How to find the root of a number. Research work on the topic: "Extracting square roots from large numbers without a calculator"

Do you want to do well in the exam in mathematics? Then you need to be able to count quickly, correctly and without a calculator. After all, the main reason for losing points on the exam in mathematics is computational errors.

According to the rules of the exam, it is forbidden to use a calculator during the exam in mathematics. The price may be too high - removal from the exam.

In fact, a calculator for the exam in mathematics is not needed. All tasks are solved without it. The main thing is attention, accuracy and some secret tricks, which we will talk about.

Let's start with the main rule. If a calculation can be simplified, simplify it.

Here, for example, is such a "devil's equation":

Seventy percent of graduates solve it head-on. The discriminant is calculated according to the formula, after which it is said that the root cannot be extracted without a calculator. But you can divide the left and right sides of the equation by . It turns out

Which way is easier? :-)

Many schoolchildren do not like multiplication in a "column". No one liked solving boring “case studies” in the fourth grade. However, in many cases it is possible to multiply numbers without a “column”, into a line. It's much faster.

Please note that we do not start with smaller digits, but with larger ones. It's comfortable.

Now, division. It is not easy to divide “in a column” into. But remember that the division sign: and the fractional line are one and the same. We write as a fraction and reduce the fraction:

Another example.

How to quickly and without any columns to square a two-digit number? We apply the abbreviated multiplication formulas:

Sometimes it is convenient to use another formula:

Numbers ending in , are squared instantly.

Suppose you need to find the square of a number (- not necessarily a digit, any natural number). Multiply by and add to the result. All!

For example: (and attributed).

(and attributed).

(and attributed).

This method is useful not only for squaring, but for taking the square root of numbers ending in .

And how do you take the square root without a calculator? We will show two ways.

The first way is to factorize the root expression.

For example, let's find
The number is divisible by (since the sum of its digits is divisible by ). Let's factor it out:

Let's find . This number is divisible by . It is also divided into. Let's factor it out.

Another example.

There is also a second way. It is convenient if the number from which the root must be extracted cannot be factored in any way.

For example, you need to find . The number under the root is odd, it is not divisible by, not divisible by, not divisible by ... You can continue to look for what it is still divisible by, or you can do it easier - find this root by selection.

Obviously, a two-digit number was squared, which is between the numbers and, since , , and the number is between them. We already know the first digit in the answer, this is .

The last digit in the number is . Since , , the last digit in the answer is either , or . Let's check:
. Happened!

Let's find .

So the first digit in the answer is five.

The last digit in the number is nine. , . So the last digit in the answer is either , or .

Let's check:

If the number from which the square root must be extracted ends with or, then the square root of it will be an irrational number. Because no integer square ends with or . Remember that in the tasks of some of the USE options in mathematics, the answer must be written as an integer or a final decimal fraction, that is, it must be a rational number.

Quadratic equations are found in tasks, and variants of the exam, as well as in part. In them, you need to consider the discriminant, and then extract the root from it. And it is not at all necessary to look for the roots of five-digit numbers. In many cases, the discriminant can be factorized.

For example, in the equation

Another situation in which the expression under the root can be factored is taken from the problem.

The hypotenuse of a right triangle is , one of the legs is equal to , find the second leg.

According to the Pythagorean theorem, it is equal to . You can count in a column for a long time, but it is easier to apply the abbreviated multiplication formula.

And now we’ll tell you the most interesting thing - because of what, after all, graduates lose precious points on the exam. After all, errors in calculations do not just happen.

1 . The surest way to lose points is sloppy calculations in which something is corrected, crossed out, one number is written on top of another. Look at your drafts. Perhaps they look the same? :-)

Write legibly! Don't skimp on paper. If something is wrong - do not correct one number for another, it is better to write again.

2. For some reason, many schoolchildren, counting in a column, try to do this 1) very, very quickly, 2) in very small numbers, in the corner of a notebook, and 3) with a pencil. The result is this:

It's impossible to parse anything. Why then be surprised that the grade for the exam is lower than expected?

3 . Many students are used to ignoring parentheses in expressions. Sometimes this also happens:

Remember that the equal sign is not placed anywhere, but only between equal values. Write well, even in draft form.

4 . A huge number of computational errors are associated with fractions. If you are dividing a fraction by a fraction, use the fact that
A "hamburger" is drawn here, that is, a multi-story fraction. It is extremely difficult to get the right answer with this method.

Let's summarize.

Checking the tasks of the first part of the profile exam in mathematics is automatic. There is no “almost right” answer here. Either he is correct or he is not. One computational error - and hello, the task does not count. Therefore, it is in your interest to learn how to count quickly, correctly and without a calculator.

The tasks of the second part of the profile exam in mathematics are checked by an expert. Take care of him! Let him understand both your handwriting and the logic of the decision.

How to extract the root from the number. In this article, we will learn how to take the square root of four and five digit numbers.

Let's take the square root of 1936 as an example.

Hence, .

The last digit in 1936 is 6. The square of 4 and 6 ends at 6. Therefore, 1936 can be the square of 44 or 46. It remains to be verified using multiplication.

Means,

Let's extract the square root of the number 15129.

Hence, .

The last digit in 15129 is 9. The 9 ends with the square of 3 and 7. Therefore, 15129 can be the square of 123 or 127. Let's check with multiplication.

Means,

How to root - video

And now I suggest you watch the video of Anna Denisova - "How to extract the root ", site author " simple physics", in which she explains how to extract square and cube roots without a calculator.

The video discusses several ways to extract roots:

1. The easiest way to extract the square root.

2. Matching using the square of the sum.

3. Babylonian way.

4. A method of extracting a square root in a column.

5. A quick way to extract the cube root.

6. The method of extracting the cube root in a column.

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Chapter first.

Extraction of the largest integer square root from a given integer.

170. Preliminary remarks.

A) Since we will be talking about extracting only the square root, for the sake of brevity in this chapter, instead of "square" root, we will simply say "root".

b) If we square the numbers of the natural series: 1,2,3,4,5. . . , then we get the following table of squares: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100,121,144. .,

Obviously, there are a lot of integers that are not in this table; from such numbers, of course, it is impossible to extract a whole root. Therefore, if you want to take the root of some integer, for example. it is required to find √4082, then we will agree to understand this requirement as follows: extract the whole root from 4082, if possible; if not, then we must find the largest integer whose square is 4082 (such a number is 63, since 63 2 \u003d 3969, and 64 2 \u003d 4090).

V) If this number is less than 100, then the root of it is in the multiplication table; so √60 would be 7, since sem 7 equals 49, which is less than 60, and 8 equals 64, which is greater than 60.

171. Extracting the root of a number less than 10,000 but greater than 100. Let it be necessary to find √4082 . Since this number is less than 10,000, then the root of it is less than √l0 000 = 100. On the other hand, this number is greater than 100; so the root of it is greater than (or equal to 10) . (If, for example, it were required to find √ 120 , then although the number 120 > 100, however √ 120 is equal to 10 because 11 2 = 121.) But any number that is greater than 10 but less than 100 has 2 digits; so the desired root is the sum:

tens + units,

and therefore its square must equal the sum:

This sum should be the largest square, consisting in 4082.

Let's take the largest of them, 36, and suppose that the square of the tens of the root will be equal to this largest square. Then the number of tens in the root must be 6. Let us now check that this must always be the case, i.e., the number of tens of the root is always equal to the largest integer root of the hundreds of the root number.

Indeed, in our example, the number of tens of the root cannot be more than 6, since (7 dec.) 2 \u003d 49 hundreds, which exceeds 4082. But it cannot be less than 6, since 5 dec. (with units) is less than 6 dess, and meanwhile (6 decs.) 2 = 36 hundreds, which is less than 4082. And since we are looking for the largest integer root, we should not take 5 dess for the root, when 6 tens is not a lot of.

So, we have found the number of tens of the root, namely 6. We write this number to the right of the = sign, remembering that it means the tens of the root. Raising it to the square, we get 36 hundreds. We subtract these 36 hundreds from the 40 hundreds of the root number and demolish the other two digits of this number. The remainder 482 must contain 2 (6 dec.) (units) + (units) 2. The product of (6 dec.) (unit) should be tens; therefore, the double product of tens by units must be sought in the tens of the remainder, i.e., in 48 (we will get their number by separating one digit from the right in the remainder 48 "2). which are not yet known), then we should get the number contained in 48. Therefore, we will divide 48 by 12.

To do this, we draw a vertical line to the left of the remainder and behind it (departing from the line one place to the left for the target that will now be found) we write the doubled first digit of the root, i.e. 12, and divide 48 into it. In the quotient we get 4.

However, one cannot guarantee in advance that the number 4 can be taken as the units of the root, since we have now divided by 12 the entire number of tens of the remainder, while some of them may not belong to the double product of tens by units, but are part of the square of units. Therefore, the number 4 may be large. You have to test her. It is obviously suitable if the sum of 2 (6 dec.) 4 + 4 2 turns out to be no more than the remainder of 482.

As a result, we immediately get the sum of both. The resulting product turned out to be 496, which is more than the remainder of 482; So 4 is big. Then we will test the next smaller number 3 in the same way.

Examples.

In the 4th example, when dividing 47 tens of the remainder by 4, we get 11 in the quotient. But since the units digit of the root cannot be a two-digit number 11 or 10, we must directly test the number 9.

In the 5th example, after subtracting 8 from the first face of the square, the remainder is 0, and the next face also consists of zeros. This shows that the desired root consists of only 8 tens, and therefore zero must be put in place of units.

172. Extracting the root of a number greater than 10000. Let it be required to find √35782 . Since the radical number is greater than 10,000, then the root of it is greater than √10000 = 100 and, therefore, it consists of 3 digits or more. No matter how many digits it consists of, we can always consider it as the sum of only tens and units. If, for example, the root turned out to be 482, then we can consider it as the sum of 48 dess. + 2 units Then the square of the root will consist of 3 terms:

(dec.) 2 + 2 (dec.) (un.) + (un.) 2 .

Now we can reason in exactly the same way as when finding √4082 (in the previous paragraph). The only difference will be that in order to find the tens of the root of 4082, we had to extract the root of 40, and this could be done using the multiplication table; now, to get tens√35782, we will have to take the root of 357, which cannot be done using the multiplication table. But we can find √357 by the trick described in the previous paragraph, since the number 357< 10 000. Наибольший целый корень из 357 оказывается 18. Значит, в √3"57"82 должно быть 18 десятков. Чтобы найти единицы, надо из 3"57"82 вычесть квадрат 18 десятков, для чего достаточно вычесть квадрат 18 из 357 сотен и к остатку снести 2 последние цифры подкоренного числа. Остаток от вычитания квадpaта 18 из 357 у нас уже есть: это 33. Значит, для получения остатка от вычитания квадрата 18 дес. из 3"57"82, достаточно к 33 приписать справа цифры 82.

Next, we proceed as we did when finding √4082, namely: to the left of the remainder of 3382 we draw a vertical line and after it we write (departing from the line by one place) twice the number of root tens found, i.e. 36 (twice 18). In the remainder, we separate one digit on the right and divide the number of tens of the remainder, i.e. 338, by 36. In the quotient we get 9. We test this number, for which we attribute it to 36 on the right and multiply it by it. The product turned out to be 3321, which is less than the remainder. So the number 9 is good, we write it at the root.

In general, to take the square root of any whole number, one must first take the root of its hundreds; if this number is more than 100, then you will have to look for the root from the number of hundreds of these hundreds, that is, from tens of thousands of a given number; if this number is more than 100, you will have to take the root from the number of hundreds of tens of thousands, that is, from millions of a given number, etc.

Examples.

In the last example, finding the first digit and subtracting its square, we get 0 in the remainder. We demolish the next 2 digits 51. Separating the tens, we get 5 dec, while the root digit found twice is 6. So, dividing 5 by 6 we get 0 We put 0 at the root in second place and demolish the next 2 digits to the remainder; we get 5110. Then we continue as usual.

In this example, the desired root consists of only 9 hundreds, and therefore zeros must be put in place of tens and units.

Rule. In order to extract the square root of a given integer, break it, from the right hand to the left, on the edge, with 2 digits in each, except for the last one, which can have one digit.
To find the first digit of the root, take the square root of the first face.
To find the second digit, the square of the first digit of the root is subtracted from the first face, the second face is demolished to the remainder, and the number of tens of the resulting number is divided by twice the first digit of the root; the resulting integer is tested.
This test is carried out as follows: behind the vertical line (to the left of the remainder) they write the twice previously found number of the root and to it, on the right side, they attribute the test figure, the resulting number, after this addition, the number is multiplied by the test figure. If, after multiplication, a number is obtained that is greater than the remainder, then the test figure is not good and the next smaller number must be tested.
The following numbers of the root are found by the same method.
If, after demolishing the face, the number of tens of the resulting number turns out to be less than the divisor, i.e., less than twice the found part of the root, then 0 is put in the root, the next face is demolished and the action continues further.

173. The number of digits of the root. From consideration of the process of finding the root, it follows that there are as many digits in the root as there are faces of 2 digits each in the root number (there may be one digit in the left side).

Chapter two.

Extracting approximate square roots from whole and fractional numbers .

Extracting the square root of polynomials, see the additions to the 2nd part of § 399 et seq.

174. Signs of an exact square root. The exact square root of a given number is a number whose square is exactly equal to the given number. Let us indicate some signs by which one can judge whether the exact root is extracted from a given number or not:

A) If the exact integer root is not extracted from a given integer (it is obtained when extracting the remainder), then a fractional exact root cannot be found from such a number, since any fraction that is not equal to an integer, when multiplied by itself, also gives a fraction in the product, not an integer.

b) Since the root of a fraction is equal to the root of the numerator divided by the root of the denominator, the exact root of an irreducible fraction cannot be found if it cannot be extracted from the numerator or from the denominator. For example, the exact root cannot be extracted from fractions 4/5, 8/9 and 11/15, since in the first fraction it cannot be extracted from the denominator, in the second - from the numerator and in the third - neither from the numerator nor from the denominator.

From such numbers, from which it is impossible to extract the exact root, only approximate roots can be extracted.

175. Approximate root up to 1. An approximate square root up to 1 of a given number (integer or fractional - it doesn't matter) is an integer that satisfies the following two requirements:

1) the square of this number is not greater than the given number; 2) but the square of this number increased by 1 is greater than the given number. In other words, the approximate square root up to 1 is the largest integer square root of a given number, that is, the root that we learned to find in the previous chapter. This root is called approximate up to 1, because in order to obtain an exact root, some fraction less than 1 would have to be added to this approximate root, so if we take this approximate root instead of an unknown exact root, we will make an error less than 1.

Rule. To extract an approximate square root with an accuracy of 1, you need to extract the largest integer root of the integer part of a given number.

The number found according to this rule is an approximate root with a disadvantage, since it lacks some fraction (less than 1) to the exact root. If we increase this root by 1, then we get another number in which there is some excess over the exact root, and this excess is less than 1. This root increased by 1 can also be called an approximate root up to 1, but with an excess. (The names: "with a lack" or "with an excess" in some mathematical books are replaced by others equivalent: "by deficiency" or "by excess".)

176. Approximate root with an accuracy of 1/10. Let it be required to find √2.35104 up to 1/10. This means that it is required to find such a decimal fraction, which would consist of whole units and tenths, and which would satisfy the following two requirements:

1) the square of this fraction does not exceed 2.35104, but 2) if we increase it by 1/10, then the square of this increased fraction exceeds 2.35104.

To find such a fraction, we first find an approximate root up to 1, that is, we extract the root only from the integer 2. We get 1 (and the remainder is 1). We write the number 1 at the root and put a comma after it. Now we will look for the number of tenths. To do this, we take down the digits 35 to the remainder of 1, to the right of the comma, and continue the extraction as if we were extracting the root from the integer 235. We write the resulting number 5 at the root in place of the tenths. We do not need the remaining digits of the root number (104). That the resulting number 1.5 will indeed be an approximate root with an accuracy of 1/10 is evident from the following. If we were to find the largest integer root of 235 with an accuracy of 1, then we would get 15. So:

15 2 < 235, but 16 2 >235.

Dividing all these numbers by 100, we get:

This means that the number 1.5 is that decimal fraction, which we called the approximate root with an accuracy of 1/10.

We also find by this method the following approximate roots with an accuracy of 0.1:

177. Approximate square root with an accuracy of 1/100 to 1/1000, etc.

Let it be required to find an approximate √248 with an accuracy of 1/100. This means: to find such a decimal fraction, which would consist of integers, tenths and hundredths and which would satisfy two requirements:

1) its square does not exceed 248, but 2) if we increase this fraction by 1/100, then the square of this increased fraction exceeds 248.

We will find such a fraction in the following sequence: first we will find an integer, then the tenths digit, then the hundredths digit. The square root of an integer will be 15 integers. To get the number of tenths, as we have seen, it is necessary to take down to the remainder 23 2 more digits to the right of the decimal point. In our example, these numbers do not exist at all, we put zeros in their place. Assigning them to the remainder and continuing the action as if we were finding the root of the integer 24,800, we will find the tenths digit 7. It remains to find the hundredths digit. To do this, we add 2 more zeros to the remainder 151 and continue the extraction, as if we were finding the root of the integer 2,480,000. We get 15.74. That this number is indeed the approximate root of 248 to within 1/100 is evident from the following. If we were to find the largest integer square root of the integer 2,480,000, we would get 1574; Means:

1574 2 < 2,480,000 but 1575 2 > 2,480,000.

Dividing all numbers by 10,000 (= 100 2), we get:

So 15.74 is that decimal fraction that we called the approximate root with an accuracy of 1/100 of 248.

Applying this technique to finding an approximate root with an accuracy of 1/1000 to 1/10000, etc., we find the following.

Rule. To extract from a given integer or from a given decimal fraction an approximate root with an accuracy of 1/10 to 1/100 to 1/100, etc., first find an approximate root with an accuracy of 1, extracting the root from the integer (if it no, they write about the root of 0 integers).

Then find the number of tenths. To do this, the remainder is demolished, 2 digits of the radical number to the right of the comma (if they are not, two zeros are attributed to the remainder), and the extraction is continued in the same way as is done when extracting the root from an integer. The resulting figure is written at the root in place of tenths.

Then find the number of hundredths. To do this, two numbers are again demolished to the remainder, to the right of those that were just demolished, etc.

Thus, when extracting the root from an integer with a decimal fraction, it is necessary to divide by 2 digits each, starting from the comma, both to the left (in the integer part of the number) and to the right (in the fractional part).

Examples.

1) Find up to 1/100 roots: a) √2; b) √0.3;

In the last example, we converted 3/7 to a decimal by calculating 8 decimal places to form the 4 faces needed to find the 4 decimal places of the root.

178. Description of the table of square roots. At the end of this book is a table of square roots calculated with four digits. Using this table, you can quickly find the square root of an integer (or decimal fraction), which is expressed in no more than four digits. Before explaining how this table is arranged, we note that we can always find the first significant digit of the desired root without the help of tables by one glance at the root number; we can also easily determine which decimal place means the first digit of the root and, therefore, where in the root, when we find its digits, we need to put a comma. Here are some examples:

1) √5"27,3 . The first digit will be 2, since the left side of the root number is 5; and the root of 5 is 2. In addition, since there are only 2 in the integer part of the radical number of all faces, then the integer part of the desired root must have 2 digits and, therefore, its first digit 2 must mean tens.

2) √9.041. Obviously, in this root, the first digit will be 3 simple units.

3) √0.00"83"4 . The first significant digit is 9, since the face from which the root would have to be extracted to obtain the first significant digit is 83, and the root of 83 is 9. Since there will be neither integers nor tenths in the desired number, the first digit 9 must mean hundredths.

4) √0.73 "85. The first significant figure is 8 tenths.

5) √0.00 "00" 35 "7. The first significant figure will be 5 thousandths.

Let's make one more remark. Suppose that it is required to extract the root from such a number, which, after discarding the occupied one in it, is depicted by a series of such numbers: 5681. This root can be one of the following:

If we take the roots that we underlined with one line, then they will all be expressed by the same series of numbers, exactly the numbers that are obtained by extracting the root from 5681 (these will be the numbers 7, 5, 3, 7). The reason for this is that the faces into which the radical number has to be divided when finding the digits of the root will be the same in all these examples, therefore the digits for each root will be the same (only the position of the comma will, of course, be different). In the same way, in all the roots, underlined by us with two lines, the same numbers should be obtained, exactly those that express √568.1 (these numbers will be 2, 3, 8, 3), and for the same reason. Thus, the digits of the roots from the numbers depicted (by discarding the comma) by the same series of digits 5681 will be of a twofold (and only twofold) kind: either this is a series of 7, 5, 3, 7, or a series of 2, 3, 8, 3. The same, obviously, can be said about any other series of numbers. Therefore, as we will now see, in the table, each row of digits of the radical number corresponds to 2 rows of digits for the roots.

Now we can explain the structure of the table and how to use it. For clarity of explanation, we have depicted here the beginning of the first page of the table.

This table spans several pages. On each of them, in the first column on the left, the numbers 10, 11, 12 ... (up to 99) are placed. These numbers express the first 2 digits of the number from which the square root is being sought. In the upper horizontal line (as well as in the bottom) there are numbers: 0, 1, 2, 3 ... 9, which are the 3rd digit of this number, and then further to the right are the numbers 1, 2, 3. . . 9, representing the 4th digit of this number. In all other horizontal lines, 2 four-digit numbers are placed, expressing the square roots of the corresponding numbers.

Let it be required to find the square root of some number, integer or expressed as a decimal fraction. First of all, we find without the help of tables the first digit of the root and its category. Then we discard the comma in the given number, if any. Suppose first that after discarding the comma, only 3 digits remain, for example. 114. We find in the tables in the leftmost column the first 2 digits, i.e. 11, and move from them to the right along the horizontal line until we reach the vertical column, at the top (and bottom) of which is the 3rd digit of the number , i.e. 4. In this place we find two four-digit numbers: 1068 and 3376. Which of these two numbers should be taken and where to put a comma in it, this is determined by the first digit of the root and its discharge, which we found earlier. So, if you need to find √0.11 "4, then the first digit of the root is 3 tenths, and therefore we must take 0.3376 for the root. If it were required to find √1.14, then the first digit of the root would be 1, and we would then take 1.068.

Thus we can easily find:

√5.30 = 2.302; √7"18 = 26.80; √0.91"6 = 0.9571, etc.

Let us now suppose that it is required to find the root of a number expressed (by discarding the comma) by 4 digits, for example √7 "45.6. Noticing that the first digit of the root is 2 tens, we find for the number 745, as it has now been explained, the numbers 2729 (we only notice this number with a finger, but do not write it down.) Then we move further from this number to the right until on the right side of the table (behind the last bold line) we meet the vertical column that is marked above (and below) 4 th digit of this number, i.e. the number 6, and we find the number 1 there. This will be the correction that must be applied (in the mind) to the previously found number 2729; we get 2730. We write this number and put a comma in it in the proper place : 27.30.

In this way we find, for example:

√44.37 = 6.661; √4.437 = 2.107; √0.04"437 \u003d 0.2107, etc.

If the radical number is expressed in only one or two digits, then we can assume that after these digits there are one or two zeros, and then proceed as was explained for the three-digit number. For example √2.7 = √2.70 =1.643; √0.13 \u003d √0.13 "0 \u003d 0.3606, etc..

Finally, if the radical number is expressed by more than 4 digits, then we will take only the first 4 of them, and discard the rest, and to reduce the error, if the first of the discarded digits is 5 or more than 5, then we will increase the fourth of the retained digits by l . So:

√357,8| 3 | = 18,91; √0,49"35|7 | = 0.7025; and so on.

Comment. The tables indicate the approximate square root, sometimes with a deficiency, sometimes with an excess, namely, one of these approximate roots that comes closer to the exact root.

179. Extraction of square roots from ordinary fractions. The exact square root of an irreducible fraction can only be extracted when both terms of the fraction are exact squares. In this case, it is enough to extract the root from the numerator and denominator separately, for example:

The approximate square root of an ordinary fraction with some decimal precision can be most easily found if we first convert the ordinary fraction to a decimal, calculating in this fraction the number of decimal places after the decimal point, which would be twice the number of decimal places in the desired root.

However, you can do otherwise. Let's explain this with the following example:

Find approximate √ 5 / 24

Let's make the denominator an exact square. To do this, it would be enough to multiply both terms of the fraction by the denominator 24; but in this example, you can do otherwise. We decompose 24 into prime factors: 24 \u003d 2 2 2 3. From this decomposition it can be seen that if 24 is multiplied by 2 and another by 3, then in the product each prime factor will be repeated an even number of times, and, therefore, the denominator will become a square:

It remains to calculate √30 with some accuracy and divide the result by 12. In this case, it must be borne in mind that the fraction showing the degree of accuracy will also decrease from dividing by 12. So, if we find √30 with an accuracy of 1/10 and divide the result by 12, then we get the approximate root of the fraction 5/24 with an accuracy of 1/120 (namely 54/120 and 55/120)

Chapter three.

Function Graphx = √ y .

180. Inverse function. Let there be an equation that defines at as a function of X , for example, this: y = x 2 . We can say that it determines not only at as a function of X , but also, conversely, determines X as a function of at , albeit in an implicit way. To make this function explicit, we need to solve this equation for X , taking at for a known number; So, from the equation we have taken, we find: y = x 2 .

The algebraic expression obtained for x after solving the equation that defines y as a function of x is called the inverse function of the one that defines y.

So the function x = √ y function inverse y = x 2 . If, as is customary, the independent variable is denoted X , and dependent at , then we can express the inverse function obtained now as follows: y = √x . Thus, in order to obtain a function that is inverse to a given (direct), it is necessary to derive from the equation that defines this given function X depending on the y and in the resulting expression, replace y on x , A X on y .

181. Graph of a function y = √x . This function is not possible with a negative value X , but it can be calculated (with any accuracy) for any positive value x , and for each such value, the function receives two different values ​​with the same absolute value, but with opposite signs. If familiar √ we denote only the arithmetic value of the square root, then these two values ​​of the function can be expressed as follows: y= ± √ x To plot this function, you must first create a table of its values. The easiest way to compile this table is from a table of direct function values:

y = x 2 .

x

y

if values at take as values X , and vice versa:

y= ± √ x

Putting all these values ​​on the drawing, we get the following graph.

In the same drawing, we depicted (dashed line) and the graph of the direct function y = x 2 . Let's compare these two charts.

182. Relationship between graphs of direct and inverse functions. To compile a table of inverse function values y= ± √ x we took for X those numbers that are in the direct function table y = x 2 served as values ​​for at , and for at took those numbers; which in this table were the values ​​for x . From this it follows that both graphs are the same, only the graph of the direct function is so located relative to the axis at - s how the graph of the inverse function is located relative to the axis X - ov. As a result, if we fold the drawing around a straight line OA bisecting a right angle xOy , so that the part of the drawing containing the semiaxis OU , fell on the part that contains the semi-axis Oh , That OU compatible with Oh , all divisions OU coincide with divisions Oh , and the points of the parabola y = x 2 coincide with the corresponding points on the graph y= ± √ x . For example, dots M And N , whose ordinate 4 , and the abscissa 2 And - 2 , coincide with the points M" And N" , whose abscissa 4 , and the ordinates 2 And - 2 . If these points coincide, then this means that the lines MM" And NN" perpendicular to OA and divide this straight line in half. The same can be said for all other relevant points on both graphs.

Thus, the graph of the inverse function should be the same as the graph of the direct function, but these graphs are located differently, namely symmetrically with each other with respect to the bisector of the angle hoy . We can say that the graph of the inverse function is a reflection (as in a mirror) of the graph of the direct function with respect to the bisector of the angle hoy .

The calculation (or extraction) of the square root can be done in several ways, but all of them are not very simple. It's easier, of course, to resort to the help of a calculator. But if this is not possible (or you want to understand the essence of the square root), I can advise you to go the following way, its algorithm is as follows:

If you don’t have the strength, desire or patience for such lengthy calculations, you can resort to rough selection, its plus is that it is incredibly fast and, with due ingenuity, accurate. Example:

When I was in school (in the early 60s), we were taught to take the square root of any number. The technique is simple, outwardly similar to division by a column, but to state it here, it will take half an hour of time and 4-5 thousand characters of text. But why do you need it? Do you have a phone or other gadget, there is a calculator in nm. There is a calculator in every computer. Personally, I prefer to do this kind of calculation in Excel.

Often in school it is required to find the square roots of different numbers. But if we are used to using a calculator all the time for this, then there will be no such opportunity in exams, so you need to learn how to look for the root without the help of a calculator. And it is in principle possible to do so.

The algorithm is:

Look first at the last digit of your number:

For example,

Now you need to determine approximately the value for the root from the leftmost group

In the case when the number has more than two groups, then you need to find the root like this:

But the next number should be exactly the largest, you need to pick it up like this:

Now we need to form a new number A by adding to the remainder that was obtained above, the next group.

In our examples:

A column of najna, and when more than fifteen characters are needed, then computers and phones with calculators most often rest. It remains to check whether the description of the methodology will take 4-5 thousand characters.

Berm any number, from a comma we count pairs of digits to the right and left

For example, 1234567890.098765432100

A pair of digits is like a two-digit number. The root of a two-digit is one-to-one. We select a single-valued one, the square of which is less than the first pair of digits. In our case it is 3.

As when dividing by a column, under the first pair we write out this square and subtract from the first pair. The result is underlined. 12 - 9 = 3. Add a second pair of digits to this difference (it will be 334). To the left of the number of berms, the doubled value of the part of the result that has already been found is supplemented with a digit (we have 2 * 6 = 6), such that when multiplied by the not received number, it does not exceed the number with the second pair of digits. We get that the figure found is five. Again we find the difference (9), demolish the next pair of digits, getting 956, again write out the doubled part of the result (70), again add the necessary digit and so on until it stops. Or to the required accuracy of calculations.

Firstly, in order to calculate the square root, you need to know the multiplication table well. The simplest examples are 25 (5 by 5 = 25) and so on. If we take numbers more complicated, then we can use this table, where there are units horizontally and tens vertically.



There is a good way to find the root of a number without the help of calculators. To do this, you will need a ruler and a compass. The bottom line is that you find on the ruler the value that you have under the root. For example, put a mark near 9. Your task is to divide this number into an equal number of segments, that is, into two lines of 4.5 cm each, and into an even segment. It is easy to guess that in the end you will get 3 segments of 3 centimeters.

The method is not easy and will not work for large numbers, but it is considered without a calculator.

without the help of a calculator, the method of extracting the square root was taught in Soviet times at school in the 8th grade.

To do this, you need to break a multi-digit number from right to left into faces of 2 digits :

The first digit of the root is the whole root of the left side, in this case 5.

Subtract 5 squared from 31, 31-25=6 and add the next face to the six, we have 678.

The next digit x is selected to double the five so that

10x*x was the maximum, but less than 678.

x=6 because 106*6=636,

now we calculate 678 - 636 = 42 and add the next face 92, we have 4292.

Again we are looking for the maximum x, such that 112x*x lt; 4292.

Answer: the root is 563

So you can continue as long as you want.

In some cases, you can try to expand the root number into two or more square factors.

It is also useful to remember the table (or at least some part of it) - the squares of natural numbers from 10 to 99.



I propose a variant of extracting the square root into a column that I invented. It differs from the well-known, except for the selection of numbers. But as I found out later, this method already existed many years before my birth. The great Isaac Newton described it in his book General Arithmetic or a book on arithmetic synthesis and analysis. So here I present my vision and rationale for the algorithm of the Newton method. You don't need to memorize the algorithm. You can simply use the diagram in the figure as a visual aid if necessary.







With the help of tables, you can not calculate, but find, the square roots only from the numbers that are in the tables. The easiest way to calculate the roots is not only square, but also other degrees, by the method of successive approximations. For example, we calculate the square root of 10739, replace the last three digits with zeros and extract the root of 10000, we get 100 with a disadvantage, so we take the number 102 and square it, we get 10404, which is also less than the specified one, we take 103*103=10609 again with a disadvantage, we take 103.5 * 103.5 \u003d 10712.25, we take even more 103.6 * 103.6 \u003d 10732, we take 103.7 * 103.7 \u003d 10753.69, which is already in excess. You can take the square root of 10739 to be approximately equal to 103.6. More precisely 10739=103.629... . . Similarly, we calculate the cube root, first from 10000 we get approximately 25 * 25 * 25 = 15625, which is in excess, we take 22 * ​​22 * ​​22 = 10.648, we take a little more than 22.06 * 22.06 * 22.06 = 10735, which is very close to the given one.