The concept of a consequence of an equation. Extraneous roots. Presentation "Equivalence of equations. Equation %U2013 consequence" Which equation is a consequence of another

To study today's topic, we need to repeat which equation is called a corollary equation, which theorems are “troubling” and what stages the solution of any equation consists of.

Definition. If each root of the equation ef from x is equal to the same from x (we denote it by the number one) is at the same time the root of the equation pe from x, equal to ax from x (we denote it by the number two), then equation two is called a consequence of equation one.

Theorem four. If both sides of the equation ef from x are equal to the same from x multiplied by the same expression ax from x, which:

Firstly, it makes sense everywhere in the domain of definition (in the domain of admissible values) of the equation eff in x, equal to in x.

Secondly, nowhere in this region does not vanish, then the equation eff from x, multiplied by ash from x is equal to the same from x, multiplied by ax from x, is equivalent to that given in its ODZ.

Consequence theorems four is another “calm” statement: if both sides of the equation are multiplied or divided by the same non-zero number, you get an equation equivalent to the given one.

Theorem five. If both sides of the equation

eff from x is equal to ix is ​​not negative in the ODS equation, then after raising both of its parts to the same even power n, the equation eff from x to the nth power is equal to the nth power is equivalent to the given equation in its o de ze.

Theorem six. Let a be greater than zero, and not equal to one, and ef of x greater than zero,

zhe from x is greater than zero, the logarithmic equation logarithm ef from x to the base a, equal to the logarithm zhe from x to the base a,

is equivalent to the equation ef from x is equal to from x .

As we have already said, solving any equations occurs in three stages:

The first stage is technical. Using a chain of transformations from the original equation, we arrive at a fairly simple equation, which we solve and find the roots.

The second stage is solution analysis. We analyze the transformations that we performed and find out whether they are equivalent.

The third stage is verification. Checking all found roots by substituting them into the original equation is mandatory when performing transformations that can lead to a corollary equation.

In this lesson we will find out, when applying what transformations, this equation turns into a corollary equation? Consider the following tasks.

Exercise 1

Which equation is a corollary of the equation x minus three equals two?

Solution

The equation x minus three equals two has a single root - x equals five. Let's multiply both sides of this equation by the expression x minus six, add similar terms and get the quadratic equation x square minus eleven x plus thirty equals zero. Let's calculate its roots: x first is equal to five; x second is equal to six. It already contains two roots. The equation x square minus eleven x plus thirty equals zero contains a single root - x equals five; the equation x minus three equals two, so x squared minus eleven x plus thirty is a consequence of the equation x minus three equals two.

Task 2

What other equation is a consequence of the equation x-3=2?

Solution

In the equation x minus three equals two, let’s square both sides, apply the formula for the square of the difference, add similar terms, and get the quadratic equation x square minus six x plus five equals zero.

Let's calculate its roots: x first is equal to five, x second is equal to one.

The root x equals one is extraneous to the equation x minus three equals two. This happened because both sides of the original equation were squared (an even power). But at the same time, its left side - x minus three - can be negative (the conditions are violated theorems five). This means that the equation x square minus six x plus five equals zero is a consequence of the equation x minus three equals two.

Task 3

Find the corollary equation for the equation

The logarithm of x plus one to base three plus the logarithm of x plus three to base three is equal to one.

Solution

Let's imagine one as the logarithm of three to base three, potentiate the logarithmic equation, perform the multiplication, add similar terms and get the quadratic equation x square plus four x equals zero. Let's calculate its roots: the first x is equal to zero, the second x is equal to minus four. The root x equals minus four is extraneous to the logarithmic equation, since when substituting it into the logarithmic equation, the expressions x plus one and x plus three take on negative values ​​- the conditions are violated theorems six.

This means that the equation x squared plus four x equals zero is a consequence of this equation.

Based on the solution of these examples, we can do conclusion:the corollary equation is obtained from this equation by expanding the domain of definition of the equation. And this is possible by performing transformations such as

1) getting rid of denominators containing a variable value;

2) raising both sides of the equation to the same even power;

3) liberation from logarithmic signs.

Remember! If in the process of solving an equation the domain of definition of the equation was expanded, then it is necessary to check all the roots found.

Task 4

Solve the equation x minus three divided by x minus five plus one divided by x is equal to x plus five divided by x times x minus five.

Solution

The first stage is technical.

Let's carry out a chain of transformations, get the simplest equation and solve it. To do this, we multiply both sides of the equation by the common denominator of the fractions, that is, by the expression x multiplied by x minus five.

We get the quadratic equation x squared minus three x minus ten equals zero. Let's calculate the roots: x first is equal to five, x second is equal to minus two.

The second stage is solution analysis.

When solving an equation, we multiplied both sides by an expression containing a variable. This means that the scope of the equation has expanded. Therefore, checking the roots is mandatory.

The third stage is verification.

When x is equal to minus two, the common denominator does not go to zero. This means that x equals minus two is the root of this equation.

When x equals five, the common denominator goes to zero. Therefore, x equals five - an extraneous root.

Answer: minus two.

Task 5

Solve the equation square root of x minus six equals square root of four minus x.

Solution

The first stage is technical .

In order to obtain a simple equation and solve it, we perform a chain of transformations.

Let's square (an even power) both sides of this equation, move the x's to the left side, and the numbers to the right side of the equation, add similar terms, we get: two x's equals ten. X equals five.

The second stage is solution analysis.

Let's check the completed transformations for equivalence.

When solving the equation, we squared both sides of it. This means that the scope of the equation has expanded. Therefore, checking the roots is mandatory.

The third stage is verification.

Let's substitute the found roots into the original equation.

If x equals five, the expression square root of four minus x is undefined, so x equals five is an extraneous root. This means that this equation has no roots.

Answer: the equation has no roots.

Task 6

Solve the equation the natural logarithm of the expression x squared plus two x minus seven equals the natural logarithm of the expression x minus one.

Solution

The first stage is technical .

Let's carry out a chain of transformations, get the simplest equation and solve it. To do this, let's potentiate

equation, we move all the terms to the left side of the equation, bring similar terms, we get a quadratic equation x square plus x minus six equals zero. Let's calculate the roots: x first is equal to two, x second is equal to minus three.

The second stage is solution analysis.

Let's check the completed transformations for equivalence.

In the process of solving this equation, we freed ourselves from the signs of logarithms. This means that the scope of the equation has expanded. Therefore, checking the roots is mandatory.

The third stage is verification.

Let's substitute the found roots into the original equation.

If x is equal to two, then we get the natural logarithm of one equal to the natural logarithm of one -

true equality.

This means that x equals two is the root of this equation.

If x is equal to minus three, then the natural logarithm of the expression x square plus two x minus seven and the natural logarithm of the expression x minus one are not defined. This means that x equals minus three is an extraneous root.

Answer: two.

Is it always necessary to distinguish three stages when solving an equation? What other way can I check?

We will get answers to these questions in the next lesson.

In the presentation we will continue to consider equivalent equations, theorems, and will dwell in more detail on the stages of solving such equations.

To begin with, let us recall the condition under which one of the equations is a consequence of the other (slide 1). The author cites once again some theorems on equivalent equations that were discussed earlier: on multiplying parts of the equation by the same value h (x); raising parts of an equation to the same even power; obtaining an equivalent equation from the equation log a f(x) = log a g (x).

The 5th slide of the presentation highlights the main steps with which it is convenient to solve equivalent equations:

Find solutions to the equivalent equation;

Analyze solutions;

Check.

Let's consider example 1. It is necessary to find a consequence of the equation x - 3 = 2. Let's find the root of the equation x = 5. We write the equivalent equation (x - 3)(x - 6) = 2(x - 6), using the method of multiplying the parts of the equation by (x - 6). Simplifying the expression to the form x 2 - 11x +30 = 0, we find the roots x 1 = 5, x 2 = 6. Because Each root of the equation x - 3 = 2 is also a solution to the equation x 2 - 11x +30 = 0, then x 2 - 11x +30 = 0 is a corollary equation.

Example 2. Find another consequence of the equation x - 3 = 2. To obtain an equivalent equation, we use the method of raising to an even power. Simplifying the resulting expression, we write x 2 - 6x +5 = 0. Find the roots of the equation x 1 = 5, x 2 = 1. Because x = 5 (the root of the equation x - 3 = 2) is also a solution to the equation x 2 - 6x +5 = 0, then the equation x 2 - 6x +5 = 0 is also a corollary equation.

Example 3. It is necessary to find a consequence of the equation log 3 (x + 1) + log 3 (x + 3) = 1.

Let us replace 1 = log 3 3 in the equation. Then, applying the statement from Theorem 6, we write the equivalent equation (x + 1)(x +3) = 3. Simplifying the expression, we obtain x 2 + 4x = 0, where the roots are x 1 = 0, x 2 = - 4. So the equation x 2 + 4x = 0 is a consequence for the given equation log 3 (x + 1) + log 3 (x + 3) = 1.

So, we can conclude: if the domain of definition of an equation is expanded, then a corollary equation is obtained. Let us highlight the standard actions when finding a corollary equation:

Getting rid of denominators that contain a variable;

Raising parts of an equation to the same even power;

Liberation from logarithmic signs.

But it is important to remember: when during the solution the domain of definition of the equation expands, it is necessary to check all the roots found - whether they will fall into the ODZ.

Example 4. Solve the equation presented on slide 12. First, let's find the roots of the equivalent equation x 1 = 5, x 2 = - 2 (first stage). It is imperative to check the roots (second stage). Checking roots (third stage): x 1 = 5 does not belong to the range of permissible values ​​of the given equation, therefore the equation has one solution only x = - 2.

In example 5, the found root of the equivalent equation is not included in the ODZ of the given equation. In Example 6, the value of one of the two roots found is undefined, so this root is not a solution to the original equation.

Class: 11

Duration: 2 lessons.

The purpose of the lesson:

• (for teacher) formation in students of a holistic understanding of methods for solving irrational equations.
• (for students) Development of the ability to observe, compare, generalize, and analyze mathematical situations (slide 2). Preparation for the Unified State Exam.

First lesson plan(slide 3)

1. Updating knowledge
2. Analysis of the theory: Raising an equation to an even power
3. Workshop on solving equations

Second lesson plan

1. Differentiated independent work in groups “Irrational equations on the Unified State Exam”
2. Summary of lessons
3. Homework

Progress of lessons

I. Updating knowledge

Target: repeat the concepts necessary to successfully master the lesson topic.

Frontal survey.

– Which two equations are called equivalent?

– What transformations of an equation are called equivalent?

– Replace this equation with an equivalent one with an explanation of the applied transformation: (slide 4)

a) x+ 2x +1; b) 5 = 5; c) 12x = -3; d) x = 32; d) = -4.

– What equation is called the corollary equation of the original equation?

– Can a corollary equation have a root that is not the root of the original equation? What are these roots called?

– What transformations of the equation lead to corollary equations?

– What is called an arithmetic square root?

Today we will dwell in more detail on the transformation “Raising an equation to an even power”.

II. Analysis of the theory: Raising an equation to an even power

Teacher's explanation with active participation of students:

Let 2m(mN) is a fixed even natural number. Then the consequence of the equationf(x) =g(x) is the equation (f(x)) = (g(x)).

Very often this statement is used when solving irrational equations.

Definition. An equation containing an unknown under the root sign is called irrational.

When solving irrational equations, the following methods are used: (slide 5)

Attention! Methods 2 and 3 require mandatory checks.

ODZ does not always help eliminate extraneous roots.

Conclusion: When solving irrational equations, it is important to go through three stages: technical, solution analysis, verification (slide 6).

III. Workshop on solving equations

Solve the equation:

After discussing how to solve an equation by squaring, solve by going to an equivalent system.

Conclusion: The simplest equations with integer roots can be solved by any familiar method.

b) = x – 2

By solving by raising both sides of the equation to the same power, students obtain roots x = 0, x = 3 -, x = 3 +, which are difficult and time-consuming to check by substitution. (Slide 7). Transition to an equivalent system

allows you to quickly get rid of foreign roots. The condition x ≥ 2 is satisfied only by x.

Answer: 3 +

Conclusion: It is better to check irrational roots by moving to an equivalent system.

c) = x – 3

In the process of solving this equation, we obtain two roots: 1 and 4. Both roots satisfy the left side of the equation, but when x = 1 the definition of an arithmetic square root is violated. The ODZ equation does not help eliminate extraneous roots. The transition to an equivalent system gives the correct answer.

Conclusion:good knowledge and understanding of all the conditions for determining the arithmetic square root helps to move on toperforming equivalent transformations.

By squaring both sides of the equation, we get the equation

x + 13 - 8 + 16 = 3 + 2x - x, placing the radical on the right side, we get

26 – x + x = 8. Application of further actions to square both sides of the equation will lead to an equation of the 4th degree. The transition to the ODZ equation gives a good result:

let's find the ODZ equation:

x = 3.

Check: - 4 = , 0 = 0 correct.

Conclusion:Sometimes it is possible to solve using the definition of the ODZ equation, but be sure to check.

Solution: ODZ equation: -2 – x ≥ 0 x ≤ -2.

For x ≤ -2,< 0, а ≥ 0.

Therefore, the left side of the equation is negative, and the right side is non-negative; therefore the original equation has no roots.

Answer: no roots.

Conclusion:Having made the correct reasoning on the limitation in the condition of the equation, you can easily find the roots of the equation, or establish that they do not exist.

Using the example of solving this equation, show the double squaring of the equation, explain the meaning of the phrase “solitude of radicals” and the need to check the roots found.

h) + = 1.

The solution of these equations is carried out using the variable replacement method until the return to the original variable. Offer the solution to those who complete the tasks of the next stage earlier.

Control questions

• How to solve the simplest irrational equations?
• What do you need to remember when raising an equation to an even power? ( foreign roots may appear)
• What is the best way to test irrational roots? ( using ODZ and conditions for the coincidence of the signs of both sides of the equation)
• Why is it necessary to be able to analyze mathematical situations when solving irrational equations? ( For the correct and quick choice of how to solve the equation).

IV. Differentiated independent work in groups “Irrational equations on the Unified State Exam”

The class is divided into groups (2-3 people) according to levels of training, each group chooses an option with a task, discusses and solves the selected tasks. If necessary, seek advice from the teacher. After completing all the tasks in their version and checking the answers by the teacher, group members individually finish solving equations g) and h) of the previous stage of the lesson. For options 4 and 5 (after checking the answers and solutions by the teacher), additional tasks are written on the board and are completed individually.

All individual solutions are submitted to the teacher for verification at the end of the lessons.

Option 1

Solve the equations:

a) = 6;
b) = 2;
c) = 2 – x;
d) (x + 1) (5 – x) (+ 2 = 4.

Option 5

1. Solve the equation:

a) = ;
b) = 3 – 2x;

2. Solve the system of equations:

Additional tasks:

V. Lesson summary

What difficulties did you experience when completing USE tasks? What is needed to overcome these difficulties?

VI. Homework

Repeat the theory of solving irrational equations, read paragraph 8.2 in the textbook (pay attention to example 3).

Solve No. 8.8 (a, c), No. 8.9 (a, c), No. 8.10 (a).

Literature:

1. Nikolsky S.M., Potapov M.K., N.N. Reshetnikov N.N., Shevkin A.V. Algebra and beginning of mathematical analysis , textbook for 11th grade of general education institutions, M.: Prosveshchenie, 2009.
2. Mordkovich A.G. On some methodological issues related to solving equations. Mathematics at school. -2006. -No. 3.
3. M. Shabunin. Equations. Lectures for high school students and applicants. Moscow, “Chistye Prudy”, 2005. (library “First of September”)
4. E.N. Balayan. Problem solving workshop. Irrational equations, inequalities and systems. Rostov-on-Don, “Phoenix”, 2006.
5. Mathematics. Preparation for the Unified State Exam 2011. Edited by F.F. Lysenko, S.Yu. Kulabukhova Legion-M, Rostov-on-Don, 2010.

Municipal educational institution

"Novoukolovskaya secondary school"

Krasnensky district, Belgorod region

Algebra lesson in 11th grade

“Application of several transformations leading to a corollary equation”

Prepared and carried out

Mathematic teacher

Kharkovskaya Valentina Grigorievna

Algebra 11th grade

Subject: Application of several transformations leading to the corollary equation.

Target: create conditions for consolidating material on the topic: “Application of several transformations leading to an equation-consequence”; Rdevelop independence, improve speech literacy; to develop students’ computing skills; complete tasks corresponding to the Unified State Examination level.

Equipment: textbook, computer, cards

Lesson type: lesson on complex application of ZUN

During the classes

Organizational moment (Slide 1)

Good afternoon guys! Look at these pictures and choose which one you liked the most. I see that you, like me, came to class in a good mood, and I think it will remain the same until the end of the lesson. I would like to wish you fruitful work.

Guys, each of you has assessment sheets on your table in which you will evaluate yourself at each stage of the lesson.

Checking homework. (Slide 2)

Highlight the solutions on the slide and the children give themselves grades

self-control sheet. No errors – “5”, if 1 error – “4”, 2

errors – “3”. If you get a lot of children who have 2

mistakes, then solve this task at the board.

Announcing the topic of the lesson (Slide 3). setting lesson goals

You can see the topic of our lesson on the slide. What do you think than

Are we going to study with you in class today?

Well, guys, let's remember the material we covered. .

Let's start with oral work :

Oral work (Slide 4)

What equations are called corollary equations? (if any root of the first equation is a root of the second, then the second equation is called a consequence of the first);

What is called the transition to a corollary equation? (replacing an equation with another equation, which is its consequence);

What transformations lead to the corollary equation? Give examples. (raising an equation to an even power; potentiating a logarithmic equation; freeing the equation from the denominator; bringing similar terms of the equation; applying formulas).

Solve the equations (Slide 5)

(equations are displayed on the screen):

1) = 6; (answer: 36)

2) = 3; (answer: 11)

3) = 4; (answer: 6)

4) = - 2; (answer: no solutions, since the left side of the equation takes only non-negative values)

5) = 9; (answer: -9 and 9)

6) = -2; (answer: no solutions, since the sum of two

non-negative numbers cannot be negative)

Guys, I think you noticed that when doing homework and oral work, we came across tasks that corresponded to the demo version, specification and Unified State Examination codifier.

4.Completing tasks

Guys, let's work in our notebooks:

№8.26 (a) – at the blackboard

№8.14 (c) – at the blackboard

Exercise for the eyes (music)

№8.8 (c)-at the board

№8.9-(e)-at the board

5.Independent work (Slide 6)

Solution for independent work (Slide 7)

6. Homework: complete No. 8.14 (d), Unified State Examination B5 task in options 21,23,25 (Slide 8)

7. Lesson summary (Slide 9)

8.Reflection (Slide 10)

Questionnaire.

1. I worked during the lesson

2. Through my work in class I

3. The lesson seemed to me

4. For the lesson I

5. My mood

6. I had the lesson material

7. Do you think you can cope with such tasks in the exam?

8. Homework seems to me

active / passive

satisfied/dissatisfied

short / long

not tired / tired

it got better/it got worse

clear / not clear

useful/useless

interesting / boring

yes/no/don't know

easy / difficult

interesting / uninteresting

Resources used:

Nikolsky S.M., Potapov K.M., . Algebra and the beginnings of mathematical analysis, 11th grade M.: Prosveshcheniye, 2010

Collection of tasks for preparing for the Unified State Exam in mathematics