2. Because -3 and 1 both belong to A, that is to say, when x=-3 or 1, y-x=0 holds. Then according to the square of y=x -ax+b, we can get

-3=9+3a+b, 1= 1-a+b, the solution is: a=-3, b=-3.

Then the original formula is y=x squared +3x-3.

Set b is y-ax=0, so y=x square +3x-3-(-3x)=x square +6x-3=0.

We can get that the two roots of X are -3+2√3 and -3-2√3, which is the set B.

You wrote it correctly. Judge δ

Get the square of 4a+a.

When δ > =0, we have

The two roots of the equation are (-(2+a)+√(4a+a square)) /2 and (-(2+a)-√(4a+a square)) /2.

Because δ >; =0, so there is 4a+ a square >; =0, get a & gt=0 or a.

Because (-(2+a)+√(4a+a squared))/2 >; (-(2+a)-√(4a+a squared)) /2

According to A∩B=? We know that as long as (-(2+a)+√(4a+a squared)) /2.

That is -(2+a)+√(4a+a squared) < =0.

Because √ (square of 4a+a) < = √ (square of 4a+a+4) = √ (square of a+2) =|a+2|

When a> =-2,2+a > 0, then (-(2+a)+√(4a+a squared)); =0。

Combining the above two is the range of a, which is -4.