It is known that the root sign is the square root of some number. However, the root sign means not only an algebraic operation, but is also used in woodworking - in the calculation of relative sizes.

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If you want to learn how to multiply roots "with" or "without" factors, then this article is for you. In it, we will consider methods for multiplying roots:

• without multipliers;
• with multipliers;
• with different indicators.

Root multiplication method without multipliers

Action algorithm:

Make sure that the root has the same exponents (degrees). Recall that the degree is written on the left above the root sign. If there is no degree designation, this means that the root is square, i.e. with degree 2, and it can be multiplied by other roots with degree 2.

Example

Example 1: 18 × 2 = ?

Example 2: 10 × 5 = ?

Example

Example 1: 18 × 2 = 36

Example 2: 10 × 5 = 50

Example 3: 3 3 × 9 3 = 27 3

Simplify root expressions. When we multiply the roots with each other, we can simplify the resulting radical expression to the product of a number (or expression) by a full square or cube:

Example

Example 1: 36 = 6 . 36 is the square root of six (6 × 6 = 36).

Example 2: 50 = (25 × 2) = (5 × 5) × 2 = 5 2 . We decompose the number 50 into the product of 25 and 2. The root of 25 is 5, so we take out 5 from under the root sign and simplify the expression.

Example 3: 27 3 = 3 . The cube root of 27 is 3: 3 × 3 × 3 = 27.

The method of multiplying indicators with multipliers

Action algorithm:

Multiply multipliers. The multiplier is the number that comes before the root sign. In the absence of a multiplier, it is, by default, considered one. Next, you need to multiply the factors:

Example

Example 1: 3 2 × 10 = 3 ? 3 x 1 = 3

Example 2: 4 3 × 3 6 = 12 ? 4 x 3 = 12

Multiply the numbers under the root sign. Once you have multiplied the factors, feel free to multiply the numbers under the root sign:

Example

Example 1: 3 2 × 10 = 3 (2 × 10) = 3 20

Example 2: 4 3 × 3 6 = 12 (3 × 6) = 12 18

Simplify the root expression. Next, you should simplify the values ​​that are under the root sign - you need to take the corresponding numbers out of the root sign. After that, you need to multiply the numbers and factors that come before the root sign:

Example

Example 1: 3 20 = 3 (4 × 5) = 3 (2 × 2) × 5 = (3 × 2) 5 = 6 5

Example 2: 12 18 = 12 (9 × 2) = 12 (3 × 3) × 2 = (12 × 3) 2 = 36 2

Root multiplication method with different exponents

Action algorithm:

Find the least common multiple (LCM) of the exponents. The least common multiple is the smallest number divisible by both exponents.

Example

It is necessary to find the LCM of indicators for the following expression:

The exponents are 3 and 2 . For these two numbers, the least common multiple is the number 6 (it is divisible without a remainder by both 3 and 2). To multiply the roots, an exponent of 6 is needed.

Write each expression with a new exponent:

Find the numbers by which you need to multiply the indicators to get the LCM.

In the expression 5 3 you need to multiply 3 by 2 to get 6 . And in the expression 2 2 - on the contrary, it is necessary to multiply by 3 to get 6.

Raise the number under the root sign to the power equal to the number found in the previous step. For the first expression, 5 needs to be raised to the power of 2, and the second - 2 to the power of 3:

2 → 5 6 = 5 2 6 3 → 2 6 = 2 3 6

Raise to the power of expression and write the result under the root sign:

5 2 6 = (5 × 5) 6 = 25 6 2 3 6 = (2 × 2 × 2) 6 = 8 6

Multiply the numbers under the root:

(8×25) 6

Write result:

(8 × 25) 6 = 200 6

If possible, simplify the expression, but in this case it is not simplified.

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Hello kitties! Last time we analyzed in detail what roots are (if you don’t remember, I recommend reading). The main conclusion of that lesson: there is only one universal definition of roots, which you need to know. The rest is nonsense and a waste of time.

Today we go further. We will learn to multiply roots, we will study some problems associated with multiplication (if these problems are not solved, then they can become fatal on the exam) and we will practice properly. So stock up on popcorn, make yourself comfortable - and we'll start. :)

You haven't smoked yet, have you?

The lesson turned out to be quite large, so I divided it into two parts:

1. First, we'll look at the rules for multiplication. The cap seems to be hinting: this is when there are two roots, there is a “multiply” sign between them - and we want to do something with it.
2. Then we will analyze the reverse situation: there is one big root, and we were impatient to present it as a product of two roots in a simpler way. With what fright it is necessary is a separate question. We will only analyze the algorithm.

For those who can't wait to jump right into Part 2, you're welcome. Let's start with the rest in order.

Basic multiplication rule

Let's start with the simplest - classical square roots. The ones that are denoted by $\sqrt(a)$ and $\sqrt(b)$. For them, everything is generally clear:

multiplication rule. To multiply one square root by another, you just need to multiply their radical expressions, and write the result under the common radical:

\[\sqrt(a)\cdot \sqrt(b)=\sqrt(a\cdot b)\]

No additional restrictions are imposed on the numbers on the right or left: if the multiplier roots exist, then the product also exists.

Examples. Consider four examples with numbers at once:

\[\begin(align) & \sqrt(25)\cdot \sqrt(4)=\sqrt(25\cdot 4)=\sqrt(100)=10; \\ & \sqrt(32)\cdot \sqrt(2)=\sqrt(32\cdot 2)=\sqrt(64)=8; \\ & \sqrt(54)\cdot \sqrt(6)=\sqrt(54\cdot 6)=\sqrt(324)=18; \\ & \sqrt(\frac(3)(17))\cdot \sqrt(\frac(17)(27))=\sqrt(\frac(3)(17)\cdot \frac(17)(27 ))=\sqrt(\frac(1)(9))=\frac(1)(3). \\ \end(align)\]

As you can see, the main meaning of this rule is to simplify irrational expressions. And if in the first example we would have extracted the roots from 25 and 4 without any new rules, then the tin begins: $\sqrt(32)$ and $\sqrt(2)$ do not count by themselves, but their product turns out to be an exact square, so the root of it is equal to a rational number.

Separately, I would like to note the last line. There, both radical expressions are fractions. Thanks to the product, many factors cancel out, and the whole expression turns into an adequate number.

Of course, not everything will always be so beautiful. Sometimes there will be complete crap under the roots - it is not clear what to do with it and how to transform after multiplication. A little later, when you start studying irrational equations and inequalities, there will be all sorts of variables and functions in general. And very often, the compilers of the problems are just counting on the fact that you will find some contracting terms or factors, after which the task will be greatly simplified.

In addition, it is not necessary to multiply exactly two roots. You can multiply three at once, four - yes even ten! This will not change the rule. Take a look:

\[\begin(align) & \sqrt(2)\cdot \sqrt(3)\cdot \sqrt(6)=\sqrt(2\cdot 3\cdot 6)=\sqrt(36)=6; \\ & \sqrt(5)\cdot \sqrt(2)\cdot \sqrt(0.001)=\sqrt(5\cdot 2\cdot 0.001)= \\ & =\sqrt(10\cdot \frac(1) (1000))=\sqrt(\frac(1)(100))=\frac(1)(10). \\ \end(align)\]

And again a small remark on the second example. As you can see, in the third multiplier, there is a decimal fraction under the root - in the process of calculations, we replace it with a regular one, after which everything is easily reduced. So: I highly recommend getting rid of decimal fractions in any irrational expressions (that is, containing at least one radical icon). This will save you a lot of time and nerves in the future.

But it was a lyrical digression. Now let's consider a more general case - when the root exponent contains an arbitrary number $n$, and not just the "classical" two.

The case of an arbitrary indicator

So, we figured out the square roots. And what to do with cubes? Or in general with roots of arbitrary degree $n$? Yes, everything is the same. The rule remains the same:

To multiply two roots of degree $n$, it is enough to multiply their radical expressions, after which the result is written under one radical.

In general, nothing complicated. Unless the volume of calculations can be more. Let's look at a couple of examples:

Examples. Calculate products:

\[\begin(align) & \sqrt(20)\cdot \sqrt(\frac(125)(4))=\sqrt(20\cdot \frac(125)(4))=\sqrt(625)= 5; \\ & \sqrt(\frac(16)(625))\cdot \sqrt(0,16)=\sqrt(\frac(16)(625)\cdot \frac(16)(100))=\sqrt (\frac(64)(((25)^(2))\cdot 25))= \\ & =\sqrt(\frac(((4)^(3)))(((25)^(3 ))))=\sqrt(((\left(\frac(4)(25) \right))^(3)))=\frac(4)(25). \\ \end(align)\]

And again attention to the second expression. We multiply the cube roots, get rid of the decimal fraction, and as a result we get the product of the numbers 625 and 25 in the denominator. This is a rather large number - personally, I won’t immediately calculate what it is equal to.

Therefore, we simply selected the exact cube in the numerator and denominator, and then used one of the key properties (or, if you like, the definition) of the root of the $n$th degree:

\[\begin(align) & \sqrt(((a)^(2n+1)))=a; \\ & \sqrt(((a)^(2n)))=\left| a\right|. \\ \end(align)\]

Such "scams" can save you a lot of time on an exam or test, so remember:

Do not rush to multiply the numbers in the radical expression. First, check: what if the exact degree of any expression is “encrypted” there?

With all the obviousness of this remark, I must admit that most unprepared students point blank do not see the exact degrees. Instead, they multiply everything ahead, and then wonder: why did they get such brutal numbers? :)

However, all this is child's play compared to what we will study now.

Multiplication of roots with different exponents

Well, now we can multiply roots with the same exponents. What if the scores are different? Say, how do you multiply an ordinary $\sqrt(2)$ by some crap like $\sqrt(23)$? Is it even possible to do this?

Yes, of course you can. Everything is done according to this formula:

Root multiplication rule. To multiply $\sqrt[n](a)$ by $\sqrt[p](b)$, just do the following transformation:

\[\sqrt[n](a)\cdot \sqrt[p](b)=\sqrt(((a)^(p))\cdot ((b)^(n)))\]

However, this formula only works if radical expressions are non-negative. This is a very important remark, to which we will return a little later.

For now, let's look at a couple of examples:

\[\begin(align) & \sqrt(3)\cdot \sqrt(2)=\sqrt(((3)^(4))\cdot ((2)^(3)))=\sqrt(81 \cdot8)=\sqrt(648); \\ & \sqrt(2)\cdot \sqrt(7)=\sqrt(((2)^(5))\cdot ((7)^(2)))=\sqrt(32\cdot 49)= \sqrt(1568); \\ & \sqrt(5)\cdot \sqrt(3)=\sqrt(((5)^(4))\cdot ((3)^(2)))=\sqrt(625\cdot 9)= \sqrt(5625). \\ \end(align)\]

As you can see, nothing complicated. Now let's figure out where the non-negativity requirement came from, and what will happen if we violate it. :)

It's easy to multiply roots.

Why do radical expressions have to be non-negative?

Of course, you can become like school teachers and quote a textbook with a smart look:

The requirement of non-negativity is associated with different definitions of roots of even and odd degrees (respectively, their domains of definition are also different).

Well, it became clearer? Personally, when I read this nonsense in the 8th grade, I understood for myself something like this: “The requirement of non-negativity is connected with *#&^@(*#@^#)~%” - in short, I didn’t understand shit at that time. :)

So now I will explain everything in a normal way.

First, let's find out where the multiplication formula above comes from. To do this, let me remind you of one important property of the root:

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

In other words, we can safely raise the root expression to any natural power $k$ - in this case, the root index will have to be multiplied by the same power. Therefore, we can easily reduce any roots to a common indicator, after which we multiply. This is where the multiplication formula comes from:

\[\sqrt[n](a)\cdot \sqrt[p](b)=\sqrt(((a)^(p)))\cdot \sqrt(((b)^(n)))= \sqrt(((a)^(p))\cdot ((b)^(n)))\]

But there is one problem that severely limits the application of all these formulas. Consider this number:

According to the formula just given, we can add any degree. Let's try adding $k=2$:

\[\sqrt(-5)=\sqrt(((\left(-5 \right))^(2)))=\sqrt(((5)^(2)))\]

We removed the minus precisely because the square burns the minus (like any other even degree). And now let's perform the reverse transformation: "reduce" the two in the exponent and degree. After all, any equality can be read both left-to-right and right-to-left:

\[\begin(align) & \sqrt[n](a)=\sqrt(((a)^(k)))\Rightarrow \sqrt(((a)^(k)))=\sqrt[n ](a); \\ & \sqrt(((a)^(k)))=\sqrt[n](a)\Rightarrow \sqrt(((5)^(2)))=\sqrt(((5)^( 2)))=\sqrt(5). \\ \end(align)\]

But then something crazy happens:

\[\sqrt(-5)=\sqrt(5)\]

This can't be because $\sqrt(-5) \lt 0$ and $\sqrt(5) \gt 0$. This means that for even powers and negative numbers, our formula no longer works. After which we have two options:

1. To fight against the wall to state that mathematics is a stupid science, where “there are some rules, but this is inaccurate”;
2. Introduce additional restrictions under which the formula will become 100% working.

In the first option, we will have to constantly catch “non-working” cases - this is difficult, long and generally fu. Therefore, mathematicians preferred the second option. :)

But don't worry! In practice, this restriction does not affect the calculations in any way, because all the described problems concern only the roots of an odd degree, and minuses can be taken out of them.

Therefore, we formulate another rule that applies in general to all actions with roots:

Before multiplying the roots, make sure that the radical expressions are non-negative.

Example. In the number $\sqrt(-5)$, you can take out the minus from under the root sign - then everything will be fine:

\[\begin(align) & \sqrt(-5)=-\sqrt(5) \lt 0\Rightarrow \\ & \sqrt(-5)=-\sqrt(((5)^(2))) =-\sqrt(25)=-\sqrt(((5)^(2)))=-\sqrt(5) \lt 0 \\ \end(align)\]

Feel the difference? If you leave a minus under the root, then when the radical expression is squared, it will disappear, and crap will begin. And if you first take out a minus, then you can even raise / remove a square until you are blue in the face - the number will remain negative. :)

Thus, the most correct and most reliable way to multiply the roots is as follows:

1. Remove all minuses from under the radicals. Minuses are only in the roots of odd multiplicity - they can be placed in front of the root and, if necessary, reduced (for example, if there are two of these minuses).
2. Perform multiplication according to the rules discussed above in today's lesson. If the indices of the roots are the same, simply multiply the root expressions. And if they are different, we use the evil formula \[\sqrt[n](a)\cdot \sqrt[p](b)=\sqrt(((a)^(p))\cdot ((b)^(n) ))\].
3. 3. We enjoy the result and good grades. :)

Well? Shall we practice?

Example 1. Simplify the expression:

\[\begin(align) & \sqrt(48)\cdot \sqrt(-\frac(4)(3))=\sqrt(48)\cdot \left(-\sqrt(\frac(4)(3 )) \right)=-\sqrt(48)\cdot \sqrt(\frac(4)(3))= \\ & =-\sqrt(48\cdot \frac(4)(3))=-\ sqrt(64)=-4; \end(align)\]

This is the simplest option: the indicators of the roots are the same and odd, the problem is only in the minus of the second multiplier. We endure this minus nafig, after which everything is easily considered.

Example 2. Simplify the expression:

\[\begin(align) & \sqrt(32)\cdot \sqrt(4)=\sqrt(((2)^(5)))\cdot \sqrt(((2)^(2)))= \sqrt(((\left(((2)^(5)) \right))^(3))\cdot ((\left(((2)^(2)) \right))^(4) ))= \\ & =\sqrt(((2)^(15))\cdot ((2)^(8)))=\sqrt(((2)^(23))) \\ \end( align)\]

Here, many would be confused by the fact that the output turned out to be an irrational number. Yes, it happens: we could not completely get rid of the root, but at least we significantly simplified the expression.

Example 3. Simplify the expression:

\[\begin(align) & \sqrt(a)\cdot \sqrt(((a)^(4)))=\sqrt(((a)^(3))\cdot ((\left((( a)^(4)) \right))^(6)))=\sqrt(((a)^(3))\cdot ((a)^(24)))= \\ & =\sqrt( ((a)^(27)))=\sqrt(((a)^(3\cdot 9)))=\sqrt(((a)^(3))) \end(align)\]

This is what I would like to draw your attention to. There are two points here:

1. Under the root is not a specific number or degree, but the variable $a$. At first glance, this is a bit unusual, but in reality, when solving mathematical problems, you will most often have to deal with variables.
2. In the end, we managed to “reduce” the root exponent and degree in the radical expression. This happens quite often. And this means that it was possible to significantly simplify the calculations if you do not use the main formula.

For example, you could do this:

\[\begin(align) & \sqrt(a)\cdot \sqrt(((a)^(4)))=\sqrt(a)\cdot \sqrt(((\left(((a)^( 4)) \right))^(2)))=\sqrt(a)\cdot \sqrt(((a)^(8))) \\ & =\sqrt(a\cdot ((a)^( 8)))=\sqrt(((a)^(9)))=\sqrt(((a)^(3\cdot 3)))=\sqrt(((a)^(3))) \ \ \end(align)\]

In fact, all transformations were performed only with the second radical. And if you do not paint in detail all the intermediate steps, then in the end the amount of calculations will significantly decrease.

In fact, we have already encountered a similar task above when solving the $\sqrt(5)\cdot \sqrt(3)$ example. Now it can be written much easier:

\[\begin(align) & \sqrt(5)\cdot \sqrt(3)=\sqrt(((5)^(4))\cdot ((3)^(2)))=\sqrt(( (\left(((5)^(2))\cdot 3 \right))^(2)))= \\ & =\sqrt(((\left(75 \right))^(2))) =\sqrt(75). \end(align)\]

Well, we figured out the multiplication of the roots. Now consider the inverse operation: what to do when there is a work under the root?

Power formulas used in the process of reducing and simplifying complex expressions, in solving equations and inequalities.

Number c is an n-th power of a number a when:

Operations with degrees.

1. Multiplying degrees with the same base, their indicators add up:

a ma n = a m + n .

2. In the division of degrees with the same base, their indicators are subtracted:

3. The degree of the product of 2 or more factors is equal to the product of the degrees of these factors:

(abc…) n = a n b n c n …

4. The degree of a fraction is equal to the ratio of the degrees of the dividend and the divisor:

(a/b) n = a n / b n .

5. Raising a power to a power, the exponents are multiplied:

(am) n = a m n .

Each formula above is correct in the directions from left to right and vice versa.

for example. (2 3 5/15)² = 2² 3² 5²/15² = 900/225 = 4.

Operations with roots.

1. The root of the product of several factors is equal to the product of the roots of these factors:

2. The root of the ratio is equal to the ratio of the dividend and the divisor of the roots:

3. When raising a root to a power, it is enough to raise the root number to this power:

4. If we increase the degree of the root in n once and at the same time raise to n th power is a root number, then the value of the root will not change:

5. If we decrease the degree of the root in n root at the same time n th degree from the radical number, then the value of the root will not change:

Degree with a negative exponent. The degree of a certain number with a non-positive (integer) exponent is defined as one divided by the degree of the same number with an exponent equal to the absolute value of the non-positive exponent:

Formula a m:a n = a m - n can be used not only for m> n, but also at m< n.

for example. a4:a 7 = a 4 - 7 = a -3.

To formula a m:a n = a m - n became fair at m=n, you need the presence of the zero degree.

Degree with zero exponent. The power of any non-zero number with a zero exponent is equal to one.

for example. 2 0 = 1,(-5) 0 = 1,(-3/5) 0 = 1.

Degree with a fractional exponent. To raise a real number a to a degree m/n, you need to extract the root n th degree of m th power of this number a.