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 - Page R-29: 20

Answer

$\sqrt{p^{7}q^{3}}-\sqrt{p^{5}q^{9}}+\sqrt{p^{9}q}=p^{2}(pq-q^{4}+p^{2})\sqrt{pq}$

Work Step by Step

$\sqrt{p^{7}q^{3}}-\sqrt{p^{5}q^{9}}+\sqrt{p^{9}q}$ Evaluate each square root: $\sqrt{p^{7}q^{3}}-\sqrt{p^{5}q^{9}}+\sqrt{p^{9}q}=...$ $...=p^{3}q\sqrt{pq}-p^{2}q^{4}\sqrt{pq}+p^{4}\sqrt{pq}=...$ The expressions inside square roots are all the same. Simplify by taking out common factor $\sqrt{pq}$: $...=(p^{3}q-p^{2}q^{4}+p^{4})\sqrt{pq}$ Take out common factor $p^{2}$ from the expression inside parentheses: $...=p^{2}(pq-q^{4}+p^{2})\sqrt{pq}$
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