Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

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:

Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12. 

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.

(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




My Personal Notes arrow_drop_up
Recommended Articles
Page :