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. :)

It's easy to multiply roots.

Why do radical expressions have to be non-negative?

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

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

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

So now I will explain everything in a normal way.

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

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

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

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

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

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

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

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

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

But then something crazy happens:

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

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

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?

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 degrees.

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!

The material of this article should be considered as part of the topic transformation of irrational expressions. Here, using examples, we will analyze all the subtleties and nuances (of which there are many) that arise when carrying out transformations based on the properties of the roots.

Page navigation.

Recall the properties of roots

Since we are going to deal with the transformation of expressions using the properties of the roots, it does not hurt to remember the main ones, or even better, write them down on paper and place them in front of you.

First, square roots and their following properties are studied (a, b, a 1, a 2, ..., a k are real numbers):

And later, the idea of ​​the root is expanded, the definition of the root of the nth degree is introduced, and such properties are considered (a, b, a 1, a 2, ..., a k are real numbers, m, n, n 1, n 2, ... , n k - natural numbers):

Converting expressions with numbers under root signs

As usual, they first learn to work with numerical expressions, and only after that they move on to expressions with variables. We will do the same, and first we will deal with the transformation of irrational expressions containing only numerical expressions under the signs of the roots, and already further in the next paragraph we will introduce variables under the signs of the roots.

How can this be used to transform expressions? Very simple: for example, we can replace an irrational expression with an expression, or vice versa. That is, if the converted expression contains an expression that matches in appearance with the expression from the left (right) part of any of the listed properties of the roots, then it can be replaced by the corresponding expression from the right (left) part. This is the transformation of expressions using the properties of the roots.

Let's take a few more examples.

Let's simplify the expression . The numbers 3 , 5 and 7 are positive, so we can safely apply the properties of the roots. Here you can act differently. For example, a property-based root can be represented as , and a property-based root with k=3 as , with this approach, the solution will look like this:

It was possible to do otherwise, replacing with , and then with , in this case the solution would look like this:

Other solutions are possible, for example:

Let's take a look at another example. Let's transform the expression. Looking at the list of properties of the roots, we select from it the properties we need to solve the example, it is clear that two of them and are useful here, which are valid for any a . We have:

Alternatively, one could first transform expressions under root signs using

and then apply the properties of the roots

Up to this point, we have converted expressions that contain only square roots. It's time to work with roots that have other indicators.

Example.

Transform Irrational Expression .

Decision.

By property the first factor of a given product can be replaced by the number −2:

Move on. By virtue of the property, the second factor can be represented as, and it does not hurt to replace 81 with the quadruple power of three, since the number 3 appears in the remaining factors under the signs of the roots:

It is advisable to replace the root of the fraction with the ratio of the roots of the form , which can be further transformed: . We have

The resulting expression after performing operations with twos will take the form , and it remains to transform the product of the roots.

To transform the products of the roots, they are usually reduced to one indicator, for which it is advisable to take the indicators of all roots. In our case, LCM(12, 6, 12)=12 , and only the root will have to be reduced to this indicator, since the other two roots already have such an indicator. To cope with this task allows equality, which is applied from right to left. So . Considering this result, we have

Now the product of the roots can be replaced by the root of the product and the remaining, already obvious, transformations can be performed:

Let's make a short version of the solution:

Answer:

.

Separately, we emphasize that in order to apply the properties of the roots, it is necessary to take into account the restrictions imposed on the numbers under the signs of the roots (a≥0, etc.). Ignoring them can lead to incorrect results. For example, we know that the property holds for non-negative a . Based on it, we can safely go, for example, from to, since 8 is a positive number. But if we take a meaningful root of a negative number, for example, , and, based on the above property, replace it with , then we will actually replace −2 with 2 . Indeed, , a . That is, for negative a, the equality may be false, just as other properties of the roots may be false without taking into account the conditions specified for them.

But what was said in the previous paragraph does not mean at all that expressions with negative numbers under the root signs cannot be transformed using the properties of the roots. They just need to be “prepared” beforehand by applying the rules of operations with numbers or using the definition of an odd degree root from a negative number, which corresponds to the equality , where −a is a negative number (while a is positive). For example, it cannot be immediately replaced by , since −2 and −3 are negative numbers, but it allows us to move from the root to , and then apply the property of the root from the product: . And in one of the previous examples, it was necessary to move from the root to the root of the eighteenth degree not like this, but like this .

So, to transform expressions using the properties of the roots, you need to

• select the appropriate property from the list,
• make sure that the numbers under the root satisfy the conditions for the selected property (otherwise, you need to perform preliminary transformations),
• and carry out the intended transformation.

Converting expressions with variables under root signs

To transform irrational expressions containing not only numbers, but also variables under the sign of the root, the properties of the roots listed in the first paragraph of this article must be applied carefully. This is due for the most part to the conditions that must be satisfied by the numbers involved in the formulas. For example, based on the formula , the expression can be replaced by an expression only for those x values ​​that satisfy the conditions x≥0 and x+1≥0 , since the specified formula is set for a≥0 and b≥0 .

What is the danger of ignoring these conditions? The answer to this question is clearly demonstrated by the following example. Let's say we need to calculate the value of an expression when x=−2 . If we immediately substitute the number −2 instead of the variable x, then we get the value we need . And now let's imagine that, based on some considerations, we converted the given expression to the form , and only after that we decided to calculate the value. We substitute the number −2 instead of x and arrive at the expression , which doesn't make sense.

Let's see what happens to the range of valid values ​​(ODV) of the x variable as we move from expression to expression. We mentioned the ODZ not by chance, since this is a serious tool for controlling the admissibility of the transformations performed, and changing the ODZ after the transformation of the expression should at least alert. It is not difficult to find the ODZ for these expressions. For the expression, the ODZ is determined from the inequality x (x+1)≥0 , its solution gives the numerical set (−∞, −1]∪∪∪)