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?)?
 A: $$\alpha +\beta =-\frac { b }{ a } \\ \alpha \beta =\frac { c }{ a } \\ a\left( x-\left( \alpha +\frac { 1 }{ \beta  }  \right)  \right) \left( x-\left( \beta +{ \frac { 1 }{ \alpha  }  } \right)  \right) =0\\ a\left( x-\frac { \alpha \beta +1 }{ \beta  }  \right) \left( x-\frac { \beta \alpha +1 }{ \alpha  }  \right) =0\\ a\left( { x }^{ 2 }-\frac { \left( \alpha \beta +1 \right) \left( \beta +\alpha  \right)  }{ \alpha \beta  } x+\frac { { \left( \beta \alpha +1 \right)  }^{ 2 } }{ \beta \alpha  }  \right) =0\\ a\left( { x }^{ 2 }-\frac { \left( \frac { c }{ a } +1 \right) \left( -\frac { b }{ a }  \right)  }{ \frac { c }{ a }  } x+\frac { { \left( \frac { c }{ a } +1 \right)  }^{ 2 } }{ \frac { c }{ a }  }  \right) =0\\ a{ x }^{ 2 }+\frac { \left( c+a \right) b }{ c } x+\frac { { \left( c+a \right)  }^{ 2 } }{ c } =0\\ c{ a }{ x }^{ 2 }+\left( bc+ba \right) x+{ \left( c+a \right)  }^{ 2 }=0$$
A: $$S=\alpha+\beta+\frac{1}{\alpha}+\frac{1}{\beta}=(\alpha+\beta)\left(1+\frac{1}{\alpha\beta}\right)$$
If I write this in terms of $a,b,c$ I get $$S=-\frac{b}{a}\left(1+\frac{a}{c}\right)=-\frac{b(a+c)}{ac}$$
A: $\alpha+\frac{1}{\beta}+\beta+\frac{1}{\alpha}=\frac{(\alpha+\beta)(\alpha\beta+1)}{\alpha\beta}=\frac{(\frac{-b}{a})(\frac{c}{a}+1)}{\frac{c}{a}}=-\frac{(b(c+a))}{ac}$
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 $x^2- (-\frac{b(c+a)}{ac})x+\frac{(c+a)^2}{ac} =x^2+\frac{b(c+a)}{ac}x+\frac{(c+a)^2}{ac}=acx^2+b(c+a)x+{(c+a)^2}$
A: Hint: Use the solution form to express $\alpha$ and $\beta$ in terms of $a$, $b$, $c$, then observe that a polynomial with roots $r$ and $s$ is of the form $f(x) = c(x - r) (x - s)$ for some constant $c$.
A: 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:
$$
\frac{ac}{(a+c)^2} x^2 + \frac{bc}{a+c} x+ c = \frac{1}{(a+c)^2}\left(ac x^2 + bc(a+c)x + (a+c)^2\right)
$$
