=\pm 1$
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