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?)?

$\alpha+\frac{1}{\beta}+\beta+\frac{1}{\alpha}=\frac{(\alpha+\beta)(\alpha\beta+1)}{\alpha\beta}=\frac{(\frac{-b}{a})(\frac{c}{a}+1)}{\frac{c}{a}}=-\frac{(b(c+a))}{ac}$
and 
$(\alpha+\frac{1}{\beta})(\beta+\frac{1}{\alpha})=\frac{(\alpha\beta+1)^2}{\alpha\beta}=\frac{(\frac{c}{a}+1)^2}{\frac{c}{a}}=\frac{(c+a)^2}{ac}$.
Sum of roots $= -$ (coefficient of $x$) and product of roots $=$ constant term.
Equation is $x^2- (-\frac{b(c+a)}{ac})x+\frac{(c+a)^2}{ac} =x^2+\frac{b(c+a)}{ac}x+\frac{(c+a)^2}{ac}=acx^2+b(c+a)x+{(c+a)^2}$
