If the solution set of inequality AX2 + bx-2 > 0 is {x │ X1 / 3}, then the value of AB is

If the solution set of inequality AX2 + bx-2 > 0 is {x │ X1 / 3}, then the value of AB is

If the solution set of inequality AX2 + bx-2 > 0 is {x │ X1 / 3}, then the value of AB is
Satisfy x1x2 = - 2 / a = - 1 / 6
a = 12
x1+x2 = -b/a = -1/2 + 1/3 = -1/6
b = a/6 = 12/6 =2
ab =12 x 2 =24

If the solution set of inequality AX2 + BX + C is less than or equal to 0 {XIX is less than or equal to - 2 or X is greater than or equal to 3}, find the value of [B + C] / A

-2. 3 is ax & # 178; + BX + C = 0
∴-b/a=-2+3 c/a=-2*3
∴b/a=-1 c/a=-6
∴(b+c)/a=b/a+c/a=-7

If there is no term containing x in the sum of 3x's Square minus 2x plus B + BX minus 1, try to find the value of B, write their sum, and prove that no matter what value x takes, its value is always positive
If there is no linear term containing x in the sum of 3x's Square minus 2x plus B + BX minus 1, find the value of B, write their sum, and prove that no matter what value x takes, its value is always positive

If there is no term containing x, then B = 2, with the time of admission, we get the square of 3x + 1. No matter what the value of X is, the square of 3x + 1 is always positive

If there is no term containing x in the sum of 3x square - 2x + B and x square + BX-1, try to find the value of B, write their sum, and prove that no matter what value x takes, their sum is correct
The value is always positive

3x²-2x+b+x²+bx-1=4x²+(b-2)x+b-1
Does not contain x items
So B-2 = 0
b=2
Sum = 4x & # 178; + 1
Because 4x & # 178; + 1 > 0
So sum is always positive