# Class 9 RD Sharma Solutions – Chapter 3 Rationalisation- Exercise 3.1

• Last Updated : 01 Dec, 2020

### Question 1: Simplify each of the following: Solution:

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(i)

Using the formula: Here, (ii)

Using the formula: Here, ### (iii) (√5 -2)( √3 – √5)

Solution:

(i) (4 + √7) (3 + √2)

= 12 + 4√2 + 3√7 + √14

(ii) (3 + √3)(5- √2)

= 15 – 3√2 + 5√3 – √6

(iii) (√5 – 2)(√3 – √5)

= √15 – √25 – 2√3 + 2√5

= √15 – 5 – 2√3 + 2√5

### (v) (√5 – √2) (√5 + √2)

Solution:

Using Identity: (a – b)(a + b) = a2 – b2

(i) (11 + √11) (11 – √11)

= 112 – (√11)2

= 121 – 11

= 110

(ii) (5 + √7) (5 –√7)

= (52 – (√7)2)

= 25 – 7 = 18

(iii) (√8 – √2) (√8 + √2)

= (√8)2 – (√2)2

= 8 – 2

= 6

(iv) (3 + √3) (3 – √3)

= (3)2 – (√3)2

= 9 – 3

= 6

(v) (√5 – √2) (√5 + √2)

= (√5)2 – (√2)2

= 5 – 2

= 3

### (iii) (2√5 + 3√2 )2

Solution:

Using identities: (a – b)2 = a2 + b2– 2ab and (a + b)2 = a2+ b2 + 2ab

(i) (√3 + √7)2

= (√3)2 + (√7)2 + 2(√3)(√7)

= 3 + 7 + 2√21

= 10 + 2√21

(ii) (√5 – √3)2

= (√5)2 + (√3)2 – 2(√5)(√3)

= 5 + 3 – 2√15

= 8 – 2√15

(iii) (2√5 + 3√2)2

= (2√5)2 + (3√2)2 + 2(2√5)( 3√2)

= 20 + 18 + 12√10

= 38 + 12√10

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