Difference between revisions of "Talk:1983 AIME Problems/Problem 3"

(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 <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 positive square root.) Thus, solution one has the correct answer.
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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 positive square root.) Thus, solution one has the correct answer.  
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--[[User:Xantos C. Guin|Xantos C. Guin]] 01:51, 22 May 2009 (UTC)

Latest revision as of 21:51, 21 May 2009

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)