Express roots of equation $acx^2-b(c+a)x+(c+a)^2=0$ in terms of $\alpha, \beta, $ 
If roots of the equation $ax^2+bx+c=0$ are $\alpha, \beta, $  find roots of equation $acx^2-b(c+a)x+(c+a)^2=0$ in terms of $\alpha, \beta$

Here's what I have tried so far,
I know that $\alpha+ \beta=\frac{-b}{a} $ and $\alpha \beta=\frac{c}{a} $
So I can express $b=-a(\alpha+\beta)$
$c=a.\alpha\beta$
Once I substitute for b and c in the equation I can get, $$\alpha\beta x^2+(\alpha+\beta)(\alpha\beta+1)x+(\alpha\beta+1)^2=0$$
I want to know whether there is any different approach other than this method?
Any hint is higly valued. thank you!
 A: As given by me in comment:
let $g(x)=acx^2-b(c+a)x+{(c+a)}^2=0$ and $f(x)=ax^2+bx+c$
we see that $$g(x)={(c+a)}^2\cdot \frac{1}{c}\cdot f(\frac{-cx}{c+a})$$ thus $g(x)=0$ implies $$\frac{-cx}{c+a}=\alpha ,\beta$$ now use vieta and rearrange .....
A: Let $\bar\alpha$, $\bar\beta$ be the roots of $acx^2-b(c+a)x+(c+a)^2=0$, which can be written as
$$\left(1+\frac ac\right)^2\frac c{x^2}
-\left(1+\frac ac\right)\frac b{x}+a=0
$$
Compare with $ax^2+bx+c=0$ written as
$$\frac c{x^2} +\frac bx +a=0$$
to establish
$$\left(1+\frac ac\right)\frac1{\bar\alpha}=-\frac1{\alpha},\>\>\>\>\>\left(1+\frac ac\right)\frac1{\bar\beta}=-\frac1{\beta}
$$
Then, with $\frac ca = \alpha \beta$
$$\bar\alpha=- \alpha -\frac1{\beta}
,\>\>\>\>\>
\bar\beta= -\beta -\frac1{\alpha}
$$
A: Solve the two equations to find
$$ay^2+by+c=0\implies y=\dfrac{-b\pm\sqrt{b^2-4ca}}{2a}\ \ \ \ (1)$$
$$cax^2-b(c+a)x+(c+a)^2=0\implies x=(c+a)\cdot\left(\dfrac{b\pm\sqrt{b^2-4ca}}{2ca}\right)\ 
 \ \ \  (2)$$
Considering the opposite signs of the roots,
$$\dfrac yx=-\dfrac c{c+a}$$
Considering the same signs of the roots,
$$\dfrac yx=-\dfrac c{c+a}\cdot\dfrac{b+m\sqrt{b^2-4ca}}{b-m\sqrt{b^2-4ca}}\text{ where }m =\pm1$$
So, the ratios are not the same!
A: Let $r_1$ and $r_2$ be the roots of the equation
$$
a c x^{2}-b(c+a) x+(c+a)^{2}=0
$$
Then $$
\begin{aligned}
r_{1}+r_{2} &=\frac{b(c+a)}{a c}=\frac{b}{a}+\frac{b}{c}\\&=-(\alpha+\beta)-\frac{\alpha+\beta}{\alpha \beta} \\
&=-(\alpha+\beta)-\left(\frac{1}{\beta}+\frac{1}{\alpha}\right)\\&=-\left(\alpha+\frac{1}{\beta}\right)-\left(\beta+\frac{1}{\alpha}\right)
\end{aligned}
$$
and
$$
\begin{aligned}
r_{1} r_{2} &=\frac{(c+a)^{2}}{a c}=\frac{c^{2}+2 a c+a^{2}}{a c} \\
&=\frac{c}{a}+2+\frac{a}{c} \\
&=\alpha \beta+2+\frac{1}{\alpha \beta} \\
&=\left[-\left(\alpha+\frac{1}{\beta}\right)\right]\left[-\left(\beta+\frac{1}{\alpha}\right)\right]
\end{aligned}
$$
Hence the required roots are $\displaystyle -\left(\alpha+\frac{1}{\beta}\right) $ and $ \displaystyle -\left(\beta+\frac{1}{\alpha}\right) .$
A: This is somewhat along the lines of the approach Quanto takes, but causes the roots to emerge in a different fashion.
Dividing the given quadratic equation $ \ ac·x^2 \ - \ b(c+a)·x \ + \ (c+a)^2 \ = \ 0 \ \ $ through by $ \ ac \ $ produces
$$ x^2 \ - \ \frac{b(c+a)}{ac}·x \ + \ \frac{(c+a)^2}{ac} \ = \ 0 \ \ \Rightarrow \ \ x^2 \ - \ \frac{b}{a} · \left(1+\frac{a}{c}\right)·x \ + \ \frac{c}{a} · \left(1+\frac{a}{c}\right)^2 \ = \ 0  \ \ . $$
Applying the Viete relations you cited, we can then express this as
$$ x^2 \ + \ (\alpha + \beta) · \left(1+\frac{1}{\alpha·\beta}\right)·x \ + \ (\alpha·\beta) · \left(1+\frac{1}{\alpha·\beta}\right)^2 \ = \ 0  \ \ . $$
Now it is a fairly familiar fact (which is simple enough to prove, though it is not often expressed this way) that if $ \ r \ $ and $ \ s \ $ are the roots of a quadratic equation $ \ x^2 + mx + n \ = \ 0 \ \ , $ then $ \ r + s \ = \ -m \ $ and $ \ r - s \ = \ \sqrt{\Delta} \ = \ \sqrt{m^2 - 4n} \ \ . $  For the equation at hand, we thus have
$$ r \ + \ s \ = \ - \left(\alpha \ + \ \beta \ + \ \frac{1}{\beta} \ + \ \frac{1}{\alpha} \right) \ \ , $$ $$ \Delta \ = \ (\alpha + \beta)^2 · \left(1+\frac{1}{\alpha·\beta}\right)^2 \ - \ \ 4  ·  (\alpha·\beta) · \left(1+\frac{1}{\alpha·\beta}\right)^2 \ \ = \ \ (\alpha - \beta)^2 · \left(1+\frac{1}{\alpha·\beta}\right)^2  $$
$$ \Rightarrow \ \ \sqrt{\Delta} \ \ = \ \ (\alpha - \beta) · \left(1+\frac{1}{\alpha·\beta}\right) \ \ = \ \ \left(\alpha \ - \ \beta \ + \ \frac{1}{\beta} \ - \ \frac{1}{\alpha} \right) \ \ . $$
Consequently,
$$ r \ \ = \ \ \frac12  ·  \left(-\alpha \ - \ \beta \ - \ \frac{1}{\beta} \ - \ \frac{1}{\alpha} \ + \ \alpha \ - \ \beta \ + \ \frac{1}{\beta} \ - \ \frac{1}{\alpha} \right) \ \ = \ \ - \left(\beta \ + \  \frac{1}{\alpha} \right) $$
and similarly,
$$ s \ \ = \ \ \frac12  ·  \left(-\alpha \ - \ \beta \ - \ \frac{1}{\beta} \ - \ \frac{1}{\alpha} \ - \ \alpha \ + \ \beta \ - \ \frac{1}{\beta} \ + \ \frac{1}{\alpha} \right) \ \ = \ \ - \left(\alpha \ + \  \frac{1}{\beta} \right) \ \ .  $$
