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: 52

Answer

$a)$ $\sqrt[3]{x^{4}}+\sqrt[3]{8x}=(x+2)\sqrt[3]{x}$ $b)$ $4\sqrt{18rt^{3}}+5\sqrt{32r^{3}t^{5}}=4(5rt^{2}+3t)\sqrt{2rt}$

Work Step by Step

$a)$ $\sqrt[3]{x^{4}}+\sqrt[3]{8x}$ Simplify each individual term: $\sqrt[3]{x^{4}}+\sqrt[3]{8x}=x\sqrt[3]{x}+2\sqrt[3]{x}=...$ Evaluate the sum by adding only the coefficients: $...=(x+2)\sqrt[3]{x}$ $b)$ $4\sqrt{18rt^{3}}+5\sqrt{32r^{3}t^{5}}$ Simplify each individual term: $4\sqrt{18rt^{3}}+5\sqrt{32r^{3}t^{5}}=4(3)t\sqrt{2rt}+5(4)rt^{2}\sqrt{2rt}=...$ $...=12t\sqrt{2rt}+20rt^{2}\sqrt{2rt}$ Evaluate the sum by adding only the coefficients: $...=(20rt^{2}+12t)\sqrt{2rt}$ Take out common factor $4$ from the expression inside the parentheses: $...=4(5rt^{2}+3t)\sqrt{2rt}$
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