AoPS Wiki talk:Problem of the Day/September 21, 2011
Revision as of 17:54, 21 September 2011 by Negativebplusorminus (talk | contribs) (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...")
We see, after substitution, that
![\[x=\sqrt{x}-2\]](http://latex.artofproblemsolving.com/e/e/4/ee42c41e09d0944aba0bf745e544f5d749972485.png)
and thus, isolating the square root and squaring,
![\[x=(x+2)^2=x^2+4x+4\]](http://latex.artofproblemsolving.com/9/f/e/9fef0ae53d1a6e05210bdf7203b3622434d4df27.png)
and therefore 
. The sum of the roots of this equation, by Vieta's formulas, are 
.
