Q1.
Using the identity (a + b)² = a² + 2ab + b², expand the following: (i) (7x + 4y)² (ii) (7x/5 + 3y/2)² (iii) (2.5p + 1.5q)² (iv) (3s/4 + 8t)² (v) (x + 1/2y)² (vi) (1/x + 1/y)²
Answer
In every part, name a and b first, then write down a² + 2ab + b² without multiplying anything out by hand.
(i) (7x + 4y)² — here a = 7x, b = 4y.
(7x)² + 2(7x)(4y) + (4y)²
= 49x² + 56xy + 16y²
= 49x² + 56xy + 16y²
(ii) (7x/5 + 3y/2)² — a = 7x/5, b = 3y/2.
(7x/5)² + 2(7x/5)(3y/2) + (3y/2)²
= 49x²/25 + (42/10)xy + 9y²/4
= 49x²/25 + 21xy/5 + 9y²/4
= 49x²/25 + (42/10)xy + 9y²/4
= 49x²/25 + 21xy/5 + 9y²/4
(iii) (2.5p + 1.5q)² — a = 2.5p, b = 1.5q.
(2.5p)² + 2(2.5p)(1.5q) + (1.5q)²
= 6.25p² + 7.5pq + 2.25q²
(as fractions: 25p²/4 + 15pq/2 + 9q²/4)
= 6.25p² + 7.5pq + 2.25q²
(as fractions: 25p²/4 + 15pq/2 + 9q²/4)
(iv) (3s/4 + 8t)² — a = 3s/4, b = 8t.
(3s/4)² + 2(3s/4)(8t) + (8t)²
= 9s²/16 + 12st + 64t²
= 9s²/16 + 12st + 64t²
(v) (x + 1/2y)² — a = x, b = 1/(2y).
x² + 2(x)(1/2y) + (1/2y)²
= x² + x/y + 1/(4y²)
= x² + x/y + 1/(4y²)
(vi) (1/x + 1/y)² — a = 1/x, b = 1/y.
(1/x)² + 2(1/x)(1/y) + (1/y)²
= 1/x² + 2/(xy) + 1/y²
= 1/x² + 2/(xy) + 1/y²
Why the identity may be used on fractions and on 1/x: an identity is true for all values of its letters. So a and b need not be whole numbers, or even numbers written without a variable in the denominator — any expression at all may be substituted, provided it is defined. In part (v) that means y ≠ 0, and in part (vi), x ≠ 0 and y ≠ 0.
Check it yourself: Put x = 1, y = 1 in (vi). The left side is (1 + 1)² = 4 and the right side is 1 + 2 + 1 = 4. A single well-chosen substitution catches most slips.