Formulas for the properties of an arithmetic root and their names. Properties of roots: formulations, evidence, examples

Lesson and presentation on the topic:
"Properties of the square root. Formulas. Examples of solutions, problems with answers"

Additional materials
Dear users, do not forget to leave your comments, reviews, wishes. All materials have been checked by an anti-virus program.

Educational aids and simulators in the Integral online store for grade 8
Interactive textbook "Geometry in 10 minutes" for 8th grade
Educational complex "1C: School. Geometry, grade 8"

Properties of square root

We continue to study square roots. Today we will look at the basic properties of roots. All the basic properties are intuitive and consistent with all the operations 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 need 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: $a=y^2$, $b=z^2$.
$a*b=x^2=y^2*z^2$, that is, $x^2=(y*z)^2$. If the squares of two non-negative numbers are equal, then the numbers themselves are equal, which is what needed to be proven.

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

Note 1. The property is also true 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 the properties of square roots

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 you don’t have a calculator at hand, what should you do 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.
Let's 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 presented 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, that is:
$\sqrt(a)*\sqrt(b)=\sqrt(a*b)$.
$\frac(\sqrt(a))(\sqrt(b))=\sqrt(\frac(a)(b))$.
Using this, let's 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 $а≥0$ and n is a natural number, then the 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 break it down into prime factors.
We received: $129600=5^2*2^6*3^4$ or $\sqrt(129600)=\sqrt(5^2*2^6*3^4)=5*2^3*3^2 =5*8*9=$360.
Answer: 360.

Problems to solve independently

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) $\frac(\sqrt(128))(\sqrt(8))$.

Properties of square roots

So far we have performed five arithmetic operations on numbers: addition, subtraction, multiplication, division and exponentiation, and in the calculations various properties of these operations were actively used, for example a + b = b + a, an-bn = (ab)n, etc.

This chapter introduces a new operation - taking the square root of a non-negative number. To use it successfully, you need to become familiar with the properties of this operation, which we will do in this section.

Proof. Let us introduce the following notation: https://pandia.ru/text/78/290/images/image005_28.jpg" alt="Equality" width="120" height="25 id=">!}.

This is exactly how we will formulate the next theorem.

(A brief formulation that is more convenient to use in practice: the root of a fraction is equal to the fraction of the roots, or the root of the quotient is equal to the quotient of the roots.)

This time we will give only a brief summary of the proof, and you try to make appropriate comments similar to those that formed the essence of the proof of Theorem 1.

Note 3. Of course, this example can be solved differently, especially if you have a microcalculator at hand: multiply the numbers 36, 64, 9, and then take the square root of the resulting product. However, you will agree that the solution proposed above looks more cultural.

Note 4. In the first method, we carried out calculations “head-on”. The second way is more elegant:
we applied formula a2 - b2 = (a - b) (a + b) and used the property of square roots.

Note 5. Some “hot heads” sometimes offer this “solution” to example 3:

This, of course, is not true: you see - the result is not the same as in example 3. The fact is that there is no property https://pandia.ru/text/78/290/images/image014_6.jpg" alt="Task" width="148" height="26 id=">!} There are only properties relating to multiplication and division of square roots. Be careful and careful, do not take wishful thinking.

To conclude this section, let us note one more quite simple and at the same time important property:
if a > 0 and n - natural number, That

Converting Expressions Containing a Square Root Operation

Until now, we have only performed transformations rational expressions, using for this the rules of operations on polynomials and algebraic fractions, abbreviated multiplication formulas, etc. In this chapter, we introduced a new operation - the operation of extracting the square root; we have established that

where, recall, a, b are non-negative numbers.

Using these formulas, you can perform various transformations on expressions that contain a square root operation. Let's look at several examples, and in all examples we will assume that the variables take only non-negative values.

Example 3. Enter the multiplier under the square root sign:

Example 6. Simplify the expression Solution. Let's perform sequential transformations:

Root formulas. Properties of square roots.

Attention!
There are additional
materials in Special Section 555.
For those who are very "not very..."
And for those who “very much…”)

In the previous lesson we figured out what a square root is. It's time to figure out which ones exist formulas for roots what are properties of roots, and what can be done with all this.

Formulas of roots, properties of roots and rules for working with roots- this is essentially the same thing. There are surprisingly few formulas for square roots. Which certainly makes me happy! Or rather, you can write a lot of different formulas, but for practical and confident work with roots, only three are enough. Everything else flows from these three. Although many people get confused in the three root formulas, yes...

Let's start with the simplest one. Here she is:

If you like this site...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)

You can get acquainted with functions and derivatives.

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!

In order to adequately solve the Unified State Exam in mathematics, you first need to study theoretical material, which introduces you to 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 Exam in mathematics is presented in an easy and understandable way for students with any level of training is in fact a rather difficult task. School textbooks cannot always be kept at hand. And finding basic formulas for the Unified State Exam in mathematics can be difficult even on the Internet.

Why is it so important to study theory in mathematics not only for those taking the Unified State Exam?

Because it broadens your horizons. Studying theoretical material in mathematics is useful for anyone who wants to get answers to a wide range of questions related to knowledge of the world around them. 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 intelligence. By studying reference materials for the Unified State Exam in mathematics, as well as solving various problems, a person learns to think and reason logically, to formulate thoughts competently and clearly. He develops the ability to analyze, generalize, and draw conclusions.

We invite you to personally evaluate all the advantages of our approach to systematization and presentation of educational materials.