## 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: 3

True statement

#### Work Step by Step

$a{{x}^{2}}+bx+c=0$ An example of equation reducible to a quadratic is ${{x}^{4}}-8{{x}^{2}}-9=0$ Let ${{x}^{2}}=u$, Then ${{x}^{4}}={{u}^{2}}$. Substitute the values of ${{x}^{2\text{ }}}\text{and }{{x}^{4}}$ into the equation ${{x}^{4}}-8{{x}^{2}}-9=0$ as follows: \begin{align} & {{x}^{4}}-8{{x}^{2}}-9=0 \\ & {{u}^{2}}-8u-9=0 \end{align} Factor ${{u}^{2}}-8u-9=0$, \begin{align} & {{u}^{2}}-8u-9=0 \\ & \left( u-9 \right)\left( u+1 \right)=0 \end{align} Thus, $\text{ }u=9\text{ or }u=-1$ Now, this is incomplete because the solution is being carried out for x and not u. Now replace $u$ with ${{x}^{2}}$. \begin{align} & \text{ }u=9\text{ or }u=-1 \\ & {{x}^{2}}\text{= 9 or }{{x}^{2}}=-1 \\ & x=\pm 3\text{ square can }\!\!'\!\!\text{ t be negative,no solution}\text{. } \\ \end{align} Thus, the final solution is $x=\pm \text{ }3$ and not $\text{ }u=9\text{ or }u=-1$.

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