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Algebraic identities · 3 marks

Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.

  1. (i) 1172
  2. (ii) 782
  3. (iii) 1982
  4. (iv) 2142
  5. (v) 11042
  6. (vi) 11202
Answer: (i) 1172 = 13689 (ii) 782 = 6084 (iii) 1982 = 39204 (iv) 2142 = 45796 (v) 11042 = 1218816 (vi) 11202 = 1254400

Step-by-step solution

Given: (a + b)2 = a2 + 2ab + b2; (a − b)2 = a2 − 2ab + b2; (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

Idea: The three identities to choose from are (a + b)2, (a − b)2 and (a + b + c)2. The easiest choice is the one that makes a a round number (100, 200, 1100…) and b as small as possible. A number just above a round number → use (a + b)2. A number just below a round number → use (a − b)2. Two terms are usually quicker than three, so (a + b + c)2 is rarely the easiest.

(i) 1172

  1. Easier identity: (a + b)2, with 117 = 100 + 17. 1002 is instant and 172 = 289 is a square most students know.
  2. 1172 = (100 + 17)2 = 1002 + 2(100)(17) + 172
  3. = 10000 + 3400 + 289 = 13689½ mark
  4. (Other routes give the same answer: (120 − 3)2 = 14400 − 720 + 9 = 13689, or (100 + 10 + 7)2 = 10000 + 100 + 49 + 2000 + 140 + 1400 = 13689. The three-term identity needs six terms, so it is the slowest.)
13689

(ii) 782

  1. Easier identity: (a − b)2, because 78 is just 2 below 80. Then b = 2 is very small.
  2. 782 = (80 − 2)2 = 802 − 2(80)(2) + 22
  3. = 6400 − 320 + 4 = 6084½ mark
6084

(iii) 1982

  1. Easier identity: (a − b)2, because 198 is just 2 below 200.
  2. 1982 = (200 − 2)2 = 2002 − 2(200)(2) + 22
  3. = 40000 − 800 + 4 = 39204½ mark
39204

(iv) 2142

  1. Easier identity: (a + b)2, because 214 is 14 above 200.
  2. 2142 = (200 + 14)2 = 2002 + 2(200)(14) + 142
  3. = 40000 + 5600 + 196 = 45796½ mark
45796

(v) 11042

  1. Easier identity: (a + b)2, with 1104 = 1100 + 4. 11002 = 112 × 1002 = 121 × 10000.
  2. 11042 = (1100 + 4)2 = 11002 + 2(1100)(4) + 42
  3. = 1210000 + 8800 + 16 = 1218816½ mark
1218816

(vi) 11202

  1. Easier identity: (a + b)2, with 1120 = 1100 + 20.
  2. 11202 = (1100 + 20)2 = 11002 + 2(1100)(20) + 202
  3. = 1210000 + 44000 + 400 = 1254400½ mark
1254400
(i) 117² = 13689 (ii) 78² = 6084 (iii) 198² = 39204 (iv) 214² = 45796 (v) 1104² = 1218816 (vi) 1120² = 1254400. Numbers just above a round number are easiest with (a + b)²; numbers just below are easiest with (a − b)².

Check: (vi) another way: 11202 = 1122 × 102, and 1122 = (100 + 12)2 = 10000 + 2400 + 144 = 12544, so 11202 = 1254400 ✓. (iii) and (ii): 8 × 8 = 64 ends in 4, so both squares must end in 4 ✓.

Answer to write in the exam

(i)

1172 = (100 + 17)2 = 1002 + 2(100)(17) + 172 [(a + b)2 = a2 + 2ab + b2]

= 10000 + 3400 + 289

∴ 1172 = 13689

(ii)

782 = (80 − 2)2 = 802 − 2(80)(2) + 22 [(a − b)2 = a2 − 2ab + b2]

= 6400 − 320 + 4

∴ 782 = 6084

(iii)

1982 = (200 − 2)2 = 2002 − 2(200)(2) + 22 [(a − b)2 = a2 − 2ab + b2]

= 40000 − 800 + 4

∴ 1982 = 39204

(iv)

2142 = (200 + 14)2 = 2002 + 2(200)(14) + 142 [(a + b)2 = a2 + 2ab + b2]

= 40000 + 5600 + 196

∴ 2142 = 45796

(v)

11042 = (1100 + 4)2 = 11002 + 2(1100)(4) + 42 [(a + b)2 = a2 + 2ab + b2]

= 1210000 + 8800 + 16

∴ 11042 = 1218816

(vi)

11202 = (1100 + 20)2 = 11002 + 2(1100)(20) + 202 [(a + b)2 = a2 + 2ab + b2]

= 1210000 + 44000 + 400

∴ 11202 = 1254400

Common mistakes that cost marks

  • Using (a + b)2 for 78 as (70 + 8)2. It is not wrong, but 80 − 2 is much quicker because the small number is only 2.
  • In (a − b)2, subtracting b2: 40000 − 800 − 4 = 39196 is wrong. The last term is always +b2.
  • Errors in counting zeros: 11002 = 1210000 (121 followed by four zeros), not 121000.

How this can come in the exam

MCQ (1 mark)

To find 992 most easily using an identity, we should write 99 as

  1. 90 + 9
  2. 100 − 1
  3. 50 + 49
  4. 33 × 3
Show answer

(B) 100 − 1
(100 − 1)2 = 10000 − 200 + 1 = 9801. Here a = 100 is round and b = 1 is as small as possible.

Short answer (2 marks)

Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca, find 1122.

Show answer112 = 100 + 10 + 2 (½ mark). 1122 = 1002 + 102 + 22 + 2(100)(10) + 2(10)(2) + 2(2)(100) (½ mark) = 10000 + 100 + 4 + 2000 + 40 + 400 (½ mark) = 12544 (½ mark).

Try one yourself

Choose the easier identity and find 3022 and 892.

Show answer

302 = 300 + 2, use (a + b)2: 90000 + 1200 + 4 = 91204. 89 = 90 − 1, use (a − b)2: 8100 − 180 + 1 = 7921.

More questions like this

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← Let us use this identity to find the square of a number, say 119:Factor using suitable identities: →