Difference between revisions of "AoPS Wiki talk:Problem of the Day/September 21, 2011"

(Created page with "We see, after substitution, that <cmath>x=\sqrt{x}-2</cmath> and thus, isolating the square root and squaring, <cmath>x=(x+2)^2=x^2+4x+4</cmath> and therefore <math>x^2+3x+4=0</m...")
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Revision as of 17:54, 21 September 2011

We see, after substitution, that \[x=\sqrt{x}-2\] and thus, isolating the square root and squaring, \[x=(x+2)^2=x^2+4x+4\] and therefore $x^2+3x+4=0$. The sum of the roots of this equation, by Vieta's formulas, are $\boxed{-3}$.