Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.5 - Rationalizing Denominators and Numerators of Radical Expressions - Exercise Set: 75

Answer

$\dfrac{3}{10+5\sqrt{7}}$

Work Step by Step

Multiplying by the conjugate of the numerator, then the rationalized-numerator form of the given expression, $ \dfrac{2-\sqrt{7}}{-5} ,$ is \begin{array}{l}\require{cancel} \dfrac{2-\sqrt{7}}{-5}\cdot\dfrac{2+\sqrt{7}}{2+\sqrt{7}} \\\\= \dfrac{(2)^2-(\sqrt{7})^2}{-5(2)-5(\sqrt{7})} \\\\= \dfrac{4-7}{-10-5\sqrt{7}} \\\\= \dfrac{-3}{-10-5\sqrt{7}} \\\\= \dfrac{-(3)}{-(10+5\sqrt{7})} \\\\= \dfrac{3}{10+5\sqrt{7}} .\end{array}
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