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Here you will learn about when radical expressions can be subtracted, how to subtract them, and when simplifying a radical is necessary in order to subtract them.

Students first learn how to subtract radicals in algebra and build upon those strategies as they progress through high school math

**Subtracting radicals** is where you can subtract **radical expressions** when the numbers or expressions under the root symbols (the **radicands**) are the same; these are called β**like radicals**β.

The strategy of subtracting like roots and radicals is very similar to the strategy of subtracting like algebraic terms (**combining like terms**).

Recall simplifying 4a-a-2a\text{:}

Gather like terms by subtracting them (combine like terms).

4 a-a-2 a=3 a-2 a=1 a=aThis strategy can be applied to radicals.

For example, simplify or subtract the radical expression:

4 \sqrt{3}-\sqrt{3}If this was 4 x-x, you would subtract these terms to get 3x.

\sqrt{3}=1 \sqrt{3}So, 4 \sqrt{3}-\sqrt{3}=3 \sqrt{3}

Subtracting like radicals simplifies the expression to be: 3 \sqrt{3}.

Notice how the number in front of the like radicals is what you are subtracting, not the number under the radical (exactly the same if you were simplifying algebraic terms).

If the radicals are not the same, you cannot simplify the expression, just like when the variables of an algebraic expression are not the same, you cannot simplify it.

- 2a-5b cannot be simplified because the variables are not alike.
- 2\sqrt{3}-5\sqrt{7} cannot be simplified because the radicals are not alike.

So, radicals can only be subtracted (simplified) if the radicals are the same.

By simplifying radicals, the radicals can become alike.

For example,

\sqrt{8}-\sqrt{32}-\sqrt{2} \begin{aligned}& \sqrt{8}=\sqrt{4 \times 2}=\sqrt{4} \times \sqrt{2}=2 \sqrt{2} \\\\ & \sqrt{32}=\sqrt{16 \times 2}=\sqrt{16} \times \sqrt{2}=4 \sqrt{2} \end{aligned}Rewrite the expression in its simplest form.

\sqrt{8}-\sqrt{32}-\sqrt{2}=2 \sqrt{2}-4 \sqrt{2}-\sqrt{2}=- \, 3 \sqrt{2}**Step-by-step guide:** How to simplify radicals

How does this relate to high school math?

**High school- The Real Number System (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 non-zero rational number and an irrational number is irrational.

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

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

DOWNLOAD FREEIn order to subtract radicals:

**Simplify any unlike radicals.****Subtract the radical terms.**

Simplify the radical expression:

5\sqrt{3}-2\sqrt{3}**Simplify any unlike radicals.**

Both radicals are the same, so you donβt need to simplify any radicals.

2**Subtract the radical terms.**

Subtract the coefficients in front of the radicals.

5\sqrt{3}-2\sqrt{3}=3\sqrt{3}Remember to only subtract the coefficients of the radical.

Subtract the radical expression 7\sqrt{5}-9\sqrt{5}-6\sqrt{5} .

**Simplify any unlike radicals.**

All the radicals are the same, so you donβt need to simplify any radicals.

**Subtract the radical terms.**

Subtract the coefficients in front of the radicals.

7 \sqrt{5}-9 \sqrt{5}-6 \sqrt{5}=- \, 8 \sqrt{5}

*Remember to apply rules of integers when subtracting.*

Subtract the radical expression 10\sqrt{2}-4\sqrt{11}.

**Simplify any unlike radicals.**

\sqrt{2} cannot be simplified because there is not a perfect square factor of 2.

\sqrt{11} cannot be simplified because there is not a perfect square factor of 11.

**Subtract the radical terms.**

In this case, the radical terms cannot be subtracted because they are not like radicals.

10\sqrt{2}-4\sqrt{11}

Subtract the radical expression.

4-6\sqrt{2}-3\sqrt{2}-\sqrt{25}**Simplify any unlike radicals.**

There are like radicals in the expression, but there is also a radical that can be simplified.

\sqrt{25} is a perfect square, so you can take the square root.

\sqrt{25}=5 (take the principal square root when simplifying)

**Subtract the radical terms.**

The expression is now, 4-6\sqrt{2}-3\sqrt{2}-5.

The like radical terms can be subtracted because the radicals match.

- \, 6 \sqrt{2}-3 \sqrt{2}=- \, 9 \sqrt{2}

The integers can also be subtracted.

4-5=- \, 1

The new expressions, 4-6 \sqrt{2}-3 \sqrt{2}-5, simplifies to be:

- \, 9 \sqrt{2}-1

Subtract the radical expression \sqrt{7}-\sqrt{28} .

**Simplify any unlike radicals.**

\sqrt{7} cannot be simplified because there is not a perfect square factor of 7.

\sqrt{28} can be simplified because 4 is the perfect square factor of 28.

\sqrt{28}=\sqrt{4\times7}=\sqrt{4}\times\sqrt{7}=2\times\sqrt{7}=2\sqrt{7}

**Subtract the radical terms.**

\sqrt{7}-\sqrt{28} can be rewritten as \sqrt{7}-2\sqrt{7}, the radicals are alike and can now be subtracted.

Subtract the coefficients in front of the radicals. Remember, there is a 1 in front of \sqrt{7}.

\sqrt{7}-2 \sqrt{7}=- \, 1 \sqrt{7}=- \, \sqrt{7}

\sqrt{7}-\sqrt{28}=- \, \sqrt{7}

Subtract the radical expression \sqrt{8}-\sqrt{72} .

**Simplify any unlike radicals.**

\sqrt{8} can be simplified; 4 is the largest perfect square factor of 8.

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

\sqrt{72} can be simplified; 36 is the largest perfect square factor of 72.

\sqrt{72}=\sqrt{36\times{2}}=\sqrt{36}\times\sqrt{2}=6\times\sqrt{2}=6\sqrt{2}

**Subtract the radical terms.**

\sqrt{8}-\sqrt{72} can now be rewritten as ββ 2 \sqrt{2}-6 \sqrt{2}, the radicals are the same and now they can be subtracted.

Subtract the coefficients in front of the radicals.

2 \sqrt{2}-6 \sqrt{2}=- \, 4 \sqrt{2}

\sqrt{8}-\sqrt{72}=- \, 4 \sqrt{2}

Subtract the radical expression \sqrt{75}-\sqrt{50}.

**Simplify any unlike radicals.**

\sqrt{75} can be simplified; the largest perfect square factor of 75 is 25.

\sqrt{75}=\sqrt{25\times{3}}=\sqrt{25}\times\sqrt{3}=5\times\sqrt{3}=5\sqrt{3}

\sqrt{50} can be simplified; the largest perfect square factor of 50 is 25.

\sqrt{50}=\sqrt{25\times{2}}=\sqrt{25}\times\sqrt{2}=5\times\sqrt{2}=5\sqrt{2}

**Subtract the radical terms.**

\sqrt{75}-\sqrt{50} can be rewritten as 5\sqrt{3}-5\sqrt{2}.

After simplifying the radical expressions, they do not have like radicals. So, leave the expression in its simplified form.

\sqrt{75}-\sqrt{50}=5\sqrt{3}-5\sqrt{2}

- Remind students that there is always a 1 in front of the radical symbol.

- Encourage students to use previously learned concepts such as gathering like terms and integer rules.

- Have students practice skills by using digital platforms such as Khan Academy which also have embedded tutorials.

**Thinking that a root with no integer coefficient has a \bf{0} in front of it**

\sqrt{10}β {0\sqrt{10}}

\sqrt{10}=1\sqrt{10}

**Forgetting to simplify each radical fully**

For example, when subtracting 5 \sqrt{8}-4 \sqrt{32} simplifying the radical expression to be 5 \times \sqrt{4 \times 2}-4 \times \sqrt{4 \times 8}=5 \times 2 \sqrt{2}-4 \times 2 \sqrt{8}=10 \sqrt{2}-8 \sqrt{8} instead of fully simplifying using the largest square factors of 4 \sqrt{32}=4 \times \sqrt{16 \times 2}=4 \times 4 \sqrt{2}=16 \sqrt{2}, which you can then subtract 10 \sqrt{2}-16 \sqrt{2}=- \, 6 \sqrt{2}.

**Trying to combine unlike radicals**

Itβs OK to leave an answer with more than one radical in it if it will not simplify further. For example, 7\sqrt{3}-4\sqrt{5} cannot be subtracted because the radicals are not the same. So, leave the most simplified answer as 7\sqrt{3}-4\sqrt{5} .

1. Simplify the expression 3 \sqrt{11}-2 \sqrt{11}-4 \sqrt{11} .

– \, 3 \sqrt{11}

3 \sqrt{11}

5 \sqrt{11}

– \, 5 \sqrt{11}

The expression 3 \sqrt{11}-2 \sqrt{11}-4 \sqrt{11} has like radicals so you can simply subtract left to right.

3 \sqrt{11}-2 \sqrt{11}-4 \sqrt{11}

(3-2-4=-3) \rightarrow Subtracting the coefficients

Remember that \sqrt{11} has a coefficient of 1.

2. Subtract the radical expression.

3\sqrt{5}-2\sqrt{3}-4\sqrt{3}

\sqrt{15}

3\sqrt{5}-6\sqrt{3}

9\sqrt{11}

3\sqrt{5}-2\sqrt{3}Β

In the expression, 3\sqrt{5}-2\sqrt{3}-4\sqrt{3}, there are two radicals that match and one that does not.

3 \sqrt{5} cannot be simplified, so leave it as is.

– \, 2 \sqrt{3}-4 \sqrt{3} have like radicals so they can be subtracted.

– \, 2 \sqrt{3}-4 \sqrt{3}=- \, 6 \sqrt{3}

*Remember to keep the negative sign with the coefficient 2. *

So the simplified answer is: 3\sqrt{5}-6\sqrt{3}

3. Subtract the radical expression.

\sqrt{11}-\sqrt{44}

5\sqrt{11}

\sqrt{55}

11\sqrt{5}

– \, \sqrt{11}Β

The expression \sqrt{11}-\sqrt{44} does not have like radicals but \sqrt{44} can be simplified.

\begin{aligned}\sqrt{44}&=\sqrt{4\times{11}} \\\\ &=\sqrt{4}\times\sqrt{11} \\\\ &=2\times\sqrt{11} \\\\ &=2\sqrt{11} \end{aligned}

So, now the radical expression can be subtracted because the radicals are the same.

\sqrt{11}-2 \sqrt{11}=- \, 1 \sqrt{11}=- \, \sqrt{11}

4. Simplify the radical expression.

\sqrt{54}-\sqrt{24}+\sqrt{6}

2\sqrt{6}

– \, 2 \sqrt{6}

\sqrt{6}

6

The expression cannot be simplified the way it is written because the radicals are not the same. But, two of them can be simplified.

\sqrt{54} can be simplified:

\begin{aligned}\sqrt{54}&=\sqrt{9\times{6}} \\\\ &=\sqrt{9}\times\sqrt{6} \\\\ &=3\times\sqrt{6} \\\\ &=3\sqrt{6} \end{aligned}

\sqrt{24} can be simplified:

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

So the expression \sqrt{54}-\sqrt{24}+\sqrt{6}=3 \sqrt{6}-2 \sqrt{6}+\sqrt{6}=2 \sqrt{6}

5. Subtract the radical expression.

3\sqrt{20}-\sqrt{50}

6\sqrt{10}\sqrt{5}

6\sqrt{5}-5\sqrt{2}

1 \sqrt{5}-1 \sqrt{6}

3\sqrt{1000}

The expression, 3\sqrt{20}-\sqrt{50} does not have like radicals, but they can be simplified.

3\sqrt{20} simplifies to be:

\begin{aligned}3\sqrt{20}&=3\times\sqrt{4\times{5}} \\\\ &=3\times\sqrt{4}\times\sqrt{5} \\\\ &=3\times{2}\times\sqrt{5} \\\\ &=6\times\sqrt{5} \\\\ &=6\sqrt{5} \end{aligned}

\sqrt{50} simplifies to be:

\begin{aligned}\sqrt{50}&=\sqrt{25\times{2}} \\\\ &=\sqrt{25}\times\sqrt{2} \\\\ &=5\sqrt{2} \end{aligned}

So, 3\sqrt{20}-\sqrt{50}=6\sqrt{5}-5\sqrt{2} which still does not have like radicals.

The simplified answer is, 6\sqrt{5}-5\sqrt{2}

6. Simplify the radical expression.

3\sqrt{16}-\sqrt{50}+\sqrt{8}

12+7 \sqrt{2}

12-3 \sqrt{2}

9 \sqrt{2}

12+3\sqrt{2}

The expression 3\sqrt{16}-\sqrt{50}+\sqrt{8} does not have like radicals, but they can be simplified.

3 \sqrt{16} simplifies to be:

\begin{aligned}3\sqrt{16}&=3\times{4} \\\\ &=12 \end{aligned}

\sqrt{50} simplifies to be:

\begin{aligned}\sqrt{50}&=\sqrt{25\times{2}} \\\\ &=\sqrt{25}\times\sqrt{2} \\\\ &=5\sqrt{2} \end{aligned}

\sqrt{8} simplifies to be:

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

So, 3\sqrt{16}-\sqrt{50}+\sqrt{8}=12-5\sqrt{2}+2\sqrt{2}

12-5\sqrt{2}+2\sqrt{2} now has like radicals.

12-5 \sqrt{2}+2 \sqrt{2}=12-3 \sqrt{2}

12-3 \sqrt{2} is the most simplified answer. 12 cannot be combined with 3 \sqrt{2} because 12 is an integer, not a radical.

Yes, adding radical expressions follow the same rules as subtracting radical expressions. Multiplying radical expressions and dividing radical expressions are the operations you learn soon after adding and subtracting.

Visit our web page on multiplying radicals and dividing radicals to learn more about the process.

Yes, you can subtract cube root expressions because they are considered radical expressions.

The process is different. To solve simple radical equations, you need to isolate the radical part to be on one side of the equation. Then apply the correct exponent to both sides of the equation to get rid of the radical sign.

For example, if the root is a cube root, then the exponent or power each side of the equation needs to be raised to is 3.

A rational expression can have a radical expression in the numerator or the denominator, although the value is irrational.

- Rational functions
- Vectors

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