If f(x)=ax^2+bx+c satisfies f(1)+2f(2)=0 and | vedclass

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(C) Given $f(x) = ax^2 + bx + c$. $f(1) = a + b + c$ and $f(2) = 4a + 2b + c$. From the first condition: $f(1) + 2f(2) = 0$ $(a + b + c) + 2(4a + 2b +

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(C) Given $f(x) = ax^2 + bx + c$. $f(1) = a + b + c$ and $f(2) = 4a + 2b + c$. From the first condition: $f(1) + 2f(2) = 0$ $(a + b + c) + 2(4a + 2b + c) = 0$ $a + b + c + 8a + 4b + 2c = 0$ $9a + 5b + 3c = 0$ ... $(i)$ From the second condition: $2f(1) + f(2) = 0$ $2(a + b + c) + (4a + 2b + c) = 0$ $2a + 2b + 2c + 4a + 2b + c = 0$ $6a + 4b + 3c = 0$ ... $(ii)$ Subtracting equation $(ii)$ from equation $(i)$: $(9a + 5b + 3c) - (6a + 4b + 3c) = 0 - 0$ $3a + b = 0$.

If $f(x)=ax^2+bx+c$ satisfies $f(1)+2f(2)=0$ and $2f(1)+f(2)=0$,then $3a+b=$

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