How to subtract a root from a root. How to add square roots

The library of Alexander Sergeevich Pushkin's works is very rich. It contains works of various genres and different topics. Literary critics divide all the work of the poet into several periods. There are five of them in total, and each of them is associated with a specific event in Pushkin's life: graduation from the Lyceum, southern exile, and others.

To the question: "What became the subject of Alexander Sergeevich's lyrics?" - it is impossible to answer unambiguously.

He wrote about love, and friendship, and about the Motherland, touched upon, among other things, philosophical themes. It is quite possible to say that everything became the subject of his lyrics.

But, probably, the main and main theme for the poet was the theme of love, which he sang, and at the very beginning of his work he elevated and elevated to the rank of the most valuable human feelings, as, for example, in his poem "Love alone is the fun of a cold life":

A hundred times blessed, who in his youth is charming

This quick moment will catch on the fly;

Who to the joys and bliss of the unknown

Bashful beauty will bow!

But gradually, with the maturation and development of his work, the poet rethinks this topic. He starts giving great attention feelings and experiences of a woman, as well as enjoy even the sadness of love:

I'm sad and easy; my sadness is light;

My sorrow is full of you...

Another direction in Pushkin's work is the theme of friendship. The works on this topic are mainly devoted to the friends of the poet's lyceum time: I. Pushchin, A. Delvig, and V. Küchelbecker. Friendship in his youth embodied carelessness and joy for Pushkin.

The theme of friendship, like the theme of love, is gradually evolving. The writer begins to see in her tragedy, sadness, disappointment from the loss of close friends. Such motifs are especially acute in his work "The Twelfth of October":

I am sad: there is no friend with me ...

I drink alone, and on the banks of the Neva

My friends are calling me...

But how many of you feast there too?

Who else have you missed?

The next important and high-profile topic in Pushkin's lyrics became the theme of freedom. In many works of the poet, one can see the motives of love of freedom, the desire for restriction absolute power king, for example, in the ode "Liberty":

Masters! you crown and throne

The law gives, not nature;

You stand above the people

But the eternal Law is above you.

Alexander Sergeevich in it addresses the authorities, in the lines there is a clear call to limit the powers of the tsar by the Law, that is, by the Constitution.

Later, the author departs from a strictly political understanding of freedom and shows interest in the freedom of a simple Russian person. That is, this topic is also evolving in its own way. This is clearly seen in the poem "The Village":

I see, my friends! oppressed people

And slavery, fallen at the behest of the king...

The apogee of the hymn to freedom, already personal, is the work "From Pindemonti", where there is a line:

Do not bend either conscience, or thoughts, or neck ...

Of course, speaking of Pushkin's work, one cannot avoid one of the deepest philosophical themes, the theme of the poet and poetry. Alexander Sergeevich was aware that the poet is alone in society and can often be misunderstood, that the noise and praise of the crowd is only periodic and fickle, temporary. This is very clear in one of his poems:

Poet! Do not value the love of the people.

enthusiastic praise will pass minute noise;

Another of the works on this topic was "Monument". It sounds the belief that the poet's work is immortal, that it will remain in the hearts of his admirers, and that the poet himself will remain alive after death thanks to his creations, which is confirmed by the lines:

No, all of me will not die - the soul is in the cherished lyre

My ashes will survive and decay will flee...

The lyrics of the great Alexander Sergeevich do not lose their relevance over the years, because the author touched on the most vital and pressing topics even for our days, eternal themes, in each of which there is a gradual evolution of thoughts, feelings lyrical hero. Creativity, Pushkin's lyrics developed along with him, with his spiritual world, his view of everything around him.

Effective preparation for the exam (all subjects) -

Attention!
There are additional
material in Special Section 555.
For those who are strong "not very. »
And for those who “very even. "")

In the previous lesson, we figured out what a square root is. It's time to figure out what are formulas for roots, what are root properties and what can be done about it all.

Root Formulas, Root Properties, and Rules for Actions with Roots are essentially the same thing. Formulas for square roots surprisingly little. Which, of course, pleases! Rather, you can write a lot of all sorts of formulas, but only three are enough for practical and confident work with roots. Everything else flows from these three. Although many stray in the three formulas of the roots, yes.

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

I remind you (from the previous lesson): a and b are non-negative numbers! Otherwise, the formula makes no sense.

This is property of roots , as you can see, simple, short and harmless. But with this root formula, you can do a lot of useful things! Let's take a look at examples all these useful things.

Useful thing first. This formula allows us multiply roots.

How to multiply roots?

Yes, very simple. Straight to the formula. For example:

It would seem that they have multiplied, so what? Is there a lot of joy? I agree, a little. But how do you like this example?

Roots are not exactly extracted from factors. And the result is great! Already better, right? Just in case, I will inform you that there can be as many multipliers as you like. The root multiplication formula still works. For example:

So, with multiplication, everything is clear why this is needed property of roots- is also understandable.

Useful thing the second. Entering a number under the sign of the root.

How to enter a number under the root?

Let's say we have this expression:

Is it possible to hide the deuce inside the root? Easily! If you make a root out of two, the formula for multiplying the roots will work. And how to make a root from a deuce? Yes, that's not a question either! The double is square root of four!

The root, by the way, can be made from any non-negative number! This will be the square root of the square of this number. 3 is the root of 9. 8 is the root of 64. 11 is the root of 121. Well, and so on.

Of course, there is no need to paint in such detail. Except, for starters. It is enough to realize that any non-negative number multiplied by the root can be brought under the root. But don't forget! - under the root this number will become square himself. This action - entering a number under the root - can also be called multiplying a number by the root. In general terms, one can write:

The process is simple, as you can see. Why is she needed?

Like any transformation, this procedure expands our possibilities. Opportunities to turn a cruel and uncomfortable expression into a soft and fluffy one). Here's a simple one for you example:

As you can see root property, which makes it possible to introduce a factor under the sign of the root, is quite suitable for simplification.

In addition, adding a multiplier under the root makes it easy and simple to compare values various roots. Without any calculation and calculator! The third useful thing.

How to compare roots?

This skill is very important in solid missions, when unlocking modules, and other cool things.

Compare these expressions. Which one is more? Without a calculator! Each with a calculator. uh-uh. In short, everyone can do it!)

You don't say so right away. And if you enter numbers under the sign of the root?

Remember (suddenly, did not know?): if the number under the sign of the root is greater, then the root itself is greater! Hence the immediately correct answer, without any complicated calculations and calculations:

It's great, right? But that's not all! Recall that all formulas work both from left to right and from right to left. We have so far used the formula for multiplying roots from left to right. Let's run this root property backwards, from right to left. Like this:

And what's the difference? Does it give you something!? Certainly! Now you will see for yourself.

Suppose we need to extract (without a calculator!) The square root of the number 6561. Some people at this stage will fall in an unequal struggle with the task. But we are stubborn, we do not give up! Useful thing fourth.

How to extract roots from large numbers?

We recall the formula for extracting roots from a product. The one I posted above. But where is our work? We have a huge number 6561 and that's it. Yes, there is no art. But if we need it, we let's do! Let's factor this number. We have the right.

First, let's figure out what this number is divisible by exactly? What, you don't know!? Did you forget the signs of divisibility!? In vain. Go to Special Section 555, the topic is “Fractions”, there they are. This number is divisible by 3 and 9. Because the sum of the digits (6+5+6+1=18) is divisible by these numbers. This is one of the signs of divisibility. We do not need to divide by three (now you will understand why), but we will divide by 9. At least in a corner. We get 729. So we found two factors! The first one is a nine (we chose it ourselves), and the second one is 729 (it turned out like that). You can already write:

Get the idea? Let's do the same with the number 729. It is also divisible by 3 and 9. Again, we don’t divide by 3, we divide by 9. We get 81. And we know this number! We write down:

Everything turned out easy and elegant! The root had to be removed piece by piece, well, okay. This can be done with any big numbers. Multiply them, and go!

By the way, why didn’t you have to divide by 3, did you guess? Yes, because the root of three is not exactly extracted! It makes sense to decompose into such factors that at least one root can be well extracted. It's 4, 9, 16 well, and so on. Divide your huge number by these numbers in turn, you see, and you're lucky!

But not necessarily. Maybe not lucky. Let's say the number 432, when factored and using the root formula for the product, will give the following result:

Well, okay. We've simplified the expression anyway. In mathematics, it is customary to leave the most small number of the possible. In the process of solving, everything depends on the example (maybe everything is reduced without simplification), but in the answer it is necessary to give a result that cannot be further simplified.

By the way, do you know what we have done with the root of 432 now?

We taken out factors from under the sign of the root ! That's what this operation is called. And then the task will fall - " take the factor out from under the sign of the root"But the men don't even know.) Here's another use for you root properties. Useful thing fifth.

How to take the multiplier out from under the root?

Easily. Factorize the root expression and extract the roots that are extracted. We look:

Nothing supernatural. It is important to choose the right multipliers. Here we have decomposed 72 as 36 2. And everything turned out well. Or they could have decomposed it differently: 72 = 6 12. So what!? Neither from 6 nor from 12 the root is extracted. What to do?!

It's OK. Or look for other decomposition options, or continue to lay out everything to the stop! Like this:

As you can see, everything worked out. This, by the way, is not the fastest, but the most reliable way. Decompose the number into the smallest factors, and then collect the same ones in piles. The method is also successfully applied when multiplying inconvenient roots. For example, you need to calculate:

Multiply everything - you get a crazy number! And then how to extract the root from it ?! Multiply again? No, we don't need extra work. We immediately decompose into factors and collect the same in piles:

That's all. Of course, it is not necessary to lay out to the stop. Everything is determined by your personal abilities. Brought the example to a state where everything is clear to you so you can already count. The main thing is not to make mistakes. Not a man for mathematics, but mathematics for a man!)

Let's apply knowledge to practice? Let's start with a simple one:

Rule for adding square roots

Properties of square roots

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

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

Proof. Let us introduce the following notation:
We need to prove that for negative numbers x, y, z, x = yz.

So x 2 = ab, y 2 = a, z 2 = b. Then x 2 \u003d y 2 z 2, i.e. x 2 \u003d (yz) 2.

If a squares two non-negative numbers are equal, then the numbers themselves are equal, which means that from the equality x 2 \u003d (yz) 2 it follows that x \u003d yz, and this was required to be proved.

Let's bring short note proof of the theorem:

Remark 1. The theorem remains valid for the case when the radical expression is the product of more than two non-negative factors.

Remark 2. Theorem 1 can be written using the “if. , then” (as is customary for theorems in mathematics). We give the corresponding formulation: if a and b are non-negative numbers, then the equality .

This is how we formulate the following theorem.

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

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

Example 1. Calculate .
Decision. Using the first property square roots(Theorem 1), we obtain

Remark 3. Of course, this example can be solved in another way, especially if you have a calculator 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.

Remark 4. In the first method, we carried out head-on calculations. The second way is more elegant:
we applied formula a 2 - b 2 \u003d (a - b) (a + b) and used the property of square roots.

Remark 5. Some "hotheads" sometimes offer the following "solution" to Example 3:

This, of course, is not true: you see - the result is not the same as in our example 3. The fact is that there is no property as no and properties There are only properties concerning the multiplication and division of square roots. Be careful and careful, do not take wishful thinking.

Example 4. Calculate: a)
Decision. Any formula in algebra is used not only "from right to left", but also "from left to right". So, the first property of square roots means that, if necessary, it can be represented as , and vice versa, which can be replaced by the expression The same applies to the second property of square roots. With this in mind, let's solve the proposed example.

Concluding the section, we note one more rather simple and at the same time important property:
if a > 0 and n - natural number , then

Example 5 Calculate , without using a table of squares of numbers and a calculator.

Decision. Let's decompose the root number into prime factors:

Remark 6. This example could be solved in the same way as the similar example in § 15. It is easy to guess that the answer will be “80 with a tail”, since 80 2 2 . Let's find the "tail", i.e. the last digit of the desired number. So far we know that if the root is extracted, then the answer can be 81, 82, 83, 84, 85, 86, 87, 88 or 89. Only two numbers need to be checked: 84 and 86, since only they, when squared, will give as a result four-digit a number ending in 6, i.e. the same digit that ends with the number 7056. We have 84 2 \u003d 7056 - this is what we need. Means,

Mordkovich A. G., Algebra. Grade 8: Proc. for general education institutions. - 3rd ed., finalized. - M.: Mnemosyne, 2001. - 223 p.: ill.

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How to add square roots

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 them simplest form- if there are similar ones between them, it is necessary 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 output common root 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 a full squares is not under the root sign, you can try to take out the multiplier of the number 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 .

In list 24 we have a multiplier 4 , it can be taken out from under the square root sign. In list 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 in the following way: 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 let's consider enough complex example 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 .

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 you add square roots, 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 final result of this mathematical calculation should be the number we were originally given.

Action with roots: addition and subtraction

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

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

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

6 50 — 2 8 + 5 12

1. 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).
2. Then you need to extract the root from square number and write the resulting value before the root sign. Please note that the second factor is entered under the root sign.
3. After the simplification process, it is necessary to underline the roots with the same radical expressions - only they can be added and subtracted.
4. 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!

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

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.

• 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 .
• 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.

As we can see, it is not possible to simplify the radical numbers, so we look for members with the same radical numbers in the example, 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 operations with fractions, you need to find a number that is divisible by each denominator, then bring the fractions to common denominator, then add the numerators and leave the denominators unchanged.

Properties of the arithmetic square root. Power of the arithmetic square root

Converting arithmetic square roots. Conversion of arithmetic square roots

To extract square root of a polynomial, it is necessary to calculate the polynomial and extract the root from the resulting number.

Attention! It is impossible to extract the root from each term (reduced and subtracted) separately.

Shchob to win square root of polynomial, the requirement is to calculate the rich term and from the subtracted number to take the root.

Respect! It is impossible to extract the root from the skin supplement (changed and visible) OKremo.

To extract the square root of the product (quotient), you can calculate the square root of each factor (dividend and divisor), and take the resulting values ​​​​by the product (quotient).

To win the square root of the dobutka (parts), you can calculate the square root of the skin multiplier (divided and dilnik), and remove the value by taking a supplementary (frequent).

To take the square root of a fraction, you need to extract the square root of the numerator and denominator separately, and leave the resulting values ​​as a fraction or calculate as a quotient (if possible by condition).

To win the square root of the fraction, you need to take the square root of the number book and the banner of the okremo, and deprive the value of the fraction with a fraction, or count it as a part (as it is possible for the mind).

A factor can be taken out from under the root sign and a factor can be introduced under the root sign. When a factor is taken out, the root is extracted from it, and when introduced, it is raised to the corresponding power.

The 3rd root sign can be multiplied and the root sign can be multiplied. With the fault of the multiplier, the roots are twisted, and with the introduction, the roots are built at the higher feet.

Examples. Apply

To convert the sum (difference) of square roots, you need to bring the root expressions to one base of the degree, if possible, extract the roots from the degrees and write them before the signs of the roots, and the remaining square roots with the same root expressions can be added, for which the coefficients are added before the sign root and add the same square root.

In order to remake the sum (cost) of square roots, it is necessary to bring the root roots to one of the bases of the step, as it is possible, to take the root of the steps and write them down before the signs of the roots, and the solution of the square roots with the same root words, which I can put together for what I can add and add the same square root.

We bring all radical expressions to base 2.

From an even degree, the root is extracted completely, from an odd degree, the root of the base in degree 1 is left under the sign of the root.

We give similar integers and add the coefficients with the same roots. We write the binomial as the product of a number and the binomial of the sum.

Bring all sub-roots of the virazi to the base 2.

From the paired stage, the roots are drawn in a row, from the unpaired stage, the roots of the base in stage 1 are filled under the sign of the root.

It is suggested that similar numbers and coefficients are added to the same roots. We write the binomial as a supplement of the number i of the sumi binomial.

We bring the radical expressions to the smallest base or the product of powers with the smallest bases. We extract the root from even degrees of radical expressions, leave the remainders in the form of a base of a degree with an indicator of 1 or the product of such bases under the sign of the root. We give similar terms (add the coefficients of the same roots).

We lead the root of the virazi to the smallest base or the addition of steps with the smallest bases. From the steamy steps under the roots of the viraz, the roots are taken, the surplus at the base of the step with the indicator 1 or the addition of such bases is filled under the sign of the root. We suggest similar terms (we add up the coefficients of the same roots).

Let's replace the division of fractions with multiplication (with the replacement of the second fraction by the reciprocal). Multiply the numerators and denominators separately. Under each sign of the root, we highlight the degrees. Let's cut same multipliers in the numerator and denominator. We extract roots from even powers.

We replace the division of fractions with a multiplication (with the replacement of another fraction with a return). Multiply okremo numbers and banners of fractions. Steps are visible under the skin sign of the root. We will speed up the same multipliers in the number book and banner. Blame the root of the twin steps.

To compare two square roots, their radical expressions must be brought to a degree with the same base, then the more the degree of the radical expression is shown, the more value square root.

In this example, radical expressions cannot be reduced to one base, since the base is 3 in the first, and 3 and 7 in the second.

The second way to compare is to add the root factor to the root expression and compare numerical values rooted expressions. For a square root, the larger the root expression, the greater the value of the root.

To match two square roots, their sub-roots must be brought to a level with the same basis, while the greater the indicator of the degree of the sub-root of the virus, the greater the value of the square root.

In this case, it is not possible to bring to one basis the root roots of the virazi, since in the first one the basis is 3, and in the other - 3 and 7.

Another way to equalize is to add the root coefficient to the root virase and equalize the numerical values ​​of the root virase. The square root has more sub-root viraz, the more value of the root.

Using the distributive law of multiplication and the rule for multiplying roots with the same exponents (in our case, square roots), we obtained the sum of two square roots with the product under the root sign. We decompose 91 into prime factors and take the root out of brackets with common radical factors (13 * 5).

We have obtained the product of a root and a binomial, in which one of the monomials is an integer (1).

Vikoristovuyuchi rozpodilny law of multiplication and the rule of multiplication of roots with the same indicators (in our case - square roots), took the sum of two square roots with an additional root under the sign of the root. We can lay out 91 multipliers in simple terms and take the root for the arches from the root multipliers (13 * 5).

We took the addition of a root and a binary, which has one of the mononomials in the whole number (1).

Example 9:

In the radical expressions, we select by factors the numbers from which we can extract the whole square root. We extract the square roots from the powers and put the numbers by the coefficients of the square roots.

The terms of this polynomial have a common factor √3, which can be taken out of the brackets. Let us present similar terms.

In sub-root virases, it is seen as multipliers of the number, from which one can take the square root. We blame the square roots of the steps and put the numbers by the coefficients of the square roots.

The terms of this polynomial have a common multiplier √3, which can be blamed for the arms. We suggest similar additions.

The product of the sum and difference of two same bases(3 and √5) using the abbreviated multiplication formula can be written as the difference of the squares of the bases.

The square root squared is always equal to the radical expression, so we will get rid of the radical (root sign) in the expression.

Dobutok sum and difference of two identical bases (3 і √5) from the formula of fast multiplication can be written as a difference of square bases.

The square root of the square zavzhd is equal to the sub-root virase, so we will call the radical (root sign) of the virase.

Back to school. Addition of roots

Nowadays, modern electronic computers calculation of the root of a number is not represented challenging task. For example, √2704=52, any calculator will calculate this for you. Fortunately, the calculator is not only in Windows, but also in an ordinary, even the simplest, phone. True, if suddenly (with a small degree of probability, the calculation of which, by the way, includes the addition of roots) you find yourself without available funds, then, alas, you will have to rely only on your brains.

Mind training never fails. Especially for those who do not work with numbers so often, and even more so with roots. Adding and subtracting roots is a good workout for a bored mind. And I will show you the addition of roots step by step. Examples of expressions can be the following.

The equation to be simplified is:

This is an irrational expression. In order to simplify it, you need to reduce all radical expressions to general view. We do it in stages:

The first number can no longer be simplified. Let's move on to the second term.

3√48 we factorize 48: 48=2×24 or 48=3×16. The square root of 24 is not an integer, i.e. has a fractional remainder. Since we need an exact value, approximate roots are not suitable for us. The square root of 16 is 4, take it out from under the root sign. We get: 3×4×√3=12×√3

Our next expression is negative, i.e. written with a minus sign -4×√(27.) Factoring 27. We get 27=3×9. We do not use fractional factors, because it is more difficult to calculate the square root from fractions. We take out 9 from under the sign, i.e. calculate the square root. We get following expression: -4×3×√3 = -12×√3

The next term √128 calculates the part that can be taken out from under the root. 128=64×2 where √64=8. If it makes it easier for you, you can represent this expression like this: √128=√(8^2×2)

We rewrite the expression with simplified terms:

Now we add the numbers with the same radical expression. You cannot add or subtract expressions with different radical expressions. The addition of roots requires compliance with this rule.

We get the following answer:

√2=1×√2 - I hope that it is customary in algebra to omit such elements will not be news to you.

Expressions can be represented not only by square roots, but also by cube or nth roots.

The addition and subtraction of roots with different exponents, but with an equivalent root expression, occurs as follows:

If we have an expression like √a+∛b+∜b, then we can simplify this expression like this:

12√b4 +12×√b3=12×√b4 + b3

We brought two such members to general indicator root. The property of the roots was used here, which says: if the number of the degree of the radical expression and the number of the root exponent are multiplied by the same number, then its calculation will remain unchanged.

Note: exponents are added only when multiplied.

Consider an example where fractions are present in an expression.

Let's solve it step by step:

5√8=5*2√2 - we take out the extracted part from under the root.

If the body of the root is represented by a fraction, then often this fraction will not change if the square root of the dividend and divisor is taken. As a result, we have obtained the equality described above.

Here is the answer.

The main thing to remember is that a root with an even exponent is not extracted from negative numbers. If an even degree radical expression is negative, then the expression is unsolvable.

The addition of the roots is possible only if the root expressions coincide, since they are like terms. The same applies to difference.

The addition of roots with different numerical exponents is carried out by reducing both terms to a common root degree. This law operates in the same way as reduction to a common denominator when adding or subtracting fractions.

If the radical expression contains a number raised to a power, then this expression can be simplified provided that there is a common denominator between the root and the exponent.

The square root of a product and a fraction

The square root of a is a number whose square is a. For example, the numbers -5 and 5 are the square roots of the number 25. That is, the roots of the equation x^2=25 are the square roots of the number 25. Now you need to learn how to work with the square root operation: study its basic properties.

The square root of the product

√(a*b)=√a*√b

The square root of the product of two non-negative numbers, is equal to the product square roots of these numbers. For example, √(9*25) = √9*√25 =3*5 =15;

It is important to understand that this property also applies to the case when the radical expression is the product of three, four, etc. non-negative multipliers.

Sometimes there is another formulation of this property. If a and b are non-negative numbers, then the following equality holds: √(a*b) =√a*√b. There is absolutely no difference between them, you can use either one or the other wording (which one is more convenient to remember).

The square root of a fraction

If a>=0 and b>0, then the following equality is true:

√(a/b)=√a/√b.

For example, √(9/25) = √9/√25 =3/5;

This property also has a different formulation, in my opinion, more convenient to remember.
The square root of the quotient is equal to the quotient of the roots.

It is worth noting that these formulas work both from left to right and from right to left. That is, if necessary, we can represent the product of the roots as the root of the product. The same goes for the second property.

As you can see, these properties are very convenient, and I would like to have the same properties for addition and subtraction:

√(a+b)=√a+√b;

√(a-b)=√a-√b;

But unfortunately such properties are square have no roots, and so cannot be done in calculations..

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Hello kitties! AT 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 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:

1. 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.
2. 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 to study irrational equations and inequalities, there will generally be all sorts of variables and functions. 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(i.e. containing at least one radical icon). This will save you a lot of time and nerves in the future.

But it was lyrical digression. Now consider more general case- when the root index 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 cube roots, get rid of decimal fraction and as a result we get the product of the numbers 625 and 25 in the denominator. This is quite big number- Personally, I don’t immediately consider what it is equal to.

Therefore, we simply selected the exact cube in the numerator and denominator, and then used one of 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 the exam or control work 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 very important note, 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 be like school teachers and cleverly quote the textbook:

The non-negativity requirement is related to different definitions roots of even and odd degree (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 radical expression to any natural degree$k$ - in this case, the root index will have to be multiplied by the same degree. 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 execute inverse transformation: "reduce" the deuce 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 $\sqrt(5) \gt 0$. So for even powers and negative numbers, our formula no longer works. After which we have two options:

1. To fight against the wall to state that mathematics is a stupid science, where “there are some rules, but this is inaccurate”;
2. Enter additional restrictions, at 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 root 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:

1. 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).
2. 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. 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 what the output turned out 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:

1. Under the root is not specific number or degree, and the variable is $a$. At first glance, this is a little unusual, but in reality, when solving math problems most often you will have to deal with variables.
2. 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?