If the equation AX2 + 2x + 1 = 0 has and only has one negative root, then the value range of a is______ ．

If the equation AX2 + 2x + 1 = 0 has and only has one negative root, then the value range of a is______ ．

(1) When a = 0, the equation becomes 2x + 1 = 0, then x = - 12, and there is only one negative root, which is consistent with the meaning of the problem; (2) when a < 0, ∵ f (0) = 1 > 0, then the equation has only one negative root, which is consistent with the meaning of the problem; (3) when a > 0, f (0) = 1 > 0, and △ = 4-4a, if △ ≥ 0, the equation has two negative roots

If the solution set of the equation ax2-2x + a = 0 is p and there is only one element in the set P at most, the value range of a is obtained

When a = 0, the original formula = 2x = 0, x = 0
So Δ = B ^ 2-4ac ≤ 0, so (- 2) ^ 2-4aa ≤ 0, a ^ 2-1 ≥ 0, (A-1) (a + 1) ≥ 0, a ≥ 1 or a ≤ - 1
So the value range of a is a ≥ 1 or a ≤ - 1 or a = 0
Do you learn imaginary numbers now? It's the square root of negative numbers. If you learn, then Δ = B ^ 2-4ac = 0, so (- 2) ^ 2-4aa = 0, a ^ 2-1 = 0, (A-1) (a + 1) = 0, a = 1 or a = 1
So the value range of a is a = 1 or a = 1 or a = 0

The equation ax2-2x-1 = 0 has at least one root. Find the value range of a and find the corresponding root

If a = 0, f (x) = - 2x-1 = 0, there is a unique solution
If a ≠ 0, △ ≥ 0, that is (- 2) &# 178; - 4 × a × (- 1) ≥ 0, the solution is a ≥ - 1
In conclusion, a ≥ - 1

How to find the intersection of the x power of 2 and the quadratic power of X + 1

y=2^x-(x^2+1)
Obviously, y (0) = 0, y (1) = 0, so x = 0,1 are two intersections
The other intersection has to be solved numerically: X3 = 4.257461914

A = {the square of X | x + px-2 = 0}, B = {the square of X | X - x + q = 0}, and the intersection of a and B = {- 2,0,1}, find P, Q

Because the intersection of a and B = {- 2,0,1}
If 0 belongs to a
So - 2 = 0, obviously impossible
So 0 belongs to B
So we get q = 0
So B is x ^ 2-x = 0
X = 1 or x = 0
So - 2 must belong to a, substituting
(-2)^2-2p-2=0
4-2p-2=0
p=1
So p = 1, q = 0

If the solution of the equation X-1 / 2x-a = 1 of X is a positive number, then the value range of a is () process

(2x-a-x+1)/(x-1)=0
(x-a+1)/(x-1)=0
x≠1 x=a-1>0
a> 1 and a ≠ 2

If the solution of the equation X-2 / 2x + a = - 1 is positive, then the value range of a is

Separation a, X ≠ 2, so, 2x + a = 2-x, x = (2-A) / 3 > 0, so a < 2

If the solution of the equation X-2 / 2x + a = negative one is a positive number, find the value range of A
In a hurry

(2x+a)/(x-2)=-1
2x+a=2-x
x=(2-a)/3
Because x > 0
So (2-A) / 3 > 0
a

If the solution of the equation (2x + a) \ \ (X-2) = - 1 about X is positive, then the value range of a is ()

If the equation (2x + a) \ \ (X-2) = - 1 about X, the solution is positive
2x+a=-x+2
3x=2-a
x=(2-a)/3>0
a

When a is a value, the solution of the equation x + 1 x − 2 − XX + 3 = x + a (x − 2) (x + 3) is positive?

By multiplying (X-2) (x + 3) on both sides of the equation, (x + 1) (x + 3) - x (X-2) = x + A, it is sorted out that 5x = A-3, the solution of the equation is x = a − 35, ∵ a − 35 > 0, and a − 35-2 ≠ 0, a − 35 + 3 ≠ 0, the solution is a > 3 and a ≠ 13. That is, when a > 3 and a ≠ 13, the solution of the equation about X is a positive number. So the answer is: a > 3 and a ≠ 13