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: 11

Answer

$x=\pm 3$ or $x=\pm 2$

Work Step by Step

$ x^{4}-13x^{2}+36=0\qquad$...substitute $x^{2}$ for $u$ $ u^{2}-13u+36=0\qquad$... solve with the Quadractic formula. $a=1,\ b=-13,\ c=36$ $ u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $-13,\ a$ for $1$ and $c$ for $36$. $ u=\displaystyle \frac{-(-13)\pm\sqrt{(-13)^{2}-4\cdot(36)\cdot 1}}{2\cdot 1}\qquad$... simplify. $u=\displaystyle \frac{13\pm\sqrt{169-144}}{2}$ $ u=\displaystyle \frac{13\pm\sqrt{25}}{2}\qquad$... the symbol $\pm$ indicates two solutions. $u=\displaystyle \frac{13+5}{2}=\frac{18}{2}=9$ or $u=\displaystyle \frac{13-5}{2}=\frac{8}{2}=4$ Bring back $x$. $x^{2}=9$ or $x^{2}=4$ $x=\pm 3$ or $x=\pm 2$
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