Extract the root 4. Extract the roots: methods, methods, solutions. Bitwise Finding the Root Value

Posted on our website. Extracting the root of a number is often used in various calculations, and our calculator is a great tool for such mathematical calculations.

An online calculator with roots will allow you to quickly and easily make any calculations containing root extraction. The third root is just as easy to calculate as the square root of a number, the root of a negative number, the root of a complex number, the root of pi, etc.

The calculation of the root of a number is possible manually. If it is possible to calculate the integer root of a number, then we simply find the value of the root expression from the table of roots. In other cases, the approximate calculation of the roots comes down to decomposing the root expression into the product of simpler factors, which are powers and can be removed from the root sign, simplifying the expression under the root as much as possible.

But you should not use such a root solution. And that's why. First, you have to spend a lot of time on such calculations. The numbers at the root, or rather, the expressions can be quite complex, and the degree is not necessarily quadratic or cubic. Secondly, the accuracy of such calculations is not always satisfied. And thirdly, there is an online root calculator that will do any root extraction for you in a matter of seconds.

To extract a root from a number means to find a number that, when raised to the power of n, will be equal to the value of the root expression, where n is the degree of the root, and the number itself is the base of the root. The root of the 2nd degree is called simple or square, and the root of the third degree is called cubic, omitting the indication of the degree in both cases.

Solving roots in an online calculator comes down to just writing a mathematical expression in the input line. Extracting from the root in the calculator is denoted as sqrt and is performed using three keys - extracting the square root of sqrt(x), extracting the cubic root of sqrt3(x) and extracting the root of n degree sqrt(x,y). More detailed information about the control panel is presented on the page.

Extracting the square root

Pressing this button will insert a square root entry into the input line: sqrt(x), you only need to enter the root expression and close the bracket.

An example of solving square roots in a calculator:

If the root is a negative number, and the degree of the root is even, then the answer will be represented as a complex number with an imaginary unit i.

The square root of a negative number:

Third root

Use this key when you need to calculate the cube root. It inserts the entry sqrt3(x) into the input line.

3rd degree root:

Root of degree n

Naturally, the online root calculator allows you to extract not only the square and cube roots of a number, but also the root of the degree n. Pressing this button will display a record of the form sqrt(x x,y).

4th degree root:

An exact nth root of a number can only be extracted if the number itself is an exact nth power. Otherwise, the calculation will turn out to be approximate, although very close to the ideal, since the accuracy of the online calculator's calculations reaches 14 decimal places.

5th root with approximate result:

The root of the fraction

The calculator can calculate the root from various numbers and expressions. Finding the root of a fraction comes down to separately extracting the root from the numerator and denominator.

The square root of a fraction:

root from root

In cases where the root of the expression is under the root, by the property of the roots, they can be replaced by one root, the degree of which will be equal to the product of the degrees of both. Simply put, to extract a root from a root, it is enough to multiply the exponents of the roots. In the example shown in the figure, the expression root of the third degree of the root of the second degree can be replaced by a single root of the 6th degree. Specify the expression as you like. In any case, the calculator will calculate everything correctly.

The nth root of a number x is a non-negative number z that, when raised to the nth power, becomes x. The definition of the root is included in the list of basic arithmetic operations that we get acquainted with in childhood.

Mathematical notation

"Root" comes from the Latin word radix and today the word "radical" is used as a synonym for this mathematical term. Since the 13th century, mathematicians have denoted the operation of extracting the root with the letter r with a horizontal bar above the radical expression. In the 16th century, the designation V was introduced, which gradually replaced the sign r, but the horizontal line was preserved. It is easy to type it in a printing house or write by hand, but the letter designation of the root - sqrt has spread in electronic publications and programming. This is how we will denote square roots in this article.

Square root

The square radical of a number x is a number z that, when multiplied by itself, becomes x. For example, if we multiply 2 by 2, we get 4. Two in this case is the square root of four. Multiply 5 by 5, we get 25 and now we already know the value of the expression sqrt(25). We can multiply and -12 by -12 and get 144, and the radical 144 will be both 12 and -12. Obviously, square roots can be both positive and negative numbers.

The peculiar dualism of such roots is important for solving quadratic equations, therefore, when searching for answers in such problems, it is required to indicate both roots. When solving algebraic expressions, arithmetic square roots are used, that is, only their positive values.

Numbers whose square roots are integers are called perfect squares. There is a whole sequence of such numbers, the beginning of which looks like:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256…

The square roots of other numbers are irrational numbers. For example, sqrt(3) = 1.73205080757... and so on. This number is infinite and not periodic, which causes some difficulties in calculating such radicals.

The school mathematics course states that you cannot take square roots from negative numbers. As we learn in the high school course of mathematical analysis, this can and should be done - this is what complex numbers are needed for. However, our program is designed to extract real values ​​of roots, so it does not calculate even radicals from negative numbers.

cube root

The cubic radical of a number x is the number z that, when multiplied by itself three times, gives the number x. For example, if we multiply 2 × 2 × 2, we get 8. Therefore, two is the cube root of eight. Multiply four times by itself and get 4 × 4 × 4 = 64. Obviously, four is the cube root of 64. There is an infinite sequence of numbers whose cubic radicals are integers. Its beginning looks like:

1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744…

For the rest of the numbers, cube roots are irrational numbers. Unlike square radicals, cube roots, like any odd roots, can be taken from negative numbers. It's all about the product of numbers less than zero. A minus by a minus gives a plus - a rule known from the school bench. A minus times a plus makes a minus. If we multiply negative numbers an odd number of times, then the result will also be negative, therefore, nothing prevents us from extracting an odd radical from a negative number.

However, the calculator program works differently. In fact, extracting a root is raising to the inverse power. The square root is considered as raising to the power of 1/2, and the cube - 1/3. The formula for raising to the power of 1/3 can be reversed and expressed as 2/6. The result is the same, but it is impossible to extract such a root from a negative number. Thus, our calculator calculates arithmetic roots only from positive numbers.

Nth root

Such an ornate way of calculating radicals allows you to determine the roots of any degree from any expression. You can extract the fifth root of the cube of a number, or the 19th radical of a number to the 12th. All this is elegantly implemented as exponentiation to the power of 3/5 or 12/19, respectively.

Consider an example

Square diagonal

The irrationality of the diagonal of a square was known to the ancient Greeks. They were faced with the problem of calculating the diagonal of a flat square, since its length is always proportional to the square root of two. The formula for determining the length of the diagonal is derived from and ultimately takes the form:

d = a × sqrt(2).

Let's determine the square radical of two using our calculator. Let's enter the value 2 in the "Number (x)" cell, and also 2 in the "Power (n)" cell. As a result, we get the expression sqrt (2) = 1.4142. Thus, for a rough estimate of the diagonal of a square, it is enough to multiply its side by 1.4142.

Conclusion

The search for a radical is a standard arithmetic operation, without which scientific or design calculations are indispensable. Of course, we do not need to determine the roots to solve everyday problems, but our online calculator will definitely come in handy for schoolchildren or students to check their homework in algebra or calculus.

Often the transformation and simplification of mathematical expressions requires the transition from roots to powers and vice versa. This article talks about how to convert a root to a power and vice versa. The theory, practical examples and the most common mistakes are considered.

Transition from powers with fractional exponents to roots

Let's say we have a number with an exponent in the form of an ordinary fraction - a m n. How to write such an expression as a root?

The answer follows from the very definition of degree!

Definition

A positive number a raised to the power m n is the n-th root of a m .

In this case, the following condition must be met:

a > 0 m ∈ ℤ ; n ∈ ℕ.

The fractional power of the number zero is defined similarly, however, in this case, the number m is taken not as an integer, but as a natural number, so that division by 0 does not occur:

0 m n = 0 m n = 0 .

According to the definition, the power a m n can be represented as a root a m n .

For example: 3 2 5 = 3 2 5 , 1 2 3 - 3 4 = 1 2 3 - 3 4 .

However, as already mentioned, we should not forget about the conditions: a > 0 ; m ∈ ℤ ; n ∈ ℕ.

So, the expression - 8 1 3 cannot be represented as - 8 1 3, since the notation - 8 1 3 simply does not make sense - the degree of negative numbers is not defined. At the same time, the root itself - 8 1 3 makes sense.

The transition from degrees with expressions in the base and fractional indicators is carried out similarly throughout the entire range of acceptable values ​​(hereinafter - ODZ) of the original expressions in the base of the degree.

For example, the expression x 2 + 2 x + 1 - 4 1 2 can be represented as the square root x 2 + 2 x + 1 - 4. The expression to the power x 2 + x y z - z 3 - 7 3 goes into the expression x 2 + x · y · z - z 3 - 7 3 for all x , y , z from the ODZ of the given expression.

The reverse replacement of roots with degrees, when instead of an expression with a root, an expression with a degree is written, is also possible. Just reverse the equality from the previous paragraph and get:

Again, the transition is obvious for positive numbers a . For example, 7 6 4 = 7 6 4 , or 2 7 - 5 3 = 2 7 - 5 3 .

For negative a, the roots make sense. For example - 4 2 6 , - 2 3 . However, it is impossible to represent these roots as powers - 4 2 6 and - 2 1 3.

Is it even possible to transform such expressions with powers? Yes, if you make some preliminary transformations. Let's see which ones.

Using the properties of degrees, you can perform transformations of the expression - 4 2 6 .

4 2 6 = - 1 2 4 2 6 = 4 2 6 .

Since 4 > 0 , we can write:

In the case of an odd root of a negative number, we can write:

A 2 m + 1 = - a 2 m + 1 .

Then the expression - 2 3 will take the form:

2 3 = - 2 3 = - 2 1 3 .

Let us now understand how the roots, under which the expressions are contained, are replaced by the degrees containing these expressions at the base.

Denote by the letter A some expression. However, let's not rush to represent A m n as A m n . Let us explain what is meant here. For example, the expression x - 3 2 3 , based on the equality from the first paragraph, would like to be represented as x - 3 2 3 . Such a replacement is possible only for x - 3 ≥ 0, and it is not suitable for the remaining x from the ODZ, since for negative a the formula a m n = a m n does not make sense.

Thus, in the considered example, a transformation of the form A m n = A m n is a transformation that narrows the ODZ, and due to the inaccurate application of the formula A m n = A m n, errors often occur.

In order to correctly move from the root A m n to the degree A m n, several points must be observed:

• If the number m is an integer and odd, and n is a natural number and even, then the formula A m n = A m n is valid for the entire ODZ of variables.
• If m is an integer and odd, and n is natural and odd, then the expression A m n can be replaced:
  - on A m n for all values ​​of variables for which A ≥ 0 ;
  - on - - A m n for all values ​​of variables for which A< 0 ;
• If m is integer and even, and n is any natural number, then A m n can be replaced by A m n .

Let's summarize all these rules in a table and give some examples of their use.

Let's return to the expression x - 3 2 3 . Here m = 2 is an integer and even number, and n = 3 is a natural number. So, the expression x - 3 2 3 will be correctly written as:

x - 3 2 3 = x - 3 2 3 .

Here is another example with roots and powers.

Example. Converting a root to a power

x + 5 - 3 5 = x + 5 - 3 5 , x > - 5 - - x - 5 - 3 5 , x< - 5

Let us substantiate the results given in the table. If the number m is an integer and odd, and n is a positive integer and even, for all variables from the ODZ in the expression A m n the value of A is positive or non-negative (for m > 0). That is why A m n = A m n .

In the second case, when m is an integer, positive and odd, and n is natural and odd, the values ​​of A m n are separated. For variables from the ODZ for which A is non-negative, A m n = A m n = A m n . For variables where A is negative, we get A m n = - A m n = - 1 m · A m n = - A m n = - A m n = - A m n .

Similarly, consider the following case, when m is an integer and even, and n is any natural number. If the value of A is positive or non-negative, then for such values ​​of variables from the ODZ A m n = A m n = A m n . For negative A we get A m n = - A m n = - 1 m · A m n = A m n = A m n .

Thus, in the third case, for all variables from the ODZ, we can write A m n = A m n .

If you notice a mistake in the text, please highlight it and press Ctrl+Enter

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Extraction of roots (square root, cubic root, as well as the root of the n-th degree);
ex (e to x power), exponent;
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inverse trigonometric functions: arcsine - sin-1, arccosine - cos-1, arctangent - tan-1;
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Congratulations: today we will analyze the roots - one of the most mind-blowing topics of the 8th grade. :)

Many people get confused about the roots not because they are complex (which is complicated - a couple of definitions and a couple more properties), but because in most school textbooks the roots are defined through such wilds that only the authors of the textbooks themselves can understand this scribbling. And even then only with a bottle of good whiskey. :)

Therefore, now I will give the most correct and most competent definition of the root - the only one that you really need to remember. And only then I will explain: why all this is necessary and how to apply it in practice.

But first, remember one important point, which for some reason many compilers of textbooks “forget” about:

Roots can be of even degree (our favorite $\sqrt(a)$, as well as any $\sqrt(a)$ and even $\sqrt(a)$) and odd degree (any $\sqrt(a)$, $\ sqrt(a)$ etc.). And the definition of the root of an odd degree is somewhat different from the even one.

Here in this fucking “somewhat different” is hidden, probably, 95% of all errors and misunderstandings associated with the roots. So let's clear up the terminology once and for all:

Definition. Even root n from the number $a$ is any non-negative a number $b$ such that $((b)^(n))=a$. And the root of an odd degree from 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 an even degree), and for $n=3$ we get a cubic root (an 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 $((1)^(2))=1$.

Cubic roots are also common - do not 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 is the difference between an even and an odd degree, reread the definition again. It is very important!

In the meantime, we will consider one unpleasant feature of the 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 did mathematicians smoke when they came up with this?” And really: why do we need all these roots?

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 were tastier, our main concern was to multiply the numbers correctly. Well, something in the spirit of "five by five - twenty-five", that's all. But after all, you can multiply numbers not in pairs, but in triplets, fours, 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 to write down the multiplication of ten fives like this:

So they came up with degrees. Why not write the number of factors as a superscript instead of a long string? Like this one:

It's very convenient! All calculations are reduced by several times, and you can not spend a bunch of parchment sheets of notebooks to write down some 5 183 . Such an entry was called the degree of a number, a bunch of properties were found in it, but happiness turned out to be short-lived.

After a grandiose booze, which was organized just about the “discovery” of degrees, some especially stoned mathematician suddenly asked: “What if we know the degree of a number, but we don’t know the number itself?” Indeed, if we know that a certain number $b$, for example, gives 243 to the 5th power, 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 the majority of “ready-made” degrees 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 you need to find a certain number, which, when multiplied by itself three times, will give us 50. But what is this number? It is clearly greater than 3 because 3 3 = 27< 50. С тем же успехом оно меньше 4, поскольку 4 3 = 64 >50. I.e. this number lies somewhere between three and four, but what it is equal to - FIG you will understand.

This is exactly why mathematicians came up with $n$-th roots. That is why the radical icon $\sqrt(*)$ was introduced. To denote the same number $b$, which, to the specified power, will give us a previously known value

\[\sqrt[n](a)=b\Rightarrow ((b)^(n))=a\]

I do not argue: often these roots are easily considered - 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 are in for a cruel 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 drive 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 instance:

\[\sqrt(2)=1.4142...\approx 1.4 \lt 1.5\]

Or here's another example:

\[\sqrt(3)=1.73205...\approx 1.7 \gt 1.5\]

But all these roundings are, firstly, rather 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 necessarily checked at the profile exam).

Therefore, in serious mathematics, one cannot do without roots - they are the same equal representatives of the set of all real numbers $\mathbb(R)$, like fractions and integers that we have long known.

The impossibility of representing the 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, degrees, limits, etc.). But more on that another time.

Consider 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, by the appearance of the root, it is almost impossible to guess which numbers will come after the decimal point. However, it is possible to calculate on a calculator, but even the most advanced date calculator gives us only the first few digits of an irrational number. Therefore, it is much more correct to write the answers as $\sqrt(5)$ and $\sqrt(-2)$.

That's what they were invented for. To make it easy to write down 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 zero. But cube roots are calmly extracted from absolutely any number - even positive, even 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 $y=4$ (marked in red) is drawn on the graph, 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, therefore it is the root:

But then what to do with the second point? Does the 4 have 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 records as if they want to eat you? :)

The trouble is that if no additional conditions are imposed, then the four will have two square roots - positive and negative. And any positive number will also have two of them. But negative numbers will not have roots at all - this can be seen from the same graph, since the parabola never falls below the axis y, i.e. does not take negative values.

A similar problem occurs for all roots with an even exponent:

1. Strictly speaking, each positive number will have two roots with an even exponent $n$;
2. From negative numbers, the root with even $n$ is not extracted at all.

That is why the definition of an even root $n$ specifically stipulates 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 take a look at the graph of the function $y=((x)^(3))$:

The cubic parabola takes on any value, so the cube root can be taken from any number

Two conclusions can be drawn from this graph:

1. The branches of a cubic parabola, unlike the usual one, go to infinity in both directions - both up and down. Therefore, at whatever height we draw a horizontal line, this line will definitely intersect with our graph. Therefore, the cube root can always be taken, absolutely from any number;
2. In addition, such an intersection will always be unique, so you don’t need to think about which number to consider the “correct” root, and which one to score. That is why the definition of roots for an odd degree is simpler than for an even one (there is no non-negativity requirement).

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 do not argue: what is an arithmetic root - you also need to know. And I will talk about this in detail in a separate lesson. Today we will also talk about it, because without it, all reflections on the roots of the $n$-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.

And all you need to understand is the difference between even and odd numbers. Therefore, once again we will collect everything that you really need to know about the roots:

1. An even root exists only from a non-negative number and is itself always a non-negative number. For negative numbers, such a root is undefined.
2. 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. Clear? Yes, it's obvious! Therefore, now we will practice a little with the calculations.

Basic properties and restrictions

Roots have a lot of strange properties and restrictions - this will be a separate lesson. Therefore, now we will consider only the most important "chip", which applies only to roots with an even exponent. We write this property in the form of 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 degree from this, we will get not the original number, but its modulus. This is a simple theorem that is easy to prove (it suffices to consider separately non-negative $x$, and then separately consider negative ones). 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 the sign of the radical), the students forget this formula together.

To understand the issue in detail, let's forget all the formulas for a minute and try to count two numbers ahead:

\[\sqrt(((3)^(4)))=?\quad \sqrt(((\left(-3 \right))^(4)))=?\]

These are very simple examples. The first example will be solved by most of the people, but on the second, many stick. To solve any such crap without problems, always consider the procedure:

1. First, the number is raised to the fourth power. Well, it's kind of easy. A new number will be obtained, which can even be found in the multiplication table;
2. And now from this new number it is necessary to extract the root of the fourth degree. Those. there is no "reduction" of roots and degrees - these are sequential actions.

Let's deal with 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, for which we need to multiply 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 pieces, and they will all cancel each other out (after all, a minus by a minus gives a plus). Next, extract the root again:

In principle, this line could not be written, since it is a no brainer that the answer will be the same. Those. an even root of the same even power "burns" the minuses, and in this sense the result is indistinguishable from the usual 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 the root of an even degree: the result is always non-negative, and the radical sign is also always a non-negative number. Otherwise, the root is not defined.

Note on the order of operations

1. 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 a non-negative number always sits under the root sign, since $((a)^(2))\ge 0$ anyway;
2. But the notation $((\left(\sqrt(a) \right))^(2))$, on the contrary, means that we first extract the root from a certain number $a$ and only then square the result. Therefore, the number $a$ in no case can be negative - this is a mandatory requirement embedded in the definition.

Thus, in no case should one thoughtlessly reduce the roots and degrees, thereby supposedly "simplifying" the original expression. Because if there is a negative number under the root, and its exponent is even, we will get a lot of problems.

However, all these problems are relevant only for even indicators.

Removing a minus sign from under the root sign

Naturally, roots with odd exponents also have their own feature, which, in principle, does not exist for even ones. Namely:

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

In short, you can take out a minus from under the sign of the roots of an odd degree. This is a very useful property that allows you to "throw" all the minuses out:

\[\begin(align) & \sqrt(-8)=-\sqrt(8)=-2; \\ & \sqrt(-27)\cdot \sqrt(-32)=-\sqrt(27)\cdot \left(-\sqrt(32) \right)= \\ & =\sqrt(27)\cdot \sqrt(32)= \\ & =3\cdot 2=6. \end(align)\]

This simple property greatly simplifies many calculations. Now you don’t need to worry: what if a negative expression got under the root, and 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 “classic” roots are guaranteed to lead us to an error.

And here another definition enters the scene - the very one with which most schools begin the study of irrational expressions. And without which our reasoning would be incomplete. Meet!

arithmetic root

Let's assume for a moment that only positive numbers or, in extreme cases, zero can be under the root sign. Let's score on even / odd indicators, score on all the definitions given above - we will work only with non-negative numbers. What then?

And then we get the arithmetic root - it partially intersects 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 you 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 already familiar to us:

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 root a negative number or not. Because negative numbers are no longer considered in principle.

You may ask: “Well, why do we need such a castrated 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 exponentiation rule:

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

Please note: we can raise the root 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 some examples:

\[\begin(align) & \sqrt(5)=\sqrt(((5)^(2)))=\sqrt(25) \\ & \sqrt(2)=\sqrt(((2)^ (4)))=\sqrt(16) \\ \end(align)\]

Well, what's wrong with that? Why couldn't we do it before? Here's why. Consider a simple expression: $\sqrt(-2)$ is a number that is quite normal in our classical sense, 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 took the minus out from under the radical (we have every right, because the indicator is odd), and in the second, we used the above formula. Those. from the point of view of mathematics, 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 exponentiation formula, which works great for positive numbers and zero, starts to give out complete heresy in the case of negative numbers.

Here, in order to get rid of such ambiguity, they came up with arithmetic roots. A separate large lesson is devoted to them, where we consider in detail all their properties. So now we will not dwell on them - the lesson turned out to be too long anyway.

Algebraic root: for those who want to know more

I thought for a long time: to make this topic in a separate paragraph or not. In the end, I decided to leave here. This material is intended for those who want to understand the roots even better - no longer at the average “school” level, but at the level close to the Olympiad.

So: in addition to the "classical" definition of the root of the $n$-th degree from a number and the associated division into even and odd indicators, there is a more "adult" definition, which does not depend on parity and other subtleties at all. This is called an algebraic root.

Definition. An algebraic $n$-th root of any $a$ is the set of all numbers $b$ such that $((b)^(n))=a$. There is no well-established designation for such roots, so 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 the algebraic root is not a specific number, but a set. And since we are working with real numbers, this set is of only three types:

1. Empty set. Occurs when it is required to find an algebraic root of an even degree from a negative number;
2. A set consisting of a single element. All roots of odd powers, as well as roots of even powers from zero, fall into this category;
3. Finally, the set can include two numbers - the same $((x)_(1))$ and $((x)_(2))=-((x)_(1))$ that we saw on the chart quadratic function. Accordingly, such an alignment 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. Compute 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 exponent of the root is odd.

Finally, the last expression:

\[\overline(\sqrt(-16))=\varnothing \]

We got an empty set. Because there is not a single real number that, when raised to the fourth (that is, even!) Power, will give us a negative number −16.

Final note. Please note: it was not by chance that I noted everywhere that we are working with real numbers. Because there are also complex numbers - it is quite possible to calculate $\sqrt(-16)$ and many other strange things there.

However, in the modern school curriculum of mathematics, complex numbers are almost never found. They have been omitted 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. :)