Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 8 - Radical Functions - 8.2 Simplifying, Adding, and Subtracting Radicals - 8.2 Exercises - Page 633: 9

Answer

$\sqrt{-9}$ is not a real number because there is no real number that when squared gives $-9$.

Work Step by Step

Note that: $\sqrt{9} = 3$ because $3^2=9$ $\sqrt{16}=4$ because $4^2=16$ The given radical is NOT A REAL NUMBER because there is no real number that when squared gives $-9$.
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