Question

What is the value of $\lambda$ for which the quadratic equation $3x^2+2({\lambda }^2+1)x+({\lambda }^2-3\lambda +2)=0$ has roots of opposite signs?

• A. $\lambda \in (-\infty ,1)\cup (2,3)$
• B. $\lambda \in (-\infty ,-1)\cup (2,\infty )$
• C. $1<\lambda <2$
• D. $-1<\lambda <-2$

Answer 1

The correct option is C $1<\lambda <2$

Since the roots are of opposite signs, so product of the roots is negative.
$\Rightarrow \frac{{\lambda }^2-3\lambda +2}{3}<0\dots (productofroots=\frac{c}{a})$
$\Rightarrow {\lambda }^2-3\lambda +2<0$
$\Rightarrow (\lambda -1)(\lambda -2)<0$
$\Rightarrow 1<\lambda <2$