# Class 8 NCERT Solutions – Chapter 9 Algebraic Expressions and Identities – Exercise 9.5 | Set 2

### Chapter 9 Algebraic Expressions and Identities – **Exercise 9.5 | Set 1 **

**Question 5. Show that:**

**(i) (3x + 7) ^{2} – 84x = (3x – 7)^{2}**

**Solution:**

L.H.S. = (3x + 7)

^{2}– 84x= 9x

^{2}+ 42x + 49 – 84x= 9x

^{2}– 42x + 49= (3x – 7)

^{2}= R.H.S.

L.H.S. = R.H.S.

**(ii) (9p – 5q) ^{2} + 180pq = (9p + 5q)^{2}**

**Solution:**

LHS = (9p – 5q)

^{2 }+ 180pq= 81p

^{2}– 90pq + 25q^{2}+ 180pq= 81p

^{2}+ 90pq + 25q^{2}RHS = (9p + 5q)

^{2}= 81p

^{2}+ 90pq + 25q^{2}LHS = RHS

**(iii) (4/3 m – 3/4 n) ^{2}+ 2mn = 16/9 m^{2}+ 9/16 n^{2}**

**Solution:**

LHS = (4/3 m – 3/4 n)

^{2 }+ 2mn= 16/9m

^{2}+ 9/16n^{2}– 2nm + 2mn=16/9 m

^{2 }+ 9/16 n^{2}= RHS

LHS = RHS

**(iv) (4pq + 3q) ^{2} – (4pq – 3q)^{2} = 48pq^{2}**

**Solution:**

LHS = (4pq + 3q)

^{2 }– (4pq – 3q)^{2}= 16p

^{2}q^{2}+ 24pq^{2}+ 9q^{2}– 16p^{2}q^{2}+ 24pq^{2}– 9q^{2}= 48pq

^{2}RHS = 48pq

^{2}LHS = RHS

**(v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0**

**Solution:**

LHS = (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)

= a

^{2}– b^{2}+ b^{2}– c^{2}+ c^{2}– a^{2}= 0

= RHS

**Question 6. Using identities, evaluate.**

**(i) 71²**

**Solution:**

71

^{2}= (70+1)^{2}Using formula (a + b)

^{ 2}= a^{2}+ b^{2}+ 2ab= 70

^{2}+ 1^{2 }+ 140= 4900 + 140 +1

= 5041

**(ii) 99²**

**Solution:**

99² = (100 -1)

^{2}Using formula (a – b)

^{2}= a^{2}+ b^{2}– 2ab= 100

^{2}+ 1^{2}– 200= 10000 – 200 + 1

= 9801

**(iii) 102 ^{2}**

**Solution:**

102

^{2 }= (100 + 2)^{2}Using formula (a + b)

^{2}= a^{2}+ b^{2}+ 2ab= 100

^{2}+ 400 + 2^{2}= 10000 + 400 + 4

= 10404

**(iv) 998 ^{2}**

**Solution:**

998

^{2 }= (1000 – 2)^{2}Using formula (a – b)

^{2}= a^{2}+ b^{2}– 2ab= 1000

^{2}– 4000 + 2^{2}= 1000000 – 4000 + 4

= 996004

**(v) 5.2²**

**Solution:**

5.2

^{2 }= (5 + 0.2)^{2}Using formula (a + b)

^{2}= a^{2}+ b^{2}+ 2ab= 5

^{2}+ 2 + 0.2^{2}= 25 + 2 + 0.4

= 27.4

**(vi) 297 × 303**

**Solution:**

297 × 303

= (300 – 3 ) (300 + 3)

Using formula (a + b) (a – b) = a

^{2}– b^{2}= 300

^{2}– 3^{2}= 90000 – 9

= 89991

**(vii) 78 × 82**

**Solution:**

78 × 82

= (80 – 2) (80 + 2)

Using formula (a + b) (a – b) = a

^{2}– b^{2}= 80

^{2}– 2^{2}= 6400 – 4

= 6396

**(viii) 8.9 ^{2}**

**Solution:**

8.9

^{2}= (9 – 0.1)^{2}Using formula (a – b)

^{2}= a^{2}+ b^{2}– 2ab= 9

^{2}– 1.8 + 0.1^{2}= 81 – 1.8 + 0.01

= 79.21

**(ix) 10.5 × 9.5**

**Solution:**

10.5 × 9.5 = (10 + 0.5) (10 – 0.5)

Using formula (a + b) (a – b) = a

^{2}– b^{2}= 10

^{2}– 0.5^{2}= 100 – 0.25

= 99.75

**Question 7. Using a**^{2} – b^{2} = (a + b) (a – b), find

^{2}– b

^{2}= (a + b) (a – b), find

**(i) 51 ^{2 }– 49^{2}**

**Solution:**

51

^{2}– 49^{2}= (51 + 49) (51 – 49)

= 100 × 2

= 200

**(ii) (1.02) ^{2 }– (0.98)^{2}**

**Solution:**

(1.02)

^{2 }– (0.98)^{2}= (1.02 + 0.98) (1.02 – 0.98)

= 2 × 0.04

= 0.08

**(iii) 153 ^{2 }– 147^{2}**

**Solution:**

153

^{2}– 147^{2}= (153 + 147) (153 – 147)

= 300 × 6

= 1800

**(iv) 12.1 ^{2}– 7.9^{2}**

**Solution:**

12.1

^{2}– 7.9^{2}= (12.1 + 7.9) (12.1 – 7.9)

= 20 × 4.2 = 84

**Question 8. Using (x + a) (x + b) = x**^{2} + (a + b) x + ab, find

^{2}+ (a + b) x + ab, find

**(i) 103 × 104**

**Solution:**

103 × 104

= (100 + 3) (100 + 4)

= 100

^{2}+ (3 + 4)100 + 12= 10000 + 700 + 12

= 10712

**(ii) 5.1 × 5.2**

**Solution:**

5.1 × 5.2

= (5 + 0.1) (5 + 0.2)

= 5

^{2}+ (0.1 + 0.2)5 + 0.1 × 0.2= 25 + 1.5 + 0.02

= 26.52

**(iii) 103 × 98**

**Solution:**

103 × 98

= (100 + 3) (100 – 2)

= 100

^{2}+ (3-2)100 – 6= 10000 + 100 – 6

= 10094

**(iv) 9.7 × 9.8**

**Solution:**

9.7 × 9.8

= (9 + 0.7) (9 + 0.8)

= 9

^{2}+ (0.7 + 0.8)9 + 0.56= 81 + 13.5 + 0.56

= 95.06