Let  $\alpha ,\beta$  be real and  $z$  be a complex number. If  $z^2+\alpha z+\beta =0$  has two distinct roots on the line  $\mathrm{Re}\left(z\right)=1$  then it is necessary that

 $z^2+\alpha z+\beta =0$             ……………(1)
$z=x+iy$  Lies on              $\mathrm{Re}\left(z\right)=1\Rightarrow x=1$
$\therefore z=1+iy$    is a root of (1)
$\Rightarrow \overline{z}=1-iy$  is also root of (1)
Hence Product of roots 
 $\begin{array}{l}\beta =(1+iy)(1-iy)\\ =1+y^2\ge 1\\ \Rightarrow \beta \ge 1\Rightarrow \beta \in [1,\infty )\end{array}$