Question

If $(\alpha +1)(\beta -1)+(\beta +1)(\alpha -1)\alpha +(\alpha -1)(\beta -1)=0$ and $\alpha (\alpha +1)(\beta +1)-(\alpha -1)(\beta -1)=0$
Also let $A=\{\frac{\alpha +1}{\alpha -1},\frac{\beta +1}{\beta -1}\}$ and $B=\{\frac{2\alpha }{\alpha -1},\frac{2\beta }{\beta +1}\}$ If $A\cap B\ne \varphi$, then find all the permissible value of parameter $"a"$.

Answer 1

1. Here,
   $\frac{\alpha +1}{\alpha -1}+\frac{\beta +1}{\beta -1}$ and $\frac{(\alpha +1)(\beta +1)}{(\alpha -1)(\beta -1)}=\frac{1}{\alpha }$
   Now the quadratic,
   $x^2-(1-\frac{1}{\alpha })x+\frac{1}{\alpha }=0$ has roots $\left(\frac{\alpha +1}{\alpha -1}and\frac{\beta +1}{\beta -1}\right)$
   $\Rightarrow$ $a{x}^2+x+1=0$
   Let, $x=\frac{2\alpha }{\alpha +1}\Rightarrow$ $\alpha =\frac{x}{2-x}\Rightarrow$ $\frac{\alpha +1}{\alpha -1}=\frac{1}{x-1}$
   Now replacing $x$ by $\frac{1}{x-1}$ in equation $\left(i\right)$, we get quadratic whose roots are $\frac{2\alpha }{\alpha +1}$ and $\frac{2\beta }{\beta +1}$
   $\Rightarrow x^2-x+a=0$
   Hence $\left(i\right)$ and $\left(ii\right)$ are the two quadratic whose roots are elements of set ${}^{\prime }{A}^{\prime }$ and set ${}^{\prime }{B}^{\prime }$ respectively.
   $\therefore$ They must have a roots in common as $A\cap B\ne \varphi$
   $\Rightarrow$ case $I$:Both roots common