Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 10 - Section 10.5 - Rationalizing Numerators and Denominators of Radical Expressions - Exercise Set: 38

Answer

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

Work Step by Step

$\dfrac{2\sqrt{a}-3}{2\sqrt{a}-\sqrt{b}}$ Multiply the numerator and the denominator of this expression by the conjugate of the denominator and simplify if possible: $\dfrac{2\sqrt{a}-3}{2\sqrt{a}-\sqrt{b}}=\dfrac{2\sqrt{a}-3}{2\sqrt{a}-\sqrt{b}}\cdot\dfrac{2\sqrt{a}+\sqrt{b}}{2\sqrt{a}+\sqrt{b}}=...$ $...=\dfrac{(2\sqrt{a}-3)(2\sqrt{a}+\sqrt{b})}{(2\sqrt{a})^{2}-(\sqrt{b})^{2}}=...$ $...=\dfrac{(2\sqrt{a})^{2}+2\sqrt{ab}-6\sqrt{a}-3\sqrt{b}}{4a-b}=\dfrac{4a+2\sqrt{ab}-6\sqrt{a}-3\sqrt{b}}{4a-b}$
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