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: 41

Answer

The solution is $s=-18$.

Work Step by Step

The given equation is $\Rightarrow \sqrt[3]{2s+9}=-3$ Cube each side of the equation. $\Rightarrow (\sqrt[3]{2s+9})^3=(-3)^3$ Simplify. $\Rightarrow 2s+9=-27$ Subtract $9$ from each side. $\Rightarrow 2s+9-9=-27-9$ Simplify. $\Rightarrow 2s=-36$ Divide each side by $2$. $\Rightarrow s=-18$ Check $s=-18$. $\Rightarrow \sqrt[3]{2s+9}=-3$ $\Rightarrow \sqrt[3]{2(-18)+9}=-3$ $\Rightarrow \sqrt[3]{-36+9}=-3$ $\Rightarrow \sqrt[3]{-27}=-3$ $\Rightarrow \sqrt[3]{(-3)^3}=-3$ $\Rightarrow -3=-3$ True. Hence, the solution is $s=-18$.
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.