# Square root of a product

π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:
$\sqrt{(a\cdot b)}=\sqrt{a}\cdot\sqrt{b}$

## Test yourself on the root of a product!

Solve the following exercise:

$$\sqrt{30}\cdot\sqrt{1}=$$

## Let's look at this in the example

$\sqrt{4\cdot400}$
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:
$\sqrt{4}\cdot\sqrt{400}$
$2*20=40$

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

Root of the Quotient

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:

$\sqrt{16}\cdot\sqrt{1}=$

### Step-by-Step Solution

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

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

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:

$\sqrt{16}\cdot\sqrt{1}= \\ \downarrow\\ \sqrt{16}\cdot\sqrt[2]{1}=\\ \sqrt{16}\cdot 1^{\frac{1}{2}}=\\ \sqrt{16} \cdot1=\\ \sqrt{16} =\\ \boxed{4}$Therefore, the correct answer is answer D.

$4$

### Exercise #2

Solve the following exercise:

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

### Step-by-Step Solution

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

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

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:

$\sqrt{1} \cdot \sqrt{2}= \\ \downarrow\\ \sqrt[2]{1}\cdot \sqrt{2}=\\ 1^{\frac{1}{2}} \cdot\sqrt{2} =\\ 1\cdot\sqrt{2}=\\ \boxed{\sqrt{2}}$Therefore, the correct answer is answer a.

$\sqrt{2}$

### Exercise #3

Solve the following exercise:

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

### Step-by-Step Solution

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

A. Defining the root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$B. The law of exponents for dividing powers with the same bases (in the opposite direction):

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

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

$\sqrt{2}\cdot\sqrt{5}= \\ \downarrow\\ 2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}=$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:

$2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}= \\ (2\cdot5)^{\frac{1}{2}}=\\ 10^{\frac{1}{2}}=\\ \boxed{\sqrt{10}}$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.

$\sqrt{10}$

### Exercise #4

Solve the following exercise:

$\sqrt{9}\cdot\sqrt{3}=$

### 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:

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

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

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

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

$9^{\frac{1}{2}}\cdot3^{\frac{1}{2}}= \\ (9\cdot3)^{\frac{1}{2}}=\\ 27^{\frac{1}{2}}=\\ \boxed{\sqrt{27}}$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.

$\sqrt{27}$

### Exercise #5

Solve the following exercise:

$\sqrt{10}\cdot\sqrt{3}=$

### Step-by-Step Solution

To simplify the given expression, we use two laws of exponents:

A. Defining the root as an exponent:

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

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

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

$\sqrt{10}\cdot\sqrt{3}= \\ \downarrow\\ 10^{\frac{1}{2}}\cdot3^{\frac{1}{2}}=$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:

$10^{\frac{1}{2}}\cdot3^{\frac{1}{2}}= \\ (10\cdot3)^{\frac{1}{2}}=\\ 30^{\frac{1}{2}}=\\ \boxed{\sqrt{30}}$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.

$\sqrt{30}$