Question

For what value of a, the roots of the equation $2x^2+6x+a=0$, satisfy the conditions $\left(\frac{\alpha }{\beta }\right)+\left(\frac{\beta }{\alpha }\right)<2$ (where $\beta ,\alpha$ are the roots of equation).

• A. $a>0$
• B. $-1<a<0$
• C. $-1<a<1$
• D. None of these

Answer 1

Answer: (D)

None of these

Given equation is $2x^2+6x+a=0$

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

$\alpha \beta =\frac{a}{2}$

According to the question,

$\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }<2$

$\frac{(\alpha +\beta {)}^2-2ab}{\alpha \beta }<2$

$\Rightarrow \frac{9-a}{\frac{a}{2}}<2$

$\therefore a>\frac{9}{2}$