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, about which 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. This 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 are roots needed 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 for a moment to primary classes. 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 unobvious errors (by the way, the skill of comparing and rounding in mandatory checked 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 given 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 consider several 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, according to appearance root it is almost impossible to guess which 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 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 $y=((x)^(2))$:

Schedule 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:

1. Strictly speaking, each positive number will have two roots with 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 of $n$ 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 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:

1. 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 definitely intersect with our graph. Consequently, the cube root can always be taken from absolutely any number;
2. 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:

1. 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.
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. 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 take the root of the same power from it, we will not get the original number, but its modulus. This simple theorem, which is easy to prove (it is enough to separately consider 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 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:

1. 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;
2. 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 because total quantity There are 4 minuses in the work, 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 under the radical sign it is also always not negative number. Otherwise, the root is undefined.

Note on procedure

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 there is always a non-negative number under the root sign, since $((a)^(2))\ge 0$ in any case;
2. 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 the root has a negative number and its exponent is even, we 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 very useful property, which allows you to “throw out” all the negatives:

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

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!

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, 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 earlier? 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 algebraic root- this is not a specific number, but a set. And since we work with real numbers, this set comes in only three types:

1. Empty set. Occurs when you need to find an algebraic root of an even degree from a negative number;
2. 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;
3. 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, in modern school course In mathematics, complex numbers are almost never encountered. They have been removed from most textbooks because our officials consider the topic “too difficult to understand.”

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

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

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

But you shouldn't use this root solution. And here's why. Firstly, you will have to spend a lot of time on such calculations. Numbers at the root, or more precisely, expressions can be quite complex, and the degree is not necessarily quadratic or cubic. Secondly, the accuracy of such calculations is not always satisfactory. 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 n, will be equal to the value of the radical expression, where n is the power 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 the roots in online calculator just comes down to writing mathematical expression in the input line. Extracting from a root in the calculator is designated as sqrt and is performed using three keys - extract square root sqrt(x), cube root sqrt3(x) and nth root sqrt(x,y). More detailed information about the control panel is presented on the page.

Square Root

Clicking this button will insert the square root entry in the input line: sqrt(x), you only need to enter the radical expression and close the parenthesis.

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 imaginary unit i.

Square root of a negative number:

Third root

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

3rd degree root:

Root of degree n

Naturally, the online roots calculator allows you to extract not only the square and cubic roots of a number, but also the root of degree n. Clicking this button will display an entry like sqrt(x x,y).

4th root:

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

5th root with approximate result:

Root of a 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 of the numerator and denominator.

Square root of a fraction:

Root from the root

In cases where the root of the expression is under the root, according to the properties of 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 indicators of the roots. In the example shown in the figure, the expression third-degree root of the second-degree root can be replaced by one 6th-degree root. Specify the expression as you wish. In any case, the calculator will calculate everything correctly.

An example of how to extract a root from a root:

Degree at the root

The root of the degree calculator allows you to calculate in one step, without first reducing the root and degree indicators.

Square root of a degree:

All functions of our free calculator are collected in one section.

Solving roots in an online calculator was last modified: March 3rd, 2016 by Admin

We have already sorted out a large number without a calculator. In this article we will look at how to extract the cube root (third degree root). Let me make a reservation that we are talking about natural numbers. How long do you think it takes to verbally calculate roots such as:

Quite a bit, and if you practice two or three times for 20 minutes, then you can extract any such root orally in 5 seconds.

*It should be noted that we are talking about numbers under the root that are the result of cubed natural numbers from 0 to 100.

We know that:

So, the number a that we will find is natural number from 0 to 100. Look at the table of cubes of these numbers (results of raising to the third power):

You can easily extract the cube root of any number in this table. What do you need to know?

1. These are cubes of numbers that are multiples of ten:

I would even say that these are “beautiful” numbers, they are easy to remember. It's easy to learn.

2. This is a property of numbers when they are producted.

Its essence lies in the fact that when raising a certain number to the third power, the result will have a peculiarity. Which one?

For example, let's cube 1, 11, 21, 31, 41, etc. You can look at the table.

1 3 = 1, 11 3 = 1331, 21 3 = 9261, 31 3 = 26791, 41 3 = 68921 …

That is, when we cube a number with a unit at the end, the result will always be a number with a unit at the end.

When you cube a number with a two at the end, the result will always be a number with an eight at the end.

Let's show the correspondence in the table for all numbers:

Knowledge of the two points presented is quite enough.

Let's look at examples:

Take the cube root of 21952.

This number is in the range from 8000 to 27000. This means that the result of the root lies in the range from 20 to 30. The number 29952 ends in 2. This option is only possible when a number with an eight at the end is cubed. Thus, the result of the root is 28.

Find the cube root of 54852.

This number is in the range from 27000 to 64000. This means that the result of the root lies in the range from 30 to 40. The number 54852 ends in 2. This option is only possible when a number with an eight at the end is cubed. Thus, the result of the root is 38.

Take the cube root of 571787.

This number is in the range from 512000 to 729000. This means that the result of the root lies in the range from 80 to 90. The number 571787 ends in 7. This option is only possible when a number with a three at the end is cubed. Thus, the result of the root is 83.

Take the cube root of 614125.

This number is in the range from 512000 to 729000. This means that the result of the root lies in the range from 80 to 90. The number 614125 ends in 5. This option is only possible when a number with a five at the end is cubed. Thus, the result of the root is 85.

I think that you can now easily extract the cube root of the number 681472.

Of course, extracting such roots orally takes a little practice. But by restoring the two indicated tablets on paper, you can easily extract such a root within a minute, in any case.

After you have found the result, be sure to check it (raise it to the third power). *Nobody canceled multiplication by column 😉

Actually Unified State Exam problems with such “ugly” roots, no. For example, you need to extract the cube root of 1728. I think this is no longer a problem for you.

If you know any interesting methods of calculations without a calculator, send them, I will publish them in due course.That's all. Good luck to you!

Sincerely, Alexander Krutitskikh.

P.S: I would be grateful if you tell me about the site on social networks.

How many angry words were spoken to him? Sometimes it seems that the cube root is incredibly different from the square root. Actually the difference is not that big. Especially if you understand that they are only special cases common root nth degree.

However, problems may arise with its extraction. But most often they are associated with the cumbersomeness of calculations.

What do you need to know about the root of an arbitrary power?

First, a definition of this concept. The nth root of some “a” is a number that, when raised to the power n, gives the original “a”.

Moreover, there are even and odd degrees at the roots. If n is even, then the radical expression can only be zero or a positive number. Otherwise there will be no real answer.

When the degree is odd, then there is a solution for any value of “a”. It may well be negative.

Secondly, the root function can always be written as a power, the exponent of which is a fraction. Sometimes this can be very convenient.

For example, “a” to the power 1/n will be the nth root of “a”. In this case, the base of the degree is always greater than zero.

Similarly, “a” to the power n/m will be represented as the mth root of “a n”.

Thirdly, all operations with powers are valid for them.

• They can be multiplied. Then the exponents add up.
• The roots can be divided. Degrees will need to be subtracted.
• And raise it to a power. Then they should be multiplied. That is, the degree that was, to the one to which they are raised.

What are the similarities and differences between square and cube roots?

They are similar, like siblings, only their degrees are different. And the principle of their calculation is the same, the only difference is how many times the number must be multiplied by itself in order to obtain the radical expression.

And the significant difference was mentioned a little higher. But it wouldn’t hurt to repeat it. The square is only extracted from a non-negative number. While calculating the cube root of a negative value is not difficult.

Extracting cube root on a calculator

Everyone has done this for square roots at least once. But what if the degree is “3”?

On a regular calculator there is only a button for square, but not for cubic. A simple search of numbers that are multiplied by themselves three times will help here. Did you get a radical expression? So this is the answer. Didn't work out? Pick again.

What is the engineering form of a calculator on a computer? Hooray, there's a cube root here. You can simply press this button, and the program will give you the answer. But that's not all. Here you can calculate not only the 2nd and 3rd degree roots, but also any arbitrary one. Because there is a button that has “y” in the root degree. That is, after pressing this key, you will need to enter another number, which will be equal to the degree of the root, and only then “=”.

Extracting cube roots manually

This method will be required when a calculator is not at hand or cannot be used. Then, in order to calculate the cube root of a number, you will need to make an effort.

First, see if a full cube is obtained from some integer value. Maybe the root is 2, 3, 5 or 10 to the third power?

1. Mentally divide the radical expression into groups of three digits from the decimal point. Most often you need a fractional part. If it is not there, then zeros must be added.
2. Determine the number whose cube is less than the integer part of the radical expression. Write it down in the intermediate answer above the root sign. And under this group place its cube.
3. Perform subtraction.
4. Add the first group of digits after the decimal point to the remainder.
5. In the draft, write down the expression: a 2 * 300 * x + a * 30 * x 2 + x 3. Here “a” is an intermediate answer, “x” is a number that is less than the resulting remainder with the numbers assigned to it.
6. The number “x” must be written after the decimal point of the intermediate answer. And write the value of this entire expression under the remainder being compared.
7. If the accuracy is sufficient, then stop the calculations. Otherwise, you need to return to point number 3.

An illustrative example of calculating a cube root

It is needed because the description may seem complicated. The figure below shows how to take the cube root of 15 to the nearest hundredth.

The only difficulty this method has is that with each step the numbers increase many times and counting in a column becomes more and more difficult.

1. 15> 2 3, means under whole part written 8, and above the root 2.
2. After subtracting eight from 15, the remainder is 7. Three zeros must be added to it.
3. a = 2. Therefore: 2 2 * 300 * x +2 * 30 * x 2 + x 3< 7000, или 1200 х + 60 х 2 + х 3 < 7000.
4. Using the selection method, it turns out that x = 4. 1200 * 4 + 60 * 16 + 64 = 5824.
5. Subtraction gives 1176, and the number 4 appears above the root.
6. Add three zeros to the remainder.
7. a = 24. Then 172800 x + 720 x 2 + x 3< 1176000.
8. x = 6. Evaluating the expression gives the result 1062936. Remainder: 113064, above root 6.
9. Add zeros again.
10. a = 246. The inequality turns out like this: 18154800x + 7380x 2 + x 3< 113064000.
11. x = 6. Calculations give the number: 109194696, Remainder: 3869304. Above root 6.

The answer is the number: 2, 466. Since the answer must be given to the nearest hundredth, it must be rounded: 2.47.

An unusual way to extract cube roots

It can be used when the answer is an integer. Then the cube root is extracted by decomposing the radical expression into odd terms. Moreover, there should be the minimum possible number of such terms.

For example, 8 is represented by the sum of 3 and 5. And 64 = 13 + 15 + 17 + 19.

The answer will be a number that is equal to the number of terms. So the cube root of 8 will be equal to two, and of 64 - four.

If the root is 1000, then its decomposition into terms will be 91 + 109 + 93 + 107 + 95 + 105 + 97 + 103 + 99 + 101. There are 10 terms in total. This is the answer.

If you have a calculator at hand, extracting the cube root of any number will not be a problem. But if you don't have a calculator or you just want to impress others, find the cube root by hand. Most people will find the process described here quite complicated, but with practice, extracting cube roots will become much easier. Before you start reading this article, remember the basic mathematical operations and calculations with cubed numbers.

Steps

Part 1

Extracting cube roots using a simple example

Write down the task. Taking cube roots by hand is similar to long division, but with some nuances. First, write down the task in a specific form.

• Write down the number from which you want to take the cube root. Divide the number into groups of three digits, starting with the decimal point. For example, you need to extract the cube root of 10. Write this number like this: 10,000,000. The additional zeros are intended to increase the accuracy of the result.
• Draw a root sign next to and above the number. Think of it as the horizontal and vertical lines you draw when dividing. The only difference is the shape of the two signs.
• Place a decimal point above the horizontal line. Do this directly above the decimal point of the original number.

1. Remember the results of cubed integers. They will be used in calculations.

   • 1 3 = 1 ∗ 1 ∗ 1 = 1 (\displaystyle 1^(3)=1*1*1=1)
   • 2 3 = 2 ∗ 2 ∗ 2 = 8 (\displaystyle 2^(3)=2*2*2=8)
   • 3 3 = 3 ∗ 3 ∗ 3 = 27 (\displaystyle 3^(3)=3*3*3=27)
   • 4 3 = 4 ∗ 4 ∗ 4 = 64 (\displaystyle 4^(3)=4*4*4=64)
   • 5 3 = 5 ∗ 5 ∗ 5 = 125 (\displaystyle 5^(3)=5*5*5=125)
   • 6 3 = 6 ∗ 6 ∗ 6 = 216 (\displaystyle 6^(3)=6*6*6=216)
   • 7 3 = 7 ∗ 7 ∗ 7 = 343 (\displaystyle 7^(3)=7*7*7=343)
   • 8 3 = 8 ∗ 8 ∗ 8 = 512 (\displaystyle 8^(3)=8*8*8=512)
   • 9 3 = 9 ∗ 9 ∗ 9 = 729 (\displaystyle 9^(3)=9*9*9=729)
   • 10 3 = 10 ∗ 10 ∗ 10 = 1000 (\displaystyle 10^(3)=10*10*10=1000)

2. Find the first digit of the answer. Choose the cube of the integer that is closest but smaller than the first group of three digits.

   • In our example, the first group of three digits is the number 10. Find the largest cube that is less than 10. This cube is 8, and the cube root of 8 is 2.
   • Above the horizontal line above the number 10, write the number 2. Then write down the value of the operation 2 3 (\displaystyle 2^(3))= 8 under 10. Draw a line and subtract 8 from 10 (as with regular long division). The result is 2 (this is the first remainder).
   • Thus, you have found the first digit of the answer. Think about whether this result quite accurate. In most cases this will be a very rough answer. Cube the result to find out how close it is to the original number. In our example: 2 3 (\displaystyle 2^(3))= 8, which is not very close to 10, so the calculations need to be continued.

3. Find the next digit of the answer. Add a second group of three digits to the first remainder, and draw a vertical line to the left of the resulting number. Using the resulting number you will find the second digit of the answer. In our example, we need to add a second group of three digits (000) to the first remainder (2) to get the number 2000.

   • To the left of the vertical line you will write three numbers, the sum of which is equal to a certain first factor. Leave empty spaces for these numbers and put plus signs between them.

4. Find the first term (out of three). In the first empty space, write the result of multiplying the number 300 by the square of the first digit of the answer (it is written above the root sign). In our example, the first digit of the answer is 2, so 300*(2^2) = 300*4 = 1200. Write 1200 in the first blank space. The first term is the number 1200 (plus two more numbers to find).

   Find the second digit of the answer. Find out what number you need to multiply 1200 by so that the result is close, but does not exceed 2000. This number can only be 1, since 2 * 1200 = 2400, which is more than 2000. Write 1 (the second digit of the answer) after 2 and the decimal point above the root sign.

   Find the second and third terms (out of three). The multiplier consists of three numbers (terms), the first of which you have already found (1200). Now we need to find the remaining two terms.

   • Multiply 3 by 10 and by each digit of the answer (they are written above the root sign). In our example: 3*10*2*1 = 60. Add this result to 1200 and get 1260.
   • Finally, square the last digit of your answer. In our example, the last digit of the answer is 1, so 1^2 = 1. Thus, the first factor is equal to the sum of the following numbers: 1200 + 60 + 1 = 1261. Write this number to the left of the vertical bar.

5. Multiply and subtract. Multiply the last digit of the answer (in our example it is 1) by the found factor (1261): 1*1261 = 1261. Write this number under 2000 and subtract it from 2000. You will get 739 (this is the second remainder).

6. Consider whether the answer you receive is accurate enough. Do this every time you complete another subtraction. After the first subtraction, the answer was 2, which is not an accurate result. After the second subtraction, the answer is 2.1.

   • To check the accuracy of your answer, cube it: 2.1*2.1*2.1 = 9.261.
   • If you think the answer is accurate enough, you don't have to continue the calculations; otherwise, do another subtraction.

7. Find the second factor. To practice your calculations and get a more accurate result, repeat the steps above.

   • To the second remainder (739) add the third group of three digits (000). You will get the number 739000.
   • Multiply 300 by the square of the number written above the root sign (21): 300 ∗ 21 2 (\displaystyle 300*21^(2)) = 132300.
   • Find the third digit of the answer. Find out what number you need to multiply 132300 by so that the result is close to, but does not exceed 739000. This number is 5: 5 * 132200 = 661500. Write 5 (the third digit of the answer) after the 1 above the root sign.
   • Multiply 3 by 10 by 21 and by the last digit of the answer (they are written above the root sign). In our example: 3 ∗ 21 ∗ 5 ∗ 10 = 3150 (\displaystyle 3*21*5*10=3150).
   • Finally, square the last digit of your answer. In our example, the last digit of the answer is 5, so 5 2 = 25. (\displaystyle 5^(2)=25.)
   • Thus, the second multiplier is: 132300 + 3150 + 25 = 135475.

8. Multiply the last digit of the answer by the second factor. Once you have found the second factor and third digit of the answer, proceed as follows:

   • Multiply the last digit of the answer by the found factor: 135475*5 = 677375.
   • Subtract: 739000-677375 = 61625.
   • Consider whether the answer you receive is accurate enough. To do this, cube it: 2 , 15 ∗ 2 , 15 ∗ 2 , 15 = 9 , 94 (\displaystyle 2.15*2.15*2.15=9.94).

9. Write down your answer. The result, written above the root sign, is the answer accurate to two decimal places. In our example, the cube root of 10 is 2.15. Check your answer by cubeing it: 2.15^3 = 9.94, which is approximately 10. If you need more precision, continue with the calculation (as described above).

   Part 2

   Extracting the cube root using the estimation method

   1. Use number cubes to determine upper and lower limits. If you need to take the cube root of almost any number, find the cubes (of some numbers) that are close to the given number.

      • For example, you need to take the cube root of 600. Since 8 3 = 512 (\displaystyle 8^(3)=512) And 9 3 = 729 (\displaystyle 9^(3)=729), then the value of the cube root of 600 lies between 8 and 9. Therefore, use the numbers 512 and 729 as the upper and lower limits of the answer.

   2. Estimate the second number. You found the first number thanks to your knowledge of cubes of integers. Now turn the integer into decimal, adding to it (after the decimal point) a certain number from 0 to 9. It is necessary to find a decimal fraction, the cube of which will be close to, but less than the original number.

      • In our example, the number 600 is located between the numbers 512 and 729. For example, add the number 5 to the first number found (8). The number you get is 8.5.
   3. • In our example: 8 , 5 ∗ 8 , 5 ∗ 8 , 5 = 614 , 1. (\displaystyle 8.5*8.5*8.5=614.1.)

   4. Compare the cube of the resulting number with the original number. If the cube of the resulting number is larger than the original number, try estimating the smaller number. If the cube of the resulting number is much smaller than the original number, estimate big numbers until the cube of one of them exceeds the original number.

      • In our example: 8 , 5 3 (\displaystyle 8.5^(3))> 600. So evaluate the smaller number to 8.4. Cube this number and compare it with the original number: 8 , 4 ∗ 8 , 4 ∗ 8 , 4 = 592 , 7 (\displaystyle 8.4*8.4*8.4=592.7). This result is less than the original number. So the cube root of 600 is between 8.4 and 8.5.

   5. Estimate the following number to improve your answer accuracy. For each number you estimated last, add a number from 0 to 9 until you get the exact answer. In each evaluation round, you need to find the upper and lower limits between which the original number lies.

      • In our example: 8 , 4 3 = 592 , 7 (\displaystyle 8.4^(3)=592.7) And 8 , 5 3 = 614 , 1 (\displaystyle 8.5^(3)=614.1). The original number 600 is closer to 592 than to 614. Therefore, to the last number you estimated, assign a figure that is closer to 0 than to 9. For example, such a number is 4. Therefore, cube the number 8.44.

   6. If necessary, estimate a different number. Compare the cube of the resulting number with the original number. If the cube of the resulting number is larger than the original number, try estimating the smaller number. In short, you need to find two numbers whose cubes are slightly larger and slightly smaller than the original number.

      • In our example 8 , 44 ∗ 8 , 44 ∗ 8 , 44 = 601 , 2 (\displaystyle 8.44*8.44*8.44=601.2). This is slightly larger than the original number, so estimate another (smaller) number, such as 8.43: 8 , 43 ∗ 8 , 43 ∗ 8 , 43 = 599 , 07 (\displaystyle 8.43*8.43*8.43=599.07). Thus, the value of the cube root of 600 lies between 8.43 and 8.44.

   7. Follow the described process until you get an answer that you are happy with. Estimate the next number, compare it with the original, then, if necessary, estimate another number, and so on. Please note that each additional digit after the decimal point increases the accuracy of the answer.

      • In our example, the cube of 8.43 is less than 1 less than the original number. If you need more precision, cube 8.434 and get: 8, 434 3 = 599, 93 (\displaystyle 8,434^(3)=599,93), that is, the result is less than 0.1 less than the original number.