Frank Solutions for Chapter 4 Expansions Class 9 Mathematics ICSE
Exercise 4.1
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)
Answer
(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
Answer
(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)
Answer
(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 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}
Answer
(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
Answer
(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
Answer
(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
Answer
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
Answer
(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
Answer
(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)
Answer
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)
Answer
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)}
Answer
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)}
Answer
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, 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)}
Answer
(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)
Answer
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
16. If a2 – 3a – 1 = 0 and a ≠ 0, find ;
(i) a – 1/a,
(ii) a + 1/a,
(iii) a2 – 1/a2
Answer
(i)
17. If 2x + 3y = 10 xy = 5: find the value of 4x2 + 9y2
Answer
18. If x + y + z = 12 and xy + yz + zx = 27; find x2 + y2 + z2.
Answer
19. If a2 + b2 + c2 = 41 and a + b + c = 9; find ab + bc + ca.
Answer:
20. If p2 + q2 + r2 = 82 and pq + qr + pr = 18; find p + q + r.
Answer
21. If x + y + z = p and xy + yz + zx = q; find x2 + y2 + z2
Answer
Exercise 4.2
1. 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
Answer
(i)
Answer
3. If 3x – 1/3x = 9: find the value of 27x3 – 1/27x3
Answer
4. If x + 1/x = 5, find the value of x2 + 1/x2, x3 + 1/x3 and x4 + 1/x4.
Answer
5. If a – 1/a = 7, find a2 + 1/a2, a2 – 1/a2 and a3 – 1/a3
Answer
6. If a2 + 1/a2 = 14; find the value of a + 1/a
(b) If a2 + 1/a2 = 14; find the value of a3 + 1/a3
Answer
(a)
(b)
7. If m2 + 1/m2 = 51; find the value of m3 – 1/m3
Answer
8. If 9a2 + 1/9a2 = 23; find the value of 27a3 + 1/27a3
Answer
9. If x2 + 1/x2 = 18; find :
(i) x – 1/x,
(ii) x3 – 1/x3
Answer
x2 + 1/x2 = 18
(i)
10. If p + 1/p = 6; find :
(i) p2 + 1/p2,
(ii) p4 + 1/p4
(iii) p3 + 1/p3
Answer
(i)
(ii)
(iii)
11. If r – 1/r = 4; find
(i) r2 + 1/r2
(ii) r4 + 1/r4,
(iii) r3 – 1/r3
Answer
(i)
Answer
13. If x + 1/x = p, x – 1/x = q; find the relation between p and q.
Answer
14. If a + 1/a = p; then show that a3 + 1/a3 = p(p2 – 3)
Answer
15. If (a + 1/a)2 = 3; then show that a3 + 1/a3 = 0
Answer
16. If a + b + c = 0; then show that a3 + b3 + c3 = 3abc.
Answer
17. If a + 2b + c = 0; then show that a3 + 8b3 + c3 = 6abc
Answer
18. If x3 + y3 = 9 and x + y = 3, find xy.
Answer
19. If a + b = 5 and ab = 2, find a3 + b3
Answer
20. If p – q = - 1 and pq = - 12, find p3 – q3
Answer:
21: If m – n = - 2 and m3 – n3 = - 26, find mn.
Answer
22: If 2a – 3b = 10 ab = 16; find the value of 8a3 – 27b3.
Answer
23: If x + 2y = 5, then show that x3 + 8y3 + 30xy = 125.
Answer
24. Simplify
(a) (4x + 5y)2 + (4x – 5y)2
(b) (7a + 5b)2 – (7a – 5b)2
(c) (a + b)3 = (a – b)3
(d) (a – 1/a)2 + (a + 1/a)2
(e) (x + y – z)2 + (x – y + z)2
(f) (a + 1/a)3 – (a – 1/a)3
(g) (2x + y)(4x2 – 2xy + y2)
(h) (x – 1/x)(x2 + 1 + 1/x2)
(i) (x + 2y + 3z)(x2 + 4y2 + 9z2 – 2xy – 6yz – 3zx)
(j) (1 + x)(1 – x)(1 – x + x2)(1 + x + x2)
(k) (3a + 2b – c)(9a2 + 4b2 + c2 – 6ab + 2bc + 3ca)
(l) (3x + 5y + 2z)(3x – 5y + 2z)
(m) (2x – 4y + 7)(2x + 4y + 7)
(n) (3a – 7b + 3)(3a – 7b + 5)
(o) (4m – 5n – 8)(4m – 5n + 5)
Answer
(a)
(b)
25: Evaluate the following:
(i) (3.29)3 + (6.71)3
(ii) (5.45)3 + (3.55)3
(iii) (8.12)3 – (3.12)3
(iv) 7.16×7.16 + 2.16×7.16 + 2.16×2.16
(v) 1.81×1.81 – 1.81×2.19 + 2.19×2.19
Answer
(i)
