There are three ways to solve a linear equation with one variable.

Solution of quadratic equation in one variable

First, the main points of knowledge:

One-dimensional quadratic equation and one-dimensional linear equation are both integral equations, which are a key content of junior high school mathematics and the basis of studying mathematics in the future.

The foundation should attract students' attention.

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

The whole equation of.

The basic idea of solving quadratic equations with one variable is to simplify them into two quadratic equations with one variable. Quadratic equation with one variable has four solutions.

Methods: 1, direct Kaiping method; 2. Matching method; 3. Formula method; 4. Factorial decomposition method.

Second, detailed methods and examples:

1, direct Kaiping method:

The direct Kaiping method is a method to solve a quadratic equation with a direct square root. Solving (x-m)2=n (n≥0) by direct Kaiping method

The solution is an equation of x = m.

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

∴(3x+ 1)2=5

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

∴x=

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

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

∴(3x-4)2= 1 1

∴3x-4=

∴x=

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

2. Matching method: solve the equation ax2+bx+c=0 (a≠0) by matching method.

First, move the constant c to the right of the equation: AX2+BX =-C.

Convert the quadratic term to 1: x2+x =-

Add half the square of the first order coefficient on both sides of the equation: x2+x+( )2=- +( )2.

The left side of the equation becomes completely flat: (x+ )2=

When b2-4ac≥0, x+=

∴x= (this is the root formula)

Example 2. Solving Equation 3x2-4x-2=0 by Matching Method

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

Transform the quadratic term into 1: x2-x =

Add half the square of the coefficient of the first order term on both sides of the equation: x2-x+( )2= +( )2.

Formula: (x-)2=

Direct square: x-=

∴x=

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

3. Formula method: convert the quadratic equation of one variable into a general form, and then calculate the value of the discriminant △=b2-4ac. B2-4ac≥0, release all items.

Substitute the values of coefficients A, B and C into the formula x=(b2-4ac≥0) to get the root of the equation.

Example 3. Solving Equation 2x2-8x=-5 by Formula Method

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

∴a=2，b=-8，c=5

B2-4ac =(-8)2-4×2×5 = 64-40 = 24 & gt； 0

∴x= = =

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

4. Factorial decomposition method: the equation is deformed into a form with one side zero, and the quadratic trinomial on the other side is decomposed into the product of two linear factors, so that,

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

Root. 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)2 x2+3x = 0

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

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

X2-3x- 10=0 (the equation has a quadratic trinomial on the left and 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 original equation.

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

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=- 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 there are two solutions to the quadratic equation of one variable.

(3) Solution: 6x2+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 1=, x2=- is the solution of the original equation.

(4) Solution: x2-2(+ )x+4 =0 (∵4 can be decomposed into 2.2, ∴ this problem can be factorized).

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

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

Summary:

Usually, factorization is the most commonly used method to solve quadratic equations with one variable. When factorization is applied, the equation should be written as a general formula first.

Form, and the quadratic coefficient should become 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,

In this method, in order to determine the coefficient, the original equation must be transformed into a general form, and the value of discriminant must be calculated before using this formula, so as to judge the equation.

Whether there is a solution.

Matching method is a tool to derive formulas. After mastering the formula method, you can directly use the formula method to solve the quadratic equation of one variable, and generally do not need to use the matching method.

Solve a quadratic equation with one variable. Collocation is widely used in the study of other mathematical knowledge, and it is three important mathematical methods that junior high school requires to master.

One of the methods must be mastered. Three important mathematical methods: method of substitution, collocation method and undetermined coefficient method.

Example 5. Solve the following equations in an appropriate way. (optional research)

( 1)4(x+2)2-9(x-3)2 = 0(2)x2+(2-)x+-3 = 0

(3)x2-2x =-(4)4x 2-4mx- 10x+m2+5m+6 = 0

Analysis: (1) First of all, we should observe whether the topic has characteristics, and don't blindly do multiplication first. After observation, it is found that the square difference is available on the left side of the equation.

This formula decomposes a factor into the product of two linear factors.

(2) The left factor of the equation can be decomposed by cross multiplication.

(3) After it is transformed into a general form, it is solved by the formula method.

(4) Transform the equation into 4x2-2(2m+5)x+(m+2)(m+3)=0, and then decompose it by cross factor.

(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.

∴x 1= 1,x2= 13

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

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

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

∴x 1=-3,x2= 1

(3) Solution: x2-2 x=-

X2-2 x+ =0 (first converted to general form)

△=(-2)2-4×= 12-8 = 4 & gt； 0

∴x=

∴x 1=,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.

∴x 1=，x2=

Example 6. Find two roots of equation 3 (x+1) 2+5 (x+1) (x-4)+2 (x-4) 2 = 0. (optional research)

Analysis: If this equation is multiplied first, then multiplied, and similar terms are merged into a common form, it will be more complicated. Watch the topic carefully, I will.

Scientists have found that if x+ 1 and x-4 are regarded as a whole respectively, the cross factorization factor can be used on the left side of the equation (in fact, method of substitution is used).

Law)

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

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

Example 7. Solving the unary quadratic equation x2+px+q=0 by collocation method.

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

X2+px=-q (constant term moved to the right of the equation)

X2+px+( )2=-q+()2 (both sides of the equation plus half the square of the first coefficient).

(x+)2= (formula)

When p2-4q≥0, ≥0 (p2-4q must be discussed separately).

∴x=- =

∴x 1=，x2=

When p2-4q

Note: this is called the letter coefficient equation, and there are no additional conditions for P and Q in the problem, so you should always pay attention to letters in the process of solving problems.

Requirements for value selection, and classified discussion when necessary.

Exercise:

(1) Solve the following equation with an appropriate method:

1.6 x2-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 equation about x.

1 . x2-ax+-B2 = 0 ^ 2。 x2-( + )ax+ a2=0

Practice reference answer:

( 1) 1.x 1 =-，x2 = 2.x 1 = 2，x2 =-2。

3.x 1=0，x2 = 4 . x 1 = x2 = 2 5 . x 1 = x2 =

6. Solution: (Take 2x+3 as a whole and decompose the factors on the left side of the equation)

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

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

* 2x+9 = 0 or 2x+2=0

∴x 1=-,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.

∴x 1= +b, x2= -b is ∴x 1= a, x2=a is.

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

test

Multiple choice

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

a、x=5 B、x=-5 C、x 1=x2=5 D、x 1=x2=-5

2. The value of polynomial a2+4a- 10 is equal to 1 1, so 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 quadratic coefficient, linear coefficient and constant term in unary quadratic equation ax2+bx+c=0 is equal to zero, then there must be an equation.

The root is ().

a、0 B、 1 C 、- 1 D、 1

4. The root of the unary quadratic equation ax2+bx+c=0 is zero, if ().

A, b≠0 and c=0 B, b=0 and c≠0.

C and b=0 and c=0 D and c=0.

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

a 、-2，5 B、2 、-5 C、2，5 D、2

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

A, B, C, D, have no real roots.

7. The solution of equation 2x2-0. 15=0 is ().

a、x= B、x=-

c、x 1=0.27，x2=-0.27

8. Equation x2-x-4=0. After matching the left sides in a completely flat way, the equation obtained is ().

a 、( x-)2= B 、( x- )2=-

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

9. It is known that the univariate quadratic equation x2-2x-m=0, and the equation after solving the formula of this equation by matching 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

Answer and analysis

Answer:1.c2.c3.b4.d5.a6.d7.d8.c9.d.

Analysis:

1. Analysis: (x-5)2=0, then x 1=x2=5,

Note: Don't easily divide the two sides of the equation with an algebraic expression. Another quadratic equation with one variable has real roots, and it must be two.

2. Analysis: According to the meaning of the question: A2+4a-10 =11,the solution is a=3 or a=-7.

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

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

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

Then ax2+bx+c must have a factor X. If only c=0, there is a common factor X, so c=0.

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

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

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

X-5=0 or x+2=0.

x 1=5，x2=-2。

6. Analysis: δ = 9-4× 3 =-3

7. Analysis: 2x2=0. 15

x2=

x=

Pay attention to the simplification of roots, square directly and don't lose roots.

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

The sort is: (x-)2=

The equation can be transformed by using the property of equation. When x2-bx is formulated, the term of the formula is the square of half the coefficient of the first term-b.

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

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

Analysis of senior high school entrance examination

Comments on examination questions

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

(A) (B) (C) or (d) or

Comments: Since the quadratic equation of one variable has two roots, we exclude options A and B by exclusion method, and then select the correct options C and D by verification method.

Options. This equation can also be solved by factorization, and the results can also be compared with options. Options a and b only consider one hand and forget one yuan.

Quadratic equation has two roots, so it is wrong, and x =- 1 in option D can't make the left and right sides of the equation equal, so it is also wrong. The correct option is

C.

In addition, students often use an algebraic expression to divide both sides of the equation at the same time, which makes the equation lose its roots. This kind of mistake should be avoided.

2. (Jilin Province) The root of the quadratic equation of one variable is _ _ _ _ _ _ _.

Comments: The train of thought can be solved by factorization or formula method according to the characteristics of the equation.

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

0(B)– 1(C)0，– 1(D)0， 1

Comments: train of thought: Because the equation is a quadratic equation with two real roots, the correct option can be selected through exclusion verification, and a,

Two options have only one root. D option a number is not the root of the equation. In addition, you can also use the method of directly finding the root of the equation.

4. (Henan Province) It is known that one root of the quadratic equation of X is–2, so k = _ _ _ _ _ _ _ _

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

5.(Xi 'an) Solve the equation (x-3)2=8 by direct Kaiping method, and the root of the equation is ().

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

x 1=3+2，x2=3-2

Comments: You can solve the equation directly, or you don't need to calculate it. If there is a solution with a quadratic equation, there must be two solutions and the square of 8.

Root, you can choose the answer.

Extracurricular development

monadic quadratic equation

A quadratic equation with one variable means that there is an unknown number, and the highest term of the unknown number is 2.

Integral equation of degree. The general form is

ax2+bx+c=0，(a≠0)

Around 2000 BC, a quadratic equation of one yuan appeared on the clay tablet of Babylon and its solution: find a number to make it and it.

The sum of the reciprocal of is equal to a given number, that is, to find the sum of such an x, so that

x= 1，x+ =b，

x2-bx+ 1=0，

They made () 2; Do it again and get the solutions:+and-. It can be seen that the Babylonians already knew that one yuan was twice.

The root formula of the equation. 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: AX2 = 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.

One.

In 628 AD, a root of quadratic equation x2+px+q=0 was obtained from the Yarlung Zangbo River correction system written by India.

Type.

In Algebra, Arab Al-Walazimi discussed the solutions of equations, and solved the first and second kinds of equations, including six categories.

In 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 of discussion are based on the practice of Diophantine. In addition to giving several special solutions of quadratic equations, Al-Hualazimi is also the first time.

The general solution of quadratic equation is given, and it is admitted that the equation has two roots and irrational roots, but there is no understanding of imaginary roots. Italians in the sixteenth century

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 ninth chapter arithmetic Pythagorean method problem 20, finding the positive root is equivalent to x2+34x-7 1000=0. China Mathematics

Economists also apply interpolation in the study of equations.