Square root formulas division subtraction multiplication addition. Operation with roots: addition and subtraction

The square root of a number x is a number a, which when multiplied by itself gives the number x: a * a = a^2 = x, ?x = a. As with any numbers, you can perform arithmetic operations of addition and subtraction with square roots.

Instructions

1. Firstly, when adding square roots try to extract these roots. This will be acceptable if the numbers under the root sign are perfect squares. Let's say the given expression is ?4 + ?9. The first number 4 is the square of the number 2. The second number 9 is the square of the number 3. Thus it turns out that: ?4 + ?9 = 2 + 3 = 5.

2. If there are no complete squares under the root sign, then try moving the multiplier of the number from under the root sign. Let's say, let's say the expression is given?24 +?54. Factor the numbers: 24 = 2 * 2 * 2 * 3, 54 = 2 * 3 * 3 * 3. The number 24 has a factor of 4, the one that can be transferred from under the sign square root. In the number 54 there is a factor of 9. Thus, it turns out that: ?24 + ?54 = ?(4 * 6) + ?(9 * 6) = 2 * ?6 + 3 * ?6 = 5 * ?6. In this example, as a result of removing the multiplier from under the root sign, it was possible to simplify the given expression.

3. Let the sum of 2 square roots be the denominator of a fraction, say A / (?a + ?b). And let your task be “to get rid of irrationality in the denominator.” Then you can use the next method. Multiply the numerator and denominator of the fraction by the expression ?a – ?b. Thus, the denominator will contain the abbreviated multiplication formula: (?a + ?b) * (?a – ?b) = a – b. By analogy, if the denominator contains the difference between the roots: ?a – ?b, then the numerator and denominator of the fraction must be multiplied by the expression ?a + ?b. For example, let the fraction 4 / (?3 + ?5) = 4 * (?3 – ?5) / ((?3 + ?5) * (?3 – ?5)) = 4 * (?3 – ?5) / (-2) = 2 * (?5 – ?3).

4. Consider a more complex example of getting rid of irrationality in the denominator. Let the fraction 12 / (?2 + ?3 + ?5) be given. You need to multiply the numerator and denominator of the fraction by the expression?2 + ?3 – ?5:12 / (?2 + ?3 + ?5) = 12 * (?2 + ?3 – ?5) / ((?2 + ?3 + ?5) * (?2 + ?3 – ?5)) = 12 * (?2 + ?3 – ?5) / (2 * ?6) = ?6 * (?2 + ?3 – ?5) = 2 * ?3 + 3 * ?2 – ?30.

5. And finally, if you only need an approximate value, you can calculate the square roots using a calculator. Calculate the values ​​separately for the entire number and write it down to the required precision (say, two decimal places). And after that, perform the required arithmetic operations, as with ordinary numbers. Let's say, let's say you need to find out the approximate value of the expression ?7 + ?5 ? 2.65 + 2.24 = 4.89.

Video on the topic

Note!
In no case can square roots be added as primitive numbers, i.e. ?3 + ?2 ? ?5!!!

Helpful advice
If you factor a number in order to move the square from under the root sign, then do reverse check– multiply all the resulting factors and get the original number.

Square root of a number X called number A, which in the process of multiplying by itself ( A*A) can give a number X.
Those. A * A = A 2 = X, And √X = A.

Above square roots ( √x), like other numbers, you can perform arithmetic operations such as subtraction and addition. To subtract and add roots, they need to be connected using signs corresponding to these actions (for example √x - √y ).
And then bring the roots to their simplest form - if there are similar ones between them, it is necessary to make a reduction. It consists in taking the coefficients of similar terms with the signs of the corresponding terms, then putting them in parentheses and deducing common root outside the multiplier brackets. The coefficient we obtained is simplified according to the usual rules.

Step 1: Extracting square roots

Firstly, to add square roots, you first need to extract these roots. This can be done if the numbers under the root sign are perfect squares. For example, take the given expression √4 + √9 . First number 4 is the square of the number 2 . Second number 9 is the square of the number 3 . Thus, we can obtain the following equality: √4 + √9 = 2 + 3 = 5 .
That's it, the example is solved. But it doesn’t always happen that easily.

Step 2. Taking out the multiplier of the number from under the root

If there are no perfect squares under the root sign, you can try to remove the multiplier of the number from under the root sign. For example, let's take the expression √24 + √54 .

Factor the numbers:
24 = 2 * 2 * 2 * 3 ,
54 = 2 * 3 * 3 * 3 .

Among 24 we have a multiplier 4 , it can be taken out from under the square root sign. Among 54 we have a multiplier 9 .

We get equality:
√24 + √54 = √(4 * 6) + √(9 * 6) = 2 * √6 + 3 * √6 = 5 * √6 .

Considering this example, we get the factor taken out from under the root sign, thereby simplifying the given expression.

Step 3: Reducing the Denominator

Consider the following situation: the sum of two square roots is the denominator of the fraction, for example, A/(√a + √b).
Now we are faced with the task of “getting rid of irrationality in the denominator.”
Let's use the following method: multiply the numerator and denominator of the fraction by the expression √a - √b.

We now get the abbreviated multiplication formula in the denominator:
(√a + √b) * (√a - √b) = a - b.

Similarly, if the denominator has a root difference: √a - √b, the numerator and denominator of the fraction are multiplied by the expression √a + √b.

Let's take the fraction as an example:
4 / (√3 + √5) = 4 * (√3 - √5) / ((√3 + √5) * (√3 - √5)) = 4 * (√3 - √5) / (-2) = 2 * (√5 - √3) .

Example of complex denominator reduction

Now we will consider a rather complex example of getting rid of irrationality in the denominator.

For example, let's take a fraction: 12 / (√2 + √3 + √5) .
You need to take its numerator and denominator and multiply by the expression √2 + √3 - √5 .

We get:

12 / (√2 + √3 + √5) = 12 * (√2 + √3 - √5) / (2 * √6) = 2 * √3 + 3 * √2 - √30.

Step 4. Calculate the approximate value on the calculator

If you only need an approximate value, this can be done on a calculator by calculating the value of the square roots. The value is calculated separately for each number and written down with the required accuracy, which is determined by the number of decimal places. Next, all the required operations are performed, as with ordinary numbers.

Example of calculating an approximate value

It is necessary to calculate the approximate value of this expression √7 + √5 .

As a result we get:

√7 + √5 ≈ 2,65 + 2,24 = 4,89 .

Please note: under no circumstances should you add square roots as prime numbers; this is completely unacceptable. That is, if we add the square root of five and the square root of three, we cannot get the square root of eight.

Helpful advice: if you decide to factor a number, in order to derive the square from under the root sign, you need to do a reverse check, that is, multiply all the factors that resulted from the calculations, and end result This mathematical calculation should result in the number that was originally given to us.

In mathematics, any action has its opposite pair - in essence, this is one of the manifestations of the Hegelian law of dialectics: “the unity and struggle of opposites.” One of the actions in such a “pair” is aimed at increasing the number, and the other, its opposite, is aimed at decreasing it. For example, the opposite of addition is subtraction, and division is the opposite of multiplication. Exponentiation also has its own dialectical opposite pair. We are talking about extracting the root.

To extract the root of such and such a power from a number means to calculate which number must be raised to the appropriate power in order to end up with a given number. The two degrees have their own separate names: the second degree is called “square”, and the third is called “cube”. Accordingly, it is nice to call the roots of these powers square and cubic roots. Actions with cube roots are a topic for a separate discussion, but now let's talk about adding square roots.

Let's start with the fact that in some cases it is easier to first extract square roots and then add the results. Suppose we need to find the value of the following expression:

After all, it’s not at all difficult to calculate that the square root of 16 is 4, and of 121 is 11. Therefore,

√16+√121=4+11=15

However, this is the simplest case - here we're talking about about perfect squares, i.e. about those numbers that are obtained by squaring integers. But this doesn't always happen. For example, the number 24 is not a perfect square (there is no integer that, when raised to the second power, would result in 24). The same applies to a number like 54... What if we need to add the square roots of these numbers?

In this case, we will receive in the answer not a number, but another expression. The maximum we can do here is to simplify the original expression as much as possible. To do this, you will have to take out the factors from under the square root. Let's see how this is done using the numbers mentioned above as an example:

To begin with, let's factor 24 into factors so that one of them can easily be extracted as a square root (i.e., so that it is a perfect square). There is such a number – it’s 4:

Now let's do the same with 54. In its composition, this number will be 9:

Thus, we get the following:

√24+√54=√(4*6)+ √(9*6)

Now let’s extract the roots from what we can extract them from: 2*√6+3*√6

There is a common factor here that we can take out of brackets:

(2+3)* √6=5*√6

This will be the result of addition - nothing more can be extracted here.

True, you can resort to using a calculator - however, the result will be approximate and with a huge number of decimal places:

√6=2,449489742783178

Gradually rounding it up, we get approximately 2.5. If we would still like to bring the solution to the previous example to its logical conclusion, we can multiply this result by 5 - and we will get 12.5. More exact result cannot be obtained with such initial data.

Fact 1.
\(\bullet\) Let's take some non-negative number \(a\) (that is, \(a\geqslant 0\) ). Then (arithmetic) square root from the number \(a\) is called such a non-negative number \(b\) , when squared we get the number \(a\) : \[\sqrt a=b\quad \text(same as )\quad a=b^2\] From the definition it follows that \(a\geqslant 0, b\geqslant 0\). These restrictions are an important condition for the existence of a square root and should be remembered!
Remember that any number when squared gives a non-negative result. That is, \(100^2=10000\geqslant 0\) and \((-100)^2=10000\geqslant 0\) .
\(\bullet\) What is \(\sqrt(25)\) equal to? We know that \(5^2=25\) and \((-5)^2=25\) . Since by definition we must find a non-negative number, then \(-5\) is not suitable, therefore, \(\sqrt(25)=5\) (since \(25=5^2\) ).
Finding the value of \(\sqrt a\) is called taking the square root of the number \(a\) , and the number \(a\) is called the radical expression.
\(\bullet\) Based on the definition, expression \(\sqrt(-25)\), \(\sqrt(-4)\), etc. don't make sense.

Fact 2.
For quick calculations it will be useful to learn the table of squares natural numbers from \(1\) to \(20\) : \[\begin(array)(|ll|) \hline 1^2=1 & \quad11^2=121 \\ 2^2=4 & \quad12^2=144\\ 3^2=9 & \quad13 ^2=169\\ 4^2=16 & \quad14^2=196\\ 5^2=25 & \quad15^2=225\\ 6^2=36 & \quad16^2=256\\ 7^ 2=49 & \quad17^2=289\\ 8^2=64 & \quad18^2=324\\ 9^2=81 & \quad19^2=361\\ 10^2=100& \quad20^2= 400\\ \hline \end(array)\]

Fact 3.
What operations can you do with square roots?
\(\bullet\) The sum or difference of square roots IS NOT EQUAL to the square root of the sum or difference, that is \[\sqrt a\pm\sqrt b\ne \sqrt(a\pm b)\] Thus, if you need to calculate, for example, \(\sqrt(25)+\sqrt(49)\) , then initially you must find the values ​​of \(\sqrt(25)\) and \(\sqrt(49)\ ) and then fold them. Hence, \[\sqrt(25)+\sqrt(49)=5+7=12\] If the values ​​\(\sqrt a\) or \(\sqrt b\) cannot be found when adding \(\sqrt a+\sqrt b\), then such an expression is not transformed further and remains as it is. For example, in the sum \(\sqrt 2+ \sqrt (49)\) we can find \(\sqrt(49)\) is \(7\) , but \(\sqrt 2\) cannot be transformed in any way, That's why \(\sqrt 2+\sqrt(49)=\sqrt 2+7\). Unfortunately, this expression cannot be simplified further\(\bullet\) The product/quotient of square roots is equal to the square root of the product/quotient, that is \[\sqrt a\cdot \sqrt b=\sqrt(ab)\quad \text(s)\quad \sqrt a:\sqrt b=\sqrt(a:b)\] (provided that both sides of the equalities make sense)
Example: \(\sqrt(32)\cdot \sqrt 2=\sqrt(32\cdot 2)=\sqrt(64)=8\); \(\sqrt(768):\sqrt3=\sqrt(768:3)=\sqrt(256)=16\); \(\sqrt((-25)\cdot (-64))=\sqrt(25\cdot 64)=\sqrt(25)\cdot \sqrt(64)= 5\cdot 8=40\). \(\bullet\) Using these properties, it is convenient to find square roots of large numbers by factoring them.
Let's look at an example. Let's find \(\sqrt(44100)\) . Since \(44100:100=441\) , then \(44100=100\cdot 441\) . According to the criterion of divisibility, the number \(441\) is divisible by \(9\) (since the sum of its digits is 9 and is divisible by 9), therefore, \(441:9=49\), that is, \(441=9\ cdot 49\) .
Thus we got: \[\sqrt(44100)=\sqrt(9\cdot 49\cdot 100)= \sqrt9\cdot \sqrt(49)\cdot \sqrt(100)=3\cdot 7\cdot 10=210\] Let's look at another example: \[\sqrt(\dfrac(32\cdot 294)(27))= \sqrt(\dfrac(16\cdot 2\cdot 3\cdot 49\cdot 2)(9\cdot 3))= \sqrt( \ dfrac(16\cdot4\cdot49)(9))=\dfrac(\sqrt(16)\cdot \sqrt4 \cdot \sqrt(49))(\sqrt9)=\dfrac(4\cdot 2\cdot 7)3 =\dfrac(56)3\]
\(\bullet\) Let's show how to enter numbers under the square root sign using the example of the expression \(5\sqrt2\) (short notation for the expression \(5\cdot \sqrt2\)). Since \(5=\sqrt(25)\) , then \ Note also that, for example,
1) \(\sqrt2+3\sqrt2=4\sqrt2\) ,
2) \(5\sqrt3-\sqrt3=4\sqrt3\)
3) \(\sqrt a+\sqrt a=2\sqrt a\) .

Why is that? Let's explain using example 1). As you already understand, we cannot somehow transform the number \(\sqrt2\). Let's imagine that \(\sqrt2\) is some number \(a\) . Accordingly, the expression \(\sqrt2+3\sqrt2\) is nothing more than \(a+3a\) (one number \(a\) plus three more of the same numbers \(a\)). And we know that this is equal to four such numbers \(a\) , that is, \(4\sqrt2\) .

Fact 4.
\(\bullet\) They often say “you can’t extract the root” when you can’t get rid of the sign \(\sqrt () \ \) of the root (radical) when finding the value of a number. For example, you can take the root of the number \(16\) because \(16=4^2\) , therefore \(\sqrt(16)=4\) . But it is impossible to extract the root of the number \(3\), that is, to find \(\sqrt3\), because there is no number that squared will give \(3\) .
Such numbers (or expressions with such numbers) are irrational. For example, numbers \(\sqrt3, \ 1+\sqrt2, \ \sqrt(15)\) and so on. are irrational.
Also irrational are the numbers \(\pi\) (the number “pi”, approximately equal to \(3.14\)), \(e\) (this number is called the Euler number, it is approximately equal to \(2.7\)) etc.
\(\bullet\) Please note that any number will be either rational or irrational. And together all rational and all irrational numbers form a set called a set of real numbers. This set is denoted by the letter \(\mathbb(R)\) .
This means that all the numbers that are on this moment we know are called real numbers.

Fact 5.
\(\bullet\) The modulus of a real number \(a\) is a non-negative number \(|a|\) equal to the distance from the point \(a\) to \(0\) on the real line. For example, \(|3|\) and \(|-3|\) are equal to 3, since the distances from the points \(3\) and \(-3\) to \(0\) are the same and equal to \(3 \) .
\(\bullet\) If \(a\) is a non-negative number, then \(|a|=a\) .
Example: \(|5|=5\) ; \(\qquad |\sqrt2|=\sqrt2\) . \(\bullet\) If \(a\) is a negative number, then \(|a|=-a\) .
Example: \(|-5|=-(-5)=5\) ; \(\qquad |-\sqrt3|=-(-\sqrt3)=\sqrt3\).
They say that for negative numbers the modulus “eats” the minus, while positive numbers, as well as the number \(0\), are left unchanged by the modulus.
BUT This rule only applies to numbers. If under your modulus sign there is an unknown \(x\) (or some other unknown), for example, \(|x|\) , about which we do not know whether it is positive, zero or negative, then get rid of the modulus we can not. In this case, this expression remains the same: \(|x|\) . \(\bullet\) The following formulas hold: \[(\large(\sqrt(a^2)=|a|))\] \[(\large((\sqrt(a))^2=a)), \text( provided ) a\geqslant 0\] Very often the following mistake is made: they say that \(\sqrt(a^2)\) and \((\sqrt a)^2\) are one and the same. This is only true if \(a\) is a positive number or zero. But if \(a\) is a negative number, then this is false. It is enough to consider this example. Let's take instead of \(a\) the number \(-1\) . Then \(\sqrt((-1)^2)=\sqrt(1)=1\) , but the expression \((\sqrt (-1))^2\) does not exist at all (after all, it is impossible to use the root sign put negative numbers!).
Therefore, we draw your attention to the fact that \(\sqrt(a^2)\) is not equal to \((\sqrt a)^2\) ! Example: 1) \(\sqrt(\left(-\sqrt2\right)^2)=|-\sqrt2|=\sqrt2\), because \(-\sqrt2<0\) ;

\(\phantom(00000)\) 2) \((\sqrt(2))^2=2\) . \(\bullet\) Since \(\sqrt(a^2)=|a|\) , then \[\sqrt(a^(2n))=|a^n|\] (the expression \(2n\) denotes an even number)
That is, when taking the root of a number that is to some degree, this degree is halved.
Example:
1) \(\sqrt(4^6)=|4^3|=4^3=64\)
2) \(\sqrt((-25)^2)=|-25|=25\) (note that if the module is not supplied, it turns out that the root of the number is equal to \(-25\) ; but we remember , that by definition of a root this cannot happen: when extracting a root, we should always get a positive number or zero)
3) \(\sqrt(x^(16))=|x^8|=x^8\) (since any number to an even power is non-negative)

Fact 6.
How to compare two square roots?
\(\bullet\) For square roots it is true: if \(\sqrt a<\sqrt b\) , то \(aExample:
1) compare \(\sqrt(50)\) and \(6\sqrt2\) . First, let's transform the second expression into \(\sqrt(36)\cdot \sqrt2=\sqrt(36\cdot 2)=\sqrt(72)\). Thus, since \(50<72\) , то и \(\sqrt{50}<\sqrt{72}\) . Следовательно, \(\sqrt{50}<6\sqrt2\) .
2) Between what integers is \(\sqrt(50)\) located?
Since \(\sqrt(49)=7\) , \(\sqrt(64)=8\) , and \(49<50<64\) , то \(7<\sqrt{50}<8\) , то есть число \(\sqrt{50}\) находится между числами \(7\) и \(8\) .
3) Let's compare \(\sqrt 2-1\) and \(0.5\) . Let's assume that \(\sqrt2-1>0.5\) : \[\begin(aligned) &\sqrt 2-1>0.5 \ \big| +1\quad \text((add one to both sides))\\ &\sqrt2>0.5+1 \ \big| \ ^2 \quad\text((squaring both sides))\\ &2>1.5^2\\ &2>2.25 \end(aligned)\] We see that we have obtained an incorrect inequality. Therefore, our assumption was incorrect and \(\sqrt 2-1<0,5\) .
Note that adding a certain number to both sides of the inequality does not affect its sign. Multiplying/dividing both sides of an inequality by a positive number also does not affect its sign, but multiplying/dividing by a negative number reverses the sign of the inequality!
You can square both sides of an equation/inequality ONLY IF both sides are non-negative. For example, in the inequality from the previous example you can square both sides, in the inequality \(-3<\sqrt2\) нельзя (убедитесь в этом сами)! \(\bullet\) It should be remembered that \[\begin(aligned) &\sqrt 2\approx 1.4\\ &\sqrt 3\approx 1.7 \end(aligned)\] Knowing the approximate meaning of these numbers will help you when comparing numbers! \(\bullet\) In order to extract the root (if it can be extracted) from some large number that is not in the table of squares, you must first determine between which “hundreds” it is located, then – between which “tens”, and then determine the last digit of this number. Let's show how this works with an example.
Let's take \(\sqrt(28224)\) . We know that \(100^2=10\,000\), \(200^2=40\,000\), etc. Note that \(28224\) is between \(10\,000\) and \(40\,000\) . Therefore, \(\sqrt(28224)\) is between \(100\) and \(200\) .
Now let’s determine between which “tens” our number is located (that is, for example, between \(120\) and \(130\)). Also from the table of squares we know that \(11^2=121\) , \(12^2=144\) etc., then \(110^2=12100\) , \(120^2=14400 \) , \(130^2=16900\) , \(140^2=19600\) , \(150^2=22500\) , \(160^2=25600\) , \(170^2=28900 \) . So we see that \(28224\) is between \(160^2\) and \(170^2\) . Therefore, the number \(\sqrt(28224)\) is between \(160\) and \(170\) .
Let's try to determine the last digit. Let's remember what single-digit numbers, when squared, give \(4\) at the end? These are \(2^2\) and \(8^2\) . Therefore, \(\sqrt(28224)\) will end in either 2 or 8. Let's check this. Let's find \(162^2\) and \(168^2\) :
\(162^2=162\cdot 162=26224\)
\(168^2=168\cdot 168=28224\) .
Therefore, \(\sqrt(28224)=168\) . Voila!

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I looked again at the sign... And, let's go!

Let's start with something simple:

Just a minute. this, which means we can write it like this:

Got it? Here's the next one for you:

Are the roots of the resulting numbers not exactly extracted? No problem - here are some examples:

What if there are not two, but more multipliers? The same! The formula for multiplying roots works with any number of factors:

Now completely on your own:

Answers: Well done! Agree, everything is very easy, the main thing is to know the multiplication table!

Root division

We've sorted out the multiplication of roots, now let's move on to the property of division.

Let me remind you that the general formula looks like this:

Which means that the root of the quotient is equal to the quotient of the roots.

Well, let's look at some examples:

That's all science is. Here's an example:

Everything is not as smooth as in the first example, but, as you can see, there is nothing complicated.

What if you come across this expression:

You just need to apply the formula in the opposite direction:

And here's an example:

You may also come across this expression:

Everything is the same, only here you need to remember how to translate fractions (if you don’t remember, look at the topic and come back!). Do you remember? Now let's decide!

I am sure that you have coped with everything, now let’s try to raise the roots to degrees.

Exponentiation

What happens if the square root is squared? It's simple, remember the meaning of the square root of a number - this is a number whose square root is equal to.

So, if we square a number whose square root is equal, what do we get?

Well, of course, !

Let's look at examples:

It's simple, right? What if the root is to a different degree? It's OK!

Follow the same logic and remember the properties and possible actions with degrees.

Read the theory on the topic “” and everything will become extremely clear to you.

For example, here is an expression:

In this example, the degree is even, but what if it is odd? Again, apply the properties of exponents and factor everything:

Everything seems clear with this, but how to extract the root of a number to a power? Here, for example, is this:

Pretty simple, right? What if the degree is greater than two? We follow the same logic using the properties of degrees:

Well, is everything clear? Then solve the examples yourself:

And here are the answers:

Entering under the sign of the root

What haven’t we learned to do with roots! All that remains is to practice entering the number under the root sign!

It's really easy!

Let's say we have a number written down

What can we do with it? Well, of course, hide the three under the root, remembering that the three is the square root of!

Why do we need this? Yes, just to expand our capabilities when solving examples:

How do you like this property of roots? Does it make life much easier? For me, that's exactly right! Only We must remember that we can only enter positive numbers under the square root sign.

Solve this example yourself -
Did you manage? Let's see what you should get:

Well done! You managed to enter the number under the root sign! Let's move on to something equally important - let's look at how to compare numbers containing a square root!

Comparison of roots

Why do we need to learn to compare numbers that contain a square root?

Very simple. Often, in large and long expressions encountered in the exam, we receive an irrational answer (remember what this is? We already talked about this today!)

We need to place the received answers on the coordinate line, for example, to determine which interval is suitable for solving the equation. And here the problem arises: there is no calculator in the exam, and without it, how can you imagine which number is greater and which is less? That's it!

For example, determine which is greater: or?

You can’t tell right away. Well, let's use the disassembled property of entering a number under the root sign?

Then go ahead:

Well, obviously, the larger the number under the root sign, the larger the root itself!

Those. if, then, .

From this we firmly conclude that. And no one will convince us otherwise!

Extracting roots from large numbers

Before this, we entered a multiplier under the sign of the root, but how to remove it? You just need to factor it into factors and extract what you extract!

It was possible to take a different path and expand into other factors:

Not bad, right? Any of these approaches is correct, decide as you wish.

Factoring is very useful when solving such non-standard problems as this:

Let's not be afraid, but act! Let's decompose each factor under the root into separate factors:

Now try it yourself (without a calculator! It won’t be on the exam):

Is this the end? Let's not stop halfway!

That's all, it's not so scary, right?

Happened? Well done, that's right!

Now try this example:

But the example is a tough nut to crack, so you can’t immediately figure out how to approach it. But, of course, we can handle it.

Well, let's start factoring? Let us immediately note that you can divide a number by (remember the signs of divisibility):

Now, try it yourself (again, without a calculator!):

Well, did it work? Well done, that's right!

Let's sum it up

1. The square root (arithmetic square root) of a non-negative number is a non-negative number whose square is equal to.
   .
2. If we simply take the square root of something, we always get one non-negative result.
3. Properties of an arithmetic root:
4. When comparing square roots, it is necessary to remember that the larger the number under the root sign, the larger the root itself.

How's the square root? All clear?

We tried to explain to you without any fuss everything you need to know in the exam about the square root.

It's your turn. Write to us whether this topic is difficult for you or not.

Did you learn something new or was everything already clear?

Write in the comments and good luck on your exams!