<s>[INST] 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!
 [/INST] 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,
$$ s \ \ = \ \ \frac12  ·  \left(-\alpha \ - \ \beta \ - \ \frac{1}{\beta} \ - \ \frac{1}{\alpha} \ - \ \alpha \ + \ \beta \ - \ \frac{1}{\beta} \ + \ \frac{1}{\alpha} \right) \ \ = \ \ - \left(\alpha \ + \  \frac{1}{\beta} \right) \ \ .  $$
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