Intermediate Algebra (6th Edition)

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

Chapter 5 - Section 5.8 - Solving Equations by Factoring and Problem Solving - Exercise Set - Page 323: 50


$x=\left\{ -6,-3 \right\}$

Work Step by Step

The factored form of the given expression, $ \dfrac{x^2}{18}+\dfrac{x}{2}+1=0 $, is \begin{array}{} 18\left( \dfrac{x^2}{18}+\dfrac{x}{2}+1 \right)=(0)18 \\\\ 1(x^2)+9(x)+18(1)=0 \\\\ x^2+9x+18=0 \\\\ (x+6)(x+3)=0 .\end{array} Equating each factor to zero, then \begin{array}{l} x+6=0 \\ x=0-6 \\ x=-6 ,\text{ OR}\\\\ x+3=0 \\ x=0-3 \\ x=-3 \end{array} Hence, $ x=\left\{ -6,-3 \right\} $.
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