Real Numbers
Class 9th Mathematics Part I MHB Solution
Practice Set 2.4
Question 1.Multiply
i. √3(√7 – √3)
ii. (√5 – √7)√2
iii. (3√2 – √3)(4√3 – √2)
Answer:i. √3(√7 – √3)
=√3 × √7 – √3 × √3
[∵√a(√b–√c)=√a×√b–√a×√c]
=√21 – 3
ii. (√5 – √7)√2
=√5 × √2 – √7 × √2
[∵√a(√b–√c)=√a×√b–√a×√c]
=√10 – √14
iii. (3√2 – √3)(4√3 – √2)
=3√2(4√3 – √2) – √3(4√3 – √2)
=3√2×4√3 – 3√2×√2 – √3×4√3 + √3×√2
[∵√a(√b–√c)=√a×√b–√a×√c]
=12√6 – 3×2 – 4×3 + √6
=12√6 – 6 – 12 + √6
=13√6 – 18
Question 2.Rationalize the denominator.
i. 
ii. 
iii. 
iv. 
Answer:i. The rationalizing factor of √7 + √2 is √7 – √2. Therefore, multiply both numerator and denominator by √7 – √2.
[∵ (a-b)(a+b) = a2 – b2]
ii. The rationalizing factor of 2√5 – 3√2 is 2√5 + 3√2. Therefore, multiply both numerator and denominator by 2√5 + 3√2.
[∵ (a-b)(a+b) = a2 – b2]
iii. The rationalizing factor of 7 + 4√3 is 7 – 4√3. Therefore, multiply both numerator and denominator by 7 – 4√3.
[∵ (a-b)(a+b) = a2 – b2]
iv. The rationalizing factor of √5 + √3 is √5 - √3. Therefore, multiply both numerator and denominator by √5 - √3.
[∵ (a-b)(a+b) = a2 – b2]
[∵ (a-b)2 = a2 + b2 – 2ab]
i. √3(√7 – √3)
ii. (√5 – √7)√2
iii. (3√2 – √3)(4√3 – √2)
i.
ii. iii.
iv. 
