Given that the solution set of inequality ax & # 178; + BX + 1 ≥ 0 is {X / - 5 ≤ x ≤ 1}, find the value of a and B

Given that the solution set of inequality ax & # 178; + BX + 1 ≥ 0 is {X / - 5 ≤ x ≤ 1}, find the value of a and B

That is - 5 and 1 are the roots of the equation AX & # 178; + BX + 1 = 0
So - 5 + 1 = - B / A
-5×1=1/a
So a = - 1 / 5
b=-4/5
The solution set of inequality ax & # 178; + BX + 1 > 0 is x ≠ 2, and the values of a and B are obtained
The solution set of inequality ax & # 178; + BX + 1 > 0 is x ≠ 2
That is: the solution set of inequality ax & # 178; + BX + 1 > 0 is x > 2 and x < 2
So, a > 0
When x = 2, ax & # 178; + BX + 1 = 0, that is, 4A + 2B + 1 = 0
Because only when x = 2, ax & # 178; + BX + 1 = 0,
We obtain B & # 178; - 4A = 0, a = 1 / 4, B = - 1 with 4A + 2B + 1 = 0
If the solution set of inequality ax & # 178; + BX + 2 > 0 is (- 1 / 2,1 / 3), what is the value of a + B
-10
The solution set of known inequality ax & # 178; + BX + 1 > 0 is {x ｜ - 1 / 2
The inequality is constructed from the solution set
(x+1/2)(x-1/3)
The square of X + 4x + 8 = 2x + 11 is solved by formula!
x^2+4x+8=2x+11
The results are as follows
x^2+2x-3=0
（x+3）*（x-1）=0
So: x = - 3 or x = 1
Factorization of square difference
2-1/2x²
=1/2(4-x^2)
=1/2(2+x)(2-x)
Do not understand welcome to ask
=1/2)4-x^2)
=1/2(2+x)(2-x)
Please write out a linear equation with x = 1 as the solution. There are two steps to merge the similar terms and change the coefficient to 1
3X=X+2
3X-X=2
Merging similar items: 2x = 2
The conversion coefficient is 1: x = 1
（3X+1）-（2X+2）=0
simple form
3X+1-2X-2=0
（3X-2X）+1-2=0
X-1=0
X=1
2x+x=3
3x-4+x=0
If the complete set u = {1,2,3,4}, a ∩ B = {2}, (CUA) ∩ B = {1,4}, then cub=
Cub = {3} or empty set. Because the intersection of a and B is 2, there are 2 in a and 2 in B. and the intersection of complement B of a is 1,4, 1,4 in a and 1,4 in B. In conclusion, there are 1,2,4 in B. whether 3 exists or not is unknown. So cub = {3} or empty set
If x = (1 / 2), y = - 1 is the solution of the system of equations ax-3y = 5,2x + by = 1, try to find the value of a + B
Put this set of solutions in
a/2+3=5
1-b=1
therefore
A=4
B=0
a+b=4
It is known that the quadratic equation x & sup2; + 2 (k-1) x + K & sup2; - 1 = 0 has two unequal real roots.
(1) Find the value range of the real number k; (2) can 0 be a follower of the equation? If not, ask for another root. If not, explain the reason
1,delta=2-2k>0,k
1. According to the discriminant of root = 8-8k > 0, K0 result K is obtained