properties of square roots. Root formulas

This article is a collection of detailed information that deals with the topic of properties of roots. Considering the topic, we will start with the properties, study all the formulations and give proofs. To consolidate the topic, we will consider the properties of the nth degree.

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Root Properties

We'll talk about properties.

Property multiplied numbers a And b, which is represented as the equality a · b = a · b . It can be represented as multipliers, positive or equal to zero a 1 , a 2 , … , a k as a 1 a 2 … a k = a 1 a 2 … a k ;
from private a: b = a: b, a ≥ 0, b > 0, it can also be written in this form a b = a b ;
Property from the power of a number a with an even exponent a 2 m = a m for any number a, for example, a property from the square of a number a 2 = a .

In any of the presented equations, you can swap the parts before and after the dash sign, for example, the equality a · b = a · b is transformed as a · b = a · b . Equality properties are often used to simplify complex equations.

The proof of the first properties is based on the definition square root and properties of powers with a natural exponent. To substantiate the third property, it is necessary to refer to the definition of the modulus of a number.

First of all, it is necessary to prove the properties of the square root a · b = a · b . According to the definition, it is necessary to consider that a b is a number, positive or equal to zero, which will be equal to a b during construction into a square. The value of the expression a · b is positive or equal to zero as a product of non-negative numbers. The property of the degree of multiplied numbers allows us to represent equality in the form (a · b) 2 = a 2 · b 2 . By the definition of the square root a 2 \u003d a and b 2 \u003d b, then a b \u003d a 2 b 2 \u003d a b.

In a similar way, one can prove that from the product k multipliers a 1 , a 2 , … , a k will be equal to the product of the square roots of these factors. Indeed, a 1 · a 2 · … · ak 2 = a 1 2 · a 2 2 · … · ak 2 = a 1 · a 2 · … · a k .

It follows from this equality that a 1 · a 2 · … · a k = a 1 · a 2 · … · a k .

Let's look at a few examples to reinforce the topic.

Example 1

3 5 2 5 = 3 5 2 5 , 4 , 2 13 1 2 = 4 , 2 13 1 2 and 2 , 7 4 12 17 0 , 2 (1) = 2 , 7 4 12 17 0 . 2 (1) .

It is necessary to prove the property of the arithmetic square root of the quotient: a: b = a: b, a ≥ 0, b > 0. The property allows you to write the equality a: b 2 = a 2: b 2 , and a 2: b 2 = a: b , while a: b is a positive number or equal to zero. This expression will be the proof.

For example, 0:16 = 0:16, 80:5 = 80:5 and 30, 121 = 30, 121.

Consider the property of the square root of the square of a number. It can be written as an equality as a 2 = a To prove this property, it is necessary to consider in detail several equalities for a ≥ 0 and at a< 0 .

Obviously, for a ≥ 0, the equality a 2 = a is true. At a< 0 the equality a 2 = - a will be true. Actually, in this case − a > 0 and (− a) 2 = a 2 . We can conclude that a 2 = a , a ≥ 0 - a , a< 0 = a . Именно это и требовалось доказать.

Let's look at a few examples.

Example 2

5 2 = 5 = 5 and - 0 . 36 2 = - 0 . 36 = 0 . 36 .

The proved property will help to justify a 2 m = a m , where a- real, and m –natural number. Indeed, the exponentiation property allows us to replace the degree a 2 m expression (am) 2, then a 2 · m = (a m) 2 = a m .

Example 3

3 8 = 3 4 = 3 4 and (- 8 , 3) 14 = - 8 , 3 7 = (8 , 3) 7 .

Properties of the nth root

First you need to consider the main properties of the roots of the nth degree:

Property from the product of numbers a And b, which are positive or equal to zero, can be expressed as the equality a b n = a n b n , this property is valid for the product k numbers a 1 , a 2 , … , a k as a 1 a 2 … a k n = a 1 n a 2 n … a k n ;
from fractional number has the property a b n = a n b n , where a- any real number, which is positive or equal to zero, and b is a positive real number;
For any a and even numbers n = 2 m a 2 m 2 m = a is true, and for odd n = 2 m − 1 the equality a 2 · m - 1 2 · m - 1 = a is fulfilled.
Extraction property from a m n = a n m , where a- any number, positive or equal to zero, n And m are natural numbers, this property can also be represented as . . a n k n 2 n 1 = a n 1 · n 2 . . . nk ;
For any non-negative a and arbitrary n And m, which are natural, one can also define the fair equality a m n · m = a n ;
degree property n from the power of a number a, which is positive or equal to zero, in kind m, defined by the equality a m n = a n m ;
Comparison property that have the same exponents: for any positive numbers a And b such that a< b , the inequality a n< b n ;
The comparison property that possess the same numbers root: if m And n- natural numbers that m > n, then at 0 < a < 1 the inequality a m > a n is valid, and for a > 1 a m< a n .

The above equations are valid if the parts before and after the equals sign are reversed. They can be used in this form as well. This is often used during simplification or transformation of expressions.

The proof of the above properties of the root is based on the definition, the properties of the degree, and the definition of the modulus of a number. These properties must be proven. But everything is in order.

First of all, we will prove the properties of the root of the nth degree from the product a · b n = a n · b n . For a And b , which are positive or zero , the value a n · b n is also positive or equal to zero, since it is a consequence of multiplication of non-negative numbers. The property of a natural power product allows us to write the equality a n · b n n = a n n · b n n . By definition of root n th degree a n n = a and b n n = b , therefore, a n · b n n = a · b . The resulting equality is exactly what was required to be proved.

This property is proved similarly for the product k factors: for non-negative numbers a 1 , a 2 , … , a n a 1 n · a 2 n · … · a k n ≥ 0 .

Here are examples of using the root property n th power from the product: 5 2 1 2 7 = 5 7 2 1 2 7 and 8 , 3 4 17 , (21) 4 3 4 5 7 4 = 8 , 3 17 , (21) 3 5 7 4 .

Let us prove the property of the root of the quotient a b n = a n b n . At a ≥ 0 And b > 0 the condition a n b n ≥ 0 is satisfied, and a n b n n = a n n b n n = a b .

Let's show examples:

Example 4

8 27 3 = 8 3 27 3 and 2 , 3 10: 2 3 10 = 2 , 3: 2 3 10 .

For the next step, it is necessary to prove the properties of the nth degree from the number to the degree n. We represent this as an equality a 2 m 2 m = a and a 2 m - 1 2 m - 1 = a for any real a and natural m. At a ≥ 0 we get a = a and a 2 m = a 2 m , which proves the equality a 2 m 2 m = a , and the equality a 2 m - 1 2 m - 1 = a is obvious. At a< 0 we get respectively a = - a and a 2 m = (- a) 2 m = a 2 m . The last transformation of the number is valid according to the property of the degree. This is what proves the equality a 2 m 2 m \u003d a, and a 2 m - 1 2 m - 1 \u003d a will be true, since - c 2 m - 1 \u003d - c 2 m is considered for an odd degree - 1 for any number c , positive or equal to zero.

In order to consolidate the received information, consider a few examples using the property:

Example 5

7 4 4 = 7 = 7 , (- 5) 12 12 = - 5 = 5 , 0 8 8 = 0 = 0 , 6 3 3 = 6 and (- 3 , 39) 5 5 = - 3 , 39 .

Let us prove the following equality a m n = a n · m . To do this, you need to change the numbers before the equal sign and after it in places a n · m = a m n . This will indicate the correct entry. For a , which is positive or equal to zero , from the form a m n is a positive number or equal to zero. Let us turn to the property of raising a power to a power and the definition. With their help, you can transform equalities in the form a m n n · m = a m n n m = a m m = a . This proves the considered property of a root from a root.

Other properties are proved similarly. Really, . . . a n k n 2 n 1 n 1 n 2 . . . nk = . . . a n k n 3 n 2 n 2 n 3 . . . nk = . . . a nk n 4 n 3 n 3 n 4 . . . nk = . . . = a n k n k = a .

For example, 7 3 5 = 7 5 3 and 0, 0009 6 = 0, 0009 2 2 6 = 0, 0009 24.

Let us prove the following property a m n · m = a n . To do this, it is necessary to show that a n is a number that is positive or equal to zero. When raised to a power n m is a m. If number a is positive or zero, then n th degree from among a is a positive number or equal to zero Moreover, a n · m n = a n n m , which was to be proved.

In order to consolidate the acquired knowledge, consider a few examples.

Let us prove the following property - the property of the root of the power of the form a m n = a n m . It is obvious that at a ≥ 0 the degree a n m is a non-negative number. Moreover, her n-th degree is equal to a m, indeed, a n m n = a n m · n = a n n m = a m . This proves the considered property of the degree.

For example, 2 3 5 3 = 2 3 3 5 .

We need to prove that for any positive numbers a and b a< b . Consider the inequality a n< b n . Воспользуемся методом от противного a n ≥ b n . Тогда, согласно свойству, о котором говорилось выше, неравенство считается верным a n n ≥ b n n , то есть, a ≥ b . Но это не соответствует условию a< b . Therefore, a n< b n при a< b .

For example, we give 12 4< 15 2 3 4 .

Consider the root property n-th degree. First, consider the first part of the inequality. At m > n And 0 < a < 1 true a m > a n . Suppose a m ≤ a n . Properties will simplify the expression to a n m · n ≤ a m m · n . Then, according to the properties of a degree with a natural exponent, the inequality a n m n m n ≤ a m m n m n is satisfied, that is, a n ≤ a m. The value obtained at m > n And 0 < a < 1 does not match the properties above.

In the same way, one can prove that m > n And a > 1 condition a m< a n .

In order to fix the above properties, consider a few concrete examples. Consider inequalities using specific numbers.

Example 6

0 , 7 3 < 0 , 7 5 и 12 > 12 7 .

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Square root properties

We continue to study square roots. Today we will consider the main properties of the roots. All the main properties are intuitive and consistent with all the operations that we have done before.

Property 1. The square root of the product of two non-negative numbers is equal to the product of the square roots of these numbers: $\sqrt(a*b)=\sqrt(a)*\sqrt(b)$.

It is customary to prove any properties, let's do it.
Let $\sqrt(a*b)=x$, $\sqrt(a)=y$, $\sqrt(b)=z$. Then we have to prove that $x=y*z$.
Let's square each expression.
If $\sqrt(a*b)=x$ then $a*b=x^2$.
If $\sqrt(a)=y$, $\sqrt(b)=z$, then squaring both expressions, we get: $a=y^2$, $b=z^2$.
$a*b=x^2=y^2*z^2$, i.e. $x^2=(y*z)^2$. If the squares of two non-negative numbers are equal, then the numbers themselves are equal, which was to be proved.

It follows from our property that, for example, $\sqrt(5)*\sqrt(3)=\sqrt(15)$.

Remark 1. The property is also valid for the case when there are more than two non-negative factors under the root.
Property 2. If $a≥0$ and $b>0$, then the following equality holds: $\sqrt(\frac(a)(b))=\frac(\sqrt(a))(\sqrt(b))$

That is, the root of the quotient is equal to the quotient of the roots.
Proof.
Let's use the table and briefly prove our property.

Examples of using square roots properties

Example 1
Calculate: $\sqrt(81*25*121)$.

Solution.
Of course, we can take a calculator, multiply all the numbers under the root and perform the operation of extracting the square root. And if there is no calculator at hand, what then?
$\sqrt(81*25*121)=\sqrt(81)*\sqrt(25)*\sqrt(121)=9*5*11=495$.
Answer: 495.

Example 2. Calculate: $\sqrt(11\frac(14)(25))$.

Solution.
We represent the radical number as an improper fraction: $11\frac(14)(25)=\frac(11*25+14)(25)=\frac(275+14)(25)=\frac(289)(25) $.
Let's use property 2.
$\sqrt(\frac(289)(25))=\frac(\sqrt(289))(\sqrt(25))=\frac(17)(5)=3\frac(2)(5)= 3.4$.
Answer: 3.4.

Example 3
Calculate: $\sqrt(40^2-24^2)$.

Solution.
We can evaluate our expression directly, but it can almost always be simplified. Let's try to do this.
$40^2-24^2=(40-24)(40+24)=16*64$.
So $\sqrt(40^2-24^2)=\sqrt(16*64)=\sqrt(16)*\sqrt(64)=4*8=32$.
Answer: 32.

Guys, please note that there are no formulas for the operations of addition and subtraction of radical expressions and the expressions below are not correct.
$\sqrt(a+b)≠\sqrt(a)+\sqrt(b)$.
$\sqrt(a-b)≠\sqrt(a)-\sqrt(b)$.

Example 4
Calculate: a) $\sqrt(32)*\sqrt(8)$; b) $\frac(\sqrt(32))(\sqrt(8))$.
Solution.
The properties presented above work both from left to right and in reverse order, i.e:
$\sqrt(a)*\sqrt(b)=\sqrt(a*b)$.
$\frac(\sqrt(a))(\sqrt(b))=\sqrt(\frac(a)(b))$.
Let's use this to solve our example.
a) $\sqrt(32)*\sqrt(8)=\sqrt(32*8)=\sqrt(256)=16.$

B) $\frac(\sqrt(32))(\sqrt(8))=\sqrt(\frac(32)(8))=\sqrt(4)=2$.

Answer: a) 16; b) 2.

Property 3. If $a≥0$ and n is a natural number, then the following equality holds: $\sqrt(a^(2n))=a^n$.

For example. $\sqrt(a^(16))=a^8$, $\sqrt(a^(24))=a^(12)$ and so on.

Example 5
Calculate: $\sqrt(129600)$.

Solution.
The number presented to us is quite large, let's decompose it into prime factors.
We got: $129600=5^2*2^6*3^4$ or $\sqrt(129600)=\sqrt(5^2*2^6*3^4)=5*2^3*3^2 =5*8*9=360$.
Answer: 360.

Tasks for independent solution

1. Calculate: $\sqrt(144*36*64)$.
2. Calculate: $\sqrt(8\frac(1)(36))$.
3. Calculate: $\sqrt(52^2-48^2)$.
4. Calculate:
a) $\sqrt(128*\sqrt(8))$;
b) $\frac(\sqrt(128))(\sqrt(8))$.

Mathematics was born when a person became aware of himself and began to position himself as an autonomous unit of the world. The desire to measure, compare, calculate what surrounds you is what underlay one of the fundamental sciences of our days. At first, these were pieces of elementary mathematics, which made it possible to associate numbers with their physical expressions, later the conclusions began to be presented only theoretically (due to their abstractness), but after a while, as one scientist put it, "mathematics reached the ceiling of complexity when all numbers." The concept of "square root" appeared at a time when it could be easily supported by empirical data, going beyond the plane of calculations.

How it all started

The first mention of the root, which on this moment denoted as √, was recorded in the writings of the Babylonian mathematicians, who laid the foundation for modern arithmetic. Of course, they looked a little like the current form - the scientists of those years first used bulky tablets. But in the second millennium BC. e. they came up with an approximate calculation formula that showed how to take the square root. The photo below shows a stone on which Babylonian scientists carved the output process √2, and it turned out to be so correct that the discrepancy in the answer was found only in the tenth decimal place.

In addition, the root was used if it was necessary to find the side of a triangle, provided that the other two were known. Well, when solving quadratic equations, there is no escape from extracting the root.

Along with the Babylonian works, the object of the article was also studied in the Chinese work "Mathematics in Nine Books", and the ancient Greeks came to the conclusion that any number from which the root is not extracted without a remainder gives an irrational result.

The origin of this term is associated with the Arabic representation of the number: ancient scientists believed that the square of an arbitrary number grows from the root, like a plant. In Latin, this word sounds like radix (one can trace a pattern - everything that has a "root" semantic load is consonant, be it radish or sciatica).

Scientists of subsequent generations picked up this idea, designating it as Rx. For example, in the 15th century, in order to indicate that the square root is taken from an arbitrary number a, they wrote R 2 a. Habitual modern look"tick" √ appeared only in the 17th century thanks to Rene Descartes.

Our days

Mathematically, the square root of y is the number z whose square is y. In other words, z 2 =y is equivalent to √y=z. However, this definition is relevant only for arithmetic root, since it implies a non-negative value of the expression. In other words, √y=z, where z is greater than or equal to 0.

In general, which is valid for determining an algebraic root, the value of an expression can be either positive or negative. Thus, due to the fact that z 2 =y and (-z) 2 =y, we have: √y=±z or √y=|z|.

Due to the fact that love for mathematics has only increased with the development of science, there are various manifestations of affection for it, not expressed in dry calculations. For example, along with such interesting events as the day of Pi, the holidays of the square root are also celebrated. They are celebrated nine times in a hundred years, and are determined according to the following principle: the numbers that denote the day and month in order must be the square root of the year. Yes, in next time This holiday will be celebrated on April 4, 2016.

Properties of the square root on the field R

Almost all mathematical expressions have a geometric basis, this fate did not pass and √y, which is defined as the side of a square with area y.

How to find the root of a number?

There are several calculation algorithms. The simplest, but at the same time quite cumbersome, is the usual arithmetic calculation, which is as follows:

1) from the number whose root we need, odd numbers are subtracted in turn - until the remainder at the output is less than the subtracted one or even zero. The number of moves will eventually become the desired number. For example, calculating the square root of 25:

Following odd number is 11, we have the following remainder: 1<11. Количество ходов - 5, так что корень из 25 равен 5. Вроде все легко и просто, но представьте, что придется вычислять из 18769?

For such cases, there is a Taylor series expansion:

√(1+y)=∑((-1) n (2n)!/(1-2n)(n!) 2 (4 n))y n , where n takes values from 0 to

+∞, and |y|≤1.

Graphic representation of the function z=√y

Consider an elementary function z=√y on the field of real numbers R, where y is greater than or equal to zero. Her chart looks like this:

The curve grows from the origin and necessarily crosses the point (1; 1).

Properties of the function z=√y on the field of real numbers R

1. The domain of definition of the considered function is the interval from zero to plus infinity (zero is included).

2. The range of values of the considered function is the interval from zero to plus infinity (zero is again included).

3. The function takes the minimum value (0) only at the point (0; 0). There is no maximum value.

4. The function z=√y is neither even nor odd.

5. The function z=√y is not periodic.

6. There is only one point of intersection of the graph of the function z=√y with the coordinate axes: (0; 0).

7. The intersection point of the graph of the function z=√y is also the zero of this function.

8. The function z=√y is continuously growing.

9. The function z=√y takes only positive values, therefore, its graph occupies the first coordinate angle.

Options for displaying the function z=√y

In mathematics, to facilitate the calculation of complex expressions, sometimes they use the power form of writing the square root: √y=y 1/2. This option is convenient, for example, in raising a function to a power: (√y) 4 =(y 1/2) 4 =y 2 . This method is also a good representation for differentiation with integration, since thanks to it the square root is represented by an ordinary power function.

And in programming, the replacement for the symbol √ is the combination of letters sqrt.

It is worth noting that in this area the square root is in great demand, as it is part of most of the geometric formulas necessary for calculations. The counting algorithm itself is quite complicated and is based on recursion (a function that calls itself).

The square root in the complex field C

By and large, it was the subject of this article that stimulated the discovery of the field of complex numbers C, since mathematicians were haunted by the question of obtaining an even degree root from a negative number. This is how the imaginary unit i appeared, which is characterized by a very interesting property: its square is -1. Thanks to this, the quadratic equations were also solved with a negative discriminant. In C, for the square root, the same properties are relevant as in R, the only thing is that the restrictions on the root expression are removed.

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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. Consequently, \[\sqrt(25)+\sqrt(49)=5+7=12\] 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 $\sqrt3, \ 1+\sqrt2, \ \sqrt(15)$ etc. 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 which 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 in an easy and understandable way for students with any level of preparation 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.