Fan Wenxuan (1) teaching objectives of the second volume of mathematics teaching plan for ninth grade published by People's Education Press.
1, knowing that the basic idea of solving a quadratic equation with one variable is to "simplify" the quadratic equation with one variable into a linear equation with one variable.
2. Learn to solve the equation in the form of (ax+b)2-k=0(k≥0) by factorization and direct Kaiping method.
3. Guide students to realize the idea of "demotion".
Important and difficult
Emphasis: Mastering factorization method and direct Kaiping method to solve the equation with the shape of (ax+b)2-k=0(k≥0).
Difficulty: The quadratic equation of one variable is simplified to the linear equation of one variable by factorization or direct square root.
teaching process
(a) review of the introduction
1, judge whether the following statement is correct.
(1) if p= 1, q= 1, pq = l (); If pq=l, p= 1, Q =1();
(2) If p=0 and g=0, then pq=0 (), if pq=0, then p=0 or Q = 0 ();
(3) If x+3=0 or x-6=0, then (x+3)(x-6)=0 (),
If (x+3)(x-6)=0, then x+3=0 or x-6 = 0 ();
(4) If x+3= or x-6=2, then (x+3)(x-6)= 1 (),
If (x+3)(x-6)= 1, then x+3= or x-6=2 ().
Answer: (1)√, ×. (2)√,√。 (3)√,√。 (4)√,×。
2. Fill in the blanks: If x2 = a；; Then x is called a and x =；; If x2=4, then x =；;
If x2=2, then x=.
Answer: square root, 2, 2.
(B) create a situation
We have learned how to solve linear equations in one variable and linear equations in two dimensions. What is the basic idea of solving two-dimensional linear equations? Binary linear equations are eliminated into univariate linear equations. Based on the basic idea of solving binary linear equation, can we get the basic idea of solving univariate quadratic equation?
Guide students to think and draw a conclusion: the basic idea of solving a quadratic equation is to "simplify" a quadratic equation into a quadratic equation.
The equation in section 1. 1 problem 1 gives: (35-2x)2-900=0.
Q: How to "simplify" this equation into a linear equation?
(3) Explore new knowledge
Let the students discuss the above questions, and then the teacher will guide the students with the content of "Introduction to Review". According to the textbook P.6, the equation (35-2x)2-900=0 will be "simplified" into two linear equations by factorization and direct Kaiping method. Let students know what factorization is and what direct Kaiping method is.
(4) give an example
Show the textbook P.7 case 1, case 2.
Instruct students to solve quadratic equations in one variable by factorization and direct Kaiping method according to textbooks.
Guide students to sum up: The equation with the shape of (ax+b)2-k=0(k≥0) can be solved by factorization or direct Kaiping method.
The basic steps of factorization method are: to change the equation into the form of the product of two linear factors with one side being 0 and the other side (this lesson mainly uses the square difference formula to decompose the factors), and then make each linear factor equal to 0, and solve two linear equations with one variable respectively, and the two solutions obtained are the solutions of the original quadratic equation with one variable.
The steps of direct Kaiping method are: transform the equation into (ax+b)2=k(k≥0), then directly square it to get ax+b= and ax+b=-, and solve these two linear equations respectively, and the solution obtained is the solution of the original quadratic equation.
Note: (1) factorization method is suitable for quadratic equations with one side being 0 and the other side being decomposed into the product of two linear factors;
(2) The direct Kaiping method is suitable for the equation in the form of (ax+b)2=k(k≥0). Since negative numbers have no square root, it is stipulated that k≥0, when k
(5) Application of new knowledge
Textbook P.8, practice.
(6) Class summary
1, what is the basic idea of solving a quadratic equation?
2. What are the methods to transform a quadratic equation with one variable into two linear equations with one variable by "reducing the order"? What are the basic steps?
3. Factorization method and direct Kaiping method are suitable for solving quadratic equations with one variable.
(7) Thinking and expanding
If you don't understand this equation, can you tell me the root of the following equation?
( 1)-4x 2+ 1 = 0； (2)x2+3 = 0； (3)(5-3x)2 = 0； (4)(2x+ 1)2+5=0 .
Answer: (1) has two unequal real roots; (2) and (4) have no real roots; (3) There are two equal real roots.
By solving this problem, students can make it clear that there are three solutions to the quadratic equation of one variable.
arrange work
Fan Wenxuan (II), the second volume of the ninth grade mathematics teaching plan of People's Education Press, I. Teaching objectives
1. Learn to find an acute trigonometric function value with a calculator through observation, guessing, comparison, concrete operation and other mathematical activities.
2. Experience the process of using trigonometric function knowledge to solve practical problems and promote the development of observation, analysis, induction and communication ability.
3. Feel the close connection between mathematics and life, enrich the successful experience of mathematics learning, stimulate students' curiosity to continue learning, and cultivate students' awareness of cooperation and communication with others.
Second, teaching material analysis
In life, we often encounter such problems, such as measuring the height of buildings, measuring the width of rivers, positioning ships and so on. To solve this kind of problems, we often need to apply trigonometric function knowledge. Last lesson, we have learned the trigonometric function values of 30, 45 and 60, and we can do some calculations under certain circumstances, but it is impossible to solve the problems in life only by relying on the trigonometric function values of these three special angles. In this lesson, students can use a calculator to find trigonometric function values, so that they can get rid of heavy calculations and experience the process of finding problems, asking questions, analyzing problems, exploring solutions and finally solving problems.
Third, the analysis of the situation of schools and students.
The age of ninth grade students is generally around 15 years old. At this stage, students take abstract logical thinking as the main development trend, but to a great extent, students still rely on concrete empirical materials and operational activities to understand abstract logical relations. In addition, using calculators can greatly reduce the burden on students. Therefore, based on the background materials provided by textbooks and supplemented by the use of calculators, students can solve problems better.
Students have been using calculators since primary school and are familiar with the operation of calculators. At the same time, in the previous courses, students have learned the definition of acute trigonometric functions, the values of trigonometric functions 30, 45 and 60 and their simple calculations, and they have the knowledge and skills to learn this lesson.
Fourthly, teaching design.
(a) review of issues
1. The ladder leans against the wall. If the angle between the ladder and the ground is 60 and the length of the ladder is 3 meters, what is the distance from the bottom of the ladder to the wall?
Student activity: Find the value according to the meaning of the question.
2. In life, is the angle between the ladder and the ground always 60?
No, there can be various angles. 60 is just a special phenomenon.
(B) creating situations and introducing topics
As shown in figure 1, when the hanging box of the mountaineering cable car passes through point A and reaches point B, it has already passed 200m m m. Given that the included angle between the cable car route and the plane is ∠ a = 16, what is the vertical ascending distance of the cable car?
Which line segment represents the vertical distance that the cable car rises?
Line segment BC
Which right triangle can I use to find BC?
In Rt△ABC, BC = absin 16, so BC = 200sin 16.
Do you know what sin 16 is? We can calculate the trigonometric function value of acute triangle with the help of scientific calculator. So, how to find trigonometric function with scientific calculator?
To find the value of trigonometric function with scientific calculator, sincos and tan keys should be used. Teacher activities: (1) Show the following table; (2) Dictation according to the table, so that students can learn to find the value of sin 16. The key sequence display result is sin16 sin16 = sin16 = 0? 275637355
Student activities: Find the values of sin 16 in the order listed in the table.
Can you calculate the values of cos42, tan85 and SIN 72 38' 25 "?
Student activities: By analogy with the method of finding sin 16, through guessing, discussing and learning from each other, we can find the corresponding trigonometric function value with a calculator (the operation steps are as follows):
Key sequence display result cos42 cos42=cos42 =0 = 0? 743 144825 tan 85 tan 85 = tan 85 = 1 1？ 4300523 sin 72 38′25″sin 72d′M′S
38D ' M ' S2
5D′M′S = sin 72 38′25″→
0? 95445032 1
Teacher: Use a scientific calculator to solve the problem at the beginning of this section.
Health: BC = 200 sin 16 ≈ 52? 12 (m).
Description: Use students' interest in learning to consolidate the calculation method of finding trigonometric function value with calculator.
(3) think about it.
Teacher: In the question at the beginning of this section, when the cable car continues to reach point D from point B, it has passed 200m, and the angle between the cable car's driving route from point B to point D and the horizontal plane is ∠ β = 42, so what else can be calculated?
Student activities:
(1) You can find the vertical distance DE of the second ascent, the sum of the vertical distances of the two ascends, the horizontal distance of the two passes, and so on.
(2) Complement each other, and deepen the understanding of trigonometric functions in this process.
Classroom practice
1. If a person wants to climb from the foot of the mountain to the top of the mountain, he needs to climb 300m on a slope of 40 and then climb 100m on a slope of 30 to find the height of the mountain (the result is accurate to 0. 1m).
2. As shown in Figure 2, ∠ DAB = 56, ∠ Cab = 50, AB=20m. Find the length of the lightning rod CD in the drawing (the result is accurate to 0.0 1m).
(5) detection
As shown in Figure 3, Wuhua Building is 60 meters away from Xiao Wei's home. Xiao Wei looked at the building from his own window. The elevation angle at the top of the building was 45, while the depression angle at the bottom of the building was 37. Find the height of the building (the result is accurate to 0? 1m).
Note: While students are practicing, teachers should patrol and guide, observe students' learning situation and give timely guidance to students' difficulties.
(6) Summary
Students talk about their feelings of learning this class, such as what new knowledge they have learned in this class, what difficulties they have encountered in the learning process, how to solve them and so on.
(7) homework
1. Use the calculator to find the following values:
( 1)tan 32； (2)cos24？ 53 ; (3)sin 62 1 1′； (4) Tan 39 39' 39 ".
Figure 42? As shown in Figure 4, in order to measure the width of a river, the surveyor measured the position of a tree T on the other side of the river at two points P and Q separated by 180m. T is in the south of P and 50 in the southwest of Q, and the river width is calculated (the result is accurate to 1m).
Reflection on the Teaching of verb (abbreviation of verb)
1. This section is the content of learning to find trigonometric function value with calculator and applying it in practice. Through the study in this section, students can fully realize that trigonometric function knowledge is widely used in the real world. There are not many knowledge points in this course, but students have improved their ability to analyze and solve problems by actively participating in the classroom, and have developed well in willpower, self-confidence and rational spirit.
Fan Wenxuan (3) teaching objectives of the second volume of mathematics teaching plan for ninth grade published by People's Education Press
1, understand the basic steps of solving a quadratic equation with one variable by collocation method.
2. The collocation method will be used to solve the quadratic equation with quadratic coefficient of 1.
3. Further understand the thinking method of transformation.
Important and difficult
Key point: I will use collocation method to solve the quadratic equation of one variable.
Difficulty: completely flatten the unknown term in the unary quadratic equation.
teaching process
(a) review of the introduction
1. Use the matching method to solve the equation x2+x- 1=0, and students will finish the textbook P. 13 after practice.
2. What are the basic steps to solve a quadratic equation with quadratic coefficient 1 by collocation method?
(B) create a situation
Now the collocation method can be used to solve the quadratic equation with quadratic coefficient 1, but can the collocation method be used to solve the quadratic equation with quadratic coefficient other than 1?
How to solve this kind of equation: 2x2-4x-6=0.
(3) Explore new knowledge
Ask the students to discuss the method of solving the equation 2x2-4x-6=0, and then sum up: for the unary quadratic equation whose quadratic coefficient is not 1, we can divide the two sides of the equation by the quadratic coefficient, change the quadratic coefficient into 1, and then solve it according to the method we learned last class. Let students further understand the idea of transformation.
(4) give an example
1. Show the textbook P. 14 Case X and explain according to the textbook.
2. Guide students to fill in the blanks in the textbook P. 14 Case X. ..
3. Summarize the basic steps of solving a quadratic equation with one variable by collocation method: firstly, transform the equation into a general form with quadratic coefficient of 1; Secondly, add the square of half the coefficient of the first term and subtract this number, so that the unknown term is in a completely flat way; Finally, factorization or direct Kaiping method is used to solve the quadratic equation of one variable after the formula.
(5) Application of new knowledge
Textbook P. 15, exercises.
(6) Class summary
1. What are the basic steps to solve a quadratic equation with one variable by collocation method?
2. Matching method is an important mathematical method, and its importance is not only manifested in the solution of quadratic equation in one variable, but also often used in the study of quadratic function and quadratic curve in senior high school.
3. Collocation method is a general method to solve the quadratic equation of one variable, but it is rarely used in solving the quadratic equation of one variable because of the complicated operation in the formula process.
4. Summarize the solutions learned before according to the block diagram of figure1-L.
Algorithm of quadratic equation in one variable.
(7) Thinking and expanding
Do not understand the equation, only through the formula to determine the solution of the following equation.
Situation.
( 1)4x 2+4x+ 1 = 0；
(2)x2-2x-5 = 0；
(3)–x2+2x-5 = 0 .
[Solution] Formulating each equation separately, we get:
( 1)(x+)2 = 0；
(2)(x- 1)2 = 6；
(3)(x- 1)2=-4 .
It can be concluded that equation (1) has two equal real roots, equation (2) has two unequal real roots, and equation (3) has no real roots.
Comments: By answering these three questions, students can use "collocation method" flexibly to strengthen their understanding of three situations of quadratic equation in one variable.