EntranceHow to bring roots to a common indicator. The nth root and its basic properties
Greetings, cats! Last time we discussed in detail what roots are (if you don’t remember, I recommend reading it). The main takeaway from that lesson: there is only one universal definition of roots, which is what you need to know. The rest is nonsense and a waste of time.
Today we go further. We will learn to multiply roots, we will study some problems associated with multiplication (if these problems are not solved, they can become fatal in the exam) and we will practice properly. So stock up on popcorn, get comfortable, and let's get started. :)
You haven't smoked it yet either, have you?
The lesson turned out to be quite long, so I divided it into two parts:
- 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 opposite situation: there is one big root, but we were eager to represent it as a product of two simpler roots. Why is this necessary, is a separate question. We will only analyze the algorithm.
For those who can’t wait to immediately move on to the second part, you are welcome. Let's start with the rest in order.
Basic Rule of Multiplication
Let's start with the simplest - classic square roots. The same ones that are denoted by $\sqrt(a)$ and $\sqrt(b)$. Everything is obvious to them:
Multiplication rule. To multiply one Square root on the other, you just need to multiply their radical expressions, and write the result under the common radical:
\[\sqrt(a)\cdot \sqrt(b)=\sqrt(a\cdot b)\]
No additional restrictions are imposed on the numbers on the right or left: if the root factors exist, then the product also exists.
Examples. Let's look at four examples with numbers at once:
\[\begin(align) & \sqrt(25)\cdot \sqrt(4)=\sqrt(25\cdot 4)=\sqrt(100)=10; \\ & \sqrt(32)\cdot \sqrt(2)=\sqrt(32\cdot 2)=\sqrt(64)=8; \\ & \sqrt(54)\cdot \sqrt(6)=\sqrt(54\cdot 6)=\sqrt(324)=18; \\ & \sqrt(\frac(3)(17))\cdot \sqrt(\frac(17)(27))=\sqrt(\frac(3)(17)\cdot \frac(17)(27 ))=\sqrt(\frac(1)(9))=\frac(1)(3). \\ \end(align)\]
As you can see, the main meaning of this rule is to simplify irrational expressions. And if in the first example we ourselves would have extracted the roots of 25 and 4 without any new rules, then things get tough: $\sqrt(32)$ and $\sqrt(2)$ are not considered by themselves, but their product turns out to be a perfect square, so its root is equal to a rational number.
I would especially like to highlight the last line. There, both radical expressions are fractions. Thanks to the product, many factors are canceled, and the entire expression turns into an adequate number.
Of course, 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 this was a lyrical digression. Now let's consider a more general case - when the root exponent contains an arbitrary number $n$, and not just the “classical” two.
The case of an arbitrary indicator
So, we've sorted out the square roots. What to do with cubic ones? Or even with roots of arbitrary degree $n$? Yes, everything is the same. The rule remains the same:
To multiply two roots of degree $n$, it is enough to multiply their radical expressions, and then write the result under one radical.
In general, nothing complicated. Except that the amount of calculations may be greater. Let's look at a couple of examples:
Examples. Calculate products:
\[\begin(align) & \sqrt(20)\cdot \sqrt(\frac(125)(4))=\sqrt(20\cdot \frac(125)(4))=\sqrt(625)= 5; \\ & \sqrt(\frac(16)(625))\cdot \sqrt(0.16)=\sqrt(\frac(16)(625)\cdot \frac(16)(100))=\sqrt (\frac(64)(((25)^(2))\cdot 25))= \\ & =\sqrt(\frac(((4)^(3)))(((25)^(3 ))))=\sqrt(((\left(\frac(4)(25) \right))^(3)))=\frac(4)(25). \\ \end(align)\]
And again, attention to the second expression. We multiply cube roots, get rid of decimal and as a result we get the product of the numbers 625 and 25 in the denominator. This is quite big number- Personally, I can’t calculate right off the bat what it equals.
Therefore, we simply isolated the exact cube in the numerator and denominator, and then used one of the key properties (or, if you prefer, definition) of the $n$th root:
\[\begin(align) & \sqrt(((a)^(2n+1)))=a; \\ & \sqrt(((a)^(2n)))=\left| a\right|. \\ \end(align)\]
Such “machinations” can save you a lot of time on the exam or test work, so remember:
Don't rush to multiply numbers using radical expressions. First, check: what if the exact degree of any expression is “encrypted” there?
Despite the obviousness of this remark, I must admit that most unprepared students do not see the exact degrees at point-blank range. Instead, they multiply everything outright, and then wonder: why did they get such brutal numbers? :)
However, all this is baby talk compared to what we will study now.
Multiplying roots with different exponents
Okay, now we can multiply roots with the same indicators. What if the indicators are different? Let's say, how to multiply an ordinary $\sqrt(2)$ by some crap like $\sqrt(23)$? Is it even possible to do this?
Yes of course you can. Everything is done according to this formula:
Rule for multiplying roots. To multiply $\sqrt[n](a)$ by $\sqrt[p](b)$, it is enough to perform the following transformation:
\[\sqrt[n](a)\cdot \sqrt[p](b)=\sqrt(((a)^(p))\cdot ((b)^(n)))\]
However, this formula only works if radical expressions are non-negative. This is a very important note that we will return to a little later.
For now, let's look at a couple of examples:
\[\begin(align) & \sqrt(3)\cdot \sqrt(2)=\sqrt(((3)^(4))\cdot ((2)^(3)))=\sqrt(81 \cdot 8)=\sqrt(648); \\ & \sqrt(2)\cdot \sqrt(7)=\sqrt(((2)^(5))\cdot ((7)^(2)))=\sqrt(32\cdot 49)= \sqrt(1568); \\ & \sqrt(5)\cdot \sqrt(3)=\sqrt(((5)^(4))\cdot ((3)^(2)))=\sqrt(625\cdot 9)= \sqrt(5625). \\ \end(align)\]
As you can see, nothing complicated. Now let's figure out where the non-negativity requirement came from, and what will happen if we violate it. :)
Multiplying roots is easy Why must radical expressions be non-negative?
Of course, you can be like school teachers and quote the textbook with a smart look:
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 one thing important property 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 imprecise”;
- Introduce additional restrictions under which the formula will become 100% working.
In the first option, we will have to constantly catch “non-working” cases - it’s difficult, time-consuming and generally ugh. Therefore, mathematicians preferred the second option. :)
But don't worry! In practice, this limitation does not affect the calculations in any way, because all the problems described concern only roots of odd degree, and minuses can be taken from them.
Therefore, let us formulate one more rule, which generally applies to all actions with roots:
Before multiplying roots, make sure that the radical expressions are non-negative.
Example. In the number $\sqrt(-5)$ you can remove the minus from under the root sign - then everything will be normal:
\[\begin(align) & \sqrt(-5)=-\sqrt(5) \lt 0\Rightarrow \\ & \sqrt(-5)=-\sqrt(((5)^(2))) =-\sqrt(25)=-\sqrt(((5)^(2)))=-\sqrt(5) \lt 0 \\ \end(align)\]
Do you feel the difference? If you leave a minus under the root, then when the radical expression is squared, it will disappear, and crap will begin. And if you first take out the minus, then you can square/remove until you’re blue in the face - the number will remain negative. :)
Thus, the most correct and most reliable way to multiply roots is as follows:
- 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 $\sqrt(5)\cdot \sqrt(3)$ example. 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?
I looked again at the sign... And, let's go!
Let's start with something simple:
Just a minute. this, which means we can write it like this:
Got it? Here's the next one for you:
Are the roots of the resulting numbers not exactly extracted? No problem - here are some examples:
What if there are not two, but more multipliers? The same! The formula for multiplying roots works with any number of factors:
Now completely on your own:
Answers: Well done! Agree, everything is very easy, the main thing is to know the multiplication table!
Root division
We've sorted out the multiplication of roots, now let's move on to the property of division.
Let me remind you that the formula in general view looks like that:
Which means that the root of the quotient is equal to the quotient of the roots.
Well, let's look at some examples:
That's all science is. 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 you come across this expression:
You just need to apply the formula in the opposite direction:
And here's an example:
You may also come across 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!). Do you remember? Now let's decide!
I am sure that you have coped with everything, now let’s try to raise the roots to degrees.
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, what do we get?
Well, of course, !
Let's look at examples:
It's simple, right? What if the root is to a different degree? It's OK!
Follow 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 is an expression:
In this example, the degree is even, but what if it is odd? Again, apply the properties of exponents and factor everything:
Everything seems clear with this, but how to extract the root of a number to a power? 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 the examples yourself:
And here are the answers:
Entering under the sign of the root
What haven’t we learned to do with roots! All that remains is to practice entering the number under the root sign!
It's really easy!
Let's say we have a number written down
What can we do with it? Well, of course, hide the three under the root, remembering that the three is the square root of!
Why do we need this? Yes, just to expand our capabilities when solving examples:
How do you like this property of roots? Does it make life much easier? For me, that's exactly right! Only We must remember that we can only enter positive numbers under the square root sign.
Solve this example yourself -
Did you manage? Let's see what you should get:
Well done! You managed to enter the number under the root sign! Let's move on to something equally important - let's look at how to compare numbers containing a square root!
Comparison of roots
Why do we need to learn to compare numbers that contain a square root?
Very simple. Often, in large and long expressions encountered in the exam, we receive an irrational answer (remember what this 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 here the problem arises: there is no calculator in the exam, and without it, how can you imagine which number is greater and which is less? That's it!
For example, determine which is greater: or?
You can’t tell right away. Well, let's use the disassembled property of entering a number under the root sign?
Then go ahead:
Well, obviously, what larger number under the sign of the root, the larger the root itself!
Those. if, then, .
From this we firmly conclude that. And no one will convince us otherwise!
Extracting roots from large numbers
Before this, we entered a multiplier under the sign of the root, but how to remove it? You just need to factor it into factors and extract what you extract!
It was possible to take a different path and expand into other factors:
Not bad, right? Any of these approaches is correct, decide as you wish.
Factoring is very useful when solving such non-standard tasks like this:
Let's not be afraid, but act! Let's decompose each factor under the root into separate factors:
Now try it yourself (without a calculator! It won’t be on the exam):
Is this the end? Let's not stop halfway!
That's all, it's not so scary, right?
Happened? Well done, that's right!
Now try this example:
But the example is a tough nut to crack, so you can’t immediately figure out how to approach it. But, of course, we can handle it.
Well, let's start factoring? Let us immediately note that you can divide a number by (remember the signs of divisibility):
Now, try it yourself (again, without a calculator!):
Well, did it work? Well done, that's right!
Let's sum it up
- The square root (arithmetic square root) of a non-negative number is a non-negative number whose square is equal to.
. - If we simply take the square root of something, we always get one non-negative result.
- Properties of an arithmetic root:
- When comparing square roots, it is necessary to remember that the larger the number under the root sign, the larger the root itself.
How's the square root? All clear?
We tried to explain to you without any fuss 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 was everything already clear?
Write in the comments and good luck on your exams!
Rootn-th degree and its basic properties
Degree real number A with natural indicator P there is a work P factors, each of which is equal A:
a1 = a; a2 =a·a; A n =
For example,
25 = 2 2 2 2 2 = 32,
5 times
(-3)4 = (-3)(-3)(-3)(-3) = 81.
4 times
Real number A called the basis of the degree, A natural number n- exponent.
The basic properties of powers with natural exponents follow directly from the definition: power of a positive number with any P e N positive; The power of a negative number with an even exponent is positive, with an odd exponent it is negative.
For example,
(-5)4 = (-5) (-5) (-5) (-5) = 625; (-5)3 = (-5)-(-5)-(-5) = -125.
Actions with degrees are performed as follows: rules.
1. To multiply powers with the same bases, it is enough to add the exponents of the powers and leave the base the same, that is
For example, p5∙ p3 = p5+3 =p8
2. To divide powers with the same bases, it is enough to subtract the exponent of the divisor from the index of the dividend and leave the base the same, that is
https://pandia.ru/text/78/410/images/image003_63.gif" width="95" height="44 src=">
2. To raise a degree to a power, it is enough to multiply the exponents, leaving the base the same, that is
(ap)m = at·p. For example, (23)2 = 26.
4. To raise a product to a power, it is enough to raise each factor to this power and multiply the results, that is
(A b)P= ap∙bP.
For example, (2у3)2= 4y6.
5. To raise a fraction to a power, it is enough to raise the numerator and denominator separately to this power and divide the first result by the second, that is
https://pandia.ru/text/78/410/images/image005_37.gif" width="87" height="53 src=">
Note that it is sometimes useful to read these formulas from right to left. In this case they become rules. For example, in case 4, apvp= (av)p we get the following rule: to To multiply powers with the same exponents, it is enough to multiply the bases, leaving the exponent the same.
Using this rule is effective, for example, when calculating the following product
(https://pandia.ru/text/78/410/images/image006_27.gif" width="25" height="23">+1)5=(( -1)( +1))5=( = 1.
Let us now give the definition of a root.
Root nth degree from a real number A called a real number X, the nth power of which is equal to A.
Obviously, in accordance with the basic properties of powers with natural exponents, from any positive number there are two opposite values of the root of an even power, for example, the numbers 4 and -4 are square roots of 16, since (-4)2 = 42 = 16, and the numbers 3 and -3 are the fourth roots of 81, since (-3)4 = 34 = 81.
Also, there is no even root of a negative number because the even power of any real number is non-negative. As for the odd root, for any real number there is only one odd root of that number. For example, 3 is the third root of 27, since 33 = 27, and -2 is the fifth root of -32, since (-2)5 = 32.
Due to the existence of two even-degree roots of a positive number, we introduce the concept of an arithmetic root to eliminate this ambiguity of the root.
The non-negative value of the nth root of a non-negative number is called arithmetic root.
For example, https://pandia.ru/text/78/410/images/image008_21.gif" width="13" height="16 src="> 0.
It should be remembered that when solving irrational equations, their roots are always considered as arithmetic.
Let us note the main property of the nth root.
The size of the root will not change if the indicators of the root and the degree of the radical expression are multiplied or divided by the same natural number, that is
Example 7. Reduce to a common denominator and