Talk:1983 AIME Problems/Problem 3

Revision as of 20:49, 21 May 2009 by Xantos C. Guin (talk | contribs) (New page: I think what the first solution meant to say is that <math>y=-6</math> is an extraneous solution since <math>-6 \neq 2\sqrt{-6+15}</math> (the <math>\sqrt{}</math> symbol is defined as the...)
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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.