Combining root laws

What is a root?

A root is the inverse operation of exponentiation, denoted by the symbol , and it is equivalent to the power of 0.50.5.
If a small number appears on the left side, it indicates the order of the root.

Things to know about roots:

Square root of a product

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

Square root of a quotient

ab=ab\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}

Square root of a square root

amn=anm\sqrt[n]{\sqrt[m]{a}}=\sqrt[n\cdot m]{a}

Practice

The following exercise combines all the rules of roots,
can you solve it?

416+64327+101+3=\sqrt{4\cdot16}+\sqrt{\frac{64}{3\cdot27}}+\sqrt{10-1}+3=

Solution:
Roots come before the order of operations, so we will first deal with the first root:
416=416\sqrt{4\cdot16}=\sqrt4\cdot\sqrt{16}
We could do this using the root formula of a product.
Let's move on to the second root:
64327=6481=6481\sqrt{\frac{64}{3\cdot27}}=\sqrt{\frac{64}{81}}=\frac{\sqrt{64}}{\sqrt{81}}
Note – in this root, there was an exercise in the denominator, we first solved it and then continued to simplify the root using the root formula of a quotient.
Let's move on to the third root:
101=9\sqrt{10-1}=\sqrt9
Here we simply solved the exercise inside the root without using a formula.
Now let's rewrite the exercise slowly and carefully without getting confused:
416+64819+3=\sqrt4\cdot\sqrt{16}+\frac{\sqrt{64}}{\sqrt{81}}\cdot\sqrt9+3=
There are still roots in the exercise, so we will need to get rid of them:
24+893+3=2\cdot4+\frac{8}{9}\cdot3+3=
Now that there are no more roots, we can solve according to the order of operations:
8+223+3=13238+2\frac{2}{3}+3=13\dfrac{2}{3}

Suggested Topics to Practice in Advance

  1. Square root of a product
  2. Square root of a quotient
  3. Square Roots

Practice Rules of Roots Combined

Examples with solutions for Rules of Roots Combined

Exercise #1

Solve the following exercise:

24= \sqrt{\frac{2}{4}}=

Video Solution

Step-by-Step Solution

Simplify the following expression:

Begin by reducing the fraction under the square root:

24=12= \sqrt{\frac{2}{4}}= \\ \sqrt{\frac{1}{2}}=

Apply two exponent laws:

A. Definition of root as a power:

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

B. The power law for powers applied to terms in parentheses:

(ab)n=anbn \big(\frac{a}{b}\big)^n=\frac{a^n}{b^n}

Let's return to the expression that we obtained. Apply the law mentioned in A and convert the square root to a power:

12=(12)12= \sqrt{\frac{1}{2}}=\\ \big(\frac{1}{2}\big)^{\frac{1}{2}}=

Next use the power law mentioned in B, apply the power separately to the numerator and denominator.

In the next step remember that raising the number 1 to any power will always result in 1.

In the fraction's denominator we'll return to the root notation, again, using the power law mentioned in A (in the opposite direction):

(12)12=112212=12 \big(\frac{1}{2}\big)^{\frac{1}{2}}= \\ \frac{1^{\frac{1}{2}}}{2^{\frac{1}{2}}}=\\ \boxed{\frac{1}{\sqrt{2}}}\\ Let's summarize the simplification of the given expression:

24=12=112212=12 \sqrt{\frac{2}{4}}= \\ \sqrt{\frac{1}{2}}= \\ \frac{1^{\frac{1}{2}}}{2^{\frac{1}{2}}}=\\ \boxed{\frac{1}{\sqrt{2}}}\\ Therefore, the correct answer is answer D.

Answer

12 \frac{1}{\sqrt{2}}

Exercise #2

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 a 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 #3

Solve the following exercise:

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

Video Solution

Step-by-Step Solution

In order to simplify the given expression, apply 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}

Begin 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:

125= \sqrt{1}\cdot\sqrt{25}=

Video Solution

Step-by-Step Solution

To solve the expression 125 \sqrt{1} \cdot \sqrt{25} , we will use the Product Property of Square Roots.

According to the property, we have:

125=125\sqrt{1} \cdot \sqrt{25} = \sqrt{1 \cdot 25}

First, calculate the product inside the square root:

125=251 \cdot 25 = 25

Now the expression simplifies to:

25\sqrt{25}

Finding the square root of 25 gives us:

55

Thus, the value of 125 \sqrt{1} \cdot \sqrt{25} is 5\boxed{5}.

After comparing this solution with the provided choices, we see that the correct answer is choice 3.

Answer

5 5

Exercise #5

Choose the largest value

Video Solution

Step-by-Step Solution

Let's begin by calculating the numerical value of each of the roots in the given options:

25=516=49=3 \sqrt{25}=5\\ \sqrt{16}=4\\ \sqrt{9}=3\\ We can determine that:

5>4>3>1 Therefore, the correct answer is option A

Answer

25 \sqrt{25}

Exercise #6

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 #7

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 #8

Solve the following exercise:

22525= \sqrt{\frac{225}{25}}=

Video Solution

Step-by-Step Solution

Let's simplify the expression. First, we'll reduce the fraction under the square root, then we'll calculate the result of the root:

22525=93 \sqrt{\frac{225}{25}}= \\ \sqrt{9}\\ \boxed{3} Therefore, the correct answer is option B.

Answer

3

Exercise #9

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 #10

Solve the following exercise:

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

Video Solution

Step-by-Step Solution

In order to simplify the given expression, apply 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}

Begin 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, once again 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 #11

Solve the following exercise:

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

Video Solution

Step-by-Step Solution

In order to simplify the given expression, apply 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.:

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 once again 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 #12

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 #13

Solve the following exercise:

55= \sqrt{5}\cdot\sqrt{5}=

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 multiplication between terms with identical bases:

aman=am+n a^m\cdot a^n=a^{m+n}

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

55=512512= \sqrt{5}\cdot\sqrt{5}= \\ \downarrow\\ 5^{\frac{1}{2}}\cdot5^{\frac{1}{2}}= We'll continue, since we are multiplying two terms with identical bases - we'll use the law of exponents mentioned in b:

512512=512+12=51=5 5^{\frac{1}{2}}\cdot5^{\frac{1}{2}}= \\ 5^{\frac{1}{2}+\frac{1}{2}}=\\ 5^1=\\ \boxed{5} Therefore, the correct answer is answer a.

Answer

5 5

Exercise #14

Solve the following exercise:

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

Video Solution

Step-by-Step Solution

In order to simplify the given expression, apply 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}

Begin 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}}=

Let's proceed by 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 #15

Solve the following exercise:

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

Video Solution

Step-by-Step Solution

In order to simplify the given expression, apply 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}}=

Due to the fact that there is multiplication between two terms with identical exponents, we can apply the law of exponents mentioned in b' and then proceed to combine them together inside of 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}

Topics learned in later sections

  1. Combining Powers and Roots
  2. Square Root Rules