A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
a+b+c=4
a^2+b^2+c^2=10
a^3+b^3+c^3=22
find the values of a, b ,and c
anonymous
 4 years ago
a+b+c=4 a^2+b^2+c^2=10 a^3+b^3+c^3=22 find the values of a, b ,and c

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have no idea, I'm commenting so that I'll get a notification when someone solves this.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If someone gets this they are the smartest person I know!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0First note that:\[(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)\Longrightarrow \]\[4^2=10+2(ab+ac+bc)\Longrightarrow ab+ac+bc=3\]Also:\[(a+b+c)^3=a^3+b^3+c^3+3(a+b+c)(ab+ac+bc)3abc\Longrightarrow\]\[4^3=22+3(4)(3)3abc\Longrightarrow abc=2\]So your system is equivalent to: a+b+c=10 ab+ac+bc=3 abc=2 Solving this system is the same as solving the polynomial: \[x^310x^2+3x+2=0\]So the solutions to this polynomial are the solutions to your system.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh, mistype, thats why things were acting weird. Your system is equivalent to:\[a+b+c=4\]\[ab+ac+bc=3\]\[abc=2\]which gives the polynomial:\[x^34x+3x2=0\]which gives solutions \[a= 2, b=1\sqrt{2}, c=1+\sqrt{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0mistype again, +2 instead of 2 >.>

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You must have had loads of practice to be as good as you are!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.