The known set a (A-1)

The known set a (A-1)

A(a-10
We get b = {x | x > 2 or X
There is B to know x2
And because a and B are not empty sets
a-12
So A1
Let a = {x-a}
A-1
A = { x| -1+a
If the quadratic equation AX2 + X-1 = 0 with respect to X has real roots, then the value range of a is ()
A. A ＞ - 14b. A ≥ - 14C. A ≥ - 14 and a ≠ 0d. A ＞ 14 and a ≠ 0
According to the meaning of the problem, the system of equations 12 − 4 · a · (− 1) ≥ 0A ≠ 0, the solution is a ≥ - 14 and a ≠ 0
If a, B and C are in an equal ratio sequence, then the number of intersections between the image of the function y = AX2 + BX + C and the x-axis is?
a. B and C are in equal proportion sequence, so B ^ 2 = AC
Discriminant B ^ 2-4ac = B ^ 2-4b ^ 2 = - 3B ^ 2 < 0
The equation AX2 + BX + C = 0 has no solution
So the function has no intersection with the X axis
If the equation (AX + 1) ^ 2 = a + 1 about X has real roots, then the value range of a is________
（ax+1)^2=a+1>=0
a>=-1
A equals zero, right
According to the left side of the original formula is a square formula, the equation must have a real root when it is ≥ 0,
If a + 1 ≥ 0, then a ≥ - 1
(ax+1)^2=a+1
a^2x^2+1+2ax=a+1
a^2x^2+2ax=a
a^2x^2+2ax-a=0
△=b^2-4ac=4a^2+4a^3
According to the meaning of the title:
4a^2+4a^3>=0
a^2+a^3>=0
a^2(a+1)>=0
A1 > = 0, A2 > = - 1 (rounding)
So a > = 0
If the real numbers a, B and C are in an equal ratio sequence, then the number of intersections between the image of the function y = ax ^ 2 + BX + C and the X axis
The real numbers a, B and C are in equal proportion sequence: B ^ 2 = AC, and a, B and C are not zero
Y=AX^2+Bx+C:△=B^2-4AC=AC-4AC=-3AC
1) When AC has the same sign, △ 0, the number of intersections between the image and X axis is 2
0=ax^2+bx+c
b=ay
c=ay^2
ax^2+ayx+ay^2=0
x^2+yx+y^2=0
delta=y^2-4y^2=-3y^2
y^2>0
-3y^2
The equation LG (x-1) + LG (3-x) = LG (1-ax) has a solution
I figured out Delta, and then what
I still don't know where the third came from
It is known that 1-x > 0,3-x > 0,1-ax > 0
We can get 1
∵lg（x-1）+lg（3-x）=lg（1-ax）
∴lg（x-1）（3-x）=lg（1-ax）
∴（x-1）（3-x）=（1-ax）
That is: x ^ 2 - (4 + a) x + 4 = 0
∴△=（4+a）^2-4*4=a^2+8a
When there are two solutions, X-1 > 0.3-x > 0 → 10 → 10.... (2)
f(3)=1-3a>0.......(3)
One
If a, B and C are in an equal ratio sequence, then the number of intersections between the image and the x-axis of the function y = ax & # 178; + BX + C is?
If a, B, C are in equal proportion sequence
Then B ^ 2 = AC > 0
And ax ^ 2 + BX + C = 0
Discriminant △ = B ^ 2-4ac = - 3aC
If the equation LG (AX) · lg10x + 1 = 0 of X has a solution, then the value range of real number a is?
From the known ∵ 10x ∵ 0 ∵ x ∵ 0 ∵ ax > 0 ∵ a > 0 ∵ the original formula is equivalent to: (LGA + lgx) (1 + lgx) + 1 = 0, let t = lgx, then the original formula becomes: T ^ 2 + (1 + LGA) t + 1 + LGA = 0. If the equation has a solution, then: Δ = (1 + LGA) ^ 2-4 (1 + LGA) ≥ 0, then LGA ≤ - 1 or LGA ≥ 3 ∪ (0,1 / 10] ∪ [1000, + ∞)
Given that a, B and C are in equal proportion sequence, then the number of intersections between the image and X axis of quadratic function f (x) = AX2 + BX + C is ()
A. 0b. 0 or 1C. 1D. 2
From a, B, C into an equal ratio sequence, we get B2 = AC, and AC > 0, let AX2 + BX + C = 0 (a ≠ 0), then △ = b2-4ac = ac-4ac = - 3aC < 0, so the number of intersections between the image of function f (x) = AX2 + BX + C and X axis is 0