Question

If $\alpha ,\beta$ are roots of the equation $2x^2+6x+b=0(b<0),$ then $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }$ is less then

• A. $-2$
• B. $2$
• C. $18$
• D. $0$

Answer 1

The correct option is D $0$

Given $2x^2+6x+b=0,b<0$

$2x^2+6x+b=0\alpha +\beta =-\frac{6}{2}=-3$

$\alpha \beta =\frac{b}{2}<(\because b<0)$

$\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{\alpha +\beta }{\alpha \beta }=\frac{-3}{b/2}=\frac{-6}{b}>0$ ($\because$ both $-6$ and $b$ less than $0$)