Expansions Class 9 OP Malhotra Exe-3A ICSE Maths Solutions Ch-3. We Provide Step by Step Solutions / Answer of Questions on Expansions OP Malhotra Maths. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics.
Expansions Class 9 OP Malhotra Exe-3A ICSE Maths Solutions Ch-3
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 9th |
| Chapter-3 | Expansions |
| Writer | OP Malhotra |
| Exe-3A | |
| Edition | 2025-2026 |
Exercise- 3A
Expansions Class 9 OP Malhotra Exe-3A ICSE Maths Solutions Ch-3
Que-1: Write down the products for each of the following:
(i) (x + 4) (x + 2)
(ii) (4a – 5) (5a + 6)
(iii) (xy + 6) (xy – 5)
(iv) (7x² – 5y) (x² – 3y)
Sol: (i) (x + 4) (x + 2)
= (x)² + (4 + 2) x + 4 x 2
= x² + 6x + 8
(ii) (4a – 5) (5a + 6)
= (4a x 5a) + (4a x 6) – (5 x 5a) – (5 x 6)
= 20a² + 24a – 25a – 30
= 20a² – a – 30
(iii) (xy + 6) (xy – 5)
= (xy)² + (6 – 5) xy + 6(- 5)
= x²y² + xy – 30
(iv) (7x² – 5y) (x² – 3y)
= 7x² × x² – 7x² × 3y – 5y × x² + 5y × 3y
= 7x4 – 21x²y – 5x²y + 15y²
= 7x4 – 26x²y + 15y²
Que-2: Write down the squares of the following expressions
(i) 3x + 5y
(ii) 5y – 2z
(iii) 5p – (1/4q)
(iv) (5x + 3y + z)²
(v) (- 3m – 5n + 2p)²
(v) (2x – (1/3p) + 3q)²
Sol: (i) (3x + 5y)²
= (3x)² + 2 x 3x × 5y + (5y)²
= 9x² + 30xy + 25y²
(ii) (5y – 2z)²
= (5y)² – 2 x 5y x 2z + (2z)²
= 25y² – 10yz + 4z²
(iii) (5p−(1/4q))²
= (5p)² − 2×5p×(1/4q) + (1/4q)²
= 25p² – (5p/2q) + (1/16q²)
(iv) (5x + 3y + z)²
= (5x)² + (3y)² + (z)² + (2 x 5x × 3y) + (2 x 3y x z) + (2 x z x 5x)
= 25x² + 9y² + z² + 30xy + 6yz + 10zx
(v) (-3m – 5n + 2p)²
= (-3m)² + (- 5n)² + {2 (-3m) (-5n)} + {2 (-5n) x 2p} + {1 x 2p x (-3m)}
= 9m² + 25 n² + 4p² + 30mn – 20np – 12pm
(vi) (2x−(1/3p)+3q)²
= (2x)²+(−1/3p)²+(3q)² + {2 × 2x × (−1/3p)} + {2×(−1/3p) (3q)} + {2 x 3q × 2x}
= 4x² + 19 p² + 9q² – 43 xp – 2pq + 12qx
Que-3: Simplify:
(2x – p + c)² – (2x + p – c)²
Sol: (2x – p + c)² – (2x + p – c)²
= (4x² + p² + c² – 4xp – 2pc + 4cx) – (4x² + p² + c² + 4px – 2pc – 4cx)
= 4x² + p² + c² – 4xp – 2pc + 4cx – 4x² – p² – c² – 4px + 2pc + 4cx
= – 8xp + 8cx.
Que-4: Write down the following products :
(i) (3b + 7) (3b – 7)
(ii) ((1/3)−5x) ((1/3)+5x)
(iii) (x³ – 3) (x³ + 3)
(iv) (a4−(1/5y)) (a4+(1/5y))
Sol: We know that (a + b) (a – b) = (a)² – (b)²
Now,
(i) (3b + 1) (3b – 7) = (3b)² – (7)² = 9b² – 49
(ii) ((1/3)−5x) ((1/3)+5x) = (1/3)² – (5x)²
= (1/9) – 25x².
(iii) (x³ – 3) (x³ + 3)
= (x³)² – (3)²
= x6 – 9.
(iv) (a4−(1/5y)) (a4+(1/5y))
= (a4)² − (1/5y)²
= a^8 – (1/25)y².
Que-5: Find the products :
(i) (x + y) (x – y) (x² + y²)
(ii) (a² + b²) (a4 + b4) (a + b) (a – b)
Sol: (i) (x + y) (x – y) (x² + y²)
= [(x)² – (y)²] (x² + y²)
= (x² – y²)(x² + y²)
= (x²)² – (y²)²
= x4 – y4
(ii) (a² + b²) (a4 + b4) (a + b) (a – b)
= (a² + b²) (a4 + b4) [(a)² – (b)²]
= (a² + b2) (a4 + b4) (a² – b2)
= (a² + b2) (a² – b2) (a4 + b4)
= [(a²)² – (b2)²] (a4 + b4)
= (a4 – b4) (a4 + b4)
= (a4)² – (b4)²
= a8 – b8
Que-6: State which of the following expressions is a perfect square :
(i) x² + 8x + 16
(ii) y² + 3y + 9
(iii) 4m² + 4m + 1
(iv) 4x² – 2 + (1/4)x²
(v) m² – 6m + 4
Sol: (i) x² + 8x + 16
= (x)² + 8 × x × 16 + (16)²
= (x + 16)²
Which is a perfect square of (x + 16)
(ii) y² + 3y + 9 = (y)² + 3y + (3)²
Since the second term 3y is not twice the
product of y and 3
∴ It is not a perfect square
(iii) 4m²+ 4m + 1
= (2m)² + 2 x 2m x 1 + (1)²
= (2m + 1)²
which is a perfect square of (2m + 1)
(iv) 4x² – 2 + (1/4)x²
= (2x – (1/2x))²
Which is a perfect square of (2x – (1/2x))
(v) m² – 6m + 4 = (m)² – 6m + (2)²
Since the second term 6m is not twice the product of m and 2
∴ It is not a perfect square
Que-7: If 4x² – 12x + k is a perfect square, find the numerical value of k.
Sol: 4x² – 12x + k
= (2x)² – 2 x 2x × 3 + (3)² {∴ 12x = 2 × 2x × 3}
Comparing we get, k = (3)² = 9
Hence k = 9
Que-8: What term should be added to each of the following expression to make it a perfect square?
(i) 4a² + 28a
(ii) 36a² + 49b²
(iii) 4a² + 81
(iv) 9a² + 2ab + b²
(v) 49a4 + 50a²b² + 16b4
Sol: (i) 4a² + 28a
= (2a)² + 2 x 2a x 7 + (7)²
In order to complete it in a perfect square, we have to add (7)² i.e., 49
∴ On adding 49, get (2a + 7)²
(ii) 36a² + 49b²
= (6a)² + (7b)² + 2 x 6a x 7b
In order to complete it to a perfect square, we have to add 2 x 6a x 7b i.e., 84ab
∴ On adding 84ab, we get (6a + 7b)²
(iii) 4a² + 81
= (2a)² + (9)² + 2 x 2a x 9
In order to complete it in a perfect square, we have to add 2 x 2a x 9 i.e., 36a
∴ On adding 36a, we get (2a + 9)²
(iv) 9a² + 2ab + b²
= (3a)² + (b)² + 2 x 3a x b
= (3a)² + (b)² + 6 ab
In order to complete it in a perfect square, we have to add 6ab – 2ab = 4ab
∴ On adding 4ab, we get (3a + b)²
(v) 49a4 + 50a²b² + 16b4
= (7a)² + 2 x 7a² x 4b² + (4b²)²
= (7a²)² + 56a²b² + (4b²)²
In order to complete it in a perfect square, we have add 56a²b² – 50a²b² i.e., 6a²b²
∴ On adding 6a²b², we get (7a² + 4b²)²
Que-9: Write down the expansion of the following
(i) (a + 1)³
(ii) (3x – 2y)³
(iii) (x² + y)³
(iv) (2x – (1/3x))³
(v) {(a/5) + (b2)}³
Sol: (i) (a + 1)³
= (a)³ + 3a² x 1 + 3a x (1)² + (1)³
= a³ + 3a² + 3a + 1.
(ii) (3x – 2y)³
= (3x)³ – 3(3x)² (2y) + 3(3x) (2y)² – (2y)³
= 27x³ – 3 x 9x² x 2y + 3 x 3x × 4y² – 8y³
= 27x³ – 54x²y + 36xy² – 8y³
(iii) (x² + y)³
= (x²)³ + 3 (x²)² (y) + 3 (x²) (y)² + (y)³
= x6 + 3 × x4 × y + 3x²y² + y³
= x6 + 3x4y + 3x²y² + y³.
(iv) (2x – (1/3x))³ = (2x)³ – 3 x (2x)² (1/3x) + 3(2x)(1/3x)² − (1/3x)³
= 8x³ – (3 x 4x² x (1/3x)) + (3 × 2x × (1/9x²)) − (1/27x³)
= 8x³ – 4x + (2/3x) – (1/27x³).
(v) {(a/5)+(b/2)}³ = (a/5)³ + 3(a/5)²(b/2) + 3(a/5)(b/2)² + (b/2)³
= (a³/125) + {3×(a²/25)×(b/2)} + {3×(a/5)×(b²/4)} + (b³8)
= (a³/125) + {(3a²b)/50} + {(3ab²)/20} + (b³/8).
— : End of Expansions Class 9 OP Malhotra Exe-3A ICSE Maths Solutions Ch-3 :–
Return to :– OP Malhotra S Chand Solutions for ICSE Class-9 Maths
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