Extracting a root from a large number. Dear friends!In this article, we will show you how to take the root of a large number without a calculator. This is necessary not only for solving certain types of USE problems (there are such problems for movement), but it is also desirable to know this analytical technique for general mathematical development.
It would seem that everything is simple: factorize and extract. There is no problem. For example, the number 291600, when expanded, will give the product:
We calculate:
There is one BUT! The method is good if divisors 2, 3, 4 and so on are easily determined. But what if the number from which we extract the root is a product of prime numbers? For example, 152881 is the product of the numbers 17, 17, 23, 23. Try to find these divisors right away.
The essence of the method we are considering- this is pure analysis. The root with the accumulated skill is found quickly. If the skill is not worked out, but the approach is simply understood, then it is a little slower, but still determined.
Let's take the root of 190969.
First, let's determine between what numbers (multiples of a hundred) our result lies.
Obviously, the result of the root of a given number lies in the range from 400 to 500, because
400 2 =160000 and 500 2 =250000
Really:
in the middle, closer to 160,000 or 250,000?
The number 190969 is somewhere in the middle, but still closer to 160000. We can conclude that the result of our root will be less than 450. Let's check:
Indeed, it is less than 450, since 190,969< 202 500.
Now let's check the number 440:
So our result is less than 440, since 190 969 < 193 600.
Checking the number 430:
We have established that the result of this root lies in the range from 430 to 440.
The product of numbers ending in 1 or 9 gives a number ending in 1. For example, 21 times 21 equals 441.
The product of numbers ending in 2 or 8 gives a number ending in 4. For example, 18 times 18 equals 324.
The product of numbers ending in 5 gives a number ending in 5. For example, 25 times 25 equals 625.
The product of numbers ending in 4 or 6 gives a number ending in 6. For example, 26 times 26 equals 676.
The product of numbers ending in 3 or 7 gives a number ending in 9. For example, 17 times 17 equals 289.
Since the number 190969 ends with the number 9, then this product is either 433 or 437.
*Only they, when squared, can give 9 at the end.
We check:
So the result of the root will be 437.
That is, we kind of "felt" the right answer.
As you can see, the maximum that is required is to carry out 5 actions in a column. Perhaps you will immediately get to the point, or you will do just three actions. It all depends on how accurately you make the initial estimate of the number.
Extract your own root from 148996
Such a discriminant is obtained in the problem:
The motor ship passes along the river to the destination 336 km and after parking returns to the point of departure. Find the speed of the ship in still water, if the speed of the current is 5 km / h, the parking lasts 10 hours, and the ship returns to the point of departure 48 hours after leaving it. Give your answer in km/h.
View Solution
The result of the root is between the numbers 300 and 400:
300 2 =90000 400 2 =160000
Indeed, 90000<148996<160000.
The essence of further reasoning is to determine how the number 148996 is located (distanced) relative to these numbers.
Calculate the differences 148996 - 90000=58996 and 160000 - 148996=11004.
It turns out that 148996 is close (much closer) to 160000. Therefore, the result of the root will definitely be greater than 350 and even 360.
We can conclude that our result is greater than 370. Further, it is clear: since 148996 ends with the number 6, this means that you must square the number ending either in 4 or 6. *Only these numbers, when squared, give in end 6.
Sincerely, Alexander Krutitskikh.
P.S: I would be grateful if you tell about the site in social networks.
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. Here she is:
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You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)
you can get acquainted with functions and derivatives.
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 must be the largest square contained 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, but 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, that is, 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 performed 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 figures. 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 .
Fact 1.
\(\bullet\) Take some non-negative number \(a\) (ie \(a\geqslant 0\) ). Then (arithmetic) square root from the number \(a\) such a non-negative number \(b\) is called, when squaring it we get the number \(a\) : \[\sqrt a=b\quad \text(same as )\quad a=b^2\] It follows from the definition that \(a\geqslant 0, b\geqslant 0\). These restrictions are an important condition for the existence of a square root and should be remembered!
Recall that any number when squared gives a non-negative result. That is, \(100^2=10000\geqslant 0\) and \((-100)^2=10000\geqslant 0\) .
\(\bullet\) What is \(\sqrt(25)\) ? We know that \(5^2=25\) and \((-5)^2=25\) . Since by definition we have to find a non-negative number, \(-5\) is not suitable, hence \(\sqrt(25)=5\) (since \(25=5^2\) ).
Finding the value \(\sqrt a\) is called taking the square root of the number \(a\) , and the number \(a\) is called the root expression.
\(\bullet\) Based on the definition, the expressions \(\sqrt(-25)\) , \(\sqrt(-4)\) , etc. don't make sense.
Fact 2.
For quick calculations, it will be useful to learn the table of squares of natural numbers from \(1\) to \(20\) : \[\begin(array)(|ll|) \hline 1^2=1 & \quad11^2=121 \\ 2^2=4 & \quad12^2=144\\ 3^2=9 & \quad13 ^2=169\\ 4^2=16 & \quad14^2=196\\ 5^2=25 & \quad15^2=225\\ 6^2=36 & \quad16^2=256\\ 7^ 2=49 & \quad17^2=289\\ 8^2=64 & \quad18^2=324\\ 9^2=81 & \quad19^2=361\\ 10^2=100& \quad20^2= 400\\ \hline \end(array)\]
Fact 3.
What can be done with square roots?
\(\bullet\) The sum or difference of square roots is NOT EQUAL to the square root of the sum or difference, i.e. \[\sqrt a\pm\sqrt b\ne \sqrt(a\pm b)\] Thus, if you need to calculate, for example, \(\sqrt(25)+\sqrt(49)\) , then initially you must find the values \(\sqrt(25)\) and \(\sqrt(49)\ ) and then add them up. Hence, \[\sqrt(25)+\sqrt(49)=5+7=12\] If the values \(\sqrt a\) or \(\sqrt b\) cannot be found when adding \(\sqrt a+\sqrt b\), then such an expression is not further converted and remains as it is. For example, in the sum \(\sqrt 2+ \sqrt (49)\) we can find \(\sqrt(49)\) - this is \(7\) , but \(\sqrt 2\) cannot be converted in any way, That's why \(\sqrt 2+\sqrt(49)=\sqrt 2+7\). Further, this expression, unfortunately, cannot be simplified in any way.\(\bullet\) The product/quotient of square roots is equal to the square root of the product/quotient, i.e. \[\sqrt a\cdot \sqrt b=\sqrt(ab)\quad \text(s)\quad \sqrt a:\sqrt b=\sqrt(a:b)\] (provided that both parts of the equalities make sense)
Example: \(\sqrt(32)\cdot \sqrt 2=\sqrt(32\cdot 2)=\sqrt(64)=8\);
\(\sqrt(768):\sqrt3=\sqrt(768:3)=\sqrt(256)=16\);
\(\sqrt((-25)\cdot (-64))=\sqrt(25\cdot 64)=\sqrt(25)\cdot \sqrt(64)= 5\cdot 8=40\). \(\bullet\) Using these properties, it is convenient to find the square roots of large numbers by factoring them.
Consider an example. Find \(\sqrt(44100)\) . Since \(44100:100=441\) , then \(44100=100\cdot 441\) . According to the criterion of divisibility, the number \(441\) is divisible by \(9\) (since the sum of its digits is 9 and is divisible by 9), therefore, \(441:9=49\) , that is, \(441=9\ cdot 49\) .
Thus, we got: \[\sqrt(44100)=\sqrt(9\cdot 49\cdot 100)= \sqrt9\cdot \sqrt(49)\cdot \sqrt(100)=3\cdot 7\cdot 10=210\] Let's look at another example: \[\sqrt(\dfrac(32\cdot 294)(27))= \sqrt(\dfrac(16\cdot 2\cdot 3\cdot 49\cdot 2)(9\cdot 3))= \sqrt( \ dfrac(16\cdot4\cdot49)(9))=\dfrac(\sqrt(16)\cdot \sqrt4 \cdot \sqrt(49))(\sqrt9)=\dfrac(4\cdot 2\cdot 7)3 =\dfrac(56)3\]
\(\bullet\) Let's show how to enter numbers under the square root sign using the example of the expression \(5\sqrt2\) (short for the expression \(5\cdot \sqrt2\) ). Since \(5=\sqrt(25)\) , then \
Note also that, for example,
1) \(\sqrt2+3\sqrt2=4\sqrt2\) ,
2) \(5\sqrt3-\sqrt3=4\sqrt3\)
3) \(\sqrt a+\sqrt a=2\sqrt a\) .
Why is that? Let's explain with example 1). As you already understood, we cannot somehow convert the number \(\sqrt2\) . Imagine that \(\sqrt2\) is some number \(a\) . Accordingly, the expression \(\sqrt2+3\sqrt2\) is nothing but \(a+3a\) (one number \(a\) plus three more of the same numbers \(a\) ). And we know that this is equal to four such numbers \(a\) , that is, \(4\sqrt2\) .
Fact 4.
\(\bullet\) It is often said “cannot extract the root” when it is not possible to get rid of the sign \(\sqrt () \ \) of the root (radical) when finding the value of some number. For example, you can root the number \(16\) because \(16=4^2\) , so \(\sqrt(16)=4\) . But to extract the root from the number \(3\) , that is, to find \(\sqrt3\) , it is impossible, because there is no such number that squared will give \(3\) .
Such numbers (or expressions with such numbers) are irrational. For example, numbers \(\sqrt3, \ 1+\sqrt2, \ \sqrt(15)\) and so on. are irrational.
Also irrational are the numbers \(\pi\) (the number “pi”, approximately equal to \(3,14\) ), \(e\) (this number is called the Euler number, approximately equal to \(2,7\) ) etc.
\(\bullet\) Please note that any number will be either rational or irrational. And together all rational and all irrational numbers form a set called set of real (real) numbers. This set is denoted by the letter \(\mathbb(R)\) .
This means that all the numbers that we currently know are called real numbers.
Fact 5.
\(\bullet\) Modulus of a real number \(a\) is a non-negative number \(|a|\) equal to the distance from the point \(a\) to \(0\) on the real line. For example, \(|3|\) and \(|-3|\) are equal to 3, since the distances from the points \(3\) and \(-3\) to \(0\) are the same and equal to \(3 \) .
\(\bullet\) If \(a\) is a non-negative number, then \(|a|=a\) .
Example: \(|5|=5\) ; \(\qquad |\sqrt2|=\sqrt2\) . \(\bullet\) If \(a\) is a negative number, then \(|a|=-a\) .
Example: \(|-5|=-(-5)=5\) ; \(\qquad |-\sqrt3|=-(-\sqrt3)=\sqrt3\).
They say that for negative numbers, the module “eats” the minus, and positive numbers, as well as the number \(0\) , the module leaves unchanged.
BUT this rule only applies to numbers. If you have an unknown \(x\) (or some other unknown) under the module sign, for example, \(|x|\) , about which we do not know whether it is positive, equal to zero or negative, then get rid of the module we can not. In this case, this expression remains so: \(|x|\) . \(\bullet\) The following formulas hold: \[(\large(\sqrt(a^2)=|a|))\] \[(\large((\sqrt(a))^2=a)), \text( provided ) a\geqslant 0\] The following mistake is often made: they say that \(\sqrt(a^2)\) and \((\sqrt a)^2\) are the same thing. This is true only when \(a\) is a positive number or zero. But if \(a\) is a negative number, then this is not true. It suffices to consider such an example. Let's take the number \(-1\) instead of \(a\). Then \(\sqrt((-1)^2)=\sqrt(1)=1\) , but the expression \((\sqrt (-1))^2\) does not exist at all (because it is impossible under the root sign put negative numbers in!).
Therefore, we draw your attention to the fact that \(\sqrt(a^2)\) is not equal to \((\sqrt a)^2\) ! Example: 1) \(\sqrt(\left(-\sqrt2\right)^2)=|-\sqrt2|=\sqrt2\), because \(-\sqrt2<0\)
;
\(\phantom(00000)\) 2) \((\sqrt(2))^2=2\) . \(\bullet\) Since \(\sqrt(a^2)=|a|\) , then \[\sqrt(a^(2n))=|a^n|\] (the expression \(2n\) denotes an even number)
That is, when extracting the root from a number that is in some degree, this degree is halved.
Example:
1) \(\sqrt(4^6)=|4^3|=4^3=64\)
2) \(\sqrt((-25)^2)=|-25|=25\) (note that if the module is not set, then it turns out that the root of the number is equal to \(-25\) ; but we remember , which, by definition of the root, this cannot be: when extracting the root, we should always get a positive number or zero)
3) \(\sqrt(x^(16))=|x^8|=x^8\) (since any number to an even power is non-negative)
Fact 6.
How to compare two square roots?
\(\bullet\) True for square roots: if \(\sqrt a<\sqrt b\)
, то \(a
1) compare \(\sqrt(50)\) and \(6\sqrt2\) . First, we transform the second expression into \(\sqrt(36)\cdot \sqrt2=\sqrt(36\cdot 2)=\sqrt(72)\). Thus, since \(50<72\)
, то и \(\sqrt{50}<\sqrt{72}\)
. Следовательно, \(\sqrt{50}<6\sqrt2\)
.
2) Between what integers is \(\sqrt(50)\) ?
Since \(\sqrt(49)=7\) , \(\sqrt(64)=8\) , and \(49<50<64\)
, то \(7<\sqrt{50}<8\)
, то есть число \(\sqrt{50}\)
находится между числами \(7\)
и \(8\)
.
3) Compare \(\sqrt 2-1\) and \(0,5\) . Suppose \(\sqrt2-1>0.5\) : \[\begin(aligned) &\sqrt 2-1>0.5 \ \big| +1\quad \text((add one to both sides))\\ &\sqrt2>0.5+1 \ \big| \ ^2 \quad\text((square both parts))\\ &2>1,5^2\\ &2>2,25 \end(aligned)\] We see that we have obtained an incorrect inequality. Therefore, our assumption was wrong and \(\sqrt 2-1<0,5\)
.
Note that adding a certain number to both sides of the inequality does not affect its sign. Multiplying/dividing both parts of the inequality by a positive number also does not affect its sign, but multiplying/dividing by a negative number reverses the sign of the inequality!
Both sides of an equation/inequality can be squared ONLY IF both sides are non-negative. For example, in the inequality from the previous example, you can square both sides, in the inequality \(-3<\sqrt2\)
нельзя (убедитесь в этом сами)!
\(\bullet\) Note that \[\begin(aligned) &\sqrt 2\approx 1,4\\ &\sqrt 3\approx 1,7 \end(aligned)\] Knowing the approximate meaning of these numbers will help you when comparing numbers! \(\bullet\) In order to extract the root (if it is extracted) from some large number that is not in the table of squares, you must first determine between which “hundreds” it is, then between which “tens”, and then determine the last digit of this number. Let's show how it works with an example.
Take \(\sqrt(28224)\) . We know that \(100^2=10\,000\) , \(200^2=40\,000\) and so on. Note that \(28224\) is between \(10\,000\) and \(40\,000\) . Therefore, \(\sqrt(28224)\) is between \(100\) and \(200\) .
Now let's determine between which “tens” our number is (that is, for example, between \(120\) and \(130\) ). We also know from the table of squares that \(11^2=121\) , \(12^2=144\) etc., then \(110^2=12100\) , \(120^2=14400 \) , \(130^2=16900\) , \(140^2=19600\) , \(150^2=22500\) , \(160^2=25600\) , \(170^2=28900 \) . So we see that \(28224\) is between \(160^2\) and \(170^2\) . Therefore, the number \(\sqrt(28224)\) is between \(160\) and \(170\) .
Let's try to determine the last digit. Let's remember what single-digit numbers when squaring give at the end \ (4 \) ? These are \(2^2\) and \(8^2\) . Therefore, \(\sqrt(28224)\) will end in either 2 or 8. Let's check this. Find \(162^2\) and \(168^2\) :
\(162^2=162\cdot 162=26224\)
\(168^2=168\cdot 168=28224\) .
Hence \(\sqrt(28224)=168\) . Voila!
In order to adequately solve the exam in mathematics, first of all, it is necessary to study the theoretical material, which introduces numerous theorems, formulas, algorithms, etc. At first glance, it may seem that this is quite simple. However, finding a source in which the theory for the Unified State Examination in mathematics is presented easily and understandably for students with any level of training is, in fact, a rather difficult task. School textbooks cannot always be kept at hand. And finding the basic formulas for the exam in mathematics can be difficult even on the Internet.
Why is it so important to study theory in mathematics, not only for those who take the exam?
- Because it broadens your horizons. The study of theoretical material in mathematics is useful for anyone who wants to get answers to a wide range of questions related to the knowledge of the world. Everything in nature is ordered and has a clear logic. This is precisely what is reflected in science, through which it is possible to understand the world.
- Because it develops the intellect. Studying reference materials for the exam in mathematics, as well as solving various problems, a person learns to think and reason logically, to formulate thoughts correctly and clearly. He develops the ability to analyze, generalize, draw conclusions.
We invite you to personally evaluate all the advantages of our approach to the systematization and presentation of educational materials.
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.
