If the roots of $9x^2-2x+7=0$ are $2$ more than the roots of $ax^2+bx+c=0$, then value of $4a-2b+c=$? 
If the roots of $9x^2-2x+7=0$ are $2$ more than the roots of $ax^2+bx+c=0$, then value of $4a-2b+c=$?

My approach is: Let the roots of $ax^2+bx+c=0$ are $\alpha$ and $\beta$. So, $$\alpha+\beta=-\frac{b}{a}\\ \alpha\beta=\frac{c}{a}$$
Therefore the roots of $9x^2-2x+7=0$ are $(\alpha+2)$ and $(\beta+2)$. So, $$\alpha+\beta+4=\frac{2}{9}\implies4-\frac{b}{a}=\frac{2}{9}\implies\frac{b}{a}=\frac{34}{9}$$ and 
\begin{align*}
(\alpha+2)(\beta+2)=\frac{7}{9}\\
\Rightarrow\alpha\beta+2(\alpha+\beta)+4=\frac{7}{9}\\
\Rightarrow\frac{c}{a}-2\frac{b}{a}+4=\frac{7}{9}\\
\Rightarrow\frac{4a-2b+c}{a}=\frac{7}{9}\\
\end{align*}
So, my final equations are: $\dfrac{b}{a}=\dfrac{34}{9}$ and $\dfrac{4a-2b+c}{a}=\dfrac{7}{9}$. My mind says I'm pretty close to the solution but I can't find it out. 
There is another similar question: If the roots of $px^2+qx+r=0$ are $2$ more than the roots of $ax^2+bx+c=0$, then what will be the expression of $r$ in terms of $a$, $b$, and $c$?

You can check the other answers for (possible faster) alternatives, but you were doing fine.

So, my final equations are: $\dfrac{b}{a}=\dfrac{34}{9}$ and $\dfrac{4a-2b+c}{a}=\dfrac{7}{9}$. My mind says I'm pretty close to the solution but I can't find it out.

Notice that the system
$$\left\{ \begin{array}{l}\dfrac{b}{a}=\dfrac{34}{9} \\[8pt] \dfrac{4a-2b+c}{a}=\dfrac{7}{9}
 \end{array}\right. \iff \left\{ \begin{array}{l}34a-9b=0 \\[8pt] 29 a - 18 b + 9 c = 0
 \end{array}\right.$$
has an infinite number of solutions, given by:
$$\left\{ \begin{array}{l} a = 9t \\ b = 34t \\ c = 39t 
 \end{array}\right. \quad t \in \mathbb{R_0}$$
This is logical because we can simply divide $ax^2+bx+c=0$ (for $a$ non-zero) by $a$ to obtain a quadratic equation with the same roots.
