It's time to sort it out root extraction methods. They are based on the properties of roots, in particular, on the equality, which is true for any non-negative number b.
Below we will look at the main methods of extracting roots one by one.
Let's start with the simplest case - extracting roots from natural numbers using a table of squares, a table of cubes, etc.
If tables of squares, cubes, etc. If you don’t have it at hand, it’s logical to use the method of extracting the root, which involves decomposing the radical number into prime factors.
It is worth special mentioning what is possible for roots with odd exponents.
Finally, let's consider a method that allows us to sequentially find the digits of the root value.
Let's get started.
Using a table of squares, a table of cubes, etc.
In the simplest cases, tables of squares, cubes, etc. allow you to extract roots. What are these tables?
The table of squares of integers from 0 to 99 inclusive (shown below) consists of two zones. The first zone of the table is located on a gray background; by selecting a specific row and a specific column, it allows you to compose a number from 0 to 99. For example, let’s select a row of 8 tens and a column of 3 units, with this we fixed the number 83. The second zone occupies the rest of the table. Each cell is located at the intersection of a certain row and a certain column, and contains the square of the corresponding number from 0 to 99. At the intersection of our chosen row of 8 tens and column 3 of ones there is a cell with the number 6,889, which is the square of the number 83.
Tables of cubes, tables of fourth powers of numbers from 0 to 99, and so on are similar to the table of squares, only they contain cubes, fourth powers, etc. in the second zone. corresponding numbers.
Tables of squares, cubes, fourth powers, etc. allow you to extract square roots, cube roots, fourth roots, etc. accordingly from the numbers in these tables. Let us explain the principle of their use when extracting roots.
Let's say we need to extract the nth root of the number a, while the number a is contained in the table of nth powers. Using this table we find the number b such that a=b n. Then 
, therefore, the number b will be the desired root of the nth degree.
As an example, let's show how to use a cube table to extract the cube root of 19,683. We find the number 19,683 in the table of cubes, from it we find that this number is the cube of the number 27, therefore, 
.
It is clear that tables of nth powers are very convenient for extracting roots. However, they are often not at hand, and compiling them requires some time. Moreover, it is often necessary to extract roots from numbers that are not contained in the corresponding tables. In these cases, you have to resort to other methods of root extraction.
Factoring a radical number into prime factors
A fairly convenient way to extract the root of a natural number (if, of course, the root is extracted) is to decompose the radical number into prime factors. His the point is this: after that it is quite easy to represent it as a power with the desired exponent, which allows you to obtain the value of the root. Let's clarify this point.
Let the nth root of a natural number a be taken and its value equal b. In this case, the equality a=b n is true. The number b, like any natural number, can be represented as the product of all its prime factors p 1 , p 2 , …, p m in the form p 1 ·p 2 ·…·p m , and the radical number a in this case is represented as (p 1 ·p 2 ·…·p m) n . Since the decomposition of a number into prime factors is unique, the decomposition of the radical number a into prime factors will have the form (p 1 ·p 2 ·…·p m) n, which makes it possible to calculate the value of the root as.
Note that if the decomposition into prime factors of a radical number a cannot be represented in the form (p 1 ·p 2 ·…·p m) n, then the nth root of such a number a is not completely extracted.
Let's figure this out when solving examples.
Example.
Take the square root of 144.
Solution.
If you look at the table of squares given in the previous paragraph, you can clearly see that 144 = 12 2, from which it is clear that the square root of 144 is equal to 12.
But in light of this point, we are interested in how the root is extracted by decomposing the radical number 144 into prime factors. Let's look at this solution.
Let's decompose 144 to prime factors:
That is, 144=2·2·2·2·3·3. Based on the resulting decomposition, the following transformations can be carried out: 144=2·2·2·2·3·3=(2·2) 2·3 2 =(2·2·3) 2 =12 2. Hence, 
.
Using the properties of the degree and the properties of the roots, the solution could be formulated a little differently: .
Answer:
To consolidate the material, consider the solutions to two more examples.
Example.
Calculate the value of the root.
Solution.
The prime factorization of the radical number 243 has the form 243=3 5 . Thus, 
.
Answer:
Example.
Is the root value an integer?
Solution.
To answer this question, let's factor the radical number into prime factors and see if it can be represented as a cube of an integer.
We have 285 768=2 3 ·3 6 ·7 2. The resulting expansion cannot be represented as a cube of an integer, since the power of the prime factor 7 is not a multiple of three. Therefore, the cube root of 285,768 cannot be extracted completely.
Answer:
No.
Extracting roots from fractional numbers
It's time to figure out how to extract the root of a fractional number. Let the fractional radical number be written as p/q. According to the property of the root of a quotient, the following equality is true. From this equality it follows rule for extracting the root of a fraction: The root of a fraction is equal to the quotient of the root of the numerator divided by the root of the denominator.
Let's look at an example of extracting a root from a fraction.
Example.
What is the square root of the common fraction 25/169?
Solution.
Using the table of squares, we find that the square root of the numerator of the original fraction is equal to 5, and the square root of the denominator is equal to 13. Then 
. This completes the extraction of the root of the common fraction 25/169.
Answer:
The root of a decimal fraction or mixed number is extracted after replacing the radical numbers with ordinary fractions.
Example.
Take the cube root of the decimal fraction 474.552.
Solution.
Let's imagine the original decimal fraction as an ordinary fraction: 474.552=474552/1000. Then 
. It remains to extract the cube roots that are in the numerator and denominator of the resulting fraction. Because 474 552=2 2 2 3 3 3 13 13 13=(2 3 13) 3 =78 3 and 1 000 = 10 3, then 
And 
. All that remains is to complete the calculations 
.
Answer:
.
Taking the root of a negative number
It is worthwhile to dwell on extracting roots from negative numbers. When studying roots, we said that when the root exponent is an odd number, then there can be a negative number under the root sign. We gave these entries the following meaning: for a negative number −a and an odd exponent of the root 2 n−1, 
. This equality gives rule for extracting odd roots from negative numbers: to extract the root of a negative number, you need to take the root of the opposite positive number, and put a minus sign in front of the result.
Let's look at the example solution.
Example.
Find the value of the root.
Solution.
Let's transform the original expression so that there is a positive number under the root sign: 
. Now replace the mixed number with an ordinary fraction: 
. We apply the rule for extracting the root of an ordinary fraction: 
. It remains to calculate the roots in the numerator and denominator of the resulting fraction: 
.
Here is a short summary of the solution: 
.
Answer:
.
Bitwise determination of the root value
In the general case, under the root there is a number that, using the techniques discussed above, cannot be represented as the nth power of any number. But in this case there is a need to know the meaning of a given root, at least up to a certain sign. In this case, to extract the root, you can use an algorithm that allows you to sequentially obtain a sufficient number of digit values of the desired number.
The first step of this algorithm is to find out what the most significant bit of the root value is. To do this, the numbers 0, 10, 100, ... are sequentially raised to the power n until the moment when a number exceeds the radical number is obtained. Then the number that we raised to the power n at the previous stage will indicate the corresponding most significant digit.
For example, consider this step of the algorithm when extracting the square root of five. Take the numbers 0, 10, 100, ... and square them until we get a number greater than 5. We have 0 2 =0<5 , 10 2 =100>5, which means the most significant digit will be the ones digit. The value of this bit, as well as the lower ones, will be found in the next steps of the root extraction algorithm.
All subsequent steps of the algorithm are aimed at sequentially clarifying the value of the root by finding the values of the next bits of the desired value of the root, starting with the highest one and moving to the lowest ones. For example, the value of the root at the first step turns out to be 2, at the second – 2.2, at the third – 2.23, and so on 2.236067977…. Let us describe how the values of the bits are found.
The digits are found by searching through their possible values 0, 1, 2, ..., 9. In this case, the nth powers of the corresponding numbers are calculated in parallel, and they are compared with the radical number. If at some stage the value of the degree exceeds the radical number, then the value of the digit corresponding to the previous value is considered found, and the transition to the next step of the root extraction algorithm is made; if this does not happen, then the value of this digit is 9.
Let us explain these points using the same example of extracting the square root of five.
First we find the value of the units digit. We will go through the values 0, 1, 2, ..., 9, calculating 0 2, 1 2, ..., 9 2, respectively, until we get a value greater than the radical number 5. It is convenient to present all these calculations in the form of a table: 
So the value of the units digit is 2 (since 2 2<5
, а 2 3 >5 ). Let's move on to finding the value of the tenths place. In this case, we will square the numbers 2.0, 2.1, 2.2, ..., 2.9, comparing the resulting values with the radical number 5: 
Since 2.2 2<5
, а 2,3 2 >5, then the value of the tenths place is 2. You can proceed to finding the value of the hundredths place: 
This is how the next value of the root of five was found, it is equal to 2.23. And so you can continue to find values: 2,236, 2,2360, 2,23606, 2,236067, … .
To consolidate the material, we will analyze the extraction of the root with an accuracy of hundredths using the considered algorithm.
First we determine the most significant digit. To do this, we cube the numbers 0, 10, 100, etc. until we get a number greater than 2,151,186. We have 0 3 =0<2 151,186 , 10 3 =1 000<2151,186 , 100 3 =1 000 000>2 151,186 , so the most significant digit is the tens digit.
Let's determine its value. 
Since 10 3<2 151,186
, а 20 3 >2 151.186, then the value of the tens place is 1. Let's move on to units. 
Thus, the value of the ones digit is 2. Let's move on to tenths. 
Since even 12.9 3 is less than the radical number 2 151.186, then the value of the tenths place is 9. It remains to perform the last step of the algorithm; it will give us the value of the root with the required accuracy. 
At this stage, the value of the root is found accurate to hundredths: 
.
In conclusion of this article, I would like to say that there are many other ways to extract roots. But for most tasks, the ones we studied above are sufficient.
Bibliography.
- Makarychev Yu.N., Mindyuk N.G., Neshkov K.I., Suvorova S.B. Algebra: textbook for 8th grade. educational institutions.
- Kolmogorov A.N., Abramov A.M., Dudnitsyn Yu.P. and others. Algebra and the beginnings of analysis: Textbook for grades 10 - 11 of general education institutions.
- Gusev V.A., Mordkovich A.G. Mathematics (a manual for those entering technical schools).
Converting expressions with roots and powers often requires going back and forth between roots and powers. In this article we will look at how such transitions are made, what underlies them, and at what points errors most often occur. We will provide all this with typical examples with a detailed analysis of solutions.
Page navigation.
Transition from powers with fractional exponents to roots
The possibility of moving from a degree with a fractional exponent to the root is dictated by the very definition of the degree. Let us recall how it is determined: the power of a positive number a with a fractional exponent m/n, where m is an integer and n is a natural number, is called the nth root of a m, that is, where a>0, m∈Z, n∈ N. The fractional power of zero is defined similarly 
, with the only difference that in this case m is no longer considered an integer, but a natural one, so that division by zero does not occur.
Thus, the degree can always be replaced by the root. For example, you can go from to, and the degree can be replaced by the root. But you should not move from the expression to the root, since the degree initially does not make sense (the degree of negative numbers is not defined), despite the fact that the root has meaning.
As you can see, there is absolutely nothing tricky in the transition from powers of numbers to roots. The transition to roots of powers with fractional exponents, based on arbitrary expressions, is carried out similarly. Note that this transition is carried out on the ODZ of variables for the original expression. For example, the expression 
on the entire ODZ of the variable x for this expression can be replaced by the root 
. And from the degree 
go to root 
, such a replacement takes place for any set of variables x, y and z from the ODZ for the original expression.
Replacing roots with powers
The reverse replacement is also possible, that is, replacing the roots with powers with fractional exponents. It is also based on the equality, which in this case is used from right to left, that is, in the form.
For positive a the indicated transition is obvious. For example, you can replace the degree with , and go from the root to the degree with a fractional exponent of the form .
And for negative a the equality does not make sense, but the root can still make sense. For example, roots make sense, but they cannot be replaced by powers. So is it even possible to convert them into expressions with powers? It is possible if you carry out preliminary transformations, which consist in going to the roots with non-negative numbers under them, which are then replaced by powers with fractional exponents. Let us show what these preliminary transformations are and how to carry them out.
In the case of a root, you can perform the following transformations: 
. And since 4 is a positive number, the last root can be replaced by a power. And in the second case determining the odd root of a negative number−a (where a is positive), expressed by the equality 
, allows you to replace the root with an expression in which the cube root of two can already be replaced by a degree, and it will take the form .
It remains to figure out how the roots under which the expressions are located are replaced by powers containing these expressions in the base. There is no need to rush to replace it with , we used the letter A to denote a certain expression. Let's give an example to explain what we mean by this. I just want to replace the root with a degree, based on equality. But such a replacement is appropriate only under the condition x−3≥0, and for other values of the variable x from the ODZ (satisfying the condition x−3<0 ) она не подходит, так как формула не имеет смысла для отрицательных a . Если обратить внимание на ОДЗ, то несложно заметить ее сужение при переходе от выражения к выражению , а помните, что мы договорились не прибегать к преобразованиям, сужающим ОДЗ.
Because of this inaccurate application of the formula, errors often occur when moving from roots to powers. For example, in the textbook the task is given to present an expression in the form of a power with a rational exponent, and the answer is given, which raises questions, since the condition does not specify the constraint b>0. And in the textbook there is a transition from the expression 
, most likely through the following transformations of the irrational expression 
to the expression. The latest transition also raises questions, as it narrows the DZ.
A logical question arises: “How can one correctly move from the root to the power for all values of variables from the ODZ?” This replacement is carried out on the basis of the following statements:
Before justifying the recorded results, we give several examples of their use for the transition from roots to powers. First, let's return to the expression. It should have been replaced not by , but by (in this case m=2 is an even integer, n=3 is a natural integer). Another example: 
.
Now the promised justification of the results.
When m is an odd integer, and n is an even natural integer, then for any set of variables from the ODZ for the expression, the value of expression A is positive (if m<0 ) или неотрицательно (если m>0). That's why, .
Let's move on to the second result. Let m be a positive odd integer and n an odd natural number. For all values of variables from the ODZ for which the value of expression A is non-negative, 
, and for which it is negative, 
The following result is proven similarly for negative and odd integers m and odd natural integers n. For all values of variables from the ODZ for which the value of expression A is positive, , and for which it is negative,
Finally, the last result. Let m be an even integer, n be any natural number. For all values of variables from the ODZ for which the value of expression A is positive (if m<0
) или неотрицательно (если m>0
), . And for which it is negative, . Thus, if m is an even integer, n is any natural number, then for any set of values of variables from the ODZ for expression it can be replaced by .
Bibliography.
- Algebra and the beginning of analysis: Proc. for 10-11 grades. general education institutions / A. N. Kolmogorov, A. M. Abramov, Yu. P. Dudnitsyn and others; Ed. A. N. Kolmogorov. - 14th ed. - M.: Education, 2004. - 384 pp.: ill. - ISBN 5-09-013651-3.
- Algebra and the beginning of mathematical analysis. 11th grade: educational. for general education institutions: basic and profile. levels / [Yu. M. Kolyagin, M. V. Tkacheva, N. E. Fedorova, M. I. Shabunin]; edited by A. B. Zhizhchenko. – M.: Education, 2009.- 336 pp.: ill.- ISBN 979-5-09-016551-8.
Excel uses built-in functions and mathematical operators to extract the root and raise a number to a power. Let's look at examples.
Examples of the SQRT function in Excel
The built-in SQRT function returns the positive square root value. In the Functions menu, it is under the Math category.
Function syntax: =ROOT(number).
The only and required argument is a positive number for which the function calculates the square root. If the argument is negative, Excel will return a #NUM! error.
You can specify a specific value or a reference to a cell with a numeric value as an argument.
Let's look at examples.
The function returned the square root of the number 36. The argument is a specific value.
The ABS function returns the absolute value of -36. Its use allowed us to avoid errors when extracting the square root of a negative number.
The function took the square root of the sum of 13 and the value of cell C1.
Exponentiation function in Excel
Function syntax: =POWER(value, number). Both arguments are required.
Value is any real numeric value. A number is an indicator of the power to which a given value must be raised.
Let's look at examples.
In cell C2 - the result of squaring the number 10.
The function returned the number 100 raised to ¾.
Exponentiation using operator
To raise a number to a power in Excel, you can use the mathematical operator “^”. To enter it, press Shift + 6 (with English keyboard layout).
In order for Excel to treat the entered information as a formula, the “=” sign is first placed. Next is the number that needs to be raised to a power. And after the “^” sign is the value of the degree.
Instead of any value of this mathematical formula, you can use references to cells with numbers.
This is convenient if you need to construct multiple values.
By copying the formula to the entire column, we quickly got the results of raising the numbers in column A to the third power.
Extracting nth roots
ROOT is the square root function in Excel. How to extract the root of the 3rd, 4th and other powers?
Let's remember one of the mathematical laws: to extract the nth root, you need to raise the number to the power 1/n.
For example, to extract the cube root, we raise the number to the power of 1/3.
Let's use the formula to extract roots of different degrees in Excel.
The formula returned the value of the cube root of the number 21. To raise to a fractional power, the “^” operator was used.
Congratulations: today we will look at roots - one of the most mind-blowing topics in 8th grade. :)
Many people get confused about roots, not because they are complex (what’s so complicated about it - a couple of definitions and a couple more properties), but because in most school textbooks roots are defined through such a jungle that only the authors of the textbooks themselves can understand this writing. And even then only with a bottle of good whiskey. :)
Therefore, now I will give the most correct and most competent definition of a root - the only one that you really should remember. And then I’ll explain: why all this is needed and how to apply it in practice.
But first, remember one important point that many textbook compilers for some reason “forget”:
Roots can be of even degree (our favorite $\sqrt(a)$, as well as all sorts of $\sqrt(a)$ and even $\sqrt(a)$) and odd degree (all sorts of $\sqrt(a)$, $\ sqrt(a)$, etc.). And the definition of a root of an odd degree is somewhat different from an even one.
Probably 95% of all errors and misunderstandings associated with roots are hidden in this fucking “somewhat different”. So let's clear up the terminology once and for all:
Definition. Even root n from the number $a$ is any non-negative the number $b$ is such that $((b)^(n))=a$. And the odd root of the same number $a$ is generally any number $b$ for which the same equality holds: $((b)^(n))=a$.
In any case, the root is denoted like this:
\(a)\]
The number $n$ in such a notation is called the root exponent, and the number $a$ is called the radical expression. In particular, for $n=2$ we get our “favorite” square root (by the way, this is a root of even degree), and for $n=3$ we get a cubic root (odd degree), which is also often found in problems and equations.
Examples. Classic examples of square roots:
\[\begin(align) & \sqrt(4)=2; \\ & \sqrt(81)=9; \\ & \sqrt(256)=16. \\ \end(align)\]
By the way, $\sqrt(0)=0$, and $\sqrt(1)=1$. This is quite logical, since $((0)^(2))=0$ and $((1)^(2))=1$.
Cube roots are also common - no need to be afraid of them:
\[\begin(align) & \sqrt(27)=3; \\ & \sqrt(-64)=-4; \\ & \sqrt(343)=7. \\ \end(align)\]
Well, a couple of “exotic examples”:
\[\begin(align) & \sqrt(81)=3; \\ & \sqrt(-32)=-2. \\ \end(align)\]
If you do not understand what the difference is between an even and an odd degree, re-read the definition again. It is very important!
In the meantime, we will consider one unpleasant feature of roots, because of which we needed to introduce a separate definition for even and odd exponents.
Why do we need roots at all?
After reading the definition, many students will ask: “What were the mathematicians smoking when they came up with this?” And really: why are all these roots needed at all?
To answer this question, let's go back to elementary school for a moment. Remember: in those distant times, when the trees were greener and the dumplings tastier, our main concern was to multiply numbers correctly. Well, something like “five by five - twenty-five”, that’s all. But you can multiply numbers not in pairs, but in triplets, quadruples and generally whole sets:
\[\begin(align) & 5\cdot 5=25; \\ & 5\cdot 5\cdot 5=125; \\ & 5\cdot 5\cdot 5\cdot 5=625; \\ & 5\cdot 5\cdot 5\cdot 5\cdot 5=3125; \\ & 5\cdot 5\cdot 5\cdot 5\cdot 5\cdot 5=15\ 625. \end(align)\]
However, this is not the point. The trick is different: mathematicians are lazy people, so they had a hard time writing down the multiplication of ten fives like this:
That's why they came up with degrees. Why not write the number of factors as a superscript instead of a long string? Something like this:
It's very convenient! All calculations are reduced significantly, and you don’t have to waste a bunch of sheets of parchment and notebooks to write down some 5,183. This record was called a power of a number; a bunch of properties were found in it, but the happiness turned out to be short-lived.
After a grandiose drinking party, which was organized just for the “discovery” of degrees, some particularly stubborn mathematician suddenly asked: “What if we know the degree of a number, but the number itself is unknown?” Now, indeed, if we know that a certain number $b$, say, to the 5th power gives 243, then how can we guess what the number $b$ itself is equal to?
This problem turned out to be much more global than it might seem at first glance. Because it turned out that for most “ready-made” powers there are no such “initial” numbers. Judge for yourself:
\[\begin(align) & ((b)^(3))=27\Rightarrow b=3\cdot 3\cdot 3\Rightarrow b=3; \\ & ((b)^(3))=64\Rightarrow b=4\cdot 4\cdot 4\Rightarrow b=4. \\ \end(align)\]
What if $((b)^(3))=$50? It turns out that we need to find a certain number that, when multiplied by itself three times, will give us 50. But what is this number? It is clearly greater than 3, since 3 3 = 27< 50. С тем же успехом оно меньше 4, поскольку 4 3 = 64 >50. That is this number lies somewhere between three and four, but you won’t understand what it is equal to.
This is precisely why mathematicians came up with $n$th roots. This is precisely why the radical symbol $\sqrt(*)$ was introduced. To designate the very number $b$, which to the indicated degree will give us a previously known value
\[\sqrt[n](a)=b\Rightarrow ((b)^(n))=a\]
I don’t argue: often these roots are easily calculated - we saw several such examples above. But still, in most cases, if you think of an arbitrary number and then try to extract the root of an arbitrary degree from it, you will be in for a terrible bummer.
What is there! Even the simplest and most familiar $\sqrt(2)$ cannot be represented in our usual form - as an integer or a fraction. And if you enter this number into a calculator, you will see this:
\[\sqrt(2)=1.414213562...\]
As you can see, after the decimal point there is an endless sequence of numbers that do not obey any logic. You can, of course, round this number to quickly compare with other numbers. For example:
\[\sqrt(2)=1.4142...\approx 1.4 \lt 1.5\]
Or here's another example:
\[\sqrt(3)=1.73205...\approx 1.7 \gt 1.5\]
But all these roundings, firstly, are quite rough; and secondly, you also need to be able to work with approximate values, otherwise you can catch a bunch of non-obvious errors (by the way, the skill of comparison and rounding is required to be tested on the profile Unified State Examination).
Therefore, in serious mathematics you cannot do without roots - they are the same equal representatives of the set of all real numbers $\mathbb(R)$, just like the fractions and integers that have long been familiar to us.
The inability to represent a root as a fraction of the form $\frac(p)(q)$ means that this root is not a rational number. Such numbers are called irrational, and they cannot be accurately represented except with the help of a radical or other constructions specially designed for this (logarithms, powers, limits, etc.). But more on that another time.
Let's look at a few examples where, after all the calculations, irrational numbers will still remain in the answer.
\[\begin(align) & \sqrt(2+\sqrt(27))=\sqrt(2+3)=\sqrt(5)\approx 2.236... \\ & \sqrt(\sqrt(-32 ))=\sqrt(-2)\approx -1.2599... \\ \end(align)\]
Naturally, from the appearance of the root it is almost impossible to guess what numbers will come after the decimal point. However, you can count on a calculator, but even the most advanced date calculator only gives us the first few digits of an irrational number. Therefore, it is much more correct to write the answers in the form $\sqrt(5)$ and $\sqrt(-2)$.
This is exactly why they were invented. To conveniently record answers.
Why are two definitions needed?
The attentive reader has probably already noticed that all the square roots given in the examples are taken from positive numbers. Well, at least from scratch. But cube roots can be calmly extracted from absolutely any number - be it positive or negative.
Why is this happening? Take a look at the graph of the function $y=((x)^(2))$:
The graph of a quadratic function gives two roots: positive and negative Let's try to calculate $\sqrt(4)$ using this graph. To do this, a horizontal line $y=4$ is drawn on the graph (marked in red), which intersects with the parabola at two points: $((x)_(1))=2$ and $((x)_(2)) =-2$. This is quite logical, since
Everything is clear with the first number - it is positive, so it is the root:
But then what to do with the second point? Like four has two roots at once? After all, if we square the number −2, we also get 4. Why not write $\sqrt(4)=-2$ then? And why do teachers look at such posts as if they want to eat you? :)
The trouble is that if you don’t impose any additional conditions, then the quad will have two square roots - positive and negative. And any positive number will also have two of them. But negative numbers will have no roots at all - this can be seen from the same graph, since the parabola never falls below the axis y, i.e. does not accept negative values.
A similar problem occurs for all roots with an even exponent:
- Strictly speaking, each positive number will have two roots with even exponent $n$;
- From negative numbers, the root with even $n$ is not extracted at all.
That is why in the definition of a root of an even degree $n$ it is specifically stipulated that the answer must be a non-negative number. This is how we get rid of ambiguity.
But for odd $n$ there is no such problem. To see this, let's look at the graph of the function $y=((x)^(3))$:
A cube parabola can take any value, so the cube root can be taken from any number Two conclusions can be drawn from this graph:
- The branches of a cubic parabola, unlike a regular one, go to infinity in both directions - both up and down. Therefore, no matter what height we draw a horizontal line, this line will certainly intersect with our graph. Consequently, the cube root can always be extracted from absolutely any number;
- In addition, such an intersection will always be unique, so you don’t need to think about which number is considered the “correct” root and which one to ignore. That is why determining roots for an odd degree is simpler than for an even degree (there is no requirement for non-negativity).
It's a pity that these simple things are not explained in most textbooks. Instead, our brains begin to soar with all sorts of arithmetic roots and their properties.
Yes, I don’t argue: you also need to know what an arithmetic root is. And I will talk about this in detail in a separate lesson. Today we will also talk about it, because without it all thoughts about roots of $n$-th multiplicity would be incomplete.
But first you need to clearly understand the definition that I gave above. Otherwise, due to the abundance of terms, such a mess will begin in your head that in the end you will not understand anything at all.
All you need to do is understand the difference between even and odd indicators. Therefore, let’s once again collect everything you really need to know about roots:
- A root of an even degree exists only from a non-negative number and is itself always a non-negative number. For negative numbers such a root is undefined.
- But the root of an odd degree exists from any number and can itself be any number: for positive numbers it is positive, and for negative numbers, as the cap hints, it is negative.
Is it difficult? No, it's not difficult. It's clear? Yes, it’s completely obvious! So now we will practice a little with the calculations.
Basic properties and limitations
Roots have many strange properties and limitations - this will be discussed in a separate lesson. Therefore, now we will consider only the most important “trick”, which applies only to roots with an even index. Let's write this property as a formula:
\[\sqrt(((x)^(2n)))=\left| x\right|\]
In other words, if we raise a number to an even power and then extract the root of the same power, we will not get the original number, but its modulus. This is a simple theorem that can be easily proven (it is enough to consider non-negative $x$ separately, and then negative ones separately). Teachers constantly talk about it, it is given in every school textbook. But as soon as it comes to solving irrational equations (i.e., equations containing a radical sign), students unanimously forget this formula.
To understand the issue in detail, let's forget all the formulas for a minute and try to calculate two numbers straight ahead:
\[\sqrt(((3)^(4)))=?\quad \sqrt(((\left(-3 \right))^(4)))=?\]
These are very simple examples. Most people will solve the first example, but many people get stuck on the second. To solve any such crap without problems, always consider the procedure:
- First, the number is raised to the fourth power. Well, it's kind of easy. You will get a new number that can be found even in the multiplication table;
- And now from this new number it is necessary to extract the fourth root. Those. no “reduction” of roots and powers occurs - these are sequential actions.
Let's look at the first expression: $\sqrt(((3)^(4)))$. Obviously, you first need to calculate the expression under the root:
\[((3)^(4))=3\cdot 3\cdot 3\cdot 3=81\]
Then we extract the fourth root of the number 81:
Now let's do the same with the second expression. First, we raise the number −3 to the fourth power, which requires multiplying it by itself 4 times:
\[((\left(-3 \right))^(4))=\left(-3 \right)\cdot \left(-3 \right)\cdot \left(-3 \right)\cdot \ left(-3 \right)=81\]
We got a positive number, since the total number of minuses in the product is 4, and they will all cancel each other out (after all, a minus for a minus gives a plus). Then we extract the root again:
In principle, this line could not have been written, since it’s a no brainer that the answer would be the same. Those. an even root of the same even power “burns” the minuses, and in this sense the result is indistinguishable from a regular module:
\[\begin(align) & \sqrt(((3)^(4)))=\left| 3 \right|=3; \\ & \sqrt(((\left(-3 \right))^(4)))=\left| -3 \right|=3. \\ \end(align)\]
These calculations are in good agreement with the definition of a root of an even degree: the result is always non-negative, and the radical sign also always contains a non-negative number. Otherwise, the root is undefined.
Note on procedure
- The notation $\sqrt(((a)^(2)))$ means that we first square the number $a$ and then take the square root of the resulting value. Therefore, we can be sure that there is always a non-negative number under the root sign, since $((a)^(2))\ge 0$ in any case;
- But the notation $((\left(\sqrt(a) \right))^(2))$, on the contrary, means that we first take the root of a certain number $a$ and only then square the result. Therefore, the number $a$ can in no case be negative - this is a mandatory requirement included in the definition.
Thus, in no case should one thoughtlessly reduce roots and degrees, thereby allegedly “simplifying” the original expression. Because if there is a negative number under the root, and its exponent is even, we will get a bunch of problems.
However, all these problems are relevant only for even indicators.
Removing the minus sign from under the root sign
Naturally, roots with odd exponents also have their own feature, which in principle does not exist with even ones. Namely:
\[\sqrt(-a)=-\sqrt(a)\]
In short, you can remove the minus from under the sign of roots of odd degree. This is a very useful property that allows you to “throw out” all the disadvantages:
\[\begin(align) & \sqrt(-8)=-\sqrt(8)=-2; \\ & \sqrt(-27)\cdot \sqrt(-32)=-\sqrt(27)\cdot \left(-\sqrt(32) \right)= \\ & =\sqrt(27)\cdot \sqrt(32)= \\ & =3\cdot 2=6. \end(align)\]
This simple property greatly simplifies many calculations. Now you don’t need to worry: what if a negative expression was hidden under the root, but the degree at the root turned out to be even? It is enough just to “throw out” all the minuses outside the roots, after which they can be multiplied by each other, divided, and generally do many suspicious things, which in the case of “classical” roots are guaranteed to lead us to an error.
And here another definition comes onto the scene - the same one with which in most schools they begin the study of irrational expressions. And without which our reasoning would be incomplete. Meet us!
Arithmetic root
Let's assume for a moment that under the root sign there can only be positive numbers or, in extreme cases, zero. Let's forget about even/odd indicators, let's forget about all the definitions given above - we will work only with non-negative numbers. What then?
And then we will get an arithmetic root - it partially overlaps with our “standard” definitions, but still differs from them.
Definition. An arithmetic root of the $n$th degree of a non-negative number $a$ is a non-negative number $b$ such that $((b)^(n))=a$.
As we can see, we are no longer interested in parity. Instead, a new restriction appeared: the radical expression is now always non-negative, and the root itself is also non-negative.
To better understand how the arithmetic root differs from the usual one, take a look at the graphs of the square and cubic parabola we are already familiar with:
Arithmetic root search area - non-negative numbers As you can see, from now on we are only interested in those pieces of graphs that are located in the first coordinate quarter - where the coordinates $x$ and $y$ are positive (or at least zero). You no longer need to look at the indicator to understand whether we have the right to put a negative number under the root or not. Because negative numbers are no longer considered in principle.
You may ask: “Well, why do we need such a neutered definition?” Or: “Why can’t we get by with the standard definition given above?”
Well, I will give just one property because of which the new definition becomes appropriate. For example, the rule for exponentiation:
\[\sqrt[n](a)=\sqrt(((a)^(k)))\]
Please note: we can raise the radical expression to any power and at the same time multiply the root exponent by the same power - and the result will be the same number! Here are examples:
\[\begin(align) & \sqrt(5)=\sqrt(((5)^(2)))=\sqrt(25) \\ & \sqrt(2)=\sqrt(((2)^ (4)))=\sqrt(16)\\ \end(align)\]
So what's the big deal? Why couldn't we do this before? Here's why. Let's consider a simple expression: $\sqrt(-2)$ - this number is quite normal in our classical understanding, but absolutely unacceptable from the point of view of the arithmetic root. Let's try to convert it:
$\begin(align) & \sqrt(-2)=-\sqrt(2)=-\sqrt(((2)^(2)))=-\sqrt(4) \lt 0; \\ & \sqrt(-2)=\sqrt(((\left(-2 \right))^(2)))=\sqrt(4) \gt 0. \\ \end(align)$
As you can see, in the first case we removed the minus from under the radical (we have every right, since the exponent is odd), and in the second case we used the above formula. Those. From a mathematical point of view, everything is done according to the rules.
WTF?! How can the same number be both positive and negative? No way. It’s just that the formula for exponentiation, which works great for positive numbers and zero, begins to produce complete heresy in the case of negative numbers.
It was in order to get rid of such ambiguity that arithmetic roots were invented. A separate large lesson is devoted to them, where we consider all their properties in detail. So we won’t dwell on them now - the lesson has already turned out to be too long.
Algebraic root: for those who want to know more
I thought for a long time whether to put this topic in a separate paragraph or not. In the end I decided to leave it here. This material is intended for those who want to understand the roots even better - no longer at the average “school” level, but at one close to the Olympiad level.
So: in addition to the “classical” definition of the $n$th root of a number and the associated division into even and odd exponents, there is a more “adult” definition that does not depend at all on parity and other subtleties. This is called an algebraic root.
Definition. The algebraic $n$th root of any $a$ is the set of all numbers $b$ such that $((b)^(n))=a$. There is no established designation for such roots, so we’ll just put a dash on top:
\[\overline(\sqrt[n](a))=\left\( b\left| b\in \mathbb(R);((b)^(n))=a \right. \right\) \]
The fundamental difference from the standard definition given at the beginning of the lesson is that an algebraic root is not a specific number, but a set. And since we work with real numbers, this set comes in only three types:
- Empty set. Occurs when you need to find an algebraic root of an even degree from a negative number;
- A set consisting of one single element. All roots of odd powers, as well as roots of even powers of zero, fall into this category;
- Finally, the set can include two numbers - the same $((x)_(1))$ and $((x)_(2))=-((x)_(1))$ that we saw on the graph quadratic function. Accordingly, such an arrangement is possible only when extracting the root of an even degree from a positive number.
The last case deserves more detailed consideration. Let's count a couple of examples to understand the difference.
Example. Evaluate the expressions:
\[\overline(\sqrt(4));\quad \overline(\sqrt(-27));\quad \overline(\sqrt(-16)).\]
Solution. The first expression is simple:
\[\overline(\sqrt(4))=\left\( 2;-2 \right\)\]
It is two numbers that are part of the set. Because each of them squared gives a four.
\[\overline(\sqrt(-27))=\left\( -3 \right\)\]
Here we see a set consisting of only one number. This is quite logical, since the root exponent is odd.
Finally, the last expression:
\[\overline(\sqrt(-16))=\varnothing \]
We received an empty set. Because there is not a single real number that, when raised to the fourth (i.e., even!) power, will give us the negative number −16.
Final note. Please note: it was not by chance that I noted everywhere that we work with real numbers. Because there are also complex numbers - it is quite possible to calculate $\sqrt(-16)$ there, and many other strange things.
However, complex numbers almost never appear in modern school mathematics courses. They have been removed from most textbooks because our officials consider the topic “too difficult to understand.”
That's all. In the next lesson we will look at all the key properties of roots and finally learn how to simplify irrational expressions. :)
Operations with powers and roots. Degree with negative ,
zero and fractional indicator. About expressions that have no meaning.
Operations with degrees.
1. When multiplying powers with the same base, their exponents add up:
a m · a n = a m + n .
2. When dividing degrees with the same base, their exponents are deducted .
3. The degree of the product of two or more factors is equal to the product of the degrees of these factors.
(abc… ) n = a n· b n · c n…
4. The degree of a ratio (fraction) is equal to the ratio of the degrees of the dividend (numerator) and divisor (denominator):
(a/b ) n = a n / b n .
5. When raising a power to a power, their exponents are multiplied:
(a m ) n = a m n .
All the above formulas are read and executed in both directions from left to right and vice versa.
EXAMPLE (2 · 3 · 5 / 15)² = 2² 3² 5² / 15² = 900 / 225 = 4 .
Operations with roots. In all the formulas below, the symbol means arithmetic root(the radical expression is positive).
1. The root of the product of several factors is equal to the product roots of these factors:
2. The root of a ratio is equal to the ratio of the roots of the dividend and the divisor:
3. When raising a root to a power, it is enough to raise to this power radical number:
4. If we increase the degree of the root in m raise to m the th power is a radical number, then the value of the root will not change:
5. If we reduce the degree of the root in m extract the root once and at the same time m th power of a radical number, then the value of the root is not will change:
Expanding the concept of degree. So far we have considered degrees only with natural exponents; but actions with degrees and roots can also lead to negative, zero And fractional indicators. All these exponents require additional definition.
Degree with a negative exponent. Power of some number c a negative (integer) exponent is defined as one divided by a power of the same number with an exponent equal to the absolute valuenegative indicator:
T now the formula a m: a n= a m - n can be used not only form, more than n, but also with m, less than n .
EXAMPLE a 4 :a 7 =a 4 - 7 =a - 3 .
If we want the formulaa m : a n= a m - nwas fair whenm = n, we need a definition of degree zero.
A degree with a zero index. The power of any non-zero number with exponent zero is 1.
EXAMPLES. 2 0 = 1, ( – 5) 0 = 1, (– 3 / 5) 0 = 1.
Degree with a fractional exponent. To raise a real number and to the power m/n , you need to extract the root nth power of m -th power of this number A :
About expressions that have no meaning. There are several such expressions. any number.
In fact, if we assume that this expression is equal to some number x, then according to the definition of the division operation we have: 0 = 0 · x. But this equality occurs when any number x, which was what needed to be proven.
Case 3.
0 0 - any number.
Really,
Solution. Let's consider three main cases:
1) x = 0 – this value does not satisfy this equation
(Why?).
2) when x> 0 we get: x/x = 1, i.e. 1 = 1, which means
What x– any number; but taking into account that in
In our case x> 0, the answer isx > 0 ;
3) when x < 0 получаем: – x/x= 1, i.e. e . –1 = 1, therefore,
In this case there is no solution.
Thus, x > 0.