Chapter 9 - Roots and Radicals - 9.5 - Solving Radical Equations - Problem Set 9.5 - Page 425: 48

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We square both sides of the equation and solve. Thus: $ \sqrt{2n-1} = 3 - \sqrt{ n -5} \\ 2n -1 = 9 +n -5 - 6\sqrt{n-5} \\ n -5 = -6 \sqrt{n-5}\\ n^2 -10n + 25 = 36(n-5) \\ n^2 - 46n +205 = 0 \\(n-41) (n-5) \\ n=5$ Note, when 41 is plugged back in, it does not get a true statement, so it is not a solution.