At what values of 'a' do all the zeroes of the function, f(x) = (a-2)x² + 2ax + a + 3 lie on the interval (-2, 1)?

Hi,

You need to consider 2 cases here.

Case 1 : Take a-2>0
Now the concavity is upwards, draw the desired graph and think about the conditions.
We need f(-2)>0
f(1)>0 \

But this is still insufficient so one last condition will be 1>-b/2a>-2
Solve for these 4 condityions and take intersection of solution set as we need all four conditions to be true in the same case.

Solve similarly for a-2<0 (Case 2)
i.e. f(-2)<0
f(1)<0
and 1>-b/2a>-2

In the end take union of solution set of case 1 and case 2.
That would be the final answer.