# Order of Operations: Roots - Examples, Exercises and Solutions

As we have learned in previous lessons, when working with combined operations the order of the basic operations must be followed in order to get the correct result. However, before performing these the parentheses and then the roots and powers must first be solved.

Roots are very important in mathematical calculations. They are present in a variety of exercises ranging from algebraic problems for solving a second degree equation using the general formula, to geometric problems like determining the length of the hypotenuse of a right-angled triangle. Therefore, it is fundamental that we learn how to solve combined operations where this operation appears.

When we have simplified the root and power operations, we can continue solving the exercise according to the order of the basic operations: multiplications and divisions first, followed by additions and subtractions.

Let's revisit the order of the operations:

### Suggested Topics to Practice in Advance

1. The Order of Basic Operations: Addition, Subtraction, and Multiplication

## examples with solutions for order of operations: roots

### Exercise #1

What is the answer to the following?

$3^2-3^3$

### Step-by-Step Solution

Remember that according to the order of operations, exponents come before multiplication and division, which come before addition and subtraction (and parentheses always before everything),

So first calculate the values of the terms in the power and then subtract between the results:

$3^2-3^3 =9-27=-18$Therefore, the correct answer is option A.

$-18$

### Exercise #2

Sovle:

$3^2+3^3$

### Step-by-Step Solution

Remember that according to the order of operations, exponents precede multiplication and division, which precede addition and subtraction (and parentheses always precede everything).

So first calculate the values of the terms in the power and then subtract between the results:

$3^2+3^3 =9+27=36$Therefore, the correct answer is option B.

36

### Exercise #3

Solve:

$5^2\cdot4+3^3$

### Step-by-Step Solution

Remember that according to the order of arithmetic operations, exponents precede multiplication and division, which precede addition and subtraction (and parentheses always precede everything).

So first calculate the values of the terms with exponents and then subtract the results:

$5^2\cdot4+3^3 =25\cdot4+27=100+27=127$Therefore, the correct answer is option B.

127

### Exercise #4

$(5-2)^2-2^3$

### Step-by-Step Solution

Remember that according to the order of arithmetic operations, exponents precede multiplication and division, which precede addition and subtraction (and parentheses always precede everything).

So first calculate the values of the terms with exponents and then subtract the results:

$(5-2)^2-2^3 =3^2-2^3=9-8=1$Therefore, the correct answer is option C.

1

### Exercise #5

Solve:

$\sqrt{4}\cdot4^2-5^2\cdot\sqrt{1}$

### Step-by-Step Solution

We simplify each term according to the order from left to right:

$\sqrt{4}=2$

$4^2=4\times4=16$

$5^2=5\times5=25$

$\sqrt{1}=1$

Now we rearrange the exercise accordingly:

$2\times16-25\times1$

Since there are two multiplication operations in the exercise, according to the order of operations we start with them and then subtract.

We put the two multiplication exercises in parentheses to avoid confusion during the solution, and solve from left to right:

$(2\times16)-(25\times1)=32-25=7$

7

## examples with solutions for order of operations: roots

### Exercise #1

$6\sqrt{4}:6\sqrt{4}=$

### Step-by-Step Solution

$4$

### Exercise #2

Indicates the corresponding sign:

$\frac{1}{5}\cdot((5+3:3-8)^2:\sqrt{4}-2)\text{ }_{\textcolor{red}{——}\text{\textcolor{red}{ }}}8-(3^2+1)\cdot\frac{1}{10}$

### Step-by-Step Solution

According to the given problem, whether it is discussed in addition or subtraction each of the terms that comes in its turn,

this is done within the framework of the order of operations, which states that multiplication precedes addition and subtraction, and that the preceding operations are performed before all others,

A. We start with the terms that are on the left in the given problem:

$\frac{1}{5}\cdot\big((5+3:3-8)^2:\sqrt{4}-2\big)$First, we simplify the terms that are in the denominators (the divisors) on which the multiplication operation is performed, this in accordance with the order of operations mentioned, we note that this term will change in terms that are in the denominators (the numerators) on which division is performed, therefore we start with simply the terms that are in these denominators, remembering that division precedes multiplication and subtraction, therefore the beginning will perform the division operation that is in this term and then perform the operations of multiplication and subtraction:

$\frac{1}{5}\cdot\big((5+3:3-8)^2:\sqrt{4}-2\big)\\ \frac{1}{5}\cdot\big((5+1-8)^2:\sqrt{4}-2\big)=\\ \frac{1}{5}\cdot\big((-2)^2:\sqrt{4}-2\big)\\$Since the results of the operations of multiplication and subtraction that are in the numerators the level will come out smoothly on these denominators, we continue and perform the strength on the term that is in these denominators, this within that we remember thatthe raising of any number (positive or negative) in a double strength will give a positive result, in contrast we will consider its numerical value of the other side that in strength he is the divisor that is in the term that within the denominators the divisors that were left (this within that we remember that in defining the root as strength, the root he is strength for everything):

$\frac{1}{5}\cdot\big((-2)^2:\sqrt{4}-2\big)=\\ \frac{1}{5}\cdot\big(4:2-2\big)\\$We continue and finish simply the term, we remember that division precedes subtraction and therefore the beginning will calculate the result of the division operation that in the term and then perform the operation of subtraction:

$\frac{1}{5}\cdot\big(2-2\big)=\\ \frac{1}{5}\cdot0=\\ 0$In the last stage we performed the doubling that was left (it is the doubling that is performed on the term that in the denominators), this within that we remember thatthe result of doubling any number (different from zero) in zero here zero,

We finished simply the term that is on the left in the given problem, we will summarize the stages of simply:

$\frac{1}{5}\cdot\big((5+3:3-8)^2:\sqrt{4}-2\big)\\ \frac{1}{5}\cdot\big((-2)^2:\sqrt{4}-2\big)\\ \frac{1}{5}\cdot\big(2-2\big)=\\ 0$

B. We continue and simplify the term that is on the right in the given problem:

$8-(3^2+1)\cdot\frac{1}{10}$In this part to do in the first part simplify the term within the framework of the order of operations,

In this term a doubling that is performed on the term that in the denominators, therefore we simplify first this term, this is done in accordance with the order of operations mentioned, therefore we start from considering its numerical value that in strength that in this term and then perform the operation of multiplication:

$8-(3^2+1)\cdot\frac{1}{10} =\\ 8-(9+1)\cdot\frac{1}{10}=\\ 8-10\cdot\frac{1}{10}\\$We continue and simplify the term that was received in the first stage, we remember in this that multiplication and division precede addition and subtraction, therefore the beginning will perform the doubling in the break, this within that we remember that the doubling in the break means the doubling in the amount of the break, then perform the operation of division that of the break, this is done by appointment, in the last stage will perform the operation of subtraction that remained:

$8-10\cdot\frac{1}{10}=\\ 8-\frac{10\cdot1}{10}=\\ 8-\frac{\not{10}}{\not{10}}=\\ 8-1=\\ 7$We finished simply the term that is on the right in the given problem, we will summarize the stages of simply:

$8-(3^2+1)\cdot\frac{1}{10} =\\ 8-10\cdot\frac{1}{10}\\ 7$We return now to the original problem, and we will present the results of simply the terms that were reported in A and in B:

$\frac{1}{5}\cdot\big((5+3:3-8)^2:\sqrt{4}-2\big)\text{ }_{\textcolor{red}{——}\text{\textcolor{red}{ }}}8-(3^2+1)\cdot\frac{1}{10} \\ \downarrow\\ 0\text{ }\text{\textcolor{red}{\_\_}}\text{ }7$Therefore the correct answer here is answer A.

$\ne$

### Exercise #3

Indicates the corresponding sign:

$-3+(\sqrt{100}-3^2-1^{14}):30+3\text{ }\text{\textcolor{red}{\_\_}}\text{ }6^2:6\cdot(3-2)-6$

### Step-by-Step Solution

According to the given problem, whether it's an addition or subtraction each of the digits that come up separately,

this is done within the framework of the order of operations, which states that multiplication precedes addition and subtraction, and that the preceding operations are done before division and subtraction, and that the preceding operations are done before all,

A. We start with the digits that are left in the given problem:

$-3+(\sqrt{100}-3^2-1^{14}):30+3$First, we simplify the digits that are in the denominators on which the division operation takes place, this is done in accordance with the order of operations mentioned, keeping in mind that multiplication precedes subtraction, therefore, we first calculate their numerical values of the numerators in the multiplication (this within the definition of the root as a power, the root of which is a power for everything), subsequently we perform the subtraction operation which is within the denominators and finally we perform the division operation that takes place on the denominators:

$-3+(\sqrt{100}-3^2-1^{14}):30+3 =\\ -3+(10-9-1):30+3 =\\ -3+0:30+3 =\\ -3+0+3 \\$In the last step we mentioned that dividing the number 0 by any number (different from zero) will yield the result 0, we continue with the simple digits we received in the last step and perform the addition operation:

$-3+0+3 =\\ 0$We finish the simple digits that are left in the given problem, we summarize the steps of the simplification:

$-3+(\sqrt{100}-3^2-1^{14}):30+3 =\\ -3+0:30+3 =\\ 0$

B. We continue and simplify the digits that are right in the given problem:

$6^2:6\cdot(3-2)-6$In this part, in the first step we simplify the digits within the framework of the order of operations,

In this digit, multiplication takes place on digits in the denominators, therefore, we first simplify this digit, in the process we calculate the numerical values of the numerator in the multiplication which is the divisor in the first digit from the left in the given digit:

$6^2:6\cdot(3-2)-6 =\\ 36:6\cdot1-6 \\$We continue and remember that multiplication and division precede addition and subtraction, keeping in mind that there is no predefined precedence between multiplication and division operations originating from the order of operations mentioned, therefore, we calculate the numerical values of the numerator in the first digit from the left (including the multiplication and division operations) within the execution of one operation after another according to the order from left to right (this is the natural order of operations), subsequently, we complete the calculation within the execution of the subtraction operation:

$36:6\cdot1-6 =\\ 6\cdot1-6 =\\ 6-6 =\\ 0$We finish the simple digits that are right in the given problem, we summarize the steps of the simplification:

$6^2:6\cdot(3-2)-6 =\\ 36:6\cdot1-6 =\\ 0$We return to the original problem, and we present the results of the simplifications reported in A and B:

$-3+(\sqrt{100}-3^2-1^{14}):30+3\text{ }\text{\textcolor{red}{\_\_}}\text{ }6^2:6\cdot(3-2)-6 \\ \downarrow\\ 0\text{ }\text{\textcolor{red}{\_\_}}\text{ }0$As a result, we have that:

$0 \text{ }\textcolor{red}{=}0$Therefore, the correct answer here is answer B.

$=$

### Exercise #4

$\frac{7^2-\sqrt{36}:6}{3+3}\cdot(5+2)=$

### Step-by-Step Solution

Before solving the exercise, let's start by simplifying the power and the root:

$7^2=7\times7=49$

$\sqrt{36}=\sqrt{6^2}=6$

Now, we arrange the exercise accordingly:

$\frac{49-6:6}{3+3}\times(5+2)=$

According to the rules of the order of operations, parentheses are solved first:

$\frac{49-6:6}{3+3}\times(7)=$

Now we focus on the fraction, we start with the division exercise in the numerator, then we add and subtract as appropriate:

$\frac{49-1}{3+3}\times(7)=\frac{48}{6}\times(7)=$

We solve the exercise from left to right, first the division exercise and finally we multiply:

$8\times7=56$

$56$

### Exercise #5

$\frac{(2^2-3)^{15}+4^2}{15+2}-\frac{3^2-2^2}{5}=$

### Step-by-Step Solution

This simple example illustrates the order of operations, which states that multiplication and division take precedence over addition and subtraction, and that operations within parentheses come first,

Let's say we have a fraction and a whole number (every whole number) between which a division operation takes place, meaning - we can relate to the fraction and the whole number as fractions in their simplest form, through which a division operation occurs, thus we can write the given fraction in the following form:

$\frac{(2^2-3)^{15}+4^2}{15+2}-\frac{3^2-2^2}{5}= \\ \downarrow\\ \big((2^2-3)^{15}+4^2\big):(15+2)-(3^2-2^2):5$We emphasize this by stating that we should relate to the fractions that are in the numerator and those in the denominator separately, as if they exist in their simplest form,

Let's return to the original fraction in question, meaning - in its given form, and simplify it, simplifying separately the different fractions that are in the numerator and those in the denominator (if simplification is needed), this is done in accordance with the order of operations mentioned above and in a systematic way,

We start with the first numerator from the left in the given fraction, noting that in this case it changes the fraction in the denominators that are in multiplication, therefore, we start with this fraction, this in accordance with the aforementioned order of operations, noting further that in this fraction in the denominators (which are in multiplication of 15) there exists a multiplication, therefore, we start calculating its numerical value in multiplication and then perform the subtraction operation that is in the denominators:

$\frac{(2^2-3)^{15}+4^2}{15+2}-\frac{3^2-2^2}{5}= \\ \frac{(4-3)^{15}+4^2}{15+2}-\frac{3^2-2^2}{5}= \\ \frac{1^{15}+4^2}{15+2}-\frac{3^2-2^2}{5} \\$ We continue with the fraction we received in the previous step and simplify the numerators and the denominators in the fraction, this is done in accordance with the order of operations mentioned above, therefore, we start calculating their numerical values in multiplication and then perform the division and subtraction operations that are in the numerators and in the denominators:

$\frac{1^{15}+4^2}{15+2}-\frac{3^2-2^2}{5}= \\ \frac{1+16}{17}-\frac{9-4}{5}= \\ \frac{17}{17}-\frac{5}{5}\\$ We continue and simplify the fraction we received in the previous step, again, in accordance with the order of operations mentioned above, therefore, we perform the division operation of the denominators, this is done systematically, and then perform the subtraction operation:

$\frac{17}{17}-\frac{5}{5}=\\ \frac{\not{17}}{\not{17}}-\frac{\not{5}}{\not{5}}=\\ 1-1=\\ 0$

We conclude with this, the steps of simplifying the given fraction, we received that:

$\frac{(2^2-3)^{15}+4^2}{15+2}-\frac{3^2-2^2}{5}= \\ \frac{1^{15}+4^2}{15+2}-\frac{3^2-2^2}{5} =\\ \frac{1+16}{17}-\frac{9-4}{5}= \\ 0$Therefore, the correct answer is answer D.

0

## examples with solutions for order of operations: roots

### Exercise #1

Complete the following exercise:

$\lbrack(3^2-4-5)\cdot(4+\sqrt{16})-5 \rbrack:(-5)=$

### Step-by-Step Solution

This simple example demonstrates the order of operations, which states that exponentiation precedes multiplication and division, which come before addition and subtraction, and that operations within parentheses come first,

In the given example, the operation of division between parentheses (the denominators) by a number (which is also in parentheses but only for clarification purposes), thus according to the order of operations mentioned we start with the parentheses that contain the denominators first, this parentheses that contain the denominators includes multiplication between two numbers which are also in parentheses, therefore according to the order of operations mentioned, we start with the numbers inside them, paying attention that each of these numbers, including the ones in strength, and therefore assuming that exponentiation precedes multiplication and division we consider their numerical values only in the first step and only then do we perform the operations of multiplication and division on these numbers:

$\lbrack(3^2-4-5)\cdot(4+\sqrt{16})-5 \rbrack:(-5)=\\ \lbrack(9-4-5)\cdot(4+4)-5 \rbrack:(-5)=\\ \lbrack0\cdot8-5 \rbrack:(-5)\\$Continuing with the simple division in parentheses ,and according to the order of operations mentioned, we proceed from the multiplication calculation and remember that the multiplication of the number 0 by any number will yield the result 0, in the first step the operation of subtraction is performed and finally the operation of division is initiated on the number in parentheses:

$\lbrack0\cdot8-5 \rbrack:(-5)= \\ \lbrack0-5 \rbrack:(-5)= \\ -5 :(-5)=\\ 1$ Therefore, the correct answer is answer c.

1

### Exercise #2

Solve the following:
$\big((3-2+4)^2-2^2\big):\frac{\sqrt{9}\cdot7}{3}=$

### Step-by-Step Solution

This simple example demonstrates the order of operations, which states that multiplication and division take precedence over addition and subtraction, and that operations within parentheses take precedence over all others,

Let's consider the numerator and the denominator separately (each separately) which between them performs a division operation, meaning- we can relate to the numerator and the denominator separately as fractions in their own right, thus we can write the given fraction and write it in the following form:

$\big((3-2+4)^2-2^2\big):\frac{\sqrt{9}\cdot7}{3}= \\ \downarrow\\ \big((25-2-16)^2+3\big):\big((\sqrt{9}\cdot7):\sqrt{3} \big)$We emphasize this by noting that the fractions in the numerator and the denominator should be treated separately, indeed as if they are in their own parentheses,

Let's consider additionally that the division operation between the parentheses implies that we are dividing by the value of the denominator (meaning the denominator as a whole, it is the result of the division between the numerator and the denominator) and therefore in the given fraction to form a division that we marked for attention, the denominator being in parentheses is additionally important,

Returning to the original fraction problem, meaning - in the given form, and proceed simply,

We will start and simplify the fraction in the numerator (meaning- the numerator fraction that we are dividing by), this is done in accordance with the order of operations mentioned above, therefore we will start by calculating the numerical values of the fraction that takes precedence (this within the context of setting the root as a priority, the root being strong for everything) and then proceed with the multiplication which is in the numerator, in contrast let's consider within the parentheses that are left, those parts in the denominator are divided by the whole, they are fractions in the stronger parentheses, therefore we will also simplify this fraction, this in accordance with the order of operations mentioned above:

$\big((3-2+4)^2-2^2\big):\frac{\sqrt{9}\cdot7}{3}= \\ \big(5^2-2^2\big):\frac{3\cdot7}{3}= \\ \big(5^2-2^2\big):\frac{21}{3}\\$We will continue and simplify the fraction we received in the previous step, continue simply the fraction found within the parentheses that are divided by the whole, they are the parentheses that are left, remembering that multiplication takes precedence over addition and subtraction, therefore we will start by calculating their numerical values that take precedence in those parentheses and then proceed with the subtraction operation, in the next step the division operation of the whole (and not the division operation in the whole) takes place, and in the last step the remaining division operation takes place:

$\big(5^2-2^2\big):\frac{21}{3}=\\ \big(25-4\big):\frac{21}{3}=\\ 21:\frac{21}{3}=\\ 21:\frac{\not{21}}{\not{3}}=\\ 21:7=\\ 3$Let's consider that we advanced the division operation of the whole over the division operation in the whole itself, and this means that the number 21 in the fraction we discussed is divided by its numerical values of the whole (in its entirety)- which is the result of the division of the numerator by the denominator, therefore it was necessary to complete first the calculation of the numerical values of the whole and only then to divide the number 21 in this value,

We will conclude thus with the steps of simplifying the given fraction:
$\big((3-2+4)^2-2^2\big):\frac{\sqrt{9}\cdot7}{3}= \\ \big(5^2-2^2\big):\frac{21}{3}=\\ 21:7=\\ 3$Therefore, the correct answer is answer d'.

3

### Exercise #3

Choose the correct answer to the following:

$\frac{(25-2-16)^2+3}{8+5}:\sqrt{9}=$

### Step-by-Step Solution

This simple rule is the emphasis on the order of operations which states that exponentiation precedes multiplication and division, which precede addition and subtraction, and that operations within parentheses precede all others,

Let's consider that the numerator is the whole and the denominator is the part which breaks (every break) into whole pieces (in their entirety) among which division operation is performed, meaning- we can relate the numerator and the denominator of the break as whole pieces in closures, thus we can express the given fraction and write it in the following form:

$\frac{(25-2-16)^2+3}{8+5}:\sqrt{9}= \\ \downarrow\\ \big((25-2-16)^2+3\big):(8+5):\sqrt{9}$We highlight this by noting that fractions in the numerator of the break and in its denominator are considered separately, as if they are in closures,

Let's return to the original fraction in question, meaning - in the given form, and simplify, separately, the fraction in the numerator of the break which causes it and the fraction in its denominator, this is done in accordance to the order of operations mentioned and in a systematic way,

Let's consider that in the numerator of the break the fraction we get changes into a fraction in closures which indicates strength, therefore we will start simplifying this fraction, given that this fraction includes only addition and subtraction operations, perform the operations in accordance to the natural order of operations, meaning- from left to right, simplifying the fraction in the numerator of the break:

$\frac{(25-2-16)^2+3}{8+5}:\sqrt{9}=\\ \frac{7^2+3}{13}:\sqrt{9}\\$We will continue and simplify the fraction we received in the previous step, this of course, in accordance to the natural order of operations (which states that exponentiation precedes multiplication and division, which precede addition and subtraction, and that operations within parentheses precede all others), therefore we will start from calculating the numerical values of the exponents in strength (while we remember that in defining the root as strength, the root itself is strength for everything), and then perform the division operation which is in the numerator of the break:

$\frac{7^2+3}{13}:\sqrt{9}=\\ \frac{49+3}{13}:3=\\ \frac{52}{13}:3\\$We will continue and simplify the fraction we received in the previous step, starting with performing the division operation of the break, this is done by approximation, and then perform the remaining division operation:

$\frac{\not{52}}{\not{13}}:3=\\ 4:3=\\ \frac{4}{3}$In the previous step, given that the outcome of the division operation is different from a whole (greater than whole for the numerator, given that the divisor is greater than the dividend) we marked its outcome as a fraction in approximation (where the numerator is greater than the denominator),

We conclude the steps of simplifying the given fraction, we found that:

$\frac{(25-2-16)^2+3}{8+5}:\sqrt{9}=\\ \frac{7^2+3}{13}:\sqrt{9}=\\ \frac{52}{13}:3=\\ \frac{4}{3}$Therefore, the correct answer is answer b'.

Note:

Let's consider that in the group of the previous steps in solving the problem, we can start recording the break and the division operation that affects it even without the break, but with the help of the division operation:

$\frac{52}{13}:3\\ \downarrow\\ 52:13:3$And from here on we will start calculating the division operation in the break and only after that we performed the division in number 3, we emphasize that in general we simplify this fraction in accordance to the natural order of operations, meaning we perform the operations one after the other from left to right, and this means that there is no precedence of one division operation in the given fraction over the other except as defined by the natural order of operations, meaning- in calculating from left to right, (Let's consider additionally that defining the order of operations mentioned at the beginning of the solution, which states that exponentiation precedes multiplication and division, which precede addition and subtraction, and that operations within parentheses precede all others, does not define precedence even among multiplication and division, and therefore the judgment between these two operations, in different closures, is in a different order, it is in calculating from left to right).

$\frac{4}{3}$

### Exercise #4

Mark the appropriate sign:

$3^3-5^2:(2^2+1)_{\textcolor{red}{_{——}}\text{ }}4_{\text{\textcolor{red}{ }}}^2(\sqrt{10}+\sqrt{2}\cdot\sqrt{5})^2+\sqrt{48}\cdot\sqrt{3}$

### Step-by-Step Solution

To solve the given problem and determine whether it is an equality or inequality, we need to simplify the expressions on either side.

We can deal with each of them separately and simplify them, however, a more efficient way of working will be to deal with the more complex parts of these expressions separately, that is, with the expressions containing roots,

It's important to emphasize that in generalwe want to learn to solve without using a calculator by using our algebraic tools and the laws of exponents. Let's begin:

a. Let's start with the first part of the expression on the right:

$\sqrt{10}+\sqrt{2}\cdot\sqrt{5}$(We'll focus on the expression inside the parentheses first and then continue outwards).

Let's recall two laws of exponents:

a.1: Defining a root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

a.2: The law for an exponent applied to parentheses containing multiplication, but in the opposite direction:

$x^n\cdot y^n= (x\cdot y)^n$

Usually we would replace roots with exponents, but for now we won't do that. In the meantime just understand that according to the law of defining a root as an exponent mentioned in a.1, the root is actually an exponent and therefore all laws of exponents apply to it, especially the law of exponents mentioned ina.2. Let's apply this understanding to the expression in question:

$\sqrt{10}+\sqrt{2}\cdot\sqrt{5} = \sqrt{10}+\sqrt{2\cdot5}= \sqrt{10}+\sqrt{10}$

In the first step, we notice that the second term (i.e., the product of roots) is actually a product between two terms raised to the same exponent (which is half the power of the square root). Therefore using to the law of exponents mentioned in a.2, we can multiply the bases of the terms under the same square root, and in the next stage we simplify the expression in the root.

From here we note that this expression can be factored by a common factor:

$\sqrt{10}+\sqrt{10}=(1+1)\sqrt{10}=2\cdot\sqrt{10}$

We used the commutative property of multiplication, that is, the common factor we took out:$\sqrt{10}$We took out to the right of the parentheses (instead of to their left),

Let's continue to the second problematic expression on the right:

$\sqrt{48}\cdot\sqrt{3}$And we'll deal with it the same way as in the previous part:

$\sqrt{48}\cdot\sqrt{3} = \sqrt{48\cdot3}=\sqrt{144}$Again, in the first step, we notice that the expression (i.e., the product of roots) is actually a product between two terms raised to the same exponent (which is half the power of the square root), therefore according to the law of exponents mentioned in a.2, we can multiply the bases of the terms under the same exponent, and in the next step we simplify the expression in the root,

From here we'll return to the original expression in the problem (i.e., the expression on the right) and calculate it in full, using the simplified expressions we got above:

$4^2(\sqrt{10}+\sqrt{2}\cdot\sqrt{5})^2+\sqrt{48}\cdot\sqrt{3} = \\ 4^2(2\cdot\sqrt{10})^2+\sqrt{144}$We just substitute what we calculated earlier in the two parts above - in place of the expression in parentheses and in place of the second term.

INext, let's recall again the law of exponents mentioned above in a.2, that is, the law for an exponent applied to parentheses containing a product but in the normal way.

Let's apply this law to the expression we got in the last stage:

$4^2(2\cdot\sqrt{10})^2+\sqrt{144} =4^2\cdot2^2\cdot(\sqrt{10})^2+\sqrt{144}=\\ 16\cdot4\cdot10+12$We apply the law of exponents mentioned in a.2 above and applied the exponent to each of the multiplication terms in the parentheses.

Next, we apply the exponent to the square root while remembering that these are actually two inverse operations and therefore cancel each other out and simultaneously simplify the other terms by applying the square root to the second term on the left, and the exponents to the first term.

Let's finish solving. We got that the expression is:

$16\cdot4\cdot10+12 =640+12=652$Let's summarize this part:

We got that the expression on the right is:

$4^2(\sqrt{10}+\sqrt{2}\cdot\sqrt{5})^2+\sqrt{48}\cdot\sqrt{3}=\\ 4^2(2\cdot\sqrt{10})^2+\sqrt{144} =16\cdot4\cdot10+12 =\\ 652$b. Let's continue to the expression on the left and start by writing it in the standard fraction notation while keeping in mind the order of operations, that is - parentheses, exponents, multiplication/ division, addition/ subtraction.

$3^3-5^2:(2^2+1)= 3^3-\frac{5^2}{2^2+1}$Since the division here refers to the entire expression in parentheses, we inserted it in its entirety into the denominator of the fraction,

Let's simplify this expression. We'll focus on the fraction, but first let's recall the law of multiplying exponents with identical bases:

b.1:

$\frac{a^m}{a^n}=a^{m-n}$Let's simplify the expression. We'll start by simplifying the numerator of the fraction and continue by applying the law of exponents mentioned above:

$3^3-\frac{5^2}{2^2+1} =3^3-\frac{5^2}{4+1} =3^3-\frac{5^2}{5} =\\ 3^3-5^{2-1}=3^3-5$In the first and second parts, we simplified the numerator of the fraction, in the third part we applied the law of exponents mentioned in b.1 and in the following parts we simplified the resulting expression.

Let's finish the calculation:

$3^3-5 =27-5=22$And to summarize:

$3^3-5^2:(2^2+1)= 3^3-\frac{5^2}{2^2+1} =\\ 3^3-\frac{5^2}{5} = 3^3-5^{2-1}=3^3-5=22$

Now let's return to the original problem and replace what we got for a' and b':

$3^3-5^2:(2^2+1)_{\textcolor{red}{_{——}}\text{ }}4_{\text{\textcolor{red}{ }}}^2(\sqrt{10}+\sqrt{2}\cdot\sqrt{5})^2+\sqrt{48}\cdot\sqrt{3} \\ \downarrow\\ 22\text{ }\text{\textcolor{red}{\_\_}}\text{ }652$Therefore it's clear that this is not an equality but an inequality and that the expression on the left is smaller than the expression on the right,meaning that:

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### Exercise #5

Fill in the missing sign:

$9\cdot(\sqrt{6}+\sqrt{2}\cdot\sqrt{3})^2+2^2:2\text{ }\text{\textcolor{red}{\_\_}}\text{ }5^2\cdot(\sqrt{2}+2\sqrt{2})^2$

### Step-by-Step Solution

To solve the given problem and determine whether it is an equation or inequality, we need to simplify each of the algebraic expressions.

We can deal with each of them separately and simplify them. However, in this case, a more efficient way is be to deal with the more complex parts of these expressions separately, that is - the expressions in parentheses - the expressions with roots.

It's important to emphasize that in general we want to solve without a calculator, using only our algebraic tools and laws of exponents. Let's begin:

$\sqrt{6}+\sqrt{2}\cdot\sqrt{3}$(We'll focus on the expression inside the parentheses first and then move outwards),

Let's recall two laws of exponents:

a.1: Defining a root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

a.2: The law of applying exponents to parentheses containing a product, but in the opposite direction:

$x^n\cdot y^n= (x\cdot y)^n$

Usually we replace roots with exponents, but for now we won't do that. We'll just understand that according to the law of defining roots as an exponent mentioned in a.1, the root is actually an exponent and therefore all laws of exponents apply to it, especially the law of exponents mentioned in a.2 , So we'll apply this understanding to the expression in question:

$\sqrt{6}+\sqrt{2}\cdot\sqrt{3} = \sqrt{6}+\sqrt{2\cdot3}= \sqrt{6}+\sqrt{6}$

In the first stage, we notice that the second term (i.e., the product of roots) is actually a product between two terms raised to the same exponent (which is half the power of the square root). Therefore, according to the law of exponents mentioned in a.2 , we can combine the bases of the terms as a product with the same exponent , and in the next stage we simplify the expression under the root,

From here we notice that we can simplify this expression by a using common factor:

$\sqrt{6}+\sqrt{6}=(1+1)\sqrt{6}=2\cdot\sqrt{6}$

We used the commutative property of multiplication to move the common factor we took out ($\sqrt{6}$ ) we put it on the right of the parentheses (instead of to their left) so our expression with be clearer.

Now, we'll return to the original expression in the problem (i.e., the expression on the left) and calculate in full, using the simplification from above:

$9\cdot(\sqrt{6}+\sqrt{2}\cdot\sqrt{3})^2+2^2:2= 9\cdot(2\cdot\sqrt{6})^2+\frac{2^2}{2^1}$

We substitute what we calculated above in place of the expression in parentheses and write the division operation for the last term on the left as a fraction.

In the next stage let's recall another law of exponents:

a.3: The law of dividing exponents with equal bases:

$\frac{a^m}{a^n}=a^{m-n}$And let's recall again the law of exponents mentioned above in a.2, that is, the law of an exponent applied to parentheses containing a product, but in the normal direction.

Let's apply these two laws of exponents to the expression we got in the last stage:

$9\cdot(2\cdot\sqrt{6})^2+\frac{2^2}{2^1} =9\cdot2^2\cdot(\sqrt{6})^2+2^{2-1}=9\cdot4\cdot6+2^1$

In the first stage we apply the law of exponents mentioned in a.2 above and apply the exponent to each of the multiplied terms in the parentheses.

Then, we apply the law of exponents mentioned in a.3 to the second term on the left. To make things clearer, we put the root in parentheses, but this just for convenience.

In the next stage we square the square root while remembering that these are actually two inverse operations and therefore cancel each other out and we simplify the rest of the terms.

Let's finish the calculation. We got that the expression is:

$9\cdot4\cdot6+2^1 =216+2=218$

Let's summarize:

We got that the left expression is:

$9\cdot(\sqrt{6}+\sqrt{2}\cdot\sqrt{3})^2+2^2:2= 9\cdot(2\cdot\sqrt{6})^2+\frac{2^2}{2^1} =\\ 9\cdot4\cdot6+2^1 =218$

b. Let's continue to the expression on the right and as before,we'll start with the problematic part, that is, the expression inside the parentheses with the roots:

$\sqrt{2}+2\sqrt{2}$Just as in the previous part, we can factor this expression:

$\sqrt{2}+2\sqrt{2} =(1+2)\sqrt{2}=3\cdot\sqrt{2}$

Again we use the commutative property of multiplication and the common factor-$\sqrt{2}$ we choose to take out outside the parentheses - to their right.

Next, we simplify the expression in parentheses.

We'll return to the full expression on the right and substitutewhat we got :

$5^2\cdot(\sqrt{2}+2\sqrt{2})^2 =5^2\cdot(3\cdot\sqrt{2})^2$

We'll continue and simplify this expression. Again, we wil apply the law of exponents mentioned earlier in a.2 (in its normal direction) and we keep in mind that the square root and the square exponent are inverse operations and therefore cancel each other out:

$5^2\cdot(3\cdot\sqrt{2})^2 =5^2\cdot3^2 \cdot(\sqrt{2})^2=25\cdot9\cdot2$

First, according to the law of exponents mentioned in a.2 we apply the exponent to each of the multiplied terms in parentheses, and then we simplify the resulting expression while we keep in mind that the square root and square exponent are inverse operations.

Let's finish the calculation:

$25\cdot9\cdot2=450$And to summarize , we got that:

$5^2\cdot(\sqrt{2}+2\sqrt{2})^2 =5^2\cdot(3\cdot\sqrt{2})^2 =\\ 5^2\cdot3^2 \cdot(\sqrt{2})^2=25\cdot9\cdot2 =450$

Now let's return to the original problem and substitute what we got in a and b:

$9\cdot(\sqrt{6}+\sqrt{2}\cdot\sqrt{3})^2+2^2:2\text{ }\text{\textcolor{red}{\_\_}}\text{ }5^2\cdot(\sqrt{2}+2\sqrt{2})^2 \\ \downarrow\\ 218\text{ }\text{\textcolor{red}{\_\_}}\text{ }450$Therefore it's clear that this is not an equality but an inequality and that the expression on the left is smaller than the expression on the right ,that is: