Intermediate Algebra (6th Edition)

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

Chapter 7 - Section 7.5 - Rationalizing Denominators and Numerators of Radical Expressions - Exercise Set - Page 445: 48

Answer

$\dfrac{4a-2\sqrt{ab}-6\sqrt{a}+3\sqrt{b}}{4a-b}$

Work Step by Step

Rationalizing the denominator of $ \dfrac{2\sqrt{a}-3}{2\sqrt{a}+\sqrt{b}} $ results to \begin{array}{l} \dfrac{2\sqrt{a}-3}{2\sqrt{a}+\sqrt{b}} \cdot \dfrac{2\sqrt{a}-\sqrt{b}}{2\sqrt{a}-\sqrt{b}} \\\\= \dfrac{(2\sqrt{a})(2\sqrt{a})+(2\sqrt{a})(-\sqrt{b})+(-3)(2\sqrt{a})+(-3)(-\sqrt{b})}{(2\sqrt{a})^2-(\sqrt{b})^2} \\\\= \dfrac{4a-2\sqrt{ab}-6\sqrt{a}+3\sqrt{b}}{4a-b} .\end{array}
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