College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Review Exercises: 115

Answer

$\dfrac{18+6\sqrt{2}}{7}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Rationalize the given expression, $ \dfrac{6}{3-\sqrt{2}} ,$ by multiplying the numerator and the denominator by the conjugate of the denominator. $\bf{\text{Solution Details:}}$ By reversing the operator between the terms of the denominator, $ 3-\sqrt{2} ,$ the conjugate is $ 3+\sqrt{2} .$ Multiplying both the numerator and the denominator by the conjugate results to \begin{array}{l}\require{cancel} \dfrac{6}{3-\sqrt{2}}\cdot\dfrac{3+\sqrt{2}}{3+\sqrt{2}} \\\\= \dfrac{6(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})} .\end{array} Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent\begin{array}{l}\require{cancel} \dfrac{6(3+\sqrt{2})}{(3)^2-(\sqrt{2})^2} \\\\= \dfrac{6(3+\sqrt{2})}{9-2} \\\\= \dfrac{6(3+\sqrt{2})}{7} .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{18+6\sqrt{2}}{7} .\end{array}
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