Exercises of equations in the second volume of junior high school mathematics

Exercises of equations in the second volume of junior high school mathematics

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) X2-2 (+) x + 4 = 0 (∵ 4 can be decomposed into 2 &; 2, ∵ this problem 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