1, growth rate problem: smaller number× (1+growth rate) 2 = larger number; Larger number ×( 1- growth rate) 2 = smaller number.

2. Area problem: There are two different algorithms to find the area of a graph, one is based on length × width, and the other is based on area addition and subtraction.

3. Sales problem: more money and less sales can be converted into 1 to solve the problem. For example, for every increase in 2 yuan money, 5 items are sold less, which can be regarded as 2.5 items are sold less for every increase of 1 yuan, so it is unknown that for every increase of X yuan, 2.5x items are sold less.

4, the trip problem: remember several commonly used formulas, the encounter problem, the distance apart is equal to the sum of the distances of two people; Tracing back to the problem, the distance is equal to the distance difference between two people.

5. Engineering problem: The total workload of Party A and Party B is equal to "1".

Practical problems and unary quadratic equations

1, enumerating the characteristics of solving application problems by quadratic equation with one variable.

Solving application problems with one-dimensional quadratic equation is the continuation and development of solving application problems with one-dimensional quadratic equation.

From the method of solving application problems by column equation, it is very similar to column quadratic equation. Because the unknowns of linear equations with one variable are one degree, most of these problems can be solved by arithmetic. If the unknown appears twice, it will be difficult to use arithmetic. Because the unknown is quadratic, the area problem can be solved by one-dimensional quadratic equation. The average growth rate after two increases involves the product in mathematical problems.

2. Enumerate the general steps of solving application problems with quadratic equation of one variable.

Just like solving application problems with linear equations of one variable, the general steps of solving application problems with quadratic equations of one variable are "review, design, list, solve and answer".

(1) "Examining a topic" refers to reading the topic, examining the meaning of the topic, and clarifying the known and unknown, as well as the quantitative relationship between them. This step is the basis for solving the problem;

(2) "Hypothesis" refers to the setting elements, which can be divided into direct setting elements and indirect setting elements. The so-called direct setting element means asking what to set. Although indirectly setting the unknown set of elements is not what we require, it is also very important because it is beneficial to the equation. Appropriate and flexible setting of elements directly affects the difficulty of solving equations.

(3) "Column" is a column equation, which is a very important step. Column equation is to find out the equivalence relation in the topic, and then list the equations with unknowns according to this equivalence relation, that is, equations. Finding out the column equation of equivalence relation is the key to solve the problem;

(4) "Solution" refers to finding the solution of the listed equation;

(5) "Answer" is to write the answer. It should be noted that the solution of the unary quadratic equation does not necessarily meet the meaning of the question, for example, the length of the line segment cannot be negative and the reduction rate cannot be greater than 100%. Therefore, after solving the root of the equation, it must be tested.

3, the relationship between number and number

Two digits = (ten digits) × 10+ digits

Three digits = (hundred digits) × 100+ (ten digits )×10+digits.

4, double

Doubling means twice the original quantity, and quadrupling means four times the original quantity.

5. Growth rate.

(1) related formula of growth rate:

Growth number = base × growth rate

Actual number = base number+growth number

(2) The basic equivalent relationship between two growth rates and equal growth rates is:

Original ×( 1+ growth rate) growth period = later period.

(1) The above equation only applies to the same growth rate;

(2) If it is a decreasing rate, the above relationship is:

×( 1- growth rate) = the number of original decline periods in the later period.

6. The general steps of solving geometric calculation problems by using quadratic equation of one variable.

(1) examine the meaning of the question as a whole and systematically;

(2) Find the equivalence relation in the problem (according to the nature of geometric figures);

(3) setting an unknown number and listing the equations according to the equivalence relation;

(4) Solve the equation correctly and test the rationality of the solution;

(5) write the answer.

7. The key to solving application problems with column equations

(1) Examination of questions is the basis for setting unknowns and equations. The so-called examination is to be good at understanding the meaning of the problem, find out the known quantity and unknown quantity in the problem, distinguish the quantitative relationship between them, and seek the implicit equality relationship;

(2) Setting unknowns can be divided into direct setting unknowns and indirect setting unknowns, which requires correct selection of methods and setting unknowns according to the quantitative relationship in the topic.

Attention should be paid to solving application problems with column equations:

(1) We should make full use of the known conditions in the problem setting, be good at analyzing the implicit conditions in the problem, and dig its implicit relations;

(2) Since a quadratic equation with one variable usually has two roots, it is necessary to test the two roots according to the meaning of the question, that is, to judge or determine whether the roots of the equation are consistent with the actual background and meaning of the question, and to discard the roots that are inconsistent with the meaning and actual meaning of the question.

Second, the important and difficult knowledge induction

Enumerating quadratic equations with one variable to solve practical problems.

Third, the typical case analysis

Example 1, the product of two consecutive odd numbers is 323. Find these two numbers.

Ideas:

(1) The way to express two consecutive odd numbers is: ① 2n+ 1, 2n-1; ②2n- 1，2n-3； ③2n+ 1，2n+3； ... (n stands for integer); (2) Set an element: ① Set the smaller odd number as X and the other odd number as X+2; ② Let the smaller odd number be X- 1 and the other odd number be X+1; ③ Let the smaller odd number be 2n- 1 and the other odd number be 2n+ 1.

Solution 1:

Let the smaller odd number be x and the other one be x+2.

According to the question, X (x+2) = 323? After completion

x2+2x-323=0，

To solve this equation: x 1= 17, x2 =- 19,

X+2 x=- 19 is obtained from x= 17, and X+2 =- 17 is obtained from x=- 19.

A: These two numbers are 17, 19 or-19,-17.

Solution 2:

Let these two odd numbers be x- 1 and x+ 1,

According to the meaning of the question, we can get (x- 1) (x+ 1) = 323.

X2=324，x x= 18。

When x= 18,1=17, 18+ 1 = 19.

When x =- 18,-18- 1 =- 19,-18+ 1 =- 17.

A: The two odd numbers are 17, 19 or-19 and-17 respectively.

Solution 3:

Let the smaller odd number be 2x- 1 and the larger odd number be 2x+ 1.

(2x- 1) (2x+ 1) = 323。

X2=8 1 after completion.

The answer is x 1=9 and X2 =-9.

When x 1=9, these two numbers are 17 and 19.

When x2 =-9, these two numbers are-19 and-17.

A: The two odd numbers are 17, 19 or-19 and-17 respectively.

Summary:

For some mathematical problems, if we can observe and analyze the relationship from different angles according to the basic characteristics and special factors of the topic, we can find a variety of thinking paths and get a variety of different answers, which will become interesting and eye-opening. This is the way to skillfully set elements, list three different equations, and get different values of x, and the result will be "all roads lead to the same goal." Comparatively speaking, which method is the best?

Example 2: A two-digit number whose sum is 5. After the two digits and the ten digits are reversed, the product of the two digits and the original two digits is 736. Find the original two digits.

Ideas:

The relationship between numbers is: two digits = (ten digits) × 10+ (one digit)

The key to solve the problem is to write the original two-digit number and the switched two-digit number correctly. To facilitate the analysis, the following table can be listed:

Ten digits, single digits, two digits.

Original x 5-x 10x+(5-x)

5-x x 10 (5-x)+x after the switch.

Solution:

Let the ten digits of the original two-digit number be X, then the single digit number is (5-X), according to the meaning of the question.

[ 10x+(5-x)][ 10(5-x)+x]= 736

After sorting, x2-5x+6 = 0.

To solve this equation, x 1=2, x2=3.

When x=2, 5-x = 3, with two digits 23;

When x=3, 5-x = 2, and the two-digit number is 32.

Summary: (1) Be good at representing multiple digits with the digits on each digit;

(2) After finding the solution of the equation, we should be good at checking whether it meets the meaning of the question, don't leave out the solution, and don't keep the solution that doesn't meet the meaning of the question.

Ex. 3: In a chess match, a single round-robin system is implemented (that is, each player plays a game with other players), and the winner of each game gets 2 points and the loser gets 0 points. If there is a tie, the two score 1 point. Today, four students counted the total scores of all the players in the game. The scores were 2005, 2004, 2070 and 2008 respectively.

Ideas:

(1) First, analyze the general number of games in the competition. Assuming that there are X players participating in this competition, then * * * games;

(2) Analyze the characteristics of the total score again. Because no matter whether a game is a win, a loss or a draw, 2 points are scored, and the total score of the game is the total score of all the contestants, that is, X (X- 1) points, and X must be a positive integer, so X and X-1are the products of two consecutive natural numbers and must be even numbers, so 2005 belongs to.

Solution:

Let * * have x players (x is a positive integer) to participate in the competition, then * * * counts as a game. Because each * * counts as 2 points, the total score of all the players is x(x- 1). Because the product of two adjacent natural numbers is even, its digits can only be 0, 2 and 6, so the total score cannot be.

Solving this equation, we get x 1=46, x2 =-45.

A: There are 46 contestants in this competition.

Summary:

(1) analyzing the total scores of all the contestants is the key to solve this problem.

(2) Choosing the correct data is the difficulty to solve this problem, which requires more understanding of the basic characteristics of integers.

Example 4: The sales of a commercial building this year 1 month is 600,000. In February, due to poor management, sales decreased by 65,438+00%. After the improvement of management, the enthusiasm of all employees was greatly stimulated, and the monthly sales increased significantly. By April, sales had soared to 960,000 yuan. What is the average monthly growth rate in March and April? (accurate to 0. 1%)

Ideas:

This is a question of growth rate. Find out the sales in February first, and then let the average growth rate in March and April be X, indicating the sales in April.

Solution:

According to the meaning of the question, let the average monthly growth rate in March and April be X.

60( 1- 10%)( 1+x)2 = 96。

Solve.

Since the percentage increase cannot be negative, it is irrelevant and discarded.

Namely.

A: In March and April, the average monthly sales growth rate of commercial buildings was 33.3%.

Summary:

The basic formula of the growth rate is: a (1 x) n, where a is the cardinal number, x is the growth rate or decline rate, and n represents the number of months that have passed.

Up to now, the national pilot project of returning farmland to forests has been expanded to 20 provinces, cities and districts, as shown in the following table: (unit: 10,000 hectares)

Basic information: afforestation area: returning farmland to forest area: afforestation area of barren hills and wasteland suitable for forest.

Completed in 2002 88.50 38.89 48. 1

In 2003, the number increased by 227,266.

(1) Complete the above table;

(2) If the newly planted area in 2005 is four times that in 2003, and the average growth rate in 2004 and 2005 is the same, find this growth rate.

Ideas:

As can be seen from the table, the afforestation area = the area of returning farmland to forest+the afforestation area of barren hills and wasteland suitable for forest. The newly-increased afforestation area in 2005 is four times that in 2003, which can be solved by equation.

Solution:

(1) The data in the table is 493;

(2) Let this growth rate be X.

493( 1+x)2=493×4

Solve this equation to get x 1 = 1, x 2 =-3.

∴x= 1= 100%.

A: This growth rate is 100%.

Summary:

Correctly understanding the significance of quadrupling is the key to solving the problem. In our daily life, we should be exposed to similar terms and understand their meanings.

Example 6: Take a rectangular iron sheet with a length of 80cm, cut out four squares with the same size at its four corners, and fold the four sides in half to make a rectangular box without a cover. If you make a rectangular box with a bottom area of 1500cm2, what is the side length of the cut small square?

Ideas:

Let the side length of the cut small square be x? Cm, the length of the bottom surface of the rectangular box without cover is (80-2x) cm and the width is (60-2x) cm, and the equation can be obtained.

Solution:

Let the side length of the cut small square be x? Cm, according to the meaning of the question

(80-2x)(60-2x)= 1500

The sorted x2-70x+825 = 0.

The solution is x 1= 15, x2=55.

However, when x=55, 80-2x =-30, which is irrelevant.

∴x= 15.

Answer: The side length of the cut small square is 15cm. ..

Summary:

(1) When solving the problem about the area, we should pay attention to divide irregular figures into regular figures, find out the relationship between the areas of each part, and then list the equations by using the area formula of regular figures;

(2) When solving practical problems with quadratic equation of one variable, the solution should be tested, sometimes the solution of quadratic equation of one variable does not necessarily meet the meaning of the question.

Example 7: As shown in the figure, A, B, C and D are known as the four vertices of a rectangle, AB= 16cm, AD=6cm, and the moving points P and Q start from points A and C respectively. Point P moves to point B at 3cm/s until point B, and point Q moves to point D at 2 cm/s. 。

Q: When (1) point P and point Q start for a few seconds, the area of quadrilateral PBCQ is 33cm2?

(2) When P and Q start for a few seconds, the distance between P and Q is 10cm?

Ideas:

(1) Because the quadrilateral PBCQ is trapezoidal and the height CB=6cm, the equation can be listed only by expressing the length of the upper and lower base;

(2) Because of the distance between two points of PQ, it is not easy to express the unknown by algebraic expression. It is necessary to construct the basic geometric figure-right triangle with auxiliary lines and solve it with Pythagorean theorem.

Solution:

(1) Let the area of quadrilateral PBCQ be 33cm2 when x seconds begin at p and q, then AP=3x, Pb = 16-3x and CQ = 2x. According to the area formula of trapezoid, the solution is x = 5.

Answer: The area of quadrilateral PBCQ is 33cm2 when P and Q are 5 seconds away from departure.

(2) Let the distance from point P to point Q be 10cm. From departure to y seconds.

As shown in the figure, if Q intersects with QH⊥AB and AB intersects with H, then AP=3y, CQ=2y, and pH = 16-3y-2y. According to Pythagorean Theorem, (16-3y-2y) 2 = 102.

Answer: The distance from point P to point Q is 10cm. ..

Example 8: When a shopping mall sells a brand-name shirt, it can sell an average of 20 shirts a day. Every shirt is profitable, 40 yuan. In order to expand sales, increase profits and reduce inventory, the mall decided to take appropriate measures to reduce prices. After investigation, it is found that if each shirt is reduced by 1 yuan, the mall can sell two more shirts on average a day. If the average daily profit of the mall is 1200 yuan, how much will each shirt be reduced?

Ideas:

Every time the price is reduced by 1 yuan, each shirt will earn (40- 1) yuan, and (20+2) pieces can be sold every day. Therefore, if the price of each shirt is reduced by X yuan, each shirt will earn (40-x) yuan, and (20+2x) pieces will be sold every day, and then the total profit = profit per piece ×

Solution:

Suppose the price of each piece is reduced by X yuan, then the profit of each piece is (40-X) yuan, and (20+2x) pieces can be sold every day. According to the meaning of the question, you can list the equations.

(40-x)(20+2x)= 1200

The sorted x2-30x+200 = 0.

The result is x 1= 10, x2=20.

Because we need to reduce the inventory as much as possible, the more we reduce the price, the faster the sales will be, so we have to reduce the price of each piece by 20 yuan.

A: Every shirt will be reduced in price, 20 yuan.

Summary:

Reducing inventory as much as possible is the meaning that the root of this equation must fit. It is not difficult to get one that meets the meaning of the question, but it is easy to miss the requirement of "reducing inventory as soon as possible" in the examination of the question, which leads to mistakes. Please note that in addition, the price of each shirt in this question is reduced by X yuan, that is, the profit of each shirt is reduced by X yuan. Therefore, the meaning of careful examination is the key to solving problems correctly.

Ex. 9: During driving, due to inertia, the car will slide forward for a certain distance before stopping. We call this distance the braking distance. On the curve where the speed limit is less than 35km/h, two cars, A and B, move in opposite directions. When they find something wrong, they brake at the same time, but they will still collide. After that, the braking distance of car A is 12m, and that of car B is 65438+. It is known that the relationship between braking distance S A (m) of A car and vehicle speed x(km/h) is S A = 0. 1x+0.0 1x2, and the relationship between braking distance S B (m) of B car and vehicle speed x(km/h) is SB = 0.05x+0.000.

Ideas:

In order to analyze the cause of the accident from the speed of two cars, it is necessary to calculate whether the speed exceeds the warning speed when passing through this curve from the known braking distance of two cars, so as to determine the main responsible person of the accident. Under the known conditions, the braking distances of two cars are 12m and 10m, respectively. By solving the equation, we can get the velocity and make a judgment.

Solution:

* The braking distance of vehicle A is 12m, ∴ 0.01x 2+0.1x =12.

That is, x2+ 10x- 1200 = 0.

The solution is x 1=30, and x2 =-40.

Because the speed can't be negative, ∴ x2 =-40 is irrelevant, so it is discarded.

So the speed of car A is 30km/h, and there is no speeding.

0.05x+0.005x2 = 10 for car B.

The result of solving this equation is x 1=40, x2 =-50.

So the speed of B car is 40 km/h, which exceeds the speed limit of 35 km/h. 。