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 $c$?
 A: My way:
If $p$ is a root of $9x^2-2x+7=0, p-2$ will be a root of $ax^2+bx+c=0$
Now writing $p-2=q\iff p=q+2$  in   $9x^2-2x+7=0,$
we get $0=9(q+2)^2-2(q+2)+7=9q^2+34q+39$
So, we need $$\dfrac a9=\dfrac b{34}=\dfrac c{39}\ \ \ \ (1)$$
So, with the given conditions, $a,b,c$ can assume any set of non-zero finite values that honor $(1)$
A: Hint: Find the roots of equation $9x^2-2x+7=0$ by solving this let say the roots be $\alpha$ and $\beta$ (You know the value of $\alpha$ and $\beta$.) then the roots of other equation must be $\alpha-2$ and $\beta-2$ form a equation using $$(x-(\alpha-2))(x-(\beta-2))$$ compare the coefficient. then put the value in $4a-2b+c.$
A: 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.
A: Since $ax^2+bx+c=0$ has the same roots as $x^2+\frac{b}{a}x+\frac{c}{a}=0$ we can notice that if $(a_1,b_1,c_1)$ satisfies the requirements then also does $(ta_1,tb_1,tb_2)$ since $4a-2b+c\neq 0$ we can get $4a-2b+c=\frac{7t}{9}$ for any $t$.
