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What happens if you square the root? Properties of the function z=√y on the field of real numbers R

It's time to sort it out root extraction methods. They are based on the properties of roots, in particular, on equality, which is true for any negative number b.

Below we will look at the main methods of extracting roots one by one.

Let's start with the simplest case - extracting roots from natural numbers using a table of squares, a table of cubes, etc.

If tables of squares, cubes, etc. If you don’t have it at hand, it’s logical to use the method of extracting the root, which involves decomposing the radical number into prime factors.

It is worth special mentioning what is possible for roots with odd exponents.

Finally, let's consider a method that allows us to sequentially find the digits of the root value.

Let's get started.

Using a table of squares, a table of cubes, etc.

In the simplest cases, tables of squares, cubes, etc. allow you to extract roots. What are these tables?

The table of squares of integers from 0 to 99 inclusive (shown below) consists of two zones. The first zone of the table is located on a gray background; by selecting a specific row and a specific column, it allows you to compose a number from 0 to 99. For example, let’s select a row of 8 tens and a column of 3 units, with this we fixed the number 83. The second zone occupies the rest of the table. Each cell is located at the intersection of a certain row and a certain column, and contains the square of the corresponding number from 0 to 99. At the intersection of our chosen row of 8 tens and column 3 of ones there is a cell with the number 6,889, which is the square of the number 83.

Tables of cubes, tables of fourth powers of numbers from 0 to 99, and so on are similar to the table of squares, only they contain cubes, fourth powers, etc. in the second zone. corresponding numbers.

Tables of squares, cubes, fourth powers, etc. allow you to extract square roots, cube roots, fourth roots, etc. accordingly from the numbers in these tables. Let us explain the principle of their use when extracting roots.

Let's say we need to extract the nth root of the number a, while the number a is contained in the table of nth powers. Using this table we find the number b such that a=b n. Then , therefore, the number b will be the desired root of the nth degree.

As an example, let's show how to use a cube table to extract the cube root of 19,683. We find the number 19,683 in the table of cubes, from it we find that this number is the cube of the number 27, therefore, .

It is clear that tables of nth powers are very convenient for extracting roots. However, they are often not at hand, and compiling them requires some time. Moreover, it is often necessary to extract roots from numbers that are not contained in the corresponding tables. In these cases, you have to resort to other methods of root extraction.

Factoring a radical number into prime factors

A fairly convenient way to extract the root of a natural number (if, of course, the root is extracted) is to decompose the radical number into prime factors. His the point is this: after that it is quite easy to represent it as a power with the desired exponent, which allows you to obtain the value of the root. Let's clarify this point.

Let the nth root of a natural number a be taken and its value equal b. In this case, the equality a=b n is true. Number b like any natural number can be represented as the product of all its prime factors p 1 , p 2 , …, p m in the form p 1 · p 2 · … · p m , and the radical number a in this case is represented as (p 1 · p 2 · … · p m) n. Since the decomposition of a number into prime factors is unique, the decomposition of the radical number a into prime factors will have the form (p 1 ·p 2 ·…·p m) n, which makes it possible to calculate the value of the root as.

Note that if the decomposition into prime factors of a radical number a cannot be represented in the form (p 1 ·p 2 ·…·p m) n, then the nth root of such a number a is not completely extracted.

Let's figure this out when solving examples.

Example.

Take the square root of 144.

Solution.

If you look at the table of squares given in the previous paragraph, you can clearly see that 144 = 12 2, from which it is clear that the square root of 144 is equal to 12.

But in light of this point, we are interested in how the root is extracted by decomposing the radical number 144 into prime factors. Let's look at this solution.

Let's decompose 144 to prime factors:

That is, 144=2·2·2·2·3·3. Based on the resulting decomposition, the following transformations can be carried out: 144=2·2·2·2·3·3=(2·2) 2·3 2 =(2·2·3) 2 =12 2. Hence, .

Using the properties of the degree and the properties of the roots, the solution could be formulated a little differently: .

Answer:

To consolidate the material, consider the solutions to two more examples.

Example.

Calculate the value of the root.

Solution.

The prime factorization of the radical number 243 has the form 243=3 5 . Thus, .

Answer:

Example.

Is the root value an integer?

Solution.

To answer this question, let's factor the radical number into prime factors and see if it can be represented as a cube of an integer.

We have 285 768=2 3 ·3 6 ·7 2. The resulting expansion is not represented as a cube of an integer, since the degree prime factor 7 is not a multiple of three. Therefore, the cube root of 285,768 cannot be extracted completely.

Answer:

No.

Extracting roots from fractional numbers

It's time to figure out how to extract the root from fractional number. Let the fractional radical number be written as p/q. According to the property of the root of a quotient, the following equality is true. From this equality it follows rule for extracting the root of a fraction: The root of a fraction is equal to the quotient of the root of the numerator divided by the root of the denominator.

Let's look at an example of extracting a root from a fraction.

Example.

What is the square root of common fraction 25/169 .

Solution.

Using the table of squares, we find that the square root of the numerator of the original fraction is equal to 5, and the square root of the denominator is equal to 13. Then . This completes the extraction of the root of the common fraction 25/169.

Answer:

The root of a decimal fraction or mixed number is extracted after replacing the radical numbers with ordinary fractions.

Example.

Take the cube root of the decimal fraction 474.552.

Solution.

Let's imagine the original decimal as a common fraction: 474.552=474552/1000. Then . It remains to extract the cube roots that are in the numerator and denominator of the resulting fraction. Because 474 552=2·2·2·3·3·3·13·13·13=(2 3 13) 3 =78 3 and 1 000 = 10 3, then And . All that remains is to complete the calculations .

Answer:

Taking the root of a negative number

It is worthwhile to dwell on extracting roots from negative numbers. When studying roots, we said that when the root exponent is an odd number, then there can be a negative number under the root sign. We gave these entries the following meaning: for a negative number −a and an odd exponent of the root 2 n−1, . This equality gives rule for extracting odd roots from negative numbers: to extract the root of a negative number, you need to take the root of the opposite positive number, and put a minus sign in front of the result.

Let's look at the example solution.

Example.

Find the value of the root.

Solution.

Let's transform the original expression so that under the root sign it appears positive number: . Now mixed number replace it with an ordinary fraction: . We apply the rule for extracting the root of an ordinary fraction: . It remains to calculate the roots in the numerator and denominator of the resulting fraction: .

Here is a short summary of the solution: .

Answer:

Bitwise determination of the root value

IN general case under the root there is a number that, using the techniques discussed above, cannot be represented as the nth power of any number. But in this case there is a need to know the meaning of a given root, at least up to a certain sign. In this case, to extract the root, you can use an algorithm that allows you to sequentially obtain a sufficient number of digit values of the desired number.

The first step of this algorithm is to find out what the most significant bit of the root value is. To do this, the numbers 0, 10, 100, ... are sequentially raised to the power n until the moment when a number exceeds the radical number is obtained. Then the number that we raised to the power n at the previous stage will indicate the corresponding most significant digit.

For example, consider this step of the algorithm when extracting square root out of five. Take the numbers 0, 10, 100, ... and square them until we get a number greater than 5. We have 0 2 =0<5 , 10 2 =100>5, which means the most significant digit will be the ones digit. The value of this bit, as well as the lower ones, will be found in the next steps of the root extraction algorithm.

All subsequent steps of the algorithm are aimed at sequentially clarifying the value of the root by finding the values of the next bits of the desired value of the root, starting with the highest one and moving to the lowest ones. For example, the value of the root at the first step turns out to be 2, at the second – 2.2, at the third – 2.23, and so on 2.236067977…. Let us describe how the values of the digits are found.

The digits are found by searching through their possible values 0, 1, 2, ..., 9. In this case, the nth powers of the corresponding numbers are calculated in parallel, and they are compared with the radical number. If at some stage the value of the degree exceeds the radical number, then the value of the digit corresponding to the previous value is considered found, and the transition to the next step of the root extraction algorithm is made; if this does not happen, then the value of this digit is 9.

Let us explain these points using the same example of extracting the square root of five.

First we find the value of the units digit. We will go through the values 0, 1, 2, ..., 9, calculating 0 2, 1 2, ..., 9 2, respectively, until we get a value greater than the radical number 5. It is convenient to present all these calculations in the form of a table:

So the value of the units digit is 2 (since 2 2<5 , а 2 3 >5 ). Let's move on to finding the value of the tenths place. In this case, we will square the numbers 2.0, 2.1, 2.2, ..., 2.9, comparing the resulting values with the radical number 5:

Since 2.2 2<5 , а 2,3 2 >5, then the value of the tenths place is 2. You can proceed to finding the value of the hundredths place:

This is how the next value of the root of five was found, it is equal to 2.23. And so you can continue to find values: 2,236, 2,2360, 2,23606, 2,236067, … .

To consolidate the material, we will analyze the extraction of the root with an accuracy of hundredths using the considered algorithm.

First we determine the most significant digit. To do this, we cube the numbers 0, 10, 100, etc. until we get a number greater than 2,151,186. We have 0 3 =0<2 151,186 , 10 3 =1 000<2151,186 , 100 3 =1 000 000>2 151,186 , so the most significant digit is the tens digit.

Let's determine its value.

Since 10 3<2 151,186 , а 20 3 >2 151.186, then the value of the tens place is 1. Let's move on to units.

Thus, the value of the ones digit is 2. Let's move on to tenths.

Since even 12.9 3 is less than the radical number 2 151.186, then the value of the tenths place is 9. It remains to perform the last step of the algorithm; it will give us the value of the root with the required accuracy.

At this stage, the value of the root is found accurate to hundredths: .

In conclusion of this article, I would like to say that there are many other ways to extract roots. But for most tasks, the ones we studied above are sufficient.

Bibliography.

Makarychev Yu.N., Mindyuk N.G., Neshkov K.I., Suvorova S.B. Algebra: textbook for 8th grade. educational institutions.
Kolmogorov A.N., Abramov A.M., Dudnitsyn Yu.P. and others. Algebra and the beginnings of analysis: Textbook for grades 10 - 11 of general education institutions.
Gusev V.A., Mordkovich A.G. Mathematics (a manual for those entering technical schools).

Fact 1.
\(\bullet\) Let's take some non-negative number \(a\) (that is, \(a\geqslant 0\) ). Then (arithmetic) square root from the number \(a\) is called such a non-negative number \(b\) , when squared we get the number \(a\) : \[\sqrt a=b\quad \text(same as )\quad a=b^2\] From the definition it follows 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 \(\sqrt(25)\) equal to? We know that \(5^2=25\) and \((-5)^2=25\) . Since by definition we must find a non-negative number, then \(-5\) is not suitable, therefore, \(\sqrt(25)=5\) (since \(25=5^2\) ).
Finding the value of \(\sqrt a\) is called taking the square root of the number \(a\) , and the number \(a\) is called the radical expression.
\(\bullet\) Based on the definition, expression \(\sqrt(-25)\), \(\sqrt(-4)\), 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 operations can you do with square roots?
\(\bullet\) The sum or difference of square roots IS NOT EQUAL to the square root of the sum or difference, that is \[\sqrt a\pm\sqrt b\ne \sqrt(a\pm b)\] Thus, if you need to calculate, for example, \(\sqrt(25)+\sqrt(49)\) , then initially you must find the values of \(\sqrt(25)\) and \(\sqrt(49)\ ) and then fold them. 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 transformed further and remains as it is. For example, in the sum \(\sqrt 2+ \sqrt (49)\) we can find \(\sqrt(49)\) is \(7\) , but \(\sqrt 2\) cannot be transformed in any way, That's why \(\sqrt 2+\sqrt(49)=\sqrt 2+7\). Unfortunately, this expression cannot be simplified further\(\bullet\) The product/quotient of square roots is equal to the square root of the product/quotient, that is \[\sqrt a\cdot \sqrt b=\sqrt(ab)\quad \text(s)\quad \sqrt a:\sqrt b=\sqrt(a:b)\] (provided that both sides 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 square roots of large numbers by factoring them.
Let's look at an example. Let's 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 notation 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 using example 1). As you already understand, we cannot somehow transform the number \(\sqrt2\). Let's imagine that \(\sqrt2\) is some number \(a\) . Accordingly, the expression \(\sqrt2+3\sqrt2\) is nothing more than \(a+3a\) (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.
\(\bullet\) They often say “you can’t extract the root” when you can’t get rid of the sign \(\sqrt () \ \) of the root (radical) when finding the value of a number. For example, you can take the root of the number \(16\) because \(16=4^2\) , therefore \(\sqrt(16)=4\) . But it is impossible to extract the root of the number \(3\), that is, to find \(\sqrt3\), because there is no 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, it is 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 a set of 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.
\(\bullet\) The 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 modulus “eats” the minus, while positive numbers, as well as the number \(0\), are left unchanged by the modulus.
BUT This rule only applies to numbers. If under your modulus sign there is an unknown \(x\) (or some other unknown), for example, \(|x|\) , about which we do not know whether it is positive, zero or negative, then get rid of the modulus we can not. In this case, this expression remains the same: \(|x|\) . \(\bullet\) The following formulas hold: \[(\large(\sqrt(a^2)=|a|))\] \[(\large((\sqrt(a))^2=a)), \text( provided ) a\geqslant 0\] Very often the following mistake is made: they say that \(\sqrt(a^2)\) and \((\sqrt a)^2\) are one and the same. This is only true if \(a\) is a positive number or zero. But if \(a\) is a negative number, then this is false. It is enough to consider this example. Let's take instead of \(a\) the number \(-1\) . Then \(\sqrt((-1)^2)=\sqrt(1)=1\) , but the expression \((\sqrt (-1))^2\) does not exist at all (after all, it is impossible to use the root sign put negative numbers!).
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 taking the root of a number that is to 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 supplied, it turns out that the root of the number is equal to \(-25\) ; but we remember , that by definition of a root this cannot happen: when extracting a 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?
\(\bullet\) For square roots it is true: if \(\sqrt a<\sqrt b\) , то \(aExample:
1) compare \(\sqrt(50)\) and \(6\sqrt2\) . First, let's 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 \(\sqrt(50)\) located?
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) Let's compare \(\sqrt 2-1\) and \(0.5\) . Let's assume that \(\sqrt2-1>0.5\) : \[\begin(aligned) &\sqrt 2-1>0.5 \ \big| +1\quad \text((add one to both sides))\\ &\sqrt2>0.5+1 \ \big| \ ^2 \quad\text((squaring both sides))\\ &2>1.5^2\\ &2>2.25 \end(aligned)\] We see that we have obtained an incorrect inequality. Therefore, our assumption was incorrect 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 sides of an inequality by a positive number also does not affect its sign, but multiplying/dividing by a negative number reverses the sign of the inequality!
You can square both sides of an equation/inequality 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)! \(\bullet\) It should be remembered 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 can be extracted) from some large number that is not in the table of squares, you must first determine between which “hundreds” it is located, then – between which “tens”, and then determine the last digit of this number. Let's show how this works with an example.
Let's take \(\sqrt(28224)\) . We know that \(100^2=10\,000\), \(200^2=40\,000\), etc. 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 located (that is, for example, between \(120\) and \(130\)). Also from the table of squares we know 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 squared, give \(4\) at the end? These are \(2^2\) and \(8^2\) . Therefore, \(\sqrt(28224)\) will end in either 2 or 8. Let's check this. Let's find \(162^2\) and \(168^2\) :
\(162^2=162\cdot 162=26224\)
\(168^2=168\cdot 168=28224\) .
Therefore, \(\sqrt(28224)=168\) . Voila!

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Students always ask: “Why can’t I use a calculator in the math exam? How to extract the square root of a number without a calculator? Let's try to answer this question.

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

Action square root inverse to the action of squaring.

√81= 9 9 2 =81

If you take the square root of a positive number and square the result, you get the same number.

From small numbers that are exact squares of natural numbers, for example 1, 4, 9, 16, 25, ..., 100, square roots can be extracted orally. Usually at school they teach a table of squares of natural numbers up to twenty. Knowing this table, it is easy to extract square roots from the numbers 121,144, 169, 196, 225, 256, 289, 324, 361, 400. From numbers greater than 400 you can extract them using the selection method using some tips. Let's try to look at this method with an example.

Example: Extract the root of the number 676.

We notice that 20 2 = 400, and 30 2 = 900, which means 20< √676 < 900.

Exact squares of natural numbers end in 0; 1; 4; 5; 6; 9.
The number 6 is given by 4 2 and 6 2.
This means that if the root is taken from 676, then it is either 24 or 26.

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

Answer: √676 = 26 .

More example: √6889 .

Since 80 2 = 6400, and 90 2 = 8100, then 80< √6889 < 90.
The number 9 is given by 3 2 and 7 2, then √6889 is equal to either 83 or 87.

Let's check: 83 2 = 6889.

Answer: √6889 = 83 .

If you find it difficult to solve using the selection method, you can factor the radical expression.

For example, find √893025.

Let's factor the number 893025, remember, you did this in the sixth grade.

We get: √893025 = √3 6 ∙5 2 ∙7 2 = 3 3 ∙5 ∙7 = 945.

More example: √20736. Let's factor the number 20736:

We get √20736 = √2 8 ∙3 4 = 2 4 ∙3 2 = 144.

Of course, factorization requires knowledge of divisibility signs and factorization skills.

And finally, there is rule for extracting square roots. Let's get acquainted with this rule with examples.

Calculate √279841.

To extract the root of a multi-digit integer, we divide it from right to left into faces containing 2 digits (the leftmost edge may contain one digit). We write it like this: 27’98’41

To obtain the first digit of the root (5), we take the square root of the largest perfect square contained in the first face on the left (27).
Then the square of the first digit of the root (25) is subtracted from the first face and the next face (98) is added to the difference (subtracted).
To the left of the resulting number 298, write the double digit of the root (10), divide by it the number of all tens of the previously obtained number (29/2 ≈ 2), test the quotient (102 ∙ 2 = 204 should be no more than 298) and write (2) after the first digit of the root.
Then the resulting quotient 204 is subtracted from 298 and the next edge (41) is added to the difference (94).
To the left of the resulting number 9441, write the double product of the digits of the root (52 ∙2 = 104), divide the number of all tens of the number 9441 (944/104 ≈ 9) by this product, test the quotient (1049 ∙9 = 9441) should be 9441 and write it down (9) after the second digit of the root.

We received the answer √279841 = 529.

Extract similarly roots of decimal fractions. Only the radical number must be divided into faces so that the comma is between the faces.

Example. Find the value √0.00956484.

Just remember that if a decimal fraction has an odd number of decimal places, the square root cannot be extracted from it.

So now you have seen three ways to extract the root. Choose the one that suits you best and practice. To learn to solve problems, you need to solve them. And if you have any questions, sign up for my lessons.

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In this article we will introduce concept of a root of a number. We will proceed sequentially: we will start with the square root, from there we will move on to the description of the cubic root, after which we will generalize the concept of a root, defining the nth root. At the same time, we will introduce definitions, notations, give examples of roots and give the necessary explanations and comments.

Square root, arithmetic square root

To understand the definition of the root of a number, and the square root in particular, you need to have . At this point we will often encounter the second power of a number - the square of a number.

Let's start with square root definitions.

Definition

Square root of a is a number whose square is equal to a.

In order to bring examples of square roots, take several numbers, for example, 5, −0.3, 0.3, 0, and square them, we get the numbers 25, 0.09, 0.09 and 0, respectively (5 2 =5·5=25, (−0.3) 2 =(−0.3)·(−0.3)=0.09, (0.3) 2 =0.3·0.3=0.09 and 0 2 =0·0=0 ). Then, by the definition given above, the number 5 is the square root of the number 25, the numbers −0.3 and 0.3 are the square roots of 0.09, and 0 is the square root of zero.

It should be noted that not for any number a there exists a whose square is equal to a. Namely, for any negative number a there is no real number b whose square is equal to a. In fact, the equality a=b 2 is impossible for any negative a, since b 2 is a non-negative number for any b. Thus, there is no square root of a negative number on the set of real numbers. In other words, on the set of real numbers the square root of a negative number is not defined and has no meaning.

This leads to a logical question: “Is there a square root of a for any non-negative a”? The answer is yes. This fact can be justified by the constructive method used to find the value of the square root.

Then the next logical question arises: “What is the number of all square roots of a given non-negative number a - one, two, three, or even more”? Here's the answer: if a is zero, then the only square root of zero is zero; if a is some positive number, then the number of square roots of the number a is two, and the roots are . Let's justify this.

Let's start with the case a=0 . First, let's show that zero is indeed the square root of zero. This follows from the obvious equality 0 2 =0·0=0 and the definition of the square root.

Now let's prove that 0 is the only square root of zero. Let's use the opposite method. Suppose there is some nonzero number b that is the square root of zero. Then the condition b 2 =0 must be satisfied, which is impossible, since for any non-zero b the value of the expression b 2 is positive. We have arrived at a contradiction. This proves that 0 is the only square root of zero.

Let's move on to cases where a is a positive number. We said above that there is always a square root of any non-negative number, let the square root of a be the number b. Let's say that there is a number c, which is also the square root of a. Then, by the definition of a square root, the equalities b 2 =a and c 2 =a are true, from which it follows that b 2 −c 2 =a−a=0, but since b 2 −c 2 =(b−c)·( b+c) , then (b−c)·(b+c)=0 . The resulting equality is valid properties of operations with real numbers possible only when b−c=0 or b+c=0 . Thus, the numbers b and c are equal or opposite.

If we assume that there is a number d, which is another square root of the number a, then by reasoning similar to those already given, it is proved that d is equal to the number b or the number c. So, the number of square roots of a positive number is two, and the square roots are opposite numbers.

For the convenience of working with square roots, the negative root is “separated” from the positive one. For this purpose, it is introduced definition of arithmetic square root.

Definition

Arithmetic square root of a non-negative number a is a non-negative number whose square is equal to a.

The notation for the arithmetic square root of a is . The sign is called the arithmetic square root sign. It is also called the radical sign. Therefore, you can sometimes hear both “root” and “radical”, which means the same object.

The number under the arithmetic square root sign is called radical number, and the expression under the root sign is radical expression, while the term “radical number” is often replaced by “radical expression”. For example, in the notation the number 151 is a radical number, and in the notation the expression a is a radical expression.

When reading, the word "arithmetic" is often omitted, for example, the entry is read as "the square root of seven point twenty-nine." The word “arithmetic” is used only when they want to emphasize that we are talking specifically about the positive square root of a number.

In light of the introduced notation, it follows from the definition of an arithmetic square root that for any non-negative number a .

Square roots of a positive number a are written using the arithmetic square root sign as and . For example, the square roots of 13 are and . The arithmetic square root of zero is zero, that is, . For negative numbers a, we will not attach meaning to the notation until we study complex numbers. For example, the expressions and are meaningless.

Based on the definition of the square root, the properties of square roots are proved, which are often used in practice.

In conclusion of this point, we note that the square roots of the number a are solutions of the form x 2 =a with respect to the variable x.

Cube root of a number

Definition of cube root of the number a is given similarly to the definition of the square root. Only it is based on the concept of a cube of a number, not a square.

Definition

Cube root of a is a number whose cube is equal to a.

Let's give examples of cube roots. To do this, take several numbers, for example, 7, 0, −2/3, and cube them: 7 3 =7·7·7=343, 0 3 =0·0·0=0, . Then, based on the definition of a cube root, we can say that the number 7 is the cube root of 343, 0 is the cube root of zero, and −2/3 is the cube root of −8/27.

It can be shown that the cube root of a number, unlike the square root, always exists, not only for non-negative a, but also for any real number a. To do this, you can use the same method that we mentioned when studying square roots.

Moreover, there is only a single cube root of a given number a. Let us prove the last statement. To do this, consider three cases separately: a is a positive number, a=0 and a is a negative number.

It is easy to show that if a is positive, the cube root of a can be neither a negative number nor zero. Indeed, let b be the cube root of a, then by definition we can write the equality b 3 =a. It is clear that this equality cannot be true for negative b and for b=0, since in these cases b 3 =b·b·b will be a negative number or zero, respectively. So the cube root of a positive number a is a positive number.

Now suppose that in addition to the number b there is another cube root of the number a, let's denote it c. Then c 3 =a. Therefore, b 3 −c 3 =a−a=0, but b 3 −c 3 =(b−c)·(b 2 +b·c+c 2)(this is the abbreviated multiplication formula difference of cubes), whence (b−c)·(b 2 +b·c+c 2)=0. The resulting equality is possible only when b−c=0 or b 2 +b·c+c 2 =0. From the first equality we have b=c, and the second equality has no solutions, since its left side is a positive number for any positive numbers b and c as the sum of three positive terms b 2, b·c and c 2. This proves the uniqueness of the cube root of a positive number a.

When a=0, the cube root of the number a is only the number zero. Indeed, if we assume that there is a number b, which is a non-zero cube root of zero, then the equality b 3 =0 must hold, which is possible only when b=0.

For negative a, arguments similar to the case for positive a can be given. First, we show that the cube root of a negative number cannot be equal to either a positive number or zero. Secondly, we assume that there is a second cube root of a negative number and show that it will necessarily coincide with the first.

So, there is always a cube root of any given real number a, and a unique one.

Let's give definition of arithmetic cube root.

Definition

Arithmetic cube root of a non-negative number a is a non-negative number whose cube is equal to a.

The arithmetic cube root of a non-negative number a is denoted as , the sign is called the sign of the arithmetic cube root, the number 3 in this notation is called root index. The number under the root sign is radical number, the expression under the root sign is radical expression.

Although the arithmetic cube root is defined only for non-negative numbers a, it is also convenient to use notations in which negative numbers are found under the arithmetic cube root sign. We will understand them as follows: , where a is a positive number. For example, .

We will talk about the properties of cube roots in the general article properties of roots.

Calculating the value of a cube root is called extracting a cube root; this action is discussed in the article extracting roots: methods, examples, solutions.

To conclude this point, let's say that the cube root of the number a is a solution of the form x 3 =a.

nth root, arithmetic root of degree n

Let us generalize the concept of a root of a number - we introduce definition of nth root for n.

Definition

nth root of a is a number whose nth power is equal to a.

From this definition it is clear that the first degree root of the number a is the number a itself, since when studying the degree with a natural exponent we took a 1 =a.

Above we looked at special cases of the nth root for n=2 and n=3 - square root and cube root. That is, a square root is a root of the second degree, and a cube root is a root of the third degree. To study roots of the nth degree for n=4, 5, 6, ..., it is convenient to divide them into two groups: the first group - roots of even degrees (that is, for n = 4, 6, 8, ...), the second group - roots odd degrees (that is, with n=5, 7, 9, ...). This is due to the fact that roots of even powers are similar to square roots, and roots of odd powers are similar to cubic roots. Let's deal with them one by one.

Let's start with the roots whose powers are the even numbers 4, 6, 8, ... As we already said, they are similar to the square root of the number a. That is, the root of any even degree of the number a exists only for non-negative a. Moreover, if a=0, then the root of a is unique and equal to zero, and if a>0, then there are two roots of even degree of the number a, and they are opposite numbers.

Let us substantiate the last statement. Let b be an even root (we denote it as 2·m, where m is some natural number) of the number a. Suppose that there is a number c - another root of degree 2·m from the number a. Then b 2·m −c 2·m =a−a=0 . But we know the form b 2 m −c 2 m = (b−c) (b+c) (b 2 m−2 +b 2 m−4 c 2 +b 2 m−6 c 4 +…+c 2 m−2), then (b−c)·(b+c)· (b 2 m−2 +b 2 m−4 c 2 +b 2 m−6 c 4 +…+c 2 m−2)=0. From this equality it follows that b−c=0, or b+c=0, or b 2 m−2 +b 2 m−4 c 2 +b 2 m−6 c 4 +…+c 2 m−2 =0. The first two equalities mean that the numbers b and c are equal or b and c are opposite. And the last equality is valid only for b=c=0, since on its left side there is an expression that is non-negative for any b and c as the sum of non-negative numbers.

As for the roots of the nth degree for odd n, they are similar to the cube root. That is, the root of any odd degree of the number a exists for any real number a, and for a given number a it is unique.

The uniqueness of a root of odd degree 2·m+1 of the number a is proved by analogy with the proof of the uniqueness of the cube root of a. Only here instead of equality a 3 −b 3 =(a−b)·(a 2 +a·b+c 2) an equality of the form b 2 m+1 −c 2 m+1 = is used (b−c)·(b 2·m +b 2·m−1 ·c+b 2·m−2 ·c 2 +… +c 2·m). The expression in the last bracket can be rewritten as b 2 m +c 2 m +b c (b 2 m−2 +c 2 m−2 + b c (b 2 m−4 +c 2 m−4 +b c (…+(b 2 +c 2 +b c)))). For example, with m=2 we have b 5 −c 5 =(b−c)·(b 4 +b 3 ·c+b 2 ·c 2 +b·c 3 +c 4)= (b−c)·(b 4 +c 4 +b·c·(b 2 +c 2 +b·c)). When a and b are both positive or both negative, their product is a positive number, then the expression b 2 +c 2 +b·c in the highest nested parentheses is positive as the sum of the positive numbers. Now, moving sequentially to the expressions in brackets of the previous degrees of nesting, we are convinced that they are also positive as sums of positive numbers. As a result, we obtain that the equality b 2 m+1 −c 2 m+1 = (b−c)·(b 2·m +b 2·m−1 ·c+b 2·m−2 ·c 2 +… +c 2·m)=0 possible only when b−c=0, that is, when the number b is equal to the number c.

It's time to understand the notation of nth roots. For this purpose it is given definition of arithmetic root of the nth degree.

Definition

Arithmetic root of the nth degree of a non-negative number a is a non-negative number whose nth power is equal to a.

Before calculators, students and teachers calculated square roots by hand. There are several ways to calculate the square root of a number manually. Some of them offer only an approximate solution, others give an exact answer.

Steps

Prime factorization

Factor the radical number into factors that are square numbers. Depending on the radical 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 factor the radical number into square factors.

For example, calculate the square root of 400 (by hand). 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 the square factors of 25 and 16, that is, 25 x 16 = 400.
This can be written as follows: √400 = √(25 x 16).

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 to take the square root of each square factor and multiply the results to find the answer.
- In our example, take the root of 25 and 16.
  - √(25 x 16)
  - √25 x √16
  - 5 x 4 = 20
If the radical number does not factor into two square factors (and this happens in most cases), you will not be able to find the exact answer in the form of a whole number. But you can simplify the problem by decomposing the radical 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 will take the root of the common 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 factorized into the following factors: 49 and 3. Solve the problem as follows:
  - = √(49 x 3)
  - = √49 x √3
  - = 7√3
If necessary, estimate the value of the root. Now you can estimate the value of the root (find an approximate value) by comparing it with the values of the roots of the square numbers that are closest (on both sides of the number line) to the radical number. You will receive the root value as a decimal fraction, which must be multiplied by the number behind the root sign.
- Let's return to our example. The radical number is 3. The square numbers closest to it will be the numbers 1 (√1 = 1) and 4 (√4 = 2). Thus, the value of √3 is located 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 = 11.9. If you do the math on a calculator, you'll get 12.13, which is pretty close to our answer.
  - This method also works with large numbers. For example, consider √35. The radical number is 35. The closest square numbers to it will be the numbers 25 (√25 = 5) and 36 (√36 = 6). Thus, the value of √35 is located between 5 and 6. Since the value of √35 is much closer to 6 than to 5 (because 35 is only 1 less than 36), we can say that √35 is slightly less than 6. Check on the calculator gives us the answer 5.92 - we were right.
Another way is to factor the radical number into prime factors. Prime factors are numbers that are divisible only by 1 and themselves. Write the prime factors in a series and find pairs of identical factors. Such factors can be taken out of the root sign.
- For example, calculate the square root of 45. We factor the radical number into prime factors: 45 = 9 x 5, and 9 = 3 x 3. Thus, √45 = √(3 x 3 x 5). 3 can be taken out as a root sign: √45 = 3√5. Now we can estimate √5.
- Let's look at another example: √88.
  - = √(2 x 44)
  - = √ (2 x 4 x 11)
  - = √ (2 x 2 x 2 x 11). You received three multipliers of 2; take a couple of them and move them beyond the root sign.
  - = 2√(2 x 11) = 2√2 x √11. Now you can evaluate √2 and √11 and find an approximate answer.
Calculating square root manually

Using long division
1. This method involves a process similar to long division and provides an accurate answer. First, draw a vertical line dividing the sheet into two halves, and then to the right and slightly below the top edge of the sheet, draw a horizontal line to the vertical line. Now divide the radical 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 given number in the form “7 80, 14” at the top left. It is normal that the first digit from the left is an unpaired digit. You will write the answer (the root of this number) at the top right.
2. For the first pair of numbers (or single number) from the left, find the largest integer n whose square is less than or equal to the pair of numbers (or single number) in question. In other words, find the square number that is closest to, but smaller 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 n you found at the top right, and write the square of n at the bottom right.
  - In our case, the first number on the left will be 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 single number) on the left. Write the result of the calculation under the subtrahend (the square of the number n).
  - In our example, subtract 4 from 7 and 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 the addition of "_×_=".
  - In our example, the second pair of numbers is "80". Write "80" after the 3. Then, double the number on the top right gives 4. Write "4_×_=" on the bottom right.
5. Fill in the blanks on the right.
  - In our case, if we put the number 8 instead of dashes, then 48 x 8 = 384, which is more than 380. Therefore, 8 is too large a number, but 7 will do. Write 7 instead of dashes and get: 47 x 7 = 329. Write 7 at 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 under the current number on the left, find the difference and write it under the subtrahend.
  - In our example, subtract 329 from 380, which equals 51.
7. Repeat step 4. If the pair of numbers being transferred is the fractional part of the original number, then put a separator (comma) between the integer and fractional parts in the required square root at the top right. On the left, bring down the next pair of numbers. Double the number at the top right and write the result at the bottom right with the addition of "_×_=".
  - In our example, the next pair of numbers to be removed will be the fractional part of the number 780.14, so place the separator of the integer and fractional parts in the desired square root in the upper right. Take down 14 and write it in the bottom left. Double the number on the top right (27) is 54, so write "54_×_=" on the bottom right.
8. Repeat steps 5 and 6. Find the largest number in place of the dashes on the right (instead of the dashes you need to substitute the same number) so that the result of the multiplication 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 couple of zeros to the left of the current number and repeat steps 4, 5, and 6. Repeat steps until you get the answer precision (number of decimal places) you need.
Understanding the Process
1. 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 desired square root value will be a digit whose square is less than or equal to S a (that is, we are looking for an A such that 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.
2. Mentally imagine a 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 equal to S. A, B, C are the numbers in the number L. You can write it differently: 10A + B = L (for a two-digit number) or 100A + 10B + C = 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 in which the digit B stands for units and the digit A stands for tens. For example, if A=1 and B=2, then 10A+B is equal to the number 12. (10A+B)² is the area of the entire square, 100A²- area of the large inner square, B²- area of the small inner square, 10A×B- the area of each of the two rectangles. By adding up the areas of the described figures, you will find the area of the original square.