Rules of Roots Combined: Number of terms

Identify the greater valueSame base and different indicatorUsing variablesSolving the equationUsing multiple rulesApplying the formula
Question 1

Solve the following exercise:

\( \sqrt{2}\cdot\sqrt{2}\cdot\sqrt{0}= \)

Question 2

Solve the following exercise:

\( \sqrt{2}\cdot\sqrt{5}\cdot\sqrt{2}\cdot\sqrt{2}= \)

Question 3

Solve the following exercise:

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

Question 4

Solve the following exercise:

\( \sqrt{4}\cdot\sqrt{2}\cdot\sqrt{2}= \)

Question 5

Complete the following exercise:

\( \sqrt{\sqrt{2}}\cdot\sqrt{\sqrt{4}}= \)

Examples with solutions for Rules of Roots Combined: Number of terms

Exercise #1

Solve the following exercise:

2⋅2⋅0= \sqrt{2}\cdot\sqrt{2}\cdot\sqrt{0}=

Video Solution

Step-by-Step Solution

Notice that in the given problem, a multiplication is performed between three terms, one of which is:

0 \sqrt{0} and let's remember that the root (of any order) of the number 0 is 0, meaning that:

0=0 \sqrt{0}=0 and since multiplying any number by 0 will always yield the result 0,

therefore the result of the multiplication in the problem is 0, meaning:

2⋅2⋅0=2⋅2⋅0=0 \sqrt{2}\cdot\sqrt{2}\cdot\sqrt{0}=\\ \sqrt{2}\cdot\sqrt{2}\cdot0=\\ \boxed{0} and thus the correct answer is answer C.

Answer

0 0

Exercise #2

Solve the following exercise:

2⋅5⋅2⋅2= \sqrt{2}\cdot\sqrt{5}\cdot\sqrt{2}\cdot\sqrt{2}=

Video Solution

Step-by-Step Solution

In order to simplify the given expression we use two laws of exponents:

A. Defining the root as an exponent:

an=a1n \sqrt[n]{a}=a^{\frac{1}{n}} B. The law of exponents for a product of numbers with the same base (in the opposite direction):

xn⋅yn=(x⋅y)n x^n\cdot y^n =(x\cdot y)^n

Let's start by definging the roots as exponents using the law of exponents shown in A:

2⋅5⋅2⋅2=↓212⋅512⋅212⋅212= \sqrt{2}\cdot\sqrt{5}\cdot\sqrt{2}\cdot\sqrt{2}= \\ \downarrow\\ 2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}= Since we are multiplying between four numbers with the same exponents we can use the law of exponents shown in B (which also applies to a product of numbers with the same base) and combine them together in a product wit the same base which is raised to the same exponent:

212⋅512⋅212⋅212=(2⋅5⋅2⋅2)12=4012=40 2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}= \\ (2\cdot5\cdot2\cdot2)^{\frac{1}{2}}=\\ 40^{\frac{1}{2}}=\\ \boxed{\sqrt{40}} In the last step we performed the product which is in the base, then we used again the definition of the root as an exponent shown earlier in A (in the opposite direction) to return to writing the root.

Therefore, note that the correct answer is answer C.

Answer

40 \sqrt{40}

Exercise #3

Solve the following exercise:

1⋅2⋅3= \sqrt{1}\cdot\sqrt{2}\cdot\sqrt{3}=

Video Solution

Answer

6 \sqrt{6}

Exercise #4

Solve the following exercise:

4⋅2⋅2= \sqrt{4}\cdot\sqrt{2}\cdot\sqrt{2}=

Video Solution

Answer

4

Exercise #5

Complete the following exercise:

2⋅4= \sqrt{\sqrt{2}}\cdot\sqrt{\sqrt{4}}=

Video Solution

Answer

84 \sqrt[4]{8}

Question 1

Complete the following exercise:

\( \sqrt{\sqrt{16}}\cdot\sqrt{\sqrt{8}}= \)

Question 2

Complete the following exercise:

\( \sqrt[3]{\sqrt{16}}\cdot\sqrt[]{\sqrt{16}}= \)

Question 3

Complete the following exercise:

\( \sqrt{25}\cdot\sqrt[3]{\sqrt{25}}= \)

Question 4

Complete the following exercise:

\( \sqrt{\sqrt{4}}\cdot\sqrt{\sqrt{2}}= \)

Question 5

Complete the following exercise:

\( \sqrt{\sqrt{49}}\cdot\sqrt{\sqrt{16}}= \)

Exercise #6

Complete the following exercise:

16⋅8= \sqrt{\sqrt{16}}\cdot\sqrt{\sqrt{8}}=

Video Solution

Answer

1284 \sqrt[4]{128}

Exercise #7

Complete the following exercise:

163⋅16= \sqrt[3]{\sqrt{16}}\cdot\sqrt[]{\sqrt{16}}=

Video Solution

Answer

16512 16^{\frac{5}{12}}

Exercise #8

Complete the following exercise:

25⋅253= \sqrt{25}\cdot\sqrt[3]{\sqrt{25}}=

Video Solution

Answer

5113 5^{1\frac{1}{3}}

Exercise #9

Complete the following exercise:

4⋅2= \sqrt{\sqrt{4}}\cdot\sqrt{\sqrt{2}}=

Video Solution

Answer

2142 2^{\frac{1}{4}}\sqrt{2}

Exercise #10

Complete the following exercise:

49⋅16= \sqrt{\sqrt{49}}\cdot\sqrt{\sqrt{16}}=

Video Solution

Answer

27 2\sqrt{7}

Question 1

Solve the following exercise:

\( \sqrt{\frac{2}{4}}\cdot\sqrt{\frac{8}{16}}= \)

Question 2

Solve the following exercise:

\( \sqrt{10}\cdot\sqrt{2}\cdot\sqrt{5}= \)

Question 3

Solve the following exercise:

\( \sqrt{5}\cdot\sqrt{10}\cdot\sqrt{2}\cdot\sqrt{4}= \)

Question 4

Solve the following exercise:

\( \sqrt{2}\cdot\sqrt{2}\cdot\sqrt{2}\cdot\sqrt{1}\cdot\sqrt{1}= \)

Question 5

Solve the following exercise:

\( \sqrt{5}\cdot\sqrt{2}\cdot\sqrt{5}\cdot\sqrt{2}= \)

Exercise #11

Solve the following exercise:

24⋅816= \sqrt{\frac{2}{4}}\cdot\sqrt{\frac{8}{16}}=

Video Solution

Answer

12 \frac{1}{2}

Exercise #12

Solve the following exercise:

10⋅2⋅5= \sqrt{10}\cdot\sqrt{2}\cdot\sqrt{5}=

Video Solution

Answer

10 10

Exercise #13

Solve the following exercise:

5⋅10⋅2⋅4= \sqrt{5}\cdot\sqrt{10}\cdot\sqrt{2}\cdot\sqrt{4}=

Video Solution

Answer

20 20

Exercise #14

Solve the following exercise:

2⋅2⋅2⋅1⋅1= \sqrt{2}\cdot\sqrt{2}\cdot\sqrt{2}\cdot\sqrt{1}\cdot\sqrt{1}=

Video Solution

Answer

8 \sqrt{8}

Exercise #15

Solve the following exercise:

5⋅2⋅5⋅2= \sqrt{5}\cdot\sqrt{2}\cdot\sqrt{5}\cdot\sqrt{2}=

Video Solution

Answer

10 10

Question 1

Complete the following exercise:

\( \sqrt[3]{\sqrt{3}}\cdot\sqrt[3]{\sqrt{4}}= \)

Question 2

Complete the following exercise:

\( \sqrt[5]{\sqrt{3}}\cdot\sqrt[5]{\sqrt{3}}= \)

Question 3

Complete the following exercise:

\( \sqrt[5]{\sqrt{3}}\cdot\sqrt[6]{\sqrt{3}}= \)

Question 4

Complete the following exercise:

\( \sqrt[3]{\sqrt{25}}\cdot\sqrt[3]{\sqrt{64}}= \)

Question 5

Solve the following exercise:

\( \frac{\sqrt{10}\cdot\sqrt{5}\cdot\sqrt{2}}{\sqrt{5}\cdot\sqrt{5}\cdot\sqrt{4}}= \)

Exercise #16

Complete the following exercise:

33⋅43= \sqrt[3]{\sqrt{3}}\cdot\sqrt[3]{\sqrt{4}}=

Video Solution

Answer

126 \sqrt[6]{12}

Exercise #17

Complete the following exercise:

35⋅35= \sqrt[5]{\sqrt{3}}\cdot\sqrt[5]{\sqrt{3}}=

Video Solution

Answer

910 \sqrt[10]{9}

Exercise #18

Complete the following exercise:

35⋅36= \sqrt[5]{\sqrt{3}}\cdot\sqrt[6]{\sqrt{3}}=

Video Solution

Answer

3110+112 3^{\frac{1}{10}+\frac{1}{12}}

Exercise #19

Complete the following exercise:

253⋅643= \sqrt[3]{\sqrt{25}}\cdot\sqrt[3]{\sqrt{64}}=

Video Solution

Answer

253 2\sqrt[3]{5}

Exercise #20

Solve the following exercise:

10⋅5⋅25⋅5⋅4= \frac{\sqrt{10}\cdot\sqrt{5}\cdot\sqrt{2}}{\sqrt{5}\cdot\sqrt{5}\cdot\sqrt{4}}=

Video Solution

Answer

1 1