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

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 = a² + 2ab + b² 
(a – b)² = a^{2 –} 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) = x² + (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)² – (2x – 5)²
 (iii) (7m – 8n)²+ (7m + 8n)²
 (iv) (4m + 5n)² + (5m + 4n)²
 (v) (2.5p – 1.5q)² – (1.5p – 2.5q)²
 (vi) (ab + bc)²– 2ab²c
 (vii) (m² – n²m)² + 2m³n²
Answer:
1) a⁴ - b⁴
2) (2x + 5)² – (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)² + (5m + 4n)²
= 16m² + 40mn + 25n² + 25m² + 40mn + 16n²
= 41m² + 80mn + 41n²
5) (2.5p – 1.5q)² – (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) (m² – n²m)² + 2m³n²
= m⁴ - 2m³n² + m²n⁴ + 2m³n²
= m⁴ + m²n⁴


Question 5
 Show that.
(i) (3x + 7)² – 84x = (3x – 7)²
 (ii) (9p – 5q)²+ 180pq = (9p + 5q)²
iv) (4pq + 3q)²– (4pq – 3q)² = 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 = a² + 2ab + b² 
(a – b)² = a^{2 –} 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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