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 - Page 453: 7

Answer

$x=7$

Work Step by Step

$\sqrt (4x-3)-5=0$ Add 5 to both sides of the equation. $\sqrt (4x-3)=5$ Square both sides. $(\sqrt (4x-3))^{2}=5^{2}$ $4x-3=25$ Add 3 to both sides. $4x=28$. Divide both sides by 4. $x=7$
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