If set of values of “a” for which f(x) =ax^2 -(3+2a)x+6, a is not equal to zero is positive for exactly three distinct negative integral values of x is (c,d) then the value of (c^2+16 d^2) is equal to ?

Hello student,
Please find the answer to your question below
Givenf(x) =ax^2 -(3+2a)x+6
=(ax-3)(x-2)
Here roots of equation f(x)=0 are 2 and 3/a and f(0)=6
f(x) should be positive for exactly 3 -ve integral values of x
So the graph of f(x) must be a downward parabola passing through x=2
and x=3/a and -4<3/a<-3
so a(-1,-3/4]
So c=-1,d=-3/4
so c²+16d²=1+9=10