The first volume of mathematics in the third grade of China Normal University

x^2+6x+30=0

The quadratic coefficient is 1.

The first coefficient is 6,

The constant term coefficient is 30.

An integral equation with only one unknown number and the highest degree of the unknown number is twice is called a quadratic equation or a quadratic equation.

Quadratic equation with one variable has three characteristics:

(1) contains an unknown number;

(2) and the maximum number of unknowns is 2;

(3) is an integral equation. To judge whether an equation is an unary quadratic equation, we should first look at whether it is an integral equation. If so, then tidy it up. If it can be arranged in the form of AX 2+BX+C = 0 (A ≠ 0), then this equation is an unary quadratic equation. There must be an equal sign in it, and there is no unknown denominator.

1, this part of knowledge is elementary mathematics knowledge, which is generally studied in grade three. (But general quadratic function and inverse proportional function will involve the solution of quadratic equation in one variable. )

2. This part is a hot spot in the senior high school entrance examination.

3. The two elements of the equation have the following relations with the numbers in the equation: X 1+X2= -b/a, x 1 x2 = c/a (also known as Vieta's theorem).

4. When the two equations are x 1 and x2, the equation is: x? -(x 1+x2)X+x 1x2=0 (from Vieta's theorem)

5, in the coefficient a>0, b? -4ac & gt； 0 has two unequal real roots, b? When -4ac=0, there are two equal real roots, b? -4ac & lt； 0 has no real root.

general formula

Axe? +bx+c=0(a, b, c are real numbers, a≠0)

For example: x? +2x+ 1=0

matched pattern

a(x+b/2a)？ =(b？ -4ac)/4a

Bipolar type (crossing point)

a(x-x 1)(x-x2)=0

general solution

1. factorization method

(Solvable partial univariate quadratic equation)

Factorization method can be divided into "increasing common factor method", "formula method" (divided into square difference formula and complete square formula) and "cross multiplication". Factorization is obtained by decomposing the left factor of the equation, and the content of factorization was learned in the last semester of Grade 8.

such as

1. Solve the equation: x? +2x+ 1=0

Solution: Use the factor of the complete square formula to get: (x+ 1﹚? =0

Solution: x 1= x2=- 1.

2. Solve the equation x(x+ 1)-3(x+ 1)=0.

Solution: Using the common factor method, we can get: (x-3)(x+ 1)=0.

X-3=0 or x+ 1=0.

∴ x 1=3，x2=- 1

3. Solve equation x 2-4 = 0

Solution: (x+2)(x-2)=0.

X+2=0 or x-2=0.

∴ x 1=-2，x2= 2

Cross multiplication formula:

x？ +(p+q)x+pq=(x+p)(x+q)

Example:

1.ab+b？ +a-b- 2

=ab+a+b？ -b-2

=a(b+ 1)+(b-2)(b+ 1)

=(b+ 1)(a+b-2)

2. Formula method

(You can solve all quadratic equations with one variable)

Root formula

First you have to pass δ = b? From the discriminant of the roots of -4ac, how many quadratic equations are there?

1. When δ = b? -4ac & lt； X has no real root at 0 (junior high school)

2. When δ = b? When -4ac=0, x has two identical real roots, namely x 1=x2.

3. When δ = b? -4ac & gt； When 0, x has two different real roots.

When the judgment is completed, if the equation has roots and can belong to two situations, the equation can be based on the formula: x = {-b √ (b? -4ac)}/2a

In order to find the root of the equation

3. Matching method

(You can solve all quadratic equations with one variable)

Such as: solving the equation: x? +2x-3=0

Solution: move the constant term to: x? +2x=3

Add 1 to both sides of the equation at the same time (to form a completely flat way) to get: x? +2x+ 1=4

Factorization: (x+ 1)? =4

Solution: x 1=-3, x2= 1.

Formula for solving quadratic equation with one variable by collocation method

The quadratic coefficient is converted into one.

Constants should be shifted to the right.

Half of the first coefficient

Add the most equal amount on both sides

4. Open method

(Solvable partial univariate quadratic equation)

Such as: x? -24= 1

Solution: x? =25

x= 5

∴x 1=5 x2=-5

5. Average substitution method

(Solvable partial univariate quadratic equation)

Axe? +bx+c=0

Divide by a at the same time to get x? +bx/a+c/a=0

Let x 1=-b/(2a)+m, x2=-b/(2a)-m (m≥0).

According to x1* x2 = c/a.

Find m.

Then find x 1, x2.

Such as: x? -70x+825=0

The average value is 35, let x 1=35+m, and x2=35-m (m≥0).

x 1*x2=825

So m=20

So x 1=55, x2= 15.

The relationship between the root and coefficient of quadratic equation (the following two formulas are very important and often used in exams)

General formula: ax? The relationship between two roots of +bx+c=0: x 1 and x2;

x 1+x2= -b/a

x 1*x2=c/a

How to choose the simplest solution?

1. See if it can be solved by factorization (in factorization, consider the common factor method first, then the square formula method, and finally the cross multiplication).

See if we can solve the problem directly.

3. Solve by formula method.

4. Finally, consider the matching method (although the matching method can solve all quadratic equations with one variable, it is sometimes too troublesome to solve problems). If you want to take part in the competition, you can follow the following order:

1. Factorization 2. Vieta theorem 3. Discriminant 4. Equation 5. Matching method 6. Square root 7. Root formula 8. Representative.

A detailed example

1, open method:

The direct Kaiping method is a method to solve a quadratic equation with a direct square root. The equation in the form of (x-m) 2 = n (n ≥ 0) is solved by direct Kaiping method, and its solution is x = m √ n.

Example 1. Solve the equation (1)(3x+ 1)? =7 (2)9x？ -24x+ 16= 1 1

Analysis: (1) This equation is obviously easy to do by direct flattening, (2) The left side of the equation is completely flat (3x-4) 2, and the right side =11>; 0, so this equation can also be solved by direct Kaiping method.

(1) Solution: (3x+ 1)? =7

3x+ 1 =ì7

x= ...

∴x 1=...，x2= ...

(2) Solution: 9x? -24x+ 16= 1 1

(3x-4)？ = 1 1

3x-4= √ 1 1

x= ...

∴x 1=...，x2= ...

2. Matching method:

Example 1 Solving Equation 3x by Matching Method? -4x-2=0

Solution: Move the constant term to the right of Equation 3x? -4x=2

Convert the quadratic term into 1: x? -4/3 times =2/3

On both sides of the equation, add half the square of the first coefficient: x? -4/3 times +( -2/3)? = 2/3+(-2/3 )?

Formula: (x-2/3)? = 10/9

Direct square: x-2/3 = √ (10)/3.

∴x 1，x2。

The solution of the original equation is x 1, x2.

3. Formula method: the quadratic equation of one variable is transformed into the general form of AX 2+BX+C, and then the values of various coefficients A, B and C are substituted into the formula for finding the root to get the root of the equation.

Formula: x = [-b √ (b? -4ac)]/2a

When δ = b? -4ac & gt； 0, the root formula is x 1=[-b+√(b? -4ac)]/2a，x2=[-b-√(b？ 4ac)]/2a (two unequal real roots)

When δ = b? When -4ac=0, the root formula is x 1=x2=-b/2a (two equal real roots).

When δ = b? -4ac & lt； 0, the root formula is x 1=[-b+√(4ac-b? )i]/2a，x2=[-b-√(4ac-b？ )i]/2a

(two imaginary roots) (junior high school understands that there is no real root)

Example 3. Formula method to solve equation 2x? -8x=-5

Solution: Turn the equation into a general form: 2x? -8x+5=0

∴a=2，b=-8，c=5

b？ -4ac=(-8)？ -4×2×5 = 64-40 = 24 & gt； 0

∴x= (4 √6)/2

∴ The solution of the original equation is x? =(4+√6)/2，x？ =(4-√6)/2.

4. Factorial decomposition method: the quadratic trinomial on one side of the equation is decomposed into the product of two linear factors, so that the two linear factors are equal to zero respectively, and two linear equations are obtained. The roots obtained by solving these two linear equations are the two roots of the original equation. This method of solving a quadratic equation with one variable is called factorization.

Example 4. Solve the following equation by factorization:

( 1) (x+3)(x-6)=-8

(2) 2x？ +3x=0

(3) 6x？ +5x-50=0 (optional)

(4)x？ -4x+4=0 (optional)

(1) Solution: (x+3)(x-6)=-8 Simplified sorting.

x？ -3x- 10=0 (quadratic trinomial on the left, zero on the right)

(x-5)(x+2)=0 (factorization factor on the left side of the equation)

∴x-5=0 or x+2=0 (converted into two linear equations)

∴x 1=5 x2=-2 is the solution of the equation.

X(2x+3)=0 (factorize the left side of the equation by increasing the common factor)

∴x=0 or 2x+3=0 (converted into two linear equations)

∴x 1=0, x2=-3/2 is the solution of the original equation.

Note: Some students easily lose the solution of x=0 when doing this kind of problem. It should be remembered that a quadratic equation with one variable usually has two solutions.

(3) Solution: 6x? +5x-50=0

(2x-5)(3x+ 10)=0 (pay special attention to symbols when factorizing by cross multiplication).

2x-5 = 0 or 3x+ 10=0.

∴x？ =5/2，x？ =- 10/3 is the solution of the original equation.

(4) solution: x? -4x+4 =0

(x-2)(x-2 )=0

∴x 1=x2=2 is the solution of the original equation.

5. Cross multiplication: the multiplication on the left of the cross is equal to the quadratic term coefficient, the multiplication on the right is equal to the constant term, and the cross multiplication and addition is equal to the linear term coefficient.

Example 5: Solving the following equation by cross multiplication:

Solution: m2+4m- 12=0.

∵ 1,-2

1,6

∴(m-2)(m+6)=0

∴m-2=0 or m+6=0

∴m 1=2； m2=-6

summary

Usually, factorization is the most commonly used method to solve quadratic equations with one variable. When factorization is applied, the equation is written in a general form and the quadratic coefficient is turned into a positive number.

Direct leveling method is the most basic method.

Formula and collocation are the most important methods. Formula method is suitable for any quadratic equation with one variable (some people call it universal method). When using the formula method, in order to determine the coefficient, the original equation must be transformed into a general form, and before using the formula, the value of the discriminant of the root must be calculated to judge whether the equation has a solution.

Matching method is a tool to derive formulas. After mastering the formula method, we can directly use the formula method to solve the quadratic equation of one variable, so we generally don't need to use the matching method to solve the quadratic equation of one variable. However, collocation method is widely used in the study of other mathematical knowledge, and it is one of the three important mathematical methods required to be mastered in junior high school, so we must master it well. Three important mathematical methods: method of substitution, collocation method and undetermined coefficient method.

Extracurricular development

monadic quadratic equation

One-dimensional quadratic equation refers to an integral equation with an unknown number, and the highest degree of the unknown number is quadratic. The general form is ax? +bx+c=0, (a≠0). Around 2000 BC, a quadratic equation with one variable appeared on the clay tablets of Babylon and its solution: it is known that the sum of a number and its reciprocal is equal to a given number, and this number is calculated so that X 1+X2 = b, x1x2 =/kloc-. -bx+ 1=0，

They will answer. It can be seen that the Babylonians already knew the formula for finding the root of the quadratic equation of one variable. But they didn't accept negative numbers at that time, so they omitted negative roots.

Egyptian papyrus literature also involves the simplest quadratic equation, for example: ax 2 = b.

In the 4th and 5th centuries BC, China had mastered the formula for finding the root of a quadratic equation with one variable.

Diophantine (246-330) in Greece only takes the positive root of a quadratic equation, even if both of them are positive roots, he only takes one of them.

In 628 ad, the Yarlung Zangbo River correction system, quadratic equation X? A root formula of +px+q=0.

In Algebra, Arabian Al-Walazimi discussed the solutions of equations, and solved the first and second equations, involving six different forms, so that A, B and C are positive numbers, such as ax? =bx、ax？ =c、ax？ +c=bx、ax？ +bx=c、ax？ =bx+c, etc. It is in line with Diophantine's practice to discuss quadratic equations in different forms. In addition to several special solutions of quadratic equation, Al-Hualazimi also gave the general solution of quadratic equation for the first time, admitting that the equation has two roots and irrational roots, but he doesn't know the imaginary root.

/kloc-in the 6th century, Italian mathematicians began to understand cubic equations with complex roots.

David (1540- 1603) not only knows that the unary equation always has a solution in the range of complex numbers, but also gives the relationship between roots and coefficients.

China Arithmetic Pythagoras "Nine Chapters Arithmetic" Chapter 20 is to find the equivalence of X? +34x-7 1000=0。 China mathematicians also applied interpolation in the study of equations.

Discrimination method of editing this paragraph

First, the analysis of teaching content

In the new textbook of China Normal University, the section "Discriminant formula of roots of quadratic equation in one variable" is used as reading material. The derivation and application of the theorem are relatively simple. But it plays an important role in the whole middle school mathematics. It can not only judge the root of a quadratic equation, but also lay the foundation for studying inequalities, quadratic trinomials, quadratic functions and quadratic curves in the future, and can be used to solve many other comprehensive problems. Through the study of this section, students' exploration spirit, observation, analysis and induction ability, logical thinking ability and reasoning ability are cultivated, and the beauty of classified mathematics thought and simplicity of mathematics is infiltrated into students.

Teaching emphasis: the correct understanding and application of the discriminant theorem and inverse theorem of roots

Teaching difficulties: the application of the discriminant theorem and inverse theorem of roots.

The key to teaching is to thoroughly understand the discriminant theorem of roots and the conditions for the use of their inverse theorems.

Second, the analysis of learning situation

Students have learned four solutions to the quadratic equation of one variable, and have some understanding of their functions. On this basis, the further study of its function is the deepening and summary of previous knowledge. In the way of thinking, students are exposed to the mathematical ideas of classified discussion and induction. Therefore, we can cultivate students' exploration spirit, observation, analysis and induction ability, logical thinking ability and reasoning ability by letting students use their hands and brains.

Third, the teaching objectives

According to the syllabus and the analysis of teaching materials, combined with students' existing knowledge base, the teaching objectives are:

Knowing the root, we usually use the symbol "△" as the discriminant of the root of a quadratic equation with one variable.

Edit this paragraph to solve the problem.

(1) Analyze the meaning of the problem and find out the equivalent relationship between the unknown in the problem and the conditions given in the problem;

monadic quadratic equation

(2) Set unknowns, using algebraic expressions of set unknowns to represent other unknowns;

(3) Find out the equation relation and use it to list the equations;

(4) Solving the equation to find the value of the unknown quantity in the problem;

(5) Check whether the answer meets the meaning of the question and answer it.

Edit this classic example.

1. For the definition of a univariate quadratic equation, we should fully consider the three characteristics of the definition, and don't ignore that the quadratic term coefficient is not 0.

2. When solving a quadratic equation with one variable, according to the characteristics of the equation, choose the solution method flexibly, first consider whether the direct Kaiping method and factorization method can be used, and then consider the formula method.

3. There are both positive and negative discriminant of the root of the unary quadratic equation (a≠0), which can be used to solve the equation and determine the root of the equation (1); (2) Determine the range of roots according to the properties of parameter coefficients; (3) Solve the problem of proof related to roots.

4. The roots and coefficients of a quadratic equation with one variable have many applications: (1) Knowing one root of the equation, finding another root and parameter coefficients without solving the equation; (2) Knowing the equation, finding the value of algebraic expression with two symmetrical expressions and related unknown coefficients; (3) Given two equations, find the root of an unary quadratic equation.

Edit this paragraph, Vieta's theorem

Vieta's theorem is essentially the relationship between roots and coefficients in a quadratic equation with one variable.

The content of Witt theorem.

Univariate quadratic equation ax? +bx+c=0 (a≠0 and △=b? -4ac≥0)

Let two roots be X 1 and X2.

Then x1+x2 =-b/a.

X 1*X2=c/a

Generalization of Vieta Theorem

Vieta's theorem can also be used for higher-order equations. Generally, for the unary equation of degree n ∑ AIX I = 0.

Its roots are expressed as X 1, X2…, Xn.

we have

∑xi=(- 1)^ 1*a(n- 1)/a(n)

∑XiXj=(- 1)？ *A(n-2)/A(n)

…

πxi=(- 1)^n*a(0)/a(n)

Where ∑ is the sum and π is the quadrature.

If the root of an unary quadratic equation in a complex set is 0, then the French mathematician Veda first discovered this relationship between the roots and coefficients of algebraic equations, so people call this relationship Vieta Theorem. History is very interesting. This theorem was obtained by David in16th century. The proof of this theorem depends on the basic theorem of algebra, but Gauss first demonstrated it in 1799.

From the basic theorem of algebra, we can deduce that any unary equation of degree n

There must be a root in a complex set. Therefore, the left end of the equation can be decomposed into the product of linear factors in the range of complex numbers:

Where is the root of the equation. Vieta's theorem is obtained by comparing the coefficients at both ends.

Vieta theorem is widely used in equation theory.

Proof of Vieta Theorem

Let x 1 and x2 be the quadratic equation ax? Two solutions of +bx+c=0.

Yes: a(x-x 1)(x-x2)=0.

So ax? -a(x 1+x2)x+ax 1x2=0

By comparing the coefficients, we can get:

-a(x 1+x2)=b ax 1x2=c

So x1+x2 =-b/ax1x2 = c/a.

Proof of generalization of Vieta's theorem

Let x 1, x2, ..., xn be n solutions of the unary equation ∑ aixi = 0.

Then: an (x-x1) (x-x2) ... (x-xn) = 0.