What is the relationship between quadratic inequality and quadratic equation? Please explain it in detail

What is the relationship between quadratic inequality and quadratic equation? Please explain it in detail

Bivariate linear inequality is to find the respective ranges of two variables under the condition of satisfying the inequality. The variable is a linear function, such as x + Y > 0, X

A group B problem in the textbook of inequality + quadratic equation in high school mathematics
If the inequality AX2 + BX + C about X

First, judge the sign of a, it is obvious that a is less than 0; then we can conclude that in such a case, m and N are the two roots of AX2 + BX + C = 0, so we can calculate the values of B / A and C / A. m + n = - B / A, Mn = C / A; so C is also less than 0, and B is also less than 0
B / C = - (M + n) / Mn; a / C = 1 / Mn, we can judge the two roots of CX2 BX + a = 0 - 1 / m, - 1 / n
Another 0

In the space quadrilateral ABCD, ab = BC, CD = Da, and point E is the midpoint of AC, then the positional relationship between plane BDE and plane ABC is linear

Plane BDE is perpendicular to plane ABC

Find the maximum value of F (x) = - X & # 178; + 2aX + 2 in [1,2]
If you can, the map ah what trouble with Oh!

f(x)=-x^2+2ax+2
Axis of symmetry x = a
① When a

As shown in the figure, ab = ad, AC = AE, ∠ DAB = ∠ CAE, be and DC intersect at point P

It is proved that: the crossing point a is am ⊥ DP, the perpendicular foot is m, an ⊥ PE, the perpendicular foot is n, ∵ DAB = ∠ CAE (known), and ∵ DAB + ∠ BAC = ∠ CAE + ∠ BAC (properties of the equation), that is, ∵ DAC = ∠ BAE

Equation solution or arithmetic
Master Zhang has to complete the task within the specified time. If he makes 8 parts per hour, he can complete the task 30 minutes in advance. If he makes 6 parts per hour, he has to work 30 minutes overtime. How many hours is the specified time

The original plan was x hours
8（x-0.5）=6（x+0.5）
8x-4=6x+3
2x=7
x=7/2=3.5
The time limit is 3.5 hours

Write out the teaching process of deducing the formula for calculating the area of parallelogram

The area formula of a rectangle is equal to length multiplied by width. Cut both sides of a parallelogram to form a rectangle, then the area of a parallelogram is equal to low multiplied by high

One person is infected with influenza. After two rounds of infection, a total of 100 people are infected with influenza. Then the average number of people infected by one person in each round of infection is ()
A. 8 people B. 9 people C. 10 people D. 11 people

Suppose that the average number of people infected by one person in each round is x, there are (1 + x) people infected after the first round, and (1 + x) + X (1 + x) people infected after the second round, then from the meaning of the question, we can know that 1 + X + X (1 + x) = 100, sort out, X2 + 2x-99 = 0, solve x = 9 or - 11, x = - 11 is not in line with the meaning of the question, then the average number of people infected by one person in each round is 9

It is known that vector a is not equal to vector E. for any t belonging to R, it is proved that e is perpendicular to (A-E) if there is always a-TE ≥ a-e

Let m = vector a · vector E
According to the theme | a-TE | ^ 2 ≥ | A-E | ^ 2
a^2-2mt+t^2≥a^2-2m+1
t^2-2mt+2m-1≥0
For any real number, Δ = (- 2m) ^ 2-4 (2m-1) ≤ 0
m^2-2m+1≤0
（m-1)^2≤0
So only m = 1
That is, vector a · vector E = 1
So only E. (A-E) = e.a-e ^ 2 = 1-1 = 0
That is vector E ⊥ vector (A-E)

Let f (x) = a ^ 3 + BX + C (a is not equal to 0) be an odd function, the tangent of the image at (1, f (1)) is perpendicular to the straight line x-6y-7 = 0, and the minimum value of derivative f (x) is - 12
(1) Find the value of a, B, C
(2) Find the monotone interval of the function f (x), and find the maximum value of the function in the closed interval - 1 to 3

Odd function
f(-x)=-ax^3-bx+c=-f(x)=-ax^3-bx-c
So C = - C
c=0
f'(x)=3ax^2+b
x=1,f'(1)=3a+b
So the tangent slope is 3A + B,
And x-6y-7 = 0, so the slope 3A + B = - 6
If f '(x) = 3ax ^ 2 + B has a minimum value, 3A > 0, and the minimum value = b = - 12
So a = 2, 3A > 0
So a = 2, B = - 12, C = 0
f(x)=2x^3-12x
f'(x)=6x^2-12=0
x=-√2,x=√2,
X = √ 2 in [- 1,3]
-10, f (x) is an increasing function
This is the monotone interval
So f (√ 2) is a minimum,
Obviously it's the minimum
So the maximum value is at the boundary
f(-1)=10,f(3)=18
So the maximum is 18