It is known that the vertex of the image of quadratic function f (x) is m (- 1.5,49), and the difference between the two equations f (x) = 0 = 7. The analytic expression of quadratic function is obtained

It is known that the vertex of the image of quadratic function f (x) is m (- 1.5,49), and the difference between the two equations f (x) = 0 = 7. The analytic expression of quadratic function is obtained

It is known that the vertex of quadratic function f (x) image is m (- 1.5,49), so f (x) = a (x + 1.5) ^ 2 + 49 = ax ^ 2 + 3ax + 9 / 4A + 49, so X1 + x2 = - B / a = - 3x1x2 = C / a = (9a + 196) / 4A and (x1-x2) ^ 2 = (x1 + x2) ^ 2-4x1x2 = 196 / A, so: x1-x2 = 14 radical a / a = 7, so A1 = 4, A2 = 0 (rounding) so f

Given the function f (x) = - A & # 178; X & # 179 / / 3 + ax & # 178 / / 2 + CX (a is not equal to 0), when a ≥ 1 / 2, if the real number equation f '(x) = 0 of X has two real number roots m, N, and | m | ≤ 1, | n | ≤ 1

f(x)=-a²x³/3+ax²/2+cx
f'(x)= -a²x²+ax+c
△=a²+4a²c≥0
c≥-1/4 ①
Let f '(x) = 0
-a²x²+ax+c=0
x=[-a±a√(1+4c)]/(-2a²)
x=[-1±√(1+4c)]/(-2a)
| [-1+√(1+4c)]/(-2a) |≤1 | 1-√(1+4c) |≤2a c≤a²+a ②
| [-1-√(1+4c)]/(-2a) |≤1 | 1+√(1+4c) |≤2a c≤a²-a ③
According to ①, ② and ③
-1/4≤c≤a²-a

Given the quadratic function f (x) = ax ^ 2 + BX + C, this paper proves that the equation f (x) = 0.5 [f (0) + F (1)] has two unequal real roots, and one root is in the interval (0,1)
It should be in the interval [0,1]
Should it be in [0,
Second floor, why a + B = 0, just a = b = 0?

F (0) + F (1) = C + (a + B + C) = a + B + 2C; f (x) = 0.5 [f (0) + F (1)] that is ax ^ 2 + BX + C = 0.5A + 0.5B + C; → ax ^ 2 + bx-0.5 (a + b) = 0; its discriminant △ = B ^ 2 - 4A × [- 0.5 (a + b)] = B ^ 2 + (2a ^ 2 + 2Ab) = (a + b) ^ 2 + A ^ 2 because f (x) is a quadratic function, so a ≠ 0; then a ^ 2 ＞ 0; then the discriminant △ =

Let f (x) = x ^ 2 + X + C, (c > 0), if f (x) = 0 has two real roots X1 and X2, (x1 < x2)
(1) Finding the value range of positive real number C
(2) Finding the value range of x2-x1
Require reasoning, calculation steps The answer is accurate thank you!

(1) if f (x) = 0 has two real roots X1 and X2,
△＞0∴1-3c＞0 ,∴0＜c＜1/3
⑵x1+x2=-1,x1·x2=c
x1＜x2,x2－x1＞0
∴（x2－x1）^2=（x1+x2）^2-4x1·x2=1-4c
∴x2－x1=√1-4c
∵2/3＜1-4c＜1
∴√6/3＜x2－x1＜1