Quadratic Equation and Inequalities Part 1

Given equation x2+ax+b=0
It has two roots (not necessarily real α and β)
⇒ Either α = β or α ≠ β
I. If α = β
⇒ α=α2
⇒ α = -1, 2
When α = - 1, then (a, b) = (2,1)
When α = 2, then (a, b) = (- 4, 4)
II. If α ≠ β, then
(a) α=α2−2‌and‌β=β2−2
Here, (α,β) = (2, - 1) or (- 1, 2)
Hence (a, b) = (- α - α, αβ) = (-1 , - 2)
(b) α=β2−2‌and‌β=α2−2
Then α - β =β2−2= (β - α) (β + α)
∵ α ≠ β
⇒ α+β=β2+α2−4
or α=β=(α+β)2−2αβ−4
⇒ -1 = 1 - 2αβ - 4
⇒ αβ = -1
⇒ (a, b) = (-α -β, αβ) = (1, -1)
(c) α=α2−2=β2−2 and α≠β
⇒ α = -β
Thus,α = 2, α = -2
or α= -1, α = 1
∴(a, b) = (0, -4) and (0, -1)
(d) β=α2−2=β2−2 and α≠β (as in (c)
⇒ We get 6 pairs of (a, b)
They are (2, 1), (-4, 4), (-1, - 2), (1, - 1), (0, - 4), and (0, - 1).