Calculus with Applications (10th Edition)

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

Chapter 2 - Nonlinear Functions - Chapter Review - Review Exercises - Page 114: 78


$p\approx 1.830\qquad $or $p\approx-6.830$

Work Step by Step

$[1+\displaystyle \frac{2p}{5}]^{2}=3\qquad \color{blue}{\text{ ...take the square root from both sides } }$ $1+\displaystyle \frac{2p}{5}=\pm\sqrt{3} $ $\displaystyle \frac{5+2p}{5}=\pm\sqrt{3}\qquad \color{blue}{\text{ ... }/\times 5 }$ $5+2p=\pm 5\sqrt{3}\qquad \color{blue}{\text{ ... }/-5 }$ $2p=-5\pm\sqrt{3}\qquad \color{blue}{\text{ ... }/\div 2 }$ $p=\displaystyle \frac{-5\pm 5\sqrt{3}}{2}$ $p=\displaystyle \frac{-5+5\sqrt{3}}{2}\approx 1.830\qquad $or $p=\displaystyle \frac{-5-5\sqrt{3}}{2}\approx-6.830$
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