Question

Minimum possible number of positive root of the quadratic equation$x^2-(1+\lambda )x+\lambda -2=0,\in R:$

• A. $2$
• B. $0$
• C. $1$
• D. can not be determined

Answer 1

The correct option is A $2$

$\begin{array}{c}{x}^2-\left(\lambda +1\right)x+\left(\lambda -2\right)=0\lambda \in R\\ D={\left(\lambda +1\right)}^2-4\left(\lambda -2\right)\\ ={\left(\lambda +1\right)}^2-4\lambda +8\\ ={\left(\lambda -1\right)}^2+8\\ D⟩0\forall \lambda \in R\end{array}$

So,both roots will be real and distinct
minimum value of $f\left(x\right)=\frac{-D}{4}=\frac{-\left[{\left(\lambda -1\right)}^2+8\right]}{4}⟨0$
minimum two root possible