Frank Solutions for Class 9 Maths Chapter 1 Irrational Numbers

Frank Solutions for Class 9 Maths Chapter 1 Irrational Numbers help students to understand the concepts covered in an effective way. The solutions are available in brief step-wise explanations and are based on the current syllabus of the ICSE Board. Students who aspire to obtain proficiency in Mathematics are advised to practise these solutions on a regular basis. Solving the textbook questions by referring to these solutions will enhance their problem-solving skills.

Chapter 1 of Frank Solutions contains problems based on Irrational Numbers. A real number which cannot be written in a simple fraction is known as an irrational number. This chapter plays a vital role in the study of Mathematics as it is continued in higher classes as well. Students can download Frank Solutions for Class 9 Maths Chapter 1 Irrational Numbers PDF from the link given below and practise offline as well, without time constraints, to clear their doubts instantly.

Frank Solutions for Class 9 Maths Chapter 1 Irrational Numbers Download PDF

Access Frank Solutions for Class 9 Maths Chapter 1 Irrational Numbers

1. State which of these fractions have a terminating decimal.

(a) (3/ 5)

(b) (5 / 7)

(c) (25 / 49)

(d) (37 / 40)

(e) (57 / 64)

(f) (59 / 75)

(g) (89 / 125)

(h) (125 / 213)

(i) (147 / 160)

Solutions:

(a) (3 / 5)

5 = 1 × 5

5 = 2⁰ × 5¹

i.e, 5 can be expressed as 2^m × 5ⁿ

Therefore,

(3 / 5) has terminating decimal representation

(b) (5 / 7)

7 = 1 × 7

i.e, 7 cannot be expressed a 2^m × 5ⁿ

Therefore,

(5 / 7) does not have terminating decimal representation

(c) (25 / 49)

49 = 7 × 7

i.e, 49 cannot be expressed as 2^m × 5ⁿ

Therefore,

(25 / 49) does not have terminating decimal representation

(d) (37 / 40)

40 = 2 × 2 × 2 × 5

40 = 2³ × 5¹

i.e, 40 can be expressed as 2^m × 5ⁿ

Therefore,

(37 / 40) has terminating decimal representation

(e) (57 / 64)

64 = 2 × 2 × 2 × 2 × 2 × 2

64 = 2⁶ × 5⁰

i.e, 64 can be expressed as 2^m × 5ⁿ

Therefore,

(57 / 64) has terminating decimal representation

(f) (59 / 75)

75 = 5 × 5 × 3

75 = 5² × 3¹

i.e, 75 cannot be expressed as 2^m × 5ⁿ

Therefore,

(59 / 75) does not have terminating decimal representation

(g) (89 / 125)

125 = 5 × 5 × 5

125 = 2⁰ × 5³

i.e, 125 can be expressed as 2^m × 5ⁿ

Therefore,

(89 / 125) has terminating decimal representation

(h) (125 / 213)

213 = 3 × 71

i.e, 213 cannot be expressed as 2^m × 5ⁿ

Therefore,

(125 / 213) does not have terminating decimal representation

(i) (147 / 160)

160 = 2 × 2 × 2 × 2 × 2 × 5

160 = 2⁵ × 5¹

i.e, 160 can be expressed as 2^m × 5ⁿ

Therefore,

(147 / 160) has terminating decimal representation

2. Express each of the following decimals as a rational number

(a) 0.93

(b) 4.56

(c) 0.614

(d) 21.025

Solution:

(a) 0.93 = 93 / 100

Hence,

The rational number of decimal 0.93 is (93 / 100)

(b) 4.56 = (456 / 100)

= (456 ÷ 4) / (100 ÷ 4)

We get,

= (114 / 25)

Hence,

The rational number of decimal 4.56 is (114 / 25)

(c) 0.614 = (614 / 1000)

= (614 ÷ 2) / (1000 ÷ 2)

We get,

= (307 / 500)

Hence,

The rational number of decimal 0.614 is (307 / 500)

(d) 21.025 = (21025 / 1000)

= (21025 ÷ 25) / (1000 ÷ 25)

We get,

= (841 / 40)

Hence,

The rational number of decimal 21.025 is (841 / 40)

3. Convert the following fractions into decimals:

(i) (3 / 5)

(ii) (8 / 11)

(iii) (-2 / 7)

(iv) (12 / 21)

(v) (13 / 25)

(vi) (2 / 3)

Solution:

(i) (3 / 5)

(3 / 5) = 0.6

Hence,

The decimal form of (3 / 5) is 0.6

(ii) (8 / 11)

(8 / 11) = 0.72727272…..

(8 / 11) =

Hence,

The decimal form of (8 / 11) is

(iii) (-2 / 7)

(-2 / 7) = -0.285714285714….

(-2 / 7) = – 

Hence,

The decimal form of (-2 / 7) is – 

(iv) (12 / 21)

(12 / 21) = 0.571428571428……….

(12 / 21) = 

Hence,

The decimal form of (12 / 21) is 

(v) (13 / 25)

(13 / 25) = 0.52

Hence,

The decimal form of (13 / 25) is 0.52

(vi) (2 / 3)

(2 / 3) = 0.6666…..

(2 / 3) = 0.6

Hence,

The decimal form of (2 / 3) is 0.6

4. Express each of the following decimals as a rational number.

(a) 0.7

(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

Solution:

(a) 0.7

Let x = 0.7

Then,

x = 0.7777….. (1)

Here, the number of digits recurring is only 1,

So, we multiply both sides of the equation (1) by 10

We get,

10x = 10 × 0.7777… (2)

10x = 7.777…..

On subtracting (1) from (2),

We get,

9x = 7

x = (7 / 9)

0.7 = (7 / 9)

Therefore,

0.7 = (7 / 9)

(b)

Let x = 

Then,

x = 0.353535…….. (1)

Here,

The number of digits recurring is 2,

So, we multiply both sides of equation (1) by 100

We get,

100x = 100 × 0.353535….

= 35.3535 ……. (2)

On subtracting (1) from (2),

We get,

99x = 35

x = (35 / 99)

Therefore,

= (35 / 99)

(c)

Let x = 

Then,

x = 0.898989………. (1)

Here,

The number of digits recurring is 2,

So we multiply both sides of the equation (1) by 100

We get,

100x = 100 × 0.898989………

= 89.8989…. (2)

On subtracting (1) from (2),

We get,

99x = 89

x = (89 / 99)

= (89 / 99)

Therefore,

= (89 / 99)

(d)

Let x = 

Then,

x = 0.057057…… (1)

Here,

The number of digits recurring is 3,

So, we multiply both sides of the equation (1) by 1000,

We get,

1000x = 1000 × 0.057057……

= 57.057………. (2)

On subtracting (1) from (2),

We get,

999x = 57

x = (57 / 999)

x = (19 / 333)

= (19 / 333)

Therefore,

= (19 / 333)

(e)

Let x = 

Then,

x = 0.763763……. (1)

Here,

The number of digits recurring is 3,

So, we multiply both sides of the equation (1) by 1000,

We get,

1000x = 1000 × 0.763763………

= 763.763…….. (2)

On subtracting (1) from (2),

We get,

999x = 763

x = (763 / 999)

= (763 / 999)

Therefore,

= (763 / 999)

(f)

Let x = 

Then,

x = 2.676767………. (1)

Here,

The number of digits recurring is 2,

So, we multiply both sides of the equation (1) by 100,

We get,

100x = 100 × 2.676767……..

= 267.6767…….. (2)

On subtracting (1) from (2),

We get,

99x = 265

x = (265 / 99)

= (265 / 99)

Therefore,

= (265 / 99)

(g)

Let x = = 4.6724724…….

Here,

Only number 724 is being repeated, so first, we need to remove 6, which proceeds 724

We multiply by 10 so that only the recurring digits remain after decimal

Thus,

10x = 10 x 4.6724724…….

10x = 46. 724724……. (1)

The number of digits recurring in equation (1) is 3

Hence, we multiply both sides of the equation (1) by 1000

10000x = 1000 x 46.724724 = 46724.724.. …(2)

On subtracting (1) from (2), we get,

9990x = 46678

x = 46678 / 9990

We get,

x = 23339 / 4995

(h)

Let x = 

= 0.01717……….

Here, only number 17 is being repeated, so first, we need to remove 0, which proceeds 17

We multiply by 10 so that only the recurring digits remain after the decimal,

Hence,

10x = 0.1717…… (1)

The number of digits recurring in equation (1) is 2,

So we multiply both sides of the equation (1) by 100,

Hence,

1000x = 100 × 0.1717……..

= 17.1717….. (2)

On subtracting (1) from (2),

We get,

990x = 17

x = (17 / 990)

= (17 / 990)

Therefore,

= (17 / 990)

(i)

Let x = 

= 17.027777……

Here, only number 7 is being repeated, so first, we need to remove 02, which proceeds 7

We multiply by 100 so that only the recurring digits remain after decimal

Hence,

100x = 1702.7777……… (1)

The number of digits recurring in equation (1) is 1,

So we multiply both sides of the equation (1) by 10

Hence,

1000x = 10 × 1702.7777……….

= 17027.777………. (2)

On subtracting (1) from (2),

We get,

900x = 15325

x = (15325 / 900)

x = (613 / 36)

= (613 / 36)

Therefore,

= (613 / 36)

5. Insert a rational number between:

(a) (2 / 5) and (3 / 4)

(b) (3 / 4) and (5 / 7)

(c) (4 / 3) and (7 / 5)

(d) (5 / 9) and (6 / 7)

Solution:

(a) (2 / 5) and (3 / 4)

= [{(2 / 5) + (3 / 4)} / 2]

On further calculation, we get,

= [{(8 + 15) / 20} / 2]

= {(23 / 20) / 2}

We get,

= (23 / 40)

Therefore,

A rational number lying between (2 / 5) and (3 / 4) is (23 / 40)

(b) (3 / 4) and (5 / 7)

= [{(3 / 4) + (5 / 7)} / 2]

On further calculation, we get,

= [{(21 + 20) / 28} / 2]

= {(41 / 28) / 2}

We get,

= (41 / 56)

Therefore,

A rational number lying between (3 / 4) and (5 / 7) is (41 / 56)

(c) (4 / 3) and (7 / 5)

= [{(4 / 3) + (7 / 5)} / 2]

On further calculation, we get,

= [{(20 + 21) / 15} / 2]

= {(41 / 15) / 2}

We get,

= (41 / 30)

Therefore,

A rational number lying between (4 / 3) and (7 / 5) is (41 / 30)

(d) (5 / 9) and (6 / 7)

= [{(5 / 9) + (6 / 7)} / 2]

On further calculation, we get,

= [{(35 + 54) / 63} / 2]

= {(89 / 63) / 2}

We get,

= (89 / 126)

Therefore,

A rational number lying between (5 / 9) and (6 / 7) is (89 / 126)

6. State whether the following numbers are rational or irrational:

(a) (3 + √3)²

(b) (5 – √5)²

(c) (2 + √2) (2 – √2)

(d) {(√5) / (3√2)}²

Solution:

(a) (3 + √3)²

= (3)² + (√3)² + 2 × 3 × √3

On calculating further, we get,

= 9 + 3 + 6√3

= 12 + 6√3

– which is a rational number

Therefore,

(3 + √3)² is a rational number

(b) (5 – √5)²

= (5)² + (√5)² – 2 × 5 × √5

On further calculation, we get,

= 25 + 5 – 10√5

= 30 – 10√5

which is an irrational number

Therefore,

(5 – √5)² is an irrational number

(c) (2 + √2) (2 – √2)

= (2)² – (√2)²

= 4 – 2

= 2

-which is a rational number

Therefore,

(2 + √2) (2 – √2) is a rational number

(d) {(√5) / (3√2)}²

= {(5) / (9 × 2)}

We get,

= (5 / 18)

which is a rational number

Therefore,

{(√5) / (3√2)}² is a rational number

7. Check whether the square of the following is rational or irrational:

(a) 3√2

(b) 3 + √2

(c) (3√2) / 2

(d) √2 + √3

Solution:

(a) 3√2

(3√2)²

= 9 × 2

= 18

which is a rational number

Hence,

The square of (3√ 2) is a rational number

(b) 3 + √2

(3 + √2)²

= (3)² + (√2)² + 2 × 3 × √2

On further calculation, we get,

= 9 + 2 + 6√2

= 11 + 6√2

which is an irrational number

Hence,

The square of (3 + √2) is an irrational number

(c) (3√2) / 2

{(3√2) / 2}²

= (9 × 2) / 4

= (9 / 2)

-which is a rational number

Hence,

The square of {(3√2) / 2} is a rational number

(d) √2 + √3

(√2 + √3)²

= (√2)² + (√3)² + 2 × √2 × √3

= 2 + 3 + 2√6

We get,

= 5 + 2√6

which is an irrational number

Hence,

The square of (√2 + √3) is an irrational number

8. Show that √5 is an irrational number. (Use division method)

Solution:

Here,

Clearly, √5 = 2.23606…… is an irrational number

Therefore,

√5 is an irrational number

9. Without using the division method, show that √7 is an irrational number

Solution:

Let √7 be a rational number

Hence,

√7 = (a / b)

On squaring both sides, we get,

7 = (a² / b²)

a² = 7b²

Since, a² is divisible by 7, a is also divisible by 7………. (1)

Let a = 7c

On squaring both sides, we get,

a² = 49c²

Substituting a² = 7b²

We get,

7b² = 49c²

b² = 7c²

Since, b² is divisible by 7, b is also divisible by 7……… (2)

From (1) and (2), we can observe that both a and b are divisible by 7

i.e, a and b have a common factor 7

This contradicts our assumption that (a / b) is a rational number

i.e, a and b do not have any common factor other than unity (1)

Hence,

(a / b) is not a rational number

√7 is not a rational number

Therefore,

√7 is an irrational number

10. Write a pair of irrational numbers

(a) (√3 + 5) and (√5 – 3), whose sum is irrational

(b) (√3 + 5) and (4 – √3), whose sum is rational

(c) (√3 + 2) and (√2 – 3), whose difference is irrational

(d) (√5 – 3) and (√5 + 3), whose difference is rational

(e) (5 + √2) and (√5 – 2), whose product is irrational

(f) (√3 + √2) and (√3 – √2), whose product is rational

Solution:

(a) Given

(√3 + 5) and (√5 – 3) are irrational numbers whose sum is irrational

Thus,

We have,

(√3 + 5) + (√5 – 3)

= √3 + 5 + √5 – 3

We get,

= √3 + √5 + 2

which is an irrational number

(b) Given

(√3 + 5) and (4 – √3) are irrational numbers whose sum is rational

Thus,

We have,

(√3 + 5) + (4 – √3)

= √3 + 5 + 4 – √3

We get,

= 9

which is a rational number

(c) Given

(√3 + 2) and (√2 – 3) are irrational numbers whose difference is irrational

Thus,

We have,

(√3 + 2) – (√2 – 3)

= √3 + 2 – √2 + 3

We get,

= √3 – √2 + 5

which is irrational

(d) Given

(√5 – 3) and (√5 + 3) are irrational numbers whose difference is rational

Thus,

We have,

(√5 – 3) – (√5 + 3)

= √5 – 3 – √5 – 3

We get,

= -6

which is a rational number

(e) Given

(5 + √2) and (√5 – 2) are irrational numbers whose product is irrational

Thus,

We have,

(5 + √2) (√5 – 2)

= 5 (√5 – 2) + √2 (√5 – 2)

We get,

= 5√5 – 10 + √10 – 2√2

which is irrational numbers

(f) Given

(√3 + √2) and (√3 – √2) are irrational numbers whose product is rational

Thus,

We have,

(√3 + √2) (√3 – √2)

= (√3)² – (√2)²

= 3 – 2

We get,

= 1

which is a rational number

11. Simplify by rationalizing the denominator in each of the following:

(a) (3√2) / √5

(b) {1 / (5 + √2)}

(c) {1 / (√3 + √2)}

(d) {2 / (3 + √7)}

(e) {5 / (√7 – √2)}

(f) {42 / (2√3 + 3√2)}

(g) {(√3 + 1) / (√3 – 1)}

(h) (√5 – √7) / √3

(i) (3 – √3) / (2 + √2)

Solution:

(a) (3√2) / √5

= {(3√2) / √5} × (√5) / (√5)

On simplification, we get,

= {(3√2) × √5} / (√5)²

We get,

= (3√10) / 5

(b) {1 / (5 + √2)}

= {1 / (5 + √2)} × {(5 – √2) / (5 – √2)}

On simplification, we get,

= (5 – √2) / (5)² – (√2)²

= (5 – √2) / (25 – 2)

We get,

= (5 – √2) / 23

(c) {1 / (√3 + √2)}

= {1 / (√3 + √2)} × (√3 – √2) / (√3 – √2)

On simplification, we get,

= (√3 – √ 2) / (√3)² – (√2)²

= (√3 – √2) / (3 – 2)

= (√3 – √2) / 1

We get,

= (√3 – √2)

(d) {2 / (3 + √7)}

= {2 / (3 + √7)} × (3 – √7) / (3 – √7)

On further calculation, we get,

= {2 (3 – √7)} / (3)² – (√7)²

= {2(3 – √7)} / (9 – 7)

= {2(3 – √7)} / 2

We get,

= (3 – √7)

(e) {5 / (√7 – √2)}

= {5 / (√7 – √2)} × (√7 + √2) / (√7 + √2)

On further calculation, we get,

= {5(√7 + √2)} / (√7)² – (√2)²

= {5(√7 + √2)} / (7 – 2)

= {5(√7 + √2)} / 5

We get,

= (√7 + √2)

(f) {42 / (2√3 + 3√2)}

= {42 / (2√3 + 3√2)} × {(2√3 – 3√2) / (2√3 – 3√2)}

= {42 (2√3 – 3√2)} / (2√3)² – (3√2)²

On further calculation, we get,

= (84√3 – 126√2) / (4 × 3) – (9 × 2)

= (84√3 – 126√2) / (12 – 18)

= (84√3 – 126√2) / -6

= -14√3 + 21√2

= 21√2 – 14√3

We get,

= 7(3√2 – 2√3)

(g) {(√3 + 1) / (√3 – 1)}

= (√3 + 1) / (√3 – 1) × (√3 + 1) / (√3 + 1)

On further calculation, we get,

= (√3 + 1)² / (√3)² – (1)²

= {(√3)² + 2 × √3 × 1 + (1)²} / (3 – 1)

= (3 + 2√3 + 1) / 2

= (4 + 2√3) / 2

We get,

= 2 + √3

(h) (√5 – √7) / √3

= {(√5 – √7) / √3} × (√3) / (√3)

= (√5 × √ 3 – √7 × √ 3) / (√3)²

We get,

= (√15 – √21) / 3

(i) (3 – √3) / (2 + √2)

= (3 – √3) / (2 + √2) × (2 – √2) / (2 – √2)

= {3(2 – √2) – √3 (2 – √2)} / (2)² – (√2)²

On further calculation, we get,

= (6 – 3√2 – 2√3 + √6) / (4 – 2)

= (6 – 3√2 – 2√3 + √6) / 2

12. Simplify by rationalizing the denominator in each of the following:

(i) (5 + √6) / (5 – √6)

(ii) (4 + √8) / (4 – √8)

(iii) (√15 + 3) / (√15 – 3)

(iv) (√7 – √5) / (√7 + √5)

(v) (3√5 + √7) / (3√5 – √7)

(vi) (2√3 – √6) / (2√3 + √6)

(vii) (5√3 – √15) / (5√3 + √15)

(viii) (2√6 – √5) / (3√5 – 2√6)

(ix) (7√3 – 5√2) / (√48 + √18)

(x) (√12 + √18) / (√75 – √50)

Solution:

(i) (5 + √6) / (5 – √6)

= (5 + √6) / (5 – √6) × (5 + √6) / (5 + √6)

On further calculation, we get,

= (5 + √6)² / (5)² – (√6)²

= {(5)² + 2 × 5 × √6 + (√6)²} / (25 – 6)

= (25 + 10√6 + 6) /19

We get,

= (31 + 10√6) / 19

(ii) (4 + √8) / (4 – √8)

= (4 + √8) / (4 – √8) × (4 + √8) / (4 + √8)

On further calculation, we get,

= (4 + √8)² / (4)² – (√8)²

= {(4)² + 2 × 4 × √8 + (√8)²} / (16 – 8)

= (16 + 8√8 + 8) / (16 – 8)

= (24 + 8√8) / 8

We get,

= 3 + √8

(iii) (√15 + 3) / (√15 – 3)

= (√15 + 3) / (√15 – 3) × (√15 + 3) / (√15 + 3)

= (√15 + 3)² / (√15)² – (3)²

= {(√15)² + 2 × √15 × 3 + (3)²} / (15 – 9)

On further calculation, we get,

= (15 + 6√15 + 9) / 6

= (24 + 6√15) / 6

We get,

= 4 + √15

(iv) (√7 – √5) / (√7 + √5)

= (√7 – √5) / (√7 + √5) × (√7 – √5) / (√7 – √5)

= (√7 – √5)² / (√7)² – (√5)²

On further calculation, we get,

= (7 + 5 – 2√35) / (7 – 5)

= (12 – 2√35) / 2

We get,

= 6 – √35

(v) (3√5 + √7) / (3√5 – √7)

= (3√5 + √7) / 3√5 – √7) × (3√5 + √7) / (3√5 + √7)

= (3√5 + √7)² / (3√5)² – (√7)²

On further calculation, we get,

= {(3√5)² + (√7)² + 2 × 3√5 × √7} / (45 – 7)

= (45 + 7 + 6√35) / 38

= (52 + 6√35) / 38

We get,

= (26 + 3√35) / 19

(vi) (2√3 – √6) / (2√3 + √6)

= (2√3 – √6) / (2√3 + √6) × (2√3 – √6) / (2√3 – √6)

= (2√3 – √6)² / (2√3)² – (√6)²

On further calculation, we get,

= {(2√3)² + (√6)² – 2 × 2√3 × √6} / (4× 3 – 6)

= (12 + 6 – 4√18) / (12 – 6)

= (18 – 4√18) / 6

= (9 – 2√18) / 3

= (9 – 6√2) / 3

We get,

= 3 – 2√2

(vii) (5√3 – √15) / (5√3 + √15)

= (5√3 – √15) / 5√3 + √15) × (5√3 – √15) / (5√3 – √15)

= (5√3 – √15)² / (5√3)² – (√15)²

On further calculation, we get,

= (75 + 15 – 10√45) / (75 – 15)

= (90 – 10√45) / 60

= (9 – 1√45) / 6

= (9 – 3√5) / 6

We get,

= (3 – √5) / 2

(viii) (2√6 – √5) / (3√5 – 2√6)

= (2√6 – √5) / (3√5 – 2√6) × (3√5 + 2√6) / (3√5 + 2√6)

On simplification, we get,

= (6√30 + 24 – 15 – 2√30) / (3√5)² – (2√6)²

= (6√30 + 9 – 2√30) / (45 – 24)

We get,

= (4√30 + 9) / 21

(ix) (7√3 – 5√2) / (√48 + √18)

= (7√3 – 5√2) / (√48 + √18) × (√48 – √18) / (√48 – √18)

On simplification, we get,

= (7√144 – 7√54 – 5√96 + 5√36) / (√48)² – (√18)²

= (84 – 21√6 – 20√6 + 30) / (48 – 18)

We get,

= (114 – 41√6) / 30

(x) (√12 + √18) / (√75 – √50)

= (√12 + √18) / (√75 – √50) × (√75 + √50) / (√75 + √50)

On further calculation, we get,

= {(2√3 + 3√2) (5√3 + 5√2)} / (√75)² – (√50)²

= (30 + 10√6 + 15√6 + 30) / (75 – 50)

= (60 + 25√6) / 25

We get,

= (12 + 5√6) / 5

13. Simplify each of the following:

(i) 3 / (5 – √3) + 2 / (5 + √3)

(ii) (4 + √5) / (4 – √5) + (4 – √5) / (4 + √5)

(iii) (√5 – 2) / (√5 + 2) – (√5 + 2) / (√5 – 2)

(iv) (√7 – √3) / (√7 + √3) – (√7 + √3) / (√7 – √3)

(v) (√5 + √3) / √5 – √3) + (√5 – √3) / √5 + √3)

Solution:

(i) 3 / (5 – √3) + 2 / (5 + √3)

= {3(5 + √3) + 2(5 – √3)} / (5 – √3) (5 + √3)

On simplification, we get,

= (15 + 3√3 + 10 – 2√3) / (5)² – (√3)²

= (25 + √3) / (25 – 3)

We get,

= (25 + √3) / 22

(ii) (4 + √5) / (4 – √5) + (4 – √5) / (4 + √5)

= {(4 + √5)² + (4 – √5)²} / (4 – √5) (4 + √5)

On simplification, we get,

= (16 + 5 + 8√5 + 16 + 5 – 8√5) / (4)² – (√5)²

= (21 + 21) / (16 – 5)

We get,

= (42 / 11)

(iii) (√5 – 2) / (√5 + 2) – (√5 + 2) / (√5 – 2)

= (√5 – 2)² – (√5 + 2)²} / (√5 + 2) (√5 – 2)

On simplification, we get,

= (5 + 4 – 4√5 – 5 – 4 – 4√5) / (√5)² – (2)²

= – 8√5 / (5 – 4)

We get,

= – 8√5

(iv) (√7 – √3) / (√7 + √3) – (√7 + √3) / (√7 – √3)

= (√7 – √3)² – (√7 + √3)² / (√7 + √3) (√7 – √3)

On simplification, we get,

= (7 + 3 – 2√21 – 7 – 3 – 2√21) / (√7)² – (√3)²

= – 4√21 / (7 – 3)

= (- 4√21) / 4

We get,

= – √21

(v) (√5 + √3) / √5 – √3) + (√5 – √3) / √5 + √3)

= (√5 + √3)² + (√5 – √3)² / (√5 – √3) (√5 + √3)

On simplification, we get,

= (5 + 3 + 2√15 + 5 + 3 – 2√15) / (5 – 3)

= 16 / 2

We get,

= 8

14. Simplify the following:

(i) √6 / (√2 + √3) + 3√2 / (√6 + √3) – 4√3 / (√6 + √2)

(ii) 3√2 / (√6 – √3) – 4√3 / (√6 – √2) + 2√3 / (√6 + 2)

(iii) 6 / (2√3 – √6) + √6 / (√3 + √2) – 4√3 / (√6 – √2)

(iv) 7√3 / (√10 + √3) – 2√5 / (√6 + √5) – 3√2 / (√15 + 3√2)

(v) 4√3 / (2 – √2) – 30 / (4√3 – 3√2) – 3√2 / (3 + 2√3)

Solution:

(i) √6 / (√2 + √3) + 3√2 / (√6 + √3) – 4√3 / (√6 + √2)

Rationalizing the denominator of each term, we have

= {√6 (√2 – √3) / (√2 + √3) (√2 – √3)} + {(3√2 (√6 – √3) / (√6 + √3) (√6 – √3)}- {(4√3 (√6 – √2) / (√6 + √2) (√6 – √2)}

On further calculation, we get,

= {(√12 – √18) / (2 – 3)} + {(3√12 – 3√6) / (6 – 3)} – {(4√18 – 4√6) / (6 – 2)}

= {(√12 – √18) / -1} + {(3√12 – 3√6) / 3} – {(4√18 – 4√6) / 4}

= √18 – √12 + √12 – √6 – √18 + √6

We get,

= 0

(ii) 3√2 / (√6 – √3) – 4√3 / (√6 – √2) + 2√3 / (√6 + 2)

Rationalizing the denominator of each term, we have,

= {3√2 (√6 + √3) / (√6 – √3) (√6 + √3)} – {(4√3 (√6 + √2) / (√6 – √2) (√6 + √2)} + {(2√3 (√6 – 2) / (√6 + 2) (√6 – 2)}

On further calculation, we get,

= {(3√12 + 3√6) / (6 – 3)} – {(4√18 + 4√6) / (6 – 2)} + {(2√18 – 4√3) / (6 – 4)}

= {(3√12 + 3√6) / 3} – {(4√18 + 4√6) / 4} + {(2√18 – 4√3) / 2}

= √12 + √6 – √18 – √6 + √18 – 2√3

= √12 – 2√3

= 2√3 – 2√3

We get,

= 0

(iii) 6 / (2√3 – √6) + √6 / (√3 + √2) – 4√3 / (√6 – √2)

Rationalizing the denominator of each term, we have

= {6 (2√3 + √6) / (2√3 – √6) (2√3 + √6)} + {(√6 (√3 – √2) / (√3 + √2) (√3 – √2)} – {(4√3 (√6 + √2) / (√6 – √2) (√6 + √2)

On simplification, we get,

= (12√3 + 6√6) / (12 – 6) + (√18 – √12) / (3 – 2) – (4√18 + 4√6) / (6 – 2)

= {(12√3 + 6√6) / 6} + {(√18 – √12) / 1} – {(4√18 + 4√6) / 4}

= 2√3 + √6 + √18 – √12 – √18 – √6

= 2√3 – √12

= 2 √3 – 2√3

We get,

= 0

(iv) 7√3 / (√10 + √3) – 2 √5 / (√6 + √5) – 3√2 / (√15 + 3√2)

Rationalizing the denominator of each term, we have,

= {7√3 (√10 – √3) / (√10 + √3) (√10 – √3)} – {2√5 (√6 – √5) / (√6 + √5) (√6 – √5)} – { 3√2 (√15 – 3√2) / (√15 + 3√2) (√15 – 3√2)

= {(7√30 – 21) / (10 – 3)} – {(2√30 – 10) / (6 – 5)} – {(3√30 – 18)/ (15 – 18)}

= (7√30 – 21) / 7 – (2√30 – 10) / 1 – (3√30 – 18) / -3

= (7√30 – 21) / 7 – (2√30 – 10) / 1 + (3√30 – 18) / 3

= √30 – 3 – 2√30 + 10 + √30 – 6

We get,

= 1

(v) 4√3 / (2 – √2) – 30 / (4√3 – 3√2) – 3√2 / (3 + 2√3)

Rationalizing the denominator of each term, we have,

= {(4√3 (2 + √2) / (2 – √2) (2 + √2)} – {30 (4√3 + 3√2) / (4√3 – 3√2) (4√3 + 3√2)} – {(3√2 (3 – 2√3) / (3 + 2√3) (3 – 2√3)

= {(8√3 + 4√6) / (4 – 2)} – {(120√3 + 90√2) / (48 – 18)} – {(9√2 – 6√6) / (9 – 12)}

= {(8√3 + 4√6) / 2} – {120√3 + 90√2) / 30} – {(9√2 – 6√6) / -3}

= {(8√3 + 4√6) / 2} – {(120√3 + 90√2) / 30} + {(9√2 – 6√6) / 3}

= 4√3 + 2√6 – 4√3 – 3√2 + 3√2 – 2√6

We get,

= 0

15. If (√2.5 – √0.75) / (√2.5 + √0.75) = p + q√30, find the values of p and q.

Solution:

Given

(√2.5 – √0.75) / (√2.5 + √0.75)

= {(√2.5 – √0.75) / (√2.5 + √0.75)} × {(√2.5 – √0.75) / (√2.5 – √0.75)}

= (√2.5 – √0.75)² / (√2.5)² – (√0.75)²

On simplification, we get,

= (2.5 – 2 × √2.5 × √0.75 + 0.75) / (2.5 – 0.75)

= (3.25 – 2 × √0.25 × 10 × √0.25 × 3) / 1.75

= (3.25 – 2 × 0.25√30) / 1.75

= (3.25 – 0.5√30) / 1.75

= (3.25) / (1.75) – (0.5) / (1.75) √30

= (325 / 175) – (50 / 175) √30

= (13 / 7) – (2 / 7) √30

= (13 / 7) + (- 2 / 7) √30

= p + q√30

Therefore,

The value of p = (13 / 7) and q = (- 2 / 7)