Question

A binary operation is defined on the set of real numbers ℝ by

x ∆ y = x²− 2xy + y
²

; x, y ∈ ℝ.

i. Find m such that 2 ∆ − 5 = √m

ii. Simplify ((n+1) ∆ y)/n


; n ≠ 0​

Answer 1

I. m = 2401

II. ((n+1) ∆ y)/n = 1/n[(n – y + 2)(n – y) + 1]

Explanation:

I. Determination of m

x ∆ y = x² − 2xy + y²

2 ∆ − 5 = √m

2² − 2(2 × –5) + (–5)² = √m

4 – 2(–10) + 25 = √m

4 + 20 + 25 = √m

49 = √m

Take the square of both side

49² = m

2401 = m

m = 2401

II. Simplify ((n+1) ∆ y)/n

We'll begin by obtaining (n+1) ∆ y. This can be obtained as follow:

x ∆ y = x² − 2xy + y²

(n+1) ∆ y = (n+1)² – 2(n+1)y + y²

(n+1) ∆ y = n² + 2n + 1 – 2ny – 2y + y²

(n+1) ∆ y = n² + 2n – 2ny – 2y + y² + 1

(n+1) ∆ y = n² – 2ny + y² + 2n – 2y + 1

(n+1) ∆ y = n² – ny – ny + y² + 2n – 2y + 1

(n+1) ∆ y = n(n – y) – y(n – y) + 2(n – y) + 1

(n+1) ∆ y = (n – y + 2)(n – y) + 1

((n+1) ∆ y)/n = [(n – y + 2)(n – y) + 1] / n

((n+1) ∆ y)/n = 1/n[(n – y + 2)(n – y) + 1]