Pupils always ask: "Why can not be used by a calculator on a mathematics exam? How to extract square root from number without a calculator? " Let's try to answer this question.

How to extract the root square from the number without the help of the calculator?

Act root extraction square back the focus on the square.

√81= 9 9 2 =81

If, from a positive number, remove the root square and the result is raised into the square, we get the same number.

Of the small numbers, which are exact squares of natural numbers, for example, 1, 4, 9, 16, 25, ..., 100 square roots can be removed orally. Usually, the school is taught a table of squares of natural numbers to twenty. Knowing this table It is easy to extract square roots from numbers 121.144, 169, 196, 225, 256, 289, 324, 361, 400. From the numbers of large 400, it is possible to remove the selection by the method using, some prompts. Let's try to consider this method on the example.

Example: Extract the root of 676.

We notice that 20 2 \u003d 400, and 30 2 \u003d 900, it means 20< √676 < 900.

The exact squares of natural numbers end in numbers 0; one; four; five; 6; nine.
Figure 6 gives 4 2 and 6 2.
So, if the root is extracted from 676, then this is either 24 or 26.

It remains to check: 24 2 \u003d 576, 26 2 \u003d 676.

Answer: √676 = 26 .

Yet example: √6889 .

Since 80 2 \u003d 6400, and 90 2 \u003d 8100, then 80< √6889 < 90.
The figure 9 is 3 2 and 7 2, then √6889 is either 83 or 87.

Check: 83 2 \u003d 6889.

Answer: √6889 = 83 .

If it is difficult to solve the selection method, then you can decompose the conditioned expression on multipliers.

For example, find √893025..

Spread the number 893025 for multipliers, remember, you did it in the sixth grade.

We obtain: √893025 \u003d √3 6 ∙ 5 2 ∙ 7 2 \u003d 3 3 ∙ 5 ∙ 7 \u003d 945.

Yet example: √20736. Spread the number 20736 for multipliers:

We obtain √20736 \u003d √2 8 ∙ 3 4 \u003d 2 4 ∙ 3 2 \u003d 144.

Of course, the decomposition of multipliers requires knowledge of the signs of divisibility and the skills of decomposition of multipliers.

And finally there is rule Extraction of square roots. Let's get acquainted with this rule on the examples.

Calculate √279841..

To extract the root from a multi-inforous integer, we divide it to the right to left on the verges containing 2 numbers (one digit can be in the left extreme face). Record so 27'98'41

To obtain the first digit of the root (5), remove the square root of the largest accurate square contained in the first left of the face (27).
Then the square of the first figure of the root (25) is subtracted from the first face, and the following line (98) is attributed to the difference.
The left of the resulting number 298 is written a double root (10) digit (10), the number of all dozens of the early number (29/2 ≈ 2) is divided, they test the private (102 ∙ 2 \u003d 204 should not be more than 298) and write (2) after The first digit root.
Then they are subtracted from 298 the obtained private 204 and the difference (94) is attributed (demolition) the next line (41).
To the left of the resulting number 9441 write a double product of the root number (52 ∙ 2 \u003d 104), they divide the number of all dozens of number 9441 (944/104 ≈ 9) to this, test (1049 ∙ 9 \u003d 9441) 9441 and write it (9) After the second root digit.

Received the answer √279841 \u003d 529.

Similarly remove Roots from decimal fractions. Only the feed number must be broken on the verge so that the comma was between the edges.

Example. Find a value √0.00956484.

Just need to remember that if decimal It has an odd number of decimal signs, it is not extracting exactly the square root from it.

So, now you met with three ways to extract the root. Choose the one that suits you more and practice. To learn to solve tasks, they must be solved. And if you have questions, sign up for my lessons.

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There are several methods for calculating a square root without a calculator.

How to find a root from among - 1 way

• One method is to decompose the factors that are under the root. These components as a result of multiplication form an inhibited value. The accuracy of the result obtained depends on the number under the root.
• For example, if you take the number 1,600 and start laying it on multipliers, then the reasoning will be built in this way: this number multiple 100, it means that it can be divided into 25; Since the root from among 25 is removed, the number is square and suitable for further computing; When divided, we get another number - 64. This number is also square, so the root is retrieved well; After these calculations, under the root, it is possible to record the number 1600 in the form of a piece of 25 and 64.

• One of the rules for extracting the root says that the root from the product of the multiplers is equal to the number that is obtained by multiplying the roots from each multiplier. This means that: √ (25 * 64) \u003d √25 * √64. If from 25 and 64 remove the roots, then we obtain such an expression: 5 * 8 \u003d 40. That is, the square root from among the number 1600 is 40.
• But it happens that the number under the root is not laid out by two multipliers, of which the whole root is extracted. Usually this can be done only for one of the multipliers. Therefore, it is not possible to find a completely accurate answer in such an equation.
• In this case, only approximate value can be calculated. Therefore, it is necessary to extract the root of the multiplier, which is square number. This value is then multiplied to the root of the second number, which is not a square member of the equation.
• It looks in this way, for example, take the number 320. It can be decomposed on 64 and 5. Of the 64, the whole root can be removed, and out of 5 - no. Therefore, the expression will look like this: √320 \u003d √ (64 * 5) \u003d √64 * √5 \u003d 8√5.
• If necessary, you can find the approximate value of this result, calculating
  √5 ≈ 2.236, therefore, √320 \u003d 8 * 2,236 \u003d 17.88 ≈ 18.
• Also, the number under the root can be decomposed into several simple multipliers, and the same can be made from under it. Example: √75 \u003d √ (5 * 5 * 3) \u003d 5√3 ≈ 8.66 ≈ 9.

How to find a root from among - 2 ways

• Another way is to divide into a column. The division occurs similarly, but only to look for square numbers, of which then retrieve the root.
• In this case, the square number is writing from above and take it in the left side, and the extracted root from the bottom.
• Now you need to double the second value and write down the right in the form: number_x_ \u003d. Skipping must be filled with a number that will be less than or equal to the required value of the left - everything is as in conventional division.
• If necessary, this result is subtracted on the left. Such calculations continue until the result is reached. Zeros can also be added until you receive the required number of semicolons.

Quite often, when solving the tasks, we face large numbers from which you need to extract square root. Many students decide that this is a mistake, and begin to cut the whole example. In no case can not do it! That is, two reasons:

1. Roots from large numbers are really found in tasks. Especially in textual;
2. There is an algorithm with which these roots are considered almost orally.

This algorithm we will look at this algorithm. Perhaps some things will seem incomprehensible to you. But if you carefully react to this lesson, you will get a powerful weapon against square roots.

So, the algorithm:

1. Limit the desired root from above and below the numbers, multiple 10. Thus, we will reduce the search range up to 10 numbers;
2. Of these 10 numbers, there are those who exactly cannot be rooted. As a result, 1-2 numbers will remain;
3. Evaluate these 1-2 numbers per square. That of them, the square of which is equal to the initial number, and will be the root.

Before applying this algorithm works in practice, let's look at each separate step.

Restriction of roots

First of all, it is necessary to find out, between which numbers is our root. It is very desirable that the numbers are more than ten:

10 2 = 100;
20 2 = 400;
30 2 = 900;
40 2 = 1600;
...
90 2 = 8100;
100 2 = 10 000.

We get a number of numbers:

100; 400; 900; 1600; 2500; 3600; 4900; 6400; 8100; 10 000.

What do we give these numbers? Everything is simple: we get borders. Take, for example, the number 1296. It lies between 900 and 1600. Consequently, its root can not be less than 30 and more 40:

[Signature to Figure]

The same thing is with any other number from which you can find a square root. For example, 3364:

[Signature to Figure]

Thus, instead of an incomprehensible number, we get a completely specific range in which the original root lies. To further narrow the search area, go to the second step.

Opening obviously unnecessary numbers

So, we have 10 numbers - candidates for the root. We got them very quickly, without complicated reflection and multiplications in the column. It's time to move on.

Do not believe, but now we will reduce the number of candidate numbers up to two - and again without any complex computing! It is enough to know the special rule. Here it is:

The last digit of the square depends only on the last digit initial number.

In other words, it is enough to look at the last digit of the square - and we immediately understand what the initial number ends.

There are only 10 digits that can stand in the last place. Let's try to find out what they turn into a square. Take a look at the table:

1	2	3	4	5	6	7	8	9	0
1	4	9	6	5	6	9	4	1	0

This table is another step towards the root calculation. As you can see, the numbers in the second line turned out to be symmetrical about the five. For example:

2 2 = 4;
8 2 = 64 → 4.

As you can see, the last digit in both cases is the same. This means that, for example, the root of 3364 must end with 2 or 8. On the other hand, we remember the restriction from the previous paragraph. We get:

[Signature to Figure]

Red squares show that we do not yet know this figure. But the root lies in the range from 50 to 60, on which there are only two numbers ending at 2 and 8:

[Signature to Figure]

That's all! Of all possible roots, we left only two options! And it is in the hard case, because the last digit can be 5 or 0. And then the only candidate in the roots will remain!

Final calculations

So, we have 2 candidate numbers left. How to find out which one is the root? The answer is obvious: build both numbers per square. That which in the square gives the original number, and will be the root.

For example, for the number of 3364, we found two candidate numbers: 52 and 58. Earbing them in a square:

52 2 \u003d (50 +2) 2 \u003d 2500 + 2 · 50 · 2 + 4 \u003d 2704;
58 2 \u003d (60 - 2) 2 \u003d 3600 - 2 · 60 · 2 + 4 \u003d 3364.

That's all! It turned out that the root is 58! At the same time, to simplify the calculations, I used the summary and difference formula. Thanks to which I did not even have to multiply the numbers in the column! This is another level of optimization of calculations, but, of course, is completely optional :)

Examples of calculating roots

Theory is, of course, well. But let's check it in practice.

[Signature to Figure]

To begin with, we find out, between which numbers is the number 576:

400 < 576 < 900
20 2 < 576 < 30 2

Now we look at the last digit. It is equal to 6. When does it happen? Only if the root ends on 4 or 6. We get two numbers:

It remains to take each number in the square and compare with the original:

24 2 = (20 + 4) 2 = 576

Excellent! The first square was equal to the initial number. So, this is the root.

A task. Calculate the square root:

[Signature to Figure]

900 < 1369 < 1600;
30 2 < 1369 < 40 2;

We look at the last digit:

1369 → 9;
33; 37.

We are erected into a square:

33 2 \u003d (30 + 3) 2 \u003d 900 + 2 · 30 · 3 + 9 \u003d 1089 ≠ 1369;
37 2 \u003d (40 - 3) 2 \u003d 1600 - 2 · 40 · 3 + 9 \u003d 1369.

Here is the answer: 37.

A task. Calculate the square root:

[Signature to Figure]

We limit the number:

2500 < 2704 < 3600;
50 2 < 2704 < 60 2;

We look at the last digit:

2704 → 4;
52; 58.

We are erected into a square:

52 2 \u003d (50 + 2) 2 \u003d 2500 + 2 · 50 · 2 + 4 \u003d 2704;

Received the answer: 52. The second number is erected into a square no longer need.

A task. Calculate the square root:

[Signature to Figure]

We limit the number:

3600 < 4225 < 4900;
60 2 < 4225 < 70 2;

We look at the last digit:

4225 → 5;
65.

As you can see, after the second step, only one option remained: 65. This is the desired root. But let's ever erect him in a square and check:

65 2 \u003d (60 + 5) 2 \u003d 3600 + 2 · 60 · 5 + 25 \u003d 4225;

That's right. Record the answer.

Conclusion

Alas, no better. Let's figure it out for reasons. There are two of them:

• On any normal exam in mathematics, whether it is a GIA or EGE, to use calculators is prohibited. And the calculator can easily be expelled from the exam.
• Do not like stupid Americans. Which is not that roots - they cannot be folded two simple numbers. And at the sight of fractions, their hysteria generally begins.

When solving various tasks from the course of mathematics and physics, students and students often face the need to extract the roots of the second, third or N-essential degree. Of course information technologies It is not difficult to solve such a task with the help of a calculator. However, situations arise when it is impossible to use an electronic assistant.

For example, for many exams it is forbidden to bring electronics. In addition, the calculator may not be at hand. In such cases, it is useful to know at least some methods for calculating radicals manually.

One of the simplest ways to calculate the roots is using a special table. What does it represent and how to use it right?

Using the table, you can find the square of any number from 10 to 99. At the same time, there are values \u200b\u200bof dozens in the tables, in the columns - the values \u200b\u200bof units. The cell on the intersection of the line and the column contains the square of the double digit number. In order to calculate the square 63, you need to find a string with a value of 6 and a column with a value 3. At the intersection, you will detect a cell with a number of 3969.

Since the recovery of the root is an operation, reverse the square, to perform this actions, it is necessary to enroll on the contrary: first to find a cell with a number, the radical of which must be calculated, then to determine the answer by the values \u200b\u200band strings. As an example, consider the calculation of the square root 169.

We find a cell with this number in the table, horizontally determine tens - 1, vertically find units - 3. Answer: √169 \u003d 13.

Similarly, it is possible to calculate the roots of the cubic and N-essential, using the corresponding tables.

The advantage of the method is its simplicity and the absence of additional computing. The shortcomings are obvious: the method can only be used for a limited range of numbers (the number for which the root is located should be between 100 to 9801). In addition, it does not suit if the specified number is not in the table.

Decomposition of simple factors

If the table of squares is missing at hand or with it it turned out to be impossible to find the root, you can try eliminate the number under the root for simple factors. Simple factors are such that can be aimed (without a residue) to share only on themselves or by one. Examples can be 2, 3, 5, 7, 11, 13, etc.

Consider the root calculation on the example √576. Spread it on simple factors. We obtain the following result: √576 \u003d √ (2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 3 \u200b\u200b∙ 3) \u003d √ (2 ∙ 2 ∙ 2) ² ∙ √ 3². Using the basic properties of the roots √a² \u003d a, we will get rid of the roots and squares, after which we will calculate the answer: 2 ∙ 2 ∙ 2 ∙ 3 \u200b\u200b\u003d 24.

What to do, if any factors have no pair? For example, consider the calculation √54. After decomposition on multipliers, we get the result in next form: √54 \u003d √ (2 ∙ 3 \u200b\u200b∙ 3 ∙ 3) \u003d √ 3² ∙ √ (2 ∙ 3) \u003d 3√6. The unknown part can be left under the root. For most tasks in geometry and algebra, such an answer will be counted as a final one. But if there is a need to calculate the approximated values, you can use the methods that will be discussed below.

Gerona method

What to do when you need to at least approximately know what is equal to the extracted root (if it is impossible to get an integer value)? Fast and pretty accurate result gives the application of the geron method. Its essence is to use an approximate formula:

√R \u003d √a + (R - a) / 2√a,

where R is the number, the root of which needs to be calculated, A is the nearest number, the value of the root of which is known.

Consider how the method works in practice, and we estimate how accurate it is. Calculate what is equal to √111. The nearest to 111 number, the root of which is known - 121. Thus, R \u003d 111, a \u003d 121. We will substitute the values \u200b\u200bin the formula:

√111 = √121 + (111 - 121) / 2 ∙ √121 = 11 - 10 / 22 ≈ 10,55.

Now check the accuracy of the method:

10.55² \u003d 111,3025.

The error of the method was approximately 0.3. If the accuracy of the method should be increased, you can repeat the steps described earlier:

√111 = √111,3025 + (111 - 111,3025) / 2 ∙ √111,3025 = 10,55 - 0,3025 / 21,1 ≈ 10,536.

Check the accuracy of the calculation:

10.536² \u003d 111.0073.

After re-use of the formula, the error became completely insignificant.

Calculation of the root division in the column

This method of finding the value of the square root is slightly more complex than the previous ones. However, it is the most accurate among other methods of calculating without a calculator..

Suppose it is necessary to find a square root with an accuracy of 4 characters after the comma. We will analyze the calculation algorithm on an example of an arbitrary number 1308,1912.

1. We split the paper sheet into 2 parts by a vertical feature, and then spend another line from it to the right, slightly below the upper edge. We write the number on the left side, dividing it into groups of 2 digits, moving into the right and left side of the comma. The very first digit on the left may be without a pair. If the sign is not enough in the right part of the number, then you should add 0. In our case, it turns out 13 08,19 12.
2. We will select the largest number, the square of which will be less or equal to the first group of numbers. In our case, this is 3. We write it from the right above; 3 - the first digit of the result. On the right below, we specify 3 × 3 \u003d 9; It will be needed for subsequent calculations. Of 13 in the column, I will read 9, we get the residue 4.
3. We assign the next couple of numbers to the residue 4; We get 408.
4. The number from above on the right, multiply by 2 and write down to the right, adding to it _ x _ \u003d. We obtain 6_ x _ \u003d.
5. Instead of stiffers, it is necessary to substitute the same number smaller or equal to 408. We will get 66 × 6 \u003d 396. We will write 6 to the right above, since it is the second digit of the result. We take 396 from 408, we get 12.
6. We repeat steps 3-6. Since the numbers demolished down are in the fractional part of the number, it is necessary to put a decimal comma on the right after 6. We write down the doublet with ductures: 72_ x _ \u003d. A suitable digit will be 1: 721 × 1 \u003d 721. We write it in response. Perform subtraction 1219 - 721 \u003d 498.
7. Let us execute the sequence of actions given in the previous paragraph three more times to get the required number of semicolons. If there are not enough signs for further computing, the current left number needs to be added to two zero.

As a result, we will get the answer: √1308,1912 ≈ 36,1689. If you check the action using the calculator, you can make sure that all the signs were defined correctly.

Discongest calculation of square root value

The method has high accuracy. In addition, it is sufficiently clear and it is not necessary to memorize the formulas or complex algorithm of actions, since the essence of the method is the selection of the right result.

I remove the root from among 781. Consider in detail the sequence of actions.

1. Thusll, what discharge the value of the square root will be elder. To do this, erected in a square 0, 10, 100, 1000, etc. and find out, between which of them is a feed number. We get 10²< 781 < 100², т. е. старшим разрядом будут десятки.
2. We pick up the value of dozens. To do this, we will be in turn to erect 10, 20, ..., 90, until we obtain a number exceeding 781. For our case, we get 10² \u003d 100, 20² \u003d 400, 30² \u003d 900. The value of the result n will be in the range of 20< n <30.
3. Similar to the previous step, the value of the discharge of units is selected. Alternately erected into a square 21.22, ..., 29: 21² \u003d 441, 22² \u003d 484, 23² \u003d 529, 24² \u003d 576, 25² \u003d 625, 26² \u003d 676, 27² \u003d 729, 28² \u003d 784. We obtain that 27< n < 28.
4. Each subsequent discharge (tenths, hundredths, etc.) is calculated in the same way as shown above. Calculations are carried out until the necessary accuracy is achieved.

Korean n.degree of natural number a. called such a number n.whose degree is equal a.. The root is indicated as follows :. The symbol is called sign root or sign radical, Number a. - forbidden, n. - root indicator.

Action by which the root is given to a given degree is called root extraction.

Since, according to the definition of the concept of root n.degree

that removing the root - The effect, the reverse of the exercise, with which the foundation of the degree is found to the extent to the extent.

Square root

Square root a. called the number whose square is equal a..

The action by which the square root is calculated is called the extraction of the square root.

Extract square root - Action reverse the construction of the square (or the erection of the number in the second degree). When erected in the square, you know the number, it is required to find its square. When the square root is removed, the square is known, it is required to find the number itself.

Therefore, to verify the correctness of the actual action, you can build a found root to the second degree and, if the degree is equal to the guided number, it means the root was found correctly.

Consider the extraction of a square root and its verification on the example. Calculate or (the root rate with the value 2 is usually not written, since 2 is the smallest indicator and it should be remembered that if there is no indicator above the root sign, then the indicator 2), for this we need to find a number when the second The degree will be 49. It is obvious that such a number is 7, since

7 · 7 \u003d 7 2 \u003d 49.

Calculation of square root

If this number is 100 or less, the square root can be calculated using the multiplication table. For example, a square root of 25 is 5, because 5 · 5 \u003d 25.

Now consider the method of finding a square root from any number without using a calculator. For example, take the number 4489 and begin to gradually calculate.

1. We define from which discharges should be the desired root. Since 10 2 \u003d 10 · 10 \u003d 100, and 100 2 \u003d 100 · 100 \u003d 10,000, it becomes clear that the desired root should be more than 10 and less than 100, i.e. consist of dozens and units.
2. We find the number of dozens of root. From multiplying dozens, hundreds are obtained, in our number of them 44, so the root must contain as many dozens so that the square dozens give approximately 44 hundred. Therefore, the root should be 6 dozen, because 60 2 \u003d 3600, and 70 2 \u003d 4900 (this is too much). Thus, we found out that our root contains 6 tens and several units, as it is in the range from 60 to 70.
3. Determine the number of units in the root will help the multiplication table. Looking at the number 4489, we see that the last figure is 9. Now we look at the multiplication table and see that 9 units may only be erected in the square of numbers 3 and 7. So the root of the number will be equal to 63 or 67.
4. We check the numbers obtained by us 63 and 67. Earring them into a square: 63 2 \u003d 3969, 67 2 \u003d 4489.