Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 9 - Roots and Radicals - 9.5 - Solving Radical Equations - Problem Set 9.5 - Page 424: 34



Work Step by Step

Using the properties of equality, the given equation, $ \sqrt{-x}-6=x ,$ is equivalent to \begin{array}{l}\require{cancel} \sqrt{-x}=x+6 .\end{array} Squaring both sides of the equation and then using the properties of equality, we obtain: \begin{array}{l}\require{cancel}\left( \sqrt{-x} \right)^2=\left( x+6 \right)^2 \\\\ -x=(x)^2+2(x)(6)+(6)^2 \\\\ -x=x^2+12x+36 \\\\ 0=x^2+(12x+x)+36 \\\\ x^2+13x+36=0 \\\\ (x+9)(x+4)=0 \\\\ x=\{-9,-4\} .\end{array} Upon checking, only $ x=-4 $ satisfies the original equation.
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