Square root. Detailed theory with examples

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 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 easily and understandably for students with any level of training is, in fact, a rather difficult task. School textbooks cannot always be kept at hand. And finding the basic formulas for the exam in mathematics can be difficult even on the Internet.

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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.

The square root of a number X called a number A, which in the process of multiplying itself by itself ( A*A) can give a number X.
Those. A * A = A 2 = X, And √X = A.

Over square roots ( √x), as with other numbers, you can perform arithmetic operations such as subtraction and addition. To subtract and add roots, they must be connected using signs corresponding to these actions (for example √x- √y ).
And then bring the roots to their simplest form - if there are similar ones between them, you need to make a cast. It consists in the fact that the coefficients of similar terms are taken with the signs of the corresponding terms, then they are enclosed in brackets and the common root is displayed outside the multiplier brackets. The coefficient that we have obtained is simplified according to the usual rules.

Step 1. Extracting square roots

First, to add square roots, you first need to extract these roots. This can be done if the numbers under the root sign are perfect squares. For example, take the given expression √4 + √9 . First number 4 is the square of the number 2 . Second number 9 is the square of the number 3 . Thus, the following equality can be obtained: √4 + √9 = 2 + 3 = 5 .
Everything, the example is solved. But it doesn't always happen that way.

Step 2. Taking out the multiplier of a number from under the root

If there are no full squares under the root sign, you can try to take the multiplier of the number out from under the root sign. For example, take the expression √24 + √54 .

Let's factorize the numbers:
24 = 2 * 2 * 2 * 3 ,
54 = 2 * 3 * 3 * 3 .

Among 24 we have a multiplier 4 , it can be taken out from under the square root sign. Among 54 we have a multiplier 9 .

We get the equality:
√24 + √54 = √(4 * 6) + √(9 * 6) = 2 * √6 + 3 * √6 = 5 * √6 .

Considering this example, we get the removal of the factor from under the root sign, thereby simplifying the given expression.

Step 3. Reducing the denominator

Consider the following situation: the sum of two square roots is the denominator of a fraction, for example, A / (√a + √b).
Now we are faced with the task of "getting rid of the irrationality in the denominator."
Let's use the following method: multiply the numerator and denominator of the fraction by the expression √a - √b.

We now get the abbreviated multiplication formula in the denominator:
(√a + √b) * (√a - √b) = a - b.

Similarly, if the denominator contains the difference of the roots: √a - √b, the numerator and denominator of the fraction are multiplied by the expression √a + √b.

Let's take a fraction as an example:
4 / (√3 + √5) = 4 * (√3 - √5) / ((√3 + √5) * (√3 - √5)) = 4 * (√3 - √5) / (-2) = 2 * (√5 - √3) .

An example of complex denominator reduction

Now we will consider a rather complicated example of getting rid of irrationality in the denominator.

Let's take a fraction as an example: 12 / (√2 + √3 + √5) .
You need to take its numerator and denominator and multiply by the expression √2 + √3 - √5 .

We get:

12 / (√2 + √3 + √5) = 12 * (√2 + √3 - √5) / (2 * √6) = 2 * √3 + 3 * √2 - √30.

Step 4. Calculate the approximate value on the calculator

If you only need an approximate value, this can be done on a calculator by calculating the value of square roots. Separately, for each number, the value is calculated and recorded with the required accuracy, which is determined by the number of decimal places. Further, all the required operations are performed, as with ordinary numbers.

Estimated Calculation Example

It is necessary to calculate the approximate value of this expression √7 + √5 .

As a result, we get:

√7 + √5 ≈ 2,65 + 2,24 = 4,89 .

Please note: under no circumstances should square roots be added as prime numbers, this is completely unacceptable. That is, if you add the square root of five and three, we cannot get the square root of eight.

Useful advice: if you decide to factorize a number, in order to derive a square from under the root sign, you need to do a reverse check, that is, multiply all the factors that resulted from the calculations, and the end result of this mathematical calculation should be the number that we were originally given.

Theory

Addition and subtraction of roots is studied in the introductory mathematics course. We will assume that the reader knows the concept of degree.

Definition 1

An $n$ root of a real number $a$ is a real number $b$ whose $n$-th power is equal to $a$: $b=\sqrt[n]a, b^n=a.$ Here $a$ is the root expression, $n$ is the exponent of the root, $b$ is the value of the root. The root sign is called the radical.

The inverse of root extraction is exponentiation.

Basic operations with arithmetic roots:

Figure 1. Basic operations with arithmetic roots. Author24 - online exchange of student papers

As we can see, in the listed actions there is no formula for addition and subtraction. These actions with roots are performed in the form of transformations. For these transformations, the abbreviated multiplication formulas should be used:

$(\sqrt a - \sqrt b)(\sqrt a + \sqrt b)=a-b;$

$(\sqrta-\sqrtb)(\sqrt(a^2)+\sqrt(ab)+\sqrt(b^2))=a-b;$

$(\sqrta+\sqrtb)(\sqrt(a^2)-\sqrt(ab)+\sqrt(b^2))=a+b;$

$a\sqrt a+b\sqrt b=(\sqrt a)^3+(\sqrt b)^3=(\sqrt a+\sqrt b)(a-\sqrt(ab)+b);$

$a\sqrt a-b\sqrt b=(\sqrt a)^3-(\sqrt b)^3=(\sqrt a-\sqrt b)(a+\sqrt(ab)+b).$

It is worth noting that the operations of addition and subtraction are found in examples of irrational expressions: $ab\sqrt(m-n); 1+\sqrt3.$

Examples

Let us consider by examples the cases when the "destruction" of irrationality in the denominator is applicable. When, as a result of transformations, an irrational expression is obtained both in the numerator and in the denominator, then it is necessary to "destroy" the irrationality in the denominator.

Example 1

$\frac(1)(\sqrt7-\sqrt6)=\frac(\sqrt7+\sqrt6)((\sqrt7-\sqrt6)(\sqrt7+\sqrt6))=\frac(\sqrt7+\sqrt6)(7-6)=\frac(\sqrt7+\sqrt6)(1)=\sqrt7+\sqrt6.$

In this example, we have multiplied the numerator and denominator of a fraction by the conjugate of the denominator. Thus, the denominator is transformed by the difference of squares formula.

Extracting the square root of a number is not the only operation that can be performed with this mathematical phenomenon. Just like ordinary numbers, square roots can be added and subtracted.

Rules for adding and subtracting square roots

Definition 1

Actions such as adding and subtracting a square root are possible only if the root expression is the same.

Example 1

You can add or subtract expressions 2 3 and 6 3, but not 5 6 And 9 4 . If it is possible to simplify the expression and bring it to roots with the same root number, then simplify, and then add or subtract.

Root Actions: The Basics

Example 2

6 50 - 2 8 + 5 12

Action algorithm:

Simplify the root expression. To do this, it is necessary to decompose the root expression into 2 factors, one of which is a square number (the number from which the whole square root is extracted, for example, 25 or 9).
Then you need to take the root of the square number and write the resulting value before the root sign. Please note that the second factor is entered under the root sign.
After the simplification process, it is necessary to underline the roots with the same radical expressions - only they can be added and subtracted.
For roots with the same radical expressions, it is necessary to add or subtract the factors that precede the root sign. The root expression remains unchanged. Do not add or subtract root numbers!

Tip 1

If you have an example with a lot of identical radical expressions, then underline such expressions with single, double and triple lines to facilitate the calculation process.

Example 3

Let's try this example:

6 50 = 6 (25 × 2) = (6 × 5) 2 = 30 2 . First you need to decompose 50 into 2 factors 25 and 2, then take the root of 25, which is 5, and take 5 out from under the root. After that, you need to multiply 5 by 6 (the multiplier at the root) and get 30 2 .

2 8 = 2 (4 × 2) = (2 × 2) 2 = 4 2 . First, you need to decompose 8 into 2 factors: 4 and 2. Then, from 4, extract the root, which is equal to 2, and take 2 out from under the root. After that, you need to multiply 2 by 2 (the factor at the root) and get 4 2 .

5 12 = 5 (4 × 3) = (5 × 2) 3 = 10 3 . First, you need to decompose 12 into 2 factors: 4 and 3. Then extract the root from 4, which is 2, and take it out from under the root. After that, you need to multiply 2 by 5 (the factor at the root) and get 10 3 .

Simplification result: 30 2 - 4 2 + 10 3

30 2 - 4 2 + 10 3 = (30 - 4) 2 + 10 3 = 26 2 + 10 3 .

As a result, we saw how many identical radical expressions are contained in this example. Now let's practice with other examples.

Example 4

Simplify (45) . We factorize 45: (45) = (9 × 5) ;
We take out 3 from under the root (9 \u003d 3): 45 \u003d 3 5;
We add the factors at the roots: 3 5 + 4 5 = 7 5 .

Example 5

6 40 - 3 10 + 5:

Simplifying 6 40 . We factorize 40: 6 40 \u003d 6 (4 × 10) ;
We take out 2 from under the root (4 \u003d 2): 6 40 \u003d 6 (4 × 10) \u003d (6 × 2) 10;
We multiply the factors that are in front of the root: 12 10;
We write the expression in a simplified form: 12 10 - 3 10 + 5;
Since the first two terms have the same root numbers, we can subtract them: (12 - 3) 10 = 9 10 + 5.

Example 6

As we can see, it is not possible to simplify the radical numbers, therefore, in the example, we look for members with the same radical numbers, perform mathematical operations (add, subtract, etc.) and write the result:

(9 - 4) 5 - 2 3 = 5 5 - 2 3 .

Adviсe:

Before adding or subtracting, it is imperative to simplify (if possible) the radical expressions.
Adding and subtracting roots with different root expressions is strictly prohibited.
Do not add or subtract an integer or square root: 3 + (2 x) 1 / 2 .
When performing actions with fractions, you need to find a number that is divisible by each denominator, then bring the fractions to a common denominator, then add the numerators, and leave the denominators unchanged.

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Hello kitties! Last time we analyzed in detail what roots are (if you don’t remember, I recommend reading). The main conclusion of that lesson: there is only one universal definition of roots, which you need to know. The rest is nonsense and a waste of time.

Today we go further. We will learn to multiply roots, we will study some problems associated with multiplication (if these problems are not solved, then they can become fatal on the exam) and we will practice properly. So stock up on popcorn, make yourself comfortable - and we'll start. :)

You haven't smoked yet, have you?

The lesson turned out to be quite large, so I divided it into two parts:

First, we'll look at the rules for multiplication. The cap seems to be hinting: this is when there are two roots, there is a “multiply” sign between them - and we want to do something with it.
Then we will analyze the reverse situation: there is one big root, and we were impatient to present it as a product of two roots in a simpler way. With what fright it is necessary is a separate question. We will only analyze the algorithm.

For those who can't wait to jump right into Part 2, you're welcome. Let's start with the rest in order.

Basic multiplication rule

Let's start with the simplest - classical square roots. The ones that are denoted by $\sqrt(a)$ and $\sqrt(b)$. For them, everything is generally clear:

multiplication rule. To multiply one square root by another, you just need to multiply their radical expressions, and write the result under the common radical:

\[\sqrt(a)\cdot \sqrt(b)=\sqrt(a\cdot b)\]

No additional restrictions are imposed on the numbers on the right or left: if the multiplier roots exist, then the product also exists.

Examples. Consider four examples with numbers at once:

\[\begin(align) & \sqrt(25)\cdot \sqrt(4)=\sqrt(25\cdot 4)=\sqrt(100)=10; \\ & \sqrt(32)\cdot \sqrt(2)=\sqrt(32\cdot 2)=\sqrt(64)=8; \\ & \sqrt(54)\cdot \sqrt(6)=\sqrt(54\cdot 6)=\sqrt(324)=18; \\ & \sqrt(\frac(3)(17))\cdot \sqrt(\frac(17)(27))=\sqrt(\frac(3)(17)\cdot \frac(17)(27 ))=\sqrt(\frac(1)(9))=\frac(1)(3). \\ \end(align)\]

As you can see, the main meaning of this rule is to simplify irrational expressions. And if in the first example we would have extracted the roots from 25 and 4 without any new rules, then the tin begins: $\sqrt(32)$ and $\sqrt(2)$ do not count by themselves, but their product turns out to be an exact square, so the root of it is equal to a rational number.

Separately, I would like to note the last line. There, both radical expressions are fractions. Thanks to the product, many factors cancel out, and the whole expression turns into an adequate number.

Of course, not everything will always be so beautiful. Sometimes there will be complete crap under the roots - it is not clear what to do with it and how to transform after multiplication. A little later, when you start studying irrational equations and inequalities, there will be all sorts of variables and functions in general. And very often, the compilers of the problems are just counting on the fact that you will find some contracting terms or factors, after which the task will be greatly simplified.

In addition, it is not necessary to multiply exactly two roots. You can multiply three at once, four - yes even ten! This will not change the rule. Take a look:

\[\begin(align) & \sqrt(2)\cdot \sqrt(3)\cdot \sqrt(6)=\sqrt(2\cdot 3\cdot 6)=\sqrt(36)=6; \\ & \sqrt(5)\cdot \sqrt(2)\cdot \sqrt(0.001)=\sqrt(5\cdot 2\cdot 0.001)= \\ & =\sqrt(10\cdot \frac(1) (1000))=\sqrt(\frac(1)(100))=\frac(1)(10). \\ \end(align)\]

And again a small remark on the second example. As you can see, in the third multiplier, there is a decimal fraction under the root - in the process of calculations, we replace it with a regular one, after which everything is easily reduced. So: I highly recommend getting rid of decimal fractions in any irrational expressions (that is, containing at least one radical icon). This will save you a lot of time and nerves in the future.

But it was a lyrical digression. Now let's consider a more general case - when the root exponent contains an arbitrary number $n$, and not just the "classical" two.

The case of an arbitrary indicator

So, we figured out the square roots. And what to do with cubes? Or in general with roots of arbitrary degree $n$? Yes, everything is the same. The rule remains the same:

To multiply two roots of degree $n$, it is enough to multiply their radical expressions, after which the result is written under one radical.

In general, nothing complicated. Unless the volume of calculations can be more. Let's look at a couple of examples:

Examples. Calculate products:

\[\begin(align) & \sqrt(20)\cdot \sqrt(\frac(125)(4))=\sqrt(20\cdot \frac(125)(4))=\sqrt(625)= 5; \\ & \sqrt(\frac(16)(625))\cdot \sqrt(0,16)=\sqrt(\frac(16)(625)\cdot \frac(16)(100))=\sqrt (\frac(64)(((25)^(2))\cdot 25))= \\ & =\sqrt(\frac(((4)^(3)))(((25)^(3 ))))=\sqrt(((\left(\frac(4)(25) \right))^(3)))=\frac(4)(25). \\ \end(align)\]

And again attention to the second expression. We multiply the cube roots, get rid of the decimal fraction, and as a result we get the product of the numbers 625 and 25 in the denominator. This is a rather large number - personally, I won’t immediately calculate what it is equal to.

Therefore, we simply selected the exact cube in the numerator and denominator, and then used one of the key properties (or, if you like, the definition) of the root of the $n$th degree:

\[\begin(align) & \sqrt(((a)^(2n+1)))=a; \\ & \sqrt(((a)^(2n)))=\left| a\right|. \\ \end(align)\]

Such "scams" can save you a lot of time on an exam or test, so remember:

Do not rush to multiply the numbers in the radical expression. First, check: what if the exact degree of any expression is “encrypted” there?

With all the obviousness of this remark, I must admit that most unprepared students point blank do not see the exact degrees. Instead, they multiply everything ahead, and then wonder: why did they get such brutal numbers? :)

However, all this is child's play compared to what we will study now.

Multiplication of roots with different exponents

Well, now we can multiply roots with the same exponents. What if the scores are different? Say, how do you multiply an ordinary $\sqrt(2)$ by some crap like $\sqrt(23)$? Is it even possible to do this?

Yes, of course you can. Everything is done according to this formula:

Root multiplication rule. To multiply $\sqrt[n](a)$ by $\sqrt[p](b)$, just do the following transformation:

\[\sqrt[n](a)\cdot \sqrt[p](b)=\sqrt(((a)^(p))\cdot ((b)^(n)))\]

However, this formula only works if radical expressions are non-negative. This is a very important remark, to which we will return a little later.

For now, let's look at a couple of examples:

\[\begin(align) & \sqrt(3)\cdot \sqrt(2)=\sqrt(((3)^(4))\cdot ((2)^(3)))=\sqrt(81 \cdot8)=\sqrt(648); \\ & \sqrt(2)\cdot \sqrt(7)=\sqrt(((2)^(5))\cdot ((7)^(2)))=\sqrt(32\cdot 49)= \sqrt(1568); \\ & \sqrt(5)\cdot \sqrt(3)=\sqrt(((5)^(4))\cdot ((3)^(2)))=\sqrt(625\cdot 9)= \sqrt(5625). \\ \end(align)\]

As you can see, nothing complicated. Now let's figure out where the non-negativity requirement came from, and what will happen if we violate it. :)

It's easy to multiply roots.

Why do radical expressions have to be non-negative?

Of course, you can become like school teachers and quote a textbook with a smart look:

The requirement of non-negativity is associated with different definitions of roots of even and odd degrees (respectively, their domains of definition are also different).

Well, it became clearer? Personally, when I read this nonsense in the 8th grade, I understood for myself something like this: “The requirement of non-negativity is associated with *#&^@(*#@^#)~%” - in short, I didn’t understand shit at that time. :)

So now I will explain everything in a normal way.

First, let's find out where the multiplication formula above comes from. To do this, let me remind you of one important property of the root:

\[\sqrt[n](a)=\sqrt(((a)^(k)))\]

In other words, we can safely raise the root expression to any natural power $k$ - in this case, the root index will have to be multiplied by the same power. Therefore, we can easily reduce any roots to a common indicator, after which we multiply. This is where the multiplication formula comes from:

\[\sqrt[n](a)\cdot \sqrt[p](b)=\sqrt(((a)^(p)))\cdot \sqrt(((b)^(n)))= \sqrt(((a)^(p))\cdot ((b)^(n)))\]

But there is one problem that severely limits the application of all these formulas. Consider this number:

According to the formula just given, we can add any degree. Let's try adding $k=2$:

\[\sqrt(-5)=\sqrt(((\left(-5 \right))^(2)))=\sqrt(((5)^(2)))\]

We removed the minus precisely because the square burns the minus (like any other even degree). And now let's perform the reverse transformation: "reduce" the two in the exponent and degree. After all, any equality can be read both left-to-right and right-to-left:

\[\begin(align) & \sqrt[n](a)=\sqrt(((a)^(k)))\Rightarrow \sqrt(((a)^(k)))=\sqrt[n ](a); \\ & \sqrt(((a)^(k)))=\sqrt[n](a)\Rightarrow \sqrt(((5)^(2)))=\sqrt(((5)^( 2)))=\sqrt(5). \\ \end(align)\]

But then something crazy happens:

\[\sqrt(-5)=\sqrt(5)\]

This can't be because $\sqrt(-5) \lt 0$ and $\sqrt(5) \gt 0$. This means that for even powers and negative numbers, our formula no longer works. After which we have two options:

To fight against the wall to state that mathematics is a stupid science, where “there are some rules, but this is inaccurate”;
Introduce additional restrictions under which the formula will become 100% working.

In the first option, we will have to constantly catch “non-working” cases - this is difficult, long and generally fu. Therefore, mathematicians preferred the second option. :)

But don't worry! In practice, this restriction does not affect the calculations in any way, because all the described problems concern only the roots of an odd degree, and minuses can be taken out of them.

Therefore, we formulate another rule that applies in general to all actions with roots:

Before multiplying the roots, make sure that the radical expressions are non-negative.

Example. In the number $\sqrt(-5)$, you can take out the minus from under the root sign - then everything will be fine:

\[\begin(align) & \sqrt(-5)=-\sqrt(5) \lt 0\Rightarrow \\ & \sqrt(-5)=-\sqrt(((5)^(2))) =-\sqrt(25)=-\sqrt(((5)^(2)))=-\sqrt(5) \lt 0 \\ \end(align)\]

Feel the difference? If you leave a minus under the root, then when the radical expression is squared, it will disappear, and crap will begin. And if you first take out a minus, then you can even raise / remove a square until you are blue in the face - the number will remain negative. :)

Thus, the most correct and most reliable way to multiply the roots is as follows:

Remove all minuses from under the radicals. Minuses are only in the roots of odd multiplicity - they can be placed in front of the root and, if necessary, reduced (for example, if there are two of these minuses).
Perform multiplication according to the rules discussed above in today's lesson. If the indices of the roots are the same, simply multiply the root expressions. And if they are different, we use the evil formula \[\sqrt[n](a)\cdot \sqrt[p](b)=\sqrt(((a)^(p))\cdot ((b)^(n) ))\].
3. We enjoy the result and good grades. :)

Well? Shall we practice?

Example 1. Simplify the expression:

\[\begin(align) & \sqrt(48)\cdot \sqrt(-\frac(4)(3))=\sqrt(48)\cdot \left(-\sqrt(\frac(4)(3 )) \right)=-\sqrt(48)\cdot \sqrt(\frac(4)(3))= \\ & =-\sqrt(48\cdot \frac(4)(3))=-\ sqrt(64)=-4; \end(align)\]

This is the simplest option: the indicators of the roots are the same and odd, the problem is only in the minus of the second multiplier. We endure this minus nafig, after which everything is easily considered.

Example 2. Simplify the expression:

\[\begin(align) & \sqrt(32)\cdot \sqrt(4)=\sqrt(((2)^(5)))\cdot \sqrt(((2)^(2)))= \sqrt(((\left(((2)^(5)) \right))^(3))\cdot ((\left(((2)^(2)) \right))^(4) ))= \\ & =\sqrt(((2)^(15))\cdot ((2)^(8)))=\sqrt(((2)^(23))) \\ \end( align)\]

Here, many would be confused by the fact that the output turned out to be an irrational number. Yes, it happens: we could not completely get rid of the root, but at least we significantly simplified the expression.

Example 3. Simplify the expression:

\[\begin(align) & \sqrt(a)\cdot \sqrt(((a)^(4)))=\sqrt(((a)^(3))\cdot ((\left((( a)^(4)) \right))^(6)))=\sqrt(((a)^(3))\cdot ((a)^(24)))= \\ & =\sqrt( ((a)^(27)))=\sqrt(((a)^(3\cdot 9)))=\sqrt(((a)^(3))) \end(align)\]

This is what I would like to draw your attention to. There are two points here:

Under the root is not a specific number or degree, but the variable $a$. At first glance, this is a bit unusual, but in reality, when solving mathematical problems, you will most often have to deal with variables.
In the end, we managed to “reduce” the root exponent and degree in the radical expression. This happens quite often. And this means that it was possible to significantly simplify the calculations if you do not use the main formula.

For example, you could do this:

\[\begin(align) & \sqrt(a)\cdot \sqrt(((a)^(4)))=\sqrt(a)\cdot \sqrt(((\left(((a)^( 4)) \right))^(2)))=\sqrt(a)\cdot \sqrt(((a)^(8))) \\ & =\sqrt(a\cdot ((a)^( 8)))=\sqrt(((a)^(9)))=\sqrt(((a)^(3\cdot 3)))=\sqrt(((a)^(3))) \ \ \end(align)\]

In fact, all transformations were performed only with the second radical. And if you do not paint in detail all the intermediate steps, then in the end the amount of calculations will significantly decrease.

In fact, we have already encountered a similar task above when solving the $\sqrt(5)\cdot \sqrt(3)$ example. Now it can be written much easier:

\[\begin(align) & \sqrt(5)\cdot \sqrt(3)=\sqrt(((5)^(4))\cdot ((3)^(2)))=\sqrt(( (\left(((5)^(2))\cdot 3 \right))^(2)))= \\ & =\sqrt(((\left(75 \right))^(2))) =\sqrt(75). \end(align)\]

Well, we figured out the multiplication of the roots. Now consider the inverse operation: what to do when there is a work under the root?