Since x1, x2, x3 are in A. P. T. Therefore, let x1 = a – d, x2 = a and x3 = a + d and x1, x2, x3 are the roots of x3 – x2 + βx + γ = 0
We have ∑α = a – d + a + a + d =1 …….. (1)
∑αβ = (a – d)a + a(a + d) + (a – d) (a + d) = β ….. (2)
α β γ = (a – d)a(a +d) = -γ ………. (3)
From (1), we get, 3a = 1 ⇒ a = 1/3
From (2), we get 3a2 – a2 = β
⇒ 3(1/3)2 – a2 = β ⇒ 1/3 – β = d2
(NOTE : In this equation we have two variables β and d but we have only one equation. So at first sight it looks that this equation cannot be solved but we know that
d2 ≥ 0 ∀ d ∈ R then ; β can be solved).
⇒ 1/3 – β ≥ 0 ∵ d2 ≥ 0
⇒ β ≤ 1/3 ⇒ β ∈ (- ∞, 1/3]
From (3), a (a2 – d2) = - γ
⇒ 1/3 (1/9 – d2) = - γ ⇒ 1/27 – 1/3d2 = - γ
⇒ γ + 1/27 = 1/3d2 ⇒ γ + 1/27 ≥ 0
⇒ γ ≥ - 1/27 ⇒ - γ ∈[ -1/27,∞)
Hence β ∈ (-∞, 1/3 ) and γ ∈ [-1/27, ∞]
