# If α and β are the roots of the equation x²-p(x+1)-q=0 then find the value of (α²+2α+1/α²+2α+q)+(β²+2β+1/β²+2β+q)

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by kirubb

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by kirubb

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Now we have to find the value of (α²+2α+1/α²+2α+q)+(β²+2β+1/β²+2β+q)

But from equation it is clear that

Sum of the roots = α+β = +p

Product of the roots = α*β = -p-q

(α²+2α+1/α²+2α+q)+(β²+2β+1/β²+2β+q) = (α²+β²)+4(α+β)+(1/α²+1/β²)+2q

(α+β)² = α²+β²+2αβ

⇒α²+β² = (α+β)² - 2αβ = p² - 2*(-p-q) = p²+2p+2q

(1/α²+1/β²) = (α²+β²)/(αβ)² = (p²+2p+2q)/(-p-q)² = (p²+2p+2q)/(p²+q²+2pq)

(α²+2α+1/α²+2α+q)+(β²+2β+1/β²+2β+q) = (α²+β²)+4(α+β)+(1/α²+1/β²)+2q =

p²+2p+2q+4p+ (p²+2p+2q)/(p²+q²+2pq)+2q = p²+6p+4q+ (p²+2p+2q)/(p²+q²+2pq)

= {(p²+6p+4q)(p²+q²+2pq)+(p²+2p+2q)}/(p²+q²+2pq)

= {p⁴+6p³+4qp²+p²q²+6pq²+4q³+2p³q+12p²q+8pq²+p²+2p+2q}/(p²+q²+2pq)

This is the answer.