Ronn Blankenship wrote:
>
> i should love the quadratic equation, since if it were
> not for the application of the Fundamental Theorem of
> Algebra to equations like x�+1=0,
> mathematicians and physicists would use i a lot less
> often (and electrical engineers would never learn the
> tenth letter of the alphabet) . . .
>
But _i_ wasn't needed to solve the quadratic equation.
IIRC, the first time someone needed _i_ was to
solve a _cubic_ equation.
Something like:
"Solve x^3 + 5 = 6 * x using Cardano"
[[I don't remember the actual historical equation that
motivated B..<something> to conjure _i_, only that the
solution was 4]]
Cardano's solution is writing x = u + v, then keeping
in reserve that we may push u and v into any 1st
degree equation:
u^3 + v^3 + 3 u v (u + v) + 5 = 6 (u + v)
So, let's cancel the terms with u + v:
3 u v = 6
Which makes:
u^3 + v^3 + 5 = 0
u v = 2 ==> u^3 v^3 = 8
So, u^3 and v^3 are the roots of the quadratic equation:
y^2 + 5 y + 8 = 0
What puzzled the above mentioned italian mathematician
whose name I can't remember now is that this equation
doesn't have any solution, because there ain't no such
thing as a square root of minus seven:
y = (-5 +/- sqrt(-7)) / 2
So it's impossible to solve x, that would be:
x = cubic root((-5 + sqrt(-7))/2) +
cubic root((-5 - sqrt(-7))/2)
His leap of faith was imagining - just for a while -
that such things could exist, and that we could take
their cubic roots. With some trial and error,
we can find an "imaginary, impossible" number that
corresponds to the cubic root of that obscenity:
(1/2 - 1/2 sqrt(-7))^3 =
1/8 - 3/8 sqrt(-7) - 3 * 7/8 + 7/8 sqrt(-7) =
- 20/8 + 4/8 sqrt(-7) = (-5 + sqrt(-7)) / 2
Of course, after mathematicians write something,
it ceases to be non-existent :-)
Alberto Monteiro
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