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alpha + beta = -b/a.Then how to find (alpha - beta)

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Q. If α and β are the zeroes of the quadratic polynomial f(x) = ax² + bx + c, then evaluate :

(i) α − β
(ii)

\frac{1}{α} - \frac{1}{β}

\frac{1}{α} + \frac{1}{β} - 2 αβ

(iv) α²β − αβ²
(v) α⁴ + β⁴
(vi)

\frac{1}{a α + b} + \frac{1}{a β + b}

\frac{β}{a α + b} + \frac{α}{a β + b}

a (\frac{α^{2}}{β} + \frac{β^{2}}{α}) + b (\frac{α}{β} + \frac{β}{α})

(α + 1) (β - 1) + (β + 1) (α - 1) α + (α - 1) (β - 1) = 0

α (α + 1) (β + 1) - (α - 1) (β - 1) = 0

A = {\frac{α + 1}{α - 1}, \frac{β + 1}{β - 1}}

B = {\frac{2 α}{α - 1}, \frac{2 β}{β + 1}}

A \cap B \neq ϕ

, then find all the permissible value of parameter

" a "

Q. The roots of the quadratic equation

(x + β) (x - α) = 0

∣ ∣ ∣ ∣ ∣ \begin{matrix} β γ & β γ^{'} + β^{'} γ & β^{'} γ^{'} γ α & γ α^{'} + γ^{'} α & γ^{'} α^{'} α β & α β^{'} + α^{'} β^{'} & α^{'} β^{'} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (α β^{'} + α^{'} β) (β γ^{'} - β^{'} γ) (γ α^{'} + γ^{'} α)

Q. If α, β are the zeros of f (x) = 2x² + 6x − 6, then

(a) α + β = αβ
(b) α + β > αβ
(c) α + β < αβ
(d) α + β + αβ = 0

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