Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Review - Exercises: 49

Answer

$\dfrac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}=3\sqrt{2}-2\sqrt{3}$

Work Step by Step

$\dfrac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}$ (rationalize the denominator) Multiply the numerator and the denominator of this expression by the conjugate of the denominator and simplify if possible: $\dfrac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\dfrac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=...$ $...=\dfrac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}=\dfrac{\sqrt{18}-\sqrt{12}}{3-2}=\sqrt{18}-\sqrt{12}=...$ $...=3\sqrt{2}-2\sqrt{3}$
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