Rationalize The Denominator

When dealing with radical expressions, we apply the technique of rationalisation. Suppose we can ‘rationalise’ the denominator to convert the denominator into a rational number. For this, we require the identities comprising square roots.

In this article, you will learn how to simplify the rational expressions containing two and three terms using the rationalisation technique.

How to Rationalize The Denominator

When the denominator of an expression contains a term with a square root (or a number under a radical sign), the process of converting it to an equivalent expression whose denominator is a rational number is called rationalising the denominator. This can be understood in a better way from the example given below:

Example 1: Rationalise the denominator of 1/√3.

Solution:

Given radical expression is 1/√3.

Now, we have to write 1/√3 as an equivalent expression in which the denominator is a rational number.

As we know, √3 is irrational and the product √3.√3 is a rational number.

Thus, by multiplying 1/√3 by √3/√3 we can get the required equivalent expression.

So,

\(\frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3}\)

Note: There can be radical terms in numerator even after rationalizing the denominator.

Also, get rationalize the denominator calculator here.

Rationalize The Denominator and Simplify

The below identities can be applied when dealing with a binomial denominator involving Radicals. Using these identities we can simplify the terms in both numerator and denominator.

Expression

Conjugate

Product

√a + b

√a – b

(√a + b)(√a – b) = a – b2

√a – b

√a + b

(√a – b)(√a + b) = a – b2

a + √b

a – √b

(a + √b)(a – √b) = a2 – b

a – √b

a + √b

(a – √b)(a + √b) = a2 – b

√a + √b

√a – √b

(√a + √b)(√a – √b) = a – b

√a – √b

√a + √b

(√a – √b)(√a + √b) = a – b

How to Rationalize The Denominator with Two Terms

Below are the steps to perform rationalisation on denominators containing two terms.

Step 1: Multiply both the numerator and the denominator by the denominator’s conjugate.

Step 2: Distribute or use the FOIL technique for both the numerator and the denominator.

Step 3: We can multiply numbers inside the radical with numbers inside the radical and numbers outside the radical with numbers outside the radical. Then, combine like terms.

Step 4: Now simplify the radical terms and combine the obtained like terms.

Step 5: Finally, convert the fraction to a simplified form if possible.

Step 6: We may need to reduce each number outside the radical by the same number for the above step’s result.

Step 7: If we cannot reduce each number outside the radical by the same number, then we cannot reduce the fraction.

Let’s understand this with the help of the below example.

Example 2:

Rationalize the denominator of \(\frac{2}{\sqrt{8}-\sqrt{7}}\).

Solution:

Given expression is: \(\frac{2}{\sqrt{8}-\sqrt{7}}\)

As we know, the conjugate of √8 – √7 is √8 + √7 {since the conjugate of √a – √b is √a + √b}

Now, multiplying the numerator and denominator with √8 + √7, we get;

\(=\frac{2}{\sqrt{8}-\sqrt{7}}\times\frac{\sqrt{8}+\sqrt{7}}{\sqrt{8}+\sqrt{7}}\)

By applying FOIL technique and suitable identities,

\(=\frac{2(\sqrt{8}+\sqrt{7})}{(\sqrt{8})^2-(\sqrt{7})^2}\)

On simplification,

\(=\frac{2(\sqrt{8}+\sqrt{7})}{(8-7)}\\=\frac{2(\sqrt{8}+\sqrt{7})}{1}\\=2(\sqrt{8}+\sqrt{7})\)

Thus, the denominator obtained here is a rational number.

How to Rationalize The Denominator with Three Terms

Let’s understand how to rationalize the denominator with three terms from the example given below:

Example 3:

Rationalize the denominator: \(\frac{1}{\sqrt{7}+\sqrt{5}-\sqrt{2}}\)

Solution:

Given expression is: \(\frac{1}{\sqrt{7}+\sqrt{5}-\sqrt{2}}\)

Here, the denominator contains three terms.

Let’s write two radical terms of the denominator in one parenthesis and the remaining term with its corresponding sign such as:

√7 + (√5 – √2)

The conjugate of √7 + (√5 – √2) is √7 – (√5 – √2).

Multiply the numerator and denominator with √7 – (√5 – √2).

\(=\frac{1}{\sqrt{7}+(\sqrt{5}-\sqrt{2})}\times \frac{\sqrt{7}-(\sqrt{5})-\sqrt{2})}{\sqrt{7}-(\sqrt{5})-\sqrt{2})}\)

Using the identity (a – b)(a + b) = a2 – b2,

\(=\frac{\sqrt{7}-(\sqrt{5}-\sqrt{2})}{(\sqrt{7})^2-(\sqrt{5}-\sqrt{2})^2}\\=\frac{\sqrt{7}-(\sqrt{5}-\sqrt{2})}{7 – 5 – 2 + 2\sqrt{10}}\\=\frac{\sqrt{7}-(\sqrt{5}-\sqrt{2})}{2\sqrt{10}}\)

Again, rationalizing the denominator,

\(=\frac{\sqrt{7}-(\sqrt{5}-\sqrt{2})}{2\sqrt{10}}\times \frac{\sqrt{10}}{\sqrt{10}}\\=\frac{\sqrt{70}-\sqrt{50}+\sqrt{20}}{20}\)

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