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: 12


$x=\pm 4$ or $x=\pm 1$

Work Step by Step

$ x^{4}-17x^{2}+16=0\qquad$...substitute $x^{2}$ for $u$ $ u^{2}-17u+16=0\qquad$... solve with the Quadractic formula. $a=1,\ b=-17,\ c=16$ $ u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $-17,\ a$ for $1$ and $c$ for $16$. $ u=\displaystyle \frac{-(-17)\pm\sqrt{(-17)^{2}-4\cdot(16)\cdot 1}}{2\cdot 1}\qquad$... simplify. $u=\displaystyle \frac{17\pm\sqrt{289-64}}{2}$ $u=\displaystyle \frac{17\pm\sqrt{225}}{2}$ $ u=\displaystyle \frac{17\pm 15}{2}\qquad$... the symbol $\pm$ indicates two solutions. $u=\displaystyle \frac{17+15}{2}=\frac{32}{2}=16$ or $u=\displaystyle \frac{17-15}{2}=\frac{2}{2}=1$ Bring back $x$. $x^{2}=16$ or $x^{2}=1$ $x=\pm 4$ or $x=\pm 1$
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