If $a$ and $b$ are the non-zero distinct roots of $x^2+ax+b=0,$then the least value of $x^2+ax+b$, is

Since $a, b$are roots of $x^2+ax+b=0$. Therefore,

$a^2+a^2+b=0$ and,$b^2+ab+b=0$

$\begin{array}{l}\Rightarrow b=-2a^2\text{ and }b+a+1=0\\ \Rightarrow -2a^2+a+1=0\\ \Rightarrow 2a^2-a-1=0\Rightarrow a=1\text{ or, }a=-1/2.\end{array}$

Now,

$a=1,\Rightarrow b=-2 [\because b+a+1=0]$

and, $a=-1/2\Rightarrow b--1/2$

But, $a\ne b\text{. }$Therefore $, a=1,b=-2\text{. }$

$\therefore$Least value of $x^2+ax+b$ is $-\frac{D}{4}$

i.e.  $-\left(\frac{a^2-4b}{4}\right)=-\left(\frac{1+8}{4}\right)=-\frac{9}{4}$