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: 13


$x=\pm 2$ or $x=\pm\sqrt{3}$

Work Step by Step

$ t^{4}-7t^{2}+12=0\qquad$...substitute $t^{2}$ for $u$ $ u^{2}-7u+12=0\qquad$... solve with the Quadractic formula. $a=1,\ b=-7,\ c=12$ $ u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $-7,\ a$ for $1$ and $c$ for $12$. $ u=\displaystyle \frac{-(-7)\pm\sqrt{(-7)^{2}-4\cdot(12)\cdot 1}}{2\cdot 1}\qquad$... simplify. $u=\displaystyle \frac{7\pm\sqrt{49-48}}{2}$ $u=\displaystyle \frac{7\pm\sqrt{1}}{2}$ $ u=\displaystyle \frac{7\pm 1}{2}\qquad$... the symbol $\pm$ indicates two solutions. $u=\displaystyle \frac{7+1}{2}=\frac{8}{2}=4$ or $u=\displaystyle \frac{7-1}{2}=\frac{6}{2}=3$ Bring back $x$. $x^{2}=4$ or $x^{2}=3$ $x=\pm 2$ or $x=\pm\sqrt{3}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.