Given, $x^{2}+a x +b=0$
For district now zero roots $D > 0$
$\Rightarrow a^{2}-4 b>0$
Now, $x^{2}+a x +b$
$=\left(x+\frac{a}{2}\right)^{2}+\left(b-\frac{a^{2}}{4}\right)$
$=\left(x+\frac{a}{2}\right)^{2}-\left(a^{2}-\frac{4 b}{4}\right)$
We know, sum of roots $a+b=-a$
$\Rightarrow 2 a +b=0$ ...(i)
Product of roots $a \times b=b$
$\Rightarrow b(a-1)=0$
$\Rightarrow a=1, b \neq 0$
From Eq. (i),
$2 a +b=0$
$2(1)+b=0$
$b=-2$
Now, $\left(x+\frac{a}{2}\right)^{2}-\left(\frac{1^{2}+4 \times 2}{4}\right)$
$=\left(x+\frac{a}{2}\right)^{2}-\frac{9}{4}$
$\therefore $ Minimum value $=-\frac{9}{4}$