|
-
Feb 26th, 2003, 05:52 PM
#1
Thread Starter
New Member
complex rearangement - please help
ive been dieing over this for the past three hours - i just cant do the damn thing. please try and solve my equasion for variable 'c'. Even if you cant, please offer some advice.
im not particulary interested in the answer.. but how to get to it. i already know the answer (various calculators on the web).
the equation is:
ab-4ca-4cb+12c^2=0
thanko mucho
-
Feb 27th, 2003, 03:47 AM
#2
The general (2) solution(s) for a 2nd degree equation such as:
mx2 + nx + p = 0
is (are):
x1 = [-n + Sqr(n2 - 4mp]/2m
x2 = [-n - Sqr(n2 - 4mp]/2m
Rearrange your equation as:
12c2 - 4c(a + b) + ab = 0
and apply the above recipe. The result(s) is(are) (after some simplifications):
c1 = [(a+b) + Sqr((a-b)2 + ab)]/6
c2 = [(a+b) - Sqr((a-b)2 + ab)]/6
-
Feb 27th, 2003, 04:02 AM
#3
-
Feb 27th, 2003, 11:44 AM
#4
Thread Starter
New Member
ahhhh yessss!!
I completely neglected the quadratic formula. I thought it couldnt be solved using the formula but now its so obvious that
- 4cb - 4ca = - 4(a+b)c
thanks for the replies, its very much appreciated
-
Feb 28th, 2003, 08:56 AM
#5
Conquistador
Posting Permissions
- You may not post new threads
- You may not post replies
- You may not post attachments
- You may not edit your posts
-
Forum Rules
|
Click Here to Expand Forum to Full Width
|