If α, β, γ are roots of the equation x³+a x²+b x+c=0 then α^{-1}+β^{-1}+γ^{-1}= [AP EAPCET 19 May 2024 Shift 2]

Correct Answer is : − bc-\;b/c−cb

Correct Answer is : − bc-\;b/c−cb

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Quadratic Equations

Section: Mathematics

Question : 47 of 51

Marks: +1, -0

α, β, γ

x^3+a x^2+b x+c=0

α^{-1}+β^{-1}+γ^{-1}=

Solution:

To find

α^{-1}+β^{-1}+γ^{-1}

, we consider that

α, β

, and

γ

are the roots of the equation:

x^3+a x^2+b x+c=0 .

From Vieta's formulas, we know:

\;α+β+γ=-a,

\;α β+β γ+γ α=b,

\;α β γ=-c .

To find the sum of the reciprocals of the roots:

α^{-1}+β^{-1}+γ^{-1}=\;\frac{1}{α}+\;\frac{1}{β}+\;\frac{1}{γ}

Utilizing the identity for sums of reciprocals, this can be rewritten as:

α^{-1}+β^{-1}+γ^{-1}=\;\frac{β γ+α γ+α β}{α β γ} .

Substituting the known values from Vieta's formulas, we get:

α^{-1}+β^{-1}+γ^{-1}=\;\frac{b}{-c}=-\;\frac{b}{c} .

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