Effective OP Malhotra Class 9 Solutions Chapter 3 Expansions Ex 3(A) can help bridge the gap between theory and application.

S Chand Class 9 ICSE Maths Solutions Chapter 3 Expansions Ex 3(A)

Question 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)
Solution:
(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

Question 2.
Write down the squares of the following expressions
(i) 3x + 5y
(ii) 5y – 2z
(iii) 5p – $$\frac { 1 }{ 4q }$$
(iv) (5x + 3y + z)²
(v) (- 3m – 5n + 2p)²
(v) (2x – $$\frac { 1 }{ 3 }$$p + 3q)²
Solution:
(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) $$\left(5 p-\frac{1}{4 q}\right)^2=(5 p)^2-2 \times 5 p \times \frac{1}{4 q}+\left(\frac{1}{4 q}\right)^2$$
= 25p² – $$\frac{5 p}{2 q}+\frac{1}{16 q^2}$$

(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) $$\left(2 x-\frac{1}{3} p+3 q\right)^2=(2 x)^2+\left(\frac{-1}{3} p\right)^2+(3 q)^2$$ + 2 × 2x × $$\left(\frac{-1}{3} p\right)+2 \times\left(\frac{-1}{3} p\right)$$ (3q) + 2 x 3q × 2x
= 4x² + $$\frac { 1 }{ 9 }$$ p² + 9q² – $$\frac { 4 }{ 3 }$$ xp – 2pq + 12qx

Question 3.
Simplify:
(2x – p + c)² – (2x + p – c)²
Solution:
(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

Question 4.
Write down the following products :
(i) (3b + 7) (3b – 7)
(ii) $$\left(\frac{1}{3}-5 x\right)\left(\frac{1}{3}+5 x\right)$$
(iii) (x³ – 3) (x³ + 3)
(iv) $$\left(a^4-\frac{1}{5 y}\right)\left(a^4+\frac{1}{5 y}\right)$$
Solution:
We know that (a + b) (a – b) = (a)² – (b)²
Now,
(i) (3b + 1) (3b – 7) = (3b)² – (7)² = 9b² – 49

(ii) $$\left(\frac{1}{3}-5 x\right)\left(\frac{1}{3}+5 x\right)$$ = ($$\frac { 1 }{ 3 }$$)² – (5x)²
= $$\frac { 1 }{ 9 }$$ – 25x²

(iii) (x³ – 3) (x³ + 3) = (x³)² – (3)² = x6 – 9

(iv) $$\left(a^4-\frac{1}{5 y}\right)\left(a^4+\frac{1}{5 y}\right)=\left(a^4\right)^2-\left(\frac{1}{5 y}\right)^2$$
= a8 – $$\frac{1}{25 y^2}$$

Question 5.
Find the products :
(i) (x + y) (x – y) (x² + y²)
(ii) (a² + b²) (a4 + b4) (a + b) (a – b)
Solution:
(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
= a8 – b8

Question 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 + $$\frac { 1 }{ 4x² }$$
(v) m² – 6m + 4
Solution:
(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 squre of (2m + 1)

(iv) 4x² – 2 + $$\frac { 1 }{ 4x² }$$
= (2x – $$\frac { 1 }{ 2x }$$)²
Which is a perfect square of (2x – $$\frac { 1 }{ 2x }$$)

(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

Question 7.
If 4x² – 12x + k is a perfect square, find the numerical value of k.
Solution:
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

Question 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
Solution:
(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

Question 9.
Write down the expansion of the following
(i) (a + 1)³
(ii) (3x – 2y)³
(iii) (x² + y)³
(iv) (2x – $$\frac { 1 }{ 3x }$$)³
(v) $$\left(\frac{a}{5}+\frac{b}{2}\right)^3$$
Solution:
(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 – $$\frac { 1 }{ 3x }$$)³ = (2x)³ – 3 x (2x)² ($$\frac { 1 }{ 3x }$$) + 3(2x)$$\left(\frac{1}{3 x}\right)^2-\left(\frac{1}{3 x}\right)^3$$
= 8x³ – 3 x 4x² x $$\frac { 1 }{ 3x }$$ + 3 × 2x × $$\frac{1}{9 x^2}-\frac{1}{27 x^3}$$
= 8x³ – 4x + $$\frac { 2 }{ 3x }$$ – $$\frac{1}{27 x^3}$$

(v) $$\left(\frac{a}{5}+\frac{b}{2}\right)^3$$ = $$\left(\frac{a}{5}\right)^3+3\left(\frac{a}{5}\right)^2\left(\frac{b}{2}\right)+3\left(\frac{a}{5}\right)$$$$\left(\frac{b}{2}\right)^2+\left(\frac{b}{2}\right)^3$$
= $$\frac{a^3}{125}+3 \times \frac{a^2}{25} \times \frac{b}{2}+3 \times \frac{a}{5} \times \frac{b^2}{4}+\frac{b^3}{8}$$
= $$\frac{a^3}{125}+\frac{3 a^2 b}{50}+\frac{3 a b^2}{20}+\frac{b^3}{8}$$