Question

If $\alpha ,\beta$ are the roots of $2x^2+6x+b=0$ and $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }<2$ then b lies in the interval

• A. ( 0, ∞)
• B. ( -1, ∞)
• C. (- ∞, 0)
• D. ( -∞ , 0]

Answer 1

The correct option is C (- ∞, 0)

Solution : $\alpha ,\beta$ are roots of $2x^2+6x+b=0$
$\alpha +\beta =-3$
$\alpha \beta =\frac{b}{2}$
$\frac{\alpha }{\beta }+\frac{\beta }{\alpha }<2$
$\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }<2$
$\frac{(\alpha +\beta {)}^2-2\alpha \beta }{\alpha \beta }<2$
$\frac{(\alpha +\beta {)}^2}{\alpha \beta }<4$
$\frac{9}{\left(\frac{b}{2}\right)}<4$
$\frac{1}{b}<\frac{2}{9}$
$b\in (-\infty ,0)\cup (\frac{9}{2},\infty )$