Articles
Register
Sign In
Search
Prescilliacrnl
@Prescilliacrnl
May 2019
1
93
Report
Bonjour, il me reste un exercice sur les nombres complexes auquel je bloque
Please enter comments
Please enter your name.
Please enter the correct email address.
Agree to
terms and service
You must agree before submitting.
Send
Lista de comentários
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
1 votes
Thanks 0
More Questions From This User
See All
prescilliacrnl
January 2021 | 0 Respostas
Bonjour! J'aurais juste besoins d'aide pour la question 2 svp!
Responda
Prescilliacrnl
May 2019 | 0 Respostas
Responda
×
Report "Bonjour, il me reste un exercice sur les nombres complexes auquel je bloque.... Pergunta de ideia de Prescilliacrnl"
Your name
Email
Reason
-Select Reason-
Pornographic
Defamatory
Illegal/Unlawful
Spam
Other Terms Of Service Violation
File a copyright complaint
Description
Helpful Links
Sobre nós
Política de Privacidade
Termos e Condições
direito autoral
Contate-Nos
Helpful Social
Get monthly updates
Submit
Copyright © 2024 ELIBRARY.TIPS - All rights reserved.
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