College Algebra (11th Edition)

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

Chapter R - Section R.7 - Radical Expressions - R.7 Exercises - Page 68: 95

Answer

$\dfrac{-7+2\sqrt{14}+\sqrt{7}-2\sqrt{2}}{2}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To rationalize the denominator of the given expression, $ \dfrac{\sqrt{7}-1}{2\sqrt{7}+4\sqrt{2}} ,$ multiply the numerator and the denominator by the conjugate of the denominator. $\bf{\text{Solution Details:}}$ Reversing the operator between the terms of the denominator in the expression above, the conjugate of the denominator is $ 2\sqrt{7}-4\sqrt{2} .$ Multiplying the conjugate to both the numerator and the denominator of the expression above results to \begin{array}{l}\require{cancel} \dfrac{\sqrt{7}-1}{2\sqrt{7}+4\sqrt{2}}\cdot\dfrac{2\sqrt{7}-4\sqrt{2}}{2\sqrt{7}-4\sqrt{2}} .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel} \dfrac{\sqrt{7}(2\sqrt{7})+\sqrt{7}(-4\sqrt{2})-1(2\sqrt{7})-1(-4\sqrt{2})}{(2\sqrt{7}+4\sqrt{2})(2\sqrt{7}-4\sqrt{2})} .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{2\sqrt{7(7)}-4\sqrt{7(2)}-2\sqrt{7}+4\sqrt{2}}{(2\sqrt{7}+4\sqrt{2})(2\sqrt{7}-4\sqrt{2})} \\\\= \dfrac{2\sqrt{7^2}-4\sqrt{14}-2\sqrt{7}+4\sqrt{2}}{(2\sqrt{7}+4\sqrt{2})(2\sqrt{7}-4\sqrt{2})} \\\\= \dfrac{2(7)-4\sqrt{14}-2\sqrt{7}+4\sqrt{2}}{(2\sqrt{7}+4\sqrt{2})(2\sqrt{7}-4\sqrt{2})} \\\\= \dfrac{14-4\sqrt{14}-2\sqrt{7}+4\sqrt{2}}{(2\sqrt{7}+4\sqrt{2})(2\sqrt{7}-4\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{14-4\sqrt{14}-2\sqrt{7}+4\sqrt{2}}{(2\sqrt{7})^2-(4\sqrt{2})^2} \\\\= \dfrac{14-4\sqrt{14}-2\sqrt{7}+4\sqrt{2}}{4(7)-16(2)} \\\\= \dfrac{14-4\sqrt{14}-2\sqrt{7}+4\sqrt{2}}{28-32} \\\\= \dfrac{14-4\sqrt{14}-2\sqrt{7}+4\sqrt{2}}{-4} \\\\= \dfrac{-2(-7+2\sqrt{14}+\sqrt{7}-2\sqrt{2})}{-4} \\\\= \dfrac{\cancel{-2}(-7+2\sqrt{14}+\sqrt{7}-2\sqrt{2})}{\cancel{-4}^2} \\\\= \dfrac{-7+2\sqrt{14}+\sqrt{7}-2\sqrt{2}}{2} .\end{array}
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