RD Sharma Solutions Ex-3.2, Rationalisation, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download

1. Rationalize the denominator of each of the following:

(i) 
(ii) 
 (iii) 
 (iv) 
 (v) 
 (vi) 
 (vii) 

Solution: (i)   For rationalizing the denominator, multiply both numerator and denominator with 

(ii)    For rationalizing the denominator, multiply both numerator and denominator with 

(iii)    For rationalizing the denominator, multiply both numerator and denominator with 

(iv)   For rationalizing the denominator, multiply both numerator and denominator with 

(v)   For rationalizing the denominator, multiply both numerator and denominator with 

(vi)   For rationalizing the denominator, multiply both numerator and denominator with    

(vii)    For rationalizing the denominator, multiply both numerator and denominator with 

2. Find the value to three places of decimals of each of the following. It is given that   = 1.414, = 1.732,  = 2.236,  = 3.162.

(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 

Solution: Given,    = 1.414, = 1.732,  = 2.236,  = 3.162.

(i)  Rationalizing the denominator by multiplying both numerator and denominator with 

(ii)  Rationalizing the denominator by multiplying both numerator and denominator with 

(iii)  Rationalizing the denominator by multiplying both numerator and denominator with 

(iv)  Rationalizing the denominator by multiplying both numerator and denominator with 

(vi)  Rationalizing the denominator by multiplying both numerator and denominator with 

3. Express each one of the following with rational denominator:

(i) 
(ii) 
 (iii) 
 (iv) 
 (v) 
 (vi) 
 (vii) 
 (viii) 
 (ix) 

Solution: (i)  Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor 

As we know, (a + b) (a - b) = (a²−b²)  

(ii)   Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor   

As we know, (a + b) (a - b) = (a² − b²)

(iii)  Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor 

As we know, (a + b) (a - b) = (a²−b²) 

(iv)   Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor 

As we know, (a + b) (a - b) = (a²−b²) = 

(v)   Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor 

As we know, (a + b) (a - b) = (a²−b²) = 

(vi)  Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor 

As we know, (a + b) (a - b) = (a²−b²) = 

(vii)   Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor 

As we know, (a + b) (a - b) = (a²−b²)  As we know , (a - b)² = (a²−2×a×b+b²)

(viii)   Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor 

As we know, (a + b) (a - b) = (a²−b²) = 

(ix)  Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor 

As we know, (a + b) (a - b) = (a²−b²) 

4. Rationalize the denominator and simplify: 

(i) 

(ii) 

(iii) 

(iv)  

(v) 

(vi)  

Solution: (i)   Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor 

As we know, (a + b) (a - b)  As we know , (a - b) 2 = (a²−2×a×b+b²)

(ii)   Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor 

As we know, (a + b) (a - b) = (a²−b²) = 

(iii)   Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor 

As we know, (a + b) (a - b) = (a²−b²) 

(iv)   Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor 

 As we know, (a + b) (a - b) = (a²−b²) = 

(v)   Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor 

As we know, (a + b) (a - b) = (a²−b²) = 

(vi)    Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor 

5. Simplify: 

(i) 

(ii) 

(iii)  

(iv) 

(v) 

 Solution: (i)  Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor   and the rationalizing factor 

 Now, (a + b) (a - b) = (a²−b²) =   

As we know , (a - b)² = (a²−2×a×b+b²) = 

(ii)   Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor   and the rationalizing factor   = 

Now as we know, (a + b) (a - b) = (a²−b²), (a - b)² = (a²−2×a×b+b²) and (a+b)² = (a²+2×a×b+b²)

(iii)    Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor   and the rationalizing factor 

Now as we know, (a + b) (a - b) = (a²−b²) = 

(iv)   Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor   the rationalizing factor  , and the rationalizing factor

Since, (a + b) (a - b) = (a²−b²)

(v)   Rationalizing the denominator by multiplying both numerator and denominator with  the rationalizing factor   the rationalizing factor   and the rationalizing factor

Since, (a + b) (a - b) = (a²−b²) = 

6. In each of the following determine rational numbers a and b:

(i) 

(ii)  

(iii) 

(iv)  

(v) 

(vi)  

Solution:   (i) Given,  Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor 

As we know, (a + b) (a - b) = (a²−b²) 

On comparing the rational and irrational parts of the above equation, we get, a = 2 and b = 1

(ii)    Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor 

As we know, (a + b) (a - b) = (

On comparing the rational and irrational parts of the above equation, we get,

a = 3 and b = 2

(iii)   Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor 

As we know, (a + b) (a - b) = (

On comparing the rational and irrational parts of the above equation, we get,

(iv)    Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor 

As we know, (a + b) (a - b) = (

On comparing the rational and irrational parts of the above equation, we get,

a = -1 and

b = 1

(v)   Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor 

As we know, (a + b) (a - b) = (a²−b²) = 

On comparing the rational and irrational parts of the above equation, we get,

(vi)    Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor 

As we know, (a + b) (a - b) = (a²−b²)

On comparing the rational and irrational parts of the above equation, we get,

7. If x = 2 +  find the value of 

Solution:  Given, x = 2 +  To find the value of 

We have,  x = 2 +  

Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor 

Since, (a + b) (a - b) = (a²−b²)

We know that, 

Putting the value of   in the above equation, we get,

8. If x = 3 +  find the value of 

Solution:   Given, x = 3 + 

To find the value of    We have, 

Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor

Since, (a + b) (a - b) = (a²−b²)

9. Find the value of  it being given that  = 1.732 and = 2.236. 

Solution. Given,   Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor

Since, (a + b) (a - b) = (a²−b²) =  = 3(2.236+1.732) = 3(3.968) = 11.904 

10. Find the values of each of the following correct to three places of decimals, it being given that  = 1.414, = 1.732, = 2.236, = 2.4495, = 3.162

Solution. (i)  Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor 

Since, (a + b) (a - b) = (a²−b²) =

(ii)  Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor 

As we know, (a + b) (a - b) = (a²−b²)  = 7 + 7.07 = 14.07 

11. If x =  find the value of 4x³+2x²−8x+7.

Solution: Given,  

and given to find the value of 4x³+2x²−8x+7

Now, squaring on both the sides, we get, (2x−1)² = 3

4x²−4x+1 = 3

4x²−4x+1−3 = 0

4x²−4x−2 = 0

2x²−2x−1 = 0

Now taking 4x³+2x²−8x+7

2x(2x²−2x−1)+ 4x²+2x+2x²−8x+7

2x(2x²−2x−1)+ 6x²−6x+7

As, 2x²−2x−1=0

2x(0)+3(2x²−2x−1))+7+3

0+3(0)+10

10