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NCERT Class 9 Mathematics Exercise Set 4.3 Solutions: Exploring Algebraic Identities

Step-by-step working for every question of Exercise Set 4.3 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.

1. Ex 4.3 Q1

   Find these squares with a suitable identity, saying which identity makes the work easier: (i) 117², (ii) 78², (iii) 198², (iv) 214², (v) 1104², (vi) 1120².

   1. (i) Write 117 as (100 + 17) and use (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 100² + 2 × 100 × 17 + 17² = 10000 + 3400 + 289.
   3. (i) Add the terms to get 13689.
   4. (ii) Write 78 as (80 − 2) and use (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.

   5. (ii) Calculate 80² − 2 × 80 × 2 + 2² = 6400 − 320 + 4.
   6. (ii) Subtract and add to get 6084.
   7. (iii) Write 198 as (200 − 2) and use (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.

   8. (iii) Calculate 200² − 2 × 200 × 2 + 2² = 40000 − 800 + 4.
   9. (iii) Subtract and add to get 39204.
   10. (iv) Write 214 as (200 + 14) and use (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.

   11. (iv) Calculate 200² + 2 × 200 × 14 + 14² = 40000 + 5600 + 196.
   12. (iv) Add the terms to get 45796.
   13. (v) Write 1104 as (1100 + 4) and use (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.

   14. (v) Calculate 1100² + 2 × 1100 × 4 + 4² = 1210000 + 8800 + 16.
   15. (v) Add the terms to get 1218816.
   16. (vi) Write 1120 as (1100 + 20) and use (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.

   17. (vi) Calculate 1100² + 2 × 1100 × 20 + 20² = 1210000 + 44000 + 400.
   18. (vi) Add the terms to get 1254400.

   Answer: (i) 13689; (ii) 6084; (iii) 39204; (iv) 45796; (v) 1218816; (vi) 1254400

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2. Ex 4.3 Q2

   Factor with suitable identities: (i) 16y² − 24y + 9, (ii) (9/4)s² + 6st + 4t², (iii) m²/9 + mk/3 + k²/4 + 3nk + 2mn + 9n², (iv) p²/16 − 2 + 16/p², (v) 9a² + 4b² + c² − 12ab + 6ac − 4bc.

   1. (i) 16y² − 24y + 9 = (4y)² − 2(4y)(3) + 3²
   2. (i) Using (a − b)² = a² − 2ab + b², this equals (4y − 3)²
      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. (ii) (9/4)s² + 6st + 4t² = ((3/2)s)² + 2((3/2)s)(2t) + (2t)²
   4. (ii) Using (a + b)² = a² + 2ab + b², this equals ((3/2)s + 2t)²
      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.

   5. (iii) m²/9 + mk/3 + k²/4 + 3nk + 2mn + 9n² = (m/3)² + 2(m/3)(k/2) + (k/2)² + 2(m/3)(3n) + 2(k/2)(3n) + (3n)²
   6. (iii) Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca, this equals (m/3 + k/2 + 3n)²
   7. (iv) p²/16 − 2 + 16/p² = (p/4)² − 2(p/4)(4/p) + (4/p)²
   8. (iv) Using (a − b)² = a² − 2ab + b², this equals (p/4 − 4/p)²
      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.

   9. (v) 9a² + 4b² + c² − 12ab + 6ac − 4bc = (3a)² + (−2b)² + c² + 2(3a)(−2b) + 2(−2b)(c) + 2(c)(3a)
   10. (v) Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca, this equals (3a − 2b + c)²

   Answer: (i) (4y − 3)²; (ii) ((3/2)s + 2t)²; (iii) (m/3 + k/2 + 3n)²; (iv) (p/4 − 4/p)²; (v) (3a − 2b + c)²

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3. Ex 4.3 Q3

   Expand with (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca: (i) (p + 3q + 7r)², (ii) (3x − 2y + 4z)².

   1. (i) Identify a = p, b = 3q, c = 7r in the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.
   2. (i) Calculate the square terms: a² = p², b² = (3q)² = 9q², c² = (7r)² = 49r².
   3. (i) Calculate the product terms: 2ab = 2(p)(3q) = 6pq, 2bc = 2(3q)(7r) = 42qr, 2ca = 2(7r)(p) = 14pr.
   4. (i) Combine all terms to get p² + 9q² + 49r² + 6pq + 42qr + 14pr.
   5. (ii) Identify a = 3x, b = −2y, c = 4z in the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.
   6. (ii) Calculate the square terms: a² = (3x)² = 9x², b² = (−2y)² = 4y², c² = (4z)² = 16z².
   7. (ii) Calculate the product terms: 2ab = 2(3x)(−2y) = −12xy, 2bc = 2(−2y)(4z) = −16yz, 2ca = 2(4z)(3x) = 24xz.
   8. (ii) Combine all terms to get 9x² + 4y² + 16z² − 12xy − 16yz + 24xz.

   Answer: (i) p² + 9q² + 49r² + 6pq + 42qr + 14pr; (ii) 9x² + 4y² + 16z² − 12xy − 16yz + 24xz

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4. Ex 4.3 Q4

   Is (a + b − c)² + (a − b + c)² + (a − b − c)² = 2a² + 2b² + 2c² an identity?

   1. Expand (a + b − c)² as a² + b² + c² + 2ab − 2ac − 2bc.
      why?

      Expanding multiplies every term in one bracket by every term in the other: (a + b)(c + d) = ac + ad + bc + bd.

   2. Expand (a − b + c)² as a² + b² + c² − 2ab + 2ac − 2bc.
      why?

      Expanding multiplies every term in one bracket by every term in the other: (a + b)(c + d) = ac + ad + bc + bd.

   3. Expand (a − b − c)² as a² + b² + c² − 2ab − 2ac + 2bc.
      why?

      Expanding multiplies every term in one bracket by every term in the other: (a + b)(c + d) = ac + ad + bc + bd.

   4. Add the three expanded expressions together.
      why?

      Expanding multiplies every term in one bracket by every term in the other: (a + b)(c + d) = ac + ad + bc + bd.

   5. Combine the like terms: a² + a² + a² = 3a², b² + b² + b² = 3b², c² + c² + c² = 3c².
   6. Combine the ab terms: 2ab − 2ab − 2ab = −2ab.
   7. Combine the ac terms: −2ac + 2ac − 2ac = −2ac.
   8. Combine the bc terms: −2bc − 2bc + 2bc = −2bc.
   9. The simplified LHS is 3a² + 3b² + 3c² − 2ab − 2bc − 2ca.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   10. Compare this with the RHS 2a² + 2b² + 2c².
   11. Since 3a² + 3b² + 3c² − 2ab − 2bc − 2ca is not the same as 2a² + 2b² + 2c² (for example a = 1, b = 0, c = 0 gives 3 on the left and 2 on the right), the equation is not true for all values.

   Answer: It is not an identity: the left side equals 3a² + 3b² + 3c² − 2ab − 2bc − 2ca, not 2a² + 2b² + 2c².

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