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Ncert Solutions for Algebraic Expressions and Identities Class 8 Maths Chapter 9 Exercise 9.5





In this page we have Ncert Solutions for Algebraic Expressions and Identities Class 8 Maths Chapter 9 for Exercise 9.5. Hope you like them and do not forget to like , social share and comment at the end of the page.
Question 1
 Use a suitable identity to get each of the following products.
(i) (x + 3) (x + 3)
 (ii) (2y + 5) (2y + 5)
 (iii) (2a – 7) (2a – 7)
iv) (3a-1/2)(3a-1/2)
(v) (1.1m – 0.4) (1.1m + 0.4)
 (vi) (a2+ b2) (– a2+ b2)
 (vii) (6x – 7) (6x + 7)
 (viii) (– a + c) (– a + c)
ix) [ (x/2)+(3y/4)] [ (x/2)+(3y/4)]
(x) (7a – 9b) (7a – 9b)
Answer:
We will be using below identities in these question
(a + b)2 = a+ 2ab + b2 
(a – b)2 = a2 – 2ab + b2 
(a – b)(a + b) = a2 – b2
1) This question can be solved using the first identify
(x + 3) (x + 3)
 =x2 + 6x + 9
2) This question can be solved using the first identify
(2y + 5) (2y + 5)
=4y2 + 20y + 25
3) This question can be solved using the second identify
(2a – 7) (2a – 7)
         = 4a2 – 28a + 49
 4) This question can be solved using the second identify
       (3a-1/2)(3a-1/2)=  9a2 -3a+(1/4)
5)  This question can be solved using the third identity
(1.1m – 0.4) (1.1m + 0.4)
= 1.21m2 – 0.16
5) This question can be solved using the third identity
(a2+ b2) (– a2+ b2)
= (b2 + a2 ) (b2 – a2)
= a4 - b4
6) This question can be solved using the third identity
(6x – 7) (6x + 7)
=36x2 - 49
7) This question can be solved using the third identity
= c2 - a2
8) This question can be solved using the first identity
[ (x/2)+(3y/4)] [ (x/2)+(3y/4)]
=(x2/4) + (9y2/16) +(3xy/4)
9) This question can be solved using second identity
= 49a2 – 126ab + 81b2
Question 2
 Use the identity (x + a) (x + b) = x+ (a + b) x + ab to find the following products.
(i) (x + 3) (x + 7)
 (ii) (4x + 5) (4x + 1)
 (iii) (4x – 5) (4x – 1)
 (iv) (4x + 5) (4x – 1)
 (v) (2x + 5y) (2x + 3y)
 (vi) (2a2+ 9) (2a2+ 5)
 (vii) (xyz – 4) (xyz – 2)
Answer:
1) x2 + (3+7)x + 21
= x2 + 10x + 21
2) (4x + 5) (4x + 1)
= 16x2 + (5 + 1)4x + 5
= 16x2 + 24x + 5
3) (4x – 5) (4x – 1)
= 16x2 + (-5-1)4x + 5
= 16x2 - 20x + 5
4) (4x + 5) (4x – 1)
= 16x2 + (5-1)4x - 5
= 16x2 +16x - 5
5) (2x + 5y) (2x + 3y)
= 4x2 + (5y + 3y)4x + 15y2
= 4x2 + 32xy + 15y2
6) (2a2+ 9) (2a2+ 5)
= 4a4 + (9+5)2a2 + 45
= 4a4 + 28a2 + 45
7) (xyz – 4) (xyz – 2)
= x2y2z2 + (-4 -2)xyz - 8
= x2y2z2 - 6xyz – 8
Question 3
 Find the following squares by using the identities.
(i) (b – 7)2
 (ii) (xy + 3z)2
 (iii) (6x2– 5y)2
 iv) [(2m/3}) + (3n/2)]2
(v) (0.4p – 0.5q)2
 (vi) (2xy + 5y)
Answer:
  1. b2 - 14b + 49
  2. x2y2 + 6xyz + 9z2
  3. 36x4 - 60x2y + 25y2
  4. (4m2/9) +(9n2/4) +2mn
  5. 0.16p2 - 0.4pq + 0.25q2
  6. 4x2y2 + 20xy2 + 25y2
Question 4
 Simplify.
(i) (a2– b2)2
 (ii) (2x + 5)– (2x – 5)2
 (iii) (7m – 8n)2+ (7m + 8n)2
 (iv) (4m + 5n)+ (5m + 4n)2
 (v) (2.5p – 1.5q)– (1.5p – 2.5q)2
 (vi) (ab + bc)2– 2ab²c
 (vii) (m– n2m)+ 2m3n2
Answer:
1) a4 - b4
2) (2x + 5)– (2x – 5)2
= 4x2 + 20x +25 - (4x2 - 20x + 25)
= 4x2 + 20x + 25 - 4x2 + 20x - 25
= 40
3) (7m – 8n)2+ (7m + 8n)2
= 49m2 - 112mn + 64n2 + 49m2 + 112mn + 49n2
= 98m2 + 128n2
4) (4m + 5n)+ (5m + 4n)2
= 16m2 + 40mn + 25n2 + 25m2 + 40mn + 16n2
= 41m2 + 80mn + 41n2
5) (2.5p – 1.5q)– (1.5p – 2.5q)2
= 6.25p2 - 7.5pq + 2.25q2 - 2.25p2 + 7.5pq - 6.25q2
= 4p2 - 4q2
6) (ab + bc)2– 2ab²c
= a2b2 + 2ab2c + b2c2 - 2ab2c
= a2b2 + b2c2
7) (m– n2m)+ 2m3n2
= m4 - 2m3n2 + m2n4 + 2m3n2
= m4 + m2n4


Question 5
 Show that.
(i) (3x + 7)– 84x = (3x – 7)2
 (ii) (9p – 5q)2+ 180pq = (9p + 5q)2
iv) (4pq + 3q)2– (4pq – 3q)= 48pq2
v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
Answer:
1) LHS = 9x2 + 42x + 49 - 84x
= 9x2 - 42x + 49
RHS = 9x2 - 42x + 49
LHS = RHS
2)  LHS = 91p2 - 90pq + 25q2 + 180pq
= 91p2 + 90pq + 25q2
RHS = 91p2 + 90pq + 25q2
4)  LHS= 16p2q2 + 24pq2 + 9q2 - 16p2q2 + 24pq2 - 9q2
= 48pq2
5) LHS= a2 - b2 + b2 - c2 + c2 - a2
= 0
Question 6
Using identities, evaluate.
(i) 71²
 (ii) 99²
 (iii) 1022
 (iv) 998²
 (v) 5.2²
 (vi) 297 x 303
 (vii) 78 x 82
 (viii) 8.92
 (ix) 10.5 x 9.5
Answer:
We will be using below identities in these question
(a + b)2 = a+ 2ab + b2 
(a – b)2 = a2 – 2ab + b2 
(a – b)(a + b) = a2 – b2
1) 712 = (70+1)2
         = 702 + 140 + 12
          = 4900 + 140 +1= 5041
2) 99²
= (100 -1)2
= 1002 - 200 + 12
= 10000 - 200 + 1
= 9801
3) 1022= (100 + 2)2
= 1002 + 400 + 22
= 10000 + 400 + 4
= 10404
4) 9982= (1000 - 2)2
= 10002 - 4000 + 22
= 1000000 - 4000 + 4
= 996004
5) 5.22= (5 + 0.2)2
= 52 + 2 + 0.22
= 25 + 2 + 0.4
= 27.4
6) = (300 - 3 )(300 + 3)
= 3002 - 32
= 90000 - 9
= 89991
7) = (80 - 2)(80 + 2)
= 802 - 22
= 6400 - 4
= 6396
8) 8.92= (9 - 0.1)2
= 92 - 1.8 + 0.12
= 81 - 1.8 + 0.01
= 79.21
9) = (10 + 0.5)(10 - 0.5)
= 102 - 0.52
= 100 - 0.25
= 99.75
Question 7
 Using a2– b2 = (a + b) (a – b), find
(i) 512– 492
 (ii) (1.02)2– (0.98)2
 (iii) 1532– 1472
 (iv) 12.12– 7.92
Answer:
1) = (51 + 49)(51 - 49)
= 100 x 2
= 200
2) = (1.02 + 0.98)(1.02 - 0.98)
= 2 x 0.04
= 0.08
3) = (153 + 147)(153 - 147)
= 300 x 6
= 1800
4) = (12.1 + 7.9)(12.1 - 7.9)
= 20 x 4.2
= 84
Question 8
 Using (x + a) (x + b) = x2+ (a + b) x + ab, find
(i) 103 x 104
 (ii) 5.1 x 5.2
 (iii) 103 x 98
 (iv) 9.7 x 9.8
Answer:
1) = (100 + 3)(100 + 4)
= 1002 + (3 + 4)100 + 12
= 10000 + 1200 + 12
= 11212
2) = (5 + 0.1)(5 + 0.2)
= 52 + (0.1+0.2)5 + 0.02
= 25 + 1.5 + 0.02
= 26.52
3) = (100 + 3)(100 - 2)
= 1002 + (3-2)100 - 6
= 10000 + 100 - 6
= 10094
4) = (9 + 0.7 )(9 + 0.8)
= 92 + (0.7 + 0.8)9 + 0.63
= 81 + 13.5 + 0.63
= 95.13

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Class 8 Maths Class 8 Science
Reference Books for class 8 Math

Given below are the links of some of the reference books for class 8 Math.

  1. Mathematics Foundation Course for JEE/Olympiad : Class 8 This book can take students maths skills further. Only buy if child is interested in Olympiad/JEE foundation courses.
  2. Mathematics for Class 8 by R S Aggarwal Detailed Mathematics book to clear basics and concepts. I would say it is a must have book for class 8 student.
  3. Pearson Foundation Series (IIT -JEE / NEET) Physics, Chemistry, Maths & Biology for Class 8 (Main Books) | PCMB Combo : These set of books could help your child if he aims to get extra knowledge of science and maths. These would be helpful if child wants to prepare for competitive exams like JEE/NEET. Only buy if you can provide help to the child while studying.
  4. Reasoning Olympiad Workbook - Class 8 :- Reasoning helps sharpen the mind of child. I would recommend students practicing reasoning even though they are not appearing for Olympiad.

You can use above books for extra knowledge and practicing different questions.







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