NCERT Class 9 Maths Ganita Manjari Chapter 4 Solutions
Exercise Set 4.1
1. Using the identity (a + b)² = a² + 2ab + b², expand the following:
(i) (7x + 4y)²
Answer:
Using (a + b)² = a² + 2ab + b²
Here, a = 7x and b = 4y
= (7x)² + 2(7x)(4y) + (4y)²
= 49x² + 56xy + 16y²
(ii) [(7/5)x + (3/2)y]²
Answer:
Here, a = (7/5)x and b = (3/2)y
= [(7/5)x]² + 2 × (7/5)x × (3/2)y + [(3/2)y]²
= (49/25)x² + 2 × (21/10)xy + (9/4)y²
= (49/25)x² + (42/10)xy + (9/4)y²
Simplify:
= (49/25)x² + (21/5)xy + (9/4)y²
(iii) (2.5p + 1.5q)²
Answer:
Here, a = 2.5p and b = 1.5q
= (2.5p)² + 2(2.5p)(1.5q) + (1.5q)²
= 6.25p² + 7.5pq + 2.25q²
(iv) [(3/4)s + 8t]²
Answer:
Here, a = (3/4)s and b = 8t
= [(3/4)s]² + 2 × (3/4)s × 8t + (8t)²
= (9/16)s² + 2 × (24/4)st + 64t²
= (9/16)s² + 12st + 64t²
(v) [x + 1/(2y)]²
Answer:
Here, a = x and b = 1/(2y)
= x² + 2 × x × [1/(2y)] + [1/(2y)]²
= x² + (2x)/(2y) + 1/(4y²)
= x² + x/y + 1/(4y²)
(vi) ( 1/x + 1/y )²
Answer:
Here, a = 1/x and b = 1/y
= (1/x)² + 2(1/x)(1/y) + (1/y)²
= 1/x² + 2/(xy) + 1/y²
2. Using the identity (a + b)² = a² + 2ab + b², find the values of the following:
(i) (64)²
Answer:
Write 64 = 60 + 4
Using (a + b)² = a² + 2ab + b²:
(60 + 4)² = 60² + 2 × 60 × 4 + 4²
= 3600 + 480 + 16
= 4096
(ii) (105)²
Answer:
Writing 105 = 100 + 5
Now (105)² = (100 + 5)²
= 100² + 2 × 100 × 5 + 5² [Using (a + b)² = a² + 2ab + b²]
= 10000 + 1000 + 25
= 11025
(iii) (205)²
Answer:
Write 205 = 200 + 5
So, (205)² = (200 + 5)²
= 200² + 2 × 200 × 5 + 5² [Using (a + b)² = a² + 2ab + b²]
= 40000 + 2000 + 25
= 42025
Class 9 Maths Ganita Manjari Chapter 4 Exercise Set 4.2 Solutions
Exercise Set 4.2
1. Factor completely:
(i) 9x² + 24xy + 16y²
Answer:
We know that:
9x² = (3x)²
16y² = (4y)²
24xy = 2 × 3x × 4y
So, 9x² + 24xy + 16y²
= (3x)² + 2 × 3x × 4y + (4y)²
= (3x + 4y)² [Using a² + 2ab + b² = (a + b)²]
(ii) 4s² + 20st + 25t²
Answer:
We can write:
4s² = (2s)²
25t² = (5t)²
20st = 2 × 2s × 5t
So, 4s² + 20st + 25t²
= (2s)² + 2 × 2s × 5t + (5t)²
= (2s + 5t)² [Using a² + 2ab + b² = (a + b)²]
(iii) 49x² + 28xy + 4y²
Answer:
49x² = (7x)²
4y² = (2y)²
28xy = 2 × 7x × 2y
So, 49x² + 28xy + 4y²
= (7x)² + 2 × 7x × 2y + (2y)²
= (7x + 2y)² [Using a² + 2ab + b² = (a + b)²]
(iv) 64p² + (32/3)pq + (4/9)q²
Answer:
64p² = (8p)²
(4/9)q² = (2/3 q)²
Middle term:
2 × 8p × (2/3 q) = 32/3 pq
So, 64p² + (32/3)pq + (4/9)q²
= (8p)² + 2 × 8p × (2/3 q) + (2/3 q)²
= (8p + 2/3 q)² [Using a² + 2ab + b² = (a + b)²]
(v) 3a² + 4ab + (4/3)b²
Answer:
Take common factor 3:
= 3[a² + (4/3)ab + (4/9)b²]
Now, a² = (a)²
(4/9)b² = (2/3 b)²
(4/3)ab = 2 × a × (2/3 b)
So, 3a² + 4ab + (4/3)b²
= 3[a² + (4/3)ab + (4/9)b²]
= 3[a² + 2 × a × (2/3 b) + (2/3 b)²]
= 3(a + 2/3 b)² [Using a² + 2ab + b² = (a + b)²]
(vi) (9/5)s² + 6sv + 5v²
Answer:
Take common factor 1/5:
= (1/5)[9s² + 30sv + 25v²]
Now, 9s² = (3s)²
25v² = (5v)²
30sv = 2 × 3s × 5v
So, (9/5)s² + 6sv + 5v²
= (1/5)[9s² + 30sv + 25v²]
= (1/5)[(3s)² + 2 × 3s × 5v + (5v)²]
= (1/5)(3s + 5v)² [Using a² + 2ab + b² = (a + b)²]
2. Find the values of the following using the identity (a − b)² = a² − 2ab + b²:
(i) (79)²
Answer:
79 = 80 − 1
= (80 − 1)²
= 80² − 2×80×1 + 1² [Using (a − b)² = a² − 2ab + b²]
= 6400 − 160 + 1
= 6241
(ii) (193)²
Answer:
193 = 200 − 7
= (200 − 7)²
= 200² − 2×200×7 + 7² [Using (a − b)² = a² − 2ab + b²]
= 40000 − 2800 + 49
= 37249
(iii) (299)²
Answer:
299 = 300 − 1
= (300 − 1)²
= 300² − 2×300×1 + 1² [Using (a − b)² = a² − 2ab + b²]
= 90000 − 600 + 1
= 89401
Class 9 Maths Ganita Manjari Chapter 4 Exercise Set 4.3 Solutions
Exercise Set 4.3
1. Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.
(i) 117²
Answer:
117 = 110 + 7
So, 117² = (110 + 7)²
= 110² + + 2(110)(7) + 7² [Using (a + b)² = a² + 2ab + b²]
= 12100 + 1540 + 49
= 13689
(ii) 78²
Answer:
78 = 80 – 2
So, 78² = (80 – 2)²
= 80² – 2(80)(2) + 2² [Using (a − b)² = a² − 2ab + b²]
= 6400 – 320 + 4
= 6084
(iii) 198²
Answer:
198 = 200 – 2
So, 198² = (200 – 2)²
= 200² – 2(200)(2) + 2² [Using (a − b)² = a² − 2ab + b²]
= 40000 – 800 + 4
= 39204
(iv) 214²
Answer:
214 = 200 + 14
So, 214² = (200 + 14)²
= 200² + 2(200)(14) + 14² [Using (a − b)² = a² − 2ab + b²]
= 40000 + 5600 + 196
= 45796
(v) 1104²
Answer:
1104 = 1100 + 4
So, 1104² = (1100 + 4)²
= 1100² + 2(1100)(4) + 4² [Using (a + b)² = a² + 2ab + b²]
= 1210000 + 8800 + 16
= 1218816
(vi) 1120²
Answer:
1120 = 1100 + 20
So, 1120² = (1100 + 20)²
= 1100² + 2(1100)(20) + 20² [Using (a + b)² = a² + 2ab + b²]
= 1210000 + 44000 + 400
= 1254400
2. Factor using suitable identities:
(i) 16y² – 24y + 9
Answer:
16y² = (4y)²
9 = 3²
-24y = -2(4y)(3)
Therefore, 16y² – 24y + 9
= (4y)² -2(4y)(3) + 3²
= (4y – 3)² [Using a² – 2ab + b² = (a – b)²]
(ii) (9/4)s² + 6st + 4t²
Answer:
(9/4)s² = [(3s)/2]²
4t² = (2t)²
6st = 2 × (3s/2) × (2t)
Therefore, (9/4)s² + 6st + 4t²
= [(3s)/2]² + 2 × (3s/2) × (2t) + (2t)²
= [(3s)/2 + 2t]² [Using a² + 2ab + b² = (a + b)²]
(iii) m²/9 + mk/3 + k²/4 + 3nk + 2mn + 9n²
Answer:
Grouping the terms as:
m²/9 + mk/3 + k²/4 + 2mn + 3nk + 9n²
We have:
m²/9 = (m/3)²
k²/4 = (k/2)²
9n² = (3n)²
Now checking the cross terms:
2 × (m/3) × (k/2) = mk/3
2 × (m/3) × (3n) = 2mn
2 × (k/2) × (3n) = 3nk
So, m²/9 + mk/3 + k²/4 + 3nk + 2mn + 9n²
= m²/9 + k²/4 + 9n² + + mk/3 + 3nk + 2mn [ Rearranging the terms]
= (m/3)² + (k/2)² + (3n)² + 2 × (m/3) × (k/2) + 2 × (m/3) × (3n) + 2 × (k/2) × (3n)
= (m/3 + k/2 + 3n)² [Using a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²]
(iv) p²/16 – 2 + 16/p²
Answer:
Writing -2 as:
-2 = -2 × (p/4) × (4/p)
Now, p²/16 = (p/4)²
16/p² = (4/p)²
So, p²/16 – 2 + 16/p²
= (p/4)² – 2(p/4)(4/p) + (4/p)²
= (p/4 – 4/p)² [Using a² – 2ab + b² = (a – b)²]
(v) 9a² + 4b² + c² – 12ab + 6ac – 4bc
Answer:
We have:
9a² = (3a)²
4b² = (- 2b)² [Here 4b² = (- 2b)² as in two negative terms – 12ab and – 4bc, b is common]
c² = c²
Now, 9a² + 4b² + c² – 12ab + 6ac – 4bc
= (3a)² + (- 2b)² + c² + 2(3a)(- 2b) + 2(3a)(c) + 2(- 2b)(c)
= (3a – 2b + c)² [Using x² + y² + z² + 2xy + 2yz + 2zx = (x + y + z)²]
3. Expand the following using the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca:
(i) (p + 3q + 7r)²
Answer:
Here, a = p, b = 3q, c = 7r
Using the identity:
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
So, (p + 3q + 7r)²
= p² + (3q)² + (7r)² + 2(p)(3q) + 2(3q)(7r) + 2(p)(7r)
= p² + 9q² + 49r² + 6pq + 42qr + 14pr
(ii) (3x − 2y + 4z)²
Answer:
Write it as:
[3x + (−2y) + 4z]²
Here,
a = 3x, b = −2y, c = 4z
Using the identity:
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
So, (3x − 2y + 4z)²
= (3x)² + (−2y)² + (4z)² + 2(3x)(−2y) + 2(−2y)(4z) + 2(3x)(4z)
= 9x² + 4y² + 16z² − 12xy − 16yz + 24xz
4. Is this an identity? (a + b − c)² + (a − b + c)² + (a − b − c)² = 2a² + 2b² + 2c²
Answer:
To check whether this is an identity, expand the left-hand side.
First Part: (a + b − c)²
= a² + b² + c² + 2ab − 2ac − 2bc
Second Part: (a − b + c)²
= a² + b² + c² − 2ab + 2ac − 2bc
Third Part: (a − b − c)²
= a² + b² + c² − 2ab − 2ac + 2bc
Now adding all three:
LHS = (a² + b² + c² + 2ab − 2ac − 2bc)
+ (a² + b² + c² − 2ab + 2ac − 2bc)
+ (a² + b² + c² − 2ab − 2ac + 2bc)
= 3a² + 3b² + 3c² − 2ab − 2ac − 2bc
This is NOT equal to 2a² + 2b² + 2c²
Hence, the given statement is not true for all values of a, b and c.
So, it is not an identity.
Class 9 Maths Ganita Manjari Chapter 4 Exercise Set 4.4 Solutions
Exercise Set 4.4
1. Fill in the blanks to complete the following identities:
(i) s² − 11s + 24 = ( ____ )( ____ )
Answer:
We need two numbers whose:
Sum = 11 and Product = 24
These are 3 and 8.
So, s² − 11s + 24
= s² − 8s – 3s + 24
= s(s − 8) – 3(s − 8)
= (s − 3)(s − 8)
So, s² − 11s + 24 = = (s − 3)(s − 8)
(ii) ( ____ )(x + 1) = (3x² − 4x − 7)
Answer:
RHS: 3x² − 4x − 7
= 3x² − 7x + 3x − 7
= x(3x − 7) + 1(3x – 7)
= (3x − 7)(x + 1)
So, (3x – 7)(x + 1) = (3x² − 4x − 7)
(iii) 10x² − 11x − 6 = (2x − ____)( ____ + 2)
Answer:
LHS = 10x² − 11x − 6
= 10x² − 15x + 4x − 6
= 5x(2x − 3) + 2(2x − 3)
= (5x + 2)(2x − 3)
(2x − 3)(5x + 2)
So, 10x² − 11x − 6 = (2x − 3)(5x + 2)
(iv) 6x² + 7x + 2 = ( ____ )( ____ )
Answer:
= 6x² + 3x + 4x + 2
= 3x(2x + 1) + 2(2x + 1)
= (3x + 2)(2x + 1)
So, 6x² + 7x + 2 = = (3x + 2)(2x + 1)
2. Select and use the identity that will help you to find the following products without multiplying directly:
(i) (41)²
Answer:
41 = 40 + 1
= (40 + 1)²
= 40² + 2 × 40 × 1 + 1² [Using (a + b)² = a² + 2ab + b²]
= 1600 + 80 + 1
= 1681
(ii) (27)²
Answer:
27 = 30 − 3
So, (27)² = (30 − 3)²
= (30)² – 2 × 30 × 3 + (3)² [Using (a – b)² = a² – 2ab + b²]
= 900 − 180 + 9
= 729
(iii) (23 × 17)
Answer:
Use identity:
23 × 17 = (20 + 3)(20 − 3)
= 20² − 3² [Using (a + b)(a − b) = a² − b²]
= 400 − 9
= 391
(iv) (135)²
Answer:
135 = 100 + 35
So, (135)²
= (100 + 35)²
= (100)² + 2 × 100 × 35 + (35)² [Using (a + b)² = a² + 2ab + b²]
= 10000 + 7000 + 1225
= 18225
(v) (97)²
Answer:
97 = 100 − 3
So, (97)²
= (100 − 3)²
= (100)² – 2 × 100 × 3 + (3)² [Using (a – b)² = a² – 2ab + b²]
= 10000 − 600 + 9
= 9409
(vi) (18 × 29)
Answer:
18 × 29 =
= (20 − 2)(20 + 9)
= 20² + (− 2 + 9) × 20 − 18 [Using (x + a)(x + b) = x² + (a + b)x + a × b]
= 400 + 140 − 18
= 522
(vii) (34 × 43)
Answer:
= (38 − 4)(38 + 5)
= 38² + (− 4 + 5) × 38 − 20 [Using (x + a)(x + b) = x² + (a + b)x + a × b]
= 38² + 38 − 20
= 1444 + 38 − 20
= 1462
(viii) (205)²
Answer:
205 = 200 + 5
= (200 + 5)²
= (200)² + 2 × 200 × 5 + (5)² [Using (a + b)² = a² + 2ab + b²]
= 40000 + 2000 + 25
= 42025
3. Factor the following:
(i) 9a² + b² + 4c² − 6ab + 12ac − 4bc
Answer:
9a² + b² + 4c² − 6ab + 12ac − 4bc
= (3a)² + (b)² + (2c)² + 2(3a)(−b) + 2(−b)(2c) + 2(3a)(2c)
= (3a − b + 2c)² [Using x² + y² + z² + 2xy + 2xz + 2yz = (x + y + z)²]
(ii) 16s² + 25t² − 40st
Answer:
Now, 16s² + 25t² − 40st
= 16s² − 40st + 25t² [Rearranging the terms]
= (4s)² −2(4s)(5t) + (5t)²
= (4s − 5t)² [Using a² − 2ab + b² = (a − b)²]
(iii) r² − r − 42
Answer:
We need two numbers whose product = −42 and Sum = −1.
These numbers are −7 and 6.
So, r² − r − 42
= r² − 7r + 6r − 42
= r(r − 7) + 6(r − 7)
= (r − 7)(r + 6)
(iv) 49g² + 14gh + h²
Answer:
49g² + 14gh + h²
= (7g)² + 2(7g)(h) + (h)²
= (7g + h)² [Using a² + 2ab + b² = (a + b)²]
(v) 64u² + 121v² + 4w² − 176uv − 32uw + 44vw
Answer:
64u² + 121v² + 4w² − 176uv − 32uw + 44vw
= 64u² + 121v² + 4w² − 176uv + 44vw − 32uw [Rearranging the terms]
= (8u)² + (11v)² + (2w)² + 2(8u)(−11v) + 2(−11v)(−2w) + 2(8u)(−2w)
= (8u − 11v − 2w)² [Using x² + y² + z² + 2xy + 2xz + 2yz = = (x + y + z)²
Class 9 Maths Ganita Manjari Chapter 4 Exercise Set 4.5 Solutions
Exercise Set 4.5
1. Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero:
(i) (3p² − 3pq − 18q²)/(p² + 3pq − 10q²)
Answer:
First factorising the numerator:
3p² − 3pq − 18q²
= 3(p² − pq − 6q²)
= 3[p² − 3pq + 2pq − 6q²]
= 3[p(p − 3q) + 2q(p – 3q)
= 3(p − 3q)(p + 2q)
Now factorising the denominator:
p² + 3pq − 10q²
= p² + 5pq – 2pq − 10q²
= p(p + 5q) – 2q(p + 5q)
= (p + 5q)(p − 2q)
So, (3p² − 3pq − 18q²)/(p² + 3pq − 10q²)
= 3(p − 3q)(p + 2q)/[(p + 5q)(p − 2q)]
No common factor cancels.
(ii) (n³ − 3n²m + 3nm² − m³)/(5m² − 10mn + 5n²)
Answer:
Factorising the numerator using identity:
n³ − 3n²m + 3nm² − m³
= (n − m)³ [Using a³ − 3a²b + 3ab² − b³ = (a − b)³]
Factorising the denominator:
5m² − 10mn + 5n²
= 5(m² − 2mn + n²)
= 5(m − n)² [Using a² – 2ab + b² = (a − b)²]
Now, (n − m)³/[5(m − n)²]
= −(m − n)³/[5(m − n)²] [Since, (n − m) = −(m − n), so, (n − m)³ = −(m − n)³]
= −(m − n)/5
(iii) (w³ − v³ + x³ + 3wvx)/(w² + v² + x² − 2wv − 2vx + 2wx)
Answer:
Factorising the numerator:
w³ − v³ + x³ + 3wvx
= w³ + (- v)³ + x³ – 3w(- v)x
= (w − v + x)(w² + v² + x² + wv + vx − wx)
[Using a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)]
Now denominator:
w² + v² + x² − 2wv − 2vx + 2wx
= w² + (- v)² + x² + 2w(-v) + 2(−v)x + 2xw)
= (w − v + x)² [Using a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²]
Therefore, (w³ − v³ + x³ + 3wvx)/(w² + v² + x² − 2wv − 2vx + 2wx)
= [(w − v + x)(w² + v² + x² + wv + vx − wx)]/(w − v + x)²
= (w² + x² + v² − wx + vx − wv) / (w + x − v) [Canceling the common factor]
(iv) (4y² − 20yz + 25z²)/(25z² − 4y²)
Answer:
Factor numerator:
4y² − 20yz + 25z²
= (2y)² − 2(2y)(5z) + (5z)²
= (2y − 5z)² [Using a² – 2ab + b² = (a − b)²]
Factorising the denominator:
25z² − 4y²
= (5z − 2y)(5z + 2y) [Using a² – b² = (a − b)(a + b)]
So, (4y² − 20yz + 25z²)/(25z² − 4y²)
= (5z − 2y)²/[(5z − 2y)(5z + 2y)] [Since (2y − 5z)² = (−1)²(5z − 2y)² = (5z − 2y)²]
= (5z − 2y)/(5z + 2y)
(v) [(x² + x − 6)(x² − 7x + 12)]/[(x² − 6x + 8)(x² − 9)]
Answer:
Factorising each polynomial:
x² + x − 6 = x² + 3x – 2x − 6 = x(x + 3) – 2(x + 3) = (x + 3)(x − 2)
x² − 7x + 12 = x² – 3x – 4x + 12 = x(x – 3) – 4(x – 3) = (x − 3)(x − 4)
x² − 6x + 8 = x² – 4x – 2x + 8 = x(x – 4) – 2(x – 4) = (x − 2)(x − 4)
x² − 9 = x² − 3 = (x − 3)(x + 3)
Substituting each polynomial, we get:
[(x + 3)(x − 2)][(x − 3)(x − 4)]/[(x − 2)(x − 4)][(x − 3)(x + 3)]
= 1 [Since all factors cancel]
(vi) (p⁴ − 16) / (p² − 4p + 4)
Answer:
Factorising numerator:
p⁴ − 16
= (p² − 4)(p² + 4) [Using a² − b² = (a + b)(a − b)]
= (p − 2)(p + 2)(p² + 4) [Again using a² − b² = (a + b)(a − b)]
Factorising denominator:
p² − 4p + 4
= p² − 2(p)(2) + 2²
= (p − 2)² [Using a² – 2ab + b² = (a – b)²]
So, [(p − 2)(p + 2)(p² + 4)]/(p − 2)²
= (p + 2)(p² + 4) / (p − 2) [Cancelling common factor (p − 2)].
Class 9 Maths Ganita Manjari Chapter 4 End-of-Chapter Exercises Solutions
End-of-Chapter Exercises
1. Use suitable identities to find the following products:
(i) (−3x + 4)²
Answer:
(−3x + 4)²
= (−3x)² + 2(−3x)(4) + 4² [Using the identity (a + b)² = a² + 2ab + b²]
= 9x² − 24x + 16
(ii) (2s + 7)(2s − 7)
Answer:
(2s + 7)(2s − 7)
= (2s)² − 7² [Using the identity (a + b)(a − b) = a² − b²]
= 4s² − 49
(iii) (p² + 1/2)(p² − 1/2)
Answer:
(p² + 1/2)(p² − 1/2)
= (p²)² − (1/2)² [Using the identity (a + b)(a − b) = a² − b²]
= p⁴ − 1/4
(iv) (2n + 7)(2n − 7)
Answer:
(2n + 7)(2n − 7)
= (2n)² − 7² [Using the identity (a + b)(a − b) = a² − b²]
= 4n² − 49
(v) (s − 2t)(s² + 2st + 4t²)
Answer:
(s − 2t)(s² + 2st + 4t²)
= s³ − (2t)³ [Using the identity (a − b)(a² + ab + b²) = a³ − b³]
= s³ − 8t³
(vi) [1/(2r) − 4r]²
Answer:
[1/(2r) − 4r]²
= (1/(2r))² − 2(1/(2r))(4r) + (4r)² [Using the identity (a − b)² = a² − 2ab + b²]
= 1/(4r²) − 4 + 16r²
(vii) (−3m + 4k − l)²
Answer:
So, (−3m + 4k − l)²
= (−3m)² + (4k)² + (−l)² + 2(−3m)(4k) + 2(4k)(−l) + 2(−3m)(−l)
[Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca]
= 9m² + 16k² + l² − 24mk − 8kl + 6ml
(viii) (x − 1/3 y)³
Answer:
(x − 1/3 y)³
= x³ − 3x²(y/3) + 3x(y/3)² − (y/3)³
[Using the identity (a − b)³ = a³ − 3a²b + 3ab² − b³]
= x³ − x²y + 3x(y²/9) − y³/27
= x³ − x²y + (1/3)xy² − y³/27
(ix) (7/2 k − 2/3 m)³
Answer:
[Using the identity (a − b)³ = a³ − 3a²b + 3ab² − b³]
Here, a = 7k/2, b = 2m/3
So, a³ = (7k/2)³ = 343k³/8
3a²b = 3 × (49k²/4) × (2m/3) = 49k²m/2
3ab² = 3 × (7k/2) × (4m²/9) = 14km²/3
b³ = (2m/3)³ = 8m³/27
Therefore, (7/2 k − 2/3 m)³
= 343k³/8 − 49k²m/2 + 14km²/3 − 8m³/27.
2. Find the values using suitable identities:
(i) 17 × 21
Answer:
Using identity: (a − b)(a + b) = a² − b²
We have 17 × 21 = (19 − 2)(19 + 2)
= 19² − 2²
= 361 − 4
= 357
(ii) 104 × 96
Answer:
Using identity: (a + b)(a − b) = a² − b²
We have 104 × 96 = (100 + 4)(100 − 4)
= 100² − 4²
= 10000 − 16
= 9984
(iii) 24 × 16
Answer:
Using identity: (a + b)(a − b) = a² − b²
We have 24 × 16 = (20 + 4)(20 − 4)
= 20² − 4²
= 400 − 16
= 384
(iv) 147³
Answer:
We know that 147 = 150 − 3
So, 147³ = (150 − 3)³
= 150³ − 3 × 150² × 3 + 3 × 150 × 3² − 3³ [Using identity (a + b)³ = a³ + 3a²b + 3ab² + b³]
= 3375000 − 202500 + 4050 − 27
= 3176523
(v) 199³
Answer:
199 = 200 − 1
So, 199³ = (200 − 1)³
= 200³ − 3 × 200² × 1 + 3 × 200 × 1² − 1 [Using identity: (a − b)³ = a³ − 3a²b + 3ab² − b³]
= 8000000 − 120000 + 600 − 1
= 7880599
(vi) 127³
Answer:
Here, 127 = 130 − 3
So, 127³ = (130 − 3)³
= 130³ − 3 × 130² × 3 + 3 × 130 × 3² − 3³ [Using identity: (a − b)³ = a³ − 3a²b + 3ab² − b³]
= 2197000 − 152100 + 3510 − 27
= 2048383
(vii) (−107)³
Answer:
(−107)³ = −(107³)
Now, 107 = 100 + 7
107³ = (100 + 7)³
= 100³ + 3 × 100² × 7 + 3 × 100 × 7² + 7³ [Using identity: (a + b)³ = a³ + 3a²b + 3ab² + b³]
= 1000000 + 210000 + 14700 + 343
= 1225043
So, (−107)³ = −1225043
(viii) (−299)³
Answer:
(−299)³ = −(299³)
Here, 299 = 300 − 1
So, 299³ = (300 − 1)³
= 300³ − 3 × 300² × 1 + 3 × 300 × 1² − 1 [Using identity: (a − b)³ = a³ − 3a²b + 3ab² − b³]
= 27000000 − 270000 + 900 − 1
= 26730900 − 1
= 26730899
So, (−299)³ = −26730899
3. Factor the following algebraic expressions:
(i) 4y² + 1 + 1/(16y²)
Answer:
Here, we have 4y² = (2y)²
1/(16y²) = (1/4y)²
and 2 × 2y × 1/(4y) = 1
So, this is of the form: a² + 2ab + b² = (a + b)²
Therefore, 4y² + 1 + 1/(16y²)
= (2y + 1/(4y))²
(ii) 9m² − 1/(25n²)
Answer:
9m² − 1/(25n²)
= (3m)² − (1/5n)²
= (3m + 1/5n)(3m − 1/5n) [Using a² − b² = (a + b)(a − b)]
(iii) 27b³ − 1/(64b³)
Answer:
27b³ − 1/(64b³)
= [3b − 1/(4b)][(3b)² + (3b)(1/4b) + (1/4b)²] [Using a³ − b³ = (a − b)(a² + ab + b²)]
= (3b − 1/4b)[9b² + 3/4 + 1/16b²]
(iv) x² + 5x/6 + 1/6
Answer:
We need two numbers whose sum is 5/6 and product is 1/6.
These numbers are 1/2 and 1/3.
So, x² + 5x/6 + 1/6
= x² + x/2 + x/3 + 1/6 [Splitting the middle term]
= x(x + 1/2) + 1/3(x + 1/2)
= (x + 1/3)(x + 1/2)
(v) 27u³ − 1/125 − 27u²/5 + 9u/25
Answer:
Given expression:
27u³ − 1/125 − 27u²/5 + 9u/25
= 27u³ − 27u²/5 + 9u/25 − 1/125 [Rearranging the terms]
= (3u)³ − 3(3u)²(1/5) + 3(3u)(1/5)² − (1/5)³
= (3u − 1/5)³ [Using a³ − 3a²b + 3ab² − b³ = (a − b)³]
(vi) 64y³ + 1/125 z³
Answer:
64y³ + 1/125 z³
= (4y)³ + (z/5)³
= (4y + z/5)[(4y)² − (4y)(z/5) + (z/5)²] [Using a³ + b³ = (a + b)(a² − ab + b²)]
= (4y + z/5)(16y² − 4yz/5 + z²/25)
(vii) p³ + 27q³ + r³ − 9pqr
Answer:
p³ + 27q³ + r³ − 9pqr
= p³ + (3q)³ + r³ − 3(p)(3q)(r)
= (p + 3q + r)[p² + (3q)² + r² − p(3q) − (3q)r − pr)]
[Using a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)]
= (p + 3q + r)(p² + 9q² + r² − 3pq − 3qr − pr)
(viii) 9m² − 12m + 4
Answer:
9m² − 12m + 4
= (3m)² − 2(3m)(2) + (2)²
= (3m – 2)² [Using a² – 2ab + b² = (a – b)²]
(ix) 9x³ − 8/3 y³ + z³/3 + 6xyz
Answer:
9x³ − 8/3 y³ + z³/3 + 6xyz
= (1/3)[9x³ − 8y³ + z³ + 18xyz]
= (1/3)[(3x)³ + (- 2y)³ + z³ – 3(3x)(-2y)z]
= (1/3)(3x – 2y + z)[(3x)² + (-2y)² + z² − (3x)(-2y) − (-2y)(z) − (z)(3x)]
[Using a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)]
= (1/3)(3x – 2y + z)[9x² + 4y² + z² + 6xy + 2yz − 3zx]
(x) 4x² + 9y² + 36z² + 12xz + 36yz + 24xy
Answer:
4x² + 9y² + 36z² + 12xz + 36yz + 24xy
= 4x² + 9y² + 36z² + 24xy + 36yz + 12xz
= (2x)² + (3y)² + (6z)² + 2(2x)(3y) + 2(3y)(6z) + 2(2x)(6z)
= (2x + 3y + 6z)² [Using a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²]
(xi) 27u³ − 1/216 − 9u²/2 + u/4
Answer:
27u³ − 1/216 − 9u²/2 + u/4
= 27u³ − 9u²/2 + u/4 − 1/216 [Rearranging the terms]
= (3u)³ − 3(3u)²(1/6) + 3(3u)(1/6)² − (1/6)³
= (3u − 1/6)³ [Using a³ − 3a²b + 3ab² − b³ = (a − b)³].
4. Simplify the following:
Note: Assume that the denominators are not equal to 0.
(i) (4x² + 4x + 1)/(4x² − 1)
Answer:
Factorising numerator:
4x² + 4x + 1
= (2x)² + 2(2x)(1) + 1²
= (2x + 1)² [Using a² + 2ab + b² = (a + b)²]
Factorising denominator:
4x² − 1 = (2x + 1)(2x − 1)
Now the expression: (4x² + 4x + 1)/(4x² − 1)
= (2x + 1)²/[(2x + 1)(2x − 1)]
= (2x + 1)/(2x − 1)
(ii) 9(3a³ − 24b³)/(9a² − 36b²)
Answer:
First simplify:
9(3a³ − 24b³) = 27(a³ − 8b³)
and 9a² − 36b² = 9(a² − 4b²)
So, 9(3a³ − 24b³)/(9a² − 36b²)
= 3(a³ − 8b³)/(a² − 4b²)
= 3(a − 2b)(a² + 2ab + 4b²)/(a² − 4b²) [Since a³ − 8b³ = a³ − (2b)³ = (a − 2b)(a² + 2ab + 4b²)]
= 3(a − 2b)(a² + 2ab + 4b²)/(a − 2b)(a + 2b) [Since a² − 4b² = (a − 2b)(a + 2b)]
= 3(a² + 2ab + 4b²)/(a + 2b) [Cancelling the common factor (a − 2b)]
(iii) (s³ + 125t³)/(s² − 2st − 35t²)
Answer:
Factorising numerator:
s³ + 125t³ = s³ + (5t)³
= (s + 5t)(s² − 5st + 25t²) [Using a³ + b³ = (a + b)(a² – ab + b²)]
Factorising denominator:
s² − 2st − 35t²
s² − 7st + 5st − 35t²
= s(s – 7t) + 5t(s − 7t)
= (s + 5t)(s − 7t)
Now the given expression:
(s³ + 125t³)/(s² − 2st − 35t²)
= (s + 5t)(s² − 5st + 25t²)/(s + 5t)(s − 7t)
= (s² − 5st + 25t²)/(s − 7t) [Cancelling common factor (s + 5t)].
5. Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units.
(i) 25a² − 30ab + 9b²
Answer:
This is a perfect square:
25a² − 30ab + 9b²
= (5a)² − 2(5a)(3b) + (3b)²
= (5a − 3b)² [Using a² – 2ab + b² = (a – b)²]
So possible length and breadth are (5a − 3b) and (5a − 3b).
(ii) 36s² − 49t²
Answer:
36s² − 49t²
= (6s)² − (7t)²
= (6s + 7t)(6s − 7t) [Using a² – b² = (a + b)(a – b)]
So possible length and breadth are (6s + 7t) and (6s − 7t).
6. Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units.
(i) 6a² − 24b²
Answer:
Take common factor:
6a² − 24b²
= 6(a² − 4b²)
= 6[a² − (2b)²]
= 6(a + 2b)(a − 2b) [Using a² – b² = (a + b)(a – b)]
So possible dimensions are 6, (a + 2b) and (a − 2b).
(ii) 3ps² − 15ps + 12p
Answer:
3ps² − 15ps + 12p
= 3p(s² − 5s + 4) [Taking common factor]
= 3p(s² − 4s – 1s + 4)
= 3p[s(s − 4) – 1(s − 4)]
= 3p(s − 1)(s − 4)
So, the possible dimensions are 3p, (s − 1) and (s − 4).
7. The village playground is shaped as a square of side 40 metres. A path of width s metres is created around the playground for people to walk. Find an expression for the area of the path in terms of s.
Answer:
Side of playground = 40 m
Since a path of width s metres is made all around the outside, the side of the outer square becomes:
40 + 2s
Area of outer square = (40 + 2s)²
Area of playground = 40² = 1600
Therefore, area of the path
= (40 + 2s)² − 1600
= (1600 + 160s + 4s²) − 1600
= 4s² + 160s
Hence, the required expression: Area of path = 4s² + 160s square metres.
8. If a number plus its reciprocal equals 10/3, find the number.
Answer:
Let the number be x.
Then, according to question: x + 1/x = 10/3
Multiply both sides by 3x:
3x² + 3 = 10x
⇒ 3x² − 10x + 3 = 0
⇒ 3x² − 9x − x + 3 = 0
⇒ 3x(x − 3) − 1(x − 3) = 0
⇒ (3x − 1)(x − 3) = 0
So, 3x − 1 = 0 or x − 3 = 0
Hence, x = 1/3 or x = 3
Therefore, the number is 3 or 1/3.
9. A rectangular pool has area 2x² + 7x + 3 square hastas. If its width is 2x + 1 hastas, find its length. Hasta was a unit used to measure length.
Answer:
Area of pool = 2x² + 7x + 3
Width = 2x + 1
Length = Area/Width
= (2x² + 7x + 3)/(2x + 1)
= (2x² + 6x + x + 3)/(2x + 1)
= [2x(x + 3) + 1(x + 3)]/(2x + 1)
= (2x + 1)(x + 3)/(2x + 1)
= x + 3
Therefore, the length of the pool is x + 3 hastas.
10. If both x − 2 and x − 1/2 are factors of px² + 5x + r, show that p = r.
Answer:
Since x − 2 and x − 1/2 are factors, the quadratic polynomial can be written as:
px² + 5x + r = k(x − 2)(x − 1/2)
⇒ (x − 2)(x − 1/2) = k[x² − (5/2)x + 1]
⇒ x² − (1/2)x − 2x + 1 = k[x² − (5/2)x + 1]
⇒ x² − (5/2)x + 1 = k[x² − (5/2)x + 1]
Comparing the coefficients on both sides, we get:
Coefficient of x² gives: k = p
Constant term gives: k = r
Therefore, p = r
Hence proved.
11. If a + b + c = 5 and ab + bc + ca = 10, then prove that a³ + b³ + c³ − 3abc = −25.
Answer:
Given:
a + b + c = 5
ab + bc + ca = 10
First finding a² + b² + c² using:
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
So, 5² = a² + b² + c² + 2(10)
⇒ 25 = a² + b² + c² + 20
⇒ a² + b² + c² = 5
Now, a² + b² + c² − ab − bc − ca = 5 − 10 = −5
Using the identity: a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)
a³ + b³ + c³ − 3abc
= (5)(−5)
= −25
Hence proved a³ + b³ + c³ − 3abc = −25.
12. By factoring the expression, check that n³ − n is always divisible by 6 for all natural numbers n. Give reasons.
Answer:
We have: n³ − n
= n(n² − 1)
= n(n − 1)(n + 1)
So, n³ − n = n(n − 1)(n + 1)
Thus, n³ − n is the product of three consecutive natural numbers: (n − 1), n, (n + 1)
Among any three consecutive natural numbers:
- one is always divisible by 3
- at least one is always even, so divisible by 2
Therefore, their product is always divisible by: 2 × 3 = 6
Hence, n³ − n is always divisible by 6 for all natural numbers n.
13. Find the value of:
(i) x³ + y³ − 12xy + 64, when x + y = −4
Answer:
Since x + y = −4, we have:
(x + y)³ = (−4)³ = −64
Using identity: (x + y)³ = x³ + y³ + 3xy(x + y)
So,
−64 = x³ + y³ + 3xy(−4)
⇒ −64 = x³ + y³ − 12xy
Therefore, x³ + y³ − 12xy = −64
Now, x³ + y³ − 12xy + 64 = −64 + 64 = 0
Hence, the value is 0.
(ii) x³ − 8y³ − 36xy − 216, when x − 2y + 6 = 0
Answer:
Given: x − 2y + 6 = 0
So, x − 2y = −6
Now use the identity: (a − b)³ = a³ − 3a²b + 3ab² − b³
We have: x³ − 8y³ − 36xy − 216
= x³ − 8y³ + 3(x)(−2y)(−6) + (−6)³
So it matches: a³ + b³ + c³ − 3abc
with a = x, b = −2y, c = −6
Using identity: a³ + b³ + c³ − 3abc
= (a + b + c)(a² + b² + c² − ab − bc − ca)
Thus, x³ − 8y³ − 36xy − 216
= x³ − 8y³ − 216 − 36xy
= x³ + (-2y)³ + (-6) − 3(x)(-2y)(-6)
= (x − 2y − 6)[x² + 4y² + 36² + 2xy − 12y + 18x]
But from the question:
x − 2y + 6 = 0, which gives x − 2y = −6, so x − 2y − 6 = −12, not zero.
So this expression does not become 0 under the given condition as written.
Using x = 2y − 6 from the condition, Substitute into the expression:
x³ − 8y³ − 36xy − 216
= (2y − 6)³ − 8y³ − 36(2y − 6)y − 216
= (8y³ − 72y² + 216y − 216) − 8y³ − 72y² + 216y − 216
= −144y² + 432y − 432
= −144(y² − 3y + 3)
Therefore, under the given condition, the value is −144(y² − 3y + 3).
So, the expression does not reduce to a constant unless there is a typo in the question.
Class 9 Maths Ganita Manjari Chapter 4 List of Formulae Used
All Algebraic Identities Covered in Class 9 Ganita Manjari Chapter 4
The following is the complete list of identities studied in this chapter:
| Identity | Expanded Form |
|---|---|
| (x + y)² | x² + 2xy + y² |
| (x − y)² | x² − 2xy + y² |
| (x + y + z)² | x² + y² + z² + 2xy + 2yz + 2zx |
| (x + y)(x − y) | x² − y² |
| (x + a)(x + b) | x² + (a + b)x + ab |
| (ax + b)(cx + d) | acx² + (ad + bc)x + bd |
| x³ − y³ | (x − y)(x² + xy + y²) |
| x³ + y³ | (x + y)(x² − xy + y²) |
| (x + y)³ | x³ + 3x²y + 3xy² + y³ |
| (x − y)³ | x³ − 3x²y + 3xy² − y³ |
| x³ + y³ + z³ − 3xyz | (x + y + z)(x² + y² + z² − xy − xz − yz) |
Key Concepts and Important Points
What is an Algebraic Identity?
An algebraic identity is an equation that is true for all values of the variables in it. For example, (x + y)² = x² + 2xy + y² is true whether x and y are positive, negative, rational, or any real number. This is the fundamental difference between an identity and an ordinary equation – an equation like x² − 1 = 24 is satisfied only for specific values (x = 5 or x = −5), making it an equation but not an identity.
Difference Between Equation and Identity:
- An equation is true only for specific values of the variable.
- An identity is true for all values of the variable.
Geometric Visualisation:
The chapter uses a particularly beautiful approach – geometric squares and rectangles – to justify why identities like (a + b)² = a² + 2ab + b² work. A square of side (a + b) units is divided into one a² square, one b² square, and two ab rectangles, confirming the identity visually.
Historical Note – ลhrฤซdharฤchฤrya’s Method (750 CE):
The textbook highlights that the identity a² = (a + b)(a − b) + b² was used by the Indian mathematician ลhrฤซdharฤchฤrya in 750 CE as a method to quickly compute squares of numbers — a proud reminder of India’s mathematical heritage.
