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: 60


Zeros: $2,-2,\sqrt{6},-\sqrt{6} $ $x$-intercepts: $2,-2,\sqrt{6},-\sqrt{6} $

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 $f(x)=0$: $$x^4-10x^2+24=0$$ Let $u=x^2$, the original equation becomes $$u^2-10u+24=0$$ By factoring $$(u-4)(u-6) = 0$$ Use the Zero-Product Property by equating each factor to zero, then solve each equation to obtain: \begin{align*} u-4 &= 0 &\text{ or }& &u-6=0\\ u &= 4 &\text{ or }& &u=6\\ \end{align*} To solve for $x$, we use $u=x^2$ $$\because u = x^2$$ $$\therefore x = \pm \sqrt{u}$$ For $u=4$ $$x=\pm \sqrt{4} $$ $$x= \pm 2$$ For $u=6$ $$x= \pm \sqrt{6}$$ $\therefore x = 2,-2,\sqrt{6},-\sqrt{6} $
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