If the equation  ${\mathrm{x}}^2+2(\mathrm{\lambda }+1)\mathrm{x}+{\mathrm{\lambda }}^2+\mathrm{\lambda }+7=0$has only negative roots, then least value of $\mathrm{\lambda }$  equals

We have,
${\mathrm{x}}^2+2(\mathrm{\lambda }+1)\mathrm{x}+{\mathrm{\lambda }}^2+\mathrm{\lambda }+7=0$
Both roots are negative, then$\mathrm{D}\ge 0$
$\begin{array}{lc}\therefore & 4(\mathrm{\lambda }+1{)}^2-4({\mathrm{\lambda }}^2+\mathrm{\lambda }+7)\ge 0\end{array}$
$\begin{array}{lc}\Rightarrow & \mathrm{\lambda }-6\ge 0\Rightarrow \mathrm{\lambda }\in [6,\mathrm{\infty })\end{array}$………………..(i)

Sum of roots $\mathrm{s}=-2(\mathrm{\lambda }+1)<0$
$\Rightarrow \mathrm{\lambda }\in (-1,\mathrm{\infty })$……………………………..(ii)

and product of roots $={\lambda }^2+\lambda +7>0$
$\Rightarrow \mathrm{\lambda }\in \mathrm{R}$………………………………………(iii)
$\therefore$From (il, (ir) and (iir), we get
$\mathrm{\lambda }\in [6,\mathrm{\infty })$

The least value of $\mathrm{\lambda }=6$