# class 8 maths exercise 8.5 solutions

In this page we have class 8 maths exercise 8.5 solutions for Algebraic Expressions.This exercise has questions about Standard identities and how to apply the identities to solve various problems. Hope you like them and do not forget to like , social share and comment at the end of the page.

## class 8 maths exercise 8.5 solutions

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)
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
(i) This question can be solved using the first identify
(x + 3) (x + 3)
=x2 + 6x + 9
(ii) This question can be solved using the first identify
(2y + 5) (2y + 5)
=4y2 + 20y + 25
(iii) This question can be solved using the second identify
(2a – 7) (2a – 7)
= 4a2 – 28a + 49
(iv) This question can be solved using the second identify
(3a-1/2)(3a-1/2)=  9a2 -3a+(1/4)
(v)  This question can be solved using the third identity
(1.1m – 0.4) (1.1m + 0.4)
= 1.21m2 – 0.16
(vi) This question can be solved using the third identity
(a2+ b2) (– a2+ b2)
= (b2 + a2 ) (b2 – a2)
= a4 - b4
(vii) This question can be solved using the third identity
(6x – 7) (6x + 7)
=36x2 - 49
(viii) This question can be solved using the third identity
= c2 - a2
(ix) This question can be solved using the first identity
[ (x/2)+(3y/4)] [ (x/2)+(3y/4)]
=(x2/4) + (9y2/16) +(3xy/4)
(x) 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)
(i) x2 + (3+7)x + 21
= x2 + 10x + 21
(ii) (4x + 5) (4x + 1)
= 16x2 + (5 + 1)4x + 5
= 16x2 + 24x + 5
(iii) (4x – 5) (4x – 1)
= 16x2 + (-5-1)4x + 5
= 16x2 - 20x + 5
(iv) (4x + 5) (4x – 1)
= 16x2 + (5-1)4x - 5
= 16x2 +16x - 5
(v) (2x + 5y) (2x + 3y)
= 4x2 + (5y + 3y)4x + 15y2
= 4x2 + 32xy + 15y2
(vi) (2a2+ 9) (2a2+ 5)
= 4a4 + (9+5)2a2 + 45
= 4a4 + 28a2 + 45
(vii) (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)
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
(i) a4 - b4
(ii) (2x + 5)– (2x – 5)2
= 4x2 + 20x +25 - (4x2 - 20x + 25)
= 4x2 + 20x + 25 - 4x2 + 20x - 25
= 40
(iii) (7m – 8n)2+ (7m + 8n)2
= 49m2 - 112mn + 64n2 + 49m2 + 112mn + 49n2
= 98m2 + 128n2
(iv) (4m + 5n)+ (5m + 4n)2
= 16m2 + 40mn + 25n2 + 25m2 + 40mn + 16n2
= 41m2 + 80mn + 41n2
(v) (2.5p – 1.5q)– (1.5p – 2.5q)2
= 6.25p2 - 7.5pq + 2.25q2 - 2.25p2 + 7.5pq - 6.25q2
= 4p2 - 4q2
(vi) (ab + bc)2– 2ab²c
= a2b2 + 2ab2c + b2c2 - 2ab2c
= a2b2 + b2c2
(vii) (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
(iii)
(iv) (4pq + 3q)2– (4pq – 3q)= 48pq2
(v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
(i) LHS = 9x2 + 42x + 49 - 84x
= 9x2 - 42x + 49
RHS = 9x2 - 42x + 49
LHS = RHS
(ii)  LHS = 91p2 - 90pq + 25q2 + 180pq
= 91p2 + 90pq + 25q2
RHS = 91p2 + 90pq + 25q2
(iii) LHS = $\frac {16}{9}m^2 + \frac {9}{16}n^2 -2mn + 2mn=\frac {16}{9)m^2 + \frac {9}{16}n^2$ RHS=$\frac {16}{9}m^2 + \frac {9}{16}n^2$ (iv)  LHS= 16p2q2 + 24pq2 + 9q2 - 16p2q2 + 24pq2 - 9q2
= 48pq2
(v) 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
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
(i)712 = (70+1)2
= 702 + 140 + 12
= 4900 + 140 +1= 5041
(ii) 99²
= (100 -1)2
= 1002 - 200 + 12
= 10000 - 200 + 1
= 9801
(iii) 1022= (100 + 2)2
= 1002 + 400 + 22
= 10000 + 400 + 4
= 10404
(iv) 9982= (1000 - 2)2
= 10002 - 4000 + 22
= 1000000 - 4000 + 4
= 996004
(v) 5.22= (5 + 0.2)2
= 52 + 2 + 0.22
= 25 + 2 + 0.4
= 27.4
(vi) = (300 - 3 )(300 + 3)
= 3002 - 32
= 90000 - 9
= 89991
(vii) = (80 - 2)(80 + 2)
= 802 - 22
= 6400 - 4
= 6396
(viii) 8.92= (9 - 0.1)2
= 92 - 1.8 + 0.12
= 81 - 1.8 + 0.01
= 79.21
(ix) = (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
(i) = (51 + 49)(51 - 49)
= 100 x 2
= 200
(ii) = (1.02 + 0.98)(1.02 - 0.98)
= 2 x 0.04
= 0.08
(iii) = (153 + 147)(153 - 147)
= 300 x 6
= 1800
(iv) = (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
(i) = (100 + 3)(100 + 4)
= 1002 + (3 + 4)100 + 12
= 10000 + 1200 + 12
= 11212
(ii) = (5 + 0.1)(5 + 0.2)
= 52 + (0.1+0.2)5 + 0.02
= 25 + 1.5 + 0.02
= 26.52
(iii) = (100 + 3)(100 - 2)
= 1002 + (3-2)100 - 6
= 10000 + 100 - 6
= 10094
(iv) = (9 + 0.7 )(9 + 0.8)
= 92 + (0.7 + 0.8)9 + 0.63
= 81 + 13.5 + 0.63
= 95.13

## Summary

1. class 8 maths exercise 8.5 solutions has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail. You can download the solutions as PDF in the below Link also
2. This chapter 8 has total 4 Exercise 8.1 ,8.2 ,8.3 ,8.4 as per latest syllabus. Exercise 9.5 is deleted as per Latest syllabus This is the deleted exercise in the chapter.You can explore previous exercise of this chapter by clicking the link below