College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.3 - Page 51: 140


(a) Answer is $>$ (b) Answer is $>$

Work Step by Step

(a) The answer can be deduced logically, since $3^{\frac{1}{2}}$ is $\sqrt{3}$ and $3^{\frac{1}{3}}$ is $\sqrt[3] 3$. A square root breaks the number into 2 factors, while the cubic root breaks the number into 3 factors; therefore, the square root must be larger. (b) The answer can be estimated by the using numbers with perfect squares as a guide: $$4 < 7 < 9$$ Therefore, $$\sqrt{4} < \sqrt{7} < \sqrt{9}$$$$= 2 \sqrt{7 + 18}$$
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.