Given the parabola y = 3ax ^ 2 + 2bx + C, if a = b = 1, and when - 1 is less than x is less than 1, there is only one common point between the parabola and the x-axis, find the value range of C If a + B + C = 0 and X1 = 0, the corresponding Y1 > 0, X2 = 1, the corresponding Y2 > 0, try to judge when 0

Given the parabola y = 3ax ^ 2 + 2bx + C, if a = b = 1, and when - 1 is less than x is less than 1, there is only one common point between the parabola and the x-axis, find the value range of C If a + B + C = 0 and X1 = 0, the corresponding Y1 > 0, X2 = 1, the corresponding Y2 > 0, try to judge when 0

(1)
Parabola: y = 3x & # 178; + 2x + C
① When △ = 0
That is △ = 4-12c = 0
c=⅓
The intersection point: x = - &; in the range (- 1,1)
So C = 1 / 3
② When △ > 0 and the left intersection point is in the range of (- 1,1), the
That is, C ＜ &; and f (- 1) ＞ 0, f (1) ＜ 0
F (- 1) = 3-2 + C = 1 + C > 0, that is, C > 1
F (1) = 3 + 2 + C = 5 + C ＜ 0, that is, C ＜ - 5
C has no solution
③ When △ > 0 and the right intersection point is in the range of (- 1,1)
That is, C ＜ &; and f (- 1) ＜ 0, f (1) ＞ 0
F (- 1) = 3-2 + C = 1 + C ＜ 0, that is, C ＜ - 1
F (1) = 3 + 2 + C = 5 + C ＞ 0, that is, C ＞ - 5
∴-5＜x＜-1
④ When C = - 5, the root of 3x & # 178; + 2x-5 = 0 is - 5 / 3,1, ｛ C = - 5 does not meet the requirements
⑤ When C = - 1, the root of 3x & # 178; + 2x-1 = 0 is - 1,1 / 3, ｛ C = - 1
To sum up, C = &; or - 5 ＜ x ≤ - 1
(2)
When X1 = 0, Y1 = C > 0
f(1)=3a+2b+c>0
3a+2b+c=a+b+c+2a+b=2a+b＞0
A > - (a + b) = C > 0, i.e. the parabolic opening is upward
b=-(a+c)0
The parabola must have an intersection with the x-axis
At the same time, f (0) > 0, f (1) < 0
When 0 ＜ x ＜ - B / 3a, the parabola decreases monotonically, and there must be a common point between the parabola and the x-axis
When 0 ＜ x ＜ 1, there must be a common point between the parabola and the x-axis

Known parabola y = 3ax ^ 2 + 2bx + C
If a + B + C = 0 and X1 = 0, the corresponding Y1 is greater than 0, and if x2 = 1, the corresponding Y2 is greater than 0
Try to judge 0

From the image, when a is less than 0, and 0

It is known that a, B and C are all positive integers, and the parabola y = AX2 + BX + C has two different intersections A and B with the x-axis. If the distance from a and B to the origin is less than 1, the minimum value of a + B + C is obtained
I have found the answer: according to the meaning of the question, the equation AX2 + BX + C = 0 has two different roots, both of which are in (｛ 61485; 1,0), so A-B + C > 0, C / A0 shows A-B + C > 1 and a > C, so a + C
>B + 1 > 2 root sign AC + 1 can get (root sign a-root sign C) ^ 2 > 1, thus root sign a > root sign C + 1, so a > 4 and b > 2 root sign AC > 2 root sign 5 * 1 > 4. It can be seen that the minimum integer of ABC is 5, 1. The minimum value of a + B + C is 11?
It is proved that ﹣ a + B + C = 11 is the smallest

b*b-4ac>0
b+sqr(b*b-4ac)sqr(b*b-4ac)
Square on both sides
a>b-c
b>2a
a>c
that will do
Minimum 1,2,7
=10

a. B and C are integers, and the parabola y = ax ^ 2 + BX + C has two different intersections A and B with the X axis. If the distance from a and B to the origin is less than 1, find the minimum value of a + B + C

The point of intersection with the x-axis, that is, ax ^ 2 + BX + C = 0, can be solved by | X|