If a and bare non-zero roots of $x^{2}+a x+b=0$ then the least value of $\mathrm{x}^{2}+a \mathrm{x}+\mathrm{b}$ is

As given a and bare the roots of the equat $x^{2}+a x+b=0$ $\Rightarrow$ sum of roots, $a+b=-a$ $\Rightarrow \quad \mathrm{b}=-2 \mathrm{a}$ and product of roots, $a b=b$ $$ \begin{array}{l} \Rightarrow \quad a b-b=0 \\ \Rightarrow \quad b(a-1)=0 \end{array} $$ if $b=0$ then $a=0$ if $b \neq 0$ then $a=1$ and $b=-2$ so, the expression will be, $f(x)=x^{2}+x-2$ $$ =x^{2}+2 \cdot \frac{1}{2} x+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}-2 $$ $\Rightarrow \mathrm{f}(\mathrm{x})=\left(\mathrm{x}+\frac{1}{2}\right)^{2}-\frac{9}{4}$ So, $\mathrm{f}(\mathrm{x})$ will be minimum, if $\left(\mathrm{x}+\frac{1}{2}\right)^{2}=0$ i.e. when $x=-\frac{1}{2}$ $\Rightarrow$ minimum value of function $=-\frac{9}{4}$