MR JOHN P BURKE wrote:
>
> -- [ From: John P. Burke * EMC.Ver #2.10P ] --
>
> Larry Boyd said:
>
> >OK, Kumar, its great and wonderful that you solved the problem so
> >quickly ;). However, for those of us who still can't remember the
> >steps, what are they? ...
>
> What a treat! A question I can answer (having spent the first 28 years
> of my life discovering that I was not the next John Von Neuman) that
> has not already been answered (the hazard of getting this list in
> digest form the day after - unlike those who seem to get and respond in
> real time). So before I get the digest for Saturday and discover
> someone else has already supplied the proof:
>
> The quadratic equation ax^2+bx+c=0 where a<>0 is a special case of the
> general second degree polynomial y=f(x)=ax^2+bx+c, which geometrically
> describes a parabola. In particular, if f(x), when graphed, touches or
> intersects the x-axis, then f(x)=0 has a solution(s) which is simply
> the value of x at the point of contact with the x-axis (obvious if you
> think about it geometrically; or, as one mathematics professor used to
> say, "it is intuitively obvious to the casual observer").
>
> Rene Descartes (Cartesian coordinates) proved the formula in question
> geometrically in Le Geometrie, which was published as an appendix to
> his monumental work Discourse in 1673. He also generally gets credit
> for the a,b,c x,y algebraic notation. However, he did not state the
> equation in the general terms we use today because he only dealt with
> solutions that were real numbers (if the parabola does not intersect
> the x-axis then, while the roots in a purely mathematical sense still
> exist, they involve the square root of -1).
>
> Anyway, a purely algebraic proof of the formula can be given as follows:
>
>    ax^2+bx+c=0               step 1: subtract "c" from each side
> => ax^2+bx=-c                step 2: multiply each side by 4a
> => 4a^2x^2+4abx=-4ac         step 3: add b^2 to each side
> => 4a^2x^2+4abx+b^2=b^2-4ac  step 4: factor the left side
> => (2ax+b)^2=b^2-4ac         step 5: take sqrt of each side
> => 2ax+b=+/- sqrt(b^2-4ac)   step 6: subtract b from each side
> => 2ax=-b +/- sqrt(b^2-4ac)  step 7: divide each side by 2a
> => x = -b +/- sqrt(b^2-4ac)          (assumes a <> 0 )
>        --------------------
>               2a
>
> The algebraic solution relies on re-structuring the equation so as to
> create something which can be "factored" (a lost art).
>
> BTW, there are general formulas for cubic and quartic equations, but it
> has been proven that no general algebraic solution exists for equations
> beyond the fourth degree.
>
> Now, where did I leave my slide rule?
>
> Regards,
> John Burke
> [log in to unmask] nicely done sir! However, if you could see me now you might
observe me running my hand quickly over my head and making a "swooshing"
sound and saying, "That went over my head!"
--
Mike Farrell
Intermountain Gas Company
Data Center Operations Supervisor
555 South Cole Road
Boise,  ID  83707