Question

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

Answer 1

Let $f\left(x\right)=x^2+2(\lambda +1)x+{\lambda }^2+\lambda +7$ If both roots of $f\left(x\right)=0$ are negative, then

$D=b^2-4ac=4(\lambda +1{)}^2-4({\lambda }^2+\lambda +.7)\ge 0\Rightarrow \lambda -6\ge 0$

$\Rightarrow \lambda \in [6,\infty )$

Sum of roots $=-2(\lambda +1)<0$

$\Rightarrow \lambda \in (-1,\infty )$

and product of roots $={\lambda }^2+\lambda +7>0\forall \lambda \in R$

$\therefore$From $\left(1\right),\left(2\right),\left(3\right),$ we get $\lambda \in [6,\infty )$

(As (1),(2),(3) must be satisfied simultaneously.)

Hence, the least value of $\lambda =6$