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: 94

Answer

$\dfrac{-7-\sqrt{21}}{4}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To rationalize the denominator of the given expression, $ \dfrac{\sqrt{7}}{\sqrt{3}-\sqrt{7}} ,$ 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 $ \sqrt{3}+\sqrt{7} .$ Multiplying the conjugate to both the numerator and the denominator of the expression above results to \begin{array}{l}\require{cancel} \dfrac{\sqrt{7}}{\sqrt{3}-\sqrt{7}}\cdot\dfrac{\sqrt{3}+\sqrt{7}}{\sqrt{3}+\sqrt{7}} .\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{\sqrt{7(3)}+\sqrt{7(7)}}{(\sqrt{3}-\sqrt{7})(\sqrt{3}+\sqrt{7})} \\\\= \dfrac{\sqrt{21}+\sqrt{7^2}}{(\sqrt{3}-\sqrt{7})(\sqrt{3}+\sqrt{7})} \\\\= \dfrac{\sqrt{21}+7}{(\sqrt{3}-\sqrt{7})(\sqrt{3}+\sqrt{7})} .\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{\sqrt{21}+7}{(\sqrt{3})^2-(\sqrt{7})^2} \\\\= \dfrac{\sqrt{21}+7}{3-7} \\\\= \dfrac{\sqrt{21}+7}{-4} \\\\= -\dfrac{\sqrt{21}+7}{4} \\\\= \dfrac{-\sqrt{21}-7}{4} \\\\= \dfrac{-7-\sqrt{21}}{4} .\end{array}
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