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Prescilliacrnl
@Prescilliacrnl
May 2019
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Bonjour, il me reste un exercice sur les nombres complexes auquel je bloque
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scoladan
Verified answer
Bonjour,
1) 4z² + 8z + 29 = 0
Δ = 8² - 4x4x29 = 64 - 464 = -400 = (20i)²
donc 2 solutions dans Z:
z₁ = (-8 - 20i)/8 = -1 -5i/2
z₂ = -1 + 5i/2
2) ci-joint
A d'affixe z₁ et B d'affixe z₂
3) AB = |z₂ - z₁| = |5i| = 5
CB = |z₂ - (2 - 1,5i)| = |-1 + 5i/2 - 2 + 1,5i| = |-3 + 4i| = √[(-3)² + (4)²] = √(25) = 5
Donc AB = CB ⇒ ABC isocèle en B
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prescilliacrnl
January 2021 | 0 Respostas
Bonjour! J'aurais juste besoins d'aide pour la question 2 svp!
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Prescilliacrnl
May 2019 | 0 Respostas
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Lista de comentários
Verified answer
Bonjour,1) 4z² + 8z + 29 = 0
Δ = 8² - 4x4x29 = 64 - 464 = -400 = (20i)²
donc 2 solutions dans Z:
z₁ = (-8 - 20i)/8 = -1 -5i/2
z₂ = -1 + 5i/2
2) ci-joint
A d'affixe z₁ et B d'affixe z₂
3) AB = |z₂ - z₁| = |5i| = 5
CB = |z₂ - (2 - 1,5i)| = |-1 + 5i/2 - 2 + 1,5i| = |-3 + 4i| = √[(-3)² + (4)²] = √(25) = 5
Donc AB = CB ⇒ ABC isocèle en B