Question Types:

Question 1

Choose the largest value

Question 2

Solve the following exercise:

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

Question 3

Solve the following exercise:

\( \sqrt{16}\cdot\sqrt{1}= \)

Question 4

Solve the following exercise:

\( \sqrt{1}\cdot\sqrt{2}= \)

Question 5

Solve the following exercise:

\( \sqrt{10}\cdot\sqrt{3}= \)

Choose the largest value

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

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

5>4>3>1 __Therefore, the correct answer is option A__

$\sqrt{25}$

Solve the following exercise:

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

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

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

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:

$\sqrt{30}\cdot\sqrt{1}= \\
\downarrow\\
\sqrt{30}\cdot\sqrt[2]{1}=\\
\sqrt{30}\cdot 1^{\frac{1}{2}}=\\
\sqrt{30} \cdot1=\\
\boxed{\sqrt{30}}$__Therefore, the correct answer is answer C.__

$\sqrt{30}$

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

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

__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

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

$\sqrt{30}$

Question 1

Solve the following exercise:

\( \sqrt{100}\cdot\sqrt{25}= \)

Question 2

Solve the following exercise:

\( \sqrt{25}\cdot\sqrt{4}= \)

Question 3

Solve the following exercise:

\( \sqrt{9}\cdot\sqrt{4}= \)

Question 4

Solve the following exercise:

\( \sqrt{\frac{225}{25}}= \)

Question 5

Solve the following exercise:

\( \sqrt{2}\cdot\sqrt{5}= \)

Solve the following exercise:

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

We can simplify the expression ** without using the laws of exponents**, because the expression

$\sqrt{100}\cdot\sqrt{25}=\\
10\cdot5=\\
\boxed{50}$__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

$\sqrt{25}\cdot\sqrt{4}=\\
5\cdot2=\\
\boxed{10}$__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

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

$6$

Solve the following exercise:

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

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

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

3

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

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

__Let's start__ **by changing the square roots to exponents** using the law of exponents shown in

$\sqrt{2}\cdot\sqrt{5}= \\
\downarrow\\
2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}=$We continue: __since we are multiplying two terms __** with equal exponents** we can use the law of exponents shown in

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

$\sqrt{10}$

Question 1

Solve the following exercise:

\( \sqrt{2}\cdot\sqrt{2}= \)

Question 2

Solve the following exercise:

\( \sqrt{9}\cdot\sqrt{3}= \)

Question 3

Solve the following exercise:

\( \sqrt{9x}= \)

Question 4

Solve the following exercise:

\( \sqrt{2}\cdot\sqrt{5}\cdot\sqrt{2}\cdot\sqrt{2}= \)

Question 5

Solve the following exercise:

\( \sqrt[6]{12}\cdot\sqrt[3]{12}= \)

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

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

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

__Let's start__ **by changing the square root into an exponent **using the law shown in

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

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

__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

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

__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

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

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

__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

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

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