2017/001) Find all values of the parameter a such that the real roots $\alpha,\beta$ of the equation $2x^2+6x+a=0$ satisfy the ineqauilty $|\frac{\alpha}{\beta} + \frac{\beta}{\alpha} |< 2$

as  for real t $|t+\frac{1}{t}| >=2$
so one root is positive and another -ve
$2x^2+6x + a = 0$
$=>4x^2 + 12x + 2a = 0$
$=>(2x+3)^2 + 2a - 9 = 0$
$=>(2x+3)^2 =  9 - 2a$
$x = -\frac{3}{2} \pm \frac{\sqrt{9-2a}}{2}$
for one root to be positive and another -ve $\frac{\sqrt{9-2a}}{2} > \frac{3}{2}$ or $ a < 0$
if we were interested in complex roots then $a > \frac{9}{2}$