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

Solution:

We will be solving the questions using the algebraic identity a² - b² = (a + b)(a - b)

(i) 51² – 49²

= (51 + 49)(51 - 49)

= (100)(2)

= 200

(ii) (1.02)² - (0.98)²

= (1.02 + 0.98)(1.02 - 0.98)

= (2)(0.04)

= 0.08

(iii) 153² - 147²

= (153 + 147)(153 - 147)

= (300)(6)

= 1800

(iv) 12.1² - 7.9²

= (12.1 + 7.9)(12.1 - 7.9)

= (20.0)(4.2)

= 84

☛ Check: NCERT Solutions for Class 8 Maths Chapter 9

Video Solution:

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

NCERT Solutions Class 8 Maths Chapter 9 Exercise 9.5 Question 7

Summary:

Using a² - b² = (a + b)(a - b), the value of the following expressions (i) 51² – 49²  (ii) (1.02)² - (0.98)² (iii) 153² -147² (iv) 12.1² - 7.9² are i) 200 ii) 0.08 iii) 1800 iv) 84

☛ Related Questions:

• 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) (1.1 m - 0.4)(1.1 m + 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)
• 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 + 3 y) (vi) (2a2 + 9)(2a2 + 5) (vii) (xyz - 4)(xyz - 2)
• Find the following squares by using the identities. (i)(b - 7)2 (ii) (xy + 3z)2 (iii) (6x2 - 5 y)2 (iv) (2m/3 + 3n/2)2 (v) (0.4 p - 0.5q)2 (vi) (2xy + 5 y)2
• 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