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ICSE Class 9 Maths Selina Solutions Chapter 4 - Expansions

ICSE Class 9 Maths Selina Solutions Chapter 4 are provided here which have been prepared by our PW experts. Get the details of the ICSE Class 9 Maths Selina Solutions Chapter 4 Overview in the below article.
authorImagePriyanka Dahima15 Jul, 2024
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ICSE Class 9 Maths Selina Solutions Chapter 4

ICSE Class 9 Maths Selina Solutions Chapter 4: Here we provide you with detailed ICSE Class 9 Maths Selina Chapter 4 Expansion. Our PW Subject experts have prepared these questions as per the latest syllabus prescribed by CISCE for ICSE Class 9 Maths Exam. Selina's solutions are considered very useful for ICSE class 9 maths exam preparation. For better and more strategic Preparation in the correct direction our blog will act as a pathfinder for the candidates going to appear for the ICSE Class 9 Maths Exam 2024.

ICSE Class 9 Maths Selina Solutions Chapter 4 Expansion Overview

In mathematics, the expansion of a product of sums is expressed as the sum of its products using the fact that multiplication divides addition - the reverse process of trying to write an expanded polynomial as a product is called polynomial factors. "Expand" means to remove ( ) ... but we have to do it right ( ) are called "parentheses" or "brackets". Anything inside ( ) should be treated as "package". So to say: tell everything inside the "package". Physics Wallah ICSE Class 9 Maths Selina Solutions Chapter 4 Expansion is prepared by PW subject experts for clear understanding. By using these ICSE Class 9 Maths Selina Solutions Chapter 4 created by experts, students can improve their math skills, deepen their understanding of the subject and achieve better results in their exams.

ICSE Class 9 Maths Selina Solutions Chapter 4 PDF

In our article we have provided the ICSE Class 9 Maths Selina Solution Chapter 4 PDF. Here you can find the ICSE Class 9 Maths Selina Solutions Chapter 4 PDF file that can be downloaded and viewed online as we have attached below the direct download link for easy and convenient access for the candidates. Students can use and download these Selina solutions for free to practice offline.

ICSE Class 9 Maths Selina Solutions Chapter 4 PDF

ICSE Class 9 Maths Selina Solutions Chapter 4 Expansion

Here we have provided ICSE Class 9 Maths Selina Solutions Chapter 4 Expansion for the ease of students so that they can prepare better for their ICSE Class 9 Maths Exam.

ICSE Class 9 Maths Selina Solutions Chapter 4 Exercise 4(A)

1. Find the square of:

(i) 2a + b

(ii) 3a + 7b

(iii) 3a – 4b

(iv) 3a/2b – 2b/3a

Solution:

Using the identities, (a + b) 2 = a 2 + b 2 + 2ab and (a – b) 2 = a 2 + b 2 – 2ab (i) (2a + b) 2 = (2a) 2 + b 2 + 2(2a)(b) = 4a 2 + b 2 + 4ab (ii) (3a + 7b) 2 = (3a) 2 + (7b) 2 + 2(3a)(7b) = 9a 2 + 49b 2 + 42ab (iii) (3a – 4b) 2 = (3a) 2 + (4b) 2 – 2(3a)(4b) = 9a 2 + 16b 2 – 24ab (iv) (3a/2b – 2b/3a) 2 = (3a/2b) 2 + (2b/3a) 2 – 2(3a/2b)(2b/3a) = 9a 2 /4b 2 + 4b 2 /9a 2 – 2

2. Use identities to evaluate:

(i) (101) 2

(ii) (502) 2

(iii) (97) 2

(iv) (998) 2

Solution:

Using the identities, (a + b) 2 = a 2 + b 2 + 2ab and (a – b) 2 = a 2 + b 2 – 2ab (i) (101) 2 = (100 + 1) 2 = 100 2 + 1 2 + 2×100×1 = 10000 + 1 + 200 = 10201 (ii) (502) 2 = (500 + 2) 2 = 500 2 + 2 2 + 2×500×2 = 250000 + 4 + 2000 = 252004

(iii) (97) 2 = (100 – 3) 2

= 100 2 + 3 2 – 2×100×3 = 10000 + 9 – 600 = 9409 (iv) (998) 2 = (1000 – 2) 2 = 1000 2 + 2 2 – 2×1000×2 = 100000 + 4 – 4000 = 996004

3. Evalute:

(i) (7x/8 + 4y/5) 2

(ii) (2x/7 – 7y/4) 2

Solution:

Using the identities, (a + b) 2 = a 2 + b 2 + 2ab and (a – b) 2 = a 2 + b 2 – 2ab (i) We have, (7x/8 + 4y/5) 2 = (7x/8) 2 + (4y/5) 2 + 2 × (7x/8) × (4y/5) = 49x 2 /64 + 16y 2 /25 + 7xy/5 (ii) We have, (2x/7 – 7y/4) 2 = (2x/7) 2 + (7y/4) 2 – 2 × (2x/7) × (7y/4) = 4x 2 /49 + 49y 2 /16 – xy

4. Evaluate:

(i) (a/2b + 2b/a) 2 – (a/2b – 2b/a) 2 – 4

(ii) (4a + 3b) 2 – (4a – 3b) 2 + 48ab

Solution:

Using the identities, (a + b) 2 = a 2 + b 2 + 2ab and (a – b) 2 = a 2 + b 2 – 2ab (i) Given expression, Selina Solutions Icse Class 9 Mathematics Chapter - Expansion 1 On expanding the term using the identity, we have (a/2b + 2b/a) 2 = (a/2b) 2 + (2b/a) 2 + 2 × (a/2b) × (2b/a) = a 2 /4b + 4b 2 /a 2 + 2 Next, expanding the second term using the identity, we have (a/2b – 2b/a) 2 = (a/2b) 2 + (2b/a) 2 – 2 × (a/2b) × (2b/a) = a 2 /4b + 4b 2 /a 2 – 2 Now, using these results in the given expression (a/2b + 2b/a) 2 – (a/2b – 2b/a) 2 – 4 = (a 2 /4b + 4b 2 /a 2 + 2) – (a 2 /4b + 4b 2 /a 2 – 2) – 4 = a 2 /4b + 4b 2 /a 2 + 2 – a 2 /4b – 4b 2 /a 2 + 2 – 4 = 0 (ii) Given expression, (4a + 3b) 2 – (4a – 3b) 2 + 48ab On expanding the term using the identity, we have (4a + 3b) 2 = (4a) 2 + (3b) 2 + 2 × (4a) × (3b) = 16a 2 + 9b 2 + 24ab Next, expanding the second term using the identity, we have (4a – 3b) 2 = (4a) 2 + (3b) 2 – 2 × (4a) × (3b) = 16a 2 + 9b 2 – 24ab Now, using these results in the given expression (4a + 3b) 2 – (4a – 3b) 2 + 48ab = (16a 2 + 9b 2 + 24ab) – (16a 2 + 9b 2 – 24ab) + 48ab = 16a 2 + 9b 2 + 24ab – 16a 2 – 9b 2 + 24ab + 48ab = 96ab

5. If a + b = 7 and ab = 10; find a – b.

Solution:

Using the identities, (a + b) 2 = a 2 + b 2 + 2ab and (a – b) 2 = a 2 + b 2 – 2ab Rewriting the above equation as (a – b) 2 = a 2 + b 2 + 2ab – 4ab = (a + b) 2 – 4ab … (i) We have, a + b = 7 and ab = 10 So, using these in equation (i), we get (a – b) 2 = (7) 2 – (4 × 10) = 49 – 40 = 9 Then, (a – b) = √9 = ±3

6. If a – b = 7 and ab = 18; find a + b.

Solution:

Using the identities, (a – b) 2 = a 2 + b 2 – 2ab and (a + b) 2 = a 2 + b 2 + 2ab Rewriting the above equation as (a + b) 2 = a 2 + b 2 – 2ab + 4ab = (a – b) 2 + 4ab … (i) We have, a – b = 7 and ab = 18 So, using these in equation (i), we get (a + b) 2 = (7) 2 + (4 × 18) = 49 + 72 = 121 Then, (a + b) = √121 = ±11

7. If x + y = 7/2 and xy = 5/2; find:

(i) x – y

(ii) x 2 – y 2

Solution:

Using the identities, (x + y) 2 = x 2 + y 2 + 2xy and (x – y) 2 = x 2 + y 2 – 2xy Rewriting the above equation as (x – b) 2 = x 2 + y 2 + 2xy – 4xy = (x + y) 2 – 4xy … (1) We have, x + y = 7/2 and xy = 5/2 So, using these in equation (1), we get (x – y) 2 = (7/2) 2 – (4 × 5/2) = 49/4 – 10 = 9/4 Then, (x – y) = √(9/4) = ± 3/2 … (2)

(ii) We know that, x 2 – y 2 = (x + y) (x – y) Substituting values in RHS using given and (2), we get x 2 – y 2 = (7/2) (± 3/2) = ± 21/4

8. If a – b = 0.9 and ab = 0.36; find:

(i) a + b

(ii) a 2 – b 2

Solution:

Using the identities, (a – b) 2 = a 2 + b 2 – 2ab and (a + b) 2 = a 2 + b 2 + 2ab Rewriting the above equation as (a + b) 2 = a 2 + b 2 – 2ab + 4ab = (a – b) 2 + 4ab … (1) We have, a – b = 0.9 and ab = 0.36 So, using these in equation (1), we get (a + b) 2 = (0.9) 2 + (4 × 0.36) = 0.81 + 1.44 = 2.25 Then, (a + b) = √2.25 = ±1.5 … (2) (ii) We know that, a 2 – b 2 = (a + b) (a – b) Substituting values in RHS using given and (2), we get a 2 – b 2 = (±1.5) (0.9) = ± 1.35

9. If a – b = 4 and a + b = 6; find

(i) a 2 + b 2

(ii) ab

Solution:

Given, a – b = 4 and a + b = 6 We know that, (a – b) 2 = a 2 + b 2 – 2ab and (a + b) 2 = a 2 + b 2 + 2ab Now, rewriting the above equation as (a + b) 2 = a 2 + b 2 – 2ab + 4ab ⇒ (a + b) 2 = (a – b) 2 + 4ab Substituting the values in the above equation, we get (6) 2 = (4) 2 + 4ab 36 = 16 + 4ab 4ab = 36 – 16 4ab = 20 ab = 20/4 (ii) Thus, ab = 5 Now, in the identity: (a + b) 2 = a 2 + b 2 + 2ab (a + b) 2 = (a 2 + b 2 ) + 2ab Let’s substitute the values of the known terms, (6) 2 = (a 2 + b 2 ) + 2 × (5) 36 = (a 2 + b 2 ) + 10 a 2 + b 2 = 36 – 10 (i) Thus, a 2 + b 2 = 26

10. If a + 1/a = 6 and a ≠ 0 find:

(i) a – 1/a

(ii) a 2 – 1/a 2

Solution:

Using the identities, (a – b) 2 = a 2 + b 2 – 2ab and (a + b) 2 = a 2 + b 2 + 2ab (i) Now, (a + 1/a) 2 = a 2 + 1/a 2 + (2 × a × 1/a) = a 2 + 1/a 2 + 2 … (1) Substituting the value of (a + 1/a) in the equation (1), we get 6 2 = a 2 + 1/a 2 + 2 36 = a 2 + 1/a 2 + 2 a 2 + 1/a 2 = 36 – 2 = 34 … (2) Similarly, (a – 1/a) 2 = a 2 + 1/a 2 – (2 × a × 1/a) = (a 2 + 1/a 2 ) – 2 = 34 – 2 … [From (2)] = 32 ⇒ (a – 1/a) 2 = 32 a – 1/a = ±√32 = ±4√2 … (3) Thus, a – 1/a = ±4√2 (ii) We know that, a 2 – 1/a 2 = (a – 1/a) (a + 1/a) Using the given and (3) in the above equation, a 2 – 1/a 2 = (±4√2) (6) = ±24√2 Thus, a 2 – 1/a 2 = ±24√2

11. If a – 1/a = 8 and a ≠ 0, find:

(i) a + 1/a

(ii) a 2 – 1/a 2

Solution:

Using the identities, (a – b) 2 = a 2 + b 2 – 2ab and (a + b) 2 = a 2 + b 2 + 2ab (i) Now, (a – 1/a) 2 = a 2 + 1/a 2 – (2 × a × 1/a) = a 2 + 1/a 2 – 2 … (1) Substituting the value of (a + 1/a) in the equation (1), we get 8 2 = a 2 + 1/a 2 – 2 64 = a 2 + 1/a 2 – 2 a 2 + 1/a 2 = 64 + 2 = 66 … (2) Similarly, (a + 1/a) 2 = a 2 + 1/a 2 + (2 × a × 1/a) = (a 2 + 1/a 2 ) + 2 = 66 + 2 … [From (2)] = 68 ⇒ (a + 1/a) 2 = 68 a + 1/a = √68 = ±2√17 … (3) Thus, a + 1/a = ±2√17 (ii) We know that, a 2 – 1/a 2 = (a – 1/a) (a + 1/a) Using the given and (3) in the above equation, a 2 – 1/a 2 = (8) (±2√17) = ±16√17 Thus, a 2 – 1/a 2 = ±16√17

12. If a 2 – 3a + 1 = 0, and a ≠ 0; find:

(i) a + 1/a

(ii) a 2 + 1/a 2

Solution:

(i) Given equation, a 2 – 3a + 1 = 0 a 2 + 1 = 3a (a 2 + 1)/a = 3 ⇒ a + 1/a = 3 … (1) (ii) We know that, (a + b) 2 = a 2 + b 2 + 2ab Now, (a + 1/a) 2 = a 2 + 1/a 2 + 2(a)(1/a) = a 2 + 1/a 2 + 2 Using (1) in the above equation, we get (3) 2 = a 2 + 1/a 2 + 2 9 = a 2 + 1/a 2 + 2 a 2 + 1/a 2 = 9 – 2 Thus, a 2 + 1/a 2 = 7

13. If a 2 – 5a – 1 = 0 and a ≠ 0; find:

(i) a – 1/a

(ii) a + 1/a

(iii) a 2 – 1/a 2

Solution:

(i) Given, a 2 – 5a – 1 = 0 Rewriting the equation, we get a 2 – 1 = 5a (a 2 – 1)/a = 5 Hence, a – 1/a = 5 … (1) (ii) We know that, (a + 1/a) 2 = a 2 + 1/a 2 + 2 Manipulating the above as, (a + 1/a) 2 = a 2 + 1/a 2 – 2 + 4 (a + 1/a) 2 = (a – 1/a) 2 + 4 Now, using (1) in the above (a + 1/a) 2 = (5) 2 + 4 (a + 1/a) 2 = 25 + 4 = 29 Hence, a + 1/a = ±√29 … (2) (iii) We know that, a 2 – 1/a 2 = (a + 1/a) (a – 1/a) Now, using (1) and (2) in the above equation, we get a 2 – 1/a 2 = (5) x (±√29) Hence, a 2 – 1/a 2 = ±5√29

14. If 3a + 4b = 16 and ab = 4; find the value of 9a 2 + 16b 2 .

Solution:

Given, 3a + 4b = 16 and ab = 4 Required to find: value of 9a 2 + 16b 2 We know that, (a + b) 2 = a 2 + b 2 + 2ab Now, the square of (3a + 4b) will be (3a + 4b) 2 = (3a) 2 + (4b) 2 + 2 × (3a) × (4b) = 9a 2 + 4b 2 + 24ab And, given 3a + 4b = 16 So, by squaring on both the sides (3a + 4b) 2 = 16 2 9a 2 + 4b 2 + 24ab = 256 9a 2 + 4b 2 + 24(4) = 256 [Given ab = 4] 9a 2 + 4b 2 = 256 – 96 ⇒ 9a 2 + 4b 2 = 160

15. The number a is 2 more than the number b. If the sum of the squares of a and b is 34, then find the product of a and b.

Solution:

Given, a is 2 more than b ⇒ a = b + 2 And, sum of squares of a and b is 34 ⇒ a 2 + b 2 = 34 Let’s replace a = (b + 2) in the above equation and solve for b Then, (b + 2) 2 + b 2 = 34 2b 2 + 4b – 30 = 0 b 2 + 2b – 15 = 0 (b + 5) (b – 3) = 0 So, b = -5 or 3 Now, For b = -5, a =-5 + 2 = -3 For b = 3, a = 3 + 2 = 5 Thus, the product of a and b is 15 in both cases.

16. The difference between two positive numbers is 5 and the sum of their squares is 73. Find the product of these numbers.

Solution:

Let’s assume the two positive numbers as a and b Given, the difference between them is 5 and the sum of their squares is 73 So, we have a – b = 5 … (i) and a 2 + b 2 = 73 … (ii) On squaring (i) on both sides, we get (a – b) 2 = 5 2 (a 2 + b 2 ) – 2ab = 25 73 – 2ab = 25 … [Using (ii), given] So, 2ab = 73 – 25 = 48 ab = 24 Therefore, the product of numbers is 24.

Exercise 4(B)

1. Find the cube of:

(i) 3a – 2b

(ii) 5a + 3b

(iii) 2a + 1/2a

(iv) 3a – 1/a (a ≠ 0)

Solution:

Using the identities, (a + b) 3 = a 3 + 3ab (a + b) + b 3 and (a – b) 3 = a 3 – 3ab (a – b) + b 3 (i) (3a – 2b) 3 = (3a) 3 – 3 × 3a × 2b (3a – 2b) – (2b) 3 = 27a 3 – 18ab (3a – 2b) – 8b 3 = 27a 3 – 54a 2 b + 36ab 2 – 8b 3 (ii) (5a + 3b) 3 = (5a) 3 + 3 × 5a × 3b (5a + 3b) + (3b) 3 = 125a 3 + 45ab (5a + 3b) + 27b 3 = 125a 3 + 225a 2 b + 135ab 2 + 27b 3 (iii) (2a + 1/2a) 3 = (2a) 3 + 3 × 2a × 1/2a (2a + 1/2a) + (1/2a) 3 = 8a 3 + 3 (2a + 1/2a) + 1/8a 3 = 8a 3 + 6a + 3/2a + 1/8a 3 (iv) (3a – 1/a) 3 = (3a) 3 – 3 × 3a × 1/a (3a – 1/a) – (1/a) 3 = 27a 3 – 9 (3a – 1/a) – 1/a 3 = 27a 3 – 27a + 9a – 1/a 3

2. If a 2 + 1/a 2 = 47 and a ≠ 0 find:

(i) a + 1/a

(ii) a 3 + 1/a 3

Solution:

(i) Given, a 2 + 1/a 2 = 47 We know that, (a + 1/a) 2 = a 2 + 1/a 2 + 2 x a x 1/a = (a 2 + 1/a 2 ) + 2 = 47 + 2 = 49 So, a + 1/a = √49 = ±7 … (1) (ii) Using the identity (a + b) 3 = a 3 + 3ab (a + b) + b 3 Now, (a + 1/a) 3 = a 3 + 1/a 3 + 3(a + 1/a) a 3 + 1/a 3 = (a + 1/a) 3 – 3(a + 1/a) = (±7) 3 – 3(±7) … [From (1)] = ±343 – ±21 Hence, a 3 + 1/a 3 = ±322

3. If a 2 + 1/a 2 = 18; a ≠ 0 find:

(i) a – 1/a

(ii) a 3 – 1/a 3

Solution:

(i) Given, a 2 + 1/a 2 = 18 Using the identity (a + b) 2 = a 2 + b 2 + 2ab Now, (a – 1/a) 2 = a 2 + 1/a 2 – 2(a)(1/a) = (a 2 + 1/a 2 ) – 2 = 18 – 2 = 16 Hence, a – 1/a = √16 = ±4 … (1) (ii) Using the identity, (a – b) 3 = a 3 – 3ab (a – b) + b 3 Now, (a – 1/a) 3 = a 3 – 3a(1/a) (a – 1/a) + (1/a) 3 = a 3 – 3 (a – 1/a) + 1/a 3 a 3 + 1/a 3 = (a – 1/a) 3 + 3 (a – 1/a) = (±4) 3 + 3(±4) = ±64 ± 12 Hence, a 3 + 1/a 3 = ±76

4. If a + 1/a = p and a ≠ 0; then show that:

a 3 + 1/a 3 = p (p 2 – 3)

Solution:

Given, a + 1/a = p … (1) Now, cubing on both sides (a + 1/a) 3 = p 3 a 3 + 1/a 3 + 3(a + 1/a) = p 3 a 3 + 1/a 3 = p 3 – 3(a + 1/a) = p 3 – 3(p) [From (1)] = p (p 2 – 3) – Hence proved

5. If a + 2b = 5; then show that:

a 3 + 8b 3 + 30ab = 125.

Solution:

Given, a + 2b = 5 Let’s cube it on both sides, (a + 2b) 3 = 5 3 a 3 + 3(a)(2b)(a + 2b) + (2b) 3 = 125 a 3 + 6ab(a + 2b) + 8b 3 = 125 a 3 + 8b 3 = 125 – 6ab (a + 2b) = 125 – 6ab (5) … [Given] = 125 – 30ab So, a 3 + 8b 3 + 30ab = 125 – Hence showed

6. If (a + 1/a) 2 = 3 and a ≠ 0, then show: a 3 + 1/a 3 = 0.

Solution:

Given, (a + 1/a) 2 = 3 ⇒ a + 1/a = ±√3 … (1) We know the identity, (a + 1/a) 3 = a 3 + 1/a 3 + 3(a + 1/a) a 3 + 1/a 3 = (a + 1/a) 3 – 3(a + 1/a) = (±√3) 3 – 3(±√3) = ±3√3 – (±3√3) = 0 Thus, a 3 + 1/a 3 = 0

7. If a + 2b + c = 0; then show that:

a 3 + 8b 3 + c 3 = 6abc

Solution:

We have, a + 2b + c = 0 a + 2b = -c Now, on cubing it on both sides we get (a + 2b) 3 = (-c) 3 a 3 + (2b) 3 + 3(a)(2b)(a + 2b) = -c 3 a 3 + 8b 3 + 6ab (a + 2b) = -c 3 a 3 + 8b 3 + 6ab (-c) = -c 3 a 3 + 8b 3 – 6abc = -c 3 Hence, a 3 + 8b 3 + c 3 = 6abc

8. Use property to evaluate:

(i) 13 3 + (-8) 3 + (-5) 3

(ii) 7 3 + 3 3 + (-10) 3

(iii) 9 3 – 5 3 – 4 3

(iv) 38 3 + (-26) 3 + (-12) 3

Solution:

The property is if a + b + c = 0 then a 3 + b 3 + c 3 = 3abc Now, (i) a = 13, b = -8 and c = -5 ⇒ 13 3 + (-8) 3 + (-5) 3 = 3(13) (-8) (-5) … [Since, 13 + (-8) + (-5) = 0] = 1560   (ii) a = 7, b = 3, c = -10 ⇒ 7 3 + 3 3 + (-10) 3 = 3(7) (3) (-10) … [Since, 7 + 3 + (-10) = 0] = -630 (iii)a = 9, b = -5, c = -4 ⇒ 9 3 – 5 3 – 4 3 = 9 3 + (-5) 3 + (-4) 3 … [Since, 9 + (-5) + (-4) = 0] = 3(9) (-5) (-4) = 540   (iv) a = 38, b = -26, c = -12 ⇒ 38 3 + (-26) 3 + (-12) 3 = 3(38) (-26) (-12) … [Since, 38 + (-26) + (-12) = 0] = 35568

9. If a ≠ 0 and a – 1/a = 3; find:

(i) a 2 + 1/a 2

(ii) a 3 – 1/a 3

Solution:

(i) We have, a – 1/a = 3 On squaring on both sides, we get (a – 1/a) 2 = 3 2 a 2 + 1/a 2 – 2 = 9 a 2 + 1/a 2 = 9 + 2 Hence, a 2 + 1/a 2 = 11   (ii) We have, a – 1/a = 3 On cubing on both sides, we get (a – 1/a) 3 = 3 3 a 2 – 1/a 3 – 3(a – 1/a) = 27 a 2 – 1/a 3 = 27 + 3(a – 1/a) = 27 + 3(3) = 27 + 9 Hence, a 3 – 1/a 3 = 36

10. If a ≠ 0 and a – 1/a = 4; find:

(i) a 2 + 1/a 2

(ii) a 4 + 1/a 4

(iii) a 3 – 1/a 3

Solution:

(i) We have, a – 1/a = 4 … (a) On squaring it on both sides, we get (a – 1/a) 2 = 4 2 a 2 + 1/a 2 – 2(a)(1/a) = 16 a 2 + 1/a 2 – 2 = 16 a 2 + 1/a 2 = 16 + 2 = 18 … (1) Hence, a 2 + 1/a 2 = 18 (ii) Now, we know that a 4 + 1/a 4 = (a 2 + 1/a 2 ) 2 – 2 = 18 2 – 2 … [From (1)] = 324 – 2 Hence, a 4 + 1/a 4 = 322 (iii) On cubing (i) on both sides, we get (a – 1/a) 3 = 4 3 a 3 – 1/a 3 – 3(a – 1/a) = 64 a 3 – 1/a 3 = 64 + 3(a – 1/a) = 64 + 3(4) … [Given] = 64 + 12 Hence, a 3 – 1/a 3 = 76

11. If x ≠ 0 and x + 1/x = 2; then show that:

x 2 + 1/x 2 = x 3 + 1/x 3 = x 4 + 1/x 4

Solution:

We have, x + 1/x = 2 We know that, (x + 1/x) 2 = x 2 + 1/x 2 + 2 (2) 2 = x 2 + 1/x 2 + 2 x 2 + 1/x 2 = 4 – 2 = 2 … (i) Next, calculating (x + 1/x) 3 = x 3 + 1/x 3 + 3(x + 1/x) (2) 3 = x 3 + 1/x 3 + 3(2) x 3 + 1/x 3 = 2 3 – 3(2) = 8 – 6 = 2 … (ii) Next, we know that x 4 + 1/x 4 = (x 2 + 1/x 2 ) – 2 = 2 2 – 2 … [From (i)] = 4 – 2 = 2 … (iii) Therefore, from (i), (ii) and (iii) we have x 2 + 1/x 2 = x 3 + 1/x 3 = x 4 + 1/x 4

12. If 2x – 3y = 10 and xy = 16; find the value of 8x 3 – 27y 3 .

Solution:

Given, 2x – 3y = 10 … (i) and xy = 16 … (ii) Now, on cubing (i) on both sides (2x – 3y) 3 = 10 3 (2x) 3 – 3(2x)(3y) (2x – 3y) – (3y) 3 = 1000 [] 8x 3 – 18(xy) (2x – 3y) – 27y 3 = 1000 8x 3 – 18 × 16 × 10 – 27y 3 = 1000 8x 3 – 2880 – 27y 3 = 1000 8x 3 – 27y 3 = 1000 + 2880 8x 3 – 27y 3 = 3880

13. Expand:

(i) (3x + 5y + 2z) (3x – 5y + 2z)

(ii) (3x – 5y – 2z) (3x – 5y + 2z)

Solution:

(i) We have, (3x + 5y + 2z) (3x – 5y + 2z) = {(3x + 2z) + (5y)} {(3x + 2z) – (5y)} … [By grouping] = (3x + 2z) 2 – (5y) 2 … [As (a + b) (a – b) = a 2 – b 2 ] = 9x 2 + 4z 2 + (2 × 3x × 2z) – 25y 2 = 9x 2 + 4z 2 + 12xz – 25y 2 = 9x 2 + 4z 2 – 25y 2 + 12xz   (ii) We have, (3x – 5y – 2z) (3x – 5y + 2z) = {(3x – 5y) – (2z)} {(3x – 5y) + (2z)} … [By grouping] = (3x – 5y) 2 – (2z) 2 … [As (a + b) (a – b) = a 2 – b 2 ] = 9x 2 + 25y 2 – 2 × 3x × 5y – 4z 2 = 9x 2 + 25y 2 – 30xy – 4z 2 = 9x 2 +25y 2 – 4z 2 – 30xy

14. The sum of two numbers is 9 and their product is 20. Find the sum of their

(i) Squares (ii) Cubes

Solution:

Given, the sum of two numbers is 9 and their product is 20 Let’s assume the numbers to ‘a’ and ‘b’ So, we have a + b = 9 … (1) and ab = 20 … (2) Now, On squaring (1) on both sides gives, we get (a + b) 2 = 9 2 a 2 + b 2 + 2ab = 81 a 2 + b 2 + 2(20) = 81 … [From (2)] a 2 + b 2 + 40 = 81 a 2 + b 2 = 81 – 40 = 41 (i) Hence, the sum of their squares is 41 Next, On cubing (1) on both sides, we get (a + b) 3 = 9 3 a 3 + b 3 + 3ab (a + b) = 729 a 3 + b 3 + 3 × (20) × (9) = 729 … [From (1) and (2)] a 3 + b 3 = 729 – 540 = 189 (ii) Hence, the sum of their cubes is 189.

15. Two positive numbers x and y are such that x > y. If the difference of these numbers is 5 and their product is 24, find:

(i) Sum of these numbers

(ii) Difference of their cubes

(iii) Sum of their cubes.

Solution:

Given x – y = 5 and xy = 24 (x>y) (x + y) 2 = (x – y) 2 + 4xy = 25 + 96 = 121 So, x + y = 11; sum of these numbers is 11. Cubing on both sides gives (x – y) 3 = 5 3 x 3 – y 3 – 3xy(x – y) = 125 x 3 – y 3 – 72(5) = 125 x 3 – y 3 = 125 + 360 = 485 So, difference of their cubes is 485. Cubing both sides, we get (x + y) 3 = 11 3 x 3 + y 3 + 3xy(x + y) = 1331 x 3 + y 3 = 1331 – 72(11) = 1331 – 792 = 539 So, sum of their cubes is 539.

16. If 4x 2 + y 2 = a and xy = b, find the value of 2x + y.

Solution:

Given, xy = b … (i) and 4x 2 + y 2 = a … (ii) Now, (2x + y) 2 = (2x) 2 + 4xy + y 2 = (4x 2 + y 2 ) + 4xy = a + 4b … [Using (i) and (ii)] Hence, 2x + y = ±√(a + 4b)

Exercise 4(C)

1. Expand:

(i) (x + 8) (x + 10)

(ii) (x + 8) (x – 10)

(iii) (x – 8) (x + 10)

(iv) (x – 8) (x – 10)

Solution:

Using the identity, (x + a) (x + b) = x 2 + (a + b) x + ab (i) We have, (x + 8) (x + 10) = x 2 + (8 + 10) x + 8 × 10 = x 2 + 18x + 80 (ii) We have, (x + 8) (x – 10) = x 2 + (8 – 10) x + 8 × (-10) = x 2 – 2x – 80 (iii) (ii) We have, (x – 8) (x + 10) = x 2 + (-8 + 10) x + (-8) × 10 = x 2 + 2x – 80 (iv) We have, (x – 8) (x – 10) = x 2 + (-8 – 10) x + (-8) × (-10) = x 2 – 18x + 80

2. Expand:

(i) (2x – 1/x) (3x + 2/x)

(ii) (3a + 2/b) (2a – 3/b)

Solution:

(i) We have, (2x – 1/x) (3x + 2/x) = (2x)(3x) + (2x)(2/x) – (1/x)(3x) – (1/x)(2/x) = 6x 2 + 4 – 3 – 2/x 2 = 6x 2 + 1 – 2/x 2 (ii) We have, (3a + 2/b) (2a – 3/b) = (3a)(2a) – (3a)(3/b) + (2/b)(2a) – (2/b)(3/b) = 6a 2 – 9a/b + 4a/b – 6/b 2 = 6a 2 – 5a/b – 6/b 2

3. Expand:

(i) (x + y – z) 2

(ii) (x – 2y + 2) 2

(iii) (5a – 3b + c) 2

(iv) (5x – 3y – 2) 2

(v) (x – 1/x + 5) 2

Solution:

(i) (x + y – z) 2 = x 2 + y 2 + z 2 + 2(x)(y) – 2(y)(z) – 2(z)(x) = x 2 + y 2 + z 2 + 2xy – 2yz – 2zx (ii) (x – 2y + 2) 2 = x 2 + (-2y) 2 + 2 2 + 2(x)(-2y) + 2(-2y)(2) + 2(2)(x) = x 2 + 4y 2 + 4 – 4xy – 8y + 4x (iii) (5a – 3b + c) 2 = (5a) 2 + (-3b) 2 + c 2 + 2(5a)(-3b) + 2(-3b)(c) + 2(c)(5a) = 25a 2 + 9b 2 + c 2 – 30ab – 6bc + 10ac (iv) (5x – 3y – 2) 2 = (5x) 2 + (-3y) 2 + (-2) 2 + 2(5x)(-3y) + 2(-3y)(-2) + 2(-2)(5x) = 25x 2 + 9y 2 + 4 – 30xy + 12y – 20x (v) (x – 1/x + 5) 2 = (x) 2 + (-1/x) 2 + (5) 2 + 2(x)(-1/x) + 2(-1/x)(5) + 2(5)(x) = x 2 + 1/x 2 + 25 – 2 – 10/x + 10x = x 2 + 1/x 2 + 23 – 10/x + 10x

4. If a + b + c = 12 and a 2 + b 2 + c 2 = 50; find ab + bc + ca.

Solution:

Given, a + b + c = 12 and a 2 + b 2 + c 2 = 50 We know that, (a + b + c) 2 = a 2 + b 2 + c 2 + 2(ab + bc + ca) 12 2 = 50 + 2(ab + bc + ca) 144 = 50 + 2(ab + bc + ca) ab + bc + ca = (144 – 50)/ 2 = 94/2 Thus, ab + bc + ca = 47

5. If a 2 + b 2 + c 2 = 35 and ab + bc + ca = 23; find a + b + c.

Solution:

Given, a 2 + b 2 + c 2 = 35 and ab + bc + ca = 23 We know that, (a + b + c) 2 = (a 2 + b 2 + c 2 ) + 2(ab + bc + ca) (a + b + c) 2 = 35 + 2(23) (a + b + c) 2 = 35 + 46 (a + b + c) 2 = 81 (a + b + c) = ±√81 Thus, a + b + c = ±9

6. If a + b + c = p and ab + bc + ca = q; find a 2 + b 2 + c 2 .

Solution:

Given, a + b + c = p and ab + bc + ca = q We know that, (a + b + c) 2 = (a 2 + b 2 + c 2 ) + 2(ab + bc + ca) (p) 2 = (a 2 + b 2 + c 2 ) + 2(q) ⇒ a 2 + b 2 + c 2 = p 2 – 2q

7. If a 2 + b 2 + c 2 = 50 and ab + bc + ca = 47, find a + b + c.

Solution:

Given, a 2 + b 2 + c 2 = 50 and ab + bc + ca = 47 We know that, (a + b + c) 2 = (a 2 + b 2 + c 2 ) + 2(ab + bc + ca) (a + b + c) 2 = 50 + 2(47) (a + b + c) 2 = 50 + 94 = 144 ⇒ (a + b + c) = √144 Thus, a + b + c = ±12

8. If x + y – z = 4 and x 2 + y 2 + z 2 = 30, then find the value of xy – yz – zx.

Solution:

Given, x + y – z = 4 and x 2 + y 2 + z 2 = 30 We know that, (x + y – z) 2 = x 2 + y 2 + z 2 + 2(xy – yz – zx) 4 2 = 30 + 2(xy – yz – zx) 16 – 30 = 2(ab + bc + ca) xy – yz – zx = -14/ 2 Thus, xy – yz – zx = -7

Exercise 4(D)

1. If x + 2y + 3z = 0 and x 3 + 4y 3 + 9z 3 = 18xyz; evaluate:

Concise Selina Solutions for Class 9 Maths Chapter 4 Expansions - 1

Solution:

Given, x 3 + 4y 3 + 9z 3 = 18xyz and x + 2y + 3z = 0 So, x + 2y = – 3z, 2y + 3z = -x and 3z + x = -2y Now, Concise Selina Solutions for Class 9 Maths Chapter 4 Expansions - 2

2. If a + 1/a = m and a ≠ 0; find in terms of ‘m’; the value of:

(i) a – 1/a

(ii) a 2 – 1/a 2

Solution:

(i) Given, a + 1/a = m On squaring on both sides, we get (a + 1/a) 2 = m 2 a 2 + 1/a 2 + 2 = m 2 a 2 + 1/a 2 = m 2 – 2 … (1) Now, consider the expansion (a – 1/a) 2 = a 2 + 1/a 2 – 2 = m 2 – 2 – 2 … [From (1)] = m 2 – 4 So, (a – 1/a) = ±√(m 2 – 4) … (2) (ii) We know that, a 2 – 1/a 2 = (a – 1/a) (a + 1/a) = m [±√(m 2 – 4)] = ±m√(m 2 – 4)

3. In the expansion of (2x 2 – 8) (x – 4) 2 ; find the value of

(i) coefficient of x 3

(ii) coefficient of x 2

(iii) constant term

Solution:

We have, (2x 2 – 8) (x – 4) 2 = (2x 2 – 8) (x 2 – 2 × 4 × x + 4 2 ) = (2x 2 – 8) (x 2 – 8x + 16) = 2x 2 (x 2 – 8x + 16) – 8(x 2 – 8x + 16) = 4x 4 – 16x 3 + 32x 2 – 8x 2 + 64x – 128 = 4x 4 – 16x 3 + 24x 2 + 64x – 128 Now, (i) coefficient of x 3 = -16 (ii) coefficient of x 2 = 24 (iii) constant term = -128

4. If x > 0 and x 2 + 1/9x 2 = 25/36. Find: x 3 + 1/27x 3

Solution:

Given, x 2 + 1/9x 2 = 25/36 … (1) Now, consider the expansion (x + 1/3x) 2 = x 2 + (1/3x) 2 + (2 × x × 1/3x) = (x 2 + 1/9x 2 ) + 2/3 = 25/36 + 2/3 … [From (1)] = 49/36 So, (x + 1/3x) = ±√(49/36) = ±7/6 … (2) Now, consider the expansion (x + 1/3x) 3 = x 3 + (1/3x) 3 + 3(x + 1/3x) (7/6) 3 = x 3 + (1/3x) 3 + 3(7/6) …… [From (2)] 343/216 = x 3 + 1/27x 3 + 21/6 x 3 + 1/27x 3 = 343/216 – 21/6 = (343 – 252)/216 = 91/216 Thus, x 3 + 1/27x 3 = 91/216

5. If 2(x 2 + 1) = 5x, find:

(i) x – 1/x

(ii) x 3 – 1/x 3

Solution:

(i) Given, 2(x 2 + 1) = 5x x 2 + 1 = 5x/2 On dividing by x on both sides, we have (x 2 + 1)/x = 5/2 ⇒ (x + 1/x) = 5/2 … (1)   Now, consider the expansion of (x + 1/x) 2 (x + 1/x) 2 = x 2 + 1/x 2 + 2 (5/2) 2 = x 2 + 1/x 2 + 2 … [From (1)] x 2 + 1/x 2 = 25/4 – 2 = (25 – 8)/4 = 17/4 … (2) Now, (x – 1/x) 2 = x 2 + 1/x 2 – 2 = 17/4 – 2 … [From (2)] = (17 – 8)/4 = 9/4 So, x – 1/x = √9/4 Thus, (i) x – 1/x = ±3/2 … (3)   Next, we know that (x 3 – 1/x 3 ) = (x – 1/x) 3 + 3(x – 1/x) = (±3/2) 3 + 3(±3/2) … [From (3)] = ± 27/8 ± 9/2 = ± (27 + 36)/8 = ± 63/8 (ii) Thus, x 3 – 1/x 3 = ±63/8

6. If a 2 + b 2 = 34 and ab = 12; find:

(i) 3(a + b) 2 + 5(a – b) 2

(ii) 7(a – b) 2 – 2(a + b) 2

Solution:

We have, a 2 + b 2 = 34 and ab= 12 We know that, (a + b) 2 = (a 2 + b 2 ) + 2ab = 34 + 2 x 12 = 34 + 24 = 58 Also, we know that (a – b) 2 = (a 2 + b 2 ) – 2ab = 34 – 2 x 12 = 34- 24 = 10 (i) 3(a + b) 2 + 5(a – b) 2 = 3 x 58 + 5 x 10 = 174 + 50 = 224   (ii) 7(a – b) 2 – 2(a + b) 2 = 7 x 10 – 2 x 58 = 70 – 116 = -46

7. If 3x – 4/x = 4 and x ≠ 0; find: 27x 3 – 64/x 3 .

Solution:

Given, 3x – 4/x = 4 Now, let’s consider the expansion of (3x – 4/x) 3 (3x – 4/x) 3 = 27x 3 – 64/x 3 – 3 × 3x × 4/x(3x – 4/x) (4) 3 = 27x 3 – 64/x 3 – 36(3x – 4/x) 64 = 27x 3 – 64/x 3 – 36(4) 64 = 27x 3 – 64/x 3 – 144 27x 3 – 64/x 3 = 144 + 64 Hence, 27x 3 – 64/x 3 = 208

8. If x 2 + 1/x 2 = 7 and x ≠ 0; find the value of: 7x 3 + 8x – 7/x 3 – 8/x.

Solution:

Given, x 2 + 1/x 2 = 7 On subtracting 2 from both sides, we get x 2 + 1/x 2 – 2 = 7 – 2 (x – 1/x) 2 = 5 x – 1/x = ±√5 … (1) Now, consider (x – 1/x) 3 = x 3 – 1/x 3 – 3(x – 1/x) (±√5) 3 = x 3 – 1/x 3 – 3(±√5) x 3 – 1/x 3 = (±√5) 3 + 3(±√5) … (2) Taking, 7x 3 + 8x – 7/x 3 – 8/x = 7x 3 – 7/x 3 + 8x – 8/x = 7(x 3 – 1/x 3 ) + 8(x – 1/x) = 7[(±√5) 3 + 3(±√5)] + 8(±√5) = ±35√5 ± 21√5 ± 8√5 = ±64√5

9. If x = 1/(x – 5) and x ≠ 5, find x 2 – 1/x 2 .

Solution:

Given, x = 1/(x – 5) By cross multiplying, we have x (x – 5) = 1 x 2 – 5x = 1 x 2 – 1 = 5x Dividing both sides by x, (x 2 – 1)/x = 5 (x – 1/x) = 5 … (1) Now, (x – 1/x) 2 = 5 2 x 2 + 1/x 2 – 2 = 25 x 2 + 1/x 2 = 25 + 2 = 27 … (2) Considering the expansion (x + 1/x) 2 (x + 1/x) 2 = x 2 + 1/x 2 + 2 (x + 1/x) 2 = 27 + 2 … [From (1)] (x + 1/x) 2 = 29 x + 1/x = ±√29 … (3) We know that, x 2 – 1/x 2 = (x + 1/x) (x – 1/x) = (±√29) (5) … [From (3)] = ±5√29

10. If x = 1/(5 – x) and x ≠ 5; find x 3 + 1/x 3 .

Solution:

Given, x = 1/(5 – x) By cross multiplying, we have x (5 – x) = 1 x 2 – 5x = -1 x 2 + 1 = 5x Dividing both sides by x, (x 2 + 1)/x = 5 x + 1/x = 5 … (1) Now, (x + 1/x) 3 = x 3 + 1/x 3 + 3(x + 1/x) x 3 + 1/x 3 = (x + 1/x) 3 – 3(x + 1/x) = 5 3 – 3(5) = 125 – 15 = 110 Thus, x 3 + 1/x 3 = 110

11. If 3a + 5b + 4c = 0,

Show that: 27a 3 + 125b 3 + 64c 3 = 180abc

Solution:

Given, 3a + 5b + 4c = 0 ⇒ 3a + 5b = -4c On cubing on both sides, we have (3a + 5b) 3 = (-4c) 3 (3a) 3 + (5b) 3 + 3 x 3a x 5b (3a + 5b) = -64c 3 27a 3 + 125b 3 + 45ab (-4c) = -64c 3 27a 3 + 125b 3 – 180abc = -64c 3 27a 3 + 125b 3 + 64c 3 = 180abc – Hence Proved.

12. The sum of two numbers is 7 and the sum of their cubes is 133, find the sum of their square.

Solution:

Let’s assume a and b to be the two numbers So, a + b = 7 and a 3 + b 3 = 133 We know that, (a + b) 3 = a 3 + b 3 + 3ab (a + b) (7) 3 = 133 + 3ab (7) 343 = 133 + 21ab 21ab = 343 – 133 = 210 ⇒ ab = 21 Now, a 2 + b 2 = (a + b) 2 – 2ab = 7 2 – 2 x 10 = 49 – 20 = 29

13. In each of the following, find the value of ‘a’:

(i) 4x 2 + ax + 9 = (2x + 3) 2

(ii) 4x 2 + ax + 9 = (2x – 3) 2

(iii) 9x 2 + (7a – 5)x + 25 = (3x + 5) 2

Solution:

(i) 4x 2 + ax + 9 = (2x + 3) 2 = 4x 2 + 12x + 9 On comparing coefficients of x terms, we get ax = 12x So, a = 12 (ii) 4x 2 + ax + 9 = (2x – 3) 2 = 4x 2 + 12x + 9 On comparing coefficients of x terms, we get ax = -12x So, a = -12 (iii) 9x 2 + (7a – 5)x + 25 = (3x + 5) 2 = 9x 2 + 30x + 25 On comparing coefficients of x terms, we get (7a – 5)x = 30x 7a – 5 = 30 7a = 35 ⇒ a = 5

14. If (x 2 + 1)/x = 3 1/3 and x > 1; find

(i) x – 1/x

(ii) x 3 – 1/x 3

Solution:

Given, (x 2 + 1)/x = 3 1/3 = 10/3 x + 1/x = 10/3 On squaring on both sides, we get (x + 1/x) 2 = (10/3) 2 x 2 + 1/x 2 + 2 = 100/9 x 2 + 1/x 2 = 100/9 – 2 = (100 – 18)/9 = 82/9 Now, (x – 1/x) 2 = x 2 + 1/x 2 – 2 = 82/9 – 2 = (82 – 18)/9 = 64/9 x – 1/x = √(64/9) = ±8/3 On cubing both sides, we get (x – 1/x) 3 = (8/3) 3 x 3 – 1/x 3 – 3(x – 1/x) = 512/27 x 3 – 1/x 3 = 3(x – 1/x) + 512/27 = 3(8/3) + 512/27 = 24/3 + 512/27 = (216 + 512)/27 = 728/27 Therefore, x 3 – 1/x 3 = 728/27

15. The difference between two positive numbers is 4 and the difference between their cubes is 316.

Find:

(i) Their product

(ii) The sum of their squares

Solution:

Given, difference between two positive numbers is 4 And, the difference between their cubes is 316 Let’s assume the positive numbers to be a and b So, a – b = 4 a 3 – b 3 = 316 On cubing both sides, we have (a – b) 3 = 64 a 3 – b 3 – 3ab(a – b) = 64 Also, Given: a 3 – b 3 = 316 So, 316 – 64 = 3ab(4) 252 = 12ab So, ab = 21 Thus, the product of numbers is 21 Now, On squaring both sides, we get (a – b) 2 = 16 a 2 + b 2 – 2ab = 16 a 2 + b 2 = 16 + 42 = 58 Thus, sum of their squares is 58.

Exercise 4(E)

1. Simplify:

(i) (x + 6)(x + 4)(x – 2)

(ii) (x – 6)(x – 4)(x + 2)

(iii) (x – 6)(x – 4)(x – 2)

(iv) (x + 6)(x – 4)(x – 2)

Solution:

Using identity: (x + a)(x + b)(x + c) = x 3 + (a + b + c)x 2 + (ab + bc + ca)x + abc (i) We have, (x + 6)(x + 4)(x – 2) = x 3 + (6 + 4 – 2)x 2 + [6 × 4 + 4 × (-2) + (-2) × 6]x + 6 × 4 × (-2) = x 3 + 8x 2 + (24 – 8 – 12)x – 48 = x 3 + 8x 2 + 4x – 48 (ii) We have, (x – 6)(x – 4)(x + 2) = x 3 + (-6 – 4 + 2)x 2 + [-6 × (-4) + (-4) × 2 + 2 × (-6)]x + (-6) × (-4) × 2 = x 3 – 8x 2 + (24 – 8 – 12)x + 48 = x 3 – 8x 2 + 4x + 48 (iii) We have, (x – 6)(x – 4)(x – 2) = x 3 + (-6 – 4 – 2)x 2 + [-6 × (-4) + (-4) × (-2) + (-2) × (-6)]x + (-6) × (-4) × (-2) = x 3 – 12x 2 + (24 + 8 + 12)x – 48 = x 3 – 12x 2 + 44x – 48   (iv) We have, (x + 6)(x – 4)(x – 2) = x 3 + (6 – 4 – 2)x 2 + [6 × (-4) + (-4) × (-2) + (-2) × 6]x + 6 × (-4) × (-2) = x 3 – 0x 2 + (-24 + 8 – 12)x + 48 = x 3 – 28x + 48

2. Simply using following identity:

(a ± b) (a 2 ∓ ab + b 2 ) = a 3 ± b 3

(i) (2x + 3y) (4x 2 – 6xy + 9y 2 )

(ii) (3x – 5/x) (9x 2 + 15 + 25/x 2 )

(iii) (a/3 – 3b) (a 2 + ab + 9b 2 )

Solution:

(i) We have, (2x + 3y) (4x 2 – 6xy + 9y 2 ) = (2x + 3y) [(2x) 2 – (2x)(3y) + (3y) 2 ] = (2x) 3 + (3y) 3 = 8x 3 + 27y 3 (ii) We have, (3x – 5/x) (9x 2 + 15 + 25/x 2 ) = (3x – 5/x) [(3x) 2 + (3x)(5/x) + (5/x) 2 ] = (3x) 3 + (5/x) 3 = 27x 3 + 125/x 3 (iii) We have, (a/3 – 3b) (a 2 /9 + ab + 9b 2 ) = (a/3 – 3b) [(a/3) 2 + (a/3)(3b) + (3b) 2 ] = (a/3) 3 – (3b) 3 = a 3 /27 – 27b 3

3. Using suitable identity, evaluate

(i) (104) 3

(ii) (97) 3

Solution:

Using identity: (a ± b) 3 = a 3 ± b 3 ± 3ab(a ± b) (i) (104) 3 = (100 + 4) 3 = (100) 3 + (4) 3 + 3 × 100 × 4(100 + 4) = 1000000 + 64 + 1200 × 104 = 1000000 + 64 + 124800 = 1124864   (ii) (97) 3 = (100 – 3) 3 = (100) 3 – (3) 3 – 3 × 100 × 3(100 – 3) = 1000000 – 27 – 900 × 97 = 1000000 – 27 – 87300 = 912673

4. Simply:

Concise Selina Solutions for Class 9 Maths Chapter 4 Expansions - 3

Solution:

We know that, If a + b + c = 0, then a 3 + b 3 + c 3 = 3abc Now, if (x 2 – y 2 ) + (y 2 – z 2 ) + (z 2 – x 2 ) = 0 Then, we have (x 2 – y 2 ) 3 + (y 2 – z 2 ) 3 + (z 2 – x 2 ) 3 = 3(x 2 – y 2 )(y 2 – z 2 )(z 2 – x 2 ) … (1) Similarly, if x – y + y – z + z – x = 0 Then, (x – y) 3 + (y – z) 3 + (z – x) 3 = 3(x – y)(y – z)(z – x) … (2) Now,

Concise Selina Solutions for Class 9 Maths Chapter 4 Expansions - 4

= (x + y)(y + z)(z + x)

5. Evaluate:

Selina Solutions Icse Class 9 Mathematics Chapter - Expansion - 5

Solution:

(i) We have, Concise Selina Solutions for Class 9 Maths Chapter 4 Expansions - 6 = a + b = 0.8 + 0.5 = 1.3 (ii) We have, Concise Selina Solutions for Class 9 Maths Chapter 4 Expansions - 7 Concise Selina Solutions for Class 9 Maths Chapter 4 Expansions - 8

6. If a – 2b + 3c = 0; state the value of a 3 – 8b 3 + 27c 3 .

Solution:

Given, a – 2b + 3c = 0 Then, a 3 – 8b 3 + 27c 3 = a 3 + (-2b) 3 + (3c) 3 = 3(a)( -2b)(3c) = -18abc

7. If x + 5y = 10; find the value of x 3 + 125y 3 + 150xy – 1000.

Solution:

Given, x + 5y = 10 On cubing both sides, we get (x + 5y) 3 = 10 3 x 3 + (5y) 3 + 3(x)(5y)(x + 5y) = 1000 x 3 + (5y) 3 + 3(x)(5y)(10) = 1000 x 3 + (5y) 3 + 150xy = 1000 Thus, x 3 + (5y) 3 + 150xy – 1000 = 0

8. If x = 3 + 2√2, find:

(i) 1/x

(ii) x – 1/x

(iii) (x – 1/x) 3

(iv) x 3 – 1/x 3

Solution:

We have, x = 3 + 2√2 (i) 1/x = 1/(3 + 2√2) = (3 – 2√2)/ [(3 + 2√2) × (3 – 2√2)] = (3 – 2√2)/ [3 2 – (2√2) 2 ] = (3 – 2√2)/ (9 – 8) = 3 – 2√2 (ii) x – 1/x = (3 + 2√2) – (3 – 2√2) … [From (i)] = (3 + 2√2 – 3 + 2√2) = 4√2 (iii) (x – 1/x) 3 = (4√2) 3 … [From (ii)] = (64 x 2√2) = 128√2 (iv) (x 3 – 1/x 3 ) = (x – 1/x) 3 – 3(x – 1/x) = 128√2 – 3(4√2) … [From (iii) and (ii)] = 128√2 – 12√2

9. If a + b = 11 and a 2 + b 2 = 65; find a 3 + b 3 .

Solution:

Given, a + b = 11 and a 2 + b 2 = 65 Now, we know that (a + b) 2 = a 2 + b 2 + 2ab (11) 2 = 65 + 2ab 121 = 65 + 2ab 2ab = 121 – 65 ab = (121 – 65)/2 = 56/2 = 28 Considering the expansion (a 3 + b 3 ) (a 3 + b 3 ) = (a + b) (a 2 + b 2 – ab) = (11) (65 – 28) = 11 × 37 = 407 Thus, a 3 + b 3 = 407

10. Prove that:

x 2 + y 2 + z 2 – xy – yz – zx is always positive.

Solution:

We have, x 2 + y 2 + z 2 – xy – yz – zx = 2(x 2 + y 2 + z 2 – xy – yz – zx) = 2x 2 + 2y 2 + 2z 2 – 2xy – 2yz – 2zx = x 2 + x 2 + y 2 + y 2 + z 2 + z 2 – 2xy – 2yz – 2zx = (x 2 + y 2 – 2xy) + (z 2 + x 2 – 2zx) + (y 2 + z 2 – 2yz) = (x – y) 2 + (z – x) 2 + (y – z) 2 As the square of any number is positive, the given equation is always positive.

11. Find:

(i) (a + b)(a + b)

(ii) (a + b)(a + b)(a + b)

(iii) (a – b)(a – b)(a – b) by using the result of part (ii)

Solution:

(i) We have, (a + b)(a + b) = (a + b) 2 = a × a + a × b + b × a + b × b = a 2 + ab + ab + b 2 = a 2 + b 2 + 2ab   (ii) We have, (a + b)(a + b)(a + b) = (a × a + a × b + b × a + b × b)(a + b) = (a 2 + ab + ab + b 2 )(a + b) = (a 2 + b 2 + 2ab)(a + b) = a 2 × a + a 2 × b + b 2 × a + b 2 × b + 2ab × a + 2ab × b = a 3 + a 2 b + ab 2 + b 3 + 2a 2 b + 2ab 2 = a 3 + b 3 + 3a 2 b + 3ab 2 (iii) We have, (a – b)(a – b)(a – b) In result (ii), replacing b by -b, we get (a – b)(a – b)(a – b) = a 3 + (-b) 3 + 3a 2 (-b) + 3a(-b) 2 = a 3 – b 3 – 3a 2 b + 3ab 2

Benefits of ICSE Class 9 Maths Selina Solutions Chapter 4 Expansion

1. Helps to plan a preparation strategy

Attempting ICSE Class 9 Maths Selina Solutions Chapter 4 is the best strategy to prepare for the exam. Candidates can plan study table for all subjects and give more time to weak sections. By this candidates can attempt all the questions during the exam. And at the closing time, they have to perfect everything they have learned.

2. Preparation Level Overview

One of the best advantages of ICSE Class 9 Maths Selina Solutions Chapter 4 Expansion is that it helps candidates know where they stand in terms of their preparation level. This helps candidates know where they are lagging behind. Candidates can learn from their mistakes and follow the test exam solutions and strategies.

3. Performance analysis

No one can learn anything unless they analyze their performance. Candidates can know their weak and strong points only by analyzing them. Analyzing the ICSE Class 9 Maths solution will make the candidates aware that they will not repeat their mistakes in the exam hall. Analyzing the solutions helps the candidate to get an idea of ​​where they stand in the competition to take the ICSE Class 9 Maths Exam.

5. Time planning aids

A candidate cannot afford to deal with a question that takes time. Some questions can be difficult and some can be solved at a glance. ICSE Class 9 Maths Selina Solutions help candidates to develop such skills. Candidates will implement the question solving pattern with right approach which will help them to crack the exam on time. Solving most of the questions in less time, they give the remaining time to the more difficult parts.

6. Improve Accuracy and Speed ​​

It helps candidates to improve accuracy and speed. The more they solve different questions, the better they explain their concepts and tricks. Accuracy and speed are important in the selection process. This will help them clear the upcoming ICSE class 9 Maths Exam.
ICSE Class 9 Maths Selina Solutions Chapter wise
Chapter 1 Rational and Irrational Numbers Chapter 11 Inequalities
Chapter 2 Compound Interest (Without Using Formula) Chapter 12 Mid-Point and Its Converse (Including Intercept Theorem)
Chapter 3 Compound Interest (Using Formula) Chapter 13  Pythagoras Theorem
Chapter 4 Expansions Chapter 14 Rectilinear Figures (Quadrilaterals: Parallelogram, Rectangles, Rhombus, Square, and Trapezium)
Chapter 5 Factorisation Chapter 15 Construction of Polygons
Chapter 6 Simultaneous (Linear) Equations (Including Problems) Chapter 16 Area Theorems
Chapter 7 Indices (Exponents) Chapter 17 Circle
Chapter 8 Logarithms Chapter 18 Statistics
Chapter 9 Triangles (Congruency in Triangles) Chapter 19 Mean and Median
Chapter 10 Isosceles Triangles Chapter 20 Area and Perimeter of Plane Figures
.

ICSE Class 9 Maths Selina Solutions Chapter 4 FAQs

Q1. What is Expansion?

Ans. The expansion of a product of sums is expressed as the sum of its products using the fact that multiplication divides addition - the reverse process of trying to write an expanded polynomial as a product is called polynomial factors. "Expand" means to remove ( ) ... but we have to do it right ( ) are called "parentheses" or "brackets". Anything inside ( ) should be treated as "package". So to say: tell everything inside the "package".

Q2. Where can I get the ICSE Class 9 Maths Selina Solutions Chapter 4 Expansion?

Ans. You can get the ICSE Class 9 Maths Selina Solutions Chapter 4 - Expansion at PW blog section.

Q3. Is ICSE Class 9 Maths Selina Solutions Chapter 4 PDF available to download?

Ans. Yes, ICSE Class 9 Maths Selina Solutions Chapter 4 PDF is available for download on our page.

Q4. What are the Benefits of ICSE Class 9 Maths Selina Solutions Chapter 4 Expansion?

Ans. The details Benefits of ICSE Class 9 Maths Selina Solutions Chapter 4 Expansion are provided in the above article.

Q5. Are ICSE Class 9 Maths Selina Solutions Chapter 4 useful for exam preparation?

Ans. Yes, these ICSE Class 9 Maths Selina Solutions Chapter 4 are extremely beneficial for exam preparation. They cover all topics and exercises prescribed in the ICSE curriculum, ensuring comprehensive revision and practice.
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