If the solution set of inequality ax & # 178; + BX + C ≤ 0 is {x | ≤ 1 or X ≥ 4}, then the trajectory equation of point m (B, c) is

If the solution set of inequality ax & # 178; + BX + C ≤ 0 is {x | ≤ 1 or X ≥ 4}, then the trajectory equation of point m (B, c) is

The solution set of inequality ax & # 178; + BX + C ≤ 0 is {x | ≤ 1 or X ≥ 4},
Then the two roots of the equation AX & # 178; + BX + C = 0 are 1 and 4, and a
If the inequality ax & # 178; + BX + C ＜ 0 (a ≠ 0) about X has no solution, then A.A ＜ 0 and B & # 178; - 4ac ＜ 0, B.A ＜ 0 and B & # 178; - 4ac ≤ 0, C.A ＞ 0 and B & # 178; - 4ac ≤ 0, D.A ＞ 0 and B & # 178; - 4ac ＞ 0, the answer is C, ax & # 178; + BX + C ＜ 0, which means y ＜ 0? How to judge the image? The teacher said that because ax & # 178; + BX + C ＜ 0, the image is below the X axis, Why? Why did the teacher draw the last image with the opening upward? In addition, I would like to ask if △ = B & # 178; - 4ac only represents the number of intersections between the image and the X axis, and whether the equation has a solution? (sorry, I really have no money, please explain in detail)
On the inequality ax & # 178; + BX + C ＜ 0 (a ≠ 0) of X, then this inequality is a quadratic inequality of variables. The solution of quadratic inequality of variables is to use the image method to solve, so that y = ax & # 178; + BX + C (a ≠ 0), so it is transformed into a quadratic function
On the solution set of inequality ax & # 178; + BX + C < 0 of X
If the equation AX & # 178; + BX + C = 0 (a > 0) of X has no real root, the solution set of the inequality ax & # 178; + BX + C < 0 is obtained
A > 0, so f (x) = ax & # 178; + BX + C opening upward
And ax & # 178; + BX + C = 0 (a > 0) has no real root, that is, f (x) has no intersection with X axis
So the solution set of inequality ax & # 178; + BX + C < 0 is empty
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If the solution of the equation AX & # 178; + BX + C = 0 (a > 0) with respect to X has no real root, then the discriminant 0)
The opening of the image is upward, and there is no intersection with the X axis
That is ax & # 178; + BX + C
If set a = x power of set y = 2, B = square of set y = x, then the relationship between set a and B?
A is really contained in B
Because a = {y | y = 2 ^ x}, y ∈ (0, + infinity)
B = {y | y = x ^ 2}, y ∈ [0, + infinity)
So B has an element of 0 more than a
So a is really contained in B
If you have studied exponential function and power function, you can easily see them by drawing their function images
If the value of 2Y2 + 3Y + 7 is 8, then the value of 4y2 + 6y-9 is ()
A. 2B. -17C. -7D. 7
The value of ∵ 2Y2 + 3Y + 7 is 8, ∵ 2Y2 + 3Y + 7 = 8, ∵ 2Y2 + 3Y = 1, ∵ 2 (2Y2 + 3Y) = 2 = 4y2 + 6y. Substituting 4y2 + 6y = 2 into 4y2 + 6y-9, 4y2 + 6y-9 = 2-9 = - 7
Say the formula for calculating the circumference and area of a circle
The letter expression of the circle perimeter calculation formula is: C = π D or C = 2 π R, and the letter expression of the circle area calculation formula is: S = π R2
If 4A ^ 2 + k = 9 is a complete square, what is k equal to
9=(±3)²
(2a±3)²
=4a²±12a+9
So k = ± 12
Solve the fractional equation x squared + 5 / X squared - 1 / x = 0
The equation has no real roots
Let a = {(x, y) x & # 178; + Y & # 178; = 4}. B = {(x, y) y = 3 to the power of X}, then the number of subsets of a intersecting B is
A is a circle
The center of the circle is the origin and the radius is 2
Draw the circle and the image of y = 3 ^ X in the same coordinate system
Two intersections, obviously
So the intersection has two elements
So the number of subsets is 2 & # 178; = 4
Given 2x-3y = 0, find the value of (2x ^ 2-3y ^ 2) / (3x ^ 2-2xy)
The solution is 2x-3y = 0, that is, 2x = 3Y to get x = 3 / 2Y, that is, (2x ^ 2-3y ^ 2) / (3x ^ 2-2xy) = [2 (3 / 2Y) ^ 2-3y ^ 2] / [3 (3 / 2Y) ^ 2-3y * y] = [9 / 2Y ^ 2-3y ^ 2] / [27 / 4Y ^ 2-3y & # 178;] = [3 / 2Y ^ 2] / [15 / 4Y ^ 2] = (3 / 2) / (15 / 4) = 3 / 2 * 4 / 15 = 2 / 5
The solution is 2x-3y = 0,
That is, 2x = 3Y
We get x = 3 / 2Y
That is, (2x ^ 2-3y ^ 2) / (3x ^ 2-2xy)
=[2（3/2y）^2-3y^2]/[3(3/2y)^2-3y*y]
=[9/2y^2-3y^2]/[27/4y^2-3y²]
=[3/2y^2]/[15/4y^2]
=(3/2)/(15/4)
=3/2*4/15
=2/5