By the way, ab is an integer, which means A is an integer and B is an integer, right?
Because the equation has two roots-1~0, let y = f (x) = ax 2+bx+2.
According to the properties of quadratic function, it is only possible to draw the image of quadratic function, a >；; 0
Both roots are at-1~0, which means that the intersection of the quadratic function and the y axis is at-1~0.
Is it equal to discriminant greater than 0? b^2-8a>； 0? ( 1)
The symmetry axis is-1 ~ 0- 1
f(- 1)>0? a-b+2 & gt； 0? (3)
f(0)>0? 2 & gt0
B>0 and b-2a are obtained from (2) and a >: 0.
Adding (3) (1) four inequalities is solved by linear programming.
Images, especially quadratic function images, are not accurate, but they are all over the mark (2, 4).
Blue is the feasible area, indicating b> IV.
B=5, a=3 meets the problem.
B the minimum value is 5.