Solve the following exercise:

\( \sqrt{\frac{2}{4}}= \)

- The result of the square root will always be positive! You will never get a negative result. We can get a result of $0$.
- For $\sqrt{}$ of a negative-number there is no answer!
- The square root is basically a half power. We can say that: $\sqrt{a}=a^{\frac{1}{2}}$
- A square root precedes the four arithmetic operations. First, perform the square root and only then order the operations of the account.

The laws of radicals are very relevant for solving exercises, and combining them with power rules can greatly help you solve exercises easily.

Test your knowledge

Question 1

Solve the following exercise:

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

Question 2

Solve the following exercise:

\( \sqrt{25x^4}= \)

Question 3

Solve the following exercise:

\( \sqrt{1}\cdot\sqrt{25}= \)

When the root appears across the entire product, we can break down each factor and apply the root to them, leaving the multiplication sign between the factors.

We formulate:

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

When the root appears over the entire quotient (over the entire fraction), we can break down each factor and apply the root to it, leaving the division sign (fraction line) between the factors.

We formulate:

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

Do you know what the answer is?

Question 1

Choose the largest value

Question 2

Solve the following exercise:

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

Question 3

Solve the following exercise:

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

The root over another root, we will multiply the order of the first root by the order of the second root and the order we obtain will be executed as a root on our number. (As the rule of power over another power)

Let's put it this way:

$\sqrt[n]{\sqrt[m]{a}}=\sqrt[n\cdot m]{a}$

A root is symbolized with the sign $√$

Indeed, when we see a number with a root, we wonder what positive number raised to $2$, will give us what is written inside the root.

A root is the opposite of a power operation. When there is no small number at the top left of the root, it denotes that it is a root of $2$, square root.

If a small number appears on the left, this will be the order of the root.

**Let's know some of the fundamental laws:**

- The result of a root will always be positive!! A negative result will never be obtained. We can get a result of $0$.
- For $√$ (negative number) there is no answer!
- The root is basically a half power. We can say that: $\sqrt a=a^{ 1 \over 2}$
- A root precedes the four arithmetic operations. First, perform the root and only then order the arithmetic operations.

**Let's see the example:**

$\sqrt {64} =8$

Let's ask, what power of $2$ will give us $64$ and the answer is $8$.

True, also $-8$ to the power of $2$ will give us $64$ but the result of the root must be positive!

Check your understanding

Question 1

Solve the following exercise:

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

Question 2

Solve the following exercise:

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

Question 3

Solve the following exercise:

\( \sqrt{100x^2}= \)

When we find a root that is in the entirety of the product, we can break down the product's factors and leave a separate root for each of them.

We will formulate this as a rule:

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

**Let's see this in an example:**$\sqrt{(64\cdot100)}$

According to the root of a product rule, 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:

$\sqrt{64}\cdot\sqrt{100}=$

$8\cdot10=80$

When we encounter a root that is over the entire quotient (fraction) we can break down the factors of the quotient and leave a separate root for each of them. We will place the division operation between the two factors: the fraction line.

Let's formulate it this way:$\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$

**Let's see this in an example:**

$\sqrt{\frac{36}{9}}$

According to the rule of the root of a quotient, 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:

$\frac{\sqrt{36}}{\sqrt{9}}$

$\frac{6}{3}=2$

Do you think you will be able to solve it?

Question 1

Solve the following exercise:

\( \sqrt{16x^2}= \)

Question 2

Solve the following exercise:

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

Question 3

Solve the following exercise:

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

When we encounter an exercise where there is a root over a root, we will multiply the order of the first root by the order of the second root and the order we obtain we will multiply as a root over our number. (As in the rule of power over power)

Let's formulate it this way:

$\sqrt[n]{\sqrt[m]{a}}=\sqrt[n\cdot m]{a}$

Let's see this in the example:

$\sqrt[2]{\sqrt[4]{100}}=\sqrt[2\cdot4]{100}=\sqrt[8]{100}$

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

**In the blog of** **Tutorela** **you will find a variety of articles on mathematics.**

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:

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

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.__

$\sqrt{10}$

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:

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

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.__

$\sqrt{30}$

Test your knowledge

Question 1

Solve the following exercise:

\( \sqrt{25x^2}= \)

Question 2

Solve the following exercise:

\( \sqrt{2}\cdot\sqrt{3}= \)

Question 3

Solve the following exercise:

\( \sqrt{\frac{2}{4}}= \)

Related Subjects

- Large Numbers
- Small Numbers
- Power of a Quotient
- Exponent of a Multiplication
- Multiplying Exponents with the Same Base
- Division of Exponents with the Same Base
- Power of a Power
- The quadratic function
- Parabola
- Symmetry in a parabola
- Plotting the Quadratic Function Using Parameters a, b and c
- Finding the Zeros of a Parabola