If b<0, then the roots x1 and x2 of the equation 2x2+6x+b=0 satisfy the condition x1/x2+x2/x1

The discriminant of the quadratic equation 2x2+6x+b=0 is  given by D=36−8b>0 . Therefore, the given equation has real roots We have x1x2+x2x1=x12+x22x1x2=x1+x22−2x1x2x1x2=(−3)2−2(b2)b2=29-bb=18b-2<−2 [∵b<0]