Let a = {x ㄧ x ^ 2 + 4x = 0, B = {x ㄧ x ^ 2 + 2 (a + 1) x + A ^ 2-1 = 0}. A ∩ B = B, find the value of real number a I'm so tired of the title - do it right for me,

Let a = {x ㄧ x ^ 2 + 4x = 0, B = {x ㄧ x ^ 2 + 2 (a + 1) x + A ^ 2-1 = 0}. A ∩ B = B, find the value of real number a I'm so tired of the title - do it right for me,

A={xㄧx^2 +4x=0,}
x^2+4x=0
=>x=0,x=-4
B={xㄧx^2 +2（a+1)x+a^2-1= 0}.
x^2 +2（a+1)x+a^2-1= 0
(x+a+1)^2=2a+3
X = - A-1 + radical (2a + 3)
X = - a-1-radical (2a + 3)
A∩B=B
There are three situations
① B has two real roots, as shown above. One is 0 and the other is - 4
-A-1 + radical (2a + 3) = 0
-A-1-radical (2a + 3) = - 4
=>A = 10, a = 0 (rounding off)
② If B has a real root, then 2A + 3 = 0
a=-3/2
Not satisfying a ∩ B = b
③2a+3

Set a = {x | x ^ 2 + 4x = 0}, B = {x | x ^ 2 + 2 (a + 1) x + A ^ 2-1 = 0} if B belongs to a, find the condition of real number a, and explain how to calculate △ by the way

A={x|x^2+4x=0}={0,-4}
B={x|x^2+2(a+1)x+a^2-1=0}
B is a subset of A
(i) If B = {0, - 4}
Then - 2 (a + 1) = 0-4, a ^ 2-1 = 0 * (- 4)
So a = 1
(II) if B = {0}
Then 0 ^ 2 + 2 (a + 1) * 0 + A ^ 2-1 = 0
Then a = ± 1
Let's look at a = - 1
B = {x | x ^ 2 = 0} = {0}
(III) if B = {- 4}
Then (- 4) ^ 2 + 2 (a + 1) * (- 4) + A ^ 2-1 = 0
Then a = 1 or a = 7
Let's look at a = 7
B = {x | x ^ 2 + 16x + 48 = 0} = {- 4, - 12} does not conform
(III) if B is an empty set
Then Δ = 4 (a + 1) ^ 2-4 (a ^ 2-1) = 8A + 8 < 0
Then a < - 1
In conclusion, the value range of a is {a | a ≤ - 1 or a = 1}
If you don't understand, please hi me, I wish you a happy study!

Given the set M = {x | X & sup2; - 2aX + A + 2 ≤ 0} set B = {x | 1 ≤ x ≤ 4} if M ∪ B = B, find the value range of real number a

Because m ∪ B = B, when m = (- 2A) & # 178; - 4 × (a + 2) < 0, that is, (A-2) (a + 1) < 0, the solution is - 1 ＜ a ＜ 2, ② m ≠ &; when △ = (- 2A) & # 178; - 4 × (a + 2) ≥ 0, the symmetry axis X = a ∈ [1,4] f (1) = 1-2a + A + 2 = - A + 3 ≥ 0

There are always two intersections between 4x-3y-2 = 0 and x ^ 2 + y ^ 2-2ax + A ^ 2-12 = 0. Find the value range of real number a

(x-a)²+y²=12
Center (a, 0), radius 2 √ 3
There are always two intersections
The distance from the center of the circle to the straight line is less than the radius
So | 4a-0-2 | / √ (4 & sup2; + 3 & sup2;)

Given the set a = {x | x2-3x + 2 = 0}, B = {x | AX-2 = 0}, if a ∪ B = a, find the set composed of the value of real number a

A = {1,2}, from a ∪ B = A: B ⊆ A. --- - (3 points) ① if a = 0, then B = ∈, satisfying the meaning of the problem. --- - (6 points) ② if a ≠ 0, then B = {2A}, from B ⊆ A: 2A = 1 or 2A = 2, ⊆ a = 1 or a = 2, -- (11 points) -- (12 points)

Let set a be the domain of F (x) = ln [(- x ^ 2) - 2x + 8), and set B be the solution set of the inequality [AX-1 / a] (x + 4) ≥ 0 about X. if B ∈ CRA, the value range of real number a is obtained

0

Let a = {x | X-2|

A = {x | 1

Given the set a = {x | x2 + ax + 12b = 0}, the set B = {x | x2-ax + B = 0}, satisfying (∁ UA) ∩ B = {2}, a ∩ (∁ UB) = {4}, u = R, find the value of real numbers a and B

Because (CUA) ∩ B = {2}, a ∩ (cub) = {4}, so 2 ∈ B, 4 ∈ a, 4 − 2A + B = 016 + 4A + 12b = 0, a = 87B = − 127

Let u = R, a = {x | X & # 178; + ax + 12b = 0}, B = {x | X & # 178; - ax + B = 0}, satisfy CUA ∩ B = {2}, Cub ∩ a = {4}, find the value of a and B

A={x|x²+ax+12b=0},B={x|x²-ax+b=0}
CuA∩B={2},CuB∩A={4}
SO 2 ∈ B, 4 ∈ a
So 4-2a + B = 0, 16 + 4A + 12b = 0
A = 8 / 7, B = - 12 / 7
If you don't understand, please hi me, I wish you a happy study!

The definition field of function y = log2 (x ^ 2-ax + 2a) is r, and the value range of a is obtained
I'll wait here

The domain of the function y = log2 (x ^ 2-ax + 2a) is r
That is, x ^ 2-ax + 2A > 0 holds for any x ∈ R
∴Δ=a²-8a