Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.5 Inequalities - R.5 Exercises - Page R-21: 54

Answer

There is no solution for $\frac{a^2+2a}{a^2-4}\leq2$

Work Step by Step

$\frac{a^2+2a}{a^2-4}\leq2$ Distribute the 2 into the brackets $2(a^2-4)\geq a^2+2a$ $a^2-2a-8\geq0$ $(z+2)(z-4)\geq0$ $p\leq -2$or $p \geq 4$ but z can not be in $(-\infty,-2) \cup (2,\infty)$ Thus, there is no solution for $\frac{a^2+2a}{a^2-4}\leq2$
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