It is known that the coordinates of a and B are (1,0) and (2,0). The image of quadratic function y = x2 + (A-3) x + 3 has only one intersection point with line AB, and the value range of a is () Why

It is known that the coordinates of a and B are (1,0) and (2,0). The image of quadratic function y = x2 + (A-3) x + 3 has only one intersection point with line AB, and the value range of a is () Why

The meaning of the title is that f (x) = x & # 178; + (A-3) x + 3 has and has only one zero point in the interval [1,2];
According to the distribution of roots, f (1) * f (2) ≤ 0
That is: (a + 1) (2a + 1) ≤ 0
So: - 1 ≤ a ≤ - 1 / 2
If you don't understand, take the limit value. When the function image and ab intersect at points a and B respectively. Get two a values. So we get the value range of A. In this way, only - 1 ≤ a ≤ - 1 / 2 can be obtained, but the answer says that there is a second result, the root sign of a = 3-2 times.???? It's starting to slip up. Drawing, parabola opening upward, to have only one intersection with line AB, there are three cases. The first two are at point a. It's at point B. In both cases, the parabola and the x-axis have two intersections. The third case: there is only one intersection point between the parabola and the x-axis
If you don't understand, take the limit value. When the function image and ab intersect at points a and B respectively. Get two a values. So we get the value range of A. Question: in this way, only - 1 ≤ a ≤ - 1 / 2 can be obtained, but the answer says that there is a second result, the root of a = 3-2 times.????
X / 2 + Y / 3 = - 3 and 7x = 8y-4,
X / 2 + Y / 3 = - 3 multiply both sides of the equation by 24 to get 12x + 8y = - 72
If 7x = 8y-4, 8y = 7x + 4 can be obtained. Substituting the above formula, 12x + 7x + 4 = - 72, 19x + 4 = - 72, x = - 4 can be obtained
If the image of the quadratic function y = - x2 + MX-1 has two different intersections with the line AB whose two ends are a (0,3), B (3,0), then the value range of M is______ .
It is known that the equation of line segment AB is y = - x + 3 (0 ≤ x ≤ 3). Because there are two different intersections between quadratic function image and line segment AB, the equations y = − x2 + MX − 1y = − x + 3, 0 ≤ x ≤ 3 have two different real solutions. The elimination result is: X2 - (M + 1) x + 4 = 0 (0 ≤ x ≤ 3), Let f (x) = X2 - (M + 1) x +
Given the system of equations three x plus two Y equals m plus one, two x plus y equals m minus one, find the minimum integer value of m that makes x greater than y
m＞4
The coordinates of points a and B are (1,0), (2,0) respectively. If there is only one intersection point between the image of quadratic function y = x2 + (A-3) x + 3 and line AB, then the value range of a is______ .
According to the meaning of the topic, it should be divided into two cases to discuss: (1) if the vertex of quadratic function is below the x-axis, if YX = 1 < 0 and YX = 2 ≥ 0, that is 1 + (a − 3) + 3 < 04 + 2 (a − 3) + 3 ≥ 0, the solution is no solution; (2) if YX = 2 < 0 and YX = 1 ≥ 0, that is 1 + (a − 3) + 3 ≥ 04 + 2 (a − 3) + 3 < 0, the solution is - 1 ≤ a ＜ - 12; (2) when the vertex of quadratic function is on the x-axis, △ = 0, that is (A-3) 2-12 = 0, the solution is a = 3 ± 23, and The axis of symmetry is x = - a − 32, so 1 ≤ - a − 32 ≤ 2, so a = 3-23
Ternary linear equation x + y + Z = 5 4x + 8y + 5Z = 300 x + y + 2Z = 67
x+y+z=5 ① 4x+8y+5z=300 ② x+y+2z=67 ③
③-①,z=62
Substituting ①, ②, x + y = - 57, ④, 4x + 8y = - 10, that is, x + 2Y = - 2.5, ⑤
⑤-④,y=54.5
Substituting 4, x = - 111.5
x+y+z=5 x+y+2z=67
By subtracting these two expressions, we get z = 62
Substituting the first formula, we get: x + y = - 57
Substituting into the second formula, we get: 4x + 8y = - 10, that is, x + 2Y = - 2.5
By subtracting these two formulas, we get y = 54.5
Substituting: x + y = - 57
The result is: x = - 111.5
So, (x, y, z) = (- 111.5, 54.5, 62)
The coordinates of points a and B are (1,0), (2,0) respectively. If there is only one intersection point between the image of quadratic function y = x2 + (A-3) x + 3 and line AB, then the value range of a is______ .
If the vertex of quadratic function is below the x-axis, if YX = 1 ＜ 0 and YX = 2 ≥ 0, that is 1 + (a − 3) + 3 ＜ 04 + 2 (a − 3) + 3 ≥ 0, the solution of the inequality is - 1 ≤ a ＜ - 12; if YX = 2 ＜ 0 and YX = 1 ≥ 0, that is 1 + (a − 3) + 3 ≥ 04 + 2 (a − 3) + 3 ＜ 0, the solution is - 1 ≤ a ＜ - 12
X + y + Z = 51, 4x + 8y + 5Z = 300, 4.5X + 9y + 7.5z = 360 to solve the ternary linear equations
① 5, 4.5X + 4.5y + 4.5z = 229.5, ⑤ - ④, 4Y + Z = 96, z = 96-4y, ⑥, ③ - ⑤, 4.5y + 3Z = 130.5, ⑦, ⑥, 4.5y + 288-12y = 130.5, 7.5Y = 157.5, y = 21, ⑥, z = 96-84 = 12, ①, x = 51-21-12 = 18, so x = 18, y
The coordinates of points a and B are (1,0), (2,0) respectively. If there is only one intersection point between the image of quadratic function y = x2 + (A-3) x + 3 and line AB, then the value range of a is______ .
If the vertex of quadratic function is below the x-axis, if YX = 1 ＜ 0 and YX = 2 ≥ 0, that is 1 + (a − 3) + 3 ＜ 04 + 2 (a − 3) + 3 ≥ 0, the solution of the inequality is - 1 ≤ a ＜ - 12; if YX = 2 ＜ 0 and YX = 1 ≥ 0, that is 1 + (a − 3) + 3 ≥ 04 + 2 (a − 3) + 3 ＜ 0, the solution is - 1 ≤ a ＜ - 12
x+y+z=51 4x+8y+5z=300 x+y+2z=67
(3) (1) get, z = 16
Substituting (1) gives x + y = 35 (4)
By substituting (2), 4x + 8y = 220 (5)
(5) - 4 × (4), 4Y = 80
The solution is y = 20
Substituting (4) gives x + 20 = 35
The solution is x = 15
So, the solution of the equations is
x=15,y=20,z=16