College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.6 - Equations and Inequalities Involving Absolute Value - 1.6 Assess Your Understanding - Page 134: 79

Answer

See below.

Work Step by Step

I use the hint in my solution: $b-a=(\sqrt b+\sqrt a)(\sqrt b-\sqrt a)$. $(\sqrt b-\sqrt a)\gt0$ and $\sqrt a\gt0,\sqrt b\gt0$, thus $(\sqrt b+\sqrt a)\gt0$, thus their product is also positive. Thus, the result has been proved.
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