Big Ideas Math - Algebra 1, A Common Core Curriculum

Published by Big Ideas Learning LLC
ISBN 10: 978-1-60840-838-2
ISBN 13: 978-1-60840-838-2

Chapter 10 - Radical Functions and Equations - 10.3 - Solving Radical Equations - Exercises - Page 565: 39

Answer

The solution is $g=27$.

Work Step by Step

The given equation is $\Rightarrow 6=\sqrt[3]{8g}$ Cube each side of the equation. $\Rightarrow 6^3=(\sqrt[3]{8g})^3$ Simplify. $\Rightarrow 216=8g$ Divide each side by $8$. $\Rightarrow 27=g$ Check $g=27$. $\Rightarrow 6=\sqrt[3]{8(27)}$ $\Rightarrow 6=\sqrt[3]{216}$ $\Rightarrow 6=\sqrt[3]{6^3}$ $\Rightarrow 6=6$ True. Hence, the solution is $g=27$.
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