Home
Class 8
MATHS
Using a^2-b^2=(a+b) (a-b), find (i) ...

Using `a^2-b^2=(a+b) (a-b),` find
(i) `51^2-49^2`
(ii) `(1.02)^2 - (0.98)^2`
(iii) `153^2 - 147^2`
(iv) `12.1^2 - 7.9^2`

Text Solution

Verified by Experts

Using the algebraic identity `a^2 - b^2 = (a + b)(a - b)`
`(i) 51^2 – 49^2`
`= (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)`
...
Promotional Banner

Topper's Solved these Questions

  • ALGEBRAIC EXPRESSIONS AND IDENTITIES

    NCERT|Exercise EXERCISE 9.4|3 Videos
  • ALGEBRAIC EXPRESSIONS AND IDENTITIES

    NCERT|Exercise EXERCISE 9.3|5 Videos
  • ALGEBRAIC EXPRESSIONS AND IDENTITIES

    NCERT|Exercise EXERCISE 9.1|4 Videos
  • COMPARING QUANTITIES

    NCERT|Exercise EXERCISE 8.1|6 Videos

Similar Questions

Explore conceptually related problems

Evaluate: (i) (1.0)^(2) (ii) (0.7)^(2) (iii) (0.04)^(2) (iv) (0.11)^(2)

Expand : (i) (a+b-c)^(2) " " (ii) (a-2b-5c)^(2) " " (iii) (3a-2b-5c)^(2) " " (iv) (2x+(1)/(x)+1)^(2) .

By using the a^(2)-b^(2), find the value of (1.02)^(2)-(.98)^(2).

(1 + i) ^ (100) = 2 ^ (49) (x + iy) thenx ^ (2) + y ^ (2) = (i) 0 (ii) 4 (iii) 8 (iv) 16

Using (x+a)(x+b)=x^(2)+(a+b)x+ab find (i)103xx104 (ii) 51xx52 (iii) 103xx98 (iv) 97xx98

Expand (i) (2a-5b-7c)^2 (ii) ( -3a+4b-5c)^2 (iii) ((1)/(2)a-(1)/(4)b+2)^2

Evaluate : (i) (a+6b)^(2) " " (ii) (3x-4y)^(2) " " (iii) (2a-b+c)^(2)

If a = 2 and b = 3, find the value of (i) a + b (ii) a^(2) + ab (iii) ab - a^(2) (iv) 2a - 3b (v) 5a^(2) - 2ab (vi) a^(3) - b^(3)

Find the value of: (i) (82)^(2) - (18)^(2) (ii) (128)^(2) - (72)^(2) (iii) 197 xx 203 (iv) (198 xx 198 - 102 xx 102)/(96) (v) (14.7 xx 15.3) (vi) (8.63)^(2) - (1.37)^(2)

tan (i log ((a-ib)/(a+ib))) = (i) ab (ii) (2ab)/(a^(2) -b^(2)) (iii) (a^( 2) -b^(2))/(ab) (iv) (2ab)/(a^(2)+b^(2))