Difference between revisions of "AoPS Wiki talk:Problem of the Day/September 18, 2011"
(Created page with "=Solution= We see that for positive <math>x</math>, <cmath>f(x)=\sqrt{(x^3+3x+0)(x^2+3x+2)+1}=\sqrt{((x^2+3x+1)-1)((x^2+3x+1)-1)+1}=\sqrt{(x^2+3x+1)^2}=\boxed{x^2+3x+1}</cmath> a...") |
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We see that for positive <math>x</math>, | We see that for positive <math>x</math>, | ||
| − | <cmath>f(x)=\sqrt{(x^3+3x+0)(x^2+3x+2)+1}=\sqrt{((x^2+3x+1)-1)((x^2+3x+1)-1)+1}=\sqrt{(x^2+3x+1)^2}=\boxed{x^2+3x+1}</cmath> | + | <cmath>\begin{align*}f(x)&=\sqrt{(x^3+3x+0)(x^2+3x+2)+1}\\&=\sqrt{((x^2+3x+1)-1)((x^2+3x+1)-1)+1}\\&=\sqrt{(x^2+3x+1)^2}=\boxed{x^2+3x+1}\end{align*}</cmath> |
and thus <math>a+b+c=\boxed{5}</math>. | and thus <math>a+b+c=\boxed{5}</math>. | ||
Revision as of 08:01, 18 September 2011
Solution
We see that for positive 
,
and thus 
.
