If a and b $(\ne 0)$ are the roots of the equation $x^2+ax+b=0$, then find the least  value of $x^2+ax+b,\forall x\in R$ 

Since a and b are the roots of the equation

$x^2+ax+b=0$

So,

$a+b=-a,ab=b$

Now, $ab=b\Rightarrow (a-1)b=0\Rightarrow a=1,(b\ne 0)$

Putting $a=1$ in  $a+b=-a$ we get  $b=-2$

Hence, $x^2+ax+b=x^2+x-2$

Least value $=\frac{4ac-b^2}{4a}$

$\begin{array}{l}=\frac{-8-1}{4}\\ =\frac{-9}{4}\end{array}$