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 than

• A. $2$
• B. $-2$
• C. $18$
• D. $0$

Answer 1

The correct option is B $-2$

As $\alpha ,\beta$ are roots of $2x^2+6x+b=0$

$\alpha +\beta =-\frac{6}{2}=-3$ and $\alpha \beta =\frac{b}{2}$

Therefore,
$\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }=\frac{{(\alpha +\beta )}^2-2\alpha \beta }{\alpha \beta }$

$=\frac{9-\frac{2b}{2}}{\frac{b}{2}}=\frac{18-2b}{b}=\frac{18}{b}-2$

we know $b<0\Rightarrow \frac{18}{b}<0$

$\frac{18}{b}-2<-2$

$\therefore \frac{\alpha }{\beta }+\frac{\beta }{\alpha }<-2$

Hence, option B is correct.