$\Delta =4b^{2}-4b+4> 0\vee b$
$\Rightarrow$ PT luôn có 2 no pb
Theo đ/l Vi-et:
$\alpha +\beta =-2b$
$\alpha .\beta =b-1$
$\Rightarrow \alpha ^{2}+\beta ^{2}=4b^{2}-2b+1$
$\Rightarrow (\alpha -\beta )^{2}=4b^{2}-4b+4=(2b-1)^{2}+3\geq 3$
$\Rightarrow Min(\alpha -\beta )^{2}=3\Leftrightarrow b=1$