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- Jan 26, 2012

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Remember to read the POTW submission guidelines to find out how to submit your answers!

- Thread starter Jameson
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- Thread starter
- Admin
- #1

- Jan 26, 2012

- 4,041

--------------------

Remember to read the POTW submission guidelines to find out how to submit your answers!

- Thread starter
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- #2

- Jan 26, 2012

- 4,041

1) MarkFL

2) Bacterius

3) caffeinemachine

4) Siron

Solution (from Siron):

$$f(x)=ax^2+bx+c = a(x+3)^2+b(x+3)+c = a(x^2+6x+9)+bx+3b + c = ax^2+6ax+9a + bx+3b + c = ax^2+(6a+b)x+(9a+3b+c)$$

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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