Question

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

• A. 1
• B. -1
• C. 2
• D. -2

Answer 1

Answer: (D)

Given that $\alpha ,\beta$ are the roots of the given equation $2x^2+6x+b=0$, then
Sum of the roots $\alpha +\beta =-\frac{6}{2}=-3$
And product of the roots $=\alpha \beta =\frac{b}{2}$
$D=B^2-4AC$
$D=6^2-4\times 2\times 6$
$D=36-8b$
$D>0\left(\because b<0\right)$
Then, $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }=\frac{{\left(\alpha +\beta \right)}^2-2\alpha \beta }{\alpha \beta }$
$=\frac{{\left(-3\right)}^2-2\times \frac{b}{2}}{\frac{b}{2}}=\frac{9-b}{\frac{b}{2}}=\frac{18-2b}{b}=\frac{18}{b}-\frac{2b}{b}$
$=\frac{18}{b}-2<-2\left(\because b<0\right)$
$\therefore \frac{\alpha }{\beta }+\frac{\beta }{\alpha }<-2$