RATIONALISING FACTOR

Solved Problems

Write the rationalising factor of each of the following irrational numbers :

Problem 1 :

3√2

Solution :

3√2 is already in simplest form. 

=  3√2 ⋅ √2

=  3(√2 ⋅ √2)

=  3 ⋅ 2

=  6

6 is a rational number. 

So, the rationalizing factor of 3√2 is √2.

Problem 2 :

4³√25

Solution :

4³√25 is already in simplest form. 

=  4³√25 ⋅ ³√5

=  4(³√25 ⋅ ³√5)

=  4³√(25 ⋅ 5)

=  4³√(5 ⋅ 5 ⋅ 5)

=  4 ⋅ 5

=  20

20 is a rational number. 

So, the rationalizing factor of 2³√25 is ³√5.

Problem 3 :

³√432

Solution :

Write ³√432 in the simplest form. 

³√432  =  ³√(2 ⋅ 6 ⋅ 6 ⋅ 6)

=  6³√2

The simplest form of ³√412 is 6³√2.

=  6³√2 ⋅ ³√4

=  6³√(2 ⋅ 4)

=  6³√(2 ⋅ 2 ⋅ 2)

=  6 ⋅ 2

=  12

12 is a rational number. 

So, the rationalizing factor of ³√432 is ³√4.

Problem 4 :

2 + √3

Solution :

(2 + √3) is already in simplest form. 

=  (2 + √3)(2 - √3)

Use the algebraic identity a² - b² = (a + b)(a - b). 

=  2² - (√3)²

=  2² - (√3)²

=  4 - 3

=  1

1 is a rational number. 

So, the rationalizing factor of (2 + √3) is (2 - 4√3).

Problem 5 :

5 - 4√3

Solution :

(5 - 4√3) is already in simplest form. 

=  (5 - 4√3)(5 + 4√3)

Use the algebraic identity a² - b² = (a + b)(a - b). 

=  5² - (4√3)²

=  25 - 4²(√3)²

  =  25 - 16(3)

=  25 - 48

=  -23

-23 is a rational number. 

So, the rationalizing factor of (5 - 4√3) is (5 + 4√3).

Problem 6 :

√3 + √2

Solution :

(√3 + √2) is already in simplest form. 

=  (√3 + √2)(√3 - √2)

Use the algebraic identity a² - b² = (a + b)(a - b). 

=  (√3)² - (√2)²

=  3 - 2

=  1

1 is a rational number. 

So, the rationalizing factor of (√2 + √3) is (√2 - √3).

Problem 7 :

√75

Solution :

Write √75 in simplest form. 

 √75  =  √(3 ⋅ 5 ⋅ 5)

The simplest form of √75 is 5√3.

=  5√3 ⋅ √3

=  5(√3 ⋅ √3)

=  5 ⋅ 3

=  15

15 is a rational number. 

So, the rationalizing factor of √75 is √3.

Problem 8 :

√125 + 2

Solution :

Write (√125 + 2) in the simplest form. 

√125 + 2  =  √(5 ⋅ 5 ⋅ 5) + 2

=  5√5 + 2

The simplest form of (√125 + 2) is (5√5 + 2).

=  (5√5 + 2)(5√5 - 2)

Use the algebraic identity a² - b² = (a + b)(a - b). 

=  (5√5)² - 2²

=  5²(√5)² - 4

=  25(5) - 4

=  125 - 4

=  121

121 is a rational number. 

So, the rationalizing factor of (√125 + 2) is (5√5 - 2)

Problem 9 :

(√3 - 1)/(√3 + 1) = a - b√3

Solution :

= (√3 - 1)/(√3 + 1)

Conjugate of the denominator is √3 - 1

= [(√3 - 1)/(√3 + 1)] [(√3 - 1)/(√3 - 1)]

Multiplying the numerator, we get

= (√3 - 1) (√3 - 1) / (√3 + 1) (√3 - 1)

= (√3² - 2√3(1) + 1²) / (√3² - 1²)

= (3 - 2√3(1) + 1) / (3 - 1)

= (4 - 2√3) / 2

Factoring 2, we get

= 2(2 - √3) / 2

= 2 - √3

Comparing with

a - b√3

we get a = 2 and b = 1

Problem 10 :

(4 + √2)/(2 + √2) = a - √b

Solution :

= (4 + √2)/(2 + √2)

Conjugate of the denominator (2 + √2) is (2 - √2)

= (4 + √2)/(2 + √2) x [(2 - √2) / (2 - √2)]

= [(4 + √2)(2 - √2)]/[(2 + √2)(2 - √2)]

Multiplying the numerators, we get

(4 + √2)(2 - √2) = 4(2) - 4(√2) + 2√2 - √2²

= 8 - 2√2 - 2

= 6 - 2√2

Multiplying the denominators, we get

(2 + √2)(2 - √2) = 2² - √2²

= 4 - 2

= 2

Dividing numerator by denominator, we get

= (6 - 2√2)/2

= 3 - √2

Comparing with a - √b, we get a = 3 and b = -1

Problem 11 :

If x = 2 + √3, find the value of x³ + 1/x³

Solution :

x³ + 1/x³ 

Comparing with a³ + b³ = (a + b)(a² - ab + b²)

x³ + (1/x)³ = (x + 1/x)(x² - x(1/x) + (1/x)²) ---(1)

Given that x = 2 + √3

1/x = 1/(2 + √3)

Multiplying both numerator and denominator by the conjugate of the denominator, we get

= [1/(2 + √3)] [(2 - √3)/(2 - √3)]

= (2 - √3)/(2 - √3)(2 - √3)

= (2 - √3)/(2² - √3²)

= (2 - √3)/(4 - 3)

1/x = 2 - √3

x + 1/x = 2 + √3 + 2 - √3 

= 4

x² = (2 + √3)²

= 2² + 2(2) √3 + √3²

= 4 + 4√3 + 3

x² = 7 + 4√3

1/x² = (1/x)²

= (2 - √3)²

= 2² - 2(2) √3 + √3²

= 4 - 4√3 + 3

1/x² = 7 - 4√3

Applying these values in (1), we get

= (x + 1/x)(x² - x(1/x) + (1/x)²)

= (2 + √3 + 2 - √3)(7 + 4√3 - 1 + 7 - 4√3)

= 4 (14 - 1)

= 4(13)

= 52

• Rationalization of surds
• Comparison of surds
• Operations with radicals
• Ascending and descending  order of surds

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