How to subtract another root from one root. Rule for adding square roots. Root formulas. Properties of square roots

Root formulas. properties of square roots.

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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- it's essentially the same thing. There are surprisingly few formulas for square roots. 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:

If you like this site...

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

You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)

you can get acquainted with functions and derivatives.

I looked again at the plate ... And, let's go!

Let's start with a simple one:

Wait a minute. this, which means we can write it like this:

Got it? Here's the next one for you:

The roots of the resulting numbers are not exactly extracted? Don't worry, here are some examples:

But what if there are not two multipliers, but more? The same! The root multiplication formula works with any number of factors:

Now completely independent:

Answers: Well done! Agree, everything is very easy, the main thing is to know the multiplication table!

Root division

We figured out the multiplication of the roots, now let's proceed to the property of division.

Let me remind you that the formula in general looks like this:

And that means that the root of the quotient is equal to the quotient of the roots.

Well, let's look at examples:

That's all science. And here's an example:

Everything is not as smooth as in the first example, but as you can see, there is nothing complicated.

What if the expression looks like this:

You just need to apply the formula in reverse:

And here's an example:

You can also see this expression:

Everything is the same, only here you need to remember how to translate fractions (if you don’t remember, look at the topic and come back!). Remembered? Now we decide!

I am sure that you coped with everything, everything, now let's try to build roots in a degree.

Exponentiation

What happens if the square root is squared? It's simple, remember the meaning of the square root of a number - this is a number whose square root is equal to.

So, if we square a number whose square root is equal, then what do we get?

Well, of course, !

Let's look at examples:

Everything is simple, right? And if the root is in a different degree? It's OK!

Stick to the same logic and remember the properties and possible actions with powers.

Read the theory on the topic "" and everything will become extremely clear to you.

For example, here's an expression:

In this example, the degree is even, but what if it is odd? Again, apply the power properties and factor everything:

With this, everything seems to be clear, but how to extract the root from a number in a degree? Here, for example, is this:

Pretty simple, right? What if the degree is greater than two? We follow the same logic using the properties of degrees:

Well, is everything clear? Then solve your own examples:

And here are the answers:

Introduction under the sign of the root

What we just have not learned to do with the roots! It remains only to practice entering the number under the root sign!

It's quite easy!

Let's say we have a number

What can we do with it? Well, of course, hide the triple under the root, while remembering that the triple is the square root of!

Why do we need it? Yes, just to expand our capabilities when solving examples:

How do you like this property of roots? Makes life much easier? For me, that's right! Only we must remember that we can only enter positive numbers under the square root sign.

Try this example for yourself:
Did you manage? Let's see what you should get:

Well done! You managed to enter a number under the root sign! Let's move on to something equally important - consider how to compare numbers containing a square root!

Root Comparison

Why should we learn to compare numbers containing a square root?

Very simple. Often, in large and long expressions encountered in the exam, we get an irrational answer (do you remember what it is? We already talked about this today!)

We need to place the received answers on the coordinate line, for example, to determine which interval is suitable for solving the equation. And this is where the snag arises: there is no calculator on the exam, and without it, how to imagine which number is larger and which is smaller? That's it!

For example, determine which is greater: or?

You won't say right off the bat. Well, let's use the parsed property of adding a number under the root sign?

Then forward:

Well, obviously, the larger the number under the sign of the root, the larger the root itself!

Those. if means .

From this we firmly conclude that And no one will convince us otherwise!

Extracting roots from large numbers

Before that, we introduced a factor under the sign of the root, but how to take it out? You just need to factor it out and extract what is extracted!

It was possible to go the other way and decompose into other factors:

Not bad, right? Any of these approaches is correct, decide how you feel comfortable.

Factoring is very useful when solving such non-standard tasks as this one:

We don't get scared, we act! We decompose each factor under the root into separate factors:

And now try it yourself (without a calculator! It will not be on the exam):

Is this the end? We don't stop halfway!

That's all, it's not all that scary, right?

Happened? Well done, you're right!

Now try this example:

And an example is a tough nut to crack, so you can’t immediately figure out how to approach it. But we, of course, are in the teeth.

Well, let's start factoring, shall we? Immediately, we note that you can divide a number by (recall the signs of divisibility):

And now, try it yourself (again, without a calculator!):

Well, did it work? Well done, you're right!

Summing up

1. The square root (arithmetic square root) of a non-negative number is a non-negative number whose square is equal.
   .
2. If we just take the square root of something, we always get one non-negative result.
3. Arithmetic root properties:
4. When comparing square roots, it must be remembered that the larger the number under the sign of the root, the larger the root itself.

How do you like the square root? All clear?

We tried to explain to you without water everything you need to know in the exam about the square root.

It's your turn. Write to us whether this topic is difficult for you or not.

Did you learn something new or everything was already so clear.

Write in the comments and good luck on the exams!

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

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

You haven't smoked yet, have you?

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

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 studying irrational equations and inequalities, there will be all sorts of variables and functions in general. And very often, the compilers of the problems are just counting on the fact that you will find some contracting terms or factors, after which the task will be greatly simplified.

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

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

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

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

The case of an arbitrary indicator

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

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

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

Examples. Calculate products:

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

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

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

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

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

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

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

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

Multiplication of roots with different exponents

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

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

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

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

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

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

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

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

Why do radical expressions have to be non-negative?

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

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

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

So now I will explain everything in a normal way.

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

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

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

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

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

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

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

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

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

But then something crazy happens:

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

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

1. To fight against the wall to state that mathematics is a stupid science, where “there are some rules, but this is inaccurate”;
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 - this is difficult, long and generally fu. Therefore, mathematicians preferred the second option. :)

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

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

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

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

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

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

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

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

Example 3. Simplify the expression:

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

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

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

In mathematics, any action has its own pair-opposite - 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, the opposite of it, is decreasing. For example, the action opposite to addition is subtraction, and division corresponds to multiplication. Raising to a power also has its own dialectical pair-opposite. It's about root extraction.

To extract the root of such and such a degree from a number means to calculate which number must be raised to the corresponding power in order to end up with this number. The two degrees have their own separate names: the second degree is called the "square", and the third - the "cube". Accordingly, it is pleasant to call the roots of these powers the square root and the cubic root. 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 extract square roots first, and then add the results. Suppose we need to find the value of such an expression:

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

√16+√121=4+11=15

However, this is the simplest case - here we are talking about full squares, i.e. about numbers that are obtained by squaring whole numbers. But this is not always the case. For example, the number 24 is not a perfect square (you cannot find an 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 get in the answer not a number, but another expression. The maximum that 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 mentioned numbers as an example:

To begin with, let's factorize 24 - in such a way that one of them can easily be taken as a square root (i.e., so that it is a perfect square). There is such a number - this 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, which we can take out of brackets:

(2+3)* √6=5*√6

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

True, you can resort to the help of 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 still would like to bring the solution of the previous example to its logical conclusion, we can multiply this result by 5 - and we get 12.5. A more accurate result with such initial data cannot be obtained.

Addition and subtraction of roots- one of the most common "stumbling blocks" for those who take a course in mathematics (algebra) in high school. However, learning how to add and subtract them correctly is very important, because examples for 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 the solution of such examples, you need two things - to understand the rules, as well as to gain practice. Having solved one or two dozen typical examples, the student will bring this skill to automatism, and then he will have nothing to fear at the 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 the example of a square root. In mathematics, there is a well-established term "square". "Square" means to multiply 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. For those who do not know this table well, you 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 the so-called third degree);
• fourth degree;
• fifth degree.

Addition rules

In order to successfully solve a typical example, it must be borne in mind that not all root numbers can be stacked with each other. In order to be able to put them together, they must be brought to a single pattern. If this is not possible, 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 assignments when the radical expressions differ from each other. This can be illustrated with an illustrative example:

• the student is faced with the task: to add the square root of 4 and of 9;
• an inexperienced student who does not know the rule usually writes: "root of 4 + root of 9 \u003d root of 13."
• it is very easy to prove that this way of solving is wrong. To do this, you need to find the square root of 13 and check if the example is solved correctly;
• using a microcalculator, you can determine that it is approximately 3.6. Now it remains to check the solution;
• root of 4=2, and of 9=3;
• The sum of two and three is 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 the sum of two radical expressions. Thus, it is already extracted from this amount.

Addition algorithm

In order to correctly solve the simplest problem, it is necessary:

1. Determine what exactly requires addition.
2. Find out if it is possible to add values ​​​​to each other, guided by the rules existing in mathematics.
3. If they cannot be added, you need to transform them in such a way that they can be added.
4. Having carried out all the necessary transformations, it is necessary to perform addition and write down the finished answer. Addition can be done mentally or with a calculator, depending on the complexity of the example.

What are similar roots

In order to correctly solve an addition example, it is necessary, first of all, to think about how it can be simplified. To do this, you need to have a basic knowledge of what similarity is.

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

1. Find similar ones and allocate them to one group (or several groups).
2. Rewrite the existing example in such a way that the roots that have the same indicator follow each other clearly (this is called "grouping").
3. Next, you should write the expression again again, this time in such a way that similar ones (which have the same indicator and the same root figure) also follow each other.

After that, a 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, and also know what a root is and how it happens.

Sometimes such tasks look very complicated 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 "click tasks like nuts." Root addition is one of the most important branches of mathematics, so teachers should allocate enough time to study it.

Video

This video will help you understand the equations with square roots.