The root of a product - Examples, Exercises and Solutions

Solve the following exercise:

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

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.

$4$

Solve the following exercise:

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

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.

$\sqrt{2}$

Solve the following exercise:

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

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.

$\sqrt{30}$

Solve the following exercise:

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

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:

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

$50$

Solve the following exercise:

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

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:

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

$10$

Solve the following exercise:

$\sqrt{9}\cdot\sqrt{4}=$

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

$\sqrt{9}\cdot\sqrt{4}=\\ 3\cdot2=\\ \boxed{6}$Therefore, the correct answer is answer B.

$6$

Solve the following exercise:

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

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.

Therefore, the correct answer is answer B.

$\sqrt{10}$

Solve the following exercise:

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

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 multiplying exponents for identical bases:

$(a^m)^n=a^{m\cdot n}$

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

$\sqrt{2}\cdot\sqrt{2}= \\ \downarrow\\ 2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}=$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:

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

$2$

Solve the following exercise:

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

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.

Therefore, the correct answer is answer C.

$\sqrt{27}$

Solve:

$\sqrt{3}\cdot\sqrt{12}+3^2$

Recall:

A. Defining a root as a power:

$\sqrt[n]{a}=a^{\frac{1}{n}}$Means that all the laws of powers apply to roots as well.

B. Therefore, we can apply the following rule of powers in which we multiply two different bases with the same exponent:

$x^n\cdot y^n=(x\cdot y)^n$The literal meaning of this law in the given direction is that we can write a multiplication between two exponents with equal powers as a multiplication between the bases within the exponents raised to the same power,

We will apply these two laws of powers in the problem.

First, we will convert all the roots to powers using the definition of a root as a power that was mentioned in A above:

$\sqrt{3}\cdot\sqrt{12}+3^2 =3^{\frac{1}{2}}\cdot12^{\frac{1}{2}}+3^2$

Next, we will note that the two exponents in the multiplication have the same power, so we will apply the law of powers mentioned in B above:

$3^{\frac{1}{2}}\cdot12^{\frac{1}{2}}+3^2 =(3\cdot12)^\frac{1}{2}+3^2=36^\frac{1}{2} +3^2$

We will now return to writing roots using the definition of a root as a power that was mentioned in A above, but in the opposite direction:

$a^{\frac{1}{n}} = \sqrt[n]{a}$We will apply this to the expression we obtained:

$36^\frac{1}{2} +3^2 =\sqrt{36}+3^2=6+9=15$For the first term we converted the half power of the first exponent to a square root, for the next we simply calculated (without a calculator!, that's the whole point here..) the numerical value of the root.

In summary:

$\sqrt{3}\cdot\sqrt{12}+3^2 =3^{\frac{1}{2}}\cdot12^{\frac{1}{2}}+3^2 =36^\frac{1}{2} +3^2=6+9=15$Therefore, the correct answer is answer C.

$15$

Solve the following exercise:

$\sqrt{9x}=$

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

A. Definition of the root as an exponent:

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

B. Law of exponents for dividing powers with the same base:

$(a\cdot b)^n=a^n\cdot b^n$

Let's start with converting the root to an exponent using the law of exponents shown in A:

$\sqrt{9x}= \\ \downarrow\\ (9x)^{\frac{1}{2}}=$Next, we will use the law of exponents shown in B and apply the exponent to each of the factors in the numerator that are in parentheses:

$(9x)^{\frac{1}{2}}= \\ 9^{\frac{1}{2}}\cdot x^{{\frac{1}{2}}}=\\ \sqrt{9}\sqrt{x}=\\ \boxed{3\sqrt{x}}$$$In the last steps, we will multiply the half exponent by each of the factors in the numerator, returning to the root form, that is, according to the definition of the root as an exponent shown in A (in the opposite direction) and then we will calculate the known fourth root result of the number 9.

Therefore, the correct answer is answer D.

$3\sqrt{x}$

Solve the following exercise:

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

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 a product of numbers with the same base (in the opposite direction):

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

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

$\sqrt{2}\cdot\sqrt{5}\cdot\sqrt{2}\cdot\sqrt{2}= \\ \downarrow\\ 2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}=$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:

$2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}= \\ (2\cdot5\cdot2\cdot2)^{\frac{1}{2}}=\\ 40^{\frac{1}{2}}=\\ \boxed{\sqrt{40}}$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.

$\sqrt{40}$

Solve the following exercise:

$\sqrt[6]{12}\cdot\sqrt[3]{12}=$

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 multiplication of terms with identical bases:

$a^m\cdot a^n=a^{m+n}$

Let's start from converting the roots to exponents using the law of exponents shown in A:

$\sqrt[\textcolor{red}{6}]{12}\cdot\sqrt[\textcolor{blue}{3}]{12}= \\ \downarrow\\ 12^{\frac{1}{\textcolor{red}{6}}}\cdot12^{\frac{1}{\textcolor{blue}{3}}}=$We continue, since a multiplication of two terms with identical bases is performed - we use the law of exponents shown in B:

$12^{\frac{1}{6}}\cdot12^{\frac{1}{3}}= \\ \boxed{12^{\frac{1}{6}+\frac{1}{3}}}$Therefore, the correct answer is answer C.

$12^{\frac{1}{6}+\frac{1}{3}}$

Solve the following exercise:

$\sqrt[4]{3}\cdot\sqrt[6]{3}=$

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 multiplication between factors with the same bases:

$a^m\cdot a^n=a^{m+n}$

Let's start with converting the roots to exponents using the law of exponents shown in A:

$\sqrt[\textcolor{red}{4}]{3}\cdot\sqrt[\textcolor{blue}{6}]{3}= \\ \downarrow\\ 3^{\frac{1}{\textcolor{red}{4}}}\cdot3^{\frac{1}{\textcolor{blue}{6}}}=$We continue, since multiplication is performed between two factors with the same bases - we use the law of exponents shown in B:

$3^{\frac{1}{4}}\cdot3^{\frac{1}{6}}= \\ \boxed{3^{\frac{1}{4}+\frac{1}{6}}}$Therefore, the correct answer is answer D.

$3^{\frac{1}{4}+\frac{1}{6}}$

Solve the following exercise:

$\sqrt[4]{8}\cdot\sqrt[7]{8}=$

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 multiplication of terms with identical bases:

$a^m\cdot a^n=a^{m+n}$

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

$\sqrt[\textcolor{red}{4}]{8}\cdot\sqrt[\textcolor{blue}{7}]{8}= \\ \downarrow\\ 8^{\frac{1}{\textcolor{red}{4}}}\cdot8^{\frac{1}{\textcolor{blue}{7}}}=$We continue, since we have a multiplication of two terms with identical bases - we use the law of exponents shown in B:

$8^{\frac{1}{4}}\cdot8^{\frac{1}{7}}= \\ \boxed{8^{\frac{1}{4}+\frac{1}{7}}}$Therefore, the correct answer is answer D.

$8^{\frac{1}{4}+\frac{1}{7}}$