Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 8 - Polar Coordinates; Vectors - Section 8.4 Vectors - 8.4 Assess Your Understanding - Page 627: 55


$-2\pm\sqrt {21}$

Work Step by Step

1. Based on the given conditions, we have $\vec v+\vec w=(x+2)i+2j$, 2. Use its magnitude, we have $\sqrt {(x+2)^2+2^2}=5$, thus $x^2+4x+8=25$ or $x^2+4x-17=0$ 3. Thus we have $x=\frac{-4\pm\sqrt {16+4(17)}}{2}=-2\pm\sqrt {21}$
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.