If the set a = {x ∈ r ax square + 2x + 1 = 0, a ∈ r}, find the sum of the elements in a

If the set a = {x ∈ r ax square + 2x + 1 = 0, a ∈ r}, find the sum of the elements in a

1. A = 0, then x = - 1 / 2
2. A > 1, no solution
3. A ≤ 1 and a ≠ 0, X1 + x2 = - 2 / A
The sum of elements is - (4 + a) / 2A
The maximum value of function f (x) = x ^ 3-6x ^ 2 + 9x-2 is?
f '(x) = 3x^2-12x+9
When f '(x) = 0, 3x ^ 2-12x + 9 = 0
That is, (x-1) (x-3) = 0
The solution is as follows
x1=1,x2 =3
When x 3, f '(x) > 0
So,
When x = 1, the original function has a maximum of 2;
When x = 3, the original function has a minimum value of - 2
Maybe 2
Given the set a = {the square of a ax + 2x + 1 = 0, a ∈ R, X ∈ r}. If there is only one element in a, find the value of A
① When a = 0, x = - 1 / 2
② When △ = 0, a = 1, x = - 1
So when a = 0 or a = - 1, there is only one element in a
Is the element in a not a? What's the matter with x? Why deduce the value of a when x can only take one value
I feel that a can take all values=
You're right.
The subject is obviously wrong
The element should be X
The maximum value of the function y = x & # 179; - 6x + 1 is?
The derivative is 3x & # 178; - 6, the extremum is ± √ 2, where the maximum value is x = - 2, and y = - 8 √ 2 + 1
Given the set a = {x | x2-3x + 2 = 0}, set B = {x | 2x2-x + 2A = 0}, if B ∪ a = a, find the set composed of the values of A
∪ set a = {x | x2-3x + 2 = 0} = {1,2}, set B = {x | 2x2-x + 2A = 0}, B ∪ a = a, ∪ B = ∈, ∪ a ＜ 0, the solution of a ＞ 116, the value of ∪ A is {a | a ＞ 116}
The maximum and minimum values of the function y = x / (X & # 178; - 3) are respectively ()
A、√3,-√3 B、√3/6,-√3/6
C、√3,-√3/3 D、√3/6,-√3
y=x/（x²+3）
Pro, it's a big joke, no problem
It is known that the solution set of {3x-2a3} is - 1
From 3x-2a (3 + b) / 2
So (3 + b) / 2 = - 1, B = - 5
(a+1)(b-1)=-12
Let z = 1 - √ (X & # 178; + Y & # 178;), then (0,0) is the maximum point of the function and the maximum point. How to prove?
Proof: √ (X & # 178; + Y & # 178; never less than 0
When x = 0, y = 0 √ (X & # 178; + Y & # 178; take the minimum value of 0
In this case, z = 1 - √ (X & # 178; + Y & # 178;) the maximum must be the maximum point
So (0,0) is the maximum point of the function and the maximum point
Given the set a = {x | x 2 + (P + 2) x + 1 = 0, X ∈ r}, and a ⊆ negative real number, the value range of real number P is obtained
⊆ a ⊆ is a negative real number set, or ⊆ A is an empty set, or in a: x2 + (P + 2) x + 1 = 0 has negative roots. If it is an empty set, then △ = (P + 2) 2-4 < 0, the solution is - 4 < p < 0. If in a: x2 + (P + 2) x + 1 = 0 has negative roots. ⊆ 1 > 0, ⊆ a = (P + 2) 2-4 ≥ 0, and - (P + 2) < 0, the solution is p ≥ 0. In conclusion, the value range of real number P is (- 4, 0) ∪ [0, + ∞)
Given the function y = x & # 178; + 6x + 5, change the function y = x & # 178; + 6x + 5 into the form of y = a (x + m) &# 178; + K, and state the axis of symmetry, vertex coordinates and maximum value of the function,
In particular, how to say the most value situation
Don't quarrel with the answer on Baidu!
Y=X²+6X+5
=(x+3)²-4;
The axis of symmetry is x = - 3;
Vertex coordinates are (- 3, - 4)
When x = - 3, the minimum value is - 4;
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