Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 10 - Section 10.5 - Rationalizing Numerators and Denominators of Radical Expressions - Exercise Set: 48

Answer

$\sqrt{\dfrac{12}{7}}=\dfrac{6}{\sqrt{21}}$

Work Step by Step

$\sqrt{\dfrac{12}{7}}$ Rewrite this expression as $\dfrac{\sqrt{4\cdot3}}{\sqrt{7}}$ and simplify it: $\sqrt{\dfrac{12}{7}}=\dfrac{\sqrt{4\cdot3}}{\sqrt{7}}=\dfrac{2\sqrt{3}}{\sqrt{7}}=...$ Multiply this expression by $\dfrac{\sqrt{3}}{\sqrt{3}}$ and simplify again if possible: $...=\dfrac{2\sqrt{3}}{\sqrt{7}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{2\sqrt{3^{2}}}{\sqrt{21}}=\dfrac{2(3)}{\sqrt{21}}=\dfrac{6}{\sqrt{21}}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.