In this article, we will discuss how to rationalize denominators with radicals.

What is Rationalization?

"The process of removing radicals or imaginary numbers from the denominator in such a way that the number in the denominator is converted to a rational number only is known as rationalization"

Removing the radicals from the denominator of the fraction allows us to add, subtract, multiply and divide the fractions in a simple way.

There are three cases when we have a radical in the denominator. We follow different rationalization procedures to remove the radicals in all these three cases. These three scenarios are:

  • When we have a fraction of the type \frac {n} {a \sqrt {m}}
  • When we have a fraction of the type \frac {a} {b \sqrt [n] {c^m}}
  • When we have the fraction \frac {p} {\sqrt{q} + \sqrt {r}}

In the next section, we will solve examples in which we will rationalize the fraction belonging to the above three categories.

 

Example 1

Solve \frac {5} {3 \sqrt {8}}.

Solution

To rationalize the above fraction, we need to multiply and divide the fraction by the root in the denominator. In the above example, the root in the denominator is \sqrt {8}. Hence, we will multiply and divide \frac {5} {3 \sqrt {8}} by \sqrt {8} like this:

= \frac {5} {3 \sqrt {8}}

= \frac {5} {3 \sqrt {8}} \cdot \frac {\sqrt {8}} {\sqrt{8}}

= \frac {5 \sqrt {8}} {3 \cdot (\sqrt {8})^2}

= \frac {5 \sqrt {8}} {3 \cdot 8}

= \frac {5 \sqrt {8}} {24}

\sqrt {8} is equal to 2 \sqrt {2}. Hence, we can simplify the above expression further like this:

= \frac {5 \sqrt 2{2}} {24}

= \frac {5 \sqrt {2}} {12}

 

 

Example 2

Solve \frac {9} {7 \sqrt {5}}.

Solution

To rationalize the above fraction, we need to multiply and divide the fraction by the root in the denominator. In the above example, the root in the denominator is \sqrt {5}. Hence, we will multiply and divide \frac {9} {7 \sqrt {5}} by \sqrt {5} like this:

= \frac {9} {7 \sqrt {5}}

= \frac {9} {7 \sqrt {5}} \cdot \frac {\sqrt {5}} {\sqrt{5}}

= \frac {9 \sqrt {5}} {7 \cdot (\sqrt {5})^2}

= \frac {9 \sqrt {5}} {7 \cdot 5}

= \frac {9 \sqrt {5}} {35}

Example 3

Rationalize the following fraction:

\frac {3} {4 \sqrt [5] {3}}

Solution

When we get the fraction of the type \frac {a} {b \sqrt [n] {c^m}}, then to rationalize, we multiply the numerator and denominator by \sqrt [n]  {c^ {n - m}}. In the above example, we have to rationalize the fraction \frac {3} {4 \sqrt [5] {3}}, hence we will multiply the numerator and denominator by \sqrt [5]  {3^ {5 - 1}} like  this:

= \frac {3} {4 \sqrt [5] {3}} \cdot \frac {\sqrt [5]  {3^ {5 - 3}}} {\sqrt [5]  {3^ {5 - 1}}}

= \frac {3} {4 \sqrt [5] {3}} \cdot \frac {\sqrt [5]  {3^ {4}}} {\sqrt [5]  {3^ {4}}}

= \frac {3 \cdot \sqrt [5] {3^4}} {4 \cdot \sqrt [5] {3^5}}

= \frac {3 \cdot \sqrt [5] {3^4}} {4 \cdot \sqrt [5] {3^5}}

= \frac {\sqrt [5] {81}} {4}

Example 4

Rationalize the following fraction:

\frac {8} {6 \sqrt [4] {2}}

Solution

When we get the fraction of the type \frac {a} {b \sqrt [n] {c^m}}, then to rationalize, we multiply the numerator and denominator by \sqrt [n]  {c^ {n - m}}. In the above example, we have to rationalize the fraction \frac {8} {6 \sqrt [4] {2}}, hence we will multiply the numerator and denominator by \sqrt [4]  {2^ {4- 1}} like  this:

= \frac {8} {6 \sqrt [4] {2}} \cdot \frac {\sqrt [4]  {2^ {4 - 1}}} {\sqrt [4]  {2^ {4 - 1}}}

= \frac {8} {6 \sqrt [4] {2}} \cdot \frac {\sqrt [4]  {2^ {3}}} {\sqrt [4]  {2^ {3}}}

= \frac {8 \cdot \sqrt [4] {2^3}} {6 \cdot \sqrt [4] {2^4}}

= \frac {8 \cdot \sqrt [4] {2^3}} {6 \cdot 2}

= \frac {8 \cdot \sqrt [4] {8}} {12}

= \frac {2 \cdot \sqrt [4] {8}} {3}

 

Example 5

Rationalize the following fraction:

\frac {9} {3 \sqrt [3] {5}}

Solution

When we get the fraction of the type \frac {a} {b \sqrt [n] {c^m}}, then to rationalize, we multiply the numerator and denominator by \sqrt [n]  {c^ {n - m}}. In the above example, we have to rationalize the fraction \frac {9} {3 \sqrt [3] {5}}, hence we will multiply the numerator and denominator by \sqrt [3]  {5^ {3- 1}} like  this:

= \frac {9} {3 \sqrt [3] {5}} \cdot \frac {\sqrt [3]  {5^ {3 - 1}}} {\sqrt [3]  {5^ {3 - 1}}}

= \frac {9} {3 \sqrt [3] {5}} \cdot \frac {\sqrt [3]  {5^ {2}}} {\sqrt [3]  {5^ {2}}}

= \frac {9 \cdot \sqrt [3] {5^2}} {3 \cdot \sqrt [3] {5^3}}

= \frac {9 \cdot \sqrt [3] {25}} {3 \cdot 5}

= \frac {9 \cdot \sqrt [3] {25}} {15}

= \frac {3 \cdot \sqrt [3] {25}} {5}

 

Example 6

Rationalize the following fraction:

\frac {4} {7 \sqrt [5] {2^4}}

Solution

When we get the fraction of the type \frac {a} {b \sqrt [n] {c^m}}, then to rationalize, we multiply the numerator and denominator by \sqrt [n]  {c^ {n - m}}. In the above example, we have to rationalize the fraction \frac {4} {7 \sqrt [5] {2^4}}, hence we will multiply the numerator and denominator by \sqrt [3]  {2^ {5- 4}} like  this:

= \frac {4} {7 \sqrt [5] {2^4}} \cdot \frac {\sqrt [5]  {2^ {5 - 4}}} {\sqrt [5]  {2^ {5 - 4}}}

= \frac {4} {7 \sqrt [5] {32}} \cdot \frac {\sqrt [5]  {2^ {1}}} {\sqrt [5]  {2^ {1}}}

= \frac {4 \cdot \sqrt [5] {2}} {7 \cdot \sqrt [5] {2^5}}

= \frac {4 \cdot \sqrt [5] {2}} {7 \cdot 2}

= \frac {4 \cdot \sqrt [5] {2}} {14}

= \frac {2 \cdot \sqrt [5] {2}} {7}

 

Example 7

Rationalize the following fraction:

\frac {3} {11 \sqrt [3] {6}}

Solution

When we get the fraction of the type \frac {a} {b \sqrt [n] {c^m}}, then to rationalize, we multiply the numerator and denominator by \sqrt [n]  {c^ {n - m}}. In the above example, we have to rationalize the fraction \frac {3} {11 \sqrt [3] {6}}, hence we will multiply the numerator and denominator by \sqrt [3]  {6^ {3- 1}} like  this:

= \frac {3} {11 \sqrt [3] {6}} \cdot \frac {\sqrt [3]  {6^ {3 - 1}}} {\sqrt [3]  {6^ {3 - 1}}}

= \frac {3} {11 \sqrt [3] {6}} \cdot \frac {\sqrt [3]  {6^ {2}}} {\sqrt [3]  {6^ {2}}}

= \frac {3 \cdot \sqrt [3] {6^2}} {11 \cdot \sqrt [3] {6^3}}

= \frac {3 \cdot \sqrt [3] {36}} {11 \cdot 6}

= \frac {3 \cdot \sqrt [3] {36}} {66}

= \frac {\sqrt [3] {36}} {22}

 

Example 8

Rationalize the following fraction:

\frac {5} {\sqrt {7} + \sqrt {9}}

Solution

When we have the fraction of the type \frac {a} {\sqrt{b} + \sqrt {c}}, then we rationalize the denominator by multiplying and dividing the numerator by \sqrt {b} - \sqrt {c}.

= \frac {5} {\sqrt {7} + \sqrt {9}} \cdot \frac {\sqrt {7} - \sqrt {9}} {\sqrt {7} - \sqrt {9}}

= \frac {5 (\sqrt {7} - 3)} {(\sqrt {7} - \sqrt {9})^2}

= \frac {5 (\sqrt {7} - 3)} {7 - 9}

= \frac {5 (\sqrt {7} - 3)} {-2}

= -\frac {5 (\sqrt {7} - 3)} {2}

 

Example 9

Rationalize the following fraction:

\frac {8} {\sqrt {2} + \sqrt {5}}

Solution

When we have the fraction of the type \frac {a} {\sqrt{b} + \sqrt {c}}, then we rationalize the denominator by multiplying and dividing the numerator by \sqrt {b} - \sqrt {c}.

= \frac {8} {\sqrt {2} + \sqrt {5}} \cdot \frac {\sqrt {2} - \sqrt {5}} {\sqrt {2} - \sqrt {5}}

= \frac {8 (\sqrt {2} - \sqrt {5})} {(\sqrt {2} - \sqrt {5})^2}

= \frac {8 (\sqrt {2} -  \sqrt {5})} {2 - 5}

= \frac {8 (\sqrt {2} - \sqrt {5})} {-3}

= -\frac {8 (\sqrt {7} -  \sqrt {5})} {3}

 

Example 10

Rationalize the following fraction:

\frac {2} {\sqrt {12} - \sqrt {8}}

Solution

When we have the fraction of the type \frac {a} {\sqrt{b} - \sqrt {c}}, then we rationalize the denominator by multiplying and dividing the numerator by \sqrt {b} + \sqrt {c}.

= \frac {2} {\sqrt {12} - \sqrt {8}} \cdot \frac {\sqrt {12} + \sqrt {8}} {\sqrt {12} + \sqrt {8}}

= \frac {2 (\sqrt {12} + \sqrt {8})} {(\sqrt {12} - \sqrt {8})^2}

= \frac {2 (\sqrt {12} +  \sqrt {8})} {12 - 8}

= \frac {2 (\sqrt {12} + \sqrt {8})} {4}

= \sqrt {12} +  \sqrt {8}

 

Example 11

Rationalize the following fraction:

\frac {8} {\sqrt {15} - \sqrt {6}}

Solution

When we have the fraction of the type \frac {a} {\sqrt{b} - \sqrt {c}}, then we rationalize the denominator by multiplying and dividing the numerator by \sqrt {b} + \sqrt {c}.

= \frac {8} {\sqrt {15} - \sqrt {6}} \cdot \frac {\sqrt {15} + \sqrt {6}} {\sqrt {15} + \sqrt {6}}

= \frac {8 (\sqrt {15} + \sqrt {6})} {(\sqrt {15} + \sqrt {6})^2}

= \frac {8 (\sqrt {15} +  \sqrt {6})} {15 - 6}

= \frac {8 (\sqrt {15} + \sqrt {6})} {9}

 

Example 12

Rationalize the following fraction:

\frac {6} {7 \sqrt [4] {3}}

Solution

When we get the fraction of the type \frac {a} {b \sqrt [n] {c^m}}, then to rationalize, we multiply the numerator and denominator by \sqrt [n]  {c^ {n - m}}. In the above example, we have to rationalize the fraction \frac {6} {7 \sqrt [4] {3}}, hence we will multiply the numerator and denominator by \sqrt [4]  {3^ {4- 1}} like  this:

= \frac {6} {7 \sqrt [4] {3}} \cdot \frac {\sqrt [4]  {3 ^ {4 - 1}}} {\sqrt [4]  {4^ {4 - 1}}}

= \frac {6} {7 \sqrt [4] {3}} \cdot \frac {\sqrt [4]  {3^ {3}}} {\sqrt [4]  {4^ {3}}}

= \frac {6 \cdot \sqrt [4] {3^3}} {7 \cdot \sqrt [4] {3 ^4}}

= \frac {6 \cdot \sqrt [4] {27}} {7 \cdot 3}

= \frac {6 \cdot \sqrt [4] {27}} {21}

= \frac {2 \cdot \sqrt [4] {27}} {7}

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Rafia Shabbir