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Class 8 NCERT Solutions – Chapter 9 Algebraic Expressions and Identities – Exercise 9.5 | Set 1
  • Last Updated : 05 Apr, 2021

Question 1. Use a suitable identity to get each of the following products.

(i) (x + 3) (x + 3) 

Solution:

(x + 3) (x + 3) 

Putting formula (a + b)2 = a2 + b2+ 2ab

Put a = x  & b = 3



(x + 3) (x +3 ) = (x + 3)2

= x2 + 6x + 9

(ii) (2y + 5) (2y + 5)

Solution:

(2y + 5) (2y + 5)

Putting formula (a + b)2 = a2 + b2 + 2ab

Put a = 2y & b = 5

(2y + 5) (2y + 5) = (2y + 5)2

= 4y2 + 20y + 25

(iii) (2a – 7) (2a – 7)

Solution:

(2a – 7) (2a – 7)

Putting formula (x – y)2 = x2 + y2 – 2xy

Put x = 2a & y = 7

 (2a – 7) (2a – 7) = (2a – 7)2

= 4a2 – 28a + 49

(iv) (3a – 1/2) (3a – 1/2)

Solution:



(3a – 1/2) (3a – 1/2)

Putting formula (x – y)2 = x2 + y2 – 2xy

Put x = 3a & y = 1/2

= (3a)2 + (1/2)2 – 2 (3a) (1/2)

= 32a2 + 1/4 – 3a

= 9a2 + 1/4 – 3a

(v) (1.1m –  0.4) (1.1m + 0.4) 

Solution:

(1.1m – 0.4) (1.1m + 0.4)

Putting formula (a + b) (a – b) = a2– b2

Put a = 1.1m & b = 0.4

(1.1m – 0.4) (1.1m + 0.4) = (1.1m)2 – (0.4)2

= 1.21m2 – 0.16

(vi) (a2 + b2) (– a2 + b2)

Solution:

(a2 + b2) (– a2 + b2)

(a2 + b2) (– a2 + b2) = (b2 + a2) (b2 – a2)

Putting formula (x + y) (x – y) = x2 – y2

Put x = b2 &  y = a2

(b2 + a2) (b2 – a2) = b2×2 – a2×2

= b4 – a4

(vii) (6x – 7) (6x + 7)

Solution:

(6x – 7) (6x + 7)

Putting formula (a – b) (a + b) = a2 – b2

Put a = 6x & b = 7

(6x – 7) (6x + 7) = (6x)2 – 72

= 36x2 – 49

(viii) (– a + c) (– a + c)

Solution:

(– a + c) (– a + c)

(– a + c) (– a + c) = (c – a) (c – a)

Putting formula (x – y)2 = x2 + y2 – 2xy

Put x = c & y = a

 (c – a) (c – a) = c2 + a2 – 2ca

= a2 + c2 – 2ac

(ix) (x/2 + 3y/4) (x/2 + 3y/4)

Solution:

(x/2 + 3y/4) (x/2 + 3y/4)

Putting formula (a + b)2 = a2 + b2 + 2ab

put a = x/2 & b = 3y/4

= (x/2)2 + (3y/4)2 + 2 (x/2) (3y/4)

= x2/4 + 9y2/16 + 3xy/4

(x) (7a – 9b) (7a – 9b)

Solution:

(7a – 9b) (7a – 9b)

Putting formula (x – y)2 = x2+ y2– 2xy

Put x = 7a & y = 9b

(7a – 9b) (7a – 9b) = (7a)2 + (9b)2 – 2(7a)(9b)

= 49a2 + 81b2 – 126ab

Question 2. Use the identity (x + a) (x + b) = x2 + (a + b) x + ab to find the following products.

(i) (x + 3) (x + 7)

Solution:

(x + 3) (x + 7)

Formula (x + a) (x + b) = x2 + (a + b) x + ab

Put a = 3 & b = 7

= x2 + (3 + 7) x + (3 * 7)

= x2 +10x + 21 

(ii) (4x + 5) (4x + 1)

Solution:

(4x + 5) (4x + 1)

Formula (y + a) (y + b) = y2 + (a + b) y + ab

Put y = 4x , a = 5 & b = 1

= (4x)2 + (5 + 1) 4x + (5 * 1)

= 16x2 + 24x + 5

(iii) (4x – 5) (4x – 1)

Solution:

(4x – 5) (4x – 1)

Formula (y + a) (y + b) = y2 + (a + b) y + ab

Put y = 4x , a = -5 & b = -1

= (4x)2 + (-5 – 1) 4x + (-5 * -1)

= 16x2 – 24x + 5

(iv) (4x + 5) (4x – 1)

Solution:

(4x + 5) (4x – 1)

Formula (y + a) (y + b) = y2 + (a + b) y + ab

Put y = 4x , a = 5 & b = -1

= (4x)2 + (5 – 1) 4x + (5 * -1)

= 16x2 + 16x – 5

(v) (2x + 5y) (2x + 3y)

Solution:

(2x + 5y) (2x + 3y)

Formula (t + a) (t + b) = t2 + (a + b) t + ab

Put t = 2x , a = 5y & b = 3y

= (2x)2 + ( 5y + 3y) 2x + (5y * 3y)

= 4x2 + 16xy + 15y2

(vi) (2a2 + 9) (2a2 + 5)

Solution:

(2a2 + 9) (2a2 + 5)

Formula (x + y) (x + z) = x2 + (y + z) x + yz

Put x = 2a2 , y = 9 & z = 5

= (2a2)2 + (9 + 5) 2a2 + (9 * 5)

= 4a4 + 28a2 + 45

(vii) (xyz – 4) (xyz – 2)

Solution:

(xyz – 4) (xyz – 2)

Formula (t + a) (t + b) = t2 + (a + b) t + ab

Put t = xyz , a = -4 & b = -2

= (xyz)2 + (-4 + (-2)) xyz + ((-4) * (-2))

= x2y2z2 – 6xyz + 8

Question 3. Find the following squares by using the identities.

(i) (b – 7)2

Solution:

(b – 7)2

Using Formula (x – y)2 = x2 + y2 – 2xy

Putting x = b & y = 7

= b2 + 72 – 2(b)(7)

= b2 – 14b  + 49

(ii) (xy + 3z)2

Solution:

(xy + 3z)2

Using Formula (a + b)2 = a2 + b2 + 2ab

Putting a = xy & b = 3z

= x2y2 + 6xyz + 9z2

(iii) (6x2 – 5y)2

Solution:

(6x2 – 5y)2

Using Formula (a – b) 2 = a2 + b2 – 2ab

Putting a = 6x2 & b = 5y

= 36x4 – 60x2y + 25y2

(iv) [(2m/3) + (3n/2)]2

Solution:

[(2m/3) + (3n/2)]2

Using Formula (a + b)2 = a2 + b2 + 2ab

Putting a = 2m/3 & b = 3n/2

= (2m/3)2 + (3n/2)2 + 2 (2m/3) (3n/2)

= (4m2/9) + (9n2/4) + 2mn

(v) (0.4p – 0.5q)2

Solution:

(0.4p – 0.5q)2

Using Formula (a – b)2 = a2 + b2 – 2ab

Putting a = 0.4p & b = 0.5q

= 0.16p2 – 0.4pq + 0.25q2

(vi) (2xy + 5y)2

Solution:

(2xy + 5y)2

Using Formula (a + b)2 = a2 + b2 + 2ab

Putting a = 2xy & b = 5y

= (2xy)2 + (5y)2 + 2 (2xy) (5y)

= 4x2y2 + 20xy2 + 25y2

Question 4. Simplify.

(i) (a2 – b2)2

Solution:

(a2 – b2)2

Putting formula (x – y)2 = x2 + y2 – 2xy

Put x = a2 & y = b2

= a4 + b4 – 2a2b2

(ii) (2x + 5)2 – (2x – 5)2

Solution:

 (2x + 5)2  – (2x – 5)2

Putting formula (a + b)2 = a2 + b2+ 2ab &  (a – b)2 = a2 + b2 – 2ab

= 4x2 + 20x + 25 – (4x2 – 20x + 25)

= 4x2 + 20x + 25 – 4x2 + 20x – 25

= 40x

(iii) (7m – 8n)2 + (7m + 8n)2

Solution:

(7m – 8n)2 + (7m + 8n)2

Putting formula   (a – b)2 = a2 + b2 – 2ab & (a + b)2 = a2 + b2+ 2ab 

= (49m2 – 112mn + 64n2 ) + (49m2 + 112mn + 49n2)

= 98m2 + 128n2

(iv) (4m + 5n)2 + (5m + 4n)2

Solution:

(4m + 5n)2 + (5m + 4n)2

Putting formula (a + b)2 = a2 + b2+ 2ab

= (16m2 + 40mn + 25n2) + (25m2 + 40mn + 16n2)

= 41m2 + 80mn + 41n2

(v) (2.5p – 1.5q)2 – (1.5p – 2.5q)2

Solution:

(2.5p – 1.5q)2 – (1.5p – 2.5q)2

Putting formula   (a – b)2 = a2 + b2 – 2ab

= (6.25p2 – 7.5pq + 2.25q2)  –  (2.25p2 + 7.5pq – 6.25q2)

= 4p2 – 4q2

(vi) (ab + bc)2 – 2ab2c

Solution:

(ab + bc)2 –  2ab²c

Putting formula (a + b)2 = a2 + b2 + 2ab

= (a2b2 + 2ab2c + b2c2) – 2ab2c

= a2b2 + b2c2

(vii) (m2 – n2m)2 + 2m3n2

Solution:

(m2 – n2m)2 + 2m3n2

Putting formula (a – b)2 = a2 + b2 – 2ab

= (m4 – 2m3n2 + m2n4) + 2m3n2

= m4 + m2n4

Chapter 9 Algebraic Expressions and Identities – Exercise 9.5 | Set 2

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