Formula solution of quadratic equation with one variable Just a formula

The solution of quadratic equation of one variable
1、 Knowledge points:
The quadratic equation of one variable and the linear equation of one variable are integral equations. It is an important content of junior high school mathematics, and also the basis of learning mathematics in the future
Students should pay attention to the foundation
The general form of quadratic equation with one variable is: AX2 + BX + C = 0, (a ≠ 0). It contains only one unknown number, and the highest degree of the unknown number is 2
The integral equation of
The basic idea of solving quadratic equation with one variable is to change it into two quadratic equations with one variable by "reducing degree". There are four kinds of solutions to quadratic equation with one variable
Methods: 1. Direct leveling method; 2. Matching method; 3. Formula method; 4. Factorization method
2、 Methods and examples:
1. Direct leveling method:
The direct flattening method is to solve the quadratic equation of one variable with the direct square root method. The direct flattening method is used to solve the quadratic equation of the form (x-m) 2 = n (n ≥ 0)
The solution of the equation is x = m ±
Solve equation (1) (3x + 1) 2 = 7 (2) 9x2-24x + 16 = 11
Analysis: (1) this equation is obviously easy to do with the direct flattening method, (2) the left side of the equation is the complete square equation (3x-4) 2, the right side = 11 > 0, so
The equation can also be solved by direct square root method
（1）(3x+1)2=7×
∴(3x+1)2=5
Ψ 3x + 1 = ± (be careful not to lose the solution)
∴x=
The solution of the original equation is X1 =, x2=
（2） 9x2-24x+16=11
∴(3x-4)2=11
∴3x-4=±
∴x=
The solution of the original equation is X1 =, x2=
2. Collocation method: solve the equation AX2 + BX + C = 0 (a ≠ 0) by collocation method
First move the constant C to the right of the equation: AX2 + BX = - C
The quadratic coefficient is reduced to 1: x2 + X=-
On both sides of the equation, add the square of half of the coefficient of the first term: x2 + X + () 2 = - + () 2
The left side of the equation becomes a complete square: (x +) 2=
When b2-4ac ≥ 0, x + = ±
X = (this is the root formula)
Solving the equation 3x2-4x-2 = 0 by the collocation method
Move the constant term to the right of the equation 3x2-4x = 2
The quadratic coefficient is reduced to 1: x2-x=
Add the square of half of the coefficient of the first term to both sides of the equation: x2-x + () 2 = + () 2
Formula: (x -) 2=
Direct square root: X - = ±
∴x=
The solution of the original equation is X1 =, X2 =
3. Formula method: change the quadratic equation of one variable into a general form, and then calculate the value of the discriminant △ = b2-4ac. When b2-4ac ≥ 0, change the
The root of the equation can be obtained by substituting the values of coefficients a, B and C into the root formula x = (b2-4ac ≥ 0)
Solving equation 2x2-8x = - 5 by formula method
The equation is reduced to a general form: 2x2-8x + 5 = 0
∴a=2, b=-8, c=5
b2-4ac=(-8)2-4×2×5=64-40=24>0
∴x= = =
The solution of the original equation is X1 =, X2 =
4. Factorization: transform the equation into zero on one side, and decompose the quadratic trinomial on the other side into the product of two first-order factors
Two first-order factors are equal to zero respectively, and two first-order equations with one variable are obtained. The roots obtained by solving the two first-order equations with one variable are the two roots of the original equation
This method of solving quadratic equation of one variable is called factorization
The following equation is solved by factorization
(1) (x+3)(x-6)=-8 (2) 2x2+3x=0
(3) 6x2 + 5x-50 = 0 (optional) (4) x2-2 (+) x + 4 = 0 (optional)
(1) (x + 3) (X-6) = - 8
X2-3x-10 = 0 (quadratic trinomial on the left and zero on the right)
(X-5) (x + 2) = 0 (factorization on the left side of the equation)
Ψ X-5 = 0 or x + 2 = 0 (converted into two linear equations with one variable)
X 1 = 5, x 2 = - 2 are the solutions of the original equation
(2)2x2+3x=0
X (2x + 3) = 0
Ψ x = 0 or 2x + 3 = 0 (converted into two linear equations with one variable)
X 1 = 0, x 2 = - are the solutions of the original equation
Note: some students are easy to lose the solution of x = 0 when doing this kind of problem
(3)6x2+5x-50=0
(2x-5) (3x + 10) = 0
Ψ 2x-5 = 0 or 3x + 10 = 0
X 1 =, x 2 = - are the solutions of the original equation
(4) X 2-2 (+) x + 4 = 0 (∵ 4 can be decomposed into 2.2, which can be solved by factorization)
(x-2)(x-2 )=0
X 1 = 2, x 2 = 2 are the solutions of the original equation
Summary:
In general, the most commonly used method to solve quadratic equation of one variable is factorization. When factorization is applied, the equation should be written as a general equation first
At the same time, the coefficient of quadratic term should be positive
Direct leveling is the most basic method
Formula method and collocation method are the most important methods. Formula method is suitable for any quadratic equation with one variable (some people call it universal method)
In order to determine the coefficient, the original equation must be transformed into a general form, and the value of the discriminant should be calculated before using the formula
Whether there is a solution
The collocation method is a tool to deduce formulas. After mastering the formula method, you can directly use the formula method to solve the quadratic equation of one variable, so the collocation method is generally not used
Solving quadratic equation with one variable. However, collocation method is widely used in learning other mathematical knowledge. It is three important mathematical methods that junior middle school requires to master
Three important mathematical methods: substitution method, collocation method and undetermined coefficient method
Solve the following equation in an appropriate way
（1）4(x+2)2-9(x-3)2=0 （2）x2+(2-)x+ -3=0
（3） x2-2 x=- （4）4x2-4mx-10x+m2+5m+6=0
Analysis: (1) first of all, we should observe the characteristics of the topic, not blindly do the multiplication operation. After observation, we found that the square difference can be used on the left side of the equation
A formula decomposes a factor into the product of two first-order factors
(2) The left side of the equation can be factorized by cross multiplication
(3) After being transformed into a general form, it is solved by formula method
(4) The equation is transformed into 4x2-2 (2m + 5) x + (M + 2) (M + 3) = 0, and then can be decomposed by cross phase multiplication
（1）4(x+2)2-9(x-3)2=0
[2(x+2)+3(x-3)][2(x+2)-3(x-3)]=0
(5x-5)(-x+13)=0
5x-5 = 0 or - x + 13 = 0
∴x1=1,x2=13
（2） x2+(2- )x+ -3=0
[x-(-3)](x-1)=0
X - (- 3) = 0 or X-1 = 0
∴x1=-3,x2=1
（3）x2-2 x=-
X2-2 x + = 0
△=(-2 )2-4 ×=12-8=4>0
∴x=
∴x1=,x2=
（4）4x2-4mx-10x+m2+5m+6=0
4x2-2(2m+5)x+(m+2)(m+3)=0
[2x-(m+2)][2x-(m+3)]=0
2X - (M + 2) = 0 or 2x - (M + 3) = 0
∴x1= ,x2=
Example 6. Find two roots of equation 3 (x + 1) 2 + 5 (x + 1) (x-4) + 2 (x-4) 2 = 0
Analysis: if this equation first do the power, multiplication, merge the similar terms into a general form, then do it will be more cumbersome, carefully observe the topic, I will
We find that if x + 1 and x-4 are regarded as a whole, then the left side of the equation can be decomposed by cross multiplication (in fact, the square of substitution is used)
(France)
[3(x+1)+2(x-4)][(x+1)+(x-4)]=0
That is, (5x-5) (2x-3) = 0
∴5(x-1)(2x-3)=0
(x-1)(2x-3)=0
Ψ X-1 = 0 or 2x-3 = 0
X 1 = 1, x 2 = is the solution of the original equation
Using the collocation method to solve the quadratic equation x2 + PX + q = 0 with respect to X
X2 + PX + q = 0
X2 + PX = - Q (the constant term is moved to the right of the equation)
X2 + PX + () 2 = - Q + () 2
(x +) 2 = (formula)
When p2-4q ≥ 0, ≥ 0 (p2-4q must be classified)
∴x=- ±=
∴x1= ,x2=
When p2-4q

The formula of solving quadratic equation of one variable
How to change x ^ 2-3x + 2 = 0 to (x-1) (X-2) = 0
What is the middle transformation formula? (all)

The integral of quadratic trinomial (quadratic coefficient is 1)
x^2+(a+b)x+ab=(x+a)(x+b)
In decomposition, cross multiplication is used
Example: x ^ 2-3x + 2 = 0
1 -1
1 -2
When writing, write 1 as X, and then write - 1 (that is, write horizontally),
We get (x-1) (X-2) = 0

How to solve quadratic equation with one variable?

http://wenku.baidu.com/view/76acd64e767f5acfa1c7cdbe.html In general, the solutions of quadratic equation of one variable are: (Note: the following ^ means square.) 1. Direct flattening method. For example: x ^ 2-4 = 0, x ^ 2 = 4, x = ± 2 (because x is the square root of 4) ∧ X1 = 2, x2 = - 2, quadratic collocation method

A farm planted 10 mu of pumpkin last year, and the yield per mu was 2000 kg. According to the market demand, this year, the farm expanded the planting area, and planted all high-yield new varieties of pumpkin. It is known that the growth rate of pumpkin planting area is twice that of the yield per mu. This year, the total yield of pumpkin is 60000 kg, and the growth rate of pumpkin yield per mu is calculated

Suppose the growth rate of pumpkin yield per mu is x, then the growth rate of planting area is 2x. According to the meaning of the question, we get 10 (1 + 2x) · 2000 (1 + x) = 60000. The solution is: X1 = 0.5, X2 = - 2 (not the meaning, rounding off). Answer: the growth rate of pumpkin yield per mu is 50%

Mathematics in the third grade of junior high school (on solving practical problems with quadratic equation of one variable)
A computer virus spreads very fast. If a computer is infected, 81 computers will be infected after two rounds of infection. Please use the knowledge you have learned to analyze how many computers will be infected by a computer in each round of infection. If the virus is not effectively controlled, will more than 700 computers be infected after three rounds of infection?
If one computer infects x, then
x＋1＋x（x＋1）＝81
The solution is: X1 = 8, X2 = - 10 (rounding off)
On average, one computer will infect 8 computers
(8 + 1) ^ 3 = 729 (sets)
After three rounds of infection, 729 computers were infected
(this is a positive solution)

Set up a round of X infection stations
x*x=81
x1=9
X2 = - 9 (shed)
So one will infect nine
Three wheel x * x * x Table = 9 * 9 * 9 = 729 > 700
So it will

Junior three mathematics (solving practical problems with quadratic equation of one variable)
Hongda car rental company has a total of 120 cars to rent out, the daily rent of each car is 160 yuan, and the rental business is in short supply every day. According to the survey, the daily rent of a car increases by 10 yuan, and the number of cars to rent decreases correspondingly by 6. How many yuan can the company increase the daily rent of each car to make the total income of the company's daily rent the highest? What is the highest?

（160+10X）*（120-6X）
=19440-60（X-2）*（X-2）
The number requires a maximum of 19440 minus the square of a number
So it must be minus the square of zero
So x = 2
That is, the rent will be increased by 10 * 2 = 20 yuan
The highest income is 19440

Quadratic equation of one variable and its solution
Given that the solution of equation (x + a) (A-3) = 0 is the same as that of equation x2-2x-3 = 0, then a =. It doesn't matter how to solve the quadratic equation of one variable to make a test paper. Except for the calculation problem, even the blank filling and multiple-choice problems need to be solved by equations. I'm not very good at these. Does anyone teach me how to solve them? For example, the root of the quadratic equation of one variable x2 + 4x + K2 + 2k-3 = 0 of X is 0, Find the value of K and another root of the equation

X ^ 2-2x-3 = 0, the solution is x = 3, - 1, the solution of the first equation is 3, - A, so a = 1
X ^ 2 + 4x + K ^ 2 + 2k-3 = 0 has a root of 0. Substitute x = 0 into the equation, K ^ + 2k-3 = 0
The results show that k = - 3, 1
The original equation is reduced to x ^ 2 + 4x = 0, so the other root is - 4

The solution of two mathematics problems in the third grade of junior high school
1. To make a cuboid box with a volume of 396cm3, a height of 6cm and a bottom length 5cm more than the width, how many centimeters should the floor be
2. It is known that the equation 4x2 - (K + 2) x + k-1 = 0 has two equal real roots
Find: 1) the value of K, 2) the root of the equation
(PS: 2 in 4x2 is a quadratic power, because I can't play it, so when you submit a question, you should explain it.)

Let the length be x, then the width be X-5, which means: X * (X-5) * 6 = 396, i.e. x ^ 2-5x-66 = 0 (X-11) (x + 6) = 0, because x > 0, so X-11 = 0, i.e. x = 11, so the length is 11cm, the width is 6cm, and the square is represented by ^ 2

What is 5m cubic - 5MN square?

5m cubic - 5MN square equals 5m (M & # 178; - N & # 178;) = 5m (M + n) (m-n)

Formula solution of quadratic equation with one variable Just a formula