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:
(ab)=ab\sqrt{(a\cdot b)}=\sqrt{a}\cdot\sqrt{b}

Practice Product Property of Square Roots

Examples with solutions for Product Property of Square Roots

Exercise #1

Solve the following exercise:

161= \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:

161=1612=16112=161=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:

12= \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:

12=122=1122=12=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:

25x4= \sqrt{25x^4}=

Video Solution

Step-by-Step Solution

To simplify the given expression, we will use the following three laws of exponents:

a. Definition of root as an exponent:

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

b. Law of exponents for an exponent applied to terms in parentheses:

(ab)n=anbn (a\cdot b)^n=a^n\cdot b^n

c. Law of exponents for an exponent raised to an exponent:

(am)n=amn (a^m)^n=a^{m\cdot n}

We'll start by converting the fourth root to an exponent using the law of exponents mentioned in a.:

25x4=(25x4)12= \sqrt{25x^4}= \\ \downarrow\\ (25x^4)^{\frac{1}{2}}=

We'll continue, using the law of exponents mentioned in b. and apply the exponent to each factor in the parentheses:

(25x4)12=2512(x4)12 (25x^4)^{\frac{1}{2}}= \\ 25^{\frac{1}{2}}\cdot(x^4)^{{\frac{1}{2}}}

We'll continue, using the law of exponents mentioned in c. and perform the exponent applied to the term with an exponent in parentheses (the second factor in the multiplication):

2512(x4)12=2512x412=2512x2=25x2=5x2 25^{\frac{1}{2}}\cdot(x^4)^{{\frac{1}{2}}} = \\ 25^{\frac{1}{2}}\cdot x^{4\cdot\frac{1}{2}}=\\ 25^{\frac{1}{2}}\cdot x^{2}=\\ \sqrt{25}\cdot x^2=\\ \boxed{5x^2}

In the final steps, we first converted the power of one-half applied to the first factor in the multiplication back to the fourth root form, again, according to the definition of root as an exponent mentioned in a. (in the reverse direction) and then calculated the known fourth root of 25.

Therefore, the correct answer is answer a.

Answer

5x2 5x^2

Exercise #4

Solve the following exercise:

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

Video Solution

Step-by-Step Solution

Let's start with a reminder of the definition of a root as a power:

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

We will then use the fact that raising the number 1 to any power always yields the result 1, particularly raising it to the power of half of the square root (which we obtain by using the definition of root as a power mentioned earlier),

In other words:

301=3012=30112=301=30 \sqrt{30}\cdot\sqrt{1}= \\ \downarrow\\ \sqrt{30}\cdot\sqrt[2]{1}=\\ \sqrt{30}\cdot 1^{\frac{1}{2}}=\\ \sqrt{30} \cdot1=\\ \boxed{\sqrt{30}} Therefore, the correct answer is answer C.

Answer

30 \sqrt{30}

Exercise #5

Solve the following exercise:

10025= \sqrt{100}\cdot\sqrt{25}=

Video Solution

Step-by-Step Solution

We can simplify the expression without using the laws of exponents, because the expression has known square roots, so let's simplify the expression and then perform the multiplication:

10025=105=50 \sqrt{100}\cdot\sqrt{25}=\\ 10\cdot5=\\ \boxed{50} Therefore, the correct answer is answer D.

Answer

50 50

Exercise #6

Solve the following exercise:

100x2= \sqrt{100x^2}=

Video Solution

Step-by-Step Solution

To simplify the given expression, we will use the following three laws of exponents:

a. Definition of root as an exponent:

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

b. Law of exponents for an exponent applied to terms in parentheses:

(ab)n=anbn (a\cdot b)^n=a^n\cdot b^n

c. Law of exponents for an exponent raised to an exponent:

(am)n=amn (a^m)^n=a^{m\cdot n}

We'll start by converting the fourth root to an exponent using the law of exponents mentioned in a.:

100x2=(100x2)12= \sqrt{100x^2}= \\ \downarrow\\ (100x^2)^{\frac{1}{2}}=

We'll continue, using the law of exponents mentioned in b. and apply the exponent to each factor in the parentheses:

(100x2)12=10012(x2)12 (100x^2)^{\frac{1}{2}}= \\ 100^{\frac{1}{2}}\cdot(x^2)^{{\frac{1}{2}}}

We'll continue, using the law of exponents mentioned in c. and perform the exponent applied to the term with an exponent in parentheses (the second factor in the multiplication):

10012(x2)12=10012x212=10012x1=100x=10x 100^{\frac{1}{2}}\cdot(x^2)^{{\frac{1}{2}}} = \\ 100^{\frac{1}{2}}\cdot x^{2\cdot\frac{1}{2}}=\\ 100^{\frac{1}{2}}\cdot x^{1}=\\ \sqrt{100}\cdot x=\\ \boxed{10x}

In the final steps, we first converted the power of one-half applied to the first factor in the multiplication back to the fourth root form, again, according to the definition of root as an exponent mentioned in a. (in reverse) and then calculated the known fourth root of 100.

Therefore, the correct answer is answer d.

Answer

10x 10x

Exercise #7

Solve the following exercise:

103= \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):

xnyn=(xy)n x^n\cdot y^n =(x\cdot y)^n

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

103=1012312= \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:

1012312=(103)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}

Exercise #8

Solve the following exercise:

1625= \sqrt{16}\cdot\sqrt{25}=

Video Solution

Step-by-Step Solution

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

a. The definition of root as an exponent:

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

b. The law of exponents for an exponent applied to a product in parentheses (in reverse direction):

xnyn=(xy)n x^n\cdot y^n =(x\cdot y)^n

Let's start by converting the square roots to exponents using the law of exponents mentioned in a':

1625=16122512= \sqrt{16}\cdot\sqrt{25}= \\ \downarrow\\ 16^{\frac{1}{2}}\cdot25^{\frac{1}{2}}=

We'll continue, since there is a multiplication between two terms with identical exponents, we can use the law of exponents mentioned in b' and combine them together in parentheses which are raised to the same exponent:

16122512=(1625)12=40012=400=20 16^{\frac{1}{2}}\cdot25^{\frac{1}{2}}= \\ (16\cdot25)^{\frac{1}{2}}=\\ 400^{\frac{1}{2}}=\\ \sqrt{400}=\\ \boxed{20}

In the final steps, we performed the multiplication within the parentheses and again used the definition of root as an exponent mentioned in a' (in reverse direction) to return to root notation.

Therefore, the correct answer is answer d.

Answer

20 20

Exercise #9

Solve the following exercise:

16x2= \sqrt{16x^2}=

Video Solution

Step-by-Step Solution

To simplify the given expression, we will use the following three laws of exponents:

a. Definition of root as an exponent:

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

b. Law of exponents for an exponent applied to terms in parentheses:

(ab)n=anbn (a\cdot b)^n=a^n\cdot b^n

c. Law of exponents for an exponent raised to an exponent:

(am)n=amn (a^m)^n=a^{m\cdot n}

We'll start with converting the fourth root to an exponent using the law of exponents mentioned in a.:

16x2=(16x2)12= \sqrt{16x^2}= \\ \downarrow\\ (16x^2)^{\frac{1}{2}}=

We'll continue, using the law of exponents mentioned in b. and apply the exponent to each factor in the parentheses:

(16x2)12=1612(x2)12 (16x^2)^{\frac{1}{2}}= \\ 16^{\frac{1}{2}}\cdot(x^2)^{{\frac{1}{2}}}

We'll continue, using the law of exponents mentioned in c. and perform the exponent applied to the term with an exponent in parentheses (the second factor in the multiplication):

1612(x2)12=1612x212=1612x1=16x=4x 16^{\frac{1}{2}}\cdot(x^2)^{{\frac{1}{2}}} = \\ 16^{\frac{1}{2}}\cdot x^{2\cdot\frac{1}{2}}=\\ 16^{\frac{1}{2}}\cdot x^{1}=\\ \sqrt{16}\cdot x=\\ \boxed{4x}

In the final steps, first we converted the power of one-half applied to the first factor in the multiplication back to the fourth root form, again, according to the definition of root as an exponent mentioned in a. (in the opposite direction) and then we calculated the known fourth root of 16.

Therefore, the correct answer is answer d.

Answer

4x 4x

Exercise #10

Solve the following exercise:

254= \sqrt{25}\cdot\sqrt{4}=

Video Solution

Step-by-Step Solution

We can simplify the expression directly without using the laws of exponents, since the expression has known square roots, so let's simplify the expression and then perform the multiplication:

254=52=10 \sqrt{25}\cdot\sqrt{4}=\\ 5\cdot2=\\ \boxed{10} Therefore, the correct answer is answer C.

Answer

10 10

Exercise #11

Solve the following exercise:

25x2= \sqrt{25x^2}=

Video Solution

Step-by-Step Solution

To simplify the given expression, we will use the following three laws of exponents:

a. Definition of root as an exponent:

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

b. Law of exponents for an exponent applied to terms in parentheses:

(ab)n=anbn (a\cdot b)^n=a^n\cdot b^n

c. Law of exponents for an exponent raised to an exponent:

(am)n=amn (a^m)^n=a^{m\cdot n}

We'll start by converting the fourth root to an exponent using the law of exponents mentioned in a.:

25x2=(25x2)12= \sqrt{25x^2}= \\ \downarrow\\ (25x^2)^{\frac{1}{2}}=

We'll continue, using the law of exponents mentioned in b. and apply the exponent to each factor in the parentheses:

(25x2)12=2512(x2)12 (25x^2)^{\frac{1}{2}}= \\ 25^{\frac{1}{2}}\cdot(x^2)^{{\frac{1}{2}}}

We'll continue, using the law of exponents mentioned in c. and perform the exponent applied to the term with an exponent in parentheses (the second factor in the multiplication):

2512(x2)12=2512x212=2512x1=25x=5x 25^{\frac{1}{2}}\cdot(x^2)^{{\frac{1}{2}}} = \\ 25^{\frac{1}{2}}\cdot x^{2\cdot\frac{1}{2}}=\\ 25^{\frac{1}{2}}\cdot x^{1}=\\ \sqrt{25}\cdot x=\\ \boxed{5x}

In the final steps, we first converted the power of one-half applied to the first factor in the multiplication back to the fourth root form, again, according to the definition of root as an exponent mentioned in a. (in the reverse direction) and then calculated the known fourth root of 25.

Therefore, the correct answer is answer a.

Answer

5x 5x

Exercise #12

Solve the following exercise:

22= \sqrt{2}\cdot\sqrt{2}=

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 multiplying exponents for identical bases:

(am)n=amn (a^m)^n=a^{m\cdot n}

Let's start from the square root of the exponents using the law shown in A:

22=212212= \sqrt{2}\cdot\sqrt{2}= \\ \downarrow\\ 2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}= We continue: note that we got a number times itself. According to the definition of the exponent we can write the expression as an exponent of that number. Then- we use the law of exponents shown in B and perform the whole exponent on the term in the parentheses:

212212=(212)2=2122=21=2 2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}= \\ (2^{\frac{1}{2}})^2=\\ 2^{\frac{1}{2}\cdot2}=\\ 2^1=\\ \boxed{2} Therefore, the correct answer is answer B.

Answer

2 2

Exercise #13

Solve the following exercise:

23= \sqrt{2}\cdot\sqrt{3}=

Video Solution

Step-by-Step Solution

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

a. The definition of root as an exponent:

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

b. The law of exponents for exponents applied to terms in parentheses (in reverse):

xnyn=(xy)n x^n\cdot y^n =(x\cdot y)^n

Let's start by converting the square roots to exponents using the law of exponents mentioned in a':

23=212312= \sqrt{2}\cdot\sqrt{3}= \\ \downarrow\\ 2^{\frac{1}{2}}\cdot3^{\frac{1}{2}}=

We'll continue, since there is multiplication between two terms with identical exponents, we can use the law of exponents mentioned in b' and combine them together in parentheses which are raised to the same exponent:

212312=(23)12=612=6 2^{\frac{1}{2}}\cdot3^{\frac{1}{2}}= \\ (2\cdot3)^{\frac{1}{2}}=\\ 6^{\frac{1}{2}}=\\ \boxed{\sqrt{6}}

In the final steps, we performed the multiplication within the parentheses and again used the definition of root as an exponent mentioned earlier in a' (in reverse) to return to root notation.

Therefore, the correct answer is answer b.

Answer

6 \sqrt{6}

Exercise #14

Solve the following exercise:

25= \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):

xnyn=(xy)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:

25=212512= \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:

212512=(25)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 #15

Solve the following exercise:

36x= \sqrt{36x}=

Video Solution

Step-by-Step Solution

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

a. The definition of root as an exponent:

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

b. The law of exponents for an exponent applied to terms in parentheses:

(ab)n=anbn (a\cdot b)^n=a^n\cdot b^n

Let's start by converting the square root to an exponent using the law of exponents mentioned in a:

36x=(36x)12= \sqrt{36x}= \\ \downarrow\\ (36x)^{\frac{1}{2}}=

Next, we'll use the law of exponents mentioned in b and apply the exponent to each factor within the parentheses:

(36x)12=3612x12=36x=6x (36x)^{\frac{1}{2}}= \\ 36^{\frac{1}{2}}\cdot x^{{\frac{1}{2}}}=\\ \sqrt{36}\sqrt{x}=\\ \boxed{6\sqrt{x}}

In the final steps, we first converted the power of one-half applied to each factor in the multiplication back to square root form, again, according to the definition of root as an exponent mentioned in a (in the opposite direction) and then calculated the known square root of 36.

Therefore, the correct answer is answer c.

Answer

6x 6\sqrt{x}