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Problem of the week #51 - March 18th, 2013

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Jameson

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Jan 26, 2012
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Jameson

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Jan 26, 2012
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Congratulations to the following members for their correct solutions:

1) MarkFL
2) Bacterius
3) caffeinemachine
4) Siron

Solution (from Siron):
We have
$$f(x)=ax^2+bx+c = a(x+3)^2+b(x+3)+c = a(x^2+6x+9)+bx+3b + c = ax^2+6ax+9a + bx+3b + c = ax^2+(6a+b)x+(9a+3b+c)$$

Since $f(x+3)=3x^2+7x+4 = ax^2+(6a+b)+(9a+3b+c)$ we obtain the following system of equations
$$\left \{ \begin{array}{lll} a = 3 \\ 6a+b = 7 \\ 9a+3b+c = 4 \end{array} \right. \Rightarrow \left \{ \begin{array}{lll} a = 3 \\b =-11 \\ c c = 10 \end{array} \right.$$

Hence $a+b+c = 3-11+10 = 2$
 
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