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: 35

Answer

$x=3$

Work Step by Step

Squaring both sides of the equation and then using the properties of equality, we obtain: \begin{array}{l}\require{cancel}\left( \sqrt{x^2+27} \right)^2=\left( x+3 \right)^2 \\\\ x^2+27=(x)^2+2(x)(3)+(3)^2 \\\\ x^2+27=x^2+6x+9 \\\\ (x^2-x^2)-6x+(27-9)=0 \\\\ -6x+18=0 \\\\ -6x=-18 \\\\ x=\dfrac{-18}{-6} \\\\ x=3 .\end{array} Upon checking, $ x=3 $ satisfies the original equation.
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