Intermediate Algebra (6th Edition)

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

Chapter 7 - Section 7.6 - Radical Equations and Problem Solving - Exercise Set: 29

Answer

$x=\dfrac{15}{4}$

Work Step by Step

Using the properties of equality, the given equation, $ \sqrt[4]{4x+1}-2=0 ,$ is equivalent to \begin{array}{l}\require{cancel} \sqrt[4]{4x+1}=2 .\end{array} Raising both sides of the equation above to the fourth power, then the solution/s is/are \begin{array}{l}\require{cancel} 4x+1=16 \\\\ 4x=16-1 \\\\ 4x=15 \\\\ x=\dfrac{15}{4} .\end{array} Upon checking, $ x=\dfrac{15}{4} $ satisfies the original equation.
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