Talk:1983 AIME Problems/Problem 3

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I think what the first solution meant to say is that $y=-6$ is an extraneous solution since $-6 \neq 2\sqrt{-6+15}$ (the $\sqrt{}$ symbol is defined as the positive square root.) Thus, solution one has the correct answer. --Xantos C. Guin 01:51, 22 May 2009 (UTC)

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Talk:1983 AIME Problems/Problem 3

From AoPSWiki