Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 8 - Review Exercises - Page 574: 3

Answer

$\left\{-\dfrac{15}{2},\dfrac{5}{2}\right\}$

Work Step by Step

Taking the square root of both sides (Square Root Property), the given equation, $ (2x+5)^2=100 ,$ is equivalent to \begin{align*} 2x+5&=\pm\sqrt{100} .\end{align*} Using concepts of simplifying radicals, the equation above is equivalent to \begin{align*} 2x+5&=\pm\sqrt{(10)^2} \\ 2x+5&=\pm10 .\end{align*} Using the properties of equality, the equation above is equivalent to \begin{align*} 2x&=-5\pm10 \\\\ x&=\dfrac{-5\pm10}{2} \end{align*}\begin{array}{c|c} x=\dfrac{-5-10}{2} & x=\dfrac{-5+10}{2} \\\\ x=-\dfrac{15}{2} & x=\dfrac{5}{2} .\end{array} Hence, the solution set of $ (2x+5)^2=100 $ is $ \left\{-\dfrac{15}{2},\dfrac{5}{2}\right\} $.
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