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AoPS Wiki talk:Problem of the Day/September 21, 2011

Revision as of 17:28, 21 September 2011 by Djmathman (talk | contribs)

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 $x^2-5x+4=0$. The sum of the roots of this equation, by Vieta's formulas, are $\boxed{5}$.

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