What is the root of the word: definition, examples, rules. How to quickly extract square roots

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

Below we will consider in turn the main methods of extracting roots.

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

If the tables of squares, cubes, etc. is not at hand, it is logical to use the method of extracting the root, which involves decomposing the root number into simple factors.

Separately, it is worth dwelling on, which is possible for roots with odd exponents.

Finally, consider a method that allows you to sequentially find the digits of the value of the root.

Let's get started.

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

In the simplest cases, tables of squares, cubes, etc. allow extracting 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 certain row and a certain column, it allows you to make 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 of its cells 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 one, there is a cell with the number 6889, 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. respectively from the numbers in these tables. Let us explain the principle of their application in extracting roots.

Let's say we need to extract the root of the nth degree from the number a, while the number a is contained in the table of nth degrees. According to 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 the cube root of 19683 is extracted using the cube table. We find the number 19 683 in the table of cubes, from it we find that this number is a cube of the number 27, therefore, .

It is clear that tables of n-th degrees are very convenient when extracting roots. However, they are often not at hand, and their compilation requires a certain amount of time. Moreover, it is often necessary to extract roots from numbers that are not contained in the corresponding tables. In these cases, one has to resort to other methods of extracting the roots.

Decomposition of the root number into prime factors

A fairly convenient way to extract the root from a natural number (if, of course, the root is extracted) is to decompose the root number into prime factors. His the essence is as follows: after it is quite easy to represent it as a degree with the desired indicator, which allows you to get the value of the root. Let's explain this point.

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

Note that if the factorization of the root number a cannot be represented in the form (p 1 ·p 2 ·…·p m) n , then the root of the nth degree from such a number a is not completely extracted.

Let's deal with this when solving examples.

Example.

Take the square root of 144 .

Solution.

If we turn to the table of squares given in the previous paragraph, it is clearly seen that 144=12 2 , from which it is clear that the square root of 144 is 12 .

But in the light of this point, we are interested in how the root is extracted by decomposing the root number 144 into prime factors. Let's take a 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 properties of the roots, the solution could be formulated a little differently: .

Answer:

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

Example.

Calculate the root value.

Solution.

The prime factorization of the root number 243 is 243=3 5 . Thus, .

Answer:

Example.

Is the value of the root an integer?

Solution.

To answer this question, let's decompose the root 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 decomposition is not represented as a cube of an integer, since the degree of the prime factor 7 is not a multiple of three. Therefore, the cube root of 285,768 is not taken completely.

Answer:

No.

Extracting roots from fractional numbers

It's time to figure out how the root is extracted from a fractional number. Let the fractional root number be written as p/q . According to the property of the root of the quotient, the following equality is true. From this equality it follows fraction root rule: The root of a fraction is equal to the quotient of dividing the root of the numerator 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 the common fraction 25/169.

Solution.

According to the table of squares, we find that the square root of the numerator of the original fraction is 5, and the square root of the denominator is 13. Then . This completes the extraction of the root from an ordinary fraction 25/169.

Answer:

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

Example.

Take the cube root of the decimal 474.552.

Solution.

Let's represent 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 . It remains only to complete the calculations .

Answer:

Extracting the root of a negative number

Separately, it is worth dwelling on extracting roots from negative numbers. When studying roots, we said that when the exponent of the root is an odd number, then a negative number can be under the sign of the root. We gave such notations the following meaning: for a negative number −a and an odd exponent of the root 2 n−1, we have . This equality gives rule for extracting odd roots from negative numbers: to extract the root of a negative number, you need to extract the root of the opposite positive number, and put a minus sign in front of the result.

Let's consider an example solution.

Example.

Find the root value.

Solution.

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

Here is a summary of the solution: .

Answer:

Bitwise Finding the Root Value

In the 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 at the same time, there is a need to know the value 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 consistently obtain a sufficient number of values of the digits of the desired number.

The first step of this algorithm is to find out what is the most significant bit of the root value. To do this, the numbers 0, 10, 100, ... are successively raised to the power n until a number exceeding the root number is obtained. Then the number that we raised to the power of n in the previous step will indicate the corresponding high order.

For example, consider this step of the algorithm when extracting the square root of five. We 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 that the most significant digit will be the units digit. The value of this bit, as well as lower ones, will be found in the next steps of the root extraction algorithm.

All the following steps of the algorithm are aimed at successive refinement of the value of the root due to the fact that the values of the next digits of the desired value of the root are found, starting from the highest and moving to the lowest. For example, the value of the root in the first step is 2 , in the second - 2.2 , in the third - 2.23 , and so on 2.236067977 ... . Let us describe how the values of the bits are found.

Finding bits is carried out by enumeration of 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 root 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 all these points using the same example of extracting the square root of five.

First, find the value of the units digit. We will iterate over the values 0, 1, 2, …, 9 , calculating respectively 0 2 , 1 2 , …, 9 2 until we get a value greater than the radical number 5 . All these calculations are conveniently presented in the form of a table:

So the value of the units digit is 2 (because 2 2<5 , а 2 3 >5 ). Let's move on to finding the value of the tenth place. In this case, we will square the numbers 2.0, 2.1, 2.2, ..., 2.9, comparing the obtained values \u200b\u200bwith the root number 5:

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

So the next value of the root of five is found, it is equal to 2.23. And so you can continue to find values further: 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 define the senior 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 define its value.

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

Thus, the value of the ones place is 2 . Let's move on to ten.

Since even 12.9 3 is less than the radical number 2 151.186 , the value of the tenth 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 up 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, those that we studied above are sufficient.

Bibliography.

Makarychev Yu.N., Mindyuk N.G., Neshkov K.I., Suvorova S.B. Algebra: textbook for 8 cells. educational institutions.
Kolmogorov A.N., Abramov A.M., Dudnitsyn Yu.P. and others. Algebra and the Beginnings of Analysis: A Textbook for Grades 10-11 of General Educational Institutions.
Gusev V.A., Mordkovich A.G. Mathematics (a manual for applicants to technical schools).

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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 \(aExample:
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!

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

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

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

So the algorithm:

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

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

Roots constraint

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

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

We get a series of numbers:

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

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

[Figure caption]

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

[Figure caption]

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

Elimination of obviously superfluous numbers

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

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

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

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

There are only 10 digits that can be in last place. Let's try to find out what they turn into when they are squared. Take a look at the table:

1	2	3	4	5	6	7	8	9	0
1	4	9	6	5	6	9	4	1	0

This table is another step towards calculating the root. As you can see, the numbers in the second line turned out to be symmetrical with respect to the five. For example:

2 2 = 4;
8 2 = 64 → 4.

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

[Figure caption]

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

[Figure caption]

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

Final Calculations

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

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

52 2 \u003d (50 +2) 2 \u003d 2500 + 2 50 2 + 4 \u003d 2704;
58 2 \u003d (60 - 2) 2 \u003d 3600 - 2 60 2 + 4 \u003d 3364.

That's all! It turned out that the root is 58! At the same time, in order to simplify the calculations, I used the formula of the squares of the sum and difference. Thanks to this, you didn’t even have to multiply the numbers in a column! This is another level of optimization of calculations, but, of course, it is completely optional :)

Root Calculation Examples

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

[Figure caption]

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

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

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

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

24 2 = (20 + 4) 2 = 576

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

Task. Calculate the square root:
[Figure caption]

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

Let's look at the last number:

1369 → 9;
33; 37.

Let's square it:

33 2 \u003d (30 + 3) 2 \u003d 900 + 2 30 3 + 9 \u003d 1089 ≠ 1369;
37 2 \u003d (40 - 3) 2 \u003d 1600 - 2 40 3 + 9 \u003d 1369.

Here is the answer: 37.

Task. Calculate the square root:
[Figure caption]

We limit the number:

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

Let's look at the last number:

2704 → 4;
52; 58.

Let's square it:

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

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

Task. Calculate the square root:
[Figure caption]

We limit the number:

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

Let's look at the last number:

4225 → 5;
65.

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

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

Everything is correct. We write down the answer.

Conclusion

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

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

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

Steps

Prime factorization

Factor the root number into factors that are square numbers. Depending on the root number, you will get an approximate or exact answer. Square numbers are numbers from which the whole square root can be taken. Factors are numbers that, when multiplied, give the original number. For example, the factors of the number 8 are 2 and 4, since 2 x 4 = 8, the numbers 25, 36, 49 are square numbers, since √25 = 5, √36 = 6, √49 = 7. Square factors are factors , which are square numbers. First, try to factorize the root number into square factors.

For example, calculate the square root of 400 (manually). First try factoring 400 into square factors. 400 is a multiple of 100, that is, divisible by 25 - this is a square number. Dividing 400 by 25 gives you 16. The number 16 is also a square number. Thus, 400 can be factored into square factors of 25 and 16, that is, 25 x 16 = 400.
This can be written as follows: √400 = √(25 x 16).

The square root of the product of some terms is equal to the product of the square roots of each term, that is, √(a x b) = √a x √b. Use this rule and take the square root of each square factor and multiply the results to find the answer.
- In our example, take the square root of 25 and 16.
  - √(25 x 16)
  - √25 x √16
  - 5 x 4 = 20
If the radical number does not factor into two square factors (and it does in most cases), you will not be able to find the exact answer as an integer. But you can simplify the problem by decomposing the root number into a square factor and an ordinary factor (a number from which the whole square root cannot be taken). Then you will take the square root of the square factor and you will take the root of the ordinary factor.
- For example, calculate the square root of the number 147. The number 147 cannot be factored into two square factors, but it can be factored into the following factors: 49 and 3. Solve the problem as follows:
  - = √(49 x 3)
  - = √49 x √3
  - = 7√3
If necessary, evaluate the value of the root. Now you can evaluate the value of the root (find an approximate value) by comparing it with the values of the roots of square numbers that are closest (on both sides of the number line) to the root number. You will get the value of the root as a decimal fraction, which must be multiplied by the number behind the root sign.
- Let's go back to our example. The root number is 3. The nearest square numbers to it are the numbers 1 (√1 = 1) and 4 (√4 = 2). Thus, the value of √3 lies between 1 and 2. Since the value of √3 is probably closer to 2 than to 1, our estimate is: √3 = 1.7. We multiply this value by the number at the root sign: 7 x 1.7 \u003d 11.9. If you do the calculations on a calculator, you get 12.13, which is pretty close to our answer.
  - This method also works with large numbers. For example, consider √35. The root number is 35. The nearest square numbers to it are the numbers 25 (√25 = 5) and 36 (√36 = 6). Thus, the value of √35 lies between 5 and 6. Since the value of √35 is much closer to 6 than it is to 5 (because 35 is only 1 less than 36), we can state that √35 is slightly less than 6. Checking with a calculator gives us the answer 5.92 - we were right.
Another way is to decompose the root number into prime factors. Prime factors are numbers that are only divisible by 1 and themselves. Write the prime factors in a row and find pairs of identical factors. Such factors can be taken out of the sign of the root.
- For example, calculate the square root of 45. We decompose the root number into prime factors: 45 \u003d 9 x 5, and 9 \u003d 3 x 3. Thus, √45 \u003d √ (3 x 3 x 5). 3 can be taken out of the root sign: √45 = 3√5. Now we can estimate √5.
- Consider another example: √88.
  - = √(2 x 44)
  - = √ (2 x 4 x 11)
  - = √ (2 x 2 x 2 x 11). You got three multiplier 2s; take a couple of them and take them out of the sign of the root.
  - = 2√(2 x 11) = 2√2 x √11. Now we can evaluate √2 and √11 and find an approximate answer.
Calculating the square root manually

Using column division
1. This method involves a process similar to long division and gives an accurate answer. First, draw a vertical line dividing the sheet into two halves, and then draw a horizontal line to the right and slightly below the top edge of the sheet to the vertical line. Now divide the root number into pairs of numbers, starting with the fractional part after the decimal point. So, the number 79520789182.47897 is written as "7 95 20 78 91 82, 47 89 70".
  - For example, let's calculate the square root of the number 780.14. Draw two lines (as shown in the picture) and write the number in the top left as "7 80, 14". It is normal that the first digit from the left is an unpaired digit. The answer (the root of the given number) will be written on the top right.
2. Given the first pair of numbers (or one number) from the left, find the largest integer n whose square is less than or equal to the pair of numbers (or one number) in question. In other words, find the square number that is closest to, but less than, the first pair of numbers (or single number) from the left, and take the square root of that square number; you will get the number n. Write the found n at the top right, and write down the square n at the bottom right.
  - In our case, the first number on the left will be the number 7. Next, 4< 7, то есть 2 2 < 7 и n = 2. Напишите 2 сверху справа - это первая цифра в искомом квадратном корне. Напишите 2×2=4 справа снизу; вам понадобится это число для последующих вычислений.
3. Subtract the square of the number n you just found from the first pair of numbers (or one number) from the left. Write the result of the calculation under the subtrahend (the square of the number n).
  - In our example, subtract 4 from 7 to get 3.
4. Take down the second pair of numbers and write it down next to the value obtained in the previous step. Then double the number at the top right and write the result at the bottom right with "_×_=" appended.
  - In our example, the second pair of numbers is "80". Write "80" after the 3. Then, doubling the number from the top right gives 4. Write "4_×_=" from the bottom right.
5. Fill in the blanks on the right.
  - In our case, if instead of dashes we put the number 8, then 48 x 8 \u003d 384, which is more than 380. Therefore, 8 is too large a number, but 7 is fine. Write 7 instead of dashes and get: 47 x 7 \u003d 329. Write 7 from the top right - this is the second digit in the desired square root of the number 780.14.
6. Subtract the resulting number from the current number on the left. Write the result from the previous step below the current number on the left, find the difference and write it below the subtracted one.
  - In our example, subtract 329 from 380, which equals 51.
7. Repeat step 4. If the demolished pair of numbers is the fractional part of the original number, then put the separator (comma) of the integer and fractional parts in the desired square root from the top right. On the left, carry down the next pair of numbers. Double the number at the top right and write the result at the bottom right with "_×_=" appended.
  - In our example, the next pair of numbers to be demolished will be the fractional part of the number 780.14, so put the separator of the integer and fractional parts in the required square root from the top right. Demolish 14 and write down at the bottom left. Double the top right (27) is 54, so write "54_×_=" at the bottom right.
8. Repeat steps 5 and 6. Find the largest number in place of dashes on the right (instead of dashes you need to substitute the same number) so that the multiplication result is less than or equal to the current number on the left.
  - In our example, 549 x 9 = 4941, which is less than the current number on the left (5114). Write 9 on the top right and subtract the result of the multiplication from the current number on the left: 5114 - 4941 = 173.
9. If you need to find more decimal places for the square root, write a pair of zeros next to the current number on the left and repeat steps 4, 5 and 6. Repeat steps until you get the accuracy of the answer you need (number of decimal places).
Understanding the process
1. Consider the first pair of digits Sa of the number S (Sa = 7 in our example) and find its square root. In this case, the first digit A of the sought value of the square root will be such a digit, the square of which is less than or equal to S a (that is, we are looking for such an A that satisfies the inequality A² ≤ Sa< (A+1)²). В нашем примере, S1 = 7, и 2² ≤ 7 < 3²; таким образом A = 2.
  - Let's say we need to divide 88962 by 7; here the first step will be similar: we consider the first digit of the divisible number 88962 (8) and select the largest number that, when multiplied by 7, gives a value less than or equal to 8. That is, we are looking for a number d for which the inequality is true: 7 × d ≤ 8< 7×(d+1). В этом случае d будет равно 1.
2. Mentally imagine the square whose area you need to calculate. You are looking for L, that is, the length of the side of a square whose area is S. A, B, C are numbers in the number L. You can write it differently: 10A + B \u003d L (for a two-digit number) or 100A + 10B + C \u003d L (for three-digit number) and so on.
  - Let (10A+B)² = L² = S = 100A² + 2×10A×B + B². Remember that 10A+B is a number whose B stands for ones and A stands for tens. For example, if A=1 and B=2, then 10A+B equals the number 12. (10A+B)² is the area of the whole square, 100A² is the area of the large inner square, B² is the area of the small inner square, 10A×B is the area of each of the two rectangles. Adding the areas of the figures described, you will find the area of the original square.