Frank Solutions for Class 9 Maths Chapter 4 Expansions consist of accurate answers designed by experienced faculty at BYJU’S, which helps students boost their exam preparation. The solutions, given in a step-by-step manner, enable them to grasp the concepts and topics covered in the chapter easily. Solving these problems on a daily basis will improve their conceptual knowledge and thus score well in exams.
Chapter 4 mainly deals with the problems based on Expansions. Students can analyse their preparation level by cross-checking their answers, referring to these solutions provided in PDF. Practising the problems with the help of these solutions will make them understand the various methods to solve complex problems effortlessly. To practise well and score good marks in the annual exam, students can access Frank Solutions for Class 9 Maths Chapter 4 Expansions PDF from the link below.
Frank Solutions for Class 9 Maths Chapter 4 Expansions Download PDF
Access Frank Solutions for Class 9 Maths Chapter 4 Expansions
1. Expand the following:
(i) (a + 4) (a + 7)
(ii) (m + 8) (m – 7)
(iii) (x – 5) (x – 4)
(iv) (3x + 4) (2x – 1)
(v) (2x – 5) (2x + 5) (2x – 3)
Solution:
(i) (a + 4) (a + 7)
= a2 + 4a + 7a + 28
We get,
= a2 + 11a + 28
(ii) (m + 8) (m – 7)
= m2 + 8m – 7m – 56
We get,
= m2 + m – 56
(iii) (x – 5) (x – 4)
= x2 – 5x – 4x + 20
We get,
= x2 – 9x + 20
(iv) (3x + 4) (2x – 1)
= 6x2 – 3x + 8x – 4
We get,
= 6x2 + 5x – 4
(v) (2x – 5) (2x + 5) (2x – 3)
= (4x2 – 25) (2x – 3)
We get,
= 8x3 – 12x2 – 50x + 75
2. Expand the following:
(i) (a + 3b)2
(ii) (2p – 3q)2
(iii) (2a + 1 / 2a)2
(iv) (x – 3y – 2z)2
Solution:
(i) (a + 3b)2
(a + 3b)2 = a2 + 2(a) (3b) + (3b)2 [Using (x + y)2 = x2 + 2xy + y2]
We get,
= a2 + 6ab + 9b2
(ii) (2p – 3q)2
= (2p)2 – 2 (2p) (3q) + (3q)2
We get,
= 4p2 – 12pq + 9q2
(iii) (2a + 1 / 2a)2
= (2a)2 + 2 (2a) (1 / 2a) + (1 / 2a)2
We get,
= {4a2 + 2 + (1 / 4a2)}
(iv) (x – 3y – 2z)2
= x2 + (3y)2 + (2z)2 + 2 (x) (-3y) + 2 (-3y) (-2z) + 2 (x) (-2z)
[Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac]We get,
= x2 + 9y2 + 4z2 – 6xy + 12yz – 4xz
3. Find the squares of the following:
(a) 9m – 2n
(b) 3p – 4q
(c) (7x / 9y) – (9y / 7x)
(d) (2a + 3b – 4c)
Solution:
(a) 9m – 2n
(9m – 2n)2 = (9m)2 + 2 (9m) (-2n) + (-2n)2
[Using (x + y)2 = x2 + 2xy + y2]We get,
= 81m2 – 36mn + 4n2
(b) 3p – 4q
(3p – 4q)2 = (3p)2 – 2 (3p) (4q) + (4q)2
On simplification, we get,
= 9p2 – 12pq + 16q2
(c) {(7x / 9y) – (9y / 7x)}2 = (7x / 9y)2 + 2 (7x / 9y) (9y / 7x) + (9y / 7x)2
On simplification, we get,
= (49x2 / 81y2) + 2 + (81y2 / 49x2)
(d) (2a + 3b – 4c)
(2a + 3b – 4c)2 = (2a)2 + (3b)2 + (-4c)2 + 2(2a) (3b) + 2 (3b) (-4c) + 2 (2a) (-4c)
Using (a+ b+ c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac
On simplification, we get,
= 4a2 + 9b2 + 16c2 + 12ab -24bc – 16ac
4. Simplify by using the formula:
(i) (5x – 9) (5x + 9)
(ii) (2x + 3y) (2x – 3y)
(iii) (a + b – c) (a – b + c)
(iv) (x + y – 3) (x + y + 3)
(v) (1 + a) (1 – a) (1 + a2)
(vi) {a + (2 / a) – 1}{a – (2 / a) – 1}
Solution:
(i) (5x – 9) (5x + 9)
[Using identity: (a + b) (a – b) = a2 – b2]= (5x)2 – (9)2
We get,
= 25x2 – 81
(ii) (2x + 3y) (2x – 3y)
[Using identity: (a + b) (a – b) = a2 – b2= (2x)2 – (3y)2
We get,
= 4x2 – 9y2
(iii) (a + b – c) (a – b + c)
On further calculation, we get,
= (a + b – c) {a – (b – c)}
[Using identity: (a + b) (a – b) = a2 – b2]= (a)2 – (b – c)2
= a2 – (b2 + c2 – 2bc)
= a2 – b2 – c2 + 2bc
(iv) (x + y – 3) (x + y + 3)
[Using identity: (a + b) (a – b) = a2 – b2]= (x + y)2 – (3)2
= x2 + y2 + 2xy – 9
(v) (1 + a) (1 – a) (1 + a2)
[Using identity: (a + b) (a – b) = a2 – b2]= {(1)2 – (a)2} (1 + a2)
= (1 – a2) (1 + a2)
= (1)2 – (a2)2
We get,
= 1 – a4
(vi) {a + (2 / a) – 1} {a – (2 / a) – 1}
[Using identity: (a + b) (a – b) = a2 – b2]= (a – 1)2 – (2 / a)2
We get,
= a2 + 1 – 2a – (4 / a2)
5. Evaluate the following without multiplying:
(i) (95)2
(ii) (103)2
(iii) (999)2
(iv) (1005)2
Solution:
(i) (95)2
Using (x + y)2 = x2 + 2xy + y2
We get,
(95)2 = (100 – 5)2
= (100)2 – 2 (100) (5) + (5)2
= 10000 – 1000 + 25
We get,
= 9025
(ii) (103)2
Using (x + y)2 = x2 + 2xy + y2
We get,
(103)2 = (100 + 3)2
= (100)2 + 2 (100) (3) + (3)2
= 10000 + 600 + 9
We get,
= 10609
(iii) (999)2
(999)2 = (1000 – 1)2
= (1000)2 – 2 (1000) (1) + (1)2
= 1000000 – 2000 + 1
We get,
= 998001
(iv) (1005)2
(1005)2 = (1000 + 5)2
= (1000)2 + 2 (1000) (5) + (5)2
= 1000000 + 10000 + 25
We get,
= 1010025
6. Evaluate, using (a + b) (a – b) = a2 – b2
(i) 399 × 401
(ii) 999 × 1001
(iii) 409 × 5.1
(iv) 15.9 × 16.1
Solution:
(i) 399 × 401
399 × 401 = (400 – 1) (400 + 1)
= (400)2 – (1)2
On further calculation, we get,
= 160000 – 1
We get,
= 159999
(ii) 999 × 1001
999 × 1001 = (1000 – 1) (1000 + 1)
= (1000)2 – (1)2
On further calculation, we get,
= 1000000 – 1
We get,
= 999999
(iii) 4.9 × 5.1
4.9 × 5.1 = (5 – 0.1) (5 + 0.1)
= (5)2 – (0.1)2
= 25 – 0.01
We get,
= 24.99
(iv) 15.9 × 16.1
15.9 × 16.1 = (16 – 0.1) (16 + 0.1)
= (16)2 – (0.1)2
= 256 – 0.01
We get,
= 255.99
7. If a – b = 10 and ab = 11; find a + b
Solution:
Given
a – b = 10 and ab = 11
We know that,
(a – b)2 = a2 – 2ab + b2
(10)2 = a2 – 2 (11) + b2
100 = a2 + b2 – 22
On further calculation, we get,
a2 + b2 = 100 + 22
a2 + b2 = 122
[Using (a + b)2 = a2 + b2 + 2ab]We get,
(a + b)2 = 122 + 2 (11)
(a + b)2 = 122 + 22
(a + b)2 = 144
(a + b) = √144
We get,
(a + b) = ± 12
8. If x + y = 9, xy = 20; find:
(i) x – y
(ii) x2 – y2
Solution:
(i) Given
x + y = 9, xy = 20
We know that,
(a + b)2 = a2 + 2ab + b2
(x + y)2 = x2 + y2 + 2xy
x2 + y2 + 2xy = 81
x2 + y2 = 81 – 2(xy)
x2 + y2 = 81 – 2 (20)
x2 + y2 = 81 – 40
x2 + y2 = 41
We know that,
(a – b)2 = a2 – 2ab + b2
(x – y)2 = x2 – 2xy + y2
(x – y)2 = x2 + y2 – 2xy
(x – y)2 = 41 – 2 (20)
(x – y)2 = 41 – 40
(x – y)2 = 1
x – y = ± 1
(ii) We know that,
(x – y) (x + y) = x2 – y2
x2 – y2 = (± 1) (9)
We get,
x2 – y2 = ± 9
9. Find the cube of:
(i) 2a – 5b
(ii) 4x + 7y
(iii) 3a + (1 / 3a)
(iv) 4p – (1 / p)
(v) (2m / 3n) + (3n / 2m)
(vi) a – (1 / a) + b
Solution:
(i) 2a – 5b
Using (a – b)3 = a3 – b3 – 3ab (a – b)
(2a – 5b)3 = (2a)3 – (5b)3 – 3 (2a) (5b) (2a – 5b)
On further calculation, we get,
= 8a3 – 125b3 – 30ab (2a – 5b)
= 8a3 – 125b3 – 60a2b + 150ab2
(ii) 4x + 7y
Using (a + b)3 = a3 + b3 + 3ab (a + b)
(4x + 7y)3 = (4x)3 + (7y)3 + 3 (4x) (7y) (4x + 7y)
On further calculation, we get,
= 64x3 + 343y3 + 84xy (4x + 7y)
= 64x3 + 343y3 + 336x2y + 588xy2
(iii) {3a + (1 / 3a)}3
Using (a + b)3 = a3 + b3 + 3ab (a + b)
{3a + (1 / 3a)}3 = (3a)3 + (1 / 3a)3 + 3 (3a) (1 / 3a) {3a + (1 / 3a)}
We get,
= 27a3 + (1 / 27a3) + 9a + (1/ a)
(iv) {4p – (1 / p)}3
Using (a – b)3 = a3 – b3 – 3ab (a – b)
{4p – (1 / p)}3 = (4p)3 – (1 / p)3 – 3 (4p) (1 / p) {4p – (1 / p)}
We get,
= 64p3 – (1 / p3) – 48p + (12 / p)
(v) {(2m / 3n) + (3n / 2m)}3
Using (a + b)3 = a3 + b3 + 3ab (a + b)
= (2m / 3n)3 + (3n / 2m)3 + {3 (2m / 3n) (3n / 2m)}{(2m / 3n) + (3n / 2m)}
We get,
= (8m3 / 27n3) + (27n3 / 8m3) + (2m / n) + (9n / 2m)
(vi) {a – (1 / a) + b}3
Using (a + b + c)3 = a3 + b3 + c3 + 3a2b + 3a2c + 3b2a + 3c2a + 6abc
{a – (1 / a) + b}3 = a3 + (-1 / a)3 + b3 + 3a2 (-1 / a) + 3a2b + 3 (-1 / a)2b + 3(-1/a)2a + 3b2a + 3b2(-1/a)+ 6a(-1 / a)b
We get,
= a3 – (1 / a3) + b3 – 3a + 3a2b + 3b/a2 + 3/a + 3b2a – 3b2/ a – 6b
10. If {5x + (1 / 5x)} = 7; find the value of 125x3 + (1 / 125x3)
Solution:
Given
{5x + (1 / 5x)} = 7
Using {a + (1 / a)}3 = a3 + (1 / a3) + 3 {a + (1 / a)}
We get,
{5x + (1 / 5x)}3 = (5x)3 + (1 / 5x)3 + 3 {5x + (1 / 5x)}
343 = 125x3 + (1 / 125x3) + 3 (7)
We get,
125x3 + (1 / 125x3) = 343 – 21
125x3 + (1 / 125x3) = 322
11. If {3x – (1 / 3x)} = 9; find the value of 27x3 – (1 / 27x3)
Solution:
Given
{3x – (1 / 3x)} = 9
Using {a – (1 / a)}3 = a3 – (1 / a3) – 3 {a – (1 / a)}
We get,
{3x – (1 / 3x)}3 = (3x)3 – (1 / 3x)3 – 3 {3x – (1 / 3x)}
729 = 27x3 – (1 / 27x3) – 3(9)
On calculating further, we get,
27x3 – (1 / 27x3) = 729 + 27
27x3 – (1 / 27x3) = 756
12. If (x + 1 / x) = 5, find the value of {x2 + (1 / x2), x3 + (1 / x3) and x4 + (1 / x4)}
Solution:
Given
(x + 1 / x) = 5 …… (1)
On squaring both sides,
We get,
(x + 1 / x)2 = (5)2
x2 + (1 / x2) + 2 = 25
x2 + (1 / x2) = 25 – 2
x2 + (1 / x2) = 23 ….. (2)
Now,
Cubing both sides of equation (1),
We get,
{x + (1 / x)}3 = (5)3
x3 + (1 / x3) + 3 {x + (1 / x)} = 125
x3 + (1 / x3) + 3 (5) = 125
On calculating further, we get,
x3 + (1 / x3) = 125 – 15
x3 + (1 / x3) = 110
Squaring on both sides of equation (2), we get,
{x2 + (1 / x2)}2 = (23)2
x4 + (1 / x4) + 2 = 529
x4 + (1 / x4) = 529 – 2
x4 + (1 / x4) = 527
13. If {a – (1 / a)} = 7, find {a2 + (1 / a2)}, {a2 – (1 / a2) and {a3 – (1 / a3)}
Solution:
Given
{a – (1 / a) = 7 …. (1)
Squaring on both sides, we get,
{a – (1 / a)}2 = (7)2
a2 + (1 / a2) – 2 = 49
a2 + (1 / a2) = 49 + 2
a2 + (1 / a2) = 51
Now,
{a + (1 / a)}2 = a2 + (1 / a2) + 2
Substitute the value of a2 + (1 / a)2, and we get,
= 51 + 2
= 53
a + (1 / a) = ± √53
Now,
a2 – (1 / a2) = {a + (1 / a)}{a – (1 / a)}
= (± √53) (7)
We get,
= (± 7√53)
Cubing on both sides of equation (1), we get,
{a – (1 / a)}3 = (7)3
a3 – (1 / a3) – 3 {a – (1 / a)} = 343
a3 – (1 / a3) – 3 (7) = 343
a3 – (1 / a3) = 343 + 21
a3 – (1 / a3) = 364
14. If {a2 + (1 / a2)} = 14; find the value of
(i) {a + (1 / a)}
(ii) {a3 + (1 / a3)}
Solution:
(i) Using (a + b)2 = a2 + 2ab + b2
{a + (1 / a)}2 = a2 + 2a (1 / a) + (1 / a)2
{a + (1 / a)}2 = a2 + 2 + (1 / a2)
{a + (1 / a)}2 = a2 + (1 / a2) + 2
Substitute the value of {a2 + (1 / a2)}, we get,
{a + (1 / a)}2 = 14 + 2
{a + (1 / a)}2 = 16
{a + (1 / a)} = ± 4
Therefore, the value of {a + (1 / a)} = ± 4
(ii)
{a3 + (1 / a3)} = {a + (1 / a)} {(a2 + (1 / a2) – 1}
[Using a3 + b3 = (a + b) (a2 + b2 – ab)]We get,
{a3 + (1 / a3)} = (± 4) (14 – 1)
{a3 + (1 / a3)}= (± 4) (13)
{a3 + (1 / a3)}= (± 52)
15. If {m2 + (1 / m2) = 51; find the value of m3 – (1 / m3)
Solution:
m2 + (1 / m2) = 51
We know that,
{m – (1 / m)}2 = m2 + (1 / m2) – 2
{m – (1 / m)}2 = 51 – 2
{m – (1 / m)}2 = 49
{m – (1 / m)}2 = (7)2
{m – (1 / m)} = 7
Cubing on both sides, we get,
{m – (1 / m)}3 = (7)3
m3 – (1 / m3) – 3 {m – (1 / m)}= 343
m3 – (1 / m3) – 3 (7) = 343
m3 – (1 / m3) – 21 = 343
m3 – (1 / m3) = 343 + 21
We get,
m3 – (1 / m3) = 364
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