**Next exercise – Surds – Exercise 7.2 – Class X**

#### Surds – Exercise 7.3

- 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)

- 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)

#### Surds – Exercise 7.3

**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) = a^{2}b

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) 2 ^{1/3} + 2 ^{-1/3}**

Solution:

2^{1/3} + 2^{-1/3}

a = 2^{1/3} ; b = 2 ^{-1/3}

Then, a^{3} = (2^{1/3})^{3} = 2 and b^{3} = (2^{-1/3})^{3} = ^{1}/_{2}

a^{3} + b^{3} = 2 + ^{1}/_{2} = ^{4+1}/_{2} = ^{5}/_{2}

But a^{3} + b^{3} = (a + b)(a^{2} – ab + b^{2})

^{5}/_{2} = (2^{1/3} + 2^{-1/3})[(2^{1/3})^{2} – 2^{1/3}2^{-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)

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

Solution:

5^{1/3} + 5^{-1/3}

a = 5^{1/3} ; b = 5 ^{-1/3}

Then, a^{3} = (5^{1/3})^{3} = 5 and b^{3} = (5^{-1/3})^{3} = ^{1}/_{5}

a^{3} + b^{3} = 5 + ^{1}/_{5} = ^{25+1}/_{5} = ^{26}/_{5}

But a^{3} + b^{3} = (a + b)(a^{2} – ab + b^{2})

^{26}/_{5} = (5^{1/3} + 5^{-1/3})[(5^{1/3})^{2} – 5^{1/3}5^{-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)

**(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**