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: 20



Work Step by Step

We solve: $$\sqrt{x+3}=(3- \sqrt{x-5})$$ We now square both sides: $$(\sqrt{x+3})^2=(3- \sqrt{x-5})^2$$ $${x+3}=(9- 6\sqrt{x-5})+x-5$$ $${-1}=(6\sqrt{x-5})$$ $${-1/6}=(\sqrt{x-5})$$ $${1/36}=({x-5})$$ $$x=181/36$$
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