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Suppose (Z, +, ·) is an ordered integral domain. Let a and b and c be...

Suppose (Z, +, ·) is an ordered integral domain. Let a and b and c be elements of Z such that c^2 + a · c + b = 0.

(a) Prove that if 4b = a^2 , then x^2 + a · x + b ≥ 0 for all x ∈ Z.

(b) Prove that if 4b =/= a^2 , then there is exactly one element d in Z such that d^2 + a · d + b = 0 and d =/= c.

(c) For which x ∈ Z is x^2 + a · x + b > 0? As usual, provide a proof

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Answer #1

a.) Let, for some x 62 xtaxt bco Then 4x + 4 ax + 46 40 => 4x² + 4 ax + a² co [a² - 46] -> (2x+2) (2x + a) co which is a contb) Let, x 62 8.t. xtaxt 6=0 & 27c Then (erac+6) - (x + ax+b) = 0 -> (e-x) (6+2) + a(con) = 0 (c-x) (carta) = 0 Then (exta) =

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