Expressions with roots how to solve. Square root. Detailed theory with examples. Algebraic root: for those who want to know more

It's time to disassemble root extraction methods. They are based on the properties of the roots, in particular, on the equality, which is true for any non-negative number b.

Below we will consider in turn the main methods of extracting roots.

Let's start with the simplest case - extracting roots from natural numbers using a table of squares, a table of cubes, etc.

If the tables of squares, cubes, etc. is not at hand, then it is logical to use the method of extracting the root, which involves decomposing the root number into prime factors.

Separately, it is worth dwelling on, which is possible for roots with odd exponents.

Finally, consider a method that allows you to sequentially find the digits of the value of the root.

Let's get started.

Using a table of squares, a table of cubes, etc.

In the simplest cases, tables of squares, cubes, etc. allow extracting roots. What are these tables?

The table of squares of integers from 0 to 99 inclusive (shown below) consists of two zones. The first zone of the table is located on a gray background; by selecting a certain row and a certain column, it allows you to make a number from 0 to 99. For example, let's select a row of 8 tens and a column of 3 units, with this we fixed the number 83. The second zone occupies the rest of the table. Each of its cells is located at the intersection of a certain row and a certain column, and contains the square of the corresponding number from 0 to 99 . At the intersection of our chosen row of 8 tens and column 3 of one, there is a cell with the number 6889, which is the square of the number 83.

Tables of cubes, tables of fourth powers of numbers from 0 to 99 and so on are similar to the table of squares, only they contain cubes, fourth powers, etc. in the second zone. corresponding numbers.

Tables of squares, cubes, fourth powers, etc. allow you to extract square roots, cube roots, fourth roots, etc. respectively from the numbers in these tables. Let us explain the principle of their application in extracting roots.

Let's say we need to extract the root of the nth degree from the number a, while the number a is contained in the table of nth degrees. According to this table, we find the number b such that a=b n . Then , therefore, the number b will be the desired root of the nth degree.

As an example, let's show how the cube root of 19683 is extracted using the cube table. We find the number 19 683 in the table of cubes, from it we find that this number is a cube of the number 27, therefore, .

It is clear that tables of n-th degrees are very convenient when extracting roots. However, they are often not at hand, and their compilation requires a certain amount of time. Moreover, it is often necessary to extract roots from numbers that are not contained in the corresponding tables. In these cases, one has to resort to other methods of extracting the roots.

Decomposition of the root number into prime factors

A fairly convenient way to extract the root from a natural number (if, of course, the root is extracted) is to decompose the root number into prime factors. His the essence is as follows: after it is quite easy to represent it as a degree with the desired indicator, which allows you to get the value of the root. Let's explain this point.

Let the root of the nth degree be extracted from a natural number a, and its value is equal to b. In this case, the equality a=b n is true. The number b as any natural number can be represented as a product of all its prime factors p 1 , p 2 , …, p m in the form p 1 p 2 … p m , and the root number a in this case is represented as (p 1 p 2 ... p m) n . Since the decomposition of the number into prime factors is unique, the decomposition of the root number a into prime factors will look like (p 1 ·p 2 ·…·p m) n , which makes it possible to calculate the value of the root as .

Note that if the factorization of the root number a cannot be represented in the form (p 1 ·p 2 ·…·p m) n , then the root of the nth degree from such a number a is not completely extracted.

Let's deal with this when solving examples.

Example.

Take the square root of 144 .

Solution.

If we turn to the table of squares given in the previous paragraph, it is clearly seen that 144=12 2 , from which it is clear that the square root of 144 is 12 .

But in the light of this point, we are interested in how the root is extracted by decomposing the root number 144 into prime factors. Let's take a look at this solution.

Let's decompose 144 to prime factors:

That is, 144=2 2 2 2 3 3 . Based on the resulting decomposition, the following transformations can be carried out: 144=2 2 2 2 3 3=(2 2) 2 3 2 =(2 2 3) 2 =12 2. Consequently, .

Using the properties of the degree and properties of the roots, the solution could be formulated a little differently: .

Answer:

To consolidate the material, consider the solutions of two more examples.

Example.

Calculate the root value.

Solution.

The prime factorization of the root number 243 is 243=3 5 . In this way, .

Answer:

Example.

Is the value of the root an integer?

Solution.

To answer this question, let's decompose the root number into prime factors and see if it can be represented as a cube of an integer.

We have 285 768=2 3 3 6 7 2 . The resulting decomposition is not represented as a cube of an integer, since the degree of the prime factor 7 is not a multiple of three. Therefore, the cube root of 285,768 is not taken completely.

Answer:

No.

Extracting roots from fractional numbers

It's time to figure out how the root is extracted from a fractional number. Let the fractional root number be written as p/q . According to the property of the root of the quotient, the following equality is true. From this equality it follows fraction root rule: The root of a fraction is equal to the quotient of dividing the root of the numerator by the root of the denominator.

Let's look at an example of extracting a root from a fraction.

Example.

What is the square root of common fraction 25/169 .

Solution.

According to the table of squares, we find that the square root of the numerator of the original fraction is 5, and the square root of the denominator is 13. Then . This completes the extraction of the root from an ordinary fraction 25/169.

Answer:

root of decimal fraction or a mixed number is extracted after replacing the root numbers with ordinary fractions.

Example.

Take the cube root of the decimal 474.552.

Solution.

Let's represent the original decimal as a common fraction: 474.552=474552/1000 . Then . It remains to extract the cube roots that are in the numerator and denominator of the resulting fraction. Because 474 552=2 2 2 3 3 3 13 13 13=(2 3 13) 3 =78 3 and 1 000=10 3 , then and . It remains only to complete the calculations .

Answer:

.

Extracting the root of a negative number

Separately, it is worth dwelling on extracting roots from negative numbers. When studying roots, we said that when the exponent of the root is an odd number, then under the sign of the root there can be a negative number. We gave such notations the following meaning: for a negative number −a and an odd exponent of the root 2 n−1, we have . This equality gives rule for extracting odd roots from negative numbers: to extract the root from a negative number, you need to extract the root from the opposite positive number, and put a minus sign in front of the result.

Let's consider an example solution.

Example.

Find the root value.

Solution.

Let's transform the original expression so that a positive number appears under the root sign: . Now we replace the mixed number with an ordinary fraction: . We apply the rule of extracting the root from an ordinary fraction: . It remains to calculate the roots in the numerator and denominator of the resulting fraction: .

Here is a summary of the solution: .

Answer:

.

Bitwise Finding the Root Value

In the general case, under the root there is a number that, using the techniques discussed above, cannot be represented as the nth power of any number. But at the same time, there is a need to know the value of a given root, at least up to a certain sign. In this case, to extract the root, you can use an algorithm that allows you to consistently obtain a sufficient number of values ​​​​of the digits of the desired number.

The first step of this algorithm is to find out what is the most significant bit of the root value. To do this, the numbers 0, 10, 100, ... are successively raised to the power n until a number exceeding the root number is obtained. Then the number that we raised to the power of n in the previous step will indicate the corresponding high order.

For example, consider this step of the algorithm when extracting the square root of five. We take the numbers 0, 10, 100, ... and square them until we get a number greater than 5 . We have 0 2 =0<5 , 10 2 =100>5 , which means that the most significant digit will be the units digit. The value of this bit, as well as lower ones, will be found in the next steps of the root extraction algorithm.

All the following steps of the algorithm are aimed at successive refinement of the value of the root due to the fact that the values ​​of the next digits of the desired value of the root are found, starting from the highest and moving to the lowest. For example, the value of the root in the first step is 2 , in the second - 2.2 , in the third - 2.23 , and so on 2.236067977 ... . Let us describe how the values ​​of the bits are found.

Finding bits is carried out by enumeration of their possible values ​​0, 1, 2, ..., 9 . In this case, the nth powers of the corresponding numbers are calculated in parallel, and they are compared with the root number. If at some stage the value of the degree exceeds the radical number, then the value of the digit corresponding to the previous value is considered found, and the transition to the next step of the root extraction algorithm is made, if this does not happen, then the value of this digit is 9 .

Let us explain all these points using the same example of extracting the square root of five.

First, find the value of the units digit. We will iterate over the values ​​0, 1, 2, …, 9 , calculating respectively 0 2 , 1 2 , …, 9 2 until we get a value greater than the radical number 5 . All these calculations are conveniently presented in the form of a table:

So the value of the units digit is 2 (because 2 2<5 , а 2 3 >5 ). Let's move on to finding the value of the tenth place. In this case, we will square the numbers 2.0, 2.1, 2.2, ..., 2.9, comparing the obtained values ​​\u200b\u200bwith the root number 5:

Since 2.2 2<5 , а 2,3 2 >5 , then the value of the tenth place is 2 . You can proceed to finding the value of the hundredths place:

So the next value of the root of five is found, it is equal to 2.23. And so you can continue to find values ​​further: 2,236, 2,2360, 2,23606, 2,236067, … .

To consolidate the material, we will analyze the extraction of the root with an accuracy of hundredths using the considered algorithm.

First, we define the senior digit. To do this, we cube the numbers 0, 10, 100, etc. until we get a number greater than 2,151.186 . We have 0 3 =0<2 151,186 , 10 3 =1 000<2151,186 , 100 3 =1 000 000>2 151.186 , so the most significant digit is the tens digit.

Let's define its value.

Since 10 3<2 151,186 , а 20 3 >2,151.186 , then the value of the tens digit is 1 . Let's move on to units.

Thus, the value of the ones place is 2 . Let's move on to ten.

Since even 12.9 3 is less than the radical number 2 151.186 , the value of the tenth place is 9 . It remains to perform the last step of the algorithm, it will give us the value of the root with the required accuracy.

At this stage, the value of the root is found up to hundredths: .

In conclusion of this article, I would like to say that there are many other ways to extract roots. But for most tasks, those that we studied above are sufficient.

Bibliography.

• Makarychev Yu.N., Mindyuk N.G., Neshkov K.I., Suvorova S.B. Algebra: textbook for 8 cells. educational institutions.
• Kolmogorov A.N., Abramov A.M., Dudnitsyn Yu.P. and others. Algebra and the Beginnings of Analysis: A Textbook for Grades 10-11 of General Educational Institutions.
• Gusev V.A., Mordkovich A.G. Mathematics (a manual for applicants to technical schools).

The presence of square roots in an expression complicates the process of division, but there are rules by which working with fractions becomes much easier.

The only thing you need to keep in mind all the time- radical expressions are divided into radical expressions, and factors into factors. In the process of dividing square roots, we simplify the fraction. Also, recall that the root can be in the denominator.

Method 1. Division of radical expressions

Action algorithm:

Write a fraction

If the expression is not represented as a fraction, it is necessary to write it like this, because it is easier to follow the principle of dividing square roots.

Example 1

144 ÷ 36 , this expression should be rewritten like this: 144 36

Use one root sign

If both the numerator and denominator contain square roots, it is necessary to write their root expressions under the same root sign to make the solution process easier.

We remind you that a radical expression (or number) is an expression under the root sign.

Example 2

144 36 . This expression should be written like this: 144 36

Split root expressions

Just divide one expression by another, and write the result under the root sign.

Example 3

144 36 = 4 , we write this expression as follows: 144 36 = 4

Simplify the radical expression (if needed)

If the root expression or one of the factors is a perfect square, simplify that expression.

Recall that a perfect square is a number that is the square of some integer.

Example 4

4 is a perfect square because 2 × 2 = 4. Therefore:

4 = 2 × 2 = 2. Therefore 144 36 = 4 = 2 .

Method 2. Decomposition of the radical expression into factors

Action algorithm:

Write a fraction

Rewrite the expression as a fraction (if it is represented as such). This greatly simplifies the process of dividing expressions with square roots, especially when factoring.

Example 5

8 ÷ 36 , rewrite like this 8 36

Factorize each of the radical expressions

Factor the number under the root, like any other integer, only write the factors under the root sign.

Example 6

8 36 = 2 x 2 x 2 6 x 6

Simplify the numerator and denominator of a fraction

To do this, it is necessary to take out factors that are full squares from under the root sign. Thus, the factor of the root expression becomes the factor before the root sign.

Example 7

2 2 6 6 × 6 2 × 2 × 2 , from which follows: 8 36 = 2 2 6

Rationalize the denominator (get rid of the root)

In mathematics, there are rules according to which leaving the root in the denominator is a sign of bad taste, i.e. it is forbidden. If there is a square root in the denominator, then get rid of it.

Multiply the numerator and denominator by the square root you want to get rid of.

Example 8

In the expression 6 2 3, you need to multiply the numerator and denominator by 3 to get rid of it in the denominator:

6 2 3 × 3 3 = 6 2 × 3 3 × 3 = 6 6 9 = 6 6 3

Simplify the resulting expression (if necessary)

If the numerator and denominator contain numbers that can and should be reduced. Simplify such expressions as you would any fraction.

Example 9

2 6 simplifies to 1 3 ; so 2 2 6 simplifies to 1 2 3 = 2 3

Method 3. Dividing square roots with factors

Action algorithm:

Simplify Multipliers

Recall that the factors are the numbers in front of the root sign. To simplify the factors, you will need to divide or reduce them. Do not touch the root expressions!

Example 10

4 32 6 16 . First, we reduce 4 6: we divide by 2 both the numerator and the denominator: 4 6 \u003d 2 3.

Simplify square roots

If the numerator is evenly divisible by the denominator, then divide. If not, then simplify the radical expressions like any other.

Example 11

32 is evenly divisible by 16, so: 32 16 = 2

Multiply simplified factors by simplified roots

Remember the rule: do not leave roots in the denominator. Therefore, we simply multiply the numerator and denominator by this root.

Example 12

2 3 × 2 = 2 2 3

Rationalize the denominator (get rid of the root in the denominator)

Example 13

4 3 2 7 . Multiply the numerator and denominator by 7 to get rid of the root in the denominator.

4 3 7 × 7 7 = 4 3 × 7 7 × 7 = 4 21 49 = 4 21 7

Method 4. Division by a binomial with a square root

Action algorithm:

Determine if the binomial (binomial) is in the denominator

Recall that a binomial is an expression that includes 2 monomials. This method takes place only in cases where the denominator is a binomial with a square root.

Example 14

1 5 + 2 - there is a binomial in the denominator, since there are two monomials.

Find expression conjugate to binomial

Recall that the conjugate binomial is a binomial with the same monomials but opposite signs. To simplify the expression and get rid of the root in the denominator, you should multiply the conjugate binomials.

Example 15

5 + 2 and 5 - 2 are conjugate binomials.

Multiply the numerator and denominator by the binomial that is conjugate to the binomial in the denominator

This option will help get rid of the root in the denominator, since the product of conjugate binomials is equal to the difference of the squares of each binomial term: (a - b) (a + b) = a 2 - b 2

Example 16

1 5 + 2 = 1 (5 - 2) (5 - 2) (5 + 2) = 5 - 2 (5 2 - (2) 2 = 5 - 2 25 - 2 = 5 - 2 23 .

It follows from this: 1 5 + 2 = 5 - 2 23 .

Tips:

1. If you are working with the square roots of mixed numbers, then convert them to an improper fraction.
2. The difference between addition and subtraction from division is that radical expressions in the case of division are not recommended to be simplified (due to full squares).
3. Never (!) leave the root in the denominator.
4. No decimals or mixed before the root - you need to convert them to an ordinary fraction, and then simplify.
5. Is the denominator the sum or difference of two monomials? Multiply such a binomial by its conjugate binomial and get rid of the root in the denominator.

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

Root Properties

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 as multipliers, 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 private 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 an even exponent a 2 m = a m for any number a, for example, a property from 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 natural indicator. To substantiate 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 a product of non-negative numbers. The property of the degree of multiplied numbers allows us to represent equality in the form (a · b) 2 = a 2 · b 2 . By the definition of the square root a 2 = a and b 2 = b, then a · b 2 = 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 · … · ak 2 = a 1 2 · a 2 2 · … · ak 2 = a 1 · a 2 · … · a k .

It follows from this equality 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 you 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 be the proof.

For example, 0:16 = 0:16, 80:5 = 80:5 and 30, 121 = 30, 121.

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. Actually, in this case − a > 0 and (− a) 2 = a 2 . We can conclude that 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 proved property will help to justify a 2 m = a m , where a- real, and m-natural number. Indeed, the exponentiation property allows us to replace the degree a 2 m expression (am) 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 you need to consider the main properties of the roots of the nth degree:

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 is any real number that is positive or equal to zero, and b is a positive real number;
3. For any a and even numbers 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 is fulfilled.
4. Extraction property from a m n = a n m , where a- any number, positive or equal to zero, n and m – integers, this property can also be represented as. . . a n k n 2 n 1 = a n 1 · n 2 . . . nk ;
5. For any non-negative a and arbitrary n and m, which are natural, one can also define the fair equality a m n · m = a n ;
6. degree property n from the power of a number a, which is positive or equal to zero, in kind m, defined by the equality a m n = a n m ;
7. Comparison property that have the same exponents: for any positive numbers a and b such that a< b , the inequality a n< b n ;
8. Property of comparisons 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 valid, and for a > 1 a m< a n .

The above equations are valid if the parts before and after the equals sign are reversed. They can be used in this form as well. This is often used during simplification or transformation of expressions.

The proof of the above properties of the root is based on the definition, the 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, we will prove the properties of the root of the nth degree from the product a · b n = a n · b n . For a and b , which are positive or zero , the value a n · b n is also positive or equal to zero, since it is a consequence of multiplication of non-negative numbers. The property of a natural power product allows us to write the equality a n · b n n = a n n · b n n . By definition of 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 was required to be proved.

This property is proved similarly for the product k factors: 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. We represent this as an 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 get respectively a = - a and a 2 m = (- a) 2 m = a 2 m . The last transformation of the number is valid according to the property of the degree. This is what proves the equality a 2 m 2 m \u003d a, and a 2 m - 1 2 m - 1 \u003d a will be true, since - c 2 m - 1 \u003d - c 2 m is considered for an odd degree - 1 for any number c , positive or equal to zero.

In order to consolidate the received information, consider a few 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 change the numbers before the equal sign and after it in places a n · m = a m n . This will indicate the correct entry. For a , which is positive or equal to zero , from the form a m n is a positive number or equal to zero. Let us turn to the property of raising a power to a power and the 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 considered property of a root from a root.

Other properties are proved similarly. Really, . . . a n k n 2 n 1 n 1 n 2 . . . nk = . . . a n k n 3 n 2 n 2 n 3 . . . nk = . . . a nk n 4 n 3 n 3 n 4 . . . nk = . . . = 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 that is positive or equal to zero. When raised to a power n m is a m. If number a is positive or zero, then n th degree from among a is a positive number or equal to zero Moreover, a n · m n = a n n m , which was to be proved.

In order to consolidate the acquired knowledge, consider a few examples.

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

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

1. We need to prove that for any positive numbers a and b 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, we give 12 4< 15 2 3 4 .

1. Consider the root property n-th degree. First, consider the first part of the inequality. At m > n and 0 < a < 1 true a m > a n . Suppose a m ≤ a n . Properties will 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 is satisfied, that is, a n ≤ a m. The value obtained at m > n and 0 < a < 1 does not match the properties above.

In the same way, one can prove that m > n and a > 1 condition a m< a n .

In order to fix the above properties, consider a few concrete examples. Consider inequalities using specific numbers.

Example 6

0 , 7 3 > 0 , 7 5 and 12 > 12 7 .

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Before the advent of calculators, students and teachers calculated square roots by hand. There are several ways to manually calculate the square root of a number. Some of them offer only an approximate solution, others give an exact answer.

Steps

Prime factorization

Factor the root number into factors that are square numbers. Depending on the root number, you will get an approximate or exact answer. Square numbers are numbers from which the whole square root can be taken. Factors are numbers that, when multiplied, give the original number. For example, the factors of the number 8 are 2 and 4, since 2 x 4 = 8, the numbers 25, 36, 49 are square numbers, since √25 = 5, √36 = 6, √49 = 7. Square factors are factors , which are square numbers. First, try to factorize the root number into square factors.

• For example, calculate the square root of 400 (manually). First try factoring 400 into square factors. 400 is a multiple of 100, that is, divisible by 25 - this is a square number. Dividing 400 by 25 gives you 16. The number 16 is also a square number. Thus, 400 can be factored into square factors of 25 and 16, that is, 25 x 16 = 400.
• This can be written as follows: √400 = √(25 x 16).

1. The square root of the product of some terms is equal to the product of the square roots of each term, that is, √(a x b) = √a x √b. Use this rule and take the square root of each square factor and multiply the results to find the answer.

   • In our example, take the square root of 25 and 16.
     • √(25 x 16)
     • √25 x √16
     • 5 x 4 = 20

2. If the radical number does not factor into two square factors (and it does in most cases), you will not be able to find the exact answer as an integer. But you can simplify the problem by decomposing the root number into a square factor and an ordinary factor (a number from which the whole square root cannot be taken). Then you will take the square root of the square factor and you will take the root of the ordinary factor.

   • For example, calculate the square root of the number 147. The number 147 cannot be factored into two square factors, but it can be factored into the following factors: 49 and 3. Solve the problem as follows:
     • = √(49 x 3)
     • = √49 x √3
     • = 7√3

3. If necessary, evaluate the value of the root. Now you can evaluate the value of the root (find an approximate value) by comparing it with the values ​​​​of the roots of square numbers that are closest (on both sides of the number line) to the root number. You will get the value of the root as a decimal fraction, which must be multiplied by the number behind the root sign.

   • Let's go back to our example. The root number is 3. The nearest square numbers to it are the numbers 1 (√1 = 1) and 4 (√4 = 2). Thus, the value of √3 lies between 1 and 2. Since the value of √3 is probably closer to 2 than to 1, our estimate is: √3 = 1.7. We multiply this value by the number at the root sign: 7 x 1.7 \u003d 11.9. If you do the calculations on a calculator, you get 12.13, which is pretty close to our answer.
     • This method also works with big numbers. For example, consider √35. The root number is 35. The nearest square numbers to it are the numbers 25 (√25 = 5) and 36 (√36 = 6). Thus, the value of √35 lies between 5 and 6. Since the value of √35 is much closer to 6 than it is to 5 (because 35 is only 1 less than 36), we can state that √35 is slightly less than 6. Checking with a calculator gives us the answer 5.92 - we were right.

4. Another way - factorize the root number into prime factors . Prime factors are numbers that are only divisible by 1 and themselves. Write the prime factors in a row and find pairs of identical factors. Such factors can be taken out of the sign of the root.

   • For example, calculate the square root of 45. We decompose the root number into prime factors: 45 \u003d 9 x 5, and 9 \u003d 3 x 3. Thus, √45 \u003d √ (3 x 3 x 5). 3 can be taken out of the root sign: √45 = 3√5. Now we can estimate √5.
   • Consider another example: √88.
     • = √(2 x 44)
     • = √ (2 x 4 x 11)
     • = √ (2 x 2 x 2 x 11). You got three multiplier 2s; take a couple of them and take them out of the sign of the root.
     • = 2√(2 x 11) = 2√2 x √11. Now we can evaluate √2 and √11 and find an approximate answer.

   Calculating the square root manually

   Using column division

   1. This method involves a process similar to long division and gives an accurate answer. First, draw a vertical line dividing the sheet into two halves, and then draw a horizontal line to the right and slightly below the top edge of the sheet to the vertical line. Now divide the root number into pairs of numbers, starting with the fractional part after the decimal point. So, the number 79520789182.47897 is written as "7 95 20 78 91 82, 47 89 70".

      • For example, let's calculate the square root of the number 780.14. Draw two lines (as shown in the picture) and write the number in the top left as "7 80, 14". It is normal that the first digit from the left is an unpaired digit. The answer (the root of the given number) will be written on the top right.

   2. Given the first pair of numbers (or one number) from the left, find the largest integer n whose square is less than or equal to the pair of numbers (or one number) in question. In other words, find the square number that is closest to, but less than, the first pair of numbers (or single number) from the left, and take the square root of that square number; you will get the number n. Write the found n at the top right, and write down the square n at the bottom right.

      • In our case, the first number on the left will be the number 7. Next, 4< 7, то есть 2 2 < 7 и n = 2. Напишите 2 сверху справа - это первая цифра в искомом квадратном корне. Напишите 2×2=4 справа снизу; вам понадобится это число для последующих вычислений.

   3. Subtract the square of the number n you just found from the first pair of numbers (or one number) from the left. Write the result of the calculation under the subtrahend (the square of the number n).

      • In our example, subtract 4 from 7 to get 3.

   4. Take down the second pair of numbers and write it down next to the value obtained in the previous step. Then double the number at the top right and write the result at the bottom right with "_×_=" appended.

      • In our example, the second pair of numbers is "80". Write "80" after the 3. Then, doubling the number from the top right gives 4. Write "4_×_=" from the bottom right.

   5. Fill in the blanks on the right.

      • In our case, if instead of dashes we put the number 8, then 48 x 8 \u003d 384, which is more than 380. Therefore, 8 is too large a number, but 7 is fine. Write 7 instead of dashes and get: 47 x 7 \u003d 329. Write 7 from the top right - this is the second digit in the desired square root of the number 780.14.

   6. Subtract the resulting number from the current number on the left. Write the result from the previous step below the current number on the left, find the difference and write it below the subtracted one.

      • In our example, subtract 329 from 380, which equals 51.

   7. Repeat step 4. If the demolished pair of numbers is the fractional part of the original number, then put the separator (comma) of the integer and fractional parts in the desired square root from the top right. On the left, carry down the next pair of numbers. Double the number at the top right and write the result at the bottom right with "_×_=" appended.

      • In our example, the next pair of numbers to be demolished will be the fractional part of the number 780.14, so put the separator of the integer and fractional parts in the desired square root from the top right. Demolish 14 and write down at the bottom left. Double the top right (27) is 54, so write "54_×_=" at the bottom right.

   8. Repeat steps 5 and 6. Find the largest number in place of dashes on the right (instead of dashes you need to substitute the same number) so that the multiplication result is less than or equal to the current number on the left.

      • In our example, 549 x 9 = 4941, which is less than the current number on the left (5114). Write 9 on the top right and subtract the result of the multiplication from the current number on the left: 5114 - 4941 = 173.

   9. If you need to find more decimal places for the square root, write a pair of zeros next to the current number on the left and repeat steps 4, 5 and 6. Repeat steps until you get the accuracy of the answer you need (number of decimal places).

   Understanding the process

   To master this method, imagine the number whose square root you need to find as the area of ​​​​the square S. In this case, you will look for the length of the side L of such a square. Calculate the value of L for which L² = S.

   Enter a letter for each digit in your answer. Denote by A the first digit in the value of L (the desired square root). B will be the second digit, C the third and so on.

   Specify a letter for each pair of leading digits. Denote by S a the first pair of digits in the value S, by S b the second pair of digits, and so on.

   Explain the connection of this method with long division. As in the division operation, where each time we are only interested in one next digit of the divisible number, when calculating the square root, we work with a pair of digits in sequence (to obtain the next one digit in the square root value).

   1. Consider the first pair of digits Sa of the number S (Sa = 7 in our example) and find its square root. In this case, the first digit A of the sought value of the square root will be such a digit, the square of which is less than or equal to S a (that is, we are looking for such an A that satisfies the inequality A² ≤ Sa< (A+1)²). В нашем примере, S1 = 7, и 2² ≤ 7 < 3²; таким образом A = 2.

      • Let's say we need to divide 88962 by 7; here the first step will be similar: we consider the first digit of the divisible number 88962 (8) and select the largest number that, when multiplied by 7, gives a value less than or equal to 8. That is, we are looking for a number d for which the inequality is true: 7 × d ≤ 8< 7×(d+1). В этом случае d будет равно 1.

Congratulations: today we will analyze the roots - one of the most mind-blowing topics of the 8th grade. :)

Many people get confused about the roots not because they are complex (which is complicated - a couple of definitions and a couple more properties), but because in most school textbooks the roots are defined through such wilds that only the authors of the textbooks themselves can understand this scribbling. And even then only with a bottle of good whiskey. :)

Therefore, now I will give the most correct and most competent definition of the root - the only one that you really need to remember. And only then I will explain: why all this is necessary and how to apply it in practice.

But first remember one important point, about which many compilers of textbooks for some reason “forget”:

Roots can be of even degree (our favorite $\sqrt(a)$, as well as any $\sqrt(a)$ and even $\sqrt(a)$) and odd degree (any $\sqrt(a)$, $\ sqrt(a)$ etc.). And the definition of the root of an odd degree is somewhat different from the even one.

Here in this fucking “somewhat different” is hidden, probably, 95% of all errors and misunderstandings associated with the roots. So let's clear up the terminology once and for all:

Definition. Even root n from the number $a$ is any non-negative a number $b$ such that $((b)^(n))=a$. And the root of an odd degree from the same number $a$ is generally any number $b$ for which the same equality holds: $((b)^(n))=a$.

In any case, the root is denoted like this:

\(a)\]

The number $n$ in such a notation is called the root exponent, and the number $a$ is called the radical expression. In particular, for $n=2$ we get our “favorite” square root (by the way, this is a root of an even degree), and for $n=3$ we get a cubic root (an odd degree), which is also often found in problems and equations.

Examples. Classic examples of square roots:

\[\begin(align) & \sqrt(4)=2; \\ & \sqrt(81)=9; \\ & \sqrt(256)=16. \\ \end(align)\]

By the way, $\sqrt(0)=0$ and $\sqrt(1)=1$. This is quite logical since $((0)^(2))=0$ and $((1)^(2))=1$.

Cubic roots are also common - do not be afraid of them:

\[\begin(align) & \sqrt(27)=3; \\ & \sqrt(-64)=-4; \\ & \sqrt(343)=7. \\ \end(align)\]

Well, a couple of "exotic examples":

\[\begin(align) & \sqrt(81)=3; \\ & \sqrt(-32)=-2. \\ \end(align)\]

If you do not understand what is the difference between an even and an odd degree, reread the definition again. It is very important!

In the meantime, we will consider one unpleasant feature of the roots, because of which we needed to introduce a separate definition for even and odd exponents.

Why do we need roots at all?

After reading the definition, many students will ask: “What did mathematicians smoke when they came up with this?” And really: why do we need all these roots?

To answer this question, let's go back to elementary school for a moment. Remember: in those distant times when the trees were greener and the dumplings tastier, our main concern was to multiply the numbers correctly. Well, something in the spirit of "five by five - twenty-five", that's all. But after all, you can multiply numbers not in pairs, but in triplets, fours, and generally whole sets:

\[\begin(align) & 5\cdot 5=25; \\ & 5\cdot 5\cdot 5=125; \\ & 5\cdot 5\cdot 5\cdot 5=625; \\ & 5\cdot 5\cdot 5\cdot 5\cdot 5=3125; \\ & 5\cdot 5\cdot 5\cdot 5\cdot 5\cdot 5=15\ 625. \end(align)\]

However, this is not the point. The trick is different: mathematicians are lazy people, so they had to write down the multiplication of ten fives like this:

So they came up with degrees. Why not write the number of factors as a superscript instead of a long string? Like this one:

It's very convenient! All calculations are reduced by several times, and you can not spend a bunch of parchment sheets of notebooks to write down some 5 183 . Such an entry was called the degree of a number, a bunch of properties were found in it, but happiness turned out to be short-lived.

After a grandiose booze, which was organized just about the “discovery” of degrees, some especially stoned mathematician suddenly asked: “What if we know the degree of a number, but we don’t know the number itself?” Indeed, if we know that a certain number $b$, for example, gives 243 to the 5th power, then how can we guess what the number $b$ itself is equal to?

This problem turned out to be much more global than it might seem at first glance. Because it turned out that for the majority of “ready-made” degrees there are no such “initial” numbers. Judge for yourself:

\[\begin(align) & ((b)^(3))=27\Rightarrow b=3\cdot 3\cdot 3\Rightarrow b=3; \\ & ((b)^(3))=64\Rightarrow b=4\cdot 4\cdot 4\Rightarrow b=4. \\ \end(align)\]

What if $((b)^(3))=50$? It turns out that you need to find a certain number, which, when multiplied by itself three times, will give us 50. But what is this number? It is clearly greater than 3 because 3 3 = 27< 50. С тем же успехом оно меньше 4, поскольку 4 3 = 64 >50. I.e. this number lies somewhere between three and four, but what it is equal to - FIG you will understand.

This is exactly why mathematicians came up with $n$-th roots. That is why the radical icon $\sqrt(*)$ was introduced. To denote the same number $b$, which, to the specified power, will give us a previously known value

\[\sqrt[n](a)=b\Rightarrow ((b)^(n))=a\]

I do not argue: often these roots are easily considered - we saw several such examples above. But still, in most cases, if you think of an arbitrary number, and then try to extract the root of an arbitrary degree from it, you are in for a cruel bummer.

What is there! Even the simplest and most familiar $\sqrt(2)$ cannot be represented in our usual form - as an integer or a fraction. And if you drive this number into a calculator, you will see this:

\[\sqrt(2)=1.414213562...\]

As you can see, after the decimal point there is an endless sequence of numbers that do not obey any logic. You can, of course, round this number to quickly compare with other numbers. For example:

\[\sqrt(2)=1.4142...\approx 1.4 \lt 1.5\]

Or here's another example:

\[\sqrt(3)=1.73205...\approx 1.7 \gt 1.5\]

But all these roundings are, firstly, rather rough; and secondly, work with approximate values you also need to be able to, otherwise you can catch a bunch of non-obvious mistakes (by the way, the skill of comparison and rounding is necessarily checked at the profile exam).

Therefore, in serious mathematics, one cannot do without roots - they are the same equal representatives of the set of all real numbers $\mathbb(R)$, like fractions and integers that we have long known.

The impossibility of representing the root as a fraction of the form $\frac(p)(q)$ means that this root is not a rational number. Such numbers are called irrational, and they cannot be accurately represented except with the help of a radical, or other constructions specially designed for this (logarithms, degrees, limits, etc.). But more on that another time.

Consider a few examples where, after all the calculations, irrational numbers will still remain in the answer.

\[\begin(align) & \sqrt(2+\sqrt(27))=\sqrt(2+3)=\sqrt(5)\approx 2,236... \\ & \sqrt(\sqrt(-32 ))=\sqrt(-2)\approx -1,2599... \\ \end(align)\]

Naturally, by appearance the root is almost impossible to guess what numbers will come after the decimal point. However, it is possible to calculate on a calculator, but even the most advanced date calculator gives us only the first few digits of an irrational number. Therefore, it is much more correct to write the answers as $\sqrt(5)$ and $\sqrt(-2)$.

That's what they were invented for. To make it easy to write down answers.

Why are two definitions needed?

The attentive reader has probably already noticed that all the square roots given in the examples are taken from positive numbers. Well, at least from zero. But cube roots are calmly extracted from absolutely any number - even positive, even negative.

Why is this happening? Take a look at the graph of the function $y=((x)^(2))$:

Schedule quadratic function gives two roots: positive and negative

Let's try to calculate $\sqrt(4)$ using this graph. To do this, a horizontal line $y=4$ (marked in red) is drawn on the graph, which intersects the parabola at two points: $((x)_(1))=2$ and $((x)_(2)) =-2$. This is quite logical, since

Everything is clear with the first number - it is positive, therefore it is the root:

But then what to do with the second point? Does the 4 have two roots at once? After all, if we square the number −2, we also get 4. Why not write $\sqrt(4)=-2$ then? And why do teachers look at such records as if they want to eat you? :)

The trouble is that if no additional conditions are imposed, then the four will have two square roots - positive and negative. And any positive number will also have two of them. But negative numbers will not have roots at all - this can be seen from the same graph, since the parabola never falls below the axis y, i.e. does not take negative values.

A similar problem occurs for all roots with an even exponent:

1. Strictly speaking, each positive number will have two roots with an even exponent $n$;
2. From negative numbers, the root with even $n$ is not extracted at all.

That is why the definition of an even root $n$ specifically stipulates that the answer must be a non-negative number. This is how we get rid of ambiguity.

But for odd $n$ there is no such problem. To see this, let's take a look at the graph of the function $y=((x)^(3))$:

The cubic parabola takes on any value, so the cube root can be taken from any number

Two conclusions can be drawn from this graph:

1. The branches of a cubic parabola, unlike the usual one, go to infinity in both directions - both up and down. Therefore, at whatever height we draw a horizontal line, this line will definitely intersect with our graph. Therefore, the cube root can always be taken, absolutely from any number;
2. In addition, such an intersection will always be unique, so you don’t need to think about which number to consider the “correct” root, and which one to score. That is why the definition of roots for an odd degree is simpler than for an even one (there is no non-negativity requirement).

It's a pity that these simple things are not explained in most textbooks. Instead, our brains begin to soar with all sorts of arithmetic roots and their properties.

Yes, I do not argue: what is an arithmetic root - you also need to know. And I will talk about this in detail in a separate lesson. Today we will also talk about it, because without it, all reflections on the roots of the $n$-th multiplicity would be incomplete.

But first you need to clearly understand the definition that I gave above. Otherwise, due to the abundance of terms, such a mess will begin in your head that in the end you will not understand anything at all.

And all you need to understand is the difference between even and odd numbers. Therefore, once again we will collect everything that you really need to know about the roots:

1. An even root exists only from a non-negative number and is itself always a non-negative number. For negative numbers, such a root is undefined.
2. But the root of an odd degree exists from any number and can itself be any number: for positive numbers it is positive, and for negative numbers, as the cap hints, it is negative.

Is it difficult? No, it's not difficult. Clear? Yes, it's obvious! Therefore, now we will practice a little with the calculations.

Basic properties and limitations

Roots have a lot of strange properties and restrictions - this will be a separate lesson. Therefore, now we will consider only the most important "chip", which applies only to roots with an even exponent. We write this property in the form of a formula:

\[\sqrt(((x)^(2n)))=\left| x\right|\]

In other words, if we raise a number to an even power, and then extract the root of the same degree from this, we will get not the original number, but its modulus. This is a simple theorem that is easy to prove (it suffices to consider separately non-negative $x$, and then separately consider negative ones). Teachers constantly talk about it, they give it in every school textbook. But as soon as it comes to solving irrational equations (i.e. equations containing the sign of the radical), the students forget this formula together.

To understand the issue in detail, let's forget all the formulas for a minute and try to count two numbers ahead:

\[\sqrt(((3)^(4)))=?\quad \sqrt(((\left(-3 \right))^(4)))=?\]

This is very simple examples. The first example will be solved by most of the people, but on the second, many stick. To solve any such crap without problems, always consider the procedure:

1. First, the number is raised to the fourth power. Well, it's kind of easy. A new number will be obtained, which can even be found in the multiplication table;
2. And now from this new number it is necessary to extract the root of the fourth degree. Those. there is no "reduction" of roots and degrees - these are sequential actions.

Let's deal with the first expression: $\sqrt(((3)^(4)))$. Obviously, you first need to calculate the expression under the root:

\[((3)^(4))=3\cdot 3\cdot 3\cdot 3=81\]

Then we extract the fourth root of the number 81:

Now let's do the same with the second expression. First, we raise the number −3 to the fourth power, for which we need to multiply it by itself 4 times:

\[((\left(-3 \right))^(4))=\left(-3 \right)\cdot \left(-3 \right)\cdot \left(-3 \right)\cdot \ left(-3 \right)=81\]

We got a positive number, since the total number of minuses in the product is 4 pieces, and they will all cancel each other out (after all, a minus by a minus gives a plus). Next, extract the root again:

In principle, this line could not be written, since it is a no brainer that the answer will be the same. Those. an even root of the same even power "burns" the minuses, and in this sense the result is indistinguishable from the usual module:

\[\begin(align) & \sqrt(((3)^(4)))=\left| 3\right|=3; \\ & \sqrt(((\left(-3 \right))^(4)))=\left| -3 \right|=3. \\ \end(align)\]

These calculations are in good agreement with the definition of the root of an even degree: the result is always non-negative, and the radical sign is also always a non-negative number. Otherwise, the root is not defined.

Note on the order of operations

1. The notation $\sqrt(((a)^(2)))$ means that we first square the number $a$, and then take the square root of the resulting value. Therefore, we can be sure that a non-negative number always sits under the root sign, since $((a)^(2))\ge 0$ anyway;
2. But the notation $((\left(\sqrt(a) \right))^(2))$, on the contrary, means that we first extract the root from a certain number $a$ and only then square the result. Therefore, the number $a$ in no case can be negative - this is a mandatory requirement embedded in the definition.

Thus, in no case should one thoughtlessly reduce the roots and degrees, thereby supposedly "simplifying" the original expression. Because if there is a negative number under the root, and its exponent is even, we will get a lot of problems.

However, all these problems are relevant only for even indicators.

Removing a minus sign from under the root sign

Naturally, roots with odd exponents also have their own feature, which, in principle, does not exist for even ones. Namely:

\[\sqrt(-a)=-\sqrt(a)\]

In short, you can take out a minus from under the sign of the roots of an odd degree. This is a very useful property that allows you to "throw" all the minuses out:

\[\begin(align) & \sqrt(-8)=-\sqrt(8)=-2; \\ & \sqrt(-27)\cdot \sqrt(-32)=-\sqrt(27)\cdot \left(-\sqrt(32) \right)= \\ & =\sqrt(27)\cdot \sqrt(32)= \\ & =3\cdot 2=6. \end(align)\]

This simple property greatly simplifies many calculations. Now you don’t need to worry: what if a negative expression got under the root, and the degree at the root turned out to be even? It is enough to “throw out” all the minuses outside the roots, after which they can be multiplied by each other, divided and generally do many suspicious things, which in the case of “classic” roots are guaranteed to lead us to an error.

And here another definition enters the scene - the very one with which most schools begin the study of irrational expressions. And without which our reasoning would be incomplete. Meet!

arithmetic root

Let's assume for a moment that only positive numbers or, in extreme cases, zero can be under the root sign. Let's score on even / odd indicators, score on all the definitions given above - we will work only with non-negative numbers. What then?

And then we get the arithmetic root - it partially intersects with our "standard" definitions, but still differs from them.

Definition. An arithmetic root of the $n$th degree of a non-negative number $a$ is a non-negative number $b$ such that $((b)^(n))=a$.

As you can see, we are no longer interested in parity. Instead, a new restriction appeared: the radical expression is now always non-negative, and the root itself is also non-negative.

To better understand how the arithmetic root differs from the usual one, take a look at the graphs of the square and cubic parabola already familiar to us:

Root search area - non-negative numbers

As you can see, from now on, we are only interested in those pieces of graphs that are located in the first coordinate quarter - where the coordinates $x$ and $y$ are positive (or at least zero). You no longer need to look at the indicator to understand whether we have the right to root a negative number or not. Because negative numbers are no longer considered in principle.

You may ask: “Well, why do we need such a castrated definition?” Or: "Why can't we get by with the standard definition given above?"

Well, I will give just one property, because of which the new definition becomes appropriate. For example, the exponentiation rule:

\[\sqrt[n](a)=\sqrt(((a)^(k)))\]

Please note: we can raise the radical expression to any power and at the same time multiply the root exponent by the same power - and the result will be the same number! Here are some examples:

\[\begin(align) & \sqrt(5)=\sqrt(((5)^(2)))=\sqrt(25) \\ & \sqrt(2)=\sqrt(((2)^ (4)))=\sqrt(16) \\ \end(align)\]

Well, what's wrong with that? Why couldn't we do it before? Here's why. Consider a simple expression: $\sqrt(-2)$ is a number that is quite normal in our classical sense, but absolutely unacceptable from the point of view of the arithmetic root. Let's try to convert it:

$\begin(align) & \sqrt(-2)=-\sqrt(2)=-\sqrt(((2)^(2)))=-\sqrt(4) \lt 0; \\ & \sqrt(-2)=\sqrt(((\left(-2 \right))^(2)))=\sqrt(4) \gt 0. \\ \end(align)$

As you can see, in the first case, we took the minus out from under the radical (we have every right, because the indicator is odd), and in the second, we used the above formula. Those. from the point of view of mathematics, everything is done according to the rules.

WTF?! How can the same number be both positive and negative? No way. It's just that the exponentiation formula, which works great for positive numbers and zero, starts to give complete heresy in the case of negative numbers.

Here, in order to get rid of such ambiguity, they came up with arithmetic roots. A separate large lesson is devoted to them, where we consider in detail all their properties. So now we will not dwell on them - the lesson turned out to be too long anyway.

Algebraic root: for those who want to know more

I thought for a long time: to make this topic in a separate paragraph or not. In the end, I decided to leave here. This material is intended for those who want to understand the roots even better - no longer at the average “school” level, but at the level close to the Olympiad.

So: in addition to the "classical" definition of the root of the $n$-th degree from a number and the associated division into even and odd indicators, there is a more "adult" definition, which does not depend on parity and other subtleties at all. This is called an algebraic root.

Definition. An algebraic $n$-th root of any $a$ is the set of all numbers $b$ such that $((b)^(n))=a$. There is no well-established designation for such roots, so just put a dash on top:

\[\overline(\sqrt[n](a))=\left\( b\left| b\in \mathbb(R);((b)^(n))=a \right. \right\) \]

The fundamental difference from the standard definition given at the beginning of the lesson is that the algebraic root is not a specific number, but a set. And since we are working with real numbers, this set is of only three types:

1. Empty set. Occurs when it is required to find an algebraic root of an even degree from a negative number;
2. A set consisting of a single element. All roots of odd powers, as well as roots of even powers from zero, fall into this category;
3. Finally, the set can include two numbers - the same $((x)_(1))$ and $((x)_(2))=-((x)_(1))$ that we saw on the chart quadratic function. Accordingly, such an alignment is possible only when extracting the root of an even degree from a positive number.

The last case deserves more detailed consideration. Let's count a couple of examples to understand the difference.

Example. Compute expressions:

\[\overline(\sqrt(4));\quad \overline(\sqrt(-27));\quad \overline(\sqrt(-16)).\]

Solution. The first expression is simple:

\[\overline(\sqrt(4))=\left\( 2;-2 \right\)\]

It is two numbers that are part of the set. Because each of them squared gives a four.

\[\overline(\sqrt(-27))=\left\( -3 \right\)\]

Here we see a set consisting of only one number. This is quite logical, since the exponent of the root is odd.

Finally, the last expression:

\[\overline(\sqrt(-16))=\varnothing \]

We got an empty set. Because there is not a single real number that, when raised to the fourth (that is, even!) Power, will give us a negative number −16.

Final note. Please note: it was not by chance that I noted everywhere that we are working with real numbers. Because there are also complex numbers - it is quite possible to calculate $\sqrt(-16)$ and many other strange things there.

However, in the modern school curriculum of mathematics, complex numbers are almost never found. They have been omitted from most textbooks because our officials consider the topic "too difficult to understand."

That's all. In the next lesson, we will look at all the key properties of roots and finally learn how to simplify irrational expressions. :)