The known set a = {x | ax + BX + C > 0, a is not equal to 0} = {x | M

The known set a = {x | ax + BX + C > 0, a is not equal to 0} = {x | M

I have come up with a solution. One thing to pay attention to is that the two ends of the solution of this quadratic inequality are often the corresponding solutions of AX2 + BX + C = 0. Therefore, we can see that: C / a = m * n, B / a = - (M + n), then as long as the required inequality is transformed into the form of C / A, C / B, then we can use M, n instead, and use the general method of solving the quadratic equation, The second point we should pay attention to here is that the positive and negative of a is not difficult to know from the combination of numbers and shapes that a is less than 0, so the required solution can actually be changed to: m * n * X2 - (M + n) * x + 1 > 0. I'll find it by myself

Let u = R, a = {x │ x ^ 2 + ax-12 = 0}, B = {x │ x ^ 2 + BX + B ^ 2-28 = 0}
If a ∩ cub = {2}, find the value of real numbers a and B
X ^ 2 is the second power of X and a ^ 2 is the second power of A

According to the meaning of a ∩ cub = {2}, if 2 is one of the elements of a and the only element of B, then x = 2 is substituted into x ^ 2 + ax-12 = 0 and x ^ 2 + BX + B ^ 2-28 = 0 respectively
A = 4, B = 4 or - 6

Given the complete set u = R, and the set a = {x | x ^ 2 + ax-12 = 0}, B = {x | x ^ 2 + BX + B ^ 2-28 ≠ 0}, if a ∩ cub = {2}, find the value of a and B
process

Because a ∩ B complement = {2}, 2 is the solution of the equation x & # 178; + ax-12 = 0, substituting 4 + 2a-12 = 0, solving a = 4, that is, the equation x & # 178; + 4x-12 = 0, solving x = 2 or x = - 6, so a = {2, - 6} is also the solution of the equation x & # 178; + BX + B & # 178; - 28 = 0, that is, B & # 178; + 2b-24 = 0