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: 57


Zeros: $2\sqrt{2}, -2 \sqrt{2}$ $x$-intercepts: $2\sqrt{2}, -2 \sqrt{2}$

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 $P(x)=0$: $$x^4-6x^2-16=0$$ Let $u=x^2$, the original equation becomes $$u^2-6u-16=0$$ By factoring $$(u+2)(u-8) = 0$$ Use the Zero-Product Property by equating each factor to zero, then solve each equation to obtain: \begin{align*} u+2 &=0 &\text{ or }& &u-8=0\\ u &= -2 &\text{ or }& &u=8\\ \end{align*} To solve for $x$, we use $u=x^2$ $$\because u = x^2$$ $$\therefore x = \pm \sqrt{u}$$ For $u=-2$ $$x=\pm \sqrt{-2} \hspace{20pt} \text{ No real solution}$$ For $u=8$ $$x= \pm \sqrt{8}$$ $$x= \pm 2\sqrt{2}$$ $\therefore x = 2\sqrt{2}, -2 \sqrt{2}$
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