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Algebraic identities · 4 marks

1. Try to evaluate the following using a suitable identity:
(i) 352 (ii) 652 (iii) 852 (iv) 1052
Do you observe any interesting pattern?
2. Observe the two rows of figures below. They represent an algebraic identity. Try to identify it.

  1. 1. Try to evaluate the following using a suitable identity: (i) 352 (ii) 652 (iii) 852 (iv) 1052 Do you observe any interesting pattern?
  2. 2. Observe the two rows of figures below. They represent an algebraic identity. Try to identify it.
Row 1a + b + c+a + b − c+a − b + c+a − b − cRow 22a+2b+2c
Answer: 1. 352 = 1225, 652 = 4225, 852 = 7225, 1052 = 11025 (using a2 = (a + 5)(a − 5) + 25). Pattern: for a number ending in 5, multiply the digits before the 5 by the next number, then write 25. 2. (a + b + c)2 + (a + b − c)2 + (a − b + c)2 + (a − b − c)2 = (2a)2 + (2b)2 + (2c)2.

Step-by-step solution

Idea: For part 1, use a2 = (a + b)(a − b) + b2 with b = 5: a number ending in 5 becomes a product of two round numbers. For part 2, the total area of the squares in each row is the same, because the pieces of one row can be rearranged into the other.

1. Try to evaluate the following using a suitable identity: (i) 352 (ii) 652 (iii) 852 (iv) 1052 Do you observe any interesting pattern?

  1. Suitable identity: a2 = (a + b)(a − b) + b2, taking b = 5 so that a + 5 and a − 5 end in 0.
  2. (i) 352 = (40)(30) + 25 = 1200 + 25 = 1225½ mark
  3. (ii) 652 = (70)(60) + 25 = 4200 + 25 = 4225½ mark
  4. (iii) 852 = (90)(80) + 25 = 7200 + 25 = 7225½ mark
  5. (iv) 1052 = (110)(100) + 25 = 11000 + 25 = 11025
  6. Pattern: every answer ends in 25, and the part before 25 is (number before the 5) × (that number + 1): 3 × 4 = 12 → 1225; 6 × 7 = 42 → 4225; 8 × 9 = 72 → 7225; 10 × 11 = 110 → 11025.½ mark
  7. Why: a number ending in 5 is 10n + 5. Then (10n + 5)2 = 100n2 + 100n + 25 = 100 × n(n + 1) + 25.
1225, 4225, 7225, 11025. Square of a number ending in 5 = n(n + 1) followed by 25.

2. Observe the two rows of figures below. They represent an algebraic identity. Try to identify it.

  1. Row 1 has four squares with sides a + b + c, a + b − c, a − b + c and a − b − c. Total area = (a + b + c)2 + (a + b − c)2 + (a − b + c)2 + (a − b − c)2.½ mark
  2. Row 2 has three squares with sides 2a, 2b, 2c. Total area = (2a)2 + (2b)2 + (2c)2 = 4a2 + 4b2 + 4c2.½ mark
  3. The pieces of the first row are cut and rearranged to make the second row, so the areas are equal.½ mark
  4. Check by algebra. Each of the four squares contains a2 + b2 + c2, giving 4a2 + 4b2 + 4c2. The cross terms: ab-terms +2 + 2 − 2 − 2 = 0; bc-terms +2 − 2 − 2 + 2 = 0; ca-terms +2 − 2 + 2 − 2 = 0. All cross terms cancel.½ mark
(a + b + c)2 + (a + b − c)2 + (a − b + c)2 + (a − b − c)2 = (2a)2 + (2b)2 + (2c)2 = 4(a2 + b2 + c2)
1. 35² = 1225, 65² = 4225, 85² = 7225, 105² = 11025; a number ending in 5 squared is n(n + 1) followed by 25. 2. (a + b + c)² + (a + b − c)² + (a − b + c)² + (a − b − c)² = (2a)² + (2b)² + (2c)².

Check: Part 2 with a = 5, b = 2, c = 1: 82 + 62 + 42 + 22 = 64 + 36 + 16 + 4 = 120, and 102 + 42 + 22 = 100 + 16 + 4 = 120 ✓.

Answer to write in the exam

1.

352 = (35 + 5)(35 − 5) + 52 = 40 × 30 + 25 = 1225 [a2 = (a + b)(a − b) + b2]

652 = 70 × 60 + 25 = 4225

852 = 90 × 80 + 25 = 7225

1052 = 110 × 100 + 25 = 11025

Pattern: (10n + 5)2 = 100n2 + 100n + 25 = 100n(n + 1) + 25

∴ 352 = 1225, 652 = 4225, 852 = 7225, 1052 = 11025; each square is n(n + 1) followed by 25

2.

Row 1 area = (a + b + c)2 + (a + b − c)2 + (a − b + c)2 + (a − b − c)2

= 4a2 + 4b2 + 4c2 (cross terms cancel)

Row 2 area = (2a)2 + (2b)2 + (2c)2 = 4a2 + 4b2 + 4c2

∴ (a + b + c)2 + (a + b − c)2 + (a − b + c)2 + (a − b − c)2 = (2a)2 + (2b)2 + (2c)2

Common mistakes that cost marks

  • In part 1, forgetting to add 25 at the end: 352 = 40 × 30 = 1200 is wrong; it is 1225.
  • In part 2, writing the right side as 2a2 + 2b2 + 2c2. The squares have sides 2a, 2b, 2c, so the areas are 4a2, 4b2, 4c2.
  • Getting a sign wrong in a cross term, e.g. (a − b − c)2 has +2bc, because (−b)(−c) is positive.

How this can come in the exam

MCQ (1 mark)

The value of 752 is

  1. 4925
  2. 5525
  3. 5625
  4. 5725
Show answer

(C) 5625
7 × 8 = 56, then write 25: 5625. (80 × 70 + 25 = 5625.)

Assertion–Reason (1 mark)

Assertion (A): (a + b + c)2 + (a + b − c)2 + (a − b + c)2 + (a − b − c)2 = 4(a2 + b2 + c2).
Reason (R): In these four expansions, every cross term (ab, bc, ca) appears twice with + and twice with −.

  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true but R is false.
  4. A is false but R is true.
Show answer

(A) Both A and R are true, and R is the correct explanation of A.
The cross terms cancel and the squares add up to 4(a2 + b2 + c2), so R explains A.

Try one yourself

Find 1152 and 9952 using the pattern.

Show answer

1152: 11 × 12 = 132 → 13225. 9952: 99 × 100 = 9900 → 990025.

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