## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 11 - Quadratic Functions and Equations - 11.5 Equations Reducible to Quadratic - 11.5 Exercise Set - Page 731: 13

#### Answer

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

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