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8th grade

The date:

Lesson number 9.

Topic: Approximate calculations of the square root.

Objectives: 1. To teach students to find approximate square roots.

2. Develop observation, the ability to analyze, compare, draw conclusions.

Cultivate a positive attitude towards learning

Lesson type: combined.

Lesson organization forms: individual, collective

Equipment: project board, mood reflection cards, microcalculator

Three paths lead to knowledge: the path of reflection

This is the noblest way

the way of imitation is the easiest way

and the way of experience is the most bitter way.

Confucius

During the classes.

Organizing time

Homework check step

No. 60 - 1 student performs at the blackboard, another student checks the correctness of the assignment on the spot

Oral work: projected on the board

a) Find the value of the root:

b) Does the expression make sense:

c) Find a number whose arithmetic square root is 0; one; 3; ten; 0.6

The stage of explaining new material

In order to calculate the approximate value of the square root, you must use a microcalculator. To do this, enter the radical expression into the calculator and press the key with the radical sign. But there is not always a calculator at hand, so you can find the approximate value of the square root as follows:

Let's find the value.

Since , then . Now, among the numbers located on the interval from 1 to 2, we take the neighboring numbers 1.4 and 1.5, we get: , then we take the numbers 1.41 and 1.42, these numbers satisfy the inequality . If we continue this process of squaring neighboring numbers, we get the following system of inequalities:

Projected onto the board.

From this system, comparing the numbers after the decimal point, we get:

Approximate values ​​of square roots can be taken in terms of excess and deficiency, i.e. by deficiency with an accuracy of 0.0001 and by excess.

Consolidation of the studied material.

Level "A"

0.2664 0.2 - by deficiency

№93 (calculator is used)

5. Valeological pause: exercises for the eyes.

Level "B"

6. Historical background on the need to find the value of square roots

(The willing student is invited in advance to prepare a message on this topic using the Internet)

A formula is proposed for finding the approximate value of the square root of an irrational number:

Level "C" No. 105

7. Reflection.

Summary of the lesson.

Homework: No. 102,

Topic: "Finding
approximate values ​​of the square root "

Lesson type: ONZ, R

Basic goals:

• learn to find approximate values ​​of the square root,
• learn methods for calculating roots.

During the classes

1. Self-determination to learning activities

Purpose of the stage: 1) include students in learning activities;

2) determine the content of the lesson: we continue to work on square roots

Organization of the educational process at stage 1:

What do we study in algebra lessons now? (Square roots)

What are square roots?

- Well done! For successful work, we will perform the following tasks.

2. Actualization of knowledge and fixation of difficulties in activities

Purpose of the stage: 1) update the educational content necessary and sufficient for the perception of new material: finding the values ​​of the square root;

2) to update the mental operations necessary and sufficient for the perception of new material: comparison, analysis, generalization;

3) fix all repeated concepts and algorithms in the form of schemes and symbols;

4) fix an individual difficulty in activity, demonstrating the lack of existing knowledge at a personally significant level: find the meaning of the expression.

Organization of the educational process at stage 2:

1. Calculate: , , , ,

4. Individual task.

Find the value of an expression..

3. Identification of the cause of the difficulty and setting the goal of the activity

Purpose of the stage: 1) organize communicative interaction, during which a distinctive property of the task that caused difficulty in educational activities is revealed and fixed: the ability to find the value of the square root;

2) agree on the purpose and topic of the lesson.

Organization of the educational process at stage 3:

– what did you need to do?

– What did you get? (Students show their options)

- What was the problem?

Is √2 extracted completely?

No.

How will we find?

What are the ways to find roots?

Guys, you see, we are not always dealing with numbers that are easily represented as a square of a number, which are extracted completely from under the root.

- What is our goal?

- Formulate the topic of the lesson.

- Write the topic in your notebook.

4. Building a project for getting out of a difficulty

Purpose of the stage: 1) organize communicative interaction to build a new mode of action that eliminates the cause of the identified difficulty;

2) fix a new mode of action in a sign, verbal form.

Organization of the educational process at stage 4:

1 METHOD to calculate √2 accurate to two decimal placesWe will argue as follows.

The number √2 is greater than 1 because 1 2 2 greater than 2. Therefore, the decimal notation of the number will begin as follows: 1, ... That is, the root of two, this is a unit with something.

Now let's try to find the number of tenths.

To do this, we will square fractions from one to two until we get a number greater than two.

Let's take a division step of 0.1, since we are looking for the number of tenths.

In other words, we will square the numbers: 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9

1,1 2 =1,21; 1,2 2 =1,44; 1,3 2 =1,69; 1,4 2 =1,96; 1,5 2 =2,25.

We got a number greater than two, the remaining numbers no longer need to be squared. Number 1.4 2 is less than 2 and 1.5 is 2 is already greater than two, then the number √2 must belong to the interval from 1.4 to 1.5. Therefore, the decimal notation of the number √2 in the tenth place must contain 4. √2=1.4….

In other words, 1.4

1,41 2 =1,9881, 1,42 2 =2,0164.

Already at 1.42 we get that its square is greater than two, further squaring numbers does not make sense.

From this we get that the number √2 will belong to the interval from 1.41 to 1.42 (1.41

Since we need to write √2 with an accuracy of two decimal places, we can already stop and not continue the calculation.

√2 ≈ 1.41. This will be the answer. If it were necessary to calculate an even more accurate value, one would have to continue the calculations, repeating the chain of reasoning over and over again.

Exercise

Calculate to two decimal places

√3 = , √5 = , √6 = , √7 =, √8 =

Conclusion This technique allows you to extract the root with any predetermined accuracy.

2 METHOD To find out the integer part of the square root of a number, you can, by subtracting from it all odd numbers in order, until the remainder is less than the next subtracted number or equal to zero, count the number of actions performed.

For example, let's find √16 like this:

1. 16 - 1 = 15
2. 15 - 3 = 12
3. 12 - 5 = 7
4. 7 - 7 =0

• 4 steps completed, so √16 = 4

Task Calculate

√1 = √6 =

√2 = √7 =

√3 = √8 =

√4 = √9 =

√5 = √10 =

Conclusion This technique is convenient when the root is completely removed.

3 METHOD The ancient Babylonians used the following method to find the approximate value of the square root of their x number. They represented the number x as the sum of a 2+b,

where a 2 is the exact square of the natural number a closest to the number x, and used the formula.

We extract the square root using the formula,

For example from number 28:

Conclusion The Babylonian method gives a good approximation to the exact value of the root.

5. Primary consolidation in external speech

Purpose of the stage: fix the studied educational content in external speech.

Organization of the educational process at stage 5:

from the textbook: Nos. 336, 337, 338,339, 343,345

6. Independent work with self-test according to the standard.

Purpose of the stage: test your ability to apply the addition and subtraction algorithm under typical conditions by comparing your solution with a standard for self-testing.

Organization of the educational process at stage 6:

Nos. 338 (a), 339 (c, d)

After checking against the standard, errors are analyzed and corrected.

7. Inclusion in the knowledge system and repetition

Purpose of the stage: 1) train the skills of using new content in conjunction with previously learned;

Organization of the educational process at stage 7:

1 group (medium) "No. ______________

Group 2 (high) №№ _________________

8. Reflection of activities in the lesson

1) fix the new content learned in the lesson;

2) evaluate their own activities in the lesson;

3) thank classmates who helped to get the result of the lesson;

4) fix unresolved difficulties as directions for future learning activities;

5) Discuss and write down homework.

Organization of the educational process at stage 8:

– What did we learn in class today?

– What have we learned to do today?

– Analyze your activities in the lesson and evaluate your work.

Homework №№ 344 , 346, 351

Now the question is: how to raise a number to an irrational power? For example, we want to know what 10 √2 is. The answer is, in principle, very simple. Let us take instead of √2 its approximation in the form of a finite decimal drdbi - this is a rational number. We can raise to a rational degree; it comes down to raising to an integer power and extracting the root. We will get the approximate value of the number. You can take a longer decimal fraction (this is again a rational number). Then you have to extract the root of a greater degree; after all, the denominator of a rational fraction will increase, but we will get a more accurate approximation. Of course, if we take the approximate value of √2 as a very long fraction, then exponentiation will be very difficult. How to cope with this task?

The calculation of square roots, cube roots, and other roots of a low degree is an arithmetic process that is quite accessible to us; calculating, we sequentially, one after the other, write decimals. But in order to raise to an irrational power or take a logarithm (to solve the inverse problem), such work is needed that it is no longer easy to apply the previous procedure. Tables come to the rescue. They are called tables of logarithms or tables of powers, depending on what they are intended for. They save time: to raise a number to an irrational power, we do not calculate, but only turn the pages.

Although the calculation of the values ​​\u200b\u200bcollected in tables is a purely technical procedure, it is nevertheless an interesting matter and has a long history. So let's see how it's done. We will calculate not only x \u003d 10 √2, but we will also solve another problem: 10 x \u003d 2, or x \u003d log 10 2. When solving these problems, we will not discover new numbers; these are just computational problems. The solution will be irrational numbers, infinite decimal fractions, and it is somehow inconvenient to declare them a new kind of numbers.

Let's think about how to solve our equations. The general idea is very simple. If we calculate 10 1 and 10 1/10 , and 10 1/100 , and 10 1/1000 , etc., and then multiply the results, we get 10 1.414 ... or l0 √ 2 By doing this, we will solve any problem of such kind. However, instead of 10 1/10, etc., we will calculate 10 1/2, and 10 1/4, etc. Before we begin, let's explain why we refer to the number 10 more often than other numbers. We know that the meaning of tables of logarithms goes far beyond the mathematical problem of computing roots, because

This is well known to anyone who has used the logarithm table to multiply numbers. On what basis b to take logarithms? It doesn't matter; for such calculations are based only on the principle, the general property of the logarithmic function. Having calculated the logarithms once for some arbitrary base, you can go to the logarithms for another base using multiplication. If you multiply equation (22.3) by 61, then it will remain true, so if you multiply all the numbers in the table of logarithms to the base b by 61, then such a table can also be used. Suppose we know the logarithms of all numbers to base b. In other words, we can solve the equation b a = c for any c; there is a table for that. The problem is how to find the logarithm of the same number c in a different base, such as x. We need to solve the equation x a' = c. This is easy to do because x can always be represented as: x = b t . Finding t given x and b is simple: t = log b x. Let us now substitute x = b t into the equation x a’ = c; it will go into this equation: (b t) a’ = b ta’ = c. In other words, the product ta' is the logarithm of c to the base b. So a' = a/t. Thus, the logarithms to the base x are equal to the products of the logarithms to the base b and the constant number l/t. Therefore, all tables of logarithms are equivalent up to multiplication by the number l/log b x. This allows us to choose any base for tabulation, but we decided that it is most convenient to use the number 10 as the base. (The question may arise: is there still some natural base that makes everything look somehow simpler? We We will try to answer this question later, while all logarithms will be calculated in base 10.)

Now let's see how the table of logarithms is compiled. The work begins with successive extractions of the square root of 10. The result can be seen in Table. 22.1. The exponents are written in its first column, and the numbers 10 s are in the third. It is clear that 10 1 \u003d 10. It is easy to raise 10 to a half power - this is the square root of 10, and everyone knows how to take the square root of any number. (The square root is best taken not in the way that is usually taught in school, but a little differently. To extract the square root of the number N, we choose the number a close enough to the answer, calculate N / a and the average a' = 1/2; this the average will be a new number a, a new approximation of the root of N. This process very quickly leads to the goal: the number of significant digits doubles after each step.) So we have found the first square root; it is equal to 3.16228. What does it give? Gives something. We can already tell what 10 0.5 is, and we know at least one logarithm.

The logarithm of 3.16228 is very close to 0.50000. However, we still need to make a little effort: we need a more detailed table. Let's take another square root and find 10 1/4, which is equal to 1.77828. Now we know another logarithm: 1.250 is the logarithm of 17.78; in addition, we can say what 10 0.75 is equal to: after all, this is 10 (0.5 + 0.25), i.e., the product of the second and third numbers from the third column of Table. 22.1. If you make the first column of the table long enough, then the table will contain almost all the numbers; multiplying the numbers from the third column, we get 10 to almost any power. This is the basic idea of ​​tables. Our table contains ten consecutive roots out of 10; the main work on compiling the table is invested in calculating these roots.

Why don't we continue to improve the accuracy of the tables further? Because we already noticed something. By raising 10 to a very small power, we get a unit with a small addition. This, of course, happens because if we raise, for example, 10 1/1000 to the 1000th power, then we again get 10; it is clear that 10 1/1000 cannot be a large number: it is very close to one. Moreover, small additions to unity behave as if they were divided by 2 each time; take a closer look at the table: 1815 goes to 903, then to 450, 225, etc. Thus, if we calculate one more, eleventh, square root, it will be equal to 1.00112 with great accuracy, and we guessed this result even before the calculation. Can you tell what the addition to one will be if you raise 10 to the power of ∆/1024 as ∆ tends to zero? Can. The addition will be approximately equal to 0.0022511∆. Of course, not exactly 0.0022511∆; to calculate this addition more precisely, they do the following trick: subtract one from 10 s and divide the difference by the exponent s. The deviations of the quotient thus obtained from its exact value are the same for any power of s. It can be seen that these ratios (Table 22.1) are approximately equal. At first they differ greatly, but then they come closer to each other, clearly striving for some number. What is this number? Let's see how the numbers of the fourth column change if we go down the column. First, the difference between two adjacent numbers is 0.0211, then 0.0104, then 0.0053, and finally 0.0026. The difference each time decreases by half. Taking one more step, we will bring it to 0.0013, then to 0.0007, 0.0003, 0.0002, and finally to about 0.0001; we must sequentially divide 26 by 2. Thus, we will go down another 26 units and find for the limit 2.3025. (Later we will see that 2.3026 would be more correct, but let's take what we have.) Using this table, you can raise 10 to any power, if its exponent is expressed in any way through I / I024.

Now it is easy to make a table of logarithms, because we have already saved up everything necessary for this. The procedure for this is shown in Table. 22.2, and the required numbers are taken from the second and third columns of Table. 22.1.

Suppose we want to know the logarithm of 2. This means that we want to know to what power 10 must be raised to get 2. Maybe raise 10 to the power of 1/2? No, it's too big. Looking at Table 22.1, we can say that the number we need lies between 1/4 and 1/2. Let's start searching for it with 1/4; divide 2 by 1.778…, we get 1.124…; when dividing, we subtracted 0.250000 from the logarithm of two, and now we are interested in the logarithm of 1.124 .... Having found it, we will add 1/4 = 256/1024 to the result. Let's find in Table 22.1 the number that, when moving along the third column from top to bottom, would immediately stand behind 1.124 .... This is 1.074607. The ratio of 1.124… to 1.074607 is 1.046598. In the end, we will represent 2 as a product of the numbers from Table. 22.1:
2 = (1,77828) (1,074607) (1,036633). (1,0090350) (1,000573).
For the last factor (1.000573) there was no place in our table; to find its logarithm, it is necessary to represent this number as 10∆/1024 ≈ 1 + 2.3025∆/1024. From here it is easy to find that ∆ = 0.254. Thus, our product can be represented as a ten raised to the power of 1/1024 (266 + 32 + 16 + 4 + 0.254). Adding and dividing, we get the desired logarithm: log 10 2 = 0.30103; this result is correct up to the fifth decimal place!

We calculated logarithms in exactly the same way as Mr. Briggs of Halifax did in 1620. When he finished, he said: "I have calculated successively 54 square roots of 10." In fact, he calculated only the first 27 roots, and then he did a trick with ∆. Computing 27 times the square root of 10 is actually a little more difficult than
10 times like we did. However, Mr. Briggs did much more: he calculated the roots to the sixteenth decimal place, and when he published his tables, he left them only 14 decimal places in order to round off errors. To compile tables of logarithms to the fourteenth decimal place by this method is very difficult. But as much as 300 years later, compilers of tables of logarithms were engaged in the fact that they reduced the tables of Mr. Briggs, throwing out of them a different number of decimal places each time. Only in recent times has it been possible with the help of electronic computers to compile tables of logarithms independently of Mr. Briggs. In this case, a more efficient calculation method was used, based on expanding the logarithm into a series.

While compiling the tables, we came across an interesting fact; if the exponent ε is very small, then it is very easy to calculate 10 ε ; it's just 1+2.3025ε. This means that 10 n/2.3025 = 1 + n for very small n. In addition, we said from the very beginning that we calculate base 10 logarithms only because we have 10 fingers on our hands and it is more convenient for us to count in tens. Logarithms to any other base are obtained from logarithms to base 10 by simple multiplication. Now it's time to find out if there is a mathematically distinguished base of logarithms, distinguished for reasons that have nothing to do with the number of fingers on the hand. In this natural scale, formulas with logarithms should look simpler. Let's make a new table of logarithms by multiplying all base 10 logarithms by 2.3025…. This corresponds to the transition to a new base - natural, or base e. Note that log e (l + n) ≈ n or e n ≈ 1 + n, when n → 0.

It is easy to find the number e itself; it is equal to 101/ 2.3025 or 10 0.4342294... That's 10 to the irrational power. To calculate e, you can use the table of roots of 10. Let's represent 0.434294 ... first as 444.73 / 1024, and the numerator of this fraction as the sum 444.73 \u003d 256 + 128 + 32 + 16 + 8 + 4 + 0.73 . The number e is therefore equal to the product of the numbers
(1,77828) (1,33352) (1,074607) (1,036633) (1,018152) (1,009035) (1,001643) = 2,7184.
(The number 0.73 is not in our table, but the corresponding result can be represented as 1 + 2.3025∆/1024 and calculated with ∆ = 0.73.) Multiplying all 7 factors, we get 2.7184 (by should actually be 2.7183, but this result is good). Using such tables, you can raise a number to an irrational power and calculate the logarithms of irrational numbers. That's how to deal with irrationality!

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 root number does not factor into two square factors (and it does in most cases), you will not be able to find the exact answer in the form of 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 large 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. Verification with a calculator gives us the answer 5.92 - we were right.

4. Another way is to decompose 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 required 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

      1. 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).

      2. 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.

      3. Mentally imagine the square whose area you need to calculate. You are looking for L, that is, the length of the side of a square whose area is S. A, B, C are numbers in the number L. You can write it differently: 10A + B \u003d L (for a two-digit number) or 100A + 10B + C \u003d L (for three-digit number) and so on.

         • Let (10A+B)² = L² = S = 100A² + 2×10A×B + B². Remember that 10A+B is a number whose B stands for ones and A stands for tens. For example, if A=1 and B=2, then 10A+B equals the number 12. (10A+B)² is the area of ​​the whole square, 100A² is the area of ​​the large inner square, B² is the area of ​​the small inner square, 10A×B is the area of ​​each of the two rectangles. Adding the areas of the figures described, you will find the area of ​​the original square.