Quadratic equation... Definition

In an equation, an integral equation that contains only one unknown number and the highest degree of the unknown is 2 is called a quadratic equation.

The quadratic equation has three characteristics: (1) it contains only one unknown; (2) the highest degree of the unknown is 2; (3) it is an integral equation. To determine whether an equation is a quadratic equation, first check whether it is an integral equation, and if so, sort it out. If it can be organized into the form of ax^2+bx+c=0(a≠0), then this equation is a quadratic equation of one variable. General solution methods

1. Combination method (can solve all quadratic equations of one variable)

2. Formula method (can solve all quadratic equations of one variable)

3. Factoring method (can solve some quadratic equations)

4. Extraction method (can solve some quadratic equations) The method of solving quadratic equations is really not good (you buy a Casio fx -500 or 991 calculators can solve equations, but in general form)

1. Key points of knowledge:

Quadratic equations of one variable and linear equations of one variable are both integral equations. It is a key content of junior high school mathematics and the basis for learning mathematics in the future. It should attract the attention of students.

The general form of a quadratic equation is: ax2+bx+c=0, (a≠0), which contains only one unknown, and the highest degree of the unknown is 2

The integral equation of .

The basic idea of ??solving a quadratic equation of one variable is to transform it into two linear equations of one variable by "reducing the degree". There are four solutions to quadratic equations

: 1. Direct square root method; 2. Combination method; 3. Formula method; 4. Factoring method.

2. In-depth explanation of methods and examples:

1. Direct square root method:

The direct square root method is a method of solving quadratic equations of one variable using direct square root. . Use the direct square root method to solve the equation of the form (x-m)2=n (n≥0)

and the solution is x=m± .

Example 1. Solve equation (1) (3x+1)2=7 (2) 9x2-24x+16=11

Analysis: (1) This equation is obviously easy to solve using the direct square root method. (2) The left side of the equation It is a perfect square form (3x-4)2, and the right side = 11>0, so

This equation can also be solved by the direct square root method.

(1) Solution: (3x+1)2=7×

∴(3x+1)2=5

∴3x+1=± (Be careful not to lose the solution)

∴x=

∴The solution of the original equation is x1=,x2=

(2) Solution: 9x2-24x+ 16=11

∴(3x-4)2=11

∴3x-4=±

∴x=

∴ The solution of the original equation is x1=,x2=

2. Combination method: Use the combination method to solve the equation ax2+bx+c=0 (a≠0)

First move the constant c to the right side of the equation: ax2+bx=-c

Move The coefficient of the quadratic term is reduced to 1: x2+x=-

Add half the square of the coefficient of the linear term to both sides of the equation: x2+x+( )2=- +( )2

< p> The left side of the equation becomes a perfect square: (x+ )2=

When b2-4ac≥0, x+ =±

∴x= (this is the root formula)

Example 2. Use the compound method to solve the equation 3x2-4x-2=0

Solution: Move the constant term to the right side of the equation 3x2-4x=2

Change the coefficient of the quadratic term to 1: x2 -x=

Add half the square of the linear coefficient 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: transform the quadratic equation into a general form, and then calculate the value of the discriminant △=b^2-4ac. When b^2-4ac≥0, put the coefficients a, b of each term

, the value of c is substituted into the root finding formula x=(b^2-4ac≥0) to get the root of the equation.

Example 3. Use the formula method to solve the equation 2x2-8x=-5

Solution: Convert the equation into the general form: 2x2-8x+5=0

∴a=2, b=-8 , c=5

b^2-4ac=(-8)2-4×2×5=64-40=24>0

∴x= = =

p>

∴The solution of the original equation is x1=,x2= .

4. Factoring method: Transform the equation so that one side is zero, decompose the quadratic trinomial on the other side into the product of two linear factors, let

the two linear factors are equal to zero respectively, Two linear equations of one variable are obtained. The roots obtained by solving these two linear equations of one variable are the two roots of the original equation.

This method of solving quadratic equations is called factoring.

Example 4. Use factoring to solve the following equations:

(1) (x+3)(x-6)=-8 (2) 2x2+3x=0

(3) 6x2+5x-50=0 (optional) (4)x2-2( + )x+4=0 (optional)

(1) Solution: (x+3)(x-6 )=-8 Simplify and get

x2-3x-10=0 (the left side of the equation is a quadratic trinomial and the right side is zero)

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

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

∴x1= 5,x2=-2 is the solution of the original equation.

(2) Solution: 2x2+3x=0

x(2x+3)=0 (Use the common factor method to factor the left side of the equation)

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

∴x1=0, x2=- is the solution of the original equation.

Note: Some students tend to lose the solution x=0 when doing this kind of question. You should remember that there are two solutions to the quadratic equation.

(3) Solution: 6x2+5x-50=0

(2x-5)(3x+10)=0 (Pay special attention to the sign when factoring by cross multiplication Don’t make mistakes)

∴2x-5=0 or 3x+10=0

∴x1=, x2=- is the solution of the original equation.

(4) Solution: x2-2(+ )x+4 =0 (∵4 can be decomposed into 2·2, ∴This question can be factored)

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

∴x1=2 ,x2=2 is the solution of the original equation.

Summary:

Generally, the most commonly used method to solve quadratic equations of one variable is the factoring method. When applying the factoring method, the equation must first be written in general

p>

form, and the coefficient of the quadratic term should be turned into a positive number.

The direct opening method is the most basic method.

The formula method and the combination method are the most important methods. The formula method is applicable to any quadratic equation of one variable (some people call it the universal method). When using the formula

method, the original equation must be transformed into a general form in order to determine the coefficients, and the formula should be used before First calculate the value of the discriminant to determine whether the equation

has a solution.

The combination method is a tool for deriving formulas. After mastering the formula method, you can directly use the formula method to solve quadratic equations of one variable, so generally there is no need to use the combination method

to solve quadratic equations of one variable. However, the matching method is widely used in learning other mathematical knowledge. It is one of the three important mathematical methods required to be mastered in junior high schools.

It must be mastered well. (Three important mathematical methods: substitution method, combination method, and undetermined coefficient method).

Example 5. Solve the following equations using appropriate methods. (Optional)

(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, you should observe whether the question has any characteristics, do not blindly Do multiplication operations. After observation, we found that the left side of the equation can be factored using the square difference

formula and turned into the product of two linear factors.

(2) The cross multiplication method can be used to factorize the left side of the equation.

(3) After converting it into a general form, use the formula method to solve it.

(4) Transform the equation into 4x2-2(2m+5)x+(m+2)(m+3)=0, and then use the cross multiplication method to factorize.

(1) Solution: 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) Solution: x2+(2- )x+ -3=0

[x-(-3)](x-1)=0

x-(-3)=0 or x-1=0

∴x1=-3, x2=1

(3) Solution: x2-2 x=-

x2-2 x+ =0 (convert to general form first)

△=( -2 )2-4 ×=12-8=4>0

∴x=

∴x1=,x2=

(4) Solution: 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 the two roots of the equation 3(x+1)2+5(x+1)(x-4)+2(x-4)2=0. (Optional)

Analysis: It will be more cumbersome to do the exponentiation, multiplication, and merging of similar terms into a general form before doing this equation. Observe the question carefully, we

It is found that if x+1 and x-4 are regarded as a whole, the left side of the equation can be factored by the cross multiplication method (actually, the method of substitution is used)

< p> Solution: [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

∴x-1=0 or 2x-3=0

∴x1=1,x2= is the solution of the original equation.

Example 7. Use the combination method to solve the quadratic equation x2+px+q=0 about x

Solution: x2+px+q=0 can be transformed into

x2+px=-q (The constant term is moved to the right side of the equation)

x2+px+( )2=-q+()2 (The square of half the coefficient of the linear term is added to both sides of the equation)

(x+) 2= ??(formula)

When p2-4q≥0, ≥0 (p2-4q must be classified and discussed)

∴x=- ±=

< p> ∴x1= ,x2=

When p2-4q<0, <0, the original equation has no real roots.

Note: This question is an equation containing letter coefficients. There are no additional conditions for p and q in the question. Therefore, during the process of solving the problem, you should always pay attention to the requirements for the values ??of letters

. Conduct classified discussions when necessary.

Exercises:

(1) Use appropriate methods to solve the following equations:

1. 6x2-x-2=0 2. (x+5) (x-5)=3

3. x2-x=0 4. x2-4x+4=0

5. 3x2+1=2x 6. (2x+3 )2+5(2x+3)-6=0

(2) Solve the following equations about x

1.x2-ax+-b2=0 2. x2-( + )ax+ a2=0

Practice reference answers:

(1) 1.x1=-,x2= 2.x1=2,x2=-2

< p> 3.x1=0,x2= 4.x1=x2=2 5.x1=x2=

6. Solution: (Consider 2x+3 as a whole, factor the left side of the equation )

[(2x+3)+6][(2x+3)-1]=0

That is (2x+9)(2x+2)=0

p>

∴2x+9=0 or 2x+2=0

∴x1=-,x2=-1 is the solution of the original equation.

(2) 1. Solution: x2-ax+( +b)( -b)=0 2. Solution: x2-(+ )ax+ a· a=0

[x-( +b)] [x-( - b)]=0 (x- a)(x-a)=0

∴x-( +b)=0 or x-( -b) =0 x- a=0 or x-a=0 < /p>

∴x1= +b, x2= -b is ∴x1= a, x2=a is

the solution of the original equation. solution of the original equation.

Test (answers are below)

Multiple choice questions

1. The root of the equation x(x-5)=5(x-5) is ( )

A. x=5 B. x=-5 C. x1=x2=5 D. x1=x2= -5

2. The value of polynomial a2+4a-10 is equal to 11, then the value of a is ( ).

A, 3 or 7 B, -3 or 7 C, 3 or -7 D, -3 or -7

3. If the sum of the quadratic term coefficient, linear term coefficient and constant term in the quadratic equation ax2+bx+c=0 is equal to zero, then the equation must have a

root ( ).

A, 0 B, 1 C, -1 D, ±1

4. The condition that the quadratic equation ax2+bx+c=0 has a root that is zero is ( ).

A, b≠0 and c=0 B, b=0 and c≠0

C, b=0 and c=0 D, c=0

5． The two roots of the equation x2-3x=10 are ( ).

A, -2, 5 B, 2, -5 C, 2, 5 D, -2, -5

6. The solution to the equation x2-3x+3=0 is ( ).

A, B, C, D, no real root

7. The solution to equation 2x2-0.15=0 is ( ).

A. x= B. x=-

C. x1=0.27, x2=-0.27 D. x1=, x2=-

8. After the left side of the equation x2-x-4=0 is arranged into a perfect square, the resulting equation is ( ).

A. (x-)2= B. (x- )2=-

C. (x- )2= D. None of the above answers are correct

< p> 9. It is known that the quadratic equation of one variable x2-2x-m=0, and the equation after solving the equation using the formula method is ( ).

A. (x-1)2=m2+1 B. (x-1)2=m-1 C. (x-1)2=1-m D. (x-1) 2=m+1

Answers and analysis

Answer: 1.C 2.C 3.B 4.D 5.A 6.D 7.D 8.C 9. D

Analysis:

1. Analysis: Shift the terms to get: (x-5)2=0, then x1=x2=5,

Note: Do not easily divide both sides of the equation by an integer. In addition, the quadratic equation has real roots, so it must It's two.

2. Analysis: According to the question: a2+4a-10=11, the solution is a=3 or a=-7.

3. Analysis: According to the meaning of the question: there is a+b+c=0, the left side of the equation is a+b+c, and when x=1, ax2+bx+c=a+b+c means that when x =1

When the equation is established, there must be a root of x=1.

4. Analysis: If one root of the quadratic equation ax2+bx+c=0 is zero,

then ax2+bx+c must have a factor x, and it exists only when c=0 The common factor is x, so c=0.

In addition, you can also substitute x=0 to get c=0, which is simpler!

5. Analysis: The original equation becomes x2-3x-10=0,

Then (x-5)(x+2)=0

x-5=0 or x+2 =0

x1=5, x2=-2.

6. Analysis: Δ=9-4×3=-3<0, then the original equation has no real roots.

7. Analysis: 2x2=0.15

x2=

x=±

Pay attention to the simplification of the radical expression, and be careful not to lose the root when taking the square root directly.

8. Analysis: Multiply both sides by 3 to get: x2-3x-12=0, and then according to the linear term coefficient formula, x2-3x+(-)2=12+(-)2,

Organize as: (x -)2=

The equation can be deformed using the equality property, and when x2-bx is formulated, the formula term is half the square of the linear term coefficient -b.

9. Analysis: x2-2x=m, then x2-2x+1=m+1

Then (x-1)2=m+1.

High School Entrance Examination Analysis

Evaluation of test questions

1. (Gansu Province) The root of the equation is ( )

(A) (B) (C) or (D) or

Comment: Because the quadratic equation has two roots, So use the elimination method to eliminate options A and B, and then use the verification method to select the correct option from options C and D.

You can also use the factoring method to solve this equation to get the result. You can also check the options. Options A and B only consider one aspect and forget about one variable

The quadratic equation has two roots, so it is wrong. In option D, x=-1 does not make the left and right sides of the equation equal, so it is also wrong. Wrong. The correct option is

C.

In addition, students often divide both sides of the equation by an integer at the same time, causing the equation to lose roots. This mistake should be avoided.

2. (Jilin Province) The root of a quadratic equation of one variable is __________.

Comment: The idea is to use the factoring method or the formula method to solve it according to the characteristics of the equation.

3. (Liaoning Province) The root of the equation is ( )

(A) 0 (B) –1 (C) 0, –1 (D) 0, 1

Comments: Idea: Because the equation is a quadratic equation, it has two real roots. Using the elimination method and verification method, the correct option can be selected as C, while the two options A and

B have only one root. Option D A number is not a root of an equation. Alternatively, you can use the method of directly finding the roots of the equation.

4. (Henan Province) It is known that one root of the quadratic equation of x is –2, then k=__________.

Comment: k=4. Substituting x=-2 into the original equation, construct a quadratic equation about k, and then solve it.

5. (Xi'an City) Using the direct square root method to solve the equation (x-3)2=8, the root of the equation is ( )

(A) x=3+2 (B) x=3-2

(C) x1=3+2, x2=3-2 (D) x1=3+2, x2=3-2

Comment: Use the method of solving equations to solve directly. Yes, or you don't need to calculate. If there is a solution to the quadratic equation of one variable, then there must be two solutions and the square root of 8

and you can choose the answer.

Extracurricular development

Quadratic equation of one variable

A quadratic equation of one variable refers to a equation that contains an unknown number and the highest degree term of the unknown number is Integral equations of degree 2

. The general form is

ax2+bx+c=0, (a≠0)

Around 2000 BC, quadratic equations and their solutions had appeared among the ancient Babylonians In the clay tablet document of , x+ =b,

x2-bx+1=0,

They make ( )2; then make it, and then get the solutions: + and -. It can be seen that the Babylonians already knew the root formula of the quadratic

equation. But they did not accept negative numbers at the time, so the negative roots were omitted.

The simplest quadratic equations are also involved in Egyptian papyrus documents, for example: ax2=b.

In the 4th and 5th centuries BC, our country had mastered the formula for finding the roots of quadratic equations of one variable.

The Greek Diophantus (246-330) only took one positive root of the quadratic equation. Even if he encountered two positive roots, he only took one of them

one.

In 628 AD, from the "Brahma Correction System" written by Brahmagupta of India, a root formula for the quadratic equation x2+px+q=0 was obtained

Mode.

In Arabia Al. Al-Khwarizmi's "Algebra" discusses the solution of equations and solves linear and quadratic equations, which involve six different forms. Let a, b, and c be positive numbers. , such as ax2=bx, ax2=c, ax2+c=bx, ax2+bx=c, ax2=bx+c, etc. Divide the quadratic equation into

different forms for discussion, following Diophantus' approach. Al. In addition to giving several special solutions to quadratic equations, Al-Khwarizmi also gave for the first time a general solution to quadratic equations, admitting that the equation has two roots and that there are irrational roots. But there is no understanding of virtual roots. In sixteenth-century Italy

Mathematicians began to use complex roots to solve cubic equations.

Veda (1540-1603), in addition to knowing that one-variable equations always have solutions in the range of complex numbers, also gave the relationship between roots and coefficients.

my country's "Nine Chapters of Arithmetic. Problem 20 in the chapter "Pythagorean" is solved by finding the positive root equivalent to x2+34x-71000=0. Chinese mathematicians also used interpolation methods in the study of equations.

[Edit this paragraph] Judgment method

Judgment formula of quadratic equation:

b^2-4ac>0 equation has two unequal real roots.

b^2-4ac=0 The equation has two equal real roots.

b^2-4ac<0 The equation has no real roots.

The above can be pushed out from the left to the right, and conversely, the left can also be pushed out from the right.

[Edit this paragraph] List the steps for solving a quadratic equation of one variable

(1) Analyze the meaning of the question and find the equality between the unknowns in the question and the conditions given in the question;

(2) Assume the unknowns and use the algebraic expression of the unknowns to express the remaining unknowns;

(3) Find the equality relationship and use it to list the equations;

(4) Solve the equation to find the value of the unknown in the question;

(5) Check whether the answer you are looking for is consistent with the meaning of the question, and answer it.

[Edit this paragraph] Lectures on classic examples

1. For questions about the definition of a quadratic equation of one variable, we must fully consider the three characteristics of the definition, and do not ignore that the coefficient of the quadratic term is not 0.

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

3. The discriminant of the roots of the quadratic equation (a≠0) is true in both positive and negative directions. It can be used to (1) determine the roots of an equation without solving the equation; (2) determine the range of the roots based on the properties of the parameter coefficients; (3) solve proof problems related to the roots.

4. There are many applications for the roots and coefficients of quadratic equations: (1) One root of the equation is known, and the other root and parameter coefficients can be found without solving the equation; (2) The equation is known, and the value of the algebraic expression containing two symmetrical roots is found. Relevant unknown coefficients; (3) Given two roots of the equation, find a quadratic equation of one variable with two roots of the equation or its algebraic formula as roots