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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We see, after substitution, that | We see, after substitution, that | ||
| − | <cmath>x=\sqrt{x} | + | <cmath>x=\sqrt{x}+2</cmath> |
and thus, isolating the square root and squaring, | and thus, isolating the square root and squaring, | ||
| − | <cmath>x=(x | + | <cmath>x=(x-2)^2=x^2-4x+4</cmath> |
| − | and therefore <math>x^2 | + | and therefore <math>x^2-5x+4=0</math>. The sum of the roots of this equation, by [[Vieta's formulas]], are <math>\boxed{5}</math>. |
Revision as of 18:28, 21 September 2011
We see, after substitution, that
![\[x=\sqrt{x}+2\]](http://latex.artofproblemsolving.com/3/6/7/367fe5cfd41e070077cabdcebc29e4c3d3181c81.png)
and thus, isolating the square root and squaring,
![\[x=(x-2)^2=x^2-4x+4\]](http://latex.artofproblemsolving.com/6/0/c/60cdcd41f3165ecf865f65a5e9b3a9b3b642bdc3.png)
and therefore 
. The sum of the roots of this equation, by Vieta's formulas, are 
.
