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: 62


Zeros:$2,-2$ $x$-intercepts: $2,-2$

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 $F(x)=0$: $$2x^4-5x^2-12=0$$ Let $u=x^2$, the original equation becomes $$2u^2-5u-12=0$$ By factoring $$(2u+3)(u-4) = 0$$ Use the Zero-Product Property by equating each factor to zero, then solve each equation to obtain:: \begin{align*} 2u+3 &=0 &\text{ or }& &u-4=0\\ 2u &= -3 &\text{ or }& &u=4\\ u &= -\dfrac{3}{2} &\text{ or }& &u=4\\ \end{align*} To solve for $x$, we use $u=x^2$ $$\because u = x^2$$ $$\therefore x = \pm \sqrt{u}$$ For $u=-\dfrac{3}{2}$ $$x=\pm \sqrt{\left(-\dfrac{3}{2}\right)} \hspace{20pt} \text{ No real solution} $$ For $u=4$ $$x= \pm \sqrt{4}$$ $$x= \pm 2$$ $\therefore x = 2,-2$
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