Class 9 - Chapter 4: Exploring Algebraic Identities

(i) 9x² + 24xy + 16y²

Compare with the identity:

(a + b)² = a² + 2ab + b²

9x² = (3x)²

16y² = (4y)²

2(3x)(4y) = 24xy

Therefore,

9x² + 24xy + 16y² = (3x + 4y)²

Answer: (3x + 4y)²

(ii) 4s² + 20st + 25t²

4s² = (2s)²

25t² = (5t)²

2(2s)(5t) = 20st

Therefore,

4s² + 20st + 25t² = (2s + 5t)²

Answer: (2s + 5t)²

(iii) 49x² + 28xy + 4y²

49x² = (7x)²

4y² = (2y)²

2(7x)(2y) = 28xy

Therefore,

49x² + 28xy + 4y² = (7x + 2y)²

Answer: (7x + 2y)²

(iv) 64p² + (32/3)pq + (4/9)q²

64p² = (8p)²

4q²/9 = (2q/3)²

2(8p)(2q/3) = 32pq/3

Therefore,

64p² + (32/3)pq + (4/9)q²

= (8p + 2q/3)²

Answer: (8p + 2q/3)²

(v) 3a² + 4ab + (4/3)b²

Take 3 as a common factor:

= 3[a² + (4/3)ab + (4/9)b²]

The expression inside the bracket is:

a² + 2(a)(2b/3) + (2b/3)²

Therefore,

= 3(a + 2b/3)²

Answer: 3(a + 2b/3)²

(vi) (9/5)s² + 6sv + 5v²

Take 1/5 as a common factor:

= (1/5)(9s² + 30sv + 25v²)

The bracket is a perfect square:

9s² + 30sv + 25v²

= (3s + 5v)²

Therefore,

(9/5)s² + 6sv + 5v²

= (1/5)(3s + 5v)²

Answer: (1/5)(3s + 5v)²

(i) (79)²

79 = 80 − 1

(79)² = (80 − 1)²

= 80² − 2(80)(1) + 1²

= 6400 − 160 + 1

= 6241

Answer: 6241

(ii) (193)²

193 = 200 − 7

(193)² = (200 − 7)²

= 200² − 2(200)(7) + 7²

= 40000 − 2800 + 49

= 37249

Answer: 37249

(iii) (299)²

299 = 300 − 1

(299)² = (300 − 1)²

= 300² − 2(300)(1) + 1²

= 90000 − 600 + 1

= 89401

Answer: 89401