Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 2 - Linear and Quadratic Functions - Section 2.3 Quadratic Functions and Their Zeros - 2.3 Assess Your Understanding - Page 146: 61


Zeros: $1,-1$ $x$-intercepts: $1,-1$

Work Step by Step

To find the zeros of a function $f$, solve the equation $f(x)=0$ The zeros of the function are also the $x-$intercepts. Let $G(x)=0$: $$3x^4-2x^2-1=0$$ Let $u=x^2$, the original equation becomes $$3u^2-2u-1=0$$ By factoring $$(3u+1)(u-1) = 0$$ Use the Zero-Product Property by equating each factor to zero, then solve each equation ton obtain: \begin{align*} 3u+1 &=0 &\text{ or }& &u-1=0\\ 3u &= -1 &\text{ or }& &u=1\\ u &= -\dfrac{1}{3} &\text{ or }& &u=1\\ \end{align*} To solve for $x$, we use $u=x^2$ $$\because u = x^2$$ $$\therefore x = \pm \sqrt{u}$$ For $u=-\dfrac{1}{3}$ $$x=\pm \sqrt{\left(-\dfrac{1}{3}\right)} \hspace{20pt} \text{ No real solution} $$ For $u=1$ $$x= \pm \sqrt{1}$$ $\therefore x = 1,-1$
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