Question

a(a+1)(a+2)(a+3)(a+4)+1 = 71² Find a?​

Answer 1

Answer:

(a - b)2 = a2 - 2ab + b2

(a + b)(a - b) = a2 - b2

(i) 712

= (70 + 1)2

= (70)2 + 2(70)(1) + (1)2 [Since, (a + b)2 = a2 + 2ab + b2]

= 4900 + 140 + 1

= 5041

(ii) 992

= (100 - 1)2

= (100)2 - 2(100)(1) + (1)2 [Since, (a - b)2 = a2 - 2ab + b2]

= 10000 - 200 + 1

= 9801

(iii) 1022

= (100 + 2)2

= (100)2 + 2(100)(2) + (2)2 [Since, (a + b)2 = a2 + 2ab + b2]

= 10000 + 400 + 4

= 10404

(iv) 9982

= (1000 - 2)2

= (1000)2 - 2(1000)(2) + (2)2 [Since, (a - b)2 = a2 - 2ab + b2]

= 1000000 - 4000 + 4

= 996004

(v) (5.2)2

= (5.0 + 0.2)2

= (5.0)2 + 2(5.0)(0.2) + (0.2)2 [Since, (a + b)2 = a2 + 2ab + b2]

= 25 + 2 + 0.04

= 27.04

(vi) 297 × 303

= (300 - 3) × (300 + 3) [Since, (a + b)(a - b) = a2 - b2]

= (300)2 - (3)2

= 90000 - 9

= 89991

(vii) 78 × 82

= (80 - 2) × (80 + 2) [Since, (a + b)(a - b) = a2 - b2]

= (80)2 - (2)2

= 6400 - 4

= 6396

(viii) 8.92

= (9.0 - 0.1)2

= (9.0)2 - 2(9.0)(0.1) + (0.1)2 [Since, (a - b)2 = a2 - 2ab + b2]

= 81 - 1.8 + 0.01

= 79.21

(ix) 10.5 × 9.5

= (10 + 0.5) × (10 - 0.5)

= 102 - 0.52 [Since, (a + b)(a - b) = a2 - b2]

= 100 - 0.25

= 99.75