Click to See Complete Forum and Search --> : complex rearangement - please help
Dbase
Feb 26th, 2003, 04:52 PM
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
krtxmrtz
Feb 27th, 2003, 02:47 AM
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
sql_lall
Feb 27th, 2003, 03:02 AM
Are a and b unknown??
If not, you have the quadratic:
12 x (c2) - 4(a+b) x (c) + ab
and you can solve it that way
c = (4(a+b) +/- 4 sqrt(4(a+b)^2-3ab) )/24
Factoring doesn't seem to help:
(rA+s)(tB+u)
=(rt)(AB)+ ru(A) + st(B) + us
= AB - 4AC - 4BC + 12 C2
=> rt = 1
=> ru = -4C
=> st = -4C
=> us = 12 C^2
=> usrt = 12 C^2
but usrt = 16 C^2
so C=0
=> ab = 0
=> a = 0 or b = 0
However, quadratic approach is best
Dbase
Feb 27th, 2003, 10:44 AM
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 ;)
da_silvy
Feb 28th, 2003, 07:56 AM
sql_lall, I have a slight disagreement with your formula.
c = (4(a+b) +/- 4 sqrt(4(a+b)^2-3ab) )/24
Because sqrt(16 ~blah~) = 4 sqrt(~blah~) not 4 sqrt(4~blah~)
?
Anyway, I've attached the way I did it, it does simplify (slightly :p)
Hope it helps.
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