Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 8 - Review - Page 531: 33

Answer

$[-5, -2]$ U $[2, 5]$

Work Step by Step

$(x^2-4)(x^2-25) \le 0$ $(x-2)(x+2)(x-5)(x+5) \le 0$ $x-2=0$ $x-2+2=0+2$ $x=2$ $x-5=0$ $x-5+5=0+5$ $x=5$ $x+2=0$ $x+2-2=0-2$ $x=-2$ $x+5=0$ $x+5-5=0-5$ $x=-5$ Five regions to test: $(-∞, -5]$, $[-5, -2]$, $[-2, 2]$, $[2, 5]$, $[5, ∞)$ Let $x=-6$, $x=-3$, $x=0$, $x=3$, $x=6$ $x=-6$ $(x^2-4)(x^2-25) \le 0$ $((-6)^2-4)((-6)^2-25) \le 0$ $(36-4)(36-25) \le 0$ $32*11 \le 0$ $352 \le 0$ (false) $x=-3$ $(x^2-4)(x^2-25) \le 0$ $((-3)^2-4)((-3)^2-25) \le 0$ $(9-4)(9-25) \le 0$ $5*-16 \le 0$ $-80 \le 0$ (true) $x=0$ $(x^2-4)(x^2-25) \le 0$ $(0^2-4)(0^2-25) \le 0$ $(0-4)(0-25) \le 0$ $-4* -25 \le 0$ $100 \le 0$ $x=3$ $(x^2-4)(x^2-25) \le 0$ $((3)^2-4)((3)^2-25) \le 0$ $(9-4)(9-25) \le 0$ $5*-16 \le 0$ $-80 \le 0$ (true) $x=-6$ $(x^2-4)(x^2-25) \le 0$ $((6)^2-4)((6)^2-25) \le 0$ $(36-4)(36-25) \le 0$ $32*11 \le 0$ $352 \le 0$ (false)
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