Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
The parabola y=ax^2+bx+c, c≠0 meets the xaxis at A(α,0) and B(β,0) and the yaxis at C. If AC and BC are perpendicular, prove ac=1
 one year ago
 one year ago
The parabola y=ax^2+bx+c, c≠0 meets the xaxis at A(α,0) and B(β,0) and the yaxis at C. If AC and BC are perpendicular, prove ac=1
 one year ago
 one year ago

This Question is Closed

akash123Best ResponseYou've already chosen the best response.3
alpha and beta are the roots of quadratic ax^2+bx+c=0
 one year ago

akash123Best ResponseYou've already chosen the best response.3
so alpha + beta = b/a and alpha* beta= c/a
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.1
sure...tag the math hater to a math problem...makes a lot of sense
 one year ago

akash123Best ResponseYou've already chosen the best response.3
slope of AC * slope of BC= 1
 one year ago

akash123Best ResponseYou've already chosen the best response.3
so alpha* beta = c^2
 one year ago

akash123Best ResponseYou've already chosen the best response.3
and put this value into... alpha* beta= c/a
 one year ago

apple_piBest ResponseYou've already chosen the best response.0
how did you get to alpha*beta=c^2?
 one year ago

akash123Best ResponseYou've already chosen the best response.3
use this...slope of AC * slope of BC= 1 c=(0, beta)
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
No one tags mathslover though :P
 one year ago

apple_piBest ResponseYou've already chosen the best response.0
since when does c=(0,beta)?
 one year ago

akash123Best ResponseYou've already chosen the best response.3
so slope of AC=  c/alpha=m1 , slope of BC= c/beta=m2 m1 *m2= 1...since AC is perpendicular to BC.
 one year ago
See more questions >>>
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.