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. $(-1,\infty )$
• B. $b\in (-\infty ,0)\cup (\frac{9}{2},\infty )$
• C. $(1,\infty )$
• D. $(-\infty ,1]$

Answer 1

The correct option is B $b\in (-\infty ,0)\cup (\frac{9}{2},\infty )$

Given: $\alpha ,\beta$ are roots of $2x^2+6x+b=0$
$\alpha +\beta =-3$
$\alpha \beta =\frac{b}{2}$
$\frac{\alpha }{\beta }+\frac{\beta }{\alpha }<2$
$\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }<2$
$\frac{(\alpha +\beta {)}^2-2\alpha \beta }{\alpha \beta }<2$
$\frac{(\alpha +\beta {)}^2}{\alpha \beta }-2<2$
$\frac{(\alpha +\beta {)}^2}{\alpha \beta }<4$
$\frac{9}{\left(\frac{b}{2}\right)}<4$
$\frac{1}{b}<\frac{2}{9}$
$\frac{1}{b}-\frac{2}{9}<0$
$\frac{9-2b}{9b}<0$
$\frac{2b-9}{9b}>0$
Using the Wavy Curve Method, we get the range of $b$ as:
$b\in (-\infty ,0)\cup (\frac{9}{2},\infty )$