<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] $$\alpha +\beta =-\frac { b }{ a } \\ \alpha \beta =\frac { c }{ a } \\ a\left( x-\left( \alpha +\frac { 1 }{ \beta  }  \right)  \right) \left( x-\left( \beta +{ \frac { 1 }{ \alpha  }  } \right)  \right) =0\\ a\left( x-\frac { \alpha \beta +1 }{ \beta  }  \right) \left( x-\frac { \beta \alpha +1 }{ \alpha  }  \right) =0\\ a\left( { x }^{ 2 }-\frac { \left( \alpha \beta +1 \right) \left( \beta +\alpha  \right)  }{ \alpha \beta  } x+\frac { { \left( \beta \alpha +1 \right)  }^{ 2 } }{ \beta \alpha  }  \right) =0\\ a\left( { x }^{ 2 }-\frac { \left( \frac { c }{ a } +1 \right) \left( -\frac { b }{ a }  \right)  }{ \frac { c }{ a }  } x+\frac { { \left( \frac { c }{ a } +1 \right)  }^{ 2 } }{ \frac { c }{ a }  }  \right) =0\\ a{ x }^{ 2 }+\frac { \left( c+a \right) b }{ c } x+\frac { { \left( c+a \right)  }^{ 2 } }{ c } =0\\ c{ a }{ x }^{ 2 }+\left( bc+ba \right) x+{ \left( c+a \right)  }^{ 2 }=0$$
</s>