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Complex Numbers

Consider fx...

Question

Consider $f (x) = 8 x^{4} - 2 x^{2} + 6 x - 5$ and $α$ , $β$ , $γ$ , $δ$ are its zeroes find the value of $α + β + γ + δ$ .

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Solution

Given $α, β, γ, δ$ are the roots of the equation $8 x^{4} + 0. x^{3} - 2 x^{2} + 6 x - 5 = 0$ .
Then from root and co-efficient relation we get, $α + β + γ + δ = - \frac{0}{8} = 0$ .

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β

γ

are the roots of

x^{3} + a x^{2} + b = 0

, then the value of determinant

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.

(α, β), (γ, δ) (α, δ)

(γ, β)

taken in order, where

α, β, γ, δ

are different real numbers, are

α

β

x^{2} + p x + 1 = 0

γ

δ

are the roots of

x^{2} + q x + 1 = 0

q^{2} - p^{2} - (α - γ) (β - γ) (α + δ) (β + δ) = 0

∣ ∣ ∣ ∣ ∣ \begin{matrix} (β + γ - α - δ)^{4} & (β + γ - α - δ)^{2} & 1 (γ + α - β - δ)^{4} & (γ + α - β - δ)^{2} & 1 (α + β - γ - δ)^{4} & (α + β - γ - δ)^{2} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = - K (α - β) (α - γ) (β - γ) (β - δ) (γ - δ)

Then the value of

K

∣ ∣ ∣ ∣ ∣ \begin{matrix} {(β + γ - α - δ)}^{4} & {(β + γ - α - δ)}^{2} & 1 {(γ + α - β - δ)}^{4} & {(γ + α - β - δ)}^{2} & 1 {(α + β - γ - δ)}^{4} & {(α + β - γ - δ)}^{2} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = - k (α - β) (α - γ) (α - δ) (β - γ) (β - δ) (γ - δ)

, then the value of

{(k)}^{\frac{1}{2}}

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