Educational project "equations in tasks oge"

Finish the sentences: 1). The equation is... 2). The root of the equation is... 3). Solving an equation means...

I. Solve the equations orally: 1). 2). 3). 4). 5). 6). 7). 8). 9). 6 x + 18=0 2 x + 5=0 5 x – 3=0 -3 x + 9=0 -5 x + 1=0 -2 x – 10=0 6 x – 7=5 x 9 x + 6 \u003d 10 x 5 x - 12 \u003d 8 x

Which of the following equations has no solutions: a). 2 x - 14 \u003d x + 7 b). 2 x - 14 \u003d 2 (x - 7) c). x - 7 \u003d 2 x + 14 g). 2 x - 14 \u003d 2 x + 7?

Which of the equations has infinitely many solutions: a). 4 x - 12 = x - 12 b). 4 x - 12 \u003d 4 x + 12 c). 4(x - 3) = 4 x - 12 g). 4 (x - 3) \u003d x - 10?

EQUATIONS OF THE VIEW kx + b = 0, where k, b are given numbers, ARE CALLED LINEAR. Algorithm for solving linear equations: 1). open brackets 2). move the terms containing the unknown to the left side, and the terms not containing the unknown to the right side (the sign of the transferred member is reversed); 3). bring like members; 4). divide both sides of the equation by the coefficient of the unknown, if it is not equal to zero.

Solve in notebooks Group I: No. 681 p. 63 6 (4 -x) + 3 x \u003d 3 Group III: No. 767 p. 67 (x + 6) 2 + (x + 3) 2 \u003d 2 x 2 equations: II group: No. 697 p. 63 x-1 + (x + 2) \u003d -4 (-5 -x) -5

An equation of the form ax2 + bx + c \u003d 0, where a ≠ 0, b, c are any real numbers, is called square. Incomplete equations: ax2 + bx =0 (c=0), ax2 + c =0 (b=0).

II. Solve verbally quadratic equations, indicating whether they are complete or incomplete: 1). x2 + 15 x=0 2). -x2 +2 x = 0 3). x2 -25=0 4). -х2 +9 =0 5). -x2 - 16 \u003d 0 6). x2 - 8 x + 15=0 7). x2 + 5 x + 6=0 8). x2 + x - 12 =0 9). (-x-5)(-x+ 6)=0 10). x2 -4 x +4 =0

QUESTIONS: 1). What property of equations was used to solve incomplete quadratic equations? 2). What methods of factoring a polynomial were used to solve incomplete quadratic equations? 3). What is the algorithm for solving complete quadratic equations?

1). The product of two factors is equal to zero if one of them is equal to zero, while the second does not lose its meaning: ab = 0 if a = 0 or b = 0. 2). Taking out a common factor and a 2 - b 2 \u003d (a - b) (a + b) - the formula for the difference of squares. 3). The complete quadratic equation ax2 + bx + c = o. D=b 2 – 4 ac, if D>0, 2 roots; D = 0, 1 root; D

Theorem converse to Vieta's theorem: If the numbers a, b, c, x 1 and x 2 are such that x 1 x 2 \u003d x 1 + x 2 \u003d, and x 2 are the roots of the equation a x 2 + bx + c \u003d 0

SOLVE THE EQUATIONS: Group I: No. 802 p. 71 x2 - 5 x- 36 = 0 Group II: No. 810 p. 71 3 x2 - x + 21 = 5 x2 Group III: x4 -5 x2 - 36 = 0

III. SOLVE THE EQUATIONS: Group I and II: No. 860 Group III: =0 =0 What are such equations called? What property is used to solve them?

A rational equation is an equation of the form =0. A fraction is zero if the numerator is zero and the denominator is not zero. =0 if a = 0, b≠ 0.

Briefly from the history of mathematics Quadratic and linear equations were able to solve even the mathematicians of ancient Egypt. The Persian medieval scientist Al-Khwarizmi (IX century) first introduced algebra as an independent science about general methods for solving linear and quadratic equations, gave a classification of these equations. A new great breakthrough in mathematics is associated with the name of the French scientist Francois Vieta (XVI century). It was he who introduced letters into algebra. He owns the well-known theorem on the roots of a quadratic equation. And we owe the tradition of denoting unknown quantities with the last letters of the Latin alphabet (x, y, z) to another French mathematician - Rene Descartes (XVII).

Homework Work with sites: - Open bank of tasks OGE (mathematics) http: //85. 142.162.126/os/xmodules/qprint/index. php? proj=DE 0 E 276 E 49 7 AB 3784 C 3 FC 4 CC 20248 DC 0 ; - “I will solve the OGE” by D. Gushchin https: //oge. sdamgia. ru/ ; - Website of A. Larin (option 119) http: //alexlarin. net/. Teaching aids: - Yu. M. Kolyagin textbook "Algebra Grade 9", M., "Enlightenment", 2014, p. 308 -310; - "3000 tasks" under. edited by I. V. Yashchenko, M., "Exam", 2017, p. 5974.

Information for parents The system of preparation for the OGE in mathematics 1). Concomitant repetition in the lessons 2). Final repetition at the end of the year 3). Elective classes (on Saturdays) 4). Homework system - work with sites DECIDE OGE, OPEN BANK FIPI, A. LARIN SITE. 5). Individual consultations (on Mondays)

The fourth task in the algebra module tests knowledge in the field of handling powers and radical expressions.

When completing task No. 4 of the OGE in mathematics, not only the skills of performing calculations and converting numerical expressions are checked, but also the ability to convert algebraic expressions. You may need to perform operations with degrees with an integer exponent, with polynomials, identical transformations of rational expressions.

In accordance with the materials of the main exam, there may be tasks that require the implementation of identical transformations of rational expressions, the decomposition of polynomials into factors, the use of percentages and proportions, and signs of divisibility.

The answer in task 4 is one of the numbers 1; 2; 3; 4 corresponding to the number of the proposed answer to the task.

Theory for task number 4

From theoretical material, we will need rules for dealing with degrees:

Rules for working with rooted expressions:

In my analyzed options, these rules are presented - in the analysis of the first option of the third task, the rules for handling degrees are presented, and in the second and third options, examples of working with radical expressions are analyzed.

Analysis of typical options for task No. 4 OGE in mathematics

The first version of the task

Which of the following expressions for any values of n is equal to the product of 121 11 n ?

121n
11n+2
112n
11n+3

Solution:

To solve this problem, remember the following degree rules :

when multiplied, the exponents are added
division degrees are subtracted
when raising a power to a power, the powers are multiplied
when extracting the root, the degrees are divided

In addition, for the solution it is necessary to represent 121 as a power of 11, namely this is 11 2 .

121 11 n = 11 2 11 n

Taking into account the multiplication rule, we add the degrees:

11 2 11 n = 11 n+2

Therefore, the second answer suits us.

The second version of the task

Which of the following expressions has the largest value?

2√11
2√10

Solution:

To solve this task, you need to bring all expressions to a common form - present the expressions in the form of radical expressions:

We move 3 under the root:

3√5 = √(3² 5) = √(9 5) = √45

We move 2 under the root:

2√11 = √(2² 11) = √(4 11) =√44

We move 2 under the root:

2√10 = √(2² 10) = √(4 10) =√40

Squaring 6.5:

6.5 = √(6.5²) = √42.25

Let's look at all the resulting options:

3√5 = √45
2√11 = √44
2√10 = √40
6,5 = √42,25

Therefore, the correct answer is the first one.

The third version of the task

Which of these numbers is rational?

√810
√8,1
√0,81
all these numbers are irrational

Solution:

To solve this problem, you need to act as follows:

First, let's figure out the degree of which number is considered in this example - this is the number 9, since its square is 81, and this is already somewhat similar to the expressions in the answers. Next, consider the forms of the number 9 - these can be:

Consider each of them:

0.9 = √(0.9)² = √0.81

90 = √(90²) = √8100

Therefore, the number √0.81 is rational, while the other numbers

although similar to a 9 squared shape, they are not rational.

So the correct answer is the third one.

The fourth option

At the request of a member of my community Subsided Diana, I give an analysis of the following task number 4:

Which of the following numbers is the value of the expression?

Solution:

Note that there is a difference (4 - √14) in the denominator, which we need to get rid of. How to do it?

To do this, we recall the formula for abbreviated multiplication, namely the difference of squares! To apply it correctly in this task, you need to remember the rules for handling fractions. In this case, we recall that the fraction does not change if the numerator and denominator are multiplied by the same number or expression. For the difference of squares, we lack the expression (4 + √14), which means we multiply the numerator and denominator by it.

After that, in the numerator we get 4 + √14, and in the denominator the difference of squares: 4² - (√14)². After that, the denominator is easily calculated:

In total, our actions look like this:

Fifth option (demo version of the OGE 2017)

The value of which expression is a rational number?

√6-3
√3 √5
(√5)²
(√6-3)²

Solution:

In this task, we test the skills of operations with irrational numbers.

Let's analyze each answer in the solution:

√6 itself is an irrational number, to solve such problems it is enough to remember that it is rational to extract the root from the squares of natural numbers, for example, 4, 9, 16, 25...

When subtracting from an irrational number any other than itself, will again lead to an irrational number, thus, in this version, an irrational number is obtained.

When multiplying roots, we can extract the root from the product of radical expressions, that is:

√3 √5 = √(3 5) = √15

But √15 is irrational, so this answer doesn't work.

When squaring a square root, we get just a radical expression (to be more precise, a modulo radical expression, but in the case of a number, as in this version, this does not matter), therefore:

This answer suits us.

This expression represents a continuation of paragraph 1, but if √6-3 is an irrational number, then it cannot be converted into a rational number by any operations known to us.

Toylonov Argymai and Toylonov Erkey

Mathematical education received in a general education school is an essential component of general education and the general culture of a modern person. Almost everything that surrounds a modern person is all connected in one way or another with mathematics. And the latest advances in physics, engineering and information technology leave no doubt that in the future the state of affairs will remain the same. Therefore, the solution of many practical problems is reduced to solving various types of equations that need to be learned to solve.

And since 2013, certification in mathematics at the end of basic school has been carried out in the form of the OGE. Like the Unified State Examination, the OGE is designed to conduct certification not only in algebra, but also in the entire course of mathematics in the main school.

The lion's share of tasks, one way or another, come down to drawing up equations and their solutions. To proceed to the study of this topic, we needed to answer the questions: “What types of equations are found in the tasks of the OGE? ” and “What are the ways to solve these equations?”

Thus, there is a need to study all types of equations that are found in the tasks of the OGE. All of the above defines

aim work is to complete all types of equations found in the tasks of the OGE by type and analyze the main ways to solve these equations.

To achieve this goal, we have set the following tasks:

1) Learn the basic resources for preparing for the main state exams.

2) Complete all equations by type.

3) Analyze the ways of solving these equations.

4) Compile a collection with all types of equations and ways to solve them.

Object of study: equations.

Subject of study: equations in the tasks of the OGE.

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Municipal budgetary educational institution

"Chibit secondary school"

EDUCATIONAL PROJECT:

"EQUATIONS IN OGE TASKS"

Toylonov Erkey

8th grade students

Supervisor: Toylonova Nadezhda Vladimirovna, teacher of mathematics.

Project implementation timeline:

from 13.12.2017 to 13.02. 2018

Introduction ………………………………………………………………..
Historical reference …………………………………………………
Chapter 1 Solving Equations …………………………………………...
1.1 Solving linear equations ……………………………………
1.2 Quadratic equations ……………………………………………
1.2.1 Incomplete quadratic equations ………………………………	9-11
1.2.2 Complete quadratic equations …………………………………	11-14
1.2.3 Particular methods for solving quadratic equations …………….	14-15
1.3 Rational Equations ………………………………………….	15-17
Chapter 2 Complex Equations ………………………………………….	18-24
Conclusions …………………………………………………………………
List of used literature …………………………………
Appendix 1 "Linear Equations" ……………………………….	26-27
Appendix 2 "Incomplete Quadratic Equations" …………………	28-30
Appendix 3 "Complete Quadratic Equations" ……………………	31-33
Appendix 4 "Rational Equations" ………………………….	34-35
Appendix 5 "Complex Equations" ………………………………..	36-40

INTRODUCTION

Thus, there is a need to study all types of equations that are found in the tasks of the OGE. All of the above definesrelevance of the problem of the work performed.

aim work is to complete all types of equations found in the tasks of the OGE by type and analyze the main ways to solve these equations.

To achieve this goal, we have set the following tasks:

1) Learn the basic resources for preparing for the main state exams.

2) Complete all equations by type.

3) Analyze the ways of solving these equations.

4) Compile a collection with all types of equations and ways to solve them.

Object of study: equations.

Subject of study:equations in the tasks of the OGE.

Project work plan:

Formulation of the theme of the project.
Selection of material from official sources on a given topic.
Processing and systematization of information.
Project implementation.
Project design.
Project protection.

Problem : deepen your understanding of equations. Show the main methods for solving the equations presented in the tasks of the OGE in the first and second parts.

This work is an attempt to generalize and systematize the studied material and to study new ones. The project includes: linear equations with the transfer of terms from one part of the equation to another and using the properties of equations, as well as problems solved by the equation, all types of quadratic equations and methods for solving rational equations.

Mathematics... reveals order, symmetry and certainty,

and these are the most important kinds of beauty.

Aristotle.

Historical reference

In those distant times, when the wise men first began to think about equalities containing unknown quantities, there probably were no coins or wallets yet. But on the other hand, there were heaps, as well as pots, baskets, which were perfect for the role of caches-stores containing an unknown number of items. "We are looking for a heap, which, together with two thirds of it, a half and one seventh, is 37 ...", the Egyptian scribe Ahmes taught in the II millennium BC. In the ancient mathematical problems of Mesopotamia, India, China, Greece, unknown quantities expressed the number of peacocks in the garden, the number of bulls in the herd, the totality of things taken into account when dividing property. Scribes, officials, and priests initiated into secret knowledge, well trained in the science of counting, coped with such tasks quite successfully.

Sources that have come down to us indicate that ancient scientists possessed some general methods for solving problems with unknown quantities. However, not a single papyrus, not a single clay tablet gives a description of these techniques. The authors only occasionally supplied their numerical calculations with mean comments like: "Look!", "Do it!", "You found it right." In this sense, the exception is the "Arithmetic" of the Greek mathematician Diophantus of Alexandria (III century) - a collection of problems for compiling equations with a systematic presentation of their solutions.

However, the work of the Baghdad scholar of the 9th century became the first manual for solving problems that became widely known. Muhammad bin Musa al-Khwarizmi. The word "al-jabr" from the Arabic title of this treatise - "Kitab al-jaber wal-muqabala" ("The Book of Restoration and Contrasting") - over time turned into the word "algebra" well known to everyone, and the work of al-Khwarizmi itself served as starting point in the development of the science of solving equations.

So what is an equation?

There is an equation in rights, an equation of time (translation of true solar time into mean solar time, accepted in the hostel and in science; aster), etc.

In mathematics is a mathematical equation that contains one or more unknown quantities and remains valid only for certain values of these unknown quantities.

In equations with one variable, the unknown is usually denoted by the letter " X ". The value of "x , which satisfies these conditions, is called the root of the equation.

The equations are different. species :

ax + b = 0. - Linear equation.
ax 2 + bx + c = 0. - Quadratic equation.
ax 4 + bx 2 + c = 0. - Biquadratic equation.

– Rational equation.

– Irrational equation.
There are suchways to solve equations How: algebraic, arithmetic and geometric. Consider the algebraic way.

solve the equationis to find such values of x that, when substituted into the original expression, will give us the correct equality or prove that there are no solutions. Solving equations, however difficult, is exciting. After all, it is really surprising when a whole stream of numbers depends on one unknown number.

In equations, to find the unknown, it is necessary to transform and simplify the original expression. And so that when changing the appearance, the essence of the expression does not change. Such transformations are called identical or equivalent.

Chapter 1 Equation Solving

1.1 Solving linear equations.

Now we will consider the solutions of linear equations. Recall that an equation of the formis called a linear equation or an equation of the first degree, since with the variable " X » the highest degree is in the first degree.

The solution to the linear equation is very simple:

Example 1: Solve Equation 3 x+3=5x

The linear equation is solved by the method of transferring terms containing unknowns to the left side of the equal sign, free coefficients to the right side of the equal sign:

3 x – 5 x = – 3

2x=-3

x=1.5

The value of a variable that turns an equation into a true equality is called the root of the equation.

After checking we get:

So 1.5 is the root of the equation.

Answer: 1.5.

Solving equations by transferring terms from one part of the equation to another, while the sign of the terms changes to the opposite and apply properties equations - both parts of the equation can be multiplied (divided) by the same non-zero number or expression, can be considered when solving the following equations.

Example 2. Solve the equations:

a) 6 x +1=− 4 x ; b) 8 + 7 x \u003d 9 x +4; c) 4(x − 8)=− 5.

Solution.

a) By the transfer method we solve

6x + 4x = -1;

10x=─ 1;

x=─ 1:10;

x=─ 0.1.

Examination:

Answer: -0.1

b) Similarly to the previous example, we solve by the transfer method:

Answer: 2.

c) In this equation, it is necessary to open the brackets, applying the distributive property of multiplication with respect to the addition operation.

Answer: 6.75.

1.2 Quadratic equations

Type equation is called a quadratic equation, where a - senior coefficient, b is the average coefficient, c is the free term.

Depending on the coefficients a, b and c - the equation can be complete or incomplete, reduced or not reduced.

1.2.1 Incomplete quadratic equations

Consider ways to solve incomplete quadratic equations:

1) Let's start to deal with the solution of the first type of incomplete quadratic equations for c=0 . Incomplete quadratic equations of the form a x 2 +b x=0 allows you to solvefactorization method. In particular, the parenthesis method.

Obviously, we can, located on the left side of the equation, for which it is enough to take the common factor out of brackets x . This allows you to go from the original incomplete quadratic equation to an equivalent equation of the form: x·(a·x+b)=0 .

And this equation is equivalent to the combination of two equations x=0 or a x+b=0 , the last of which is linear and has a root x=− .

a x 2 +b x=0 has two roots

x=0 and x=− .

2) Now consider how incomplete quadratic equations are solved in which the coefficient b is zero and c≠0 , that is, equations of the form a x 2 +c=0 . We know that the transfer of a term from one side of the equation to the other with the opposite sign, as well as the division of both sides of the equation by a non-zero number, give an equivalent equation. Therefore, we can carry out the following equivalent transformations of the incomplete quadratic equation a x 2 +c=0 :

move c to the right side, which gives the equation a x 2 =−c ,
and divide both parts into a , we get.

The resulting equation allows us to draw conclusions about its roots.

If number is negative, then the equation has no roots. This statement follows from the fact that the square of any number is a non-negative number.

If is a positive number, then the situation with the roots of the equation is different. In this case, you need to remember that there is a root of the equation, it is a number. The root of the equation is calculated according to the scheme:

It is known that substitution into the equation instead of x its roots turns the equation into a true equality.

Let's summarize the information in this paragraph. Incomplete quadratic equation a x 2 +c=0 is equivalent to the equation, which

3) Solutions of incomplete quadratic equations in which the coefficients b and c are equal to zero, that is, from equations of the form a x 2 \u003d 0. The equation a x 2 =0 follows x 2 =0 , which is obtained from the original by dividing both parts of it by a non-zero number a . Obviously, the root of the equation x2=0 is zero because 0 2 =0 . This equation has no other roots.

So the incomplete quadratic equation a x 2 \u003d 0 has a single root x=0 .

Example 3 Solve the equations: a) x 2 \u003d 5x, if the equation has several roots, then in the answer indicate the smaller of them;

b) , if the equation has several roots, then in the answer indicate the largest of them;

c) x 2 −9=0, if the equation has several roots, then indicate the smaller one in your answer.

Solution.

We got an incomplete quadratic equation for which there is no free term. We solve by the method of taking out of brackets.

At The equation can have two roots, the smaller of which is 0.

Answer: 0.

b) . Similarly to the previous example, we apply the bracketing method

In the answer, you must indicate the largest of the roots. That is the number 2.

Answer: 2.

V) . This equation is an incomplete quadratic equation that does not have an average coefficient.

The smallest of these roots is the number - 3.

Answer: -3.

1.2.2 Complete quadratic equations.

1. Discriminant, the basic formula for the roots of a quadratic equation

There is a root formula.

Let's write down the formula of the roots of the quadratic equation step by step:

1) D=b 2 −4 a c - so-called.

a) if D

b) if D>0, then the equationdoes not have one root:

c) if D does not have two roots:

Algorithm for solving quadratic equations using root formulas

In practice, when solving a quadratic equation, you can immediately use the root formula, with which to calculate their values. But this is more about finding complex roots.

However, in a school algebra course, we usually talk not about complex, but about real roots of a quadratic equation. In this case, it is advisable to first find the discriminant before using the formulas for the roots of the quadratic equation, make sure that it is non-negative (otherwise, we can conclude that the equation has no real roots), and after that calculate the values of the roots.

The above reasoning allows us to writealgorithm for solving a quadratic equation. To solve a quadratic equation a x 2 +b x+c=0 , you need:

by the discriminant formula D=b 2 −4 a c calculate its value;
conclude that the quadratic equation has no real roots if the discriminant is negative;
calculate the only root of the equation by the formula if D=0 ;
find two real roots of a quadratic equation using the root formula if the discriminant is positive.

2. Discriminant, the second formula of the roots of the quadratic equation (for an even second coefficient).

To solve quadratic equations of the form, with an even coefficient b=2k there is another formula.

Let's write a new the formula for the roots of the quadratic equation for:

1) D’=k 2 −a c - so-calleddiscriminant of a quadratic equation.

a) if D' has no real roots;

b) if D'>0, then the equationdoes not have one root:

c) if D' does not have two roots:

Example 4 Solve the equation 2x 2 −3x+1=0.. If the equation has more than one root, write down the larger of the roots in response.

Solution. In the first case, we have the following coefficients of the quadratic equation: a=2 , b=-3 and c=1 D=b 2 −4 a c=(-3) 2 −4 2 1=9-8=1 . Since 1>0

We have got two roots, the largest of which is the number 1.

Answer: 1.

Example 5 Solve equation x 2 −21=4x.

If the equation has more than one root, write down the larger of the roots in response.

Solution. By analogy with the previous example, we move 4h to the left of the equal sign and get:

In this case, we have the following coefficients of the quadratic equation: a=1 , k=-2 and c=−21 . According to the algorithm, you first need to calculate the discriminant D'=k 2 −a c=(-2) 2 −1 (−21)=4+21=25 . Number 25>0 , that is, the discriminant is greater than zero, then the quadratic equation has two real roots. Let's find them by the root formula

Answer: 7.

1.2.3 Particular methods for solving quadratic equations.

1) Relationship between roots and coefficients of a quadratic equation. Vieta's theorem.

The formula for the roots of a quadratic equation expresses the roots of an equation in terms of its coefficients. Based on the formula of the roots, you can get other relationships between the roots and coefficients.

The most famous and applicable formula is called Vieta's Theorem.

Theorem: Let - roots of the reduced quadratic equation. Then the product of the roots is equal to the free term, and the sum of the roots is equal to the opposite value of the second coefficient:

Using the already written formulas, you can get a number of other relationships between the roots and coefficients of the quadratic equation. For example, you can express the sum of the squares of the roots of a quadratic equation in terms of its coefficients.

Example 6 a) Solve the equation x 2

b) Solve the equation x 2

c) Solve the equation x 2

Solution.

a) Solve the equation x 2 −6x+5=0. If the equation has more than one root, write down the smaller of the roots in response.

Choose the smallest of the roots

Answer: 1

b) Solve the equation x 2 +7x+10=0. If the equation has more than one root, write down the larger of the roots in response.

Applying the Vieta theorem, we write formulas for the roots

Logically, we conclude that. Choose the largest of the roots

Answer: ─2.

c) Solve the equation x 2 ─5x─14=0. If the equation has more than one root, write down the larger of the roots in response.

Applying the Vieta theorem, we write formulas for the roots

Logically, we conclude that. Choose the smallest of the roots

Answer: ─2.

1.3 Rational equations

If you are given an equation with fractions of the formwith a variable in the numerator or denominator, then such an expression is called a rational equation. A rational equation is any equation that includes at least one rational expression. Rational equations are solved in the same way as any equations: the same operations are performed on both sides of the equation until the variable is isolated on one side of the equation. However, there are 2 methods for solving rational equations.

1) Multiplication crosswise.If necessary, rewrite the equation given to you so that on each side there is one fraction (one rational expression); only then can you use the cross-multiplication method.

Multiply the numerator of the left fraction by the denominator of the right. Repeat this with the numerator of the right fraction and the denominator of the left.

Crosswise multiplication is based on basic algebraic principles. In rational expressions and other fractions, you can get rid of the numerator by multiplying the numerators and denominators of the two fractions accordingly.
Equate the resulting expressions and simplify them.
Solve the resulting equation, that is, find "x". If "x" is on both sides of the equation, isolate it on one side of the equation.

2) The least common denominator (LCD) is used to simplify this equation.This method is used when you cannot write the given equation with one rational expression on each side of the equation (and use the cross multiplication method). This method is used when you are given a rational equation with 3 or more fractions (in the case of two fractions, cross multiplication is better).

Find the least common denominator of fractions (or least common multiple).NOZ is the smallest number that is evenly divisible by each denominator.
Multiply both the numerator and denominator of each fraction by a number equal to the result of dividing the NOZ by the corresponding denominator of each fraction.
Find x. Now that you've reduced the fractions to a common denominator, you can get rid of the denominator. To do this, multiply each side of the equation by a common denominator. Then solve the resulting equation, that is, find "x". To do this, isolate the variable on one side of the equation.

Example 7 Solve the equations: a); b) c).

Solution.

A) . We use the cross multiplication method.

Open the brackets and add like terms.

got a linear equation with one unknown

Answer: ─10.

b) , similarly to the previous example, we apply the method of multiplication cross by cross.

Answer: ─1.9.

V) , we use the least common denominator (LCD) method.

In this example, the common denominator would be 12.

Answer: 5.

Chapter 2 Complex Equations

Equations belonging to the category of complex equations can combine various methods and techniques for solving. But, one way or another, all equations by the method of logical reasoning and equivalent actions lead to equations that were previously studied.

Example 7 Solve the equation( x +3) 2 =(x +8) 2 .

Solution. According to the formulas for abbreviated multiplication, we will open the brackets:

We transfer all terms beyond the equal sign and give similar ones,

Answer: 5.5.

Example 8 Solve the equations: a)(− 5 x +3)(− x +6)=0, b) (x +2)(− x +6)=0.

Solution.

a)(− 5 x +3)(− x +6)=0; open the brackets and give like terms

obtained a complete quadratic equation, which we will solve through the first formula of the discriminant

the equation has two roots

Answer: 0.6 and 6.

b) (x +2)(− x +6)=0, for this equation we will make logical reasoning (the product is equal to zero when one of the factors is equal to zero). Means

Answer: ─2 and 6.

Example 9 Solve the equations:, b).

Solution. Finding the lowest common denominator

We write in descending order of the powers of the variable

; obtained a complete quadratic equation with an even second coefficient

The equation has two real roots

Answer: .

b) . The reasoning is similar a). Finding NOZ

Open the brackets and give like terms

we solve the complete quadratic equation through the general formula

Answer: .

Example 10 Solve the equations:

Solution.

A) , We notice that on the left side the expression inside the brackets is a reduced multiplication formula, more precisely the square of the sum of two expressions. Let's transform it

; move the terms of this equation in one direction

take it out of brackets

The product is zero when one of the factors is zero. Means,

Answer: ─2, ─1 and 1.

b) We argue in the same way as for example a)

, by Vieta's theorem

Answer:

Example 11. Solve the equations a)

Solution.

A) ; [on the left and right sides of the equation, we can apply the bracketing method, and on the left side we will take out, and on the right side we take out the number 16.]

[Let's move everything to one side and once again apply the bracketing method. We will take out the common factor]

[the product is zero when one of the factors is zero.]

Answer:

b) . [This equation is similar to equation a). Therefore, in this case, the grouping method is applicable]

Answer:

Example 12. Solve the Equation=0.

Solution.

0 [biquadratic equation. Solved by the change of variable method].

0; [Applying the Vieta theorem we get the roots]

. [back to previous variables]

Answer:

Example 13 Solve the Equation

Solution. [biquadratic equation, get rid of the even degree by applying modulo signs.]

[we got two quadratic equations, which we solve through the basic formula of the roots of the quadratic equation]

there are no real roots the equation has two roots

Answer:

Example 14 Solve the Equation

Solution.

ODZ:

[we transfer all terms of the equation to the left side and bring like terms]

[we got the reduced quadratic equation, which is easily solved by the Vieta theorem]

The number - 1 does not satisfy the ODZ of the given equation, therefore it cannot be the root of this equation. So the root is only the number 7.

Answer: 7.

Example 15 Solve the Equation

Solution.

The sum of the squares of two expressions can only be equal to zero if the expressions are equal to zero at the same time. Namely

[Solve each equation separately]

According to Vieta's theorem

The coincidence of the roots equal to -5 will be the root of the equation.

Answer: - 5.

CONCLUSION

Summing up the results of the work done, we can conclude that equations play a huge role in the development of mathematics. We systematized the acquired knowledge, summarized the material covered. This knowledge can prepare us for the upcoming exams.

Our work makes it possible to take a different look at the problems that mathematics sets before us.

at the end of the project, we systematized and generalized the previously studied methods for solving equations;
got acquainted with new ways of solving equations and properties of equations;

considered all types of equations that are in the tasks of the OGE both in the first part and in the second part.
Created a methodical collection "Equations in the tasks of the OGE".

We believe that we have achieved the goal set before us - to consider all types of equations in the tasks of the main state exam in mathematics.

List of used literature:

1. B.V. Gnedenko "Mathematics in the Modern World". Moscow "Enlightenment" 1980

2. Ya.I. Perelman "Entertaining Algebra". Moscow "Science" 1978

6. http://tutorial.math.lamar.edu

7. http://www.regentsprep.org

8. http://www.fipi.ru

Annex 1

Linear equations

1. Find the root of the equation

2. Find the root of the equation

3. Find the root of the equation

Annex 2

Incomplete quadratic equations

1. Solve the equation x 2 =5x. If the equation has more than one root, write down the smaller of the roots in response.

2. Solve the equation 2x 2 =8x. If the equation has more than one root, write down the smaller of the roots in response.

3. Solve the equation 3x 2 =9x. If the equation has more than one root, write down the smaller of the roots in response.

4. Solve the equation 4x 2 =20x. If the equation has more than one root, write down the smaller of the roots in response.

5. Solve the equation 5x 2 =35x. If the equation has more than one root, write down the smaller of the roots in response.

6. Solve the equation 6x 2 =36x. If the equation has more than one root, write down the smaller of the roots in response.

7. Solve the equation 7x 2 =42x. If the equation has more than one root, write down the smaller of the roots in response.

8. Solve the equation 8x 2 =72x. If the equation has more than one root, write down the smaller of the roots in response.

9. Solve the equation 9x 2 =54x. If the equation has more than one root, write down the smaller of the roots in response.

10. Solve the equation 10x2 =80x. If the equation has more than one root, write down the smaller of the roots in response.

11. Solve the equation 5x2 −10x=0. If the equation has more than one root, write down the larger of the roots in response.

12. Solve the equation 3x2 −9x=0. If the equation has more than one root, write down the larger of the roots in response.

13. Solve the equation 4x2 −16x=0. If the equation has more than one root, write down the larger of the roots in response.

14. Solve the equation 5x2 +15x=0. If the equation has more than one root, write down the smaller of the roots in response.

15. Solve the equation 3x2 +18x=0. If the equation has more than one root, write down the smaller of the roots in response.

16. Solve the equation 6x2 +24x=0. If the equation has more than one root, write down the smaller of the roots in response.

17. Solve the equation 4x2 −20x=0. If the equation has more than one root, write down the larger of the roots in response.

18. Solve the equation 5x2 +20x=0. If the equation has more than one root, write down the smaller of the roots in response.

19. Solve the equation 7x2 −14x=0. If the equation has more than one root, write down the larger of the roots in response.

20. Solve the equation 3x2 +12x=0. If the equation has more than one root, write down the smaller of the roots in response.

21. Solve the equation x2 −9=0. If the equation has more than one root, write down the smaller of the roots in response.

22. Solve the equation x2 −121=0. If the equation has more than one root, write down the smaller of the roots in response.

23. Solve the equation x2 −16=0. If the equation has more than one root, write down the smaller of the roots in response.

24. Solve the equation x2 −25=0. If the equation has more than one root, write down the smaller of the roots in response.

25. Solve equation x2 −49=0. If the equation has more than one root, write down the smaller of the roots in response.

26. Solve the equation x2 −81=0. If the equation has more than one root, write down the smaller of the roots in response.

27. Solve the equation x2 −4=0. If the equation has more than one root, write down the smaller of the roots in response.

28. Solve equation x2 −64=0. If the equation has more than one root, write down the smaller of the roots in response.

29. Solve the equation x2 −36=0. If the equation has more than one root, write down the smaller of the roots in response.

30. Solve equation x2 −144=0. If the equation has more than one root, write down the smaller of the roots in response.

31. Solve the equation x2 −9=0. If the equation has more than one root, write down the larger of the roots in response.

32. Solve the equation x2 −121=0. If the equation has more than one root, write down the larger of the roots in response.

33. Solve the equation x2 −16=0. If the equation has more than one root, write down the larger of the roots in response.

34. Solve the equation x2 −25=0. If the equation has more than one root, write down the larger of the roots in response.

35. Solve equation x2 −49=0. If the equation has more than one root, write down the larger of the roots in response.

36. Solve equation x2 −81=0. If the equation has more than one root, write down the larger of the roots in response.

37. Solve the equation x2 −4=0. If the equation has more than one root, write down the larger of the roots in response.

38. Solve the equation x2 −64=0. If the equation has more than one root, write down the larger of the roots in response.

39. Solve the equation x2 −36=0. If the equation has more than one root, write down the larger of the roots in response.

40. Solve equation x2 −144=0. If the equation has more than one root, write down the larger of the roots in response.

Appendix 3

Complete quadratic equations

1. Solve the equation x2 +3x=10. If the equation has more than one root, write down the larger of the roots in response.

2. Solve equation x2 +7x=18. If the equation has more than one root, write down the larger of the roots in response.

3. Solve equation x2 +2x=15. If the equation has more than one root, write down the smaller of the roots in response.

4. Solve equation x2 −6x=16. If the equation has more than one root, write down the smaller of the roots in response.

5. Solve equation x2 −3x=18. If the equation has more than one root, write down the larger of the roots in response.

6. Solve equation x2 −18=7x. If the equation has more than one root, write down the larger of the roots in response.

7. Solve equation x2 +4x=21. If the equation has more than one root, write down the smaller of the roots in response.

8. Solve equation x2 −21=4x. If the equation has more than one root, write down the larger of the roots in response.

9. Solve equation x2 −15=2x. If the equation has more than one root, write down the smaller of the roots in response.

10. Solve equation x2 −5x=14. If the equation has more than one root, write down the larger of the roots in response.

11. Solve equation x2 +6=5x. If the equation has more than one root, write down the smaller of the roots in response.

12. Solve equation x2 +4=5x. If the equation has more than one root, write down the larger of the roots in response.

13. Solve equation x2 −x=12. If the equation has more than one root, write down the larger of the roots in response.

14. Solve equation x2 +4x=5. If the equation has more than one root, write down the smaller of the roots in response.

15. Solve equation x2 −7x=8. If the equation has more than one root, write down the larger of the roots in response.

16. Solve equation x2 +7=8x. If the equation has more than one root, write down the smaller of the roots in response.

17. Solve equation x2 +18=9x. If the equation has more than one root, write down the smaller of the roots in response.

18. Solve equation x2 +10=7x. If the equation has more than one root, write down the larger of the roots in response.

19. Solve the equation x2 −20=x. If the equation has more than one root, write down the larger of the roots in response.

20. Solve the equation x2 −35=2x. If the equation has more than one root, write down the smaller of the roots in response.

21. Solve the equation 2x2 −3x+1=0. If the equation has more than one root, write down the smaller of the roots in response.

22. Solve the equation 5x2 +4x−1=0. If the equation has more than one root, write down the larger of the roots in response.

23. Solve the equation 2x2 +5x−7=0. If the equation has more than one root, write down the smaller of the roots in response.

24. Solve the equation 5x2 −12x+7=0. If the equation has more than one root, write down the larger of the roots in response.

25. Solve the equation 5x2 −9x+4=0. If the equation has more than one root, write down the smaller of the roots in response.

26. Solve the equation 8x2 −12x+4=0. If the equation has more than one root, write down the smaller of the roots in response.

27. Solve the equation 8x2 −10x+2=0. If the equation has more than one root, write down the smaller of the roots in response.

28. Solve the equation 6x2 −9x+3=0. If the equation has more than one root, write down the smaller of the roots in response.

29. Solve the equation 5x2 +9x+4=0. If the equation has more than one root, write down the larger of the roots in response.

30. Solve the equation 5x2 +8x+3=0. If the equation has more than one root, write down the larger of the roots in response.

31. Solve the equation x2 −6x+5=0. If the equation has more than one root, write down the smaller of the roots in response.

32. Solve the equation x2 −7x+10=0. If the equation has more than one root, write down the smaller of the roots in response.

33. Solve the equation x2 −9x+18=0. If the equation has more than one root, write down the smaller of the roots in response.

34. Solve the equation x2 −10x+24=0. If the equation has more than one root, write down the smaller of the roots in response.

35. Solve equation x2 −11x+30=0. If the equation has more than one root, write down the smaller of the roots in response.

36. Solve equation x2 −8x+12=0. If the equation has more than one root, write down the larger of the roots in response.

37. Solve the equation x2 −10x+21=0. If the equation has more than one root, write down the larger of the roots in response.

38. Solve the equation x2 −9x+8=0. If the equation has more than one root, write down the larger of the roots in response.

39. Solve the equation x2 −11x+18=0. If the equation has more than one root, write down the larger of the roots in response.

40. Solve equation x2 −12x+20=0. If the equation has more than one root, write down the larger of the roots in response.

Appendix 4

Rational equations.

1. Find the root of the equation

2. Find the root of the equation

3. Find the root of the equation

4. Find the root of the equation

5. Find the root of the equation

6. Find the root of the equation.

7. Find the root of the equation

8. Find the root of the equation

9. Find the root of the equation.

10. Find the root of the equation

11. Find the root of the equation.

12. Find the root of the equation

13. Find the root of the equation

14. Find the root of the equation

15. Find the root of the equation

16. Find the root of the equation

17. Find the root of the equation

18. Find the root of the equation

19. Find the root of the equation

20. Find the root of the equation

21. Find the root of the equation

22. Find the root of the equation

23. Find the root of the equation

Appendix 5

Complex equations.

1. Find the root of the equation (x+3)2 =(x+8)2 .

2. Find the root of the equation (x−5)2 =(x+10)2 .

3. Find the root of the equation (x+9)2 =(x+6)2 .

4. Find the root of the equation (x+10)2 =(x−9)2 .

5. Find the root of the equation (x−5)2 =(x−8)2 .

6. Find the root of the equation.

7. Find the root of the equation.

8. Find the root of the equation.

9. Find the root of the equation.

10. Find the root of the equation.

11. Solve the equation (x+2)(− x+6)=0. If the equation has more than one root, write down the smaller of the roots in response.

12. Solve the equation (x+3)(− x−2)=0. If the equation has more than one root, write down the smaller of the roots in response.

13. Solve the equation (x−11)(− x+9)=0. If the equation has more than one root, write down the smaller of the roots in response.

14. Solve the equation (x−1)(− x−4)=0. If the equation has more than one root, write down the smaller of the roots in response.

15. Solve the equation (x−2)(− x−1)=0. If the equation has more than one root, write down the smaller of the roots in response.

16. Solve the equation (x+20)(− x+10)=0. If the equation has more than one root, write down the larger of the roots in response.

17. Solve the equation (x−2)(− x−3)=0. If the equation has more than one root, write down the larger of the roots in response.

18. Solve the equation (x−7)(− x+2)=0. If the equation has more than one root, write down the larger of the roots in response.

19. Solve the equation (x−5)(− x−10)=0. If the equation has more than one root, write down the larger of the roots in response.

20. Solve the equation (x+10)(− x−8)=0. If the equation has more than one root, write down the larger of the roots in response.

21. Solve the equation (− 5x+3)(− x+6)=0. If the equation has more than one root, write down the smaller of the roots in response.

22. Solve the equation (− 2x+1)(− 2x−7)=0. If the equation has more than one root, write down the smaller of the roots in response.

23. Solve the equation (− x−4)(3x+3)=0. If the equation has more than one root, write down the larger of the roots in response.

24. Solve the equation (x−6)(4x−6)=0. If the equation has more than one root, write down the smaller of the roots in response.

25. Solve the equation (− 5x−3)(2x−1)=0. If the equation has more than one root, write down the smaller of the roots in response.

26. Solve the equation (x−2)(− 2x−3)=0. If the equation has more than one root, write down the smaller of the roots in response.

27. Solve the equation (5x+2)(− x−4)=0. If the equation has more than one root, write down the larger of the roots in response.

28. Solve the equation (x−6)(− 5x−9)=0. If the equation has more than one root, write down the smaller of the roots in response.

29. Solve the equation (6x−3)(− x+3)=0. If the equation has more than one root, write down the larger of the roots in response.

30. Solve the equation (5x−2)(− x+3)=0. If the equation has more than one root, write down the smaller of the roots in response.

31. Solve the equation

32. Solve the equation

33. Solve the equation

34. Solve the equation

35. Solve the equation

36. Solve the equation

37. Solve the equation

38. Solve the equation

39. Solve the equation

40 Solve the equation

41. Solve the equation x(x2 +2x+1)=2(x+1).

42. Solve the equation (x−1)(x2 +4x+4)=4(x+2).

43. Solve the equation x(x2 +6x+9)=4(x+3).

44. Solve the equation (x−1)(x2 +8x+16)=6(x+4).

45. Solve the equation x(x2 +2x+1)=6(x+1).

46. Solve the equation (x−1)(x2 +6x+9)=5(x+3).

47. Solve the equation (x−2)(x2 +8x+16)=7(x+4).

48. Solve the equation x(x2 +4x+4)=3(x+2).

49. Solve the equation (x−2)(x2 +2x+1)=4(x+1).

50. Solve the equation (x−2)(x2 +6x+9)=6(x+3).

51. Solve the equation (x+2)4 −4(x+2)2 −5=0.

52. Solve the equation (x+1)4 +(x+1)2 −6=0.

53. Solve the equation (x+3)4 +2(x+3)2 −8=0.

54. Solve the equation (x−1)4 −2(x−1)2 −3=0.

55. Solve the equation (x−2)4 −(x−2)2 −6=0.

56. Solve equation (x−3)4 −3(x−3)2 −10=0.

57. Solve the equation (x+4)4 −6(x+4)2 −7=0.
58. Solve the equation (x−4)4 −4(x−4)2 −21=0.

59. Solve the equation (x+2)4 +(x+2)2 −12=0.

60. Solve the equation (x−2)4 +3(x−2)2 −10=0.

61. Solve equation x3 +3x2 =16x+48.

62. Solve equation x3 +4x2 =4x+16.

63. Solve equation x3 +6x2 =4x+24.

64. Solve equation x3 +6x2 =9x+54.

65. Solve equation x3 +3x2 =4x+12.

66. Solve equation x3 +2x2 =9x+18.

67. Solve equation x3 +7x2 =4x+28.

68. Solve equation x3 +4x2 =9x+36.

69. Solve equation x3 +5x2 =4x+20.

70. Solve equation x3 +5x2 =9x+45.

71. Solve equation x3 +3x2 −x−3=0.

72. Solve equation x3 +4x2 −4x−16=0.

73. Solve equation x3 +5x2 −x−5=0.

74. Solve equation x3 +2x2 −x−2=0.

75. Solve equation x3 +3x2 −4x−12=0.

76. Solve equation x3 +2x2 −9x−18=0.

77. Solve equation x3 +4x2 −x−4=0.

78. Solve equation x3 +4x2 −9x−36=0.

79. Solve equation x3 +5x2 −4x−20=0.
80. Solve equation x3 +5x2 −9x−45=0.

81. Solve equation x4 =(x−20)2 .

82. Solve equation x4 =(2x−15)2 .

83. Solve equation x4 =(3x−10)2 .

84. Solve equation x4 =(4x−5)2 .

85. Solve equation x4 =(x−12)2 .

86. Solve equation x4 =(2x−8)2 .

87. Solve equation x4 =(3x−4)2 .

88. Solve equation x4 =(x−6)2 .

89. Solve equation x4 =(2x−3)2 .

90. Solve equation x4 =(x−2)2 .

91. Solve the equation

92. Solve the equation

93. Solve the equation

94. Solve the equation

95. Solve the equation

96. Solve the equation

97. Solve the equation

98. Solve the equation

99. Solve the equation

100. Solve the equation

101. Solve the equation.

102. Solve the equation

103. Solve the equation

104. Solve the equation

105. Solve the equation

106. Solve the equation

107. Solve the equation

108. Solve the equation

109. Solve the equation

110. Solve the equation

! From theory to practice;

! From simple to complex

MAOU "Platoshinskaya secondary school",

mathematics teacher, Melekhina G.V.

General view of the linear equation: ax + b = 0 ,

Where a And b– numbers (coefficients).

If a = 0 And b = 0, That 0x+ 0 = 0 - infinitely many roots;
If a = 0 And b ≠ 0, That 0x+ b = 0- no solutions
If a ≠ 0 And b = 0 , That ax + 0 = 0 – one root, x = 0;
If a ≠ 0 And b ≠ 0 , That ax + b = 0 - one root

! If X is to the first power and is not contained in the denominator, then this is a linear equation

! What if the linear equation is complex :

! Terms with X to the left, without X to the right.

! These equations are also linear .

! The main property of proportion (crosswise).

! Open brackets, with X to the left, without X to the right.

if the coefficient a = 1, then the equation is called given :

if the coefficient b = 0 or (and) c = 0, then the equation is called incomplete :

! Basic formulas

! More formulas

Biquadratic equation is called an equation of the form ax 4 +bx 2 + c = 0 .

The biquadratic equation is reduced to quadratic equation by substitution, then

We get a quadratic equation:

Let's find the roots and return to the replacement:

Example 1:

Solve equation x 4 + 5x 2 – 36 = 0.

Solution:

Substitution: x 2 = t.

t 2 + 5t - 36 = 0. The roots of the equation t 1 = -9 and t 2 = 4.

x 2 \u003d -9 or x 2 \u003d 4.

Answer: There are no roots in the first equation, from the second: x \u003d ± 2.

Example 2:

solve the equation (2x - 1) 4 - 25 (2x - 1) 2 + 144 = 0.

Solution:

Substitution: (2x - 1) 2 = t.

t 2 - 25t + 144 = 0. The roots of the equation t 1 = 9 and t 2 = 16.

(2x - 1) 2 = 9 or (2x - 1) 2 = 16.

2x - 1 = ±3 or 2x - 1 = ±4.

From the first equation there are two roots: x \u003d 2 and x \u003d -1, from the second there are also two roots: x \u003d 2.5 and x \u003d -1.5.

Answer: -1.5; -1; 2; 2.5.

1) X 4 - 9 X 2 = 0; 2) 4 X 4 - x 2 \u003d 0;

1) X 4 + x 2 - 2 = 0;

2) X 4 - 3 X 2 - 4 = 0; 3) 9 X 4 + 8 X 2 - 1 = 0; 4) 20 X 4 - X 2 - 1 = 0.

Solve equations by extracting from the left side full square :

1) X 4 - 20 X 2 + 64 = 0; 2) X 4 - 13 X 2 + 36 = 0; 3) X 4 - 4 X 2 + 1 = 0; 4) X 4 + 2 X 2 +1 = 0.

! Remember the square of the sum and the square of the difference

rational expression is an algebraic expression made up of numbers and a variable x using the operations of addition, subtraction, multiplication, division and exponentiation with a natural exponent.

If r(x) is a rational expression, then the equation r(x)=0 called a rational equation.

Algorithm for solving a rational equation:

1. Transfer all terms of the equation to one part.

2. Convert this part of the equation to the form of an algebraic fraction p(x)/q(x)

3. solve the equation p(x)=0

4. For each root of the equation p(x)=0 check if it satisfies the condition q(x)≠0 or not. If yes, then this is the root of the given equation; if not, it is an extraneous root and should not be included in the response.

! Recall the solution of the fractional rational equation:

! To solve equations, it is useful to recall the formulas for abbreviated multiplication:

If the equation contains a variable under the square root sign, then the equation is called irrational .

Method for squaring both sides of an equation- the main method for solving irrational equations.

Having solved the resulting rational equation, it is necessary to make a check , sifting out possible extraneous roots.

Answer: 5; 4

Another example:

Examination:

The expression doesn't make sense.

Answer: there are no solutions.

SOLUTION OF EQUATIONS

preparation for the OGE

Grade 9

prepared by a teacher of mathematics at GBOU school No. 14 of the Nevsky district of St. Petersburg Putrova Marina Nikolaevna

Complete the sentences:

1). The equation is...

2). The root of the equation is...

3). Solving an equation means...

I. Solve the equations orally:

1). 6x + 18=0
2). 2x + 5=0
3). 5x - 3=0
4). -3x + 9=0
5). -5x + 1=0
6). -2x - 10=0
7). 6x - 7=5x
8). 9x + 6=10x
9). 5x - 12=8x

Which of the following equations has no solutions:

A). 2x - 14 = x + 7

b). 2x - 14 = 2(x - 7)

V). x - 7 \u003d 2x + 14

G). 2x-14 = 2x + 7?

Which equation has infinitely many solutions?

A). 4x - 12 = x - 12

b). 4x - 12 = 4x + 12

V). 4(x - 3) = 4x - 12

G). 4 (x - 3) \u003d x - 10?

EQUATIONS OF THE VIEW

kx + b = 0

CALLED LINEAR.

Algorithm for solving linear equations :

1). move the terms containing the unknown to the left side, and the terms not containing the unknown to the right side (the sign of the transferred member is reversed);

2). bring like members;

3). Divide both sides of the equation by the coefficient of the unknown, if it is not equal to zero.

Solve equations in notebooks :

II group: No. 697 p.63

x-1 +(x+2) = -4(-5-x)-5

I group:

№ 681 p.63

6(4x)+3x=3

III group: No. 767 p. 67

(x + 6) 2 + (x + 3) 2 = 2 x 2

Type equation

ah 2 + bx + c \u003d 0,

where a≠0, b, c – any real numbers is called square.

Incomplete Equations:

ah 2 + bх =0 (c=0),

ah 2 + c=0 (b=0).

II. Orally solve quadratic equations, indicating whether they are complete or incomplete:

1). 5x 2 + 15x=0

2). -X 2 +2x = 0

3). X 2 -25=0

4). -X 2 +9 =0

5). -X 2 - 16 =0

6). X 2 - 8x + 15=0

7 ) . X 2 + 5x + 6=0

8). X 2 + x - 12 = 0

9).(-x-5)(-x+ 6)=0

QUESTIONS:

1). What property of equations was used to solve incomplete quadratic equations?

2). What methods of factoring a polynomial were used to solve incomplete quadratic equations?

3). What is the algorithm for solving complete quadratic equations ?

0.2 roots; D = 0, 1 root; D X 1.2 = "width =" 640"

1). The product of two factors is equal to zero if one of them is equal to zero, while the second does not lose its meaning: ab = 0 , If a = 0 or b = 0 .

2). Taking out the common factor and

a 2 -b 2 =(a - b)(a + b) - the formula for the difference of squares.

3). Full quadratic equation ah 2 + bx + c = o.

D=b 2 – 4ac if D0, 2 roots;

D = 0, 1 root;

X 1,2 =

SOLVE EQUATIONS :

Group I: No. 802 p. 71 X 2 - 5x- 36 = 0

II group: No. 810 p. 71 3x 2 - x + 21=5x 2

III group: X 4 -5x 2 - 36 =0

III. SOLVE EQUATIONS :

I and II group: No. 860 = 0

III group: =0

What are such equations called? What property is used to solve them?

A rational equation is an equation of the form

A fraction is zero if the numerator is zero and the denominator is not zero. =0 if a = 0, b≠0.

Brief history of mathematics

The mathematicians of ancient Egypt knew how to solve quadratic and linear equations.

The Persian medieval scientist Al-Khwarizmi (IX century) first introduced algebra as an independent science about general methods for solving linear and quadratic equations, gave a classification of these equations.

A new great breakthrough in mathematics is associated with the name of the French scientist Francois Vieta (XVI century). It was he who introduced letters into algebra. He owns the well-known theorem on the roots of a quadratic equation.

And we owe the tradition of denoting unknown quantities with the last letters of the Latin alphabet (x, y, z) to another French mathematician - Rene Descartes (XVII).