College Algebra (11th Edition)

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

Chapter R - Test: 28

Answer

$\dfrac{7\sqrt{11}+7\sqrt{7}}{2}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To rationalize the given expression, $ \dfrac{14}{\sqrt{11}-\sqrt{7}} ,$ multiply the numerator and the denominator by the conjugate of the denominator. $\bf{\text{Solution Details:}}$ Reversing the operator between the expression, $ \sqrt{11}-\sqrt{7} ,$ then the conjugate is $ \sqrt{11}+\sqrt{7} .$ Multiplying both the numerator and the denominator by the conjugate results to \begin{array}{l}\require{cancel} \dfrac{14}{\sqrt{11}-\sqrt{7}}\cdot\dfrac{\sqrt{11}+\sqrt{7}}{\sqrt{11}+\sqrt{7}} \\\\= \dfrac{14(\sqrt{11}+\sqrt{7})}{(\sqrt{11}-\sqrt{7})(\sqrt{11}+\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{14(\sqrt{11}+\sqrt{7})}{(\sqrt{11})^2-(\sqrt{7})^2} \\\\= \dfrac{14(\sqrt{11}+\sqrt{7})}{11-7} \\\\= \dfrac{14(\sqrt{11}+\sqrt{7})}{4} \\\\= \dfrac{\cancel2(7)(\sqrt{11}+\sqrt{7})}{\cancel2(2)} \\\\= \dfrac{7(\sqrt{11}+\sqrt{7})}{2} .\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{7\sqrt{11}+7\sqrt{7}}{2} .\end{array}
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