<s>[INST] Finding the quadratic equation from its given roots. 
If $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c
=0$ , then form an equation whose roots are:
$\alpha+\dfrac{1}{\beta},\beta+\dfrac{1}{\alpha}$

Now, using Vieta's formula, 
For new equation, 
Product of roots ($P$) = $\dfrac{a^2+c^2+2ac}{ac}$
Sum of roots ($S$) = $\dfrac{2c-b}{c}$
Hence, 
the required equation is:
$acx^2 - ax(2c-b)+(a+c)^2=0$                       // as quadratic equation = $x^2-Sx +Px=0$
But the answer key states that the answer is: 
$acx^2 +b(a+c)x
+(a+c)^2=0$ 
I am really doubtful of this answer. Where have I gone wrong (or is the answer in the key wrong?)?
 [/INST] For an alternative approach, note that $\alpha \beta=c/a\,$, so the two new roots are $\alpha+1/\beta= \alpha(1+a/c)$ and $\beta+1/\alpha= \beta(1+a/c)$.
If $P(x)=ax^2+bx+c\,$ has roots $\alpha,\beta\,$, then the polynomial with roots $\lambda \alpha, \lambda\beta\,$ is:
$$
P\left(\frac{x}{\lambda}\right) = \frac{a}{\lambda^2} x^2 + \frac{b}{\lambda } x+ c
$$
For $\displaystyle \lambda =1+ \frac{a}{c} = \frac{a+c}{c}$ the above gives:
$$
\frac{ac}{(a+c)^2} x^2 + \frac{bc}{a+c} x+ c = \frac{1}{(a+c)^2}\left(ac x^2 + bc(a+c)x + (a+c)^2\right)
$$
</s>