Square root of a product

πŸ†Practice the root of a product

The square root of a product

When we encounter a root that encompasses the entirety of the product, we can decompose the factors of the products and leave a separate root for each of them. Let's not forget to leave the multiplication sign between the factors we have extracted.

Let's put it this way:
(aβ‹…b)=aβ‹…b\sqrt{(a\cdot b)}=\sqrt{a}\cdot\sqrt{b}

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Test yourself on the root of a product!

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Solve the following exercise:

\( \sqrt{30}\cdot\sqrt{1}= \)

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Let's look at this in the example

4β‹…400\sqrt{4\cdot400}
According to the rule of the root of a product, we can break down the factors and leave the root of each factor separately while maintaining the multiplication operation between them:
We will break it down and obtain:
4β‹…400\sqrt{4}\cdot\sqrt{400}
2βˆ—20=402*20=40

If you are interested in this article, you might also be interested in the following articles:

Laws of Radicals

Root of the Quotient

Radication

Combining Powers and Roots

In the blog of Tutorela you will find a variety of articles about mathematics.


Examples and exercises with solutions of the root of a product

Exercise #1

Solve the following exercise:

16β‹…1= \sqrt{16}\cdot\sqrt{1}=

Video Solution

Step-by-Step Solution

Let's start by recalling how to define a root as a power:

an=a1n \sqrt[n]{a}=a^{\frac{1}{n}}

Next, we will remember that raising 1 to any power will always yield the result 1, even the half power of the square root.

In other words:

16β‹…1=↓16β‹…12=16β‹…112=16β‹…1=16=4 \sqrt{16}\cdot\sqrt{1}= \\ \downarrow\\ \sqrt{16}\cdot\sqrt[2]{1}=\\ \sqrt{16}\cdot 1^{\frac{1}{2}}=\\ \sqrt{16} \cdot1=\\ \sqrt{16} =\\ \boxed{4} Therefore, the correct answer is answer D.

Answer

4 4

Exercise #2

Solve the following exercise:

1β‹…2= \sqrt{1}\cdot\sqrt{2}=

Video Solution

Step-by-Step Solution

Let's start by recalling how to define a square root as a power:

an=a1n \sqrt[n]{a}=a^{\frac{1}{n}}

Next, we remember that raising 1 to any power always gives us 1, even the half power we got from converting the square root.

In other words:

1β‹…2=↓12β‹…2=112β‹…2=1β‹…2=2 \sqrt{1} \cdot \sqrt{2}= \\ \downarrow\\ \sqrt[2]{1}\cdot \sqrt{2}=\\ 1^{\frac{1}{2}} \cdot\sqrt{2} =\\ 1\cdot\sqrt{2}=\\ \boxed{\sqrt{2}} Therefore, the correct answer is answer a.

Answer

2 \sqrt{2}

Exercise #3

Solve the following exercise:

2β‹…5= \sqrt{2}\cdot\sqrt{5}=

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 dividing powers with the same bases (in the opposite direction):

xnβ‹…yn=(xβ‹…y)n x^n\cdot y^n =(x\cdot y)^n

Let's start by changing the square roots to exponents using the law of exponents shown in A:

2β‹…5=↓212β‹…512= \sqrt{2}\cdot\sqrt{5}= \\ \downarrow\\ 2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}= We continue: since we are multiplying two terms with equal exponents we can use the law of exponents shown in B and combine them together as the same base raised to the same power:

212β‹…512=(2β‹…5)12=1012=10 2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}= \\ (2\cdot5)^{\frac{1}{2}}=\\ 10^{\frac{1}{2}}=\\ \boxed{\sqrt{10}} In the last steps wemultiplied the bases and then used the definition of the root as an exponent shown earlier in A (in the opposite direction) to return to the root notation.

Therefore, the correct answer is answer B.

Answer

10 \sqrt{10}

Exercise #4

Solve the following exercise:

9β‹…3= \sqrt{9}\cdot\sqrt{3}=

Video Solution

Step-by-Step Solution

Although the square root of 9 is known (3) , in order to get a single expression we will use the laws of parentheses:

So- in order to simplify the given expression we will use two exponents laws:

A. Defining the root as a an exponent:

an=a1n \sqrt[n]{a}=a^{\frac{1}{n}} B. Multiplying different bases with the same power (in the opposite direction):

xnβ‹…yn=(xβ‹…y)n x^n\cdot y^n =(x\cdot y)^n

Let's start by changing the square root into an exponent using the law shown in A:

9β‹…3=↓912β‹…312= \sqrt{9}\cdot\sqrt{3}= \\ \downarrow\\ 9^{\frac{1}{2}}\cdot3^{\frac{1}{2}}= Since a multiplication is performed between two bases with the same exponent we can use the law shown in B.

912β‹…312=(9β‹…3)12=2712=27 9^{\frac{1}{2}}\cdot3^{\frac{1}{2}}= \\ (9\cdot3)^{\frac{1}{2}}=\\ 27^{\frac{1}{2}}=\\ \boxed{\sqrt{27}} In the last steps we performed the multiplication, and then used the law of defining the root as an exponent shown earlier in A (in the opposite direction) in order to return to the root notation.

Therefore, the correct answer is answer C.

Answer

27 \sqrt{27}

Exercise #5

Solve the following exercise:

10β‹…3= \sqrt{10}\cdot\sqrt{3}=

Video Solution

Step-by-Step Solution

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 dividing powers with the same base (in the opposite direction):

xnβ‹…yn=(xβ‹…y)n x^n\cdot y^n =(x\cdot y)^n

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

10β‹…3=↓1012β‹…312= \sqrt{10}\cdot\sqrt{3}= \\ \downarrow\\ 10^{\frac{1}{2}}\cdot3^{\frac{1}{2}}= We continue, since we have a multiplication between two terms with equal exponents, we can use the law of exponents shown in B and combine them under the same base which is raised to the same exponent:

1012β‹…312=(10β‹…3)12=3012=30 10^{\frac{1}{2}}\cdot3^{\frac{1}{2}}= \\ (10\cdot3)^{\frac{1}{2}}=\\ 30^{\frac{1}{2}}=\\ \boxed{\sqrt{30}} In the last steps, we performed the multiplication of the bases and used the definition of the root as an exponent shown earlier in A (in the opposite direction) to return to the root notation.

Therefore, the correct answer is B.

Answer

30 \sqrt{30}

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