Sitbatalova Alma Kaparovna

mathematic teacher

Lyceum No. 15

Astana

“Argue, be mistaken, make mistakes, but, for God’s sake, think, and, even if it’s crooked, do it yourself.”

G. Lessing.

To develop schoolchildren’s ability to work with information, teach them to think independently, and be able to work in a team, you can use various educational technologies. The author gives preference to the group form of work.

Grade 11

Lesson topic: The nth root and its properties.

The purpose of the lesson:

Formation in students of a holistic understanding of the rootn -th degree, skills of conscious and rational use of the properties of the root when solving various problems;understanding the principles of simplifying expressions containing a radical. Check the level of students' understanding of the topic's questions.

Lesson objectives:

1. Update the necessary knowledge and skills.Give the concept of a rootn-th degree, consider its properties.

2. Organize students’ mental activity to solve the problem (build the necessary communication).Contribute to the development of algorithmic, creative thinking, develop self-control skills. To promote the development of interest in the subject and activity.

3. Cultivate respect for other people’s opinions and other people’s work through analysis and appropriation of a new way of activity,ability to work in a team, express one’s own opinion, give recommendations.

Equipment:

Computer, projector and screen for demonstrating the presentation; task cards for group work; cards with a table for assessing the assignment of a new type of activity; blank double sheets for students to do multi-level independent work; cards with multi-level tasks.

Lesson type:

Combined (systematization and generalization, assimilation of new knowledge, testing and evaluation of knowledge).

Forms of organization educational activities :

Individual, polylogue, dialogue, work with the text of a slide, textbook.

Methods :

Visual, verbal, graphic, conditionally symbolic, research.

Motivation cognitive activity students:

Tell students thatstudying the properties of the rootnThe th degree is a generalization of the properties of the degree already known to students.

Lesson plan:

Organizational and motivational ( teacher's greeting , acceptance of the topic, lesson goals , inclusion in work ).

Updating knowledge (systematization and generalization, assimilation of new knowledge).

Application of what has been learned ( establishing the correctness and awareness of learning new things educational material; identifying gaps and misconceptions and correcting them).

Control and self-control (Check of knowledge).

Reflection (Mobilizing students to reflect on their behavior (motivation, methods of activity, communication).

Summarizing (Give an analysis and assessment of the success of achieving the goal and outline the prospects for further work).

Homework (Ensuring an understanding of the purpose, content and methods of implementation homework).

During the classes:

Organizational and motivational ( teacher's greeting , acceptance of the topic, lesson goals, inclusion in work, 1-2 min ). Greeting students, message topic “Rootn– th degree and its properties”, communication of the purpose and method of activity.

Updating knowledge (systematization and generalization, assimilation of new knowledge, 15 min).

Repetition background knowledge(systematization and generalization):

The class is divided into three groups.

Teacher's activities: asking questions:

Definition of arithmetic square root.

Properties of the arithmetic square root.

Properties of a degree with a natural exponent.

Write properties on the sheet,

,

Answer questions,

Complete tasks.

Assimilation of new knowledge:

Teacher's activities: New concepts are introduced:

DEFINITION. Rootn th degree from amonga this number is calledn the th degree of which is equal toa .

DEFINITION. Arithmetic rootn th degree from amongA call a non-negative numbern the th degree of which is equal toa .

Basic properties of arithmetic rootsn -th degree.

When evenn there are two rootsn -th power of any positive numbera , rootn The th power of the number 0 is equal to the rudder; the even power root of negative numbers does not exist. For oddnthere is a rootn -noy from any numbera and only one at that.

For any numbers the following equalities hold:

1) ; 3) ;

2) 4) ;

5) ; 6) .

Examples of tasks are given on the slide:

Student activities in groups:

Write properties on the sheet yourself,

Check the slide for accuracy,

Answer questions,

Complete tasks.

Application of what has been learned ( establishing the correctness and awareness of mastering new educational material; identifying gaps and misconceptions and correcting them, 15 min).

Teacher's activities: Gives a comment on further actions:

Work in groups by stages,

In front of each group there is a piece of paper with the same task, but with different conditions (on the slide “Simplify the expression”):

- Stage 1 “Generation of ideas”.

1 stage:

Enter number 1.

Write down the order of expected actions required to complete the task.

Managing group activities (to ensure that all students are involved in the work).

- Stage 2 “Analysis of ideas”.

Familiarization with the activity instructions on the slide:

Stage:

Enter number 2.

Complete the task using the proposed algorithm, improving it if necessary.

Draw and write down a conclusion about whether the task can be completed using the proposed algorithm.

Management of group activities.

- Stage 3 “Examination”.

:

Stage:

Enter number 3.

Check the correctness of the task according to the algorithm.

Draw and write down a conclusion whether it was possible to create the necessary algorithm, and complete the task correctly.

- Stage 4 “Presentation of results”.

Introducing the activity instructions on the slide:

Stage:

Evaluate the activities of all groups at each stage.

Individually choose the stage at which it was easier to work and the stage at which difficulties arose.

Student activities in groups:

at stage 1:analyze tasks, perform the necessary actions,

at stage 2:analyze an algorithm proposed by another group, atmake adjustments as necessary,complete tasks,

at stage 3: analyze the workprevious groups, conclude,

at stage 4:analyze the conclusion drawn, check the correctness of the solution with the answer on the slide, fill out cards with a table, choosing the role in which they are more successful.

A minute of health (gymnastics for the eyes).

Control and self-control (Knowledge test, 7 min.)

Teacher's activities: Gives instructions for doing independent work:

All students complete tasks of level 1 (at “3”) tasks on the cards on the slide:

Independent work. Rating "3".

I option.

A)

b)

2). Compare numbers:

II option.

1). Find the value of a numeric expression:

A)

b)

2). Compare numbers:

:

Independent work. Rating "3".

Answers :

I option

1). a) 11

b) 15

2). <

II option

1). a) 7

b) 15

2. >

3. Who completed the level 1 task?

4. Students who have coped with level 1 move on to level 2 tasks (at “4”), those who have not coped with them remain at level 1 of the task on the slide, on cards:

Independent work.

Rating "3".

1). Find the value of a numeric expression:

A)

b)

2). Compare numbers:

Rating "4".

1). Solve the equation:

A)

b)

2). Simplify the expression:

Self-test based on the answers on the slide:

Independent work.

Answers :

Rating "3".

1). a) 13

b) 6

2). <

Rating "4".

1). A)

b)

2). 2a

6. Who moved to level 3?

Who stayed on level 2?

Who moved to level 2?

Who stayed at level 1?

7. Students who received a “4” complete tasks of level 3 (at “5”).

Students who do not receive a “4” and have completed Level 1 complete Level 2 assignments.

Students who do not receive a “3” complete tasks of level 1 on the cards on the slide:

Independent work.

Rating "4".

Rating "5".

Rating "4"?

Rating "3"?

10. Who failed the level 1 tasks?

Student activities in groups:

Complete tasks.

Perform a self-test, giving a mark of “3” if all tasks are completed.

Present the results.

Complete tasks.

Perform a self-test: put “3” if all tasks of level 1 are completed; put “4” if 2 out of 3 tasks of level 2 are completed.

Present the results.

Complete tasks.

Perform a self-test: put “3” if all tasks of level 1 are completed; put “4” if 2 tasks of level 2 are completed; give a rating of “5” if at least 1 task out of 2 is completed.

Present the results.

Reflection (Mobilizing students to reflect on their behavior (motivation, methods of activity, communication, 3 min).

Teacher's activities: Gives comments on writing “Cinquain”, instructions on the slide:

Sinkwine.

Line 1 – the topic or subject is stated (one noun);

Line 2 – description of the subject (two adjectives or participles);

Line 3 – characterizes the actions of the subject (three verbs);

Line 4 – expression of the author’s attitude to the subject (four words);

Line 5 – a synonym that generalizes or expands the meaning of the subject (one word).

Student activities in groups:

Get acquainted with the algorithm for writing Sinkwine,

Write cinquain on sheets of paper independent work,

Sinkwine is read out if desired,

Submit sheets for verification.

Summarizing (Give an analysis and assessment of the success of achieving the goal and outline the prospects for subsequent work, 1-2 min).

Teacher's activities: Analysis of performance assessment on different stages lesson: Why was it easier (more difficult) for you in this or that role? Each student's work is assessed.

Student activities in groups: answer the question.

Homework (Ensuring an understanding of the purpose, content and methods of completing homework, 1-2 min).

Teacher's activities: Gives instructions for execution homework: (A. Abylkasimova, natural-mathematical eg.)
§ 5, No. 83 (2; 4), No. 84 (2; 3), No. 86, 87 (3; 4), No. 89.

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We need to get acquainted with the properties of this operation, which we will do in this section.

All properties are formulated and proven only for non-negative values ​​of the variables contained under the signs of the roots.

Proof. Let us introduce the following notation: We need to prove that for non-negative numbers x, y, z the equality x-yz holds.
Because
So, But if the powers of two non-negative numbers are equal and the exponents are equal, then the bases are equal degrees; This means that from the equality x n =(уz) n it follows that x-yz, and this was what needed to be proven.

Let us give a brief summary of the proof of the theorem.

Notes:

1. Theorem 1 remains valid for the case when the radical expression is the product of more than two non-negative numbers.
2. Theorem 1 can be formulated using the “if...then” construction (as is customary for theorems in mathematics). Let us give the corresponding formulation: if a and b are non-negative numbers, then the equality holds. This is exactly how we will formulate the next theorem.

A short (albeit imprecise) formulation that is more convenient to use in practice: the root of fractions is equal to the fraction of the roots.

Proof. We will give a brief summary of the proof of Theorem 2, and you can try to make appropriate comments similar to those given in the proof of Theorem 1.

YOU, of course, noticed that the proven two properties roots nth degrees are a generalization of properties known to you from the 8th grade algebra course square roots. And if there were no other properties of roots of the nth degree, then everything would be simple (and not very interesting). In fact, there are several more interesting and important properties that we will discuss in this paragraph. But first, let's look at a few examples of using Theorems 1 and 2.

Example 1. Calculate
Solution. Using the first property of roots (Theorem 1), we obtain:

Note 3. You can, of course, solve this example differently, especially if you have a microcalculator at hand: multiply the numbers 125, 64 and 27, and then take the cube root of the resulting product. But, you see, the proposed solution is “more intelligent.”
Example 2. Calculate
Solution. Let's reverse mixed number into an improper fraction.
We have Using the second property of roots (Theorem 2), we obtain:

Example 3. Calculate:
Solution. Any formula in algebra, as you well know, is used not only “from left to right”, but also “from right to left”. Thus, the first property of roots means that they can be represented in the form and, conversely, can be replaced by the expression. The same applies to the second property of roots. Taking this into account, let's perform the calculations:

Example 4. Follow these steps:
Solution, a) We have:
b) Theorem 1 allows us to multiply only roots to the same degree, i.e. only roots with the same indicator. Here it is proposed to multiply the 2nd root of the number a by the 3rd root of the same number. We don’t know how to do this yet. Let's return to this problem later.
Let's continue studying the properties of radicals.

In other words, to build a root in natural degree, it is enough to raise the radical expression to this power.
This is a consequence of Theorem 1. In fact, for example, for k = 3 we obtain: We can reason in exactly the same way in the case of any other natural value of the exponent k.

In other words, to extract a root from a root, it is enough to multiply the indicators of the roots.
For example,
Proof. As in Theorem 2, we will give a brief summary of the proof, and you can try to make the appropriate comments on your own, similar to those given in the proof of Theorem 1.

Note 4. Let's take a breath. What have we learned from the theorems we have proven? We learned that four operations can be performed on roots: multiplication, division, exponentiation, and root extraction (from the root). But what about adding and subtracting roots? No way. We talked about this back in 8th grade about the operation of extracting a square root.

For example, you can’t write Really instead, But it’s obvious that Be careful!
Perhaps the most interesting property of roots is the one that will be discussed in the next theorem. Considering the special significance of this property, we will allow ourselves to break the certain style of formulations and proofs developed in this section, so that the formulation of Theorem 5 is a little “softer” and its proof is clearer.

For example:

(the indicators of the root and radical expression were divided by 4);

(indicators of root and radical expression were divided by 3);

(the indicators of the root and radical expression were multiplied by 2).

Proof. Let us denote the left side of the equality being proved by the letter. Then, by the definition of a root, the equality must be satisfied

Let us denote the right-hand side of the identity being proved by the letter y:

Then, by definition of a root, the equality

Let us raise both sides of the last equality to the same power p; we get:

So (see equalities (1) and (2)),

Comparing these two equalities, we come to the conclusion that x nр = y nр, and therefore x = y, which is what needed to be proved.
The proven theorem will allow us to solve the problem that we encountered above when solving example 5, where it was necessary to multiply roots with different exponents:

This is how they usually reason in such cases.
1) According to Theorem 5, in the expression it is possible to multiply both the index of the root (i.e. the number 2) and the exponent of the radical expression (i.e. the number 1) by the same thing natural number. Taking advantage of this, we multiply both indicators by 3; we get:
2) According to Theorem 5, in an expression it is possible to multiply both the exponent of the root (i.e. the number 3) and the exponent of the radical expression (i.e. the number 1) by the same natural number. Taking advantage of this, we multiply both indicators by 2; we get:

3) Since we received roots of the same 6th degree, we can multiply them:

Note 5. Have you forgotten that all the properties of roots that we discussed in this section were considered by us only for the case when the variables take only non-negative values? Why did such a restriction have to be made? Because nth root powers of a negative number does not always make sense - it is defined only for odd values ​​of n. For such values ​​of the root exponent, the considered properties of roots are also true in the case of negative radical expressions.

A.G. Mordkovich Algebra 10th grade

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This article is a collection of detailed information that relates to the topic of the properties of roots. Considering the topic, we will start with the properties, study all the formulations and provide evidence. To consolidate the topic, we will consider properties of the nth degree.

Yandex.RTB R-A-339285-1

Properties of roots

We'll talk about properties.

1. Property multiplied numbers a And b, which is represented as the equality a · b = a · b. It can be represented in the form of factors, positive or equal to zero a 1 , a 2 , … , a k as a 1 · a 2 · … · a k = a 1 · a 2 · … · a k ;
2. from the quotient a: b = a: b, a ≥ 0, b > 0, it can also be written in this form a b = a b;
3. Property from the power of a number a with even exponent a 2 m = a m for any number a, for example, the property of the square of a number a 2 = a.

In any of the presented equations, you can swap the parts before and after the dash sign, for example, the equality a · b = a · b is transformed as a · b = a · b. Equality properties are often used to simplify complex equations.

The proof of the first properties is based on the definition of the square root and the properties of powers with a natural exponent. To justify the third property, it is necessary to refer to the definition of the modulus of a number.

First of all, it is necessary to prove the properties of the square root a · b = a · b. According to the definition, it is necessary to consider that a b is a number, positive or equal to zero, which will be equal to a b during construction into a square. The value of the expression a · b is positive or equal to zero as the product of non-negative numbers. The property of powers of multiplied numbers allows us to represent equality in the form (a · b) 2 = a 2 · b 2 . By definition of the square root, a 2 = a and b 2 = b, then a · b = a 2 · b 2 = a · b.

In a similar way one can prove that from the product k multipliers a 1 , a 2 , … , a k will be equal to the product of the square roots of these factors. Indeed, a 1 · a 2 · … · a k 2 = a 1 2 · a 2 2 · … · a k 2 = a 1 · a 2 · … · a k .

From this equality it follows that a 1 · a 2 · … · a k = a 1 · a 2 · … · a k.

Let's look at a few examples to reinforce the topic.

Example 1

3 5 2 5 = 3 5 2 5, 4, 2 13 1 2 = 4, 2 13 1 2 and 2, 7 4 12 17 0, 2 (1) = 2, 7 4 12 17 · 0 , 2 (1) .

It is necessary to prove the property of the arithmetic square root of the quotient: a: b = a: b, a ≥ 0, b > 0. The property allows us to write the equality a: b 2 = a 2: b 2, and a 2: b 2 = a: b, while a: b is a positive number or equal to zero. This expression will become proof.

For example, 0:16 = 0:16, 80:5 = 80:5 and 30.121 = 30.121.

Let's consider the property of the square root of the square of a number. It can be written as an equality as a 2 = a To prove this property, it is necessary to consider in detail several equalities for a ≥ 0 and at a< 0 .

Obviously, for a ≥ 0 the equality a 2 = a is true. At a< 0 the equality a 2 = - a will be true. In fact, in this case − a > 0 and (− a) 2 = a 2 . We can conclude, a 2 = a, a ≥ 0 - a, a< 0 = a . Именно это и требовалось доказать.

Let's look at a few examples.

Example 2

5 2 = 5 = 5 and - 0.36 2 = - 0.36 = 0.36.

The proven property will help to justify a 2 m = a m, where a– real, and m-natural number. Indeed, the property of raising a power allows us to replace the power a 2 m expression (a m) 2, then a 2 m = (a m) 2 = a m.

Example 3

3 8 = 3 4 = 3 4 and (- 8 , 3) ​​14 = - 8 , 3 7 = (8 , 3) ​​7 .

Properties of the nth root

First, we need to consider the basic properties of nth roots:

1. Property from the product of numbers a And b, which are positive or equal to zero, can be expressed as the equality a · b n = a n · b n , this property is valid for the product k numbers a 1 , a 2 , … , a k as a 1 · a 2 · … · a k n = a 1 n · a 2 n · … · a k n ;
2. from a fractional number has the property a b n = a n b n , where a– any real number, which is positive or equal to zero, and b– positive real number;
3. For any a and even indicators n = 2 m a 2 · m 2 · m = a is true, and for odd n = 2 m − 1 the equality a 2 · m - 1 2 · m - 1 = a holds.
4. Property of extraction from a m n = a n m , where a– any number, positive or equal to zero, n And m are natural numbers, this property can also be represented in the form. . . a n k n 2 n 1 = a n 1 · n 2 . . . · n k ;
5. For any non-negative a and arbitrary n And m, which are natural, we can also define the fair equality a m n · m = a n ;
6. Property of degree n from the power of a number a, which is positive or equal to zero, to the natural power m, defined by the equality a m n = a n m ;
7. Comparison property that have the same indicators: for any positive numbers a And b such that a< b , the inequality a n< b n ;
8. Comparison property that have the same numbers under the root: if m And n – natural numbers that m > n, then at 0 < a < 1 the inequality a m > a n is true, and when a > 1 executed a m< a n .

The equalities given above are valid if the parts before and after the equal sign are swapped. They can also be used in this form. This is often used when simplifying or transforming expressions.

The proof of the above properties of a root is based on the definition, properties of the degree and the definition of the modulus of a number. These properties must be proven. But everything is in order.

1. First of all, let's prove the properties of the nth root of the product a · b n = a n · b n . For a And b , which are positive or equal to zero , the value a n · b n is also positive or equal to zero, since it is a consequence of multiplying non-negative numbers. The property of a product to the natural power allows us to write the equality a n · b n n = a n n · b n n . By definition of a root n-th degree a n n = a and b n n = b , therefore, a n · b n n = a · b . The resulting equality is exactly what needed to be proven.

This property can be proved similarly for the product k multipliers: for non-negative numbers a 1, a 2, …, a n, a 1 n · a 2 n · … · a k n ≥ 0.

Here are examples of using the root property n-th power from the product: 5 2 1 2 7 = 5 7 2 1 2 7 and 8, 3 4 17, (21) 4 3 4 5 7 4 = 8, 3 17, (21) 3 · 5 7 4 .

1. Let us prove the property of the root of the quotient a b n = a n b n . At a ≥ 0 And b > 0 the condition a n b n ≥ 0 is satisfied, and a n b n n = a n n b n n = a b .

Let's show examples:

Example 4

8 27 3 = 8 3 27 3 and 2, 3 10: 2 3 10 = 2, 3: 2 3 10.

1. For the next step it is necessary to prove the properties of the nth degree from the number to the degree n. Let's imagine this as the equality a 2 m 2 m = a and a 2 m - 1 2 m - 1 = a for any real a and natural m. At a ≥ 0 we get a = a and a 2 m = a 2 m, which proves the equality a 2 m 2 m = a, and the equality a 2 m - 1 2 m - 1 = a is obvious. At a< 0 we obtain, respectively, a = - a and a 2 m = (- a) 2 m = a 2 m. The last transformation of a number is valid according to the power property. This is precisely what proves the equality a 2 m 2 m = a, and a 2 m - 1 2 m - 1 = a will be true, since the odd degree is considered - c 2 m - 1 = - c 2 m - 1 for any number c , positive or equal to zero.

In order to consolidate the information received, let's consider several examples using the property:

Example 5

7 4 4 = 7 = 7, (- 5) 12 12 = - 5 = 5, 0 8 8 = 0 = 0, 6 3 3 = 6 and (- 3, 39) 5 5 = - 3, 39.

1. Let us prove the following equality a m n = a n m . To do this, you need to swap the numbers before and after the equal sign a n · m = a m n . This will mean the entry is correct. For a, which is positive or equal to zero , of the form a m n is a number positive or equal to zero. Let us turn to the property of raising a power to a power and its definition. With their help, you can transform equalities in the form a m n n · m = a m n n m = a m m = a. This proves the property of the root of the root under consideration.

Other properties are proved in a similar way. Really, . . . a n k n 2 n 1 n 1 · n 2 · . . . · n k = . . . a n k n 3 n 2 n 2 · n 3 · . . . · n k = . . . a n k n 4 n 3 n 3 · n 4 · . . . · n k = . . . = a n k n k = a .

For example, 7 3 5 = 7 5 3 and 0.0009 6 = 0.0009 2 2 6 = 0.0009 24.

1. Let us prove the following property a m n · m = a n . To do this, it is necessary to show that a n is a number, positive or equal to zero. When raised to the power n m is equal to a m. If the number a is positive or equal to zero, then n-th degree from among a is a positive number or equal to zero. In this case, a n · m n = a n n m , which is what needed to be proven.

In order to consolidate the knowledge gained, let's look at a few examples.

1. Let us prove the following property – the property of a root of a power of the form a m n = a n m . It is obvious that when a ≥ 0 the degree a n m is a non-negative number. Moreover, her n the th power is equal to a m, indeed, a n m n = a n m · n = a n n m = a m . This proves the property of the degree under consideration.

For example, 2 3 5 3 = 2 3 3 5.

1. It is necessary to prove that for any positive numbers a and b the condition is satisfied a< b . Consider the inequality a n< b n . Воспользуемся методом от противного a n ≥ b n . Тогда, согласно свойству, о котором говорилось выше, неравенство считается верным a n n ≥ b n n , то есть, a ≥ b . Но это не соответствует условию a< b . Therefore, a n< b n при a< b .

For example, let's give 12 4< 15 2 3 4 .

1. Consider the property of the root n-th degree. It is necessary to first consider the first part of the inequality. At m > n And 0 < a < 1 true a m > a n . Let's assume that a m ≤ a n. The properties will allow you to simplify the expression to a n m · n ≤ a m m · n . Then, according to the properties of a degree with a natural exponent, the inequality a n m · n m · n ≤ a m m · n m · n holds, that is, a n ≤ a m. The obtained value at m > n And 0 < a < 1 does not correspond to the properties given above.

In the same way it can be proven that when m > n And a > 1 the condition a m is true< a n .

In order to consolidate the above properties, let's consider several specific examples. Let's look at inequalities using specific numbers.

Example 6

0 , 7 3 < 0 , 7 5 и 12 > 12 7 .

If you notice an error in the text, please highlight it and press Ctrl+Enter

To successfully use the root extraction operation in practice, you need to become familiar with the properties of this operation.
All properties are formulated and proven only for non-negative values ​​of the variables contained under the signs of the roots.

Theorem 1. Root nth degree(n=2, 3, 4,...) from the product of two non-negative chips is equal to the product nth roots powers of these numbers:

Comment:

1. Theorem 1 remains valid for the case when the radical expression is the product of more than two non-negative numbers.

Theorem 2.If, and n is a natural number greater than 1, then the equality is true

Brief(albeit inaccurate) formulation, which is more convenient to use in practice: the root of a fraction is equal to the fraction of the roots.

Theorem 1 allows us to multiply t only roots of the same degree , i.e. only roots with the same index.

Theorem 3.If ,k is a natural number and n is a natural number greater than 1, then the equality is true

In other words, to raise a root to a natural power, it is enough to raise the radical expression to this power.
This is a consequence of Theorem 1. In fact, for example, for k = 3 we obtain: We can reason in exactly the same way in the case of any other natural value of the exponent k.

Theorem 4.If ,k, n are natural numbers greater than 1, then the equality is true

In other words, to extract a root from a root, it is enough to multiply the indicators of the roots.
For example,

Be careful! We learned that four operations can be performed on roots: multiplication, division, exponentiation, and root extraction (from the root). But what about adding and subtracting roots? No way.
For example, instead of writing Really, But it’s obvious that

Theorem 5.If the indicators of the root and radical expression are multiplied or divided by the same natural number, then the value of the root will not change, i.e.

Examples of problem solving

Example 1. Calculate

Solution. Using the first property of roots (Theorem 1), we obtain:

Example 2. Calculate
Solution. Convert a mixed number to an improper fraction.
We have Using the second property of roots ( Theorem 2 ), we get:

Example 3. Calculate:

Solution. Any formula in algebra, as you well know, is used not only “from left to right”, but also “from right to left”. Thus, the first property of roots means that they can be represented in the form and, conversely, can be replaced by the expression. The same applies to the second property of roots. Taking this into account, let's perform the calculations.

The nth root, its properties.

An arithmetic root of the nth degree of a number is a non-negative number, nth degree which is equal to a.

Denotes the arithmetic root of the nth degree of the number a

where n is the root exponent,

a is a radical expression.

The sign is also called a radical.

The arithmetic root of the second degree is called the square root and is denoted √, the arithmetic root of the third degree is called the cube root o is denoted

For example:

A) and 2≥0;

b) and 3≥0;

V)

From the definition arithmetic root nth degree it follows that for even n the radical expression must be greater than or equal to zero, which means the value of such a root is also non-negative, for example: the arithmetic root of the 4th degree of the number -81 does not exist, since not a single number in the fourth degree will not give -81 (when raised to an even power, the value of the expression is always non-negative).

If the root exponent is odd, the radical expression can be negative, and then the minus can be taken out beyond the knight sign.

For example:

Equation x n =a.

The equation x n =a for odd n has a unique solution x = .

For example: x 3 = -125;

x= ;

x=- ;

For clarity, let's check:

125=-125- correct.

Answer: x=-5.

The equation x n =a for even n has and positive a has two roots

For example:

x 1 = ; x 2 =-;

x 1 =2; x 2 = -2.

You can verify by checking that 2 4 =16 and (-2) 4 =16.

Answer: ±2.

Sometimes you need to apply this property of arithmetic nth root degrees:

|x|, if n is even;

x if n is odd.

x, if x≥0;

Recall that |x|= -x if x<0.

For example:

.

Because <0, следовательно

Basic properties of roots.

For an arithmetic root of the nth degree, as for a square root, there are operations of adding a factor under the root sign and removing the factor from under the root sign.

For example:

From the example it is clear that to enter a multiplier under the sign of the nth root, it must be

raise to the nth power. We must remember that under a sign with an even indicator we have the right to enter only a positive multiplier, for example:

The multiplier is removed from the root sign in the same way, for example:

formula of the nth term an: a n = a 1 + d · (n - 1)

formula for the nth term:

Functions y=kx (where k is any natural number). Direct proportionality, straight graph.
Properties:
domain of definition - R
range - R
odd
for k >0 the function increases, for k<0 –убывает

Root of a quadratic equation (formula)

Roots of the equation ax 2 + bx + c = 0 (a¹ 0) are found using the formula . Expression = D 2 – 4b ac .

is called the discriminant of a quadratic equation. A quadratic equation has real roots (or root) if and only if its discriminant is non-negative.