Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 9 - Roots and Radicals - 9.5 - Solving Radical Equations - Problem Set 9.5: 49

Answer

4, 12

Work Step by Step

We square both sides of the equation and solve. Thus: $ \sqrt{2n+1} = 2 + \sqrt{n-3} \\ 2n+1 = 4 + n -3 + 4\sqrt{n-3} \\ n = 4 \sqrt{n-3} \\ n^2 = 16(n-3) \\ n^2 - 16n + 48 = 0 \\ (n-4)(n-12) \\ n = 4, 12$
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