College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter P, Prerequisites - Chapter P Review - Exercises: 90

Answer

$\displaystyle6+2\sqrt{2}$

Work Step by Step

We multiply the numerator and denominator by $3+\sqrt{2}$ and use the fact that $(a-b)(a+b)=a^2-b^2$ to simplify: $\displaystyle \frac{14}{3-\sqrt{2}}=\frac{14}{3-\sqrt{2}}\cdot\frac{3+\sqrt{2}}{3+\sqrt{2}}=\frac{42+14\sqrt{2}}{3^{2}-(\sqrt{2})^{2}}=\frac{42+14\sqrt{2}}{9-2}=\frac{42+14\sqrt{2}}{7}=6+2\sqrt{2}$
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