If a and bare non-zero roots of x^ 2 +a x+b=0 then the least value...

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, $a b=b$ $$\begin{array}{l} \Rightarrow \quada b-b=0 \\ \Rightarrow \quad b(a-1)=0 \end{array}$$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}