# Surds –  Exercise 7.3 – Class X

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#### Surds –  Exercise 7.3

1. Write the rationalizing factor for the following surds.

1) √a

2) 2√x

3) 7√y

4) √(xy)

5) 4√(x + y)

6) 8√(x – y)

7) 1/2√p

8) a√(ab)

9) x√(mn)

10)5p√(a+b)

1. Write the conjugate of the following binomial surds

(i) √a + √b

(ii) √x – 2√y

(iii) 3√p – 2√q

(iv) x + 3√y

(v) 10√2 + 3√5

(vi) 5 + √3

(vii) √8 – 5

(viii) 3√7 + 7√3

(ix) 1/2 + √2

(x) 1/2 x + 1/2√y

(xi) x√a + y√b

(xii) xy√z + yz√x

III. Find the rationalising factor of the following binomial surds

(i) 21/3 + 21/3

(ii) 51/3 + 51/3

(iii) √(1+y) – √(1-y)

#### Surds –  Exercise 7.3

1. Write the rationalizing factor for the following surds.

1) √a

Solution:

√a x √a = a

Therefore, rationalizing factor of √a is √a

2) 2√x

Solution:

2√x x √x = 2x

Therefore, rationalizing factor of  2√x is √x

3) 7√y

Solution:

7√y x √y = 7y

Therefore, rationalizing factor of  7√y is √y

4) √(xy)

Solution:

√(xy) x √(xy) = xy

Therefore, rationalizing factor of  √(xy) is √xy

5) 4√(x + y)

Solution:

4√(x+y) x √(x+y) = 4(x+y)

Therefore, rationalizing factor of  4√(x+y) is √(x+y)

6) 8√(x – y)

Solution:

8√(x – y) x √(x – y) = 8(x – y)

Therefore, rationalizing factor of  8√(x – y) is √(x-y)

7) 1/2√p

Solution:

1/2√p x √p = 1/2 p

Therefore, rationalizing factor of  1/2√p is √p

8) a√(ab)

Solution:

a√(ab) x √(ab) = a(ab) = a2b

Therefore, rationalizing factor of  a√(ab) is √(ab)

9) x√(mn)

Solution:

x√(mn) x √(mn) = xmn

Therefore, rationalizing factor of  x√(mn) is xmn

10)5p√(a+b)

Solution:

5p√(a+b) x √(a+b) = 5p(a+b)

Therefore, rationalizing factor of  5p√(a+b) is √(a+b)

II. Write the conjugate of the following binomial surds

(i) √a + √b

Solution:

The conjugate of  binomial surd √a + √b is √a – √b

(ii) √x – 2√y

Solution:

The conjugate of  binomial surd √x – 2√y is √x + 2√y

(iii) 3√p – 2√q

Solution:

The conjugate of  binomial surd  3√p – 2√q is  3√p + 2√q

(iv) x + 3√y

Solution:

The conjugate of  binomial surd x + 3√y is x – 3√y

(v) 10√2 + 3√5

Solution:

The conjugate of  binomial surd 10√2 + 3√5 is 10√2  – 3√5

(vi) 5 + √3

Solution:

The conjugate of  binomial surd 5 + √3 is 5 – √3

(vii) √8 – 5

Solution:

The conjugate of  binomial surd √8 – 5 is √8 + 5

(viii) 3√7 + 7√3

Solution:

The conjugate of  binomial surd 3√7 + 7√3 is 3√7 – 7√3

(ix) 1/2 + √2

Solution:

The conjugate of  binomial surd 1/2 + √2 is 1/2 – √2

(x) 1/2 x + 1/2√y

Solution:

The conjugate of  binomial surd 1/2 x + 1/2√y  is 1/2 x – 1/2√y

(xi) x√a + y√b

Solution:

The conjugate of  binomial surd x√a + y√b is x√a – y√b

(xii) xy√z + yz√x

Solution:

The conjugate of  binomial surd xy√z + yz√x is xy√z – yz√x

III. Find the rationalizing factor of the following binomial surds

(i) 21/3 + 2 -1/3

Solution:

21/3 + 2-1/3

a = 21/3  ; b = 2 -1/3

Then, a3 = (21/3)3 = 2 and b3 = (2-1/3)3 = 1/2

a3 + b3 = 2 + 1/2 = 4+1/2 = 5/2

But a3 + b3 = (a + b)(a2 – ab + b2)

5/2 = (21/3 + 2-1/3)[(21/3)2 – 21/32-1/3 + (2-1/3)2]

5/2 = (21/3 + 2-1/3)(22/3 + 2-2/3 – 1)

Since 3/2 is a rational number, rationalizing factor of (21/3 – 2-1/3) is (22/3 + 2-2/3 + 1)

(ii) 51/3 + 5-1/3

Solution:

51/3 + 5-1/3

a = 51/3  ; b = 5 -1/3

Then, a3 = (51/3)3 = 5 and b3 = (5-1/3)3 = 1/5

a3 + b3 = 5 + 1/5 = 25+1/5 = 26/5

But a3 + b3 = (a + b)(a2 – ab + b2)

26/5 = (51/3 + 5-1/3)[(51/3)2 – 51/35-1/3 + (5-1/3)2]

26/5 = (51/3 + 5-1/3)(52/3 + 5-2/3 – 1)

Since 24/5 is a rational number, rationalizing factor of (51/3 + 5-1/3) is (52/3 + 5-2/3 – 1)

(iii) √(1+y) – √(1-y)

Solution:

[√(1+y) – √(1-y)]x[√(1+y) + √(1-y)]  =

= [√(1+y)]2 – [√(1-y)]2

= (1+ y) – (1 – y)

= 1 + y – 1 + y

= 2y

⸫√(1+y) – √(1-y) is rationalised by  using its conjugate √(1+y) + √(1-y).

Therefore, rationalizing factor of √(1+y) – √(1-y) is √(1+y) + √(1-y).

Next Exercise – Surds – Exercise 7.4 – Class X