# If alpha and beta are the zeros of th polynomial f(x) =x^2-p(x+1)-c, show that (alpha+beta)(beta+1)=1-c

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by Sivesh

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by Sivesh

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comparing with ax^2 + bx + c, we have,

Therefore,

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f(x) = x² - p x + q

α and β are the roots of the above equation.

to find α² / β² + β² / α² = ?

α = [ p + √(p² - 4q) ] / 2 and β = [ p - √(p² - 4 q) ] / 2

so, α + β = p and α β = q and α² β² = q²

=> α² + β² = (α+β)² - 2 αβ = p² - 2 q

=> α⁴ + β⁴ = (α² + β²)² - 2 α²β² = (p² - 2q)² - 2 q²

= p⁴ - 4 p² q + 4 q² - 2 q²

= p⁴ - 4 p² q + 2 q²

NOW , α² / β² + β² / α² = [ α⁴ + β⁴ ] / α² β² =

= [ p⁴ - 4 p² q + 2 q² ] / q²

α and β are the roots of the above equation.

to find α² / β² + β² / α² = ?

α = [ p + √(p² - 4q) ] / 2 and β = [ p - √(p² - 4 q) ] / 2

so, α + β = p and α β = q and α² β² = q²

=> α² + β² = (α+β)² - 2 αβ = p² - 2 q

=> α⁴ + β⁴ = (α² + β²)² - 2 α²β² = (p² - 2q)² - 2 q²

= p⁴ - 4 p² q + 4 q² - 2 q²

= p⁴ - 4 p² q + 2 q²

NOW , α² / β² + β² / α² = [ α⁴ + β⁴ ] / α² β² =

= [ p⁴ - 4 p² q + 2 q² ] / q²