Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.2 - Exponents and Radicals - 1.2 Exercises - Page 22: 54

Answer

$a)$ $3\sqrt{3a^{2}+7a}$ $b)$ $5\sqrt{3t+4t^{2}}$

Work Step by Step

$a)$ $\sqrt{27a^{2}+63a}$ Take out common factor $9$ from the expression inside the square root: $\sqrt{27a^{2}+63a}=\sqrt{9(3a^{2}+7a)}=...$ Take the square root of each factor: $...=(\sqrt{9})(\sqrt{3a^{2}+7a})=3\sqrt{3a^{2}+7a}$ $b)$ $\sqrt{75t+100t^{2}}$ Take out common factor $25$ from the expression inside the square root: $\sqrt{75t+100t^{2}}=\sqrt{25(3t+4t^{2})}=...$ Take the square root of each factor: $...=(\sqrt{25})(\sqrt{3t+4t^{2}})=5\sqrt{3t+4t^{2}}$
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