let alpha and beta be the zeros of the cubic polynomial x^3 + ax^2 + bx + c satisfying the equation alpha*beta + 1 = 0. prove that c^2 + ac + b +1 = 0
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agrawal:
i think u r quite good at maths
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given αβ + 1 = 0 => αβ = - 1 -- (1)
α , β, δ are the roots of the polynomial: then it is equal to the product:
(x - α ) (x - β) (X - δ)
= x³ - (α+β+δ) x² + ( αβ + βδ + δα) x - αβδ
so we have
αβ δ = - c
by (1) => c = δ --- (2)
α + β + δ = - a
α + β = - (a + δ) ---- (3)
αβ + βδ + δα = b
=> -1 + βδ + δα = b
=> βδ + δα = b + 1 --- (4)
Now to prove
c² + ac + b + 1
= δ² + a δ + βδ + δα using the above (2) , (4)
= δ (δ + a + β + α)
= δ (δ + a - a - δ ) using (3)
= δ ( 0 )
= 0
α , β, δ are the roots of the polynomial: then it is equal to the product:
(x - α ) (x - β) (X - δ)
= x³ - (α+β+δ) x² + ( αβ + βδ + δα) x - αβδ
so we have
αβ δ = - c
by (1) => c = δ --- (2)
α + β + δ = - a
α + β = - (a + δ) ---- (3)
αβ + βδ + δα = b
=> -1 + βδ + δα = b
=> βδ + δα = b + 1 --- (4)
Now to prove
c² + ac + b + 1
= δ² + a δ + βδ + δα using the above (2) , (4)
= δ (δ + a + β + α)
= δ (δ + a - a - δ ) using (3)
= δ ( 0 )
= 0
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