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Algebraic identities · 3 marks

Using the identity (a + b)2 = a2 + 2ab + b2, expand the following:

  1. (i) (7x + 4y)2
  2. (ii) (75x + 32y)2
  3. (iii) (2.5p + 1.5q)2
  4. (iv) (34s + 8t)2
  5. (v) (x + 12y)2
  6. (vi) (1x + 1y)2
Answer: (i) 49x2 + 56xy + 16y2 (ii) 4925x2 + 215xy + 94y2 (iii) 6.25p2 + 7.5pq + 2.25q2 (iv) 916s2 + 12st + 64t2 (v) x2 + xy + 14y2 (vi) 1x2 + 2xy + 1y2

Step-by-step solution

Idea: In each bracket, call the first term a and the second term b. Then write a2, 2ab and b2 and simplify. Square the whole term, number and letter together.

(i) (7x + 4y)2

  1. Here a = 7x and b = 4y.
  2. (7x + 4y)2 = (7x)2 + 2(7x)(4y) + (4y)2½ mark
  3. = 49x2 + 56xy + 16y2
49x2 + 56xy + 16y2

(ii) (75x + 32y)2

  1. Here a = 75x and b = 32y.
  2. = (75x)2 + 2(75x)(32y) + (32y)2½ mark
  3. Square top and bottom of each fraction: (75)2 = 4925 and (32)2 = 94. Middle term: 2 × 75 × 32 = 4210 = 215.
  4. = 4925x2 + 215xy + 94y2
4925x2 + 215xy + 94y2

(iii) (2.5p + 1.5q)2

  1. Here a = 2.5p and b = 1.5q.
  2. = (2.5p)2 + 2(2.5p)(1.5q) + (1.5q)2½ mark
  3. 2.5 × 2.5 = 6.25; 2 × 2.5 × 1.5 = 5 × 1.5 = 7.5; 1.5 × 1.5 = 2.25
  4. = 6.25p2 + 7.5pq + 2.25q2
6.25p2 + 7.5pq + 2.25q2

(iv) (34s + 8t)2

  1. Here a = 34s and b = 8t.
  2. = (34s)2 + 2(34s)(8t) + (8t)2½ mark
  3. Middle term: 2 × 34 × 8 = 484 = 12.
  4. = 916s2 + 12st + 64t2
916s2 + 12st + 64t2

(v) (x + 12y)2

  1. Here a = x and b = 12y.
  2. = x2 + 2(x)(12y) + (12y)2½ mark
  3. The 2 in front cancels the 2 in the denominator: 2 × x × 12y = xy. And (2y)2 = 4y2.
  4. = x2 + xy + 14y2
x2 + xy + 14y2

(vi) (1x + 1y)2

  1. Here a = 1x and b = 1y.
  2. = (1x)2 + 2(1x)(1y) + (1y)2½ mark
  3. = 1x2 + 2xy + 1y2
1x2 + 2xy + 1y2
(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

Check: Put x = 1, y = 1 in (i): (7 + 4)2 = 121 and 49 + 56 + 16 = 121 ✓. Put p = q = 1 in (iii): 42 = 16 and 6.25 + 7.5 + 2.25 = 16 ✓. Put x = y = 1 in (vi): 22 = 4 and 1 + 2 + 1 = 4 ✓.

Answer to write in the exam

(i)

(7x + 4y)2 = (7x)2 + 2(7x)(4y) + (4y)2 [(a + b)2 = a2 + 2ab + b2]

∴ (7x + 4y)2 = 49x2 + 56xy + 16y2

(ii)

(75x + 32y)2 = (75x)2 + 2(75x)(32y) + (32y)2 [(a + b)2 = a2 + 2ab + b2]

∴ (75x + 32y)2 = 4925x2 + 215xy + 94y2

(iii)

(2.5p + 1.5q)2 = (2.5p)2 + 2(2.5p)(1.5q) + (1.5q)2 [(a + b)2 = a2 + 2ab + b2]

∴ (2.5p + 1.5q)2 = 6.25p2 + 7.5pq + 2.25q2

(iv)

(34s + 8t)2 = (34s)2 + 2(34s)(8t) + (8t)2 [(a + b)2 = a2 + 2ab + b2]

∴ (34s + 8t)2 = 916s2 + 12st + 64t2

(v)

(x + 12y)2 = x2 + 2(x)(12y) + (12y)2 [(a + b)2 = a2 + 2ab + b2]

∴ (x + 12y)2 = x2 + xy + 14y2

(vi)

(1x + 1y)2 = (1x)2 + 2(1x)(1y) + (1y)2 [(a + b)2 = a2 + 2ab + b2]

∴ (1x + 1y)2 = 1x2 + 2xy + 1y2

Common mistakes that cost marks

  • Squaring only the letter: writing (7x)2 = 7x2. The number is squared too, so (7x)2 = 49x2.
  • Forgetting the 2 in the middle term 2ab, e.g. writing 28xy instead of 56xy in (i).
  • In (v), writing (12y)2 = 12y2. The 2 in the denominator must also be squared, giving 14y2.

How this can come in the exam

MCQ (1 mark)

The coefficient of xy in the expansion of (3x + 5y)2 is

  1. 15
  2. 30
  3. 8
  4. 34
Show answer

(B) 30
Middle term = 2 × 3x × 5y = 30xy.

Short answer (2 marks)

Expand (23m + 32n)2.

Show answera = 23m, b = 32n. (23m)2 + 2(23m)(32n) + (32n)2 = 49m2 + 2mn + 94n2. (½ mark for the identity, 1 mark for the three terms, ½ for simplifying.)

Try one yourself

Expand (6a + 13b)2.

Show answer

36a2 + 4ab + 19b2 (middle term: 2 × 6a × 13b = 4ab).

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