What will be the root. Extracting the square root of a multi-digit number

When solving various problems from the course of mathematics and physics, pupils and students are often faced with the need to extract roots of the second, third or nth degree. Of course, in the century information technologies It will not be difficult to solve such a problem using a calculator. However, there are situations when it is impossible to use an electronic assistant.

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

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

Using the table, you can find the square of any number from 10 to 99. At the same time, the rows of the table contain tens values, and the columns contain unit values. The cell at the intersection of a row and a column contains the square of a two-digit number. In order to calculate the square of 63, you need to find a row with a value of 6 and a column with a value of 3. At the intersection, we find a cell with the number 3969.

Since extracting the root is the inverse operation of squaring, to perform this action, you must do the opposite: first find the cell with the number whose radical you want to calculate, then determine the answer from the column and row values. As an example, consider the calculation of the square root of 169.

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

Similarly, you can calculate the roots of the cubic and n-th degree, using the appropriate tables.

The advantage of the method is its simplicity and the absence of additional calculations. The disadvantages are obvious: the method can only be used for a limited range of numbers (the number for which the root is found must be between 100 and 9801). In addition, it will not work if the given number is not in the table.

Prime factorization

If the table of squares is not at hand or with its help it was impossible to find the root, you can try decompose the number under the root into prime factors. Prime factors are those that can be completely (without a remainder) divided only by itself or by one. Examples would be 2, 3, 5, 7, 11, 13, etc.

Consider the calculation of the root using the example √576. Let's decompose it into simple factors. We get the following result: √576 = √(2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 3 ​​∙ 3) = √(2 ∙ 2 ∙ 2)² ∙ √3². Using the main property of the roots √a² = a, we get rid of the roots and squares, after which we calculate the answer: 2 ∙ 2 ∙ 2 ∙ 3 ​​= 24.

What to do if any of the factors does not have its own pair? For example, consider the calculation of √54. After factoring, we get the result in the following form: The non-removable part can be left under the root. For most problems in geometry and algebra, such an answer will be counted as the final one. But if there is a need to calculate approximate values, you can use the methods that will be discussed below.

Heron's method

What to do when you need to know at least approximately what the extracted root is (if it is impossible to get an integer value)? A quick and fairly accurate result is obtained by applying the Heron method.. Its essence lies in the use of an approximate formula:

√R = √a + (R - a) / 2√a,

where R is the number whose root is to be calculated, a is the nearest number whose root value is known.

Let's see how the method works in practice and evaluate how accurate it is. Let's calculate what √111 is equal to. The number closest to 111, the root of which is known, is 121. Thus, R = 111, a = 121. Substitute the values ​​in the formula:

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

Now let's check the accuracy of the method:

10.55² = 111.3025.

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

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

Let's check the accuracy of the calculation:

10.536² = 111.0073.

After repeated application of the formula, the error became quite insignificant.

Calculation of the root by division into a column

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

Let's say that you need to find the square root with an accuracy of 4 decimal places. Let's analyze the calculation algorithm using the example of an arbitrary number 1308.1912.

1. Divide the sheet of paper into 2 parts with a vertical line, and then draw another line from it to the right, slightly below the top edge. We write the number on the left side, dividing it into groups of 2 digits, moving to the right and left side from a comma. The very first digit on the left can be without a pair. If the sign is missing on the right side of the number, then 0 should be added. In our case, we get 13 08.19 12.
2. Let's pick the most big number, whose square will be less than or equal to the first group of digits. In our case, this is 3. Let's write it on the top right; 3 is the first digit of the result. At the bottom right, we indicate 3 × 3 = 9; this will be needed for subsequent calculations. Subtract 9 from 13 in a column, we get the remainder 4.
3. Let's add the next pair of numbers to the remainder 4; we get 408.
4. Multiply the number on the top right by 2 and write it on the bottom right, adding _ x _ = to it. We get 6_ x _ =.
5. Instead of dashes, you need to substitute the same number, less than or equal to 408. We get 66 × 6 \u003d 396. Let's write 6 on the top right, since this is the second digit of the result. Subtract 396 from 408, we get 12.
6. Let's repeat steps 3-6. Since the digits carried down are in the fractional part of the number, it is necessary to put a decimal point on the top right after 6. Let's write the doubled result with dashes: 72_ x _ =. A suitable number would be 1: 721 × 1 = 721. Let's write it down as an answer. Let's subtract 1219 - 721 = 498.
7. Let's perform the sequence of actions given in the previous paragraph three more times to get the required number of decimal places. If there are not enough signs for further calculations, two zeros must be added to the current number on the left.

As a result, we get the answer: √1308.1912 ≈ 36.1689. If you check the action with a calculator, you can make sure that all the characters were determined correctly.

Bitwise calculation of the square root value

The method is highly accurate. In addition, it is quite understandable and it does not require memorizing formulas or a complex algorithm of actions, since the essence of the method is to select the correct result.

Let's extract the root from the number 781. Let's consider in detail the sequence of actions.

1. Find out which digit of the square root value will be the highest. To do this, let's square 0, 10, 100, 1000, etc. and find out between which of them the root number is located. We get that 10²< 781 < 100², т. е. старшим разрядом будут десятки.
2. Let's take the value of tens. To do this, we will take turns raising to the power of 10, 20, ..., 90, until we get a number greater than 781. For our case, we get 10² = 100, 20² = 400, 30² = 900. The value of the result n will be within 20< n <30.
3. Similarly to the previous step, the value of the units digit is selected. We alternately square 21.22, ..., 29: 21² = 441, 22² = 484, 23² = 529, 24² = 576, 25² = 625, 26² = 676, 27² = 729, 28² = 784. We get that 27< n < 28.
4. Each subsequent digit (tenths, hundredths, etc.) is calculated in the same way as shown above. Calculations are carried out until the required accuracy is achieved.

Let's consider this algorithm with an example. Let's find

1st step. We divide the number under the root into two digits (from right to left):

2nd step. We extract the square root from the first face, that is, from the number 65, we get the number 8. Under the first face, we write the square of the number 8 and subtract. We attribute the second face (59) to the remainder:

(the number 159 is the first remainder).

3rd step. We double the found root and write the result on the left:

4th step. We separate in the remainder (159) one digit on the right, on the left we get the number of tens (it is equal to 15). Then we divide 15 by the doubled first digit of the root, that is, by 16, since 15 is not divisible by 16, then in the quotient we get zero, which we write as the second digit of the root. So, in the quotient we got the number 80, which we double again, and demolish the next face

(the number 15901 is the second remainder).

5th step. We separate one digit from the right in the second remainder and divide the resulting number 1590 by 160. The result (number 9) is written as the third digit of the root and assigned to the number 160. The resulting number 1609 is multiplied by 9 and we find the following remainder (1420):

Further actions are performed in the sequence indicated in the algorithm (the root can be extracted with the required degree of accuracy).

Comment. If the root expression is a decimal fraction, then its integer part is divided into two digits from right to left, the fractional part is divided into two digits from left to right, and the root is extracted according to the specified algorithm.

DIDACTIC MATERIAL

1. Take the square root of the number: a) 32; b) 32.45; c) 249.5; d) 0.9511.

Quite often, when solving problems, we are faced with large numbers from which we need to extract Square root. Many students decide that this is a mistake and start resolving the whole example. Under no circumstances should this be done! There are two reasons for this:

1. The roots of large numbers do occur in problems. Especially in text;
2. There is an algorithm by which these roots are considered almost verbally.

We will consider this algorithm today. Perhaps some things will seem incomprehensible to you. But if you pay attention to this lesson, you will get the most powerful weapon against square roots.

So the algorithm:

1. Limit the desired root above and below to multiples of 10. Thus, we will reduce the search range to 10 numbers;
2. From these 10 numbers, weed out those that definitely cannot be roots. As a result, 1-2 numbers will remain;
3. Square these 1-2 numbers. That of them, the square of which is equal to the original number, will be the root.

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

Roots constraint

First of all, we need to find out between which numbers our root is located. It is highly desirable that the numbers be a multiple of ten:

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

We get a series of numbers:

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

What do these numbers give us? It's simple: we get boundaries. Take, for example, the number 1296. It lies between 900 and 1600. Therefore, its root cannot be less than 30 and greater than 40:

[Figure caption]

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

[Figure caption]

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

Elimination of obviously superfluous numbers

So, we have 10 numbers - candidates for the root. We received them very quickly, without complex thinking and multiplication in a column. It's time to move on.

Believe it or not, now we will reduce the number of candidate numbers to two - and again without any complicated calculations! It is enough to know the special rule. Here it is:

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

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

There are only 10 digits that can be in last place. Let's try to find out what they turn into when they are squared. 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 calculating the root. As you can see, the numbers in the second line turned out to be symmetrical with respect to the five. For instance:

2 2 = 4;
8 2 = 64 → 4.

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

[Figure caption]

The red squares show that we don't know this figure yet. But after all, the root lies between 50 and 60, on which there are only two numbers ending in 2 and 8:

[Figure caption]

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

Final Calculations

So, we have 2 candidate numbers left. How do you know which one is the root? The answer is obvious: square both numbers. The one that squared will give the original number, and will be the root.

For example, for the number 3364, we found two candidate numbers: 52 and 58. Let's square them:

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, in order to simplify the calculations, I used the formula of the squares of the sum and difference. Thanks to this, you didn’t even have to multiply the numbers in a column! This is another level of optimization of calculations, but, of course, it is completely optional :)

Root Calculation Examples

Theory is good, of course. But let's test it in practice.

[Figure caption]

First, let's find out between which numbers the number 576 lies:

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

Now let's look at the last number. It is equal to 6. When does this happen? Only if the root ends in 4 or 6. We get two numbers:

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

24 2 = (20 + 4) 2 = 576

Fine! The first square turned out to be equal to the original number. So this is the root.

Task. Calculate the square root:

[Figure caption]

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

Let's look at the last number:

1369 → 9;
33; 37.

Let's square it:

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.

Task. Calculate the square root:

[Figure caption]

We limit the number:

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

Let's look at the last number:

2704 → 4;
52; 58.

Let's square it:

52 2 = (50 + 2) 2 = 2500 + 2 50 2 + 4 = 2704;

We got the answer: 52. The second number will no longer need to be squared.

Task. Calculate the square root:

[Figure caption]

We limit the number:

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

Let's look at the last number:

4225 → 5;
65.

As you can see, after the second step, only one option remains: 65. This is the desired root. But let's still square it and check:

65 2 = (60 + 5) 2 = 3600 + 2 60 5 + 25 = 4225;

Everything is correct. We write down the answer.

Conclusion

Alas, no better. Let's take a look at the reasons. There are two of them:

• It is forbidden to use calculators at any normal math exam, be it the GIA or the Unified State Examination. And for carrying a calculator into the classroom, they can easily be kicked out of the exam.
• Don't be like stupid Americans. Which are not like roots - they cannot add two prime numbers. And at the sight of fractions, they generally get hysterical.

In mathematics, the question of how to take a root is considered relatively easy. If we square numbers from the natural series: 1, 2, 3, 4, 5 ... n, then we get the following series of squares: 1, 4, 9, 16 ... n 2. The series of squares is infinite, and if you look closely at it, you will see that there are not very many integers in it. Why this is so will be explained a little later.

The root of the number: calculation rules and examples

So, we squared the number 2, that is, we multiplied it by itself and got 4. But how to take the root of the number 4? Let's say right away that the roots can be square, cubic, and any degree to infinity.

The degree of the root is always a natural number, that is, it is impossible to solve such an equation: the root to the power of 3.6 of n.

Square root

Let's return to the question of how to extract the square root of 4. Since we squared the number 2, we will also extract the square root. In order to correctly take the root of 4, you just need to choose the right number that, when squared, would give the number 4. And this, of course, is 2. Look at the example:

• 2 2 =4
• Root of 4 = 2

This example is pretty simple. Let's try to extract the square root of 64. What number, when multiplied by itself, gives 64? Obviously it's 8.

• 8 2 =64
• Root of 64=8

cube root

As mentioned above, the roots are not only square, using an example we will try to explain more clearly how to extract a cube root or a root of the third degree. The principle of extracting a cube root is the same as that of a square root, the only difference is that the desired number was initially multiplied by itself not once, but twice. So, let's say we take the following example:

• 3x3x3=27
• Naturally, the cube root of the number 27 will be three:
• Root 3 of 27 = 3

Suppose you need to find the cube root of 64. To solve this equation, it is enough to find a number that, when raised to the third power, would give 64.

• 4 3 =64
• Root 3 of 64 = 4

Extract the root of a number on a calculator

Of course, it is best to learn to extract square, cube and other degrees by practice, by solving many examples and memorizing a table of squares and cubes of small numbers. In the future, this will greatly facilitate and reduce the time for solving equations. Although, it should be noted that sometimes it is required to extract the root of such a large number that it will cost a lot of work, if at all, to find the correct squared number. An ordinary calculator will come to the rescue in extracting the square root. How to take a root on a calculator? It is very simple to enter the number from which you want to find the result. Now take a close look at the calculator buttons. Even on the simplest of them, there is a key with a root icon. By clicking on it, you will immediately get the finished result.

Not every number can be taken as a whole root, consider the following example:

Root of 1859 = 43.116122…

You can try to solve this example on a calculator in parallel. As you can see, the resulting number is not an integer; moreover, the set of digits after the decimal point is not finite. A more accurate result can be given by special engineering calculators, but the full result simply does not fit on the display of ordinary ones. And if you continue the series of squares that you started earlier, you will not find the number 1859 in it, precisely because the number that you squared to get it is not an integer.

If you need to extract the root of the third degree on a simple calculator, then you need to double-click on the button with the root sign. For example, let's take the number 1859 used above and extract the cube root from it:

Root 3 of 1859 = 6.5662867…

That is, if the number 6.5662867 ... is raised to the third power, then we will get approximately 1859. Thus, extracting roots from numbers is not difficult, just remember the above algorithms.

Root formulas. properties of square roots.

Attention!
There are additional
material in Special Section 555.
For those who strongly "not very..."
And for those who "very much...")

In the previous lesson, we figured out what a square root is. It's time to figure out what are formulas for roots, what are root properties and what can be done about it all.

Root Formulas, Root Properties, and Rules for Actions with Roots- it's essentially the same thing. There are surprisingly few formulas for square roots. Which, of course, pleases! Rather, you can write a lot of all sorts of formulas, but only three are enough for practical and confident work with roots. Everything else flows from these three. Although many stray in the three formulas of the roots, yes ...

Let's start with the simplest. There she is:

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