"Question 5 Rationalise the denominators of the following:
(i) 1/7^(1/2)
(ii) 1/(7^(1/2) - 6^(1/2))
(iii) 1/(5^(1/2) + 2^(1/2))
(iv) 1/(7^(1/2) - 2)
Class 9 - Math - Number Systems Page 24"
Answers
If the denominator of an expression contains a term with the square root ( or a number under a radical sign ) , then the process of converting it to an equivalent expression whose denominator is a rational number is called rationalising the denominator..
For rationalising the denominator we will multiply the numerator and denominator by conjugate of denominator to remove the radical sign from the denominator..
Conjugate of (√a+b) is (√a-b) & conjugate of (√a-√b) is (√a+√b).
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Solution:
I) 1/√7
On multiplying both numerator and denominator by √7
1×√7 / √7×√7
√7 / 7
ii) 1 /(√7-√6)
On multiplying the numerator and denominator by (√7+√6)
1× (√7+√6) / (√7-√6)× (√7+√6)
= (√7+√6) / ((√7)²- (√6)²)
[ (a+b)(a-b) = a² - b²]
= (√7+√6) / 7 - 6
= (√7+√6) / 1 = (√7+ √6)
= (√7+√6)
iii) 1 / (√5+√2)
On multiplying both numerator and denominator by (√5 - √2).
1× (√5-√2) / (√5+√2) (√5-√2)
= (√5- √2) / ((√5)²+(√2)²)
[ (a+b)(a-b) = a² - b²]
=(√5 - √2) / 5- 2 = (√5- √2) / 3
=(√5- √2) / 3
iv) 1/ (√7-2)
On Multiplying both numerator and denominator by (√7+2).
1× (√7+2) / (√7-2)× (√7+2)
= (√7+2) / ((√7)²- (2)²)
[ (a+b)(a-b) = a² - b²]
= (√7+2) / 7 - 4
= (√7+2) / 3 = (√7+ 2) /3
= (√7+2) / 3
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Hope this will help you....
