Quadratic Equation and Inequalities Part 1

Given equation $x^{2}+a x+b=0$ It has two roots (not necessarily real α and β) ⇒ Either α = β or α ≠ β I. If α = β ⇒ $\alpha=\alpha^2$ ⇒ 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) $\alpha=\alpha^2-2 \text{ and } \beta=\beta^2-2$ Here, (α,β) = (2, - 1) or (- 1, 2) Hence (a, b) = (- α - α, αβ) = (-1 , - 2) (b) $\alpha=\beta^2-2 \text{ and } \beta=\alpha^2-2$ Then α - β $= \beta^2-2 =$ (β - α) (β + α) ∵ α ≠ β ⇒ $\alpha+\beta=\beta^2+\alpha^2-4$ or $\alpha=\beta=(\alpha+\beta)^2-2\alpha\beta-4$ ⇒ -1 = 1 - 2αβ - 4 ⇒ αβ = -1 ⇒ (a, b) = (-α -β, αβ) = (1, -1) (c) $\alpha=\alpha^{2}-2=\beta^{2}-2$ and $\alpha\neq\beta$ ⇒ α = -β Thus,α = 2, α = -2 or α= -1, α = 1 ∴(a, b) = (0, -4) and (0, -1) (d) $\beta=\alpha^{2}-2=\beta^{2}-2$ and $\alpha\neq\beta$ (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).