Answer
$-7, 1$
Work Step by Step
1. Based on the given conditions, we have $\vec {PQ}=(x+3)i+3j$,
2. Use its magnitude, we have $\sqrt {(x+3)^2+3^2}=5$, thus $x^2+6x+9+9=25$ or $x^2+6x-7=0$
3. Thus we have $(x+7)(x-1)=0$ or $x=-7, 1$
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
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: 56
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