Shubham Swastik Behera
Last Activity: 9 Years ago
I guess B.
As a nd b are roots of the equation, 2x2+6x+b=0, so we have two equatins:
a+b = -(coeff of x)/(coeff of x2) = -3 …...(i)
ab = (constant)/(coeff of x2) = b/2 …....(ii)
So, from (ii) we have b(a-1/2)=0, as b cannot be zero (since root and putting x=0 we donot get LHS=RHS) so a-1/2=0 and thus a=1/2.
Putting this value in (i) we get b = -3.5
Now, (a/b)+(b/a) = (a2+b2)/ab = [(a+b)2-2ab]/ab ….......(iii)
Putting in values of (a+b) and ab we get (iii)= -7.14, less than -2.
Cheers!!