Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.6 - Rational Expressions - Concept and Vocabulary Check - Page 84: 12

Answer

Fill the blank with $(\sqrt{x+7}+\sqrt{x})$

Work Step by Step

The numerator is rationalized by applying the difference of squares formula, $(a-b)(a+b)=a^{2}-b^{2}$ Multiplying with the conjugate of the numerator, $\sqrt{x+7}+\sqrt{x}$, the numerator becomes $(x+7)-x=7\qquad $... no radicals. and the denominator becomes $7(\sqrt{x+7}+\sqrt{x})$. Fill the blank with $(\sqrt{x+7}+\sqrt{x})$
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