Difference between revisions of "AoPS Wiki talk:Problem of the Day/September 18, 2011"

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=Solution=
 
=Solution=
 
We see that for positive <math>x</math>,
 
We see that for positive <math>x</math>,
<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>
+
<cmath>\begin{align*}f(x)&=\sqrt{(x^2+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>.

Latest revision as of 13:32, 18 September 2011

Solution

We see that for positive $x$, \begin{align*}f(x)&=\sqrt{(x^2+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*} and thus $a+b+c=\boxed{5}$.