Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.7 Radicals - R.7 Exercises: 42

Answer

$\dfrac{3-\sqrt{3}}{6}=\dfrac{1}{3+\sqrt{3}}$

Work Step by Step

$\dfrac{3-\sqrt{3}}{6}$ Multiply the numerator and the denominator of the given expression by $3+\sqrt{3}$, which is the conjugate of the numerator: $\dfrac{3-\sqrt{3}}{6}\cdot\dfrac{3+\sqrt{3}}{3+\sqrt{3}}=...$ Evaluate the products: $...=\dfrac{3^{2}-(\sqrt{3})^{2}}{6(3+\sqrt{3})}=\dfrac{9-3}{6(3+\sqrt{3})}=\dfrac{6}{6(3+\sqrt{3})}=\dfrac{1}{3+\sqrt{3}}$
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