A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
Please help :)
Describe the nature of the roots for this equation.
3x^2+x5=0
A. Two complex roots
B. Two real, rational roots
C. One real, double root
D. Two real, irrational roots
anonymous
 one year ago
Please help :) Describe the nature of the roots for this equation. 3x^2+x5=0 A. Two complex roots B. Two real, rational roots C. One real, double root D. Two real, irrational roots

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well first we must find its roots. Do you know how?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0idk i think the first step is the quadratic formula?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah sure lets do the quadratic equation have you tried the problem already?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[b+or  \sqrt{b ^{2} 4ac} / 2a\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01+\[1+ \sqrt{1  4(3) (5)} / 6\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the entire thing is divided by 6 right??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01+dw:1436851541331:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the sqrt of 61 has no factors it is an irrational root

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can you answer your question now?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you!! what if it's 2x^2x+1? I got 1+(Sqrt 9)/4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sqrt 9 = 3i that is a non real number

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so would it be two complex numbers?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes it is real numbers + or  an imaginary number are complex numbers

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in this case they are complex roots

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay thanks so much! :)
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.