Rationalise the Denominator (How to Rationalise with Examples) (2024)

In Mathematics, we rationalise the denominator, when the given fraction contains a radical term or a surd in the denominator. These radical terms include square root and cube root. If the denominator of a mathematical expression with two terms includes radical, then we need to multiply both numerator and denominator by the conjugate of the denominator. This method is called rationalization.

In other words, we can say, rationalizing the denominator means moving the radical term (square root or cube root) to the numerator, such that a denominator is a whole number. When we rationalise the denominator, then it becomes easy to find the sum or difference of given fractions.

For example, 2/√2 is a fraction that has an irrational denominator. If we rationalise it, then it becomes √2. Thus, the denominator is a whole number, i.e. 1. Let us learn in this article how to make the denominator rational with examples.

Learn more:

• Fractions
• Irrational Numbers
• Rational Numbers

How to Rationalise the Denominator?

Here we will learn how to rationalise the irrational denominators that contain a single term and two terms. Also, we will learn to rationalise the denominator with variables.

Denominator with Single Term

For the fractions that include simple irrational denominators like √2, √3, √5, etc., it is easy to rationalise such denominators. We need to multiply numerator and denominator by the same radical term or by the same roots. Thus, we will get the denominator as a whole number.

Example 1: 1/√2

Multiply and divide by √2

⇒ (1/√2) x (√2/√2)

⇒ √2/(√2)²

⇒ √2/2

Example 2: 1/√3

Multiply and divide by √3

⇒ (1/√3) x (√3/√3)

⇒ √3/(√3)²

⇒ √3/3

Example 3: 1/√5

Multiplying and dividing by √5, we get

⇒ (1/√5) x (√5/√5)

⇒ √5/(√5)²

⇒ √5/5

Thus, we have learned here how to rationalise the denominator with single terms including simple roots.

Denominator with Two Terms

If the denominator of a fraction, contains two terms along with a surd, then we need to multiply both numerator and denominator by the conjugate of the denominator.

Conjugate of an expression ‘x + y’ will be ‘x – y’ and vice versa. Thus, we change the sign of the given expression. For example, the conjugate of ‘1 + √5’ is ‘1 – √5’. So, if we multiply both the expressions, we get;

(1 + √5) x (1 – √5)

= 1² – (√5)² [By the algebraic identity, we know, (a – b) (a + b) = a² – b²]

= 1 – 5

= -4

Hence, our denominator gets rationalised in this way.

After rationalization, combine the like terms and simplify to get the equivalent fraction.

Example: Rationalise the denominator for 2/(√3+5)

In the given example, the denominator has one radical and a whole number added to it.

Thus, the conjugate of √3 + 5 is √3 – 5.

Multiplying numerator and denominator by the conjugate of √3 + 5.

\(\begin{array}{l}\Rightarrow \frac{2}{\sqrt{3} + 5} \times \frac{\sqrt{3} – 5}{\sqrt{3} + 5}\end{array} \)

By the formula (a+b) (a-b) = a² – b² , we can write;

\(\begin{array}{l}\Rightarrow \frac{2\times \sqrt{3}-5}{\sqrt{3}^{2}-5^{2}}\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{2\times \sqrt{3}-5}{9-25}\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{2\times \sqrt{3}-5}{-16}\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{ \sqrt{3}-5}{-8}\end{array} \)

Multiplying the numerator and denominator by (-1) we get;

\(\begin{array}{l}\Rightarrow \frac{ 5 – \sqrt{3}}{8}\end{array} \)

This is the simplified form of 2/(√3+5)

Denominator with Variables

Rationalization of the denominator is necessary when the denominator is a radical or contains a term with a square root or a cube root (with a radical sign). Now, when the given irrational denominator is converted into a rational number to get the equivalent expression, then the process is called rationalizing the denominator.

Let us solve an example with variables.

Example: Rationalise the denominator and find the value of x and y. (5 + 4√3)/(4 + 5√3) = x + y √3.

Solution: Given, (5 + 4√3)/(4 + 5√3) = x + y √3

First we will rationalise the denominator at the LHS

\(\begin{array}{l}\frac{5+4 \sqrt{3}}{4+5 \sqrt{3}} \times \frac{4-5 \sqrt{3}}{4-5 \sqrt{3}}\end{array} \)

\(\begin{array}{l}\frac{(5+4 \sqrt{3})(4-5 \sqrt{3})}{(4+5 \sqrt{3})(4-5 \sqrt{3})}\end{array} \)

\(\begin{array}{l}\frac{5(4)-5(5 \sqrt{3})+4(4 \sqrt{3})-(4 \sqrt{3})(5 \sqrt{3})}{4^{2}-(5 \sqrt{3})^{2}}\end{array} \)

\(\begin{array}{l}\frac{20-25 \sqrt{3}+16 \sqrt{3}-20(3)}{16-25(3)}\end{array} \)

\(\begin{array}{l}\frac{20-60-9 \sqrt{3}}{16-75}\end{array} \)

\(\begin{array}{l}\frac{-40-9 \sqrt{3}}{-59}\end{array} \)

\(\begin{array}{l}\frac{40+9 \sqrt{3}}{59}\end{array} \)

We know, RHS = x + y √3

Thus, if we compare the LHS and RHS, we get;

x = 40/59 and y = 9/59

Solved Examples

Let us solve more examples based on the rationalization of the denominator for a fraction.

Q.1: Simplify 1/√252

Solution: Given,

1/√252

Prime factorisation of 252 = 2 x 2 x 3 x 3 x 7

⇒ 1/√( 2 x 2 x 3 x 3 x 7)

Taking the square values out of the root.

⇒ 1/[2 x 3√7]

⇒ 1/6√7

Multiply and divide by √7 to rationalise.

⇒ 1/6√7 x (√7 x √7)

⇒ √7/(6 x 7)

⇒ √7/42

Q.2: Simplify:

\(\begin{array}{l}\frac{3 \sqrt{10}-5 \sqrt{6}}{4 \sqrt{10}+2 \sqrt{6}}\end{array} \)

Solution: Given,

\(\begin{array}{l}\frac{3 \sqrt{10}-5 \sqrt{6}}{4 \sqrt{10}+2 \sqrt{6}}\end{array} \)

Multiplying numerator and denominator by the conjugate of denominator.

\(\begin{array}{l}\Rightarrow \frac{3 \sqrt{10}-5 \sqrt{6}}{4 \sqrt{10}+2 \sqrt{6}} \cdot \frac{4 \sqrt{10}-2 \sqrt{6}}{4 \sqrt{10}-2 \sqrt{6}}\end{array} \)

Multiply the terms

\(\begin{array}{l}\Rightarrow \frac{12 \sqrt{100}-6 \sqrt{60}-20 \sqrt{60}+10 \sqrt{36}}{16 \sqrt{100}-8 \sqrt{60}+8 \sqrt{60}-4 \sqrt{36}}\end{array} \)

Simplify the like terms

\(\begin{array}{l}\Rightarrow \frac{12(10)-26(2 \sqrt{15})+10(6)}{16(10)-4(6)}\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{120-52 \sqrt{15}+60}{160-24}\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{180-52 \sqrt{15}}{136}\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{45-13 \sqrt{15}}{34}\end{array} \)

Practice Questions

1. Simplify by rationalizing the denominator of the following:

• 1/(√3 + 4)
• 2/√17
• 4/√6
• (√3+1)/(√2 + 1)
• √8/(2√6 – 3√2)

2. Rationalise the denominator of the given expression: (1+2√3)/(2-√3) = a+b√3 and find the value of a and b.

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