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 $= \alpha\beta = \frac{b}{2}$
$D = B^{2} - 4AC$
$D = 6^{2} - 4 × 2 × 6$
$D = 36 - 8b$
$D > 0 \quad\left(\because b < 0\right)$
Then, $\frac{\alpha}{\beta}+\frac{\beta}{\alpha} = \frac{\alpha^{2}+\beta^{2}}{\alpha\beta} = \frac{\left( \alpha + \beta \right)^{2} - 2 \alpha\beta }{ \alpha \beta }$
$= \frac{\left(-3\right)^{2}-2\times\frac{b}{2}}{\frac{b}{2}} = \frac{9-b}{\frac{b}{2}} = \frac{18-2b}{b} = \frac{18}{b} - \frac{2b}{b}$
$= \frac{18}{b} - 2 < -2\quad\left(\because b < 0\right)$
$\therefore \quad\frac{\alpha }{\beta }+\frac{\beta }{\alpha } < -2$