## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter R - Algebra Reference - R.5 Inequalities - R.5 Exercises - Page R-21: 52

#### Answer

$\frac{8}{p^2+2p}\gt1$ $8\gt1(p^2+2p)$ $8\gt p^2+2p$ $0\gt p^2+2p-8$ $0\gt(p+4)(p-2)$ $-4\lt p\lt 2$ But $p$ can not be$-2\lt p\lt0$ Thus: $-4\lt p\lt -2$ $0\lt p\lt2$

#### Work Step by Step

Step 1: Multiply with $(p^2+2p)$ on both sides of the inequality Step 2: Distribute the $1$ into the brackets Step 3: Subtract $8$ from both sides of the inequality Step 4: Factorize inequality (quadratic trinomial) Step 5: substitute $p$ values in original inequality to test. Step 6: notice the restriction, where inequality will not be true is $-2\lt p\lt0$

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