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NCERT Class 9 Mathematics Exercise Set 4.1 Solutions: Exploring Algebraic Identities

AI-written, checked on 23 Oct 2026

Solutions written by Shishya, checked — not the board's official answers

Step-by-step working for every question of Exercise Set 4.1 in NCERT's Class 9 Mathematics book, chapter “Exploring Algebraic Identities”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

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  1. Ex 4.1 Q1

    Expand with (a + b)² = a² + 2ab + b²: (i) (7x + 4y)², (ii) ((7/5)x + (3/2)y)², (iii) (2.5p + 1.5q)², (iv) ((3/4)s + 8t)², (v) (x + 1/(2y))², (vi) (1/x + 1/y)².

    1. (i) Identify a = 7x and b = 4y in the identity (a + b)² = a² + 2ab + b².
      why?

      Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

    2. (i) Calculate a² = (7x)² = 49x², 2ab = 2(7x)(4y) = 56xy, and b² = (4y)² = 16y².
    3. (i) Combine the terms to get 49x² + 56xy + 16y².
    4. (ii) Identify a = (7/5)x and b = (3/2)y.
    5. (ii) Calculate a² = (49/25)x², 2ab = 2((7/5)x)((3/2)y) = (21/5)xy, and b² = (9/4)y².
    6. (ii) Combine the terms to get (49/25)x² + (21/5)xy + (9/4)y².
    7. (iii) Identify a = 2.5p and b = 1.5q.
    8. (iii) Calculate a² = (2.5)²p² = 6.25p², 2ab = 2(2.5p)(1.5q) = 7.5pq, and b² = (1.5)²q² = 2.25q².
    9. (iii) Combine the terms to get 6.25p² + 7.5pq + 2.25q².
    10. (iv) Identify a = (3/4)s and b = 8t.
    11. (iv) Calculate a² = (9/16)s², 2ab = 2((3/4)s)(8t) = 12st, and b² = (8t)² = 64t².
    12. (iv) Combine the terms to get (9/16)s² + 12st + 64t².
    13. (v) Identify a = x and b = 1/(2y).
    14. (v) Calculate a² = x², 2ab = 2(x)(1/(2y)) = x/y, and b² = (1/(2y))² = 1/(4y²).
    15. (v) Combine the terms to get x² + x/y + 1/(4y²).
    16. (vi) Identify a = 1/x and b = 1/y.
    17. (vi) Calculate a² = 1/x², 2ab = 2(1/x)(1/y) = 2/(xy), and b² = 1/y².
    18. (vi) Combine the terms to get 1/x² + 2/(xy) + 1/y².

    Answer: (i) 49x² + 56xy + 16y²; (ii) (49/25)x² + (21/5)xy + (9/4)y²; (iii) 6.25p² + 7.5pq + 2.25q²; (iv) (9/16)s² + 12st + 64t²; (v) x² + x/y + 1/(4y²); (vi) 1/x² + 2/(xy) + 1/y²

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  2. Ex 4.1 Q2

    Use (a + b)² to find (i) 64², (ii) 105², (iii) 205².

    1. (i) Write 64 as the sum of 60 and 4.
    2. (i) Apply the identity (a + b)² = a² + 2ab + b² with a = 60 and b = 4.
      why?

      Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

    3. (i) Calculate 60² + 2 × 60 × 4 + 4².
    4. (i) Add 3600 + 480 + 16 to get 4096.
    5. (ii) Write 105 as the sum of 100 and 5.
    6. (ii) Apply the identity (a + b)² = a² + 2ab + b² with a = 100 and b = 5.
      why?

      Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

    7. (ii) Calculate 100² + 2 × 100 × 5 + 5².
    8. (ii) Add 10000 + 1000 + 25 to get 11025.
    9. (iii) Write 205 as the sum of 200 and 5.
    10. (iii) Apply the identity (a + b)² = a² + 2ab + b² with a = 200 and b = 5.
      why?

      Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

    11. (iii) Calculate 200² + 2 × 200 × 5 + 5².
    12. (iii) Add 40000 + 2000 + 25 to get 42025.

    Answer: (i) 4096; (ii) 11025; (iii) 42025

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