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Thread: complex rearangement - please help

  1. #1

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    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

  2. #2
    vbuggy krtxmrtz's Avatar
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    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

  3. #3
    Fanatic Member sql_lall's Avatar
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    Talking Quadratic

    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
    sql_lall

  4. #4

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    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

  5. #5
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    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 )

    Hope it helps.
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