How to fold squares. Rule for adding square roots. Example of calculating an approximate value

Root formulas. Properties of square roots.

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IN 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. Formulas for square roots surprisingly little. 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:

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Greetings, cats! Last time we discussed in detail what roots are (if you don’t remember, I recommend reading it). The main takeaway from that lesson: there is only one universal definition of roots, which is what 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, they can become fatal in the exam) and we will practice properly. So stock up on popcorn, get comfortable, and let's get started. :)

You haven't smoked it yet either, have you?

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

1. First we will look at the rules of multiplication. Cap seems to be hinting: this is when there are two roots, between them there is a “multiply” sign - and we want to do something with it.
2. Then let's look at the opposite situation: there is one big root, but we were eager to represent it as a product of two simpler roots. Why is this necessary, is a separate question. We will only analyze the algorithm.

For those who can’t wait to immediately move on to the second part, you are welcome. Let's start with the rest in order.

Basic Rule of Multiplication

Let's start with the simplest thing - classic square roots. The same ones that are denoted by $\sqrt(a)$ and $\sqrt(b)$. Everything is obvious to them:

Multiplication rule. To multiply one Square root on the other, 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 root factors exist, then the product also exists.

Examples. Let's look at 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 ourselves would have extracted the roots of 25 and 4 without any new rules, then things get tough: $\sqrt(32)$ and $\sqrt(2)$ are not considered by themselves, but their product turns out to be a perfect square, so its root is equal to a rational number.

I would especially like to highlight the last line. There, both radical expressions are fractions. Thanks to the product, many factors are canceled, and the entire expression turns into an adequate number.

Of course, everything won't always be so beautiful. Sometimes there will be a complete mess under the roots - it’s not clear what to do with it and how to transform it after multiplication. A little later, when you start studying irrational equations and inequalities, there will be all sorts of variables and functions. And very often, problem writers count on the fact that you will discover some canceling terms or factors, after which the problem will be simplified many times over.

In addition, it is not at all necessary to multiply exactly two roots. You can multiply three, four, or even ten at once! 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 note on the second example. As you can see, in the third factor under the root there is a decimal fraction - 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 symbol). This will save you a lot of time and nerves in the future.

But this 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've sorted out the square roots. What to do with cubic ones? Or even 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, and then write the result under one radical.

In general, nothing complicated. Except that the amount of calculations may be greater. 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 and as a result we get the product of the numbers 625 and 25 in the denominator. This is quite big number- Personally, I can’t calculate right off the bat what it equals.

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

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

Such “machinations” can save you a lot of time on the exam or test work, so remember:

Don't rush to multiply numbers using radical expressions. First, check: what if the exact degree of any expression is “encrypted” there?

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

However, all this is baby talk compared to what we will study now.

Multiplying roots with different exponents

Okay, now we can multiply roots with the same indicators. What if the indicators are different? Let's say, how to 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:

Rule for multiplying roots. To multiply $\sqrt[n](a)$ by $\sqrt[p](b)$, it is enough to perform 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 note that we will return to a little later.

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

\[\begin(align) & \sqrt(3)\cdot \sqrt(2)=\sqrt(((3)^(4))\cdot ((2)^(3)))=\sqrt(81 \cdot 8)=\sqrt(648); \\ & \sqrt(2)\cdot \sqrt(7)=\sqrt(((2)^(5))\cdot ((7)^(2)))=\sqrt(32\cdot 49)= \sqrt(1568); \\ & \sqrt(5)\cdot \sqrt(3)=\sqrt(((5)^(4))\cdot ((3)^(2)))=\sqrt(625\cdot 9)= \sqrt(5625). \\ \end(align)\]

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. :)

Why must radical expressions be non-negative?

Of course you can be like school teachers and smartly quote the textbook:

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

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

So now I’ll 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 easily raise the radical expression to any natural degree$k$ - in this case, the root exponent will have to be multiplied by the same power. Therefore, we can easily reduce any roots to a common exponent, and then multiply them. 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 sharply limits the use 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). Now let’s perform the reverse transformation: “reduce” the two in the exponent and power. After all, any equality can be read both from left to right and from 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 it turns out to be some kind of crap:

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

This cannot happen, 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:

1. To hit the wall and state that mathematics is a stupid science, where “there are some rules, but these are imprecise”;
2. Introduce additional restrictions under which the formula will become 100% working.

In the first option, we will have to constantly catch “non-working” cases - it’s difficult, time-consuming and generally ugh. Therefore, mathematicians preferred the second option. :)

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

Therefore, let us formulate one more rule, which generally applies to all actions with roots:

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

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

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

Do you 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 the minus, then you can square/remove until you’re blue in the face - the number will remain negative. :)

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

1. Remove all the negatives from the radicals. Minuses exist only in 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 indicators of the roots are the same, we simply multiply the radical 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.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 roots are the same and odd, the only problem is that the second factor is negative. We take this minus out of the picture, after which everything is easily calculated.

Example 2: Simplify the expression:

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

Many here would be confused by what happened at the end irrational number. Yes, it happens: we couldn’t completely get rid of the root, but at least we significantly simplified the expression.

Example 3: Simplify the expression:

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

I would like to draw your attention to this task. There are two points here:

1. The root is not a specific number or power, but the variable $a$. At first glance, this is a little unusual, but in reality, when solving mathematical problems, you most often have to deal with variables.
2. In the end, we managed to “reduce” the radical indicator and the degree in radical expression. This happens quite often. And this means that it was possible to significantly simplify the calculations if you did not use the basic 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 describe in detail all the intermediate steps, then in the end the amount of calculations will be significantly reduced.

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

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

Well, we've sorted out the multiplication of roots. Now let's consider the reverse operation: what to do when there is a product under the root?

The square root of a number x is a number a, which when multiplied by itself gives the number x: a * a = a^2 = x, ?x = a. As with any numbers, you can perform arithmetic operations of addition and subtraction with square roots.

Instructions

1. First, when adding square roots, try to extract those roots. This will be acceptable if the numbers under the root sign are perfect squares. Let's say the given expression is ?4 + ?9. The first number 4 is the square of the number 2. The second number 9 is the square of the number 3. Thus it turns out that: ?4 + ?9 = 2 + 3 = 5.

2. If there are no complete squares under the root sign, then try moving the multiplier of the number from under the root sign. Let's say, let's say the expression is given?24 +?54. Factor the numbers: 24 = 2 * 2 * 2 * 3, 54 = 2 * 3 * 3 * 3. The number 24 has a factor of 4, the one that can be transferred from under the square root sign. In the number 54 there is a factor of 9. Thus, it turns out that: ?24 + ?54 = ?(4 * 6) + ?(9 * 6) = 2 * ?6 + 3 * ?6 = 5 * ?6. In this example, as a result of removing the multiplier from under the root sign, it was possible to simplify the given expression.

3. Let the sum of 2 square roots be the denominator of a fraction, say A / (?a + ?b). And let your task be “to get rid of irrationality in the denominator.” Then you can use the next method. Multiply the numerator and denominator of the fraction by the expression ?a – ?b. Thus, the denominator will contain the abbreviated multiplication formula: (?a + ?b) * (?a – ?b) = a – b. By analogy, if the denominator contains the difference between the roots: ?a – ?b, then the numerator and denominator of the fraction must be multiplied by the expression ?a + ?b. For example, let the fraction 4 / (?3 + ?5) = 4 * (?3 – ?5) / ((?3 + ?5) * (?3 – ?5)) = 4 * (?3 – ?5) / (-2) = 2 * (?5 – ?3).

4. Consider a more complex example of getting rid of irrationality in the denominator. Let the fraction 12 / (?2 + ?3 + ?5) be given. You need to multiply the numerator and denominator of the fraction by the expression?2 + ?3 – ?5:12 / (?2 + ?3 + ?5) = 12 * (?2 + ?3 – ?5) / ((?2 + ?3 + ?5) * (?2 + ?3 – ?5)) = 12 * (?2 + ?3 – ?5) / (2 * ?6) = ?6 * (?2 + ?3 – ?5) = 2 * ?3 + 3 * ?2 – ?30.

5. And finally, if you only need an approximate value, you can calculate the square roots using a calculator. Calculate the values ​​separately for the entire number and write it down to the required precision (say, two decimal places). And after that, perform the required arithmetic operations, as with ordinary numbers. Let's say, let's say you need to find out the approximate value of the expression?7 + ?5? 2.65 + 2.24 = 4.89.

Video on the topic

Note!
In no case can square roots be added as primitive numbers, i.e. ?3 + ?2 ? ?5!!!

Helpful advice
If you are factoring a number in order to move the square from under the root sign, then perform the reverse check - multiply all the resulting factors and get the original number.

In mathematics, any action has its opposite pair - in essence, this is one of the manifestations of the Hegelian law of dialectics: “the unity and struggle of opposites.” One of the actions in such a “pair” is aimed at increasing the number, and the other, its opposite, is aimed at decreasing it. For example, the opposite of addition is subtraction, and division is the opposite of multiplication. Exponentiation also has its own dialectical opposite pair. We are talking about extracting the root.

To extract the root of such and such a power from a number means to calculate which number must be raised to the appropriate power in order to end up with a given number. The two degrees have their own separate names: the second degree is called “square”, and the third is called “cube”. Accordingly, it is nice to call the roots of these powers square and cubic roots. Actions with cube roots are a topic for a separate discussion, but now let's talk about adding square roots.

Let's start with the fact that in some cases it is easier to first extract square roots and then add the results. Suppose we need to find the value of the following expression:

After all, it’s not at all difficult to calculate that the square root of 16 is 4, and of 121 is 11. Therefore,

√16+√121=4+11=15

However, this is the simplest case - here we're talking about about perfect squares, i.e. about those numbers that are obtained by squaring integers. But this doesn't always happen. For example, the number 24 is not a perfect square (there is no integer that, when raised to the second power, would result in 24). The same applies to a number like 54... What if we need to add the square roots of these numbers?

In this case, we will receive in the answer not a number, but another expression. The maximum we can do here is to simplify the original expression as much as possible. To do this, you will have to take out the factors from under the square root. Let's see how this is done using the numbers mentioned above as an example:

To begin with, let's factor 24 into factors so that one of them can easily be extracted as a square root (i.e., so that it is a perfect square). There is such a number - it is 4:

Now let's do the same with 54. In its composition, this number will be 9:

Thus, we get the following:

√24+√54=√(4*6)+ √(9*6)

Now let’s extract the roots from what we can extract them from: 2*√6+3*√6

There is a common factor here that we can take out of brackets:

(2+3)* √6=5*√6

This will be the result of addition - nothing more can be extracted here.

True, you can resort to using a calculator - however, the result will be approximate and with a huge number of decimal places:

√6=2,449489742783178

Gradually rounding it up, we get approximately 2.5. If we would still like to bring the solution to the previous example to its logical conclusion, we can multiply this result by 5 - and we will get 12.5. It is impossible to obtain a more accurate result with such initial data.

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

Above square roots ( √x), like other numbers, you can perform arithmetic operations such as subtraction and addition. To subtract and add roots, they need to 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, it is necessary to make a reduction. It consists in taking the coefficients of similar terms with the signs of the corresponding terms, then putting them in brackets and deducing the common root outside the brackets of the factor. The coefficient we obtained is simplified according to the usual rules.

Step 1: Extracting Square Roots

Firstly, 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, we can obtain the following equality: √4 + √9 = 2 + 3 = 5 .
That's it, the example is solved. But it doesn’t always happen that easily.

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

If there are no perfect squares under the root sign, you can try to remove the multiplier of the number from under the root sign. For example, let's take the expression √24 + √54 .

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

Considering this example, we obtain the removal of the multiplier 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 the fraction, for example, A/(√a + √b).
Now we are faced with the task of “getting rid of irrationality in the denominator.”
Let's take advantage 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 has a root difference: √a - √b, the numerator and denominator of the fraction are multiplied by the expression √a + √b.

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

Example of complex denominator reduction

Now let's consider enough complex example getting rid of irrationality in the denominator.

For example, let's take a fraction: 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 the square roots. The value is calculated separately for each number and written down with the required accuracy, which is determined by the number of decimal places. Next, all the required operations are performed, as with ordinary numbers.

Example of calculating an approximate value

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 like prime numbers, this is completely unacceptable. That is, if we add the square root of five and the square root of three, we cannot get the square root of eight.

Helpful advice: if you decide to factor a number, in order to derive the 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 that was originally given to us.