Surds –  Exercise 7.3 – Class X

Next exercise – Surds – Exercise 7.2 – 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) 2^1/3 + 2^1/3

(ii) 5^1/3 + 5^1/3

(iii) √(1+y) – √(1-y)

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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

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2) 2√x

Solution:

2√x x √x = 2x

Therefore, rationalizing factor of  2√x is √x

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3) 7√y

Solution:

7√y x √y = 7y

Therefore, rationalizing factor of  7√y is √y

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4) √(xy)

Solution:

√(xy) x √(xy) = xy

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

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5) 4√(x + y)

Solution:

4√(x+y) x √(x+y) = 4(x+y)

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

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6) 8√(x – y)

Solution:

8√(x – y) x √(x – y) = 8(x – y)

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

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7) ¹/₂√p

Solution:

¹/₂√p x √p = ¹/₂ p

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

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8) a√(ab)

Solution:

a√(ab) x √(ab) = a(ab) = a²b

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

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9) x√(mn)

Solution:

x√(mn) x √(mn) = xmn

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

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10)5p√(a+b)

Solution:

5p√(a+b) x √(a+b) = 5p(a+b)

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

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II. Write the conjugate of the following binomial surds

(i) √a + √b

Solution:

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

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(ii) √x – 2√y

Solution:

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

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(iii) 3√p – 2√q

Solution:

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

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(iv) x + 3√y

Solution:

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

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(v) 10√2 + 3√5

Solution:

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

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(vi) 5 + √3

Solution:

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

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(vii) √8 – 5

Solution:

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

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(viii) 3√7 + 7√3

Solution:

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

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(ix) ¹/₂ + √2

Solution:

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

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(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

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(xi) x√a + y√b

Solution:

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

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 (xii) xy√z + yz√x

Solution:

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

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III. Find the rationalizing factor of the following binomial surds

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

Solution:

2^1/3 + 2^-1/3

a = 2^1/3  ; b = 2 ^-1/3

Then, a³ = (2^1/3)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 = (2^1/3 + 2^-1/3)[(2^1/3)² – 2^1/32^-1/3 + (2^-1/3)²]

⁵/₂ = (2^1/3 + 2^-1/3)(2^2/3 + 2^-2/3 – 1)

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

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

Solution:

5^1/3 + 5^-1/3

a = 5^1/3  ; b = 5 ^-1/3

Then, a³ = (5^1/3)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 = (5^1/3 + 5^-1/3)[(5^1/3)² – 5^1/35^-1/3 + (5^-1/3)²]

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

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

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(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).

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Next Exercise – Surds – Exercise 7.4 – Class X

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