Using (x + a)(x + b) = x² + (a + b)x + ab
(i) 103 × 104   (ii) 5.1 × 5.2   (iii) 103 × 98   (iv) 9.7 × 9.8

Solution:

We will be using the algebraic Identity (x + a)(x + b) = x² + (a + b)x + ab

(i) 103 × 104

= (100 + 3)(100 + 4)

= (100)² + (3 + 4)(100) + (3)(4)

= 10000 + 700 + 12

= 10712

(ii) 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

= (100 + 3)(100 - 2)

= (100)² + [3 + (-2)](100) + (3)(-2)

= 10000 + 100 - 6

= 10094

(iv) 9.7 × 9.8

= (10 - 0.3)(10 - 0.2)

= (10)² + [(-0.3) + (-0.2)](10) + (-0.3)(-0.2)

= 100 + (- 0.5)10 + 0.06

= 100 - 5 + 0.06

= 95.06

☛ Check: NCERT Solutions for Class 8 Maths Chapter 9

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Video Solution:

Using (x + a)(x + b) = x² + (a + b)x + ab (i) 103 × 104 (ii) 5.1 × 5.2 (iii) 103 × 98 (iv) 9.7 × 9.8

NCERT Solutions Class 8 Maths Chapter 9 Exercise 9.5 Question 8

Summary:

Using (x + a)(x + b) = x² + (a + b)x + ab the following expressions (i) 103 × 104 (ii) 5.1 × 5.2 (iii) 103 × 98 (iv) 9.7 × 9.8 are evaluated as follows i) 10712 ii) 26.52 iii) 10094 iv) 95.06

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☛ Related Questions:

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