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Here we will learn about radicals, including simplifying radicals, adding and subtracting radicals, multiplying radicals, dividing radicals and rationalizing radicals.

Students are first exposed to a radical, or square root in 8 th grade when taking the square root of numbers. However, they greatly expand that knowledge in algebra class.

**Radicals** (or sometimes referred to as surds) are represented by \sqrt{\;\;} and are used to calculate the square root or the nth root of numbers and expressions. Expressions with \sqrt{\;\;} are called radical expressions.

For example, \sqrt{16}=4 because 16 is a perfect square number. Perfect square numbers have whole number roots.

However, \sqrt{5} \approx 2.23606 \ldots is an irrational number because it’s a non-terminating, non-repeating decimal that cannot be expressed as a fraction.

You can simplify the number under the radical to an exact calculation. For example, let’s apply the Pythagorean Theorem to find the missing length of the triangle.

\begin{aligned}& 4^2+6^2=x^2 \\\\ & 16+36=x^2 \\\\ & 52=x^2 \\\\ & \sqrt{52}=x \end{aligned}Notice how 52 is not a perfect square number. One way to find the answer is to type \sqrt{52} into a calculator.

Another strategy is to simplify the number under the radical. That means to find the largest square root factor of 52.

Here are some perfect square numbers: 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, …

4 is the largest perfect square factor of 52.

\begin{aligned}& \sqrt{52}=\sqrt{4} \times \sqrt{13} \\\\ & \sqrt{52}=2 \times \sqrt{13} \end{aligned}\sqrt{52}=2 \sqrt{13} \rightarrow simplified radical expression or exact form

You can also simplify radical expressions that contain variables.

For example, let’s simplify the expression, \sqrt{4 x^3}

4 is a perfect square number so that is simple enough to do. However, how do you simplify x^3? Variables raised to even exponents are considered perfect square expressions.

\begin{aligned}& x^2=x \cdot x \\\\ & x^4=x^2 \cdot x^2 \\\\ & x^6=x^3 \cdot x^3\end{aligned}If you notice, x^3 does not have an even exponent. However, you can break it apart so that it does.

\begin{aligned}& \sqrt{4 x^3}=\sqrt{4 \cdot x^2 \cdot x}=\sqrt{4} \cdot \sqrt{x^2} \cdot \sqrt{x} \\\\ & =2 x \sqrt{x} \end{aligned}\sqrt{4 x^3} simplified is 2 x \sqrt{x}

**Step-by-step guide:** How to simplify radicals

In order to add and subtract radical expressions, the number or expression under the radical symbol must be the same.

For example, 14 \sqrt{6}-6 \sqrt{6} can be subtracted because both radicals are identical.

So, 14 \sqrt{6}-6 \sqrt{6}=8 \sqrt{6}

**Do not add or subtract the number under the radical. It remains the same.**

If the number or expression under the radical is not the same, you must use strategies for simplifying the radical in order to add or subtract them.

For example, 5 \sqrt{12}+2 \sqrt{27} cannot be added the way it is written because the radicals do not match.

\begin{aligned}5 \sqrt{12} & =5 \cdot \sqrt{4} \cdot \sqrt{3} \\\\ & =5 \cdot 2 \sqrt{3} \\\\ & =10 \sqrt{3} \\\\ 2 \sqrt{27} & =2 \cdot \sqrt{9} \cdot \sqrt{3} \\\\ & =2 \cdot 3 \sqrt{3} \\\\ & =6 \sqrt{3} \end{aligned}In this case, the radicals simplify to be the same. So, they can now be added.

10 \sqrt{3}+6 \sqrt{3}=16 \sqrt{3}**Step-by-step guide:** Adding radicals

**Step-by-step guide: **Subtracting radical

Unlike addition and subtraction, the radicals do not have to match when multiplying or dividing. You can also multiply and divide the numbers under the radicals.

For example, 6 \sqrt{7} \cdot 9 \sqrt{2}=54 \sqrt{14} \rightarrow The numbers outside the radicals multiply together and the numbers under the radicals multiply together.

Always remember to look at the final product to see if it can be simplified. In this case, the product cannot be simplified any further because there is not a perfect square factor of 14.

Let’s look at another example. \cfrac{16 \sqrt{24}}{4 \sqrt{2}}=4 \sqrt{12}

When dividing, the numbers outside of the radicals can divide one another and the numbers inside the radicals can divide one another. In this case, notice how the quotient can be simplified.

\begin{aligned}4 \sqrt{12} & =4 \cdot \sqrt{4} \cdot \sqrt{3} \\\\ & =4 \cdot 2 \sqrt{3} \\\\ & =8 \sqrt{3}\end{aligned}**Step-by-step guide:** Multiplying radicals

**Step-by-step guide:** Dividing radicals

Expressions that have radicals in the denominator must be rationalized. This means that an equivalent expression has to be created so that it does not have a radical in the denominator.

Let’s look at the expression \cfrac{\sqrt{15}}{\sqrt{6}}, notice how 6 does not divide evenly into 15.

So, division does not work in this case. The radical cannot remain in the denominator so there is another strategy you can use to rationalize the expression.

Multiply the numerator and denominator by the radical in the denominator, which is \sqrt{6}. The goal is to find an equivalent expression that does not have a radical in the denominator.

Multiplying the numerator and denominator by \sqrt{6} is like multiplying the expression by \cfrac{\sqrt{6}}{\sqrt{6}} which is 1.

\cfrac{\sqrt{15}}{\sqrt{6}} \cdot \cfrac{\sqrt{6}}{\sqrt{6}}=\cfrac{\sqrt{90}}{\sqrt{36}}=\cfrac{\sqrt{9} \cdot \sqrt{10}}{6}=\cfrac{3 \sqrt{10}}{6}\cfrac{\sqrt{15}}{\sqrt{6}}=\cfrac{3 \sqrt{10}}{6} \rightarrow the radical is no longer in the denominator

However, the expression \cfrac{3 \sqrt{10}}{6} can be simplified further. Notice how the 3 in the numerator and the 6 in the denominator can be simplified.

So, \cfrac{\sqrt{15}}{\sqrt{6}}=\cfrac{3 \sqrt{10}}{6}=\cfrac{\sqrt{10}}{2}

**Step-by-step guide:** Rationalize the denominator

How does this relate to high school math?

**Number and Quantity – High School: (HSN-RN.B.3)**Explain why the sum or product of two rational numbers is rational; that the sum of a rational number and an irrational number is irrational; and that the product of a nonzero rational number and an irrational number is irrational.

Use this quiz to check your grade 6 to 8 students’ understanding of algebra. 10+ questions with answers covering a range of 6th and 8th grade algebra topics to identify areas of strength and support!

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DOWNLOAD FREEThere are lots of different ways we can calculate with radicals. We can:

- Simplify radicals and use them in exact calculations
- Add and subtract radical expressions
- Multiply and divide radicals
- Rationalize the denominator of radicals

Let’s have a look at each one in more detail.

Simplify the expression: \sqrt{60}

**Find the largest square factor(s) of the term under the root.**

Square numbers are 1, 4, 9, 16, 25, …

The largest square factor of 60 is 4 because 4 \times 15=60

2**Rewrite the radical as a product of the square factor(s) and the other factors.**

\begin{aligned}\sqrt{60} & =\sqrt{4 \times 15} \\\\ & =\sqrt{4} \times \sqrt{15}\end{aligned}

3**Simplify the radical expression.**

\sqrt{60} =\sqrt{4} \times \sqrt{15} \;\; (4 is a perfect square so take the square root of 4 to simplify )

\hspace{0.8cm} =2 \sqrt{15}

\sqrt{60} simplifies to be 2 \sqrt{15}

Simplify the expression: \sqrt{16 x^5}

**Find the largest square factor(s) of the term under the root.**

16 is a perfect square and x^4 is the largest square factor of x^5.

**Rewrite the radical as a product of the square factor(s) and the other factors.**

\begin{aligned}\sqrt{16 x^5} & =\sqrt{16 x^4 x^1} \\\\ & =\sqrt{16} \times \sqrt{x^4} \times \sqrt{x^1}\end{aligned}

**Simplify the radical expression.**

\begin{aligned} \sqrt{16 x^5}&=\sqrt{16} \times \sqrt{x^4} \times \sqrt{x^1} \\\\ & =4 x^2 \sqrt{x^1} \end{aligned}

\sqrt{16 x^5} simplifies to be 4 x^2 \sqrt{x}

Add the expressions: 2\sqrt{5}+\sqrt{45}

**If the radicals are not alike, simplify them to be the same.**

\sqrt{5} and \sqrt{45} are not the same.

\sqrt{5} cannot be simplified.

\sqrt{45} can be simplified because 9 is a perfect square factor of 45.

\begin{aligned} \sqrt{45}&=\sqrt{9} \times \sqrt{5} \\\\ \sqrt{45}&=3 \sqrt{5} \end{aligned}

**Rewrite the expressions to have like radicals.**

2 \sqrt{5}+\sqrt{45}=2 \sqrt{5}+3 \sqrt{5}

**Add or subtract the expressions.**

2 \sqrt{5}+3 \sqrt{5}=5 \sqrt{5}

Subtract the expressions: \sqrt{18}-10 \sqrt{2}

**If the radicals are not alike, simplify them to be the same.**

\sqrt{18} and \sqrt{2} are not the same.

\sqrt{18} can be simplified because 9 is a perfect square factor of 18.

\begin{aligned}\sqrt{18} & =\sqrt{9} \times \sqrt{2} \\\\ & =3 \sqrt{2}\end{aligned}

\sqrt{2} cannot be simplified

**Rewrite the expressions to have like radicals.**

\sqrt{18}-10 \sqrt{2}=3 \sqrt{2}-10 \sqrt{2}

**Add or subtract the expressions.**

3 \sqrt{2}-10 \sqrt{2}=-7 \sqrt{2}

Divide the radical expression.

\cfrac{2 \sqrt{10}-\sqrt{90}}{\sqrt{2}}**Divide the numbers in the radicals and the numbers outside the radicals.**

\cfrac{2 \sqrt{10}-\sqrt{90}}{\sqrt{2}}

In this case, you are going to divide 2 \sqrt{10} \div \sqrt{2} and \sqrt{90} \div \sqrt{2}

So, 2 \sqrt{10} \div \sqrt{2}=2 \sqrt{5} and \sqrt{90} \div \sqrt{2}=\sqrt{45}

**Simplify radicals that can be simplified and write the new expression.**

\cfrac{2 \sqrt{10}-\sqrt{90}}{\sqrt{2}}=2 \sqrt{5}-\sqrt{45}

\sqrt{5} cannot be simplified.

\sqrt{45} can be simplified because 45 has a perfect square factor which is 9.

\begin{aligned}\sqrt{45} & =\sqrt{9} \times \sqrt{5} \\\\ & =3 \times \sqrt{5} \\\\ & =3 \sqrt{5}\end{aligned}

\sqrt{45} simplifies to be 3 \sqrt{5}

\begin{aligned}& \cfrac{2 \sqrt{10}-\sqrt{90}}{\sqrt{2}}=2 \sqrt{5}-\sqrt{45} \\\\ & \cfrac{2 \sqrt{10}-\sqrt{90}}{\sqrt{2}}=2 \sqrt{5}-3 \sqrt{5}\end{aligned}

In this example, you can go even further to simplify because the radicals in both terms match. So, 2 \sqrt{5}-3 \sqrt{5}=-1 \sqrt{5} or -\sqrt{5}

Multiply the radical expression.

\sqrt{3}(6+\sqrt{8})**Multiply the numbers in the radicals and the numbers outside the radicals.**

In this case, you will have to use the distributive property.

\begin{aligned}\sqrt{3}(6+\sqrt{8}) & =\sqrt{3} \times 6+\sqrt{3} \times \sqrt{8} \\\\
& =6 \sqrt{3}+\sqrt{24}\end{aligned}

**Simplify radicals that can be simplified and write the new expression.**

6 \sqrt{3}+\sqrt{24} \rightarrow \sqrt{3} cannot be simplified and \sqrt{24} can be simplified because 24 has a perfect square factor which is 4.

\begin{aligned}&\begin{aligned}
\sqrt{24} & =\sqrt{4} \times \sqrt{6} \\\\ & =2 \times \sqrt{6}\\\\ & =2 \sqrt{6} \end{aligned} \\\\ &\begin{gathered}6 \sqrt{3}+\sqrt{24}=6\sqrt{3}+2 \sqrt{6}\end{gathered}\end{aligned}

Simplify the radical expression.

\cfrac{\sqrt{5}}{\sqrt{12}}**Multiply the numerator and the denominator of the expression by the radical in the denominator.**

The radical in the denominator is \sqrt{12}.

So, \cfrac{\sqrt{5}}{\sqrt{12}} \times \cfrac{\sqrt{12}}{\sqrt{12}}=\cfrac{\sqrt{60}}{\sqrt{144}}

**Simplify the expression completely.**

\cfrac{\sqrt{60}}{\sqrt{144}}=\cfrac{\sqrt{60}}{12}

\sqrt{60} can be simplified further because 60 has a perfect square factor which is 4. So, \sqrt{60}=\sqrt{4} \times \sqrt{15} which simplifies to be 2 \sqrt{15}

\cfrac{2 \sqrt{15}}{12} can then be simplified further because 2 is a factor of 12.

\cfrac{2 \sqrt{15}}{12}=\cfrac{\sqrt{15}}{6}

\cfrac{\sqrt{15}}{6} is simplified completely.

Simplify the expression by rationalizing the denominator.

\cfrac{6}{\sqrt{11}-3}**Write the conjugate.**

The conjugate of \sqrt{11}-3 is \sqrt{11}+3

**Multiply the numerator and the denominator by the conjugate.**

\begin{aligned} &\cfrac{6}{\sqrt{11}-3} \times \cfrac{\sqrt{11}+3}{\sqrt{11}+3} \\\\
&=\cfrac{6 \times \sqrt{11}+6 \times 3}{\sqrt{11} \times \sqrt{11}+\sqrt{11} \times 3-(3 \times \sqrt{11}-3 \times 3} \\\\
&=\cfrac{6 \sqrt{11}+18}{\sqrt{121}-9} \\\\
&=\cfrac{6 \sqrt{11}+18}{11-9} \\\\
&=\cfrac{6 \sqrt{11}+18}{2} \end{aligned}

**Simplify the expression completely.**

\cfrac{6 \sqrt{11}+18}{2} can be simplified because 2 is a factor of 6 and 18

\cfrac{6 \sqrt{11}+18}{2}=9+3 \sqrt{11}

- Reinforce perfect square numbers with students.
- Review with students how to break numbers down into their factors.
- Use digital platforms such as Desmos to help students develop conceptual understanding.
- Have students practice problems by using interactive digital platforms such as Khan Academy.

**Thinking that the square root is dividing a number by \bf{2}**

For example, \sqrt{16}≠ 8

\sqrt{16}=4 because 4 \times 4=16

**Rewriting the number under the square root sign (the radicand) as a product of any two factors**

In order to simplify a radical, it has to be a perfect square or has to have a perfect square factor. So, one of these factors must be a square number.

To simplify \sqrt{40} , break 40 so that one of the factors is a perfect square. The largest square factor of 40 is 4.

\begin{aligned}\sqrt{40} & =\sqrt{4} \times \sqrt{10} \\\\ & =2 \times \sqrt{10} \\\\ & =2 \sqrt{10}\end{aligned}

**Not simplifying fully**

Always check that there are no square factors of the number under the root.

For example,

\begin{aligned}\sqrt{32} & =\sqrt{4} \times \sqrt{8} \\\\ & =2 \times \sqrt{8} \\\\ & =2 \sqrt{8}\end{aligned}

This expression is not fully simplified because the number under the radical, 8, can be simplified further. 4 is a perfect square factor of 8.

So, 2 \sqrt{8}=2 \times \sqrt{4} \times \sqrt{2}=2 \times 2 \times \sqrt{2}=4 \sqrt{2}

**Remembering that a radical or root with no integer coefficient is ‘ \bf{1} lot’ of that surd**

As in algebra, we understand that a actually means 1a. So, \sqrt{2} means 1 \times \sqrt{2}

**Mixing up addition and multiplication laws**

With addition, the radicals have to match and only the numbers in front of the radicals can be added \rightarrow \sqrt{3}+\sqrt{3}=2 \sqrt{3}

With multiplication, both the numbers inside and outside the radicals are multiplied together \rightarrow \sqrt{3} \times \sqrt{3}=\sqrt{9}=3

Refer back to knowledge of algebra to help: a+a=2 a and a \times a=a^2 \text {.}

**Trying to combine unlike radicals**

For example, \sqrt{3}+\sqrt{5}≠ \sqrt{8} , just leave the expression as \sqrt{3}+\sqrt{5}

1. Simplify the radical expression fully.

\sqrt{450}

3 \sqrt{50}

8 \sqrt{2}

15 \sqrt{2}

9 \sqrt{50}

Look for the largest perfect square factor of 450.

225 is a perfect square number and a factor of 450.

So,

\begin{aligned}\sqrt{450} & =\sqrt{225 \times 2} \\\\ & =\sqrt{225} \times \sqrt{2} \\\\ & =15 \sqrt{2} \end{aligned}

2. Subtract the radical expression.

\sqrt{50}-2 \sqrt{2}

2 \sqrt{25}

3 \sqrt{2}

2 \sqrt{5}

2 \sqrt{48}

Simplify the radicals so that the radicals of both terms match.

\sqrt{50} can be simplified. Rewrite \sqrt{50}=\sqrt{25 \times 2}=5 \sqrt{2}

Both radicals match so the terms can be subtracted.

5 \sqrt{2}-2 \sqrt{2}=3 \sqrt{2}

3. Multiply the radical expressions and simplify the answer fully.

\sqrt{7} \times 2 \sqrt{8}

2 \sqrt{56}

14 \sqrt{8}

28 \sqrt{2}

4 \sqrt{14}

Multiply the numbers inside the radicals and outside the radicals.

\sqrt{7} \times 2 \sqrt{8}=2 \sqrt{56}

\sqrt{56} can be simplified because 56 has a perfect square factor which is 4.

\begin{aligned}\sqrt{56} & =\sqrt{4} \times \sqrt{14} \\\\ & =2 \times \sqrt{14} \\\\ & =2 \sqrt{14}\end{aligned}

So,

\begin{aligned}2 \sqrt{56} & =2 \times 2 \sqrt{14} \\\\ & =4 \sqrt{14}\end{aligned}

4. Divide the radical expression and simplify the answer fully.

\cfrac{\sqrt{60}}{2 \sqrt{3}}

\cfrac{\sqrt{5}}{2}

\cfrac{\sqrt{20}}{2}

2 \sqrt{20}

\sqrt{5}

Divide the number under the radicals.

\begin{aligned}& \cfrac{\sqrt{60}}{2 \sqrt{3}} \rightarrow \sqrt{60} \div \sqrt{3}=\sqrt{20} \\\\ & \cfrac{\sqrt{60}}{2 \sqrt{3}}=\cfrac{\sqrt{20}}{2}\end{aligned}

Simplify the radical because 20 has a perfect square factor of 4.

\cfrac{\sqrt{20}}{2}=\cfrac{\sqrt{4} \times \sqrt{5}}{2}=\cfrac{2 \sqrt{5}}{2}=\sqrt{5}

5. Simplify the expression.

\sqrt{100 x^7}

50 x^3 \sqrt{x}

50 \sqrt{x^7}

10 x^3 \sqrt{x}

10 \sqrt{x^7}

\sqrt{100 x^7} can be simplified because 100 is a perfect square number.

\sqrt{x^7} can be rewritten as \sqrt{x^6 \times x} and x^6 is a perfect square term.

So,

\begin{aligned}\sqrt{100 x^7} & =\sqrt{100} \times \sqrt{x^6} \times \sqrt{x^1} \\\\ & =10 x^3 \sqrt{x}\end{aligned}

6. Expand the expression and simplify.

4(3+\sqrt{28})

12+2\sqrt{7}

12+4\sqrt{28}

12+4\sqrt{7}

12+8\sqrt{7}

Use the distributive property to multiply.

\begin{aligned}4(3+\sqrt{28}) & =4 \times 3+4 \times \sqrt{28} \\\\ & =12+4 \sqrt{28}\end{aligned}

\sqrt{28} can be simplified because 28 has a perfect square factor which is 4.

So, \sqrt{28}=\sqrt{4 \times 7}=2 \sqrt{7}.

\begin{aligned}12+4 \sqrt{28} & =12+4(2 \sqrt{7}) \\\\ & =12+8 \sqrt{7}\end{aligned}

So the final answer is 12+8 \sqrt{7}.

7. Rationalize the denominator of the radical expression.

\cfrac{4}{\sqrt{5}}

\cfrac{4 \sqrt{5}}{5}

\cfrac{\sqrt{5}}{5}

4 \sqrt{5}

\cfrac{\sqrt{20}}{5}

Multiply the numerator and denominator by \sqrt{5}.

A rational exponent is a fractional exponent that represents a radical expression. For example, the irrational number, \sqrt{3} can be written as (3)\frac{{1}}{{2}}.

When you get into algebra 2 you will learn how to work with the square root of negative numbers. The square root of negative numbers produce imaginary numbers.

The absolute value of a complex number is in the form of a+b i=\sqrt{a^2+b^2}.

You can take the cube root of perfect cube numbers similarly to the way you can take the square root of perfect square numbers.

For example, \sqrt[3]{8}=2. Notice how the index of the radical is 3, that’s because you are taking a cube root and not a square root.

Radical equations are equations that have radical expressions. Solving radical equations requires undoing the radical. For example, squaring a square root expression undoes the square root and cubing a cube root expression undoes the cube root.

In algebra 1 and algebra 2, you will analyze the graph of rational expressions/equations, logarithm equations, polynomial equations, quadratic equations, and linear equations. You will learn about their features and what their graphs look like.

- Radical functions
- Vectors
- Quadratic equations
- Trigonometry

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