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Addition and subtraction Quadratic equations Equivalent fractions Rational numbers Irrational numbersSimplifying square roots

Adding and subtracting square roots

Multiplying and dividing square roots

Here we will learn about rationalizing the denominator of expressions containing square roots and look at how to rationalize the denominator for simple expressions.

Students will first learn about rationalizing the denominator as part of expressions and equations in 8 th grade.

**Rationalizing the denominator** is where we convert the denominator of a fraction from an irrational number to a rational number. This means you eliminate any radical expressions in the denominator (such as square roots).

For example,

\cfrac{8}{\sqrt{2}}=\cfrac{8\times\sqrt{2}}{\sqrt{2}\times\sqrt{2}}=\cfrac{8\sqrt{2}}{2}=4\sqrt{2}A number that can be written as an integer (whole number) or a simple fraction is called a rational number.

For example,

2, 100, - \; 30 are all rational numbers.

So are numbers like \cfrac{1}{2}, \cfrac{3}{4} and \cfrac{1}{9}.

Rational numbers can also be terminating decimals like 0.5 or recurring decimals like 0.111β¦

Any number that canβt be written in this form is called **irrational –** in decimal form, these are infinite, with no recurring or repeating pattern.

If a number is not a perfect square (meaning its square root is not an integer), then its square root is likely to be an irrational number. This can also be known as a surd.

All division equations can be written as fractions.

For example,

4 \div 2 can be written as \cfrac{4}{2}.

In a similar way:

4 \div \sqrt{2}=\cfrac{4}{\sqrt{2}}However, itβs much easier in algebra to divide by an integer where possible, so it is useful to be able to convert fractions with irrational denominators (square roots on the bottom) to fractions with rational denominators.

We do this by using the ideas associated with equivalent fractions:

If the numerator and denominator are both **multiplied by the same number or expression**, the fraction remains **equivalent** to the original.

The process of changing the **denominator of a fraction to a rational number** in this way is called **rationalizing** the denominator.

On this page, we will look at cases where the denominator is a single square root.

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

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DOWNLOAD FREEHow does this relate to 8 th grade math?

**Grade 8 – Equations and Expressions (8.EEA.2)**Use square root and cube root symbols to represent solutions to equations of the form x^2=p and x^3=p, where p is a positive rational number. Evaluate square roots of small perfect squares and cube roots of small perfect cubes. Know that \sqrt{}2 is irrational.

In order to rationalize the denominator:

**Simplify any square roots, if necessary.****Multiply both the numerator and the denominator by the square root in the denominator.****Simplify the answer fully.**

Rationalize the denominator:

\cfrac{4}{\sqrt{2}}**Simplify any square roots, if necessary.**

In this example, this is already done.

2**Multiply both the numerator and the denominator by the square root in the denominator.**

So here we multiply the top and the bottom of the fraction by the square root of 2.

\cfrac{4 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}}Numerator:

4 \times \sqrt{2}=4 \sqrt{2}Denominator:

\sqrt{2} \times \sqrt{2}=2So the full expression becomes:

\cfrac{4 \sqrt{2}}{2}The denominator is now rationalized, because 2 is a rational number.

3**Simplify the answer fully.**

So the final answer simplifies to 2 \sqrt{2}.

Rationalize the denominator:

\cfrac{6}{\sqrt{45}}**Simplify any square roots, if necessary.**

The square root of 45 will simplify:

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

**Multiply both the numerator and the denominator by the square root in the denominator.**

So here we multiply the top and the bottom of the fraction by the square root of 5.

\cfrac{6 \times \sqrt{5}}{3 \sqrt{5} \times \sqrt{5}}

Numerator:

6\times\sqrt{5}=6\sqrt{5}

Denominator:

3 \sqrt{5} \times \sqrt{5}=3 \times 5=15

So the full expression becomes:

\cfrac{6 \sqrt{5}}{15}

The denominator is now rationalized, because 15 is a rational number.

**Simplify the answer fully.**

6 \div 15=\cfrac{6}{15}=\cfrac{2}{5}

So the final answer simplifies to \cfrac{2 \sqrt{5}}{5}.

Rationalize the denominator:

\cfrac{4\sqrt{3}}{\sqrt{7}}**Simplify any square roots, if necessary.**

Both square roots are already in their simplest forms.

**Multiply both the numerator and the denominator by the square root in the denominator.**

So here we multiply the top and the bottom of the fraction by the square root of 7.

\cfrac{4 \sqrt{3} \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}}

Numerator:

4 \sqrt{3} \times \sqrt{7}=4 \sqrt{21}

Denominator:

\sqrt{7} \times \sqrt{7}=7

So the full expression becomes:

\cfrac{4 \sqrt{21}}{7}

The denominator is now rationalized, because 7 is a rational number.

**Simplify the answer fully.**

\cfrac{4 \sqrt{21}}{7} \, is already in its simplest form; there are no common factors of the integers, and there is no square factor of 21 to simplify the square root.

Rationalize the denominator:

\cfrac{6\sqrt{8}}{\sqrt{3}}**Simplify any square roots, if necessary.**

The square root of 8 will simplify:

\begin{aligned}6 \sqrt{8} &=6 \times(\sqrt{4 \times 2}) \\\\
&=6 \times(\sqrt{4} \times \sqrt{2}) \\\\
&=6\times(2 \sqrt{2}) \\\\
&=12 \sqrt{2}\end{aligned}

**Multiply both the numerator and the denominator by the square root in the denominator.**

So here we multiply the top and the bottom of the fraction by the square root of 3.

\cfrac{12 \sqrt{2} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}

Numerator:

12 \sqrt{2} \times \sqrt{3}=12 \sqrt{6}

Denominator:

\sqrt{3} \times \sqrt{3}=3

So the full expression becomes:

\cfrac{12 \sqrt{6}}{3}

The denominator is now rationalized, because 3 is a rational number.

**Simplify the answer fully.**

Find the quotient of the whole numbers: 12 \div 3=4

So the final answer simplifies to 4 \sqrt{6}.

Rationalize the denominator:

\cfrac{3 \sqrt{12}+\sqrt{108}}{\sqrt{6}}**Simplify any square roots, if necessary.**

The square root of 12 will simplify:

\begin{aligned}3 \sqrt{12} &=3 \times(\sqrt{4 \times 3}) \\\\ &=3 \times(\sqrt{4} \times \sqrt{3}) \\\\ &=3 \times(2 \sqrt{3}) \\\\ &=6 \sqrt{3}\end{aligned}

The square root of 108 will also simplify:

\begin{aligned}\sqrt{108} &=\sqrt{36 \times 3} \\\\ &=\sqrt{36} \times \sqrt{3} \\\\ &=6 \sqrt{3}\end{aligned}

So the numerator is 6 \sqrt{3}+6 \sqrt{3} and because these are like square roots, it will simplify to 12 \sqrt{3}.

**Multiply both the numerator and the denominator by the square root in the denominator.**

So here we multiply the top and the bottom of the fraction by the square root of 6.

\cfrac{12 \sqrt{3} \times \sqrt{6}}{\sqrt{6} \times \sqrt{6}}

Numerator:

12 \sqrt{3} \times \sqrt{6}=12 \sqrt{18}

Denominator:

\sqrt{6} \times \sqrt{6}=6

So the full expression becomes:

\cfrac{12 \sqrt{18}}{6}

The denominator is now rationalized because 6 is a rational number.

**Simplify the answer fully.**

The square root of 18 simplifies further:

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

So the expression becomes:

\cfrac{12 \times 3 \sqrt{2}}{6}=\cfrac{36 \sqrt{2}}{6}

Finally, 36 \div 6=6, so the fully simplified answer is 6\sqrt{2}.

- Before introducing rationalizing the denominator, ensure that students have a solid understanding of square roots. Review how to simplify square roots and basic operations involving radicals.

- Once students are comfortable with rationalizing single term denominators, provide them with practice problems where the denominator is a binomial.

- Incorporate interactive activities, such as online simulations, calculators, or other manipulative tools, that allow students to experiment with radicals and understand the impact of rationalizing the denominator.

**Multiplying by the incorrect square root, if there are square roots in both numerator and denominator**

Always make sure you multiply both top and bottom of the fraction by the square root in the denominator of the fraction.

**If dividing integers to simplify, not checking that the factor you want to divide by is common to all of the integers in the numerator, as well as the denominator**

For example, \frac{9+6 \sqrt{2}}{3}=3+2 \sqrt{2}

This can be simplified, because 3 is a factor of 3, 9 and 6. However, \frac{4+6 \sqrt{2}}{3} cannot be simplified further, because 3 isnβt a factor of 4.

1. Rationalize the denominator:

\cfrac{1}{\sqrt{5}}

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

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

\cfrac{1}{5}

\sqrt{5}

Multiply the numerator and denominator by the square root of 5.

\cfrac{1 \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}}Β

Numerator:

1 \times \sqrt{5}=\sqrt{5}

Denominator:

\sqrt{5} \times \sqrt{5}=5

So the full expression becomes:

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

The denominator is now rationalized because 5 is a rational number.

2. Rationalize the denominator:

\cfrac{7}{\sqrt{3}}

\cfrac{\sqrt{3}}{7}

\cfrac{7}{3}

\sqrt{21}

\cfrac{7 \sqrt{3}}{3}

Multiply the numerator and denominator by the square root of 3.

\cfrac{7 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}

Numerator:

7 \times \sqrt{3}=7 \sqrt{3}

Denominator:

\sqrt{3} \times \sqrt{3}=3

So the full expression becomes:

\cfrac{7 \sqrt{3}}{3}

The denominator is now rationalized because 3 is a rational number.

3. Rationalize the denominator:

\cfrac{20}{\sqrt{40}}

\cfrac{20 \sqrt{40}}{40}

\cfrac{\sqrt{40}}{2}

\sqrt{20}

\sqrt{10}

Simplify the square root of 40 first, then multiply the top and bottom by the square root of 10.

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

So the expression becomes:

\cfrac{20}{2 \sqrt{10}}

Multiply the numerator and denominator by the square root of 10 :

\cfrac{20 \times \sqrt{10}}{2 \sqrt{10} \times \sqrt{10}}

Numerator:

20 \times \sqrt{10}=20 \sqrt{10}

Denominator:

2 \sqrt{10} \times \sqrt{10}=2 \times 10=20

So the full expression becomes:

\cfrac{20 \sqrt{10}}{20}

The denominator is now rationalized because 20 is a rational number.

Finally, 20 \div 20=1, so the fully simplified answer is \sqrt{10}.

Alternatively, you could multiply the top and bottom by the square root of 40, but remember to simplify your answer fully at the end.

4. Rationalize the denominator:

\cfrac{5 \sqrt{3}}{\sqrt{10}}

\cfrac{\sqrt{30}}{2}

\sqrt{15}

\cfrac{5 \sqrt{30}}{10}

\cfrac{\sqrt{30}}{5}

Multiply the top and bottom by the square root of 10.

\cfrac{5 \sqrt{3} \times \sqrt{10}}{\sqrt{10} \times \sqrt{10}}Β

Numerator:

5 \sqrt{3} \times \sqrt{10}=5 \sqrt{30}

Denominator:

\sqrt{10} \times \sqrt{10}=10

So the full expression becomes:

\cfrac{5 \sqrt{30}}{10}

The denominator is now rationalized because 10 is a rational number.

Finally, simplify the integers:

5 \div 10=\cfrac{5}{10}=\cfrac{1}{2}

So the fully simplified answer is \cfrac{\sqrt{30}}{2}.

5. Rationalize the denominator:

\cfrac{2\sqrt{48}}{\sqrt{10}}

2\sqrt{38}

\cfrac{4\sqrt{30}}{5}

\cfrac{8\sqrt{30}}{10}

\cfrac{2\sqrt{30}}{5}

Simplify the square root of 48 first, then multiply the top and bottom by the square root of 10.

\begin{aligned}2 \times \sqrt{48} & =2 \times \sqrt{16 \times 3} \\\\ & =2 \times \sqrt{16} \times \sqrt{3} \\\\ & =2 \times 4 \sqrt{3} \\\\ & =8 \sqrt{3}\end{aligned}

So the expression becomes:

\cfrac{8 \sqrt{3}}{\sqrt{10}}

Multiply the numerator and denominator by the square root of 10 :

\cfrac{8 \sqrt{3} \times \sqrt{10}}{\sqrt{10} \times \sqrt{10}}

Numerator:

8 \sqrt{3} \times \sqrt{10}=8 \sqrt{30}

Denominator:

\sqrt{10} \times \sqrt{10}=10

So the full expression becomes:

\cfrac{8 \sqrt{30}}{10}

The denominator is now rationalized because 10 is a rational number.

Finally, 8 \div 10=\cfrac{8}{10}=\cfrac{4}{5}

So the fully simplified answer is \cfrac{4 \sqrt{30}}{5}.

To rationalize a denominator, multiply both the numerator and the denominator by the square root in the denominator.

If the denominator is a binomial, you rationalize it by multiplying both the numerator and denominator by the conjugate of the denominator.

Rationalizing a denominator is done to simplify expressions and make them easier to work with. It helps avoid having square roots or other radicals in the denominator, making mathematical expressions look cleaner and facilitating calculations.

- Rational functions
- Vectors
- Quadratic equations
- Functions in algebra
- Laws of exponents
- Math formulas

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