College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.4 - Radical Equations; Equations Quadratic in Form; Factorable Equations - 1.4 Assess Your Understanding - Page 118: 35



Work Step by Step

We solve: $(3x+1)^{1/2}=4$ $[(3x+1)^{1/2}]^{2}=(4)^{2}$ $3x+1=16$ $3x=15$ $x=15/3$ $x=5$
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