Rationalization of Denominator

In this section we will discuss rationalization of denominator.

Sometimes we come across expressions containing square roots in their denominators. In such an expression if the denominator is free from square roots then it will be easier to add/subtract/multiply or divide. To make the denominators free from square roots, we multiply the numerator and denominator by an irrational number. Such a number is called rationalization factor.

Example : (1 - √3 ) / √2
As there is √2 in the denominator and we know that √2 x √2 = 2
So, multiply top and bottom by √2 .
[(1-√3) x √2] / (√2 x √2)
= [1√2 - √2 x √3] / 2 [ use distributive property]
=[ √2 - √6] /2

Note : Rationalization factor for : 1) 1/√a ------> √a 2) a + √b -------> a - √b 3) a - √b ---------> a + √b 4) √a + √b ----------> √a - √b 5) √a - √b ------------> √a + √b

Examples on rationalization of denominator

1) Rationalise the denominator of 2/√3
Solution :
We know that the rationalization factor for 1/√a is √a .
∴ 2/√3 = (2 x √3)/ (√3 x √3)
= 2√3/3
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2) Rationalise the denominator of 1/(3 - √2)
Solution :
We have,
1/(3 - √2) = 1(3 + √2) /(3 - √2)(3 + √2)
= (3 + √2)/( 9 – 2) [ use the identity of (a+b)(a-b) = a² - b²]
= (3 + √2 )/ 7
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3) Solve : 3/(√3 + 1) + 5/(√3 – 1)
Solution :
3/(√3 + 1) + 5/(√3 – 1)
Rationalize the each term and then solve.
3/(√3 + 1) = 3 (√3 -1)/( √3 +1)( √3 -1)
=(3√3 – 3)/(3-1) = (3√3 – 3)/2 --------> (1)
5/(√3 – 1) = 5(√3 + 1)/( √3 -1)( √3 +1)
= (5√3 + 5)/(3-1) = (5√3 + 5)/2 --------> (2)
Add equation (1) and (2) we get,
(3√3 – 3)/ 2 + (5√3 + 5 )/2
= ( 3√3 – 3 + 5√3 + 5)/2
= (8√3 + 2)/2
= 2(4√3 + 1)/2
= 4√3 + 1
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4) √7(√35 - √7) = a + b√5 , find the value of a and b.
Solution :
√7(√35 - √7) = a + b√5
√7(√(7 x5) - √7) = a + b√5
√7(√7 x √5 - √7) = a + b√5
√7 x √7 x √5 - √7 x√7 = a + b√5 [ use a distributive law]
7√5 – 7 = a + b√5
-7 + 7√5 = a +b√5
∴ a = - 7 and b = 7.

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Real-Numbers

• Real Numbers
• Representation of real-numbers on number line
• Operations on Real Numbers
• Rationalization of denominator

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