A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
Determine the number and type of solutions
anonymous
 3 years ago
Determine the number and type of solutions

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Determine the number and type of solutions for x2 + 4x + 3=0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Quadratic equations has two solutions

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Unless it is perfect square

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Since , b^2 4ac >0 two rational solution

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2Compare your Quadratic equation with ax^2+bx+c=0 find a,b,c. calculate b^24ac if b^24ac is positive > 2 real roots. if b^24ac is 0>1 repeated real root. if b^24ac is negative >2 imaginary roots..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what about this one? What is the range of the graph of y = (x – 4)^2 + 3?

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0@sauravshakya b^24ac>0 doesnt mean roots are rational.

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2if b^24ac is perfect square > 2 rational roots.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I mean real solutions

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok so its what i though, 2 rational can someone help me on my other question
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.