Is it possible to add identical roots? Square root. The Comprehensive Guide (2019)

28.09.2019

Addition and subtraction of roots- one of the most common “stumbling blocks” for those taking mathematics (algebra) courses in high school. However, learning to correctly add and subtract them is very important, because examples on the sum or difference of roots are included in the program of the basic Unified State Exam in the discipline “mathematics”.

In order to master solving such examples, you need two things - to understand the rules, and also to gain practice. Having solved one or two dozen typical examples, the student will bring this skill to automatism, and then he will no longer have anything to fear on the Unified State Exam. It is recommended to start mastering arithmetic operations with addition, because adding them is a little easier than subtracting them.

What is a root

The easiest way to explain this is with an example square root. In mathematics there is a well-established term “squaring”. “Squaring” means multiplying a specific number by itself once.. For example, if you square 2, you get 4. If you square 7, you get 49. The square of 9 is 81. So the square root of 4 is 2, of 49 is 7, and of 81 is 9.

As a rule, teaching this topic in mathematics begins with square roots. In order to immediately determine it, the student high school must know the multiplication table by heart. Those who do not know this table firmly have to use hints. Usually the process of extracting the root square from a number is given in the form of a table on the covers of many school mathematics notebooks.

Roots are of the following types:

square;
cubic (or so-called third degree);
fourth degree;
fifth degree.

Addition rules

In order to successfully solve a typical example, it is necessary to keep in mind that not all root numbers can be stacked with each other. In order for them to be put together, they must be brought to a single pattern. If this is impossible, then the problem has no solution. Such problems are also often found in mathematics textbooks as a kind of trap for students.

Addition is not allowed in tasks when the radical expressions differ from each other. This can be illustrated with a clear example:

The student is faced with the task: add the square root of 4 and 9;
inexperienced student knowledgeable of the rules, usually writes: “root of 4 + root of 9 = root of 13.”
It is very easy to prove that this solution is incorrect. To do this, you need to find the square root of 13 and check whether the example is solved correctly;
using a microcalculator you can determine that it is approximately 3.6. Now all that remains is to check the solution;
root of 4=2, and root of 9=3;
The sum of the numbers “two” and “three” equals five. Thus, this solution algorithm can be considered incorrect.

If the roots have the same degree but different numeric expressions, it is taken out of brackets, and put into brackets sum of two radical expressions. Thus, it is already extracted from this amount.

Addition algorithm

In order to decide correctly simplest task, necessary:

Determine what exactly requires addition.
Find out whether it is possible to add values to each other, guided by existing rules in mathematics.
If they are not foldable, you need to transform them so that they can be folded.
Having carried out all the necessary transformations, you need to perform the addition and write down the finished answer. You can perform addition in your head or using a microcalculator, depending on the complexity of the example.

What are similar roots

To correctly solve an addition example, you must first think about how you can simplify it. To do this, you need to have basic knowledge of what similarity is.

The ability to identify similar ones helps to quickly solve similar addition examples, bringing them into a simplified form. To simplify a typical addition example, you need to:

Find similar ones and separate them into one group (or several groups).
Rewrite the existing example in such a way that the roots that have the same indicator follow each other clearly (this is called “grouping”).
Next, you should once again write the expression again, this time in such a way that similar ones (which have the same indicator and the same radical figure) also follow each other.

Once this is done, the simplified example is usually easy to solve.

In order to correctly solve any addition example, you need to clearly understand the basic rules of addition, as well as know what a root is and what it can be.

Sometimes such problems look very difficult at first glance, but usually they are easily solved by grouping similar ones. The most important thing is practice, and then the student will begin to “crack problems like nuts.” Addition of roots is one of the most important sections mathematics, so teachers should allocate enough time to study it.

Now at school curriculum something is happening that is not entirely clear. One good thing is that in mathematics everything remains unchanged. Working with roots, namely addition and subtraction, is not a very difficult operation. But some students experience certain difficulties.

And in this article we will look at the rules of how to add and subtract square roots.

You can subtract and add square roots if the condition is met that these roots have the same radical expressions. In other words, we can perform actions with 2√3 and 4√3, but not with 2√3 and 2√7. But you can carry out actions to simplify the radical expression in order to then lead them to roots that will have the same radical expressions. And only after that start adding or subtracting.

Theory of adding and subtracting square roots

The principle itself is very simple. And it will consist of three actions. We need to simplify the radical expression. Find the resulting identical radical expressions and add or subtract the roots.

How to simplify a radical expression

To do this, you need to expand the radical number so that it consists of two factors. The main condition. One of these numbers must be a square number (example: 25 or 9). After this action we extract the root of the given square number. And we write this number in front of our root, and under the root we have a second factor left.

For example, 6√50 – 2√8 + 5√12

6√50 = 6√(25 x 2) = (6 x 5)√2 = 30√2. Here we decompose 50 into two factors 25 and 2. Then we take the square root of 25 (we get the number 5) and take it out from under the root. Next we multiply 5 by 6 and get 30√2

2√8 = 2√(4 x 2) = (2 x 2)√2 = 4√2. IN given examples We decompose 8 into two numbers 4 and 2. We take the root from 4 and take the resulting number as the root and multiply it by the number that was already behind the root.

5√12 = 5√(4 x 3) = (5 x 2)√3 = 10√3. Here, as before, we decompose the number under the root into two numbers 4 and 3. We extract the root from 4. We take the resulting number as the root and multiply it by the number that was behind the root.

As a result, we transformed the equation 6√50 - 2√8 + 5√12 into this form 30√2 - 4√2 + 10√3

We emphasize roots that have the same radical expressions

In our example 30√2 - 4√2 + 10√3, we highlight 30√2 and 4√2 since these numbers have the same radical number 2.
If in your example there are several identical radical expressions. Underline the same ones with different lines.

Add or subtract our roots

Now we add or subtract numbers that have the same radical expressions. And what we leave at the root remains unchanged. The point is to show how many roots with certain radical expressions there are in a given equation.

In our example 30√2 - 4√2 + 10√3 we subtract 4 from 30 and get 26√2

The answer in our example will be like this. 26√2 + 10√3

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What is a mathematical root?

This action arose in opposition to exponentiation. Mathematics suggests two opposing operations. There is subtraction for addition. Multiplication is opposed to division. The inverse action of a degree is to extract the corresponding root.

If the degree is two, then the root will be square. It is the most common in school mathematics. It does not even have an indication that it is square, that is, the number 2 is not assigned next to it. The mathematical notation of this operator (radical) is presented in the figure.

Its definition flows smoothly from the action described. To extract the square root of a number, you need to find out what the radical expression will give when multiplied by itself. This number will be the square root. If we write this down mathematically, we get the following: x*x=x 2 =y, which means √y=x.

What actions can you perform with them?

At its core, a root is a fractional power with one in the numerator. And the denominator can be anything. For example, the square root has two. Therefore, all actions that can be performed with powers will also be valid for roots.

And the requirements for these actions are the same. If multiplication, division and exponentiation do not encounter difficulties for students, then adding roots, like subtracting them, sometimes leads to confusion. And all because I want to perform these operations without regard to the sign of the root. And this is where the mistakes begin.

What are the rules for adding and subtracting?

First you need to remember two categorical “don’ts”:

you cannot perform addition and subtraction of roots like prime numbers, that is, it is impossible to write radical expressions of the sum under one sign and perform mathematical operations with them;
You can't add and subtract roots from different indicators, such as square and cubic.

A clear example of the first prohibition: √6 + √10 ≠ √16, but √(6 + 10) = √16.

In the second case, it is better to limit ourselves to simplifying the roots themselves. And leave their amount in the answer.

Now to the rules

Find and group similar roots. That is, those who not only have the same numbers under the radical, but they themselves have the same indicator.
Perform the addition of the roots combined into one group in the first action. It is easy to implement because you only need to add the values that appear in front of the radicals.
Extract the roots of those terms in which the radical expression forms a whole square. In other words, do not leave anything under the sign of a radical.
Simplify radical expressions. To do this, you need to decompose them into prime factors and see if they give the square of any number. It is clear that this is true when we are talking about the square root. When the exponent is three or four, then the prime factors must give the cube or the fourth power of the number.
Remove from under the sign of the radical the factor that gives the whole power.
See if similar terms appear again. If yes, then perform the second step again.

In a situation where the task does not require exact value root, it can be calculated on a calculator. Round off the endless decimal fraction that appears in its window. Most often this is done to hundredths. And then perform all operations for decimal fractions.

This is all the information about how to add roots. The examples below will illustrate the above.

First task

Calculate the value of expressions:

a) √2 + 3√32 + ½ √128 – 6√18;

b) √75 - √147 + √48 - 1/5 √300;

c) √275 - 10√11 + 2√99 + √396.

a) If you follow the above algorithm, you can see that there is nothing for the first two actions in this example. But you can simplify some radical expressions.

For example, decompose 32 into two factors 2 and 16; 18 will be equal to the product of 9 and 2; 128 is 2 over 64. Given this, the expression will be written like this:

√2 + 3√(2 * 16) + ½ √(2 * 64) - 6 √(2 * 9).

Now you need to remove from under the radical sign those factors that give the square of the number. This is 16=4 2, 9=3 2, 64=8 2. The expression will take the form:

√2 + 3 * 4√2 + ½ * 8 √2 – 6 * 3√2.

We need to simplify the recording a little. To do this, multiply the coefficients before the root signs:

√2 + 12√2 + 4 √2 — 12√2.

In this expression, all terms turned out to be similar. Therefore, you just need to fold them. The answer will be: 5√2.

b) Similar to the previous example, adding roots begins with simplifying them. The radical expressions 75, 147, 48 and 300 will be represented in the following pairs: 5 and 25, 3 and 49, 3 and 16, 3 and 100. Each of them contains a number that can be taken out from under the root sign:

5√5 — 7√3 + 4√3 — 1/5 * 10√3.

After simplification, the answer is: 5√5 - 5√3. It can be left in this form, but it is better to take the common factor 5 out of brackets: 5 (√5 - √3).

c) And again factorization: 275 = 11 * 25, 99 = 11 * 9, 396 = 11 * 36. After removing the factors from under the root sign, we have:

5√11 - 10√11 + 2 * 3√11 + 6√11. After bringing similar terms we get the result: 7√11.

Example with fractional expressions

√(45/4) — √20 — 5√(1/18) — 1/6 √245 + √(49/2).

You will need to factor the following numbers: 45 = 5 * 9, 20 = 4 * 5, 18 = 2 * 9, 245 = 5 * 49. Similar to those already discussed, you need to remove the factors from under the root sign and simplify the expression:

3/2 √5 - 2√5 - 5/ 3 √(½) - 7/6 √5 + 7 √(½) = (3/2 - 2 - 7/6) √5 - (5/3 - 7 ) √(½) = - 5/3 √5 + 16/3 √(½).

This expression requires getting rid of irrationality in the denominator. To do this, you need to multiply the second term by √2/√2:

— 5/3 √5 + 16/3 √(½) * √2/√2 = — 5/3 √5 + 8/3 √2.

To complete the actions, you need to select the whole part of the factors in front of the roots. For the first one it is 1, for the second it is 2.

Greetings, cats! Last time we discussed in detail what roots are (if you don’t remember, I recommend reading it). Main conclusion 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, make yourself 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:

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.
Then let's look at the reverse situation: there is one big root, but we wanted to present it in the form of a simpler product of two 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 by another, you simply 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 would have extracted the roots of 25 and 4 ourselves 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, things won't always be so beautiful. Sometimes there will be complete crap 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 it was 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 large number- Personally, I can’t calculate right off the bat what it equals.

So 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. :)

Multiplying roots is easy

Why must radical expressions be non-negative?

Of course you can be like school teachers and with a smart look 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 power $k$ - in this case, the exponent of the root 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:

To hit the wall and state that mathematics is a stupid science, where “there are some rules, but these are 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 - 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 multiplying the roots is as follows:

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).
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.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:

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

Here, many would be confused by the fact that the output turned out to be an 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:

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 Most often you will have to deal with variables.
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 a similar task above when we solved 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 easiest way to explain this is using the square root as an example. In mathematics there is a well-established term “squaring”. “Squaring” means multiplying a specific number by itself once.. For example, if you square 2, you get 4. If you square 7, you get 49. The square of 9 is 81. So the square root of 4 is 2, of 49 is 7, and of 81 is 9.

As a rule, teaching this topic in mathematics begins with square roots. In order to immediately determine it, a high school student must know the multiplication table by heart. Those who do not know this table firmly have to use hints. Usually the process of extracting the root square from a number is given in the form of a table on the covers of many school mathematics notebooks.

Roots are of the following types:

square;
cubic (or so-called third degree);
fourth degree;
fifth degree.

Addition rules

Addition is not allowed in tasks when the radical expressions differ from each other. This can be illustrated with a clear example:

The student is faced with the task: add the square root of 4 and 9;
an inexperienced student who does not know the rule usually writes: “root of 4 + root of 9 = root of 13.”
It is very easy to prove that this solution is incorrect. To do this, you need to find the square root of 13 and check whether the example is solved correctly;
using a microcalculator you can determine that it is approximately 3.6. Now all that remains is to check the solution;
root of 4=2, and root of 9=3;
The sum of the numbers “two” and “three” equals five. Thus, this solution algorithm can be considered incorrect.

If the roots have the same degree, but different numerical expressions, it is taken out of brackets, and placed in brackets sum of two radical expressions. Thus, it is already extracted from this amount.

Addition algorithm

In order to correctly solve the simplest problem, you need to:

Determine what exactly requires addition.
Find out whether it is possible to add values to each other, guided by existing rules in mathematics.
If they are not foldable, you need to transform them so that they can be folded.
Having carried out all the necessary transformations, you need to perform the addition and write down the finished answer. You can perform addition in your head or using a microcalculator, depending on the complexity of the example.

What are similar roots

To correctly solve an addition example, you must first think about how you can simplify it. To do this, you need to have basic knowledge of what similarity is.

The ability to identify similar ones helps to quickly solve similar addition examples, bringing them into a simplified form. To simplify a typical addition example, you need to:

Find similar ones and separate them into one group (or several groups).
Rewrite the existing example in such a way that the roots that have the same indicator follow each other clearly (this is called “grouping”).
Next, you should once again write the expression again, this time in such a way that similar ones (which have the same indicator and the same radical figure) also follow each other.

Once this is done, the simplified example is usually easy to solve.

In order to correctly solve any addition example, you need to clearly understand the basic rules of addition, as well as know what a root is and what it can be.

Sometimes such problems look very difficult at first glance, but usually they are easily solved by grouping similar ones. The most important thing is practice, and then the student will begin to “crack problems like nuts.” Adding roots is one of the most important parts of mathematics, so teachers should spend enough time studying it.

Video

This video will help you understand equations with square roots.

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Fact 1.
$\bullet$ Let's take some non-negative number $a$ (that is, $a\geqslant 0$ ). Then (arithmetic) square root from the number $a$ is called such a non-negative number $b$ , when squared we get the number $a$ : \[\sqrt a=b\quad \text(same as )\quad a=b^2\] From the definition it follows that $a\geqslant 0, b\geqslant 0$. These restrictions are an important condition the existence of a square root and they should be remembered!
Remember that any number when squared gives a non-negative result. That is, $100^2=10000\geqslant 0$ and $(-100)^2=10000\geqslant 0$ .
$\bullet$ What is $\sqrt(25)$ equal to? We know that $5^2=25$ and $(-5)^2=25$ . Since by definition we must find a non-negative number, then $-5$ is not suitable, therefore, $\sqrt(25)=5$ (since $25=5^2$ ).
Finding the value of $\sqrt a$ is called taking the square root of the number $a$ , and the number $a$ is called the radical expression.
$\bullet$ Based on the definition, expression $\sqrt(-25)$, $\sqrt(-4)$, etc. don't make sense.

Fact 2.
For quick calculations it will be useful to learn the table of squares natural numbers from $1$ to $20$ : \[\begin(array)(|ll|) \hline 1^2=1 & \quad11^2=121 \\ 2^2=4 & \quad12^2=144\\ 3^2=9 & \quad13 ^2=169\\ 4^2=16 & \quad14^2=196\\ 5^2=25 & \quad15^2=225\\ 6^2=36 & \quad16^2=256\\ 7^ 2=49 & \quad17^2=289\\ 8^2=64 & \quad18^2=324\\ 9^2=81 & \quad19^2=361\\ 10^2=100& \quad20^2= 400\\ \hline \end(array)\]

Fact 3.
What operations can you do with square roots?
$\bullet$ The sum or difference of square roots is NOT EQUAL to the square root of the sum or difference, that is \[\sqrt a\pm\sqrt b\ne \sqrt(a\pm b)\] Thus, if you need to calculate, for example, $\sqrt(25)+\sqrt(49)$ , then initially you must find the values of $\sqrt(25)$ and $\sqrt(49)\ ) and then fold them. Hence, \[\sqrt(25)+\sqrt(49)=5+7=12\] If the values \(\sqrt a$ or $\sqrt b$ cannot be found when adding $\sqrt a+\sqrt b$, then such an expression is not transformed further and remains as it is. For example, in the sum $\sqrt 2+ \sqrt (49)$ we can find $\sqrt(49)$ is $7$ , but $\sqrt 2$ cannot be transformed in any way, That's why $\sqrt 2+\sqrt(49)=\sqrt 2+7$. Unfortunately, this expression cannot be simplified further$\bullet$ The product/quotient of square roots is equal to the square root of the product/quotient, that is \[\sqrt a\cdot \sqrt b=\sqrt(ab)\quad \text(s)\quad \sqrt a:\sqrt b=\sqrt(a:b)\] (provided that both sides of the equalities make sense)
Example: $\sqrt(32)\cdot \sqrt 2=\sqrt(32\cdot 2)=\sqrt(64)=8$; $\sqrt(768):\sqrt3=\sqrt(768:3)=\sqrt(256)=16$; $\sqrt((-25)\cdot (-64))=\sqrt(25\cdot 64)=\sqrt(25)\cdot \sqrt(64)= 5\cdot 8=40$. $\bullet$ Using these properties, it is convenient to find the square roots of large numbers by factoring them.
Let's look at an example. Let's find $\sqrt(44100)$ . Since $44100:100=441$ , then $44100=100\cdot 441$ . According to the criterion of divisibility, the number $441$ is divisible by $9$ (since the sum of its digits is 9 and is divisible by 9), therefore, $441:9=49$, that is, $441=9\ cdot 49$ .
Thus we got: \[\sqrt(44100)=\sqrt(9\cdot 49\cdot 100)= \sqrt9\cdot \sqrt(49)\cdot \sqrt(100)=3\cdot 7\cdot 10=210\] Let's look at another example: \[\sqrt(\dfrac(32\cdot 294)(27))= \sqrt(\dfrac(16\cdot 2\cdot 3\cdot 49\cdot 2)(9\cdot 3))= \sqrt( \ dfrac(16\cdot4\cdot49)(9))=\dfrac(\sqrt(16)\cdot \sqrt4 \cdot \sqrt(49))(\sqrt9)=\dfrac(4\cdot 2\cdot 7)3 =\dfrac(56)3\]
$\bullet$ Let's show how to enter numbers under the square root sign using the example of the expression $5\sqrt2$ (short notation for the expression $5\cdot \sqrt2$). Since $5=\sqrt(25)$ , then \ Note also that, for example,
1) $\sqrt2+3\sqrt2=4\sqrt2$ ,
2) $5\sqrt3-\sqrt3=4\sqrt3$
3) $\sqrt a+\sqrt a=2\sqrt a$ .

Why is this so? Let's explain using example 1). As you already understand, we cannot somehow transform the number $\sqrt2$. Let's imagine that $\sqrt2$ is some number $a$ . Accordingly, the expression $\sqrt2+3\sqrt2$ is nothing more than $a+3a$ (one number $a$ plus three more of the same numbers $a$). And we know that this is equal to four such numbers $a$ , that is, $4\sqrt2$ .

Fact 4.
$\bullet$ They often say “you can’t extract the root” when you can’t get rid of the sign $\sqrt () \ $ of the root (radical) when finding the value of a number. For example, you can take the root of the number $16$ because $16=4^2$ , therefore $\sqrt(16)=4$ . But it is impossible to extract the root of the number $3$, that is, to find $\sqrt3$, because there is no number that squared will give $3$ .
Such numbers (or expressions with such numbers) are irrational. For example, numbers $\sqrt3, \ 1+\sqrt2, \ \sqrt(15)$ etc. are irrational.
Also irrational are the numbers $\pi$ (the number “pi”, approximately equal to $3.14$), $e$ (this number is called the Euler number, it is approximately equal to $2.7$) etc.
$\bullet$ Please note that any number will be either rational or irrational. And together all rational and all irrational numbers form a set called a set of real numbers. This set is denoted by the letter $\mathbb(R)$ .
This means that all the numbers that are on at the moment we know are called real numbers.

Fact 5.
$\bullet$ The modulus of a real number $a$ is a non-negative number $|a|$ equal to the distance from the point $a$ to $0$ on the real line. For example, $|3|$ and $|-3|$ are equal to 3, since the distances from the points $3$ and $-3$ to $0$ are the same and equal to $3 $ .
$\bullet$ If $a$ is a non-negative number, then $|a|=a$ .
Example: $|5|=5$ ; $\qquad |\sqrt2|=\sqrt2$ . $\bullet$ If $a$ is a negative number, then $|a|=-a$ .
Example: $|-5|=-(-5)=5$ ; $\qquad |-\sqrt3|=-(-\sqrt3)=\sqrt3$.
They say that for negative numbers the modulus “eats” the minus, while positive numbers, as well as the number $0$, are left unchanged by the modulus.
BUT This rule only applies to numbers. If under your modulus sign there is an unknown $x$ (or some other unknown), for example, $|x|$ , about which we do not know whether it is positive, zero or negative, then get rid of the modulus we can't. In this case, this expression remains the same: $|x|$ . $\bullet$ The following formulas hold: \[(\large(\sqrt(a^2)=|a|))\] \[(\large((\sqrt(a))^2=a)), \text( provided ) a\geqslant 0\] Very often the following mistake is made: they say that $\sqrt(a^2)$ and $(\sqrt a)^2$ are one and the same. This is only true if $a$ is a positive number or zero. But if $a$ is a negative number, then this is false. It is enough to consider this example. Let's take instead of $a$ the number $-1$ . Then $\sqrt((-1)^2)=\sqrt(1)=1$ , but the expression $(\sqrt (-1))^2$ does not exist at all (after all, it is impossible to use the root sign put negative numbers!).
Therefore, we draw your attention to the fact that $\sqrt(a^2)$ is not equal to $(\sqrt a)^2$ ! Example: 1) $\sqrt(\left(-\sqrt2\right)^2)=|-\sqrt2|=\sqrt2$, because $-\sqrt2<0$ ;

$\phantom(00000)$ 2) $(\sqrt(2))^2=2$ . $\bullet$ Since $\sqrt(a^2)=|a|$ , then \[\sqrt(a^(2n))=|a^n|\] (the expression $2n$ denotes an even number)
That is, when taking the root of a number that is to some degree, this degree is halved.
Example:
1) $\sqrt(4^6)=|4^3|=4^3=64$
2) $\sqrt((-25)^2)=|-25|=25$ (note that if the module is not supplied, it turns out that the root of the number is equal to $-25$ ; but we remember , that by definition of a root this cannot happen: when extracting a root, we should always get a positive number or zero)
3) $\sqrt(x^(16))=|x^8|=x^8$ (since any number to an even power is non-negative)

Fact 6.
How to compare two square roots?
$\bullet$ For square roots it is true: if $\sqrt a<\sqrt b$ , то $aExample:
1) compare \(\sqrt(50)$ and $6\sqrt2$ . First, let's transform the second expression into $\sqrt(36)\cdot \sqrt2=\sqrt(36\cdot 2)=\sqrt(72)$. Thus, since $50<72$ , то и $\sqrt{50}<\sqrt{72}$ . Следовательно, $\sqrt{50}<6\sqrt2$ .
2) Between what integers is $\sqrt(50)$ located?
Since $\sqrt(49)=7$ , $\sqrt(64)=8$ , and $49<50<64$ , то $7<\sqrt{50}<8$ , то есть число $\sqrt{50}$ находится между числами $7$ и $8$ .
3) Let's compare $\sqrt 2-1$ and $0.5$ . Let's assume that $\sqrt2-1>0.5$ : \[\begin(aligned) &\sqrt 2-1>0.5 \ \big| +1\quad \text((add one to both sides))\\ &\sqrt2>0.5+1 \\big| \ ^2 \quad\text((squaring both sides))\\ &2>1.5^2\\ &2>2.25 \end(aligned)\] We see that we have obtained an incorrect inequality. Therefore, our assumption was incorrect and $\sqrt 2-1<0,5$ .
Note that adding a certain number to both sides of the inequality does not affect its sign. Multiplying/dividing both sides of an inequality by a positive number also does not affect its sign, but multiplying/dividing by a negative number reverses the sign of the inequality!
You can square both sides of an equation/inequality ONLY IF both sides are non-negative. For example, in the inequality from the previous example you can square both sides, in the inequality $-3<\sqrt2$ нельзя (убедитесь в этом сами)! $\bullet$ It should be remembered that \[\begin(aligned) &\sqrt 2\approx 1.4\\ &\sqrt 3\approx 1.7 \end(aligned)\] Knowing the approximate meaning of these numbers will help you when comparing numbers! $\bullet$ In order to extract the root (if it can be extracted) from some large number that is not in the table of squares, you must first determine between which “hundreds” it is located, then – between which “tens”, and then determine the last digit of this number. Let's show how this works with an example.
Let's take $\sqrt(28224)$ . We know that $100^2=10\,000$, $200^2=40\,000$, etc. Note that $28224$ is between $10\,000$ and $40\,000$ . Therefore, $\sqrt(28224)$ is between $100$ and $200$ .
Now let’s determine between which “tens” our number is located (that is, for example, between $120$ and $130$). Also from the table of squares we know that $11^2=121$ , $12^2=144$ etc., then $110^2=12100$ , $120^2=14400 $ , $130^2=16900$ , $140^2=19600$ , $150^2=22500$ , $160^2=25600$ , $170^2=28900 $ . So we see that $28224$ is between $160^2$ and $170^2$ . Therefore, the number $\sqrt(28224)$ is between $160$ and $170$ .
Let's try to determine the last digit. Let's remember what single-digit numbers, when squared, give $4$ at the end? These are $2^2$ and $8^2$ . Therefore, $\sqrt(28224)$ will end in either 2 or 8. Let's check this. Let's find $162^2$ and $168^2$ :
$162^2=162\cdot 162=26224$
$168^2=168\cdot 168=28224$ .
Therefore, $\sqrt(28224)=168$ . Voila!

In order to adequately solve the Unified State Exam in mathematics, you first need to study theoretical material, which introduces you to numerous theorems, formulas, algorithms, etc. At first glance, it may seem that this is quite simple. However, finding a source in which the theory for the Unified State Exam in mathematics is presented in an easy and understandable way for students with any level of training is, in fact, a rather difficult task. School textbooks cannot always be kept at hand. And finding basic formulas for the Unified State Exam in mathematics can be difficult even on the Internet.

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