Re: Satisfying Cardons' impossible quadratic equation.

William,

I rewrote this in my second post to be consistant
with the other quadratics in that second post.

x^2 - 10x = - 40
-----------------------------------
x^2 - 10x = 40 -  ((10/2)^2) = 15+ 40 = 55

-x = ((sqrt((55*4)+2)) -2)/2 = 6.4498322128... + (10/2) = -
11.4498322128...
-x^2 = ((((11.4498322128...*2)+2)^2) -2)/4 =  - 154.49832212...

- 11.449832212... *  10  =  - 114.49832212..
Therefore
- 154.49832212... - ( - 114.49832212..) = - 40
-------------------------------------------------------

This is unconventional I know where many algorithms
can do the same plot but in truth not in the negative (-x\-y)
third quadrant domain duplicating a mirror of x + y = x^2
in the fist quadrant.
This is just an exersize to show it can be done by using
these special quadratics, if you want to call them that.

Wheather (i) in the complex plain has a roll here I am not sure
because this is Cartisan.

Thanks for your input.

Dan

No.  The second equation does not follow from the first.

What are you talking about?
The solution to the (first) equation is
    x = (10 +- sqr(100 - 160))/2 = 5 +- 15i

x^2 - 10x = - 40
-----------------------------------
x^2 - 10x + (10/2)+10 =15 + 40 = 55

When solving for -x and -x^2 in the third negative quadrant of the
Cartisan coordinate system where both x and y are negative values
you need to derive -x and -x^2 this way --

-x = ((sqrt((55*4)+2)) -2)/2 = 6.4498322128... + (10/2) = -
11.4498322128...
-x^2 = ((((11.4498322128...*2)+2)^2) -2)/4 =  - 154.49832212...

Where the full negative values are --
10x =  -
114.498322128756698596659080230619755701172608310906236690287015059644675373511228185049178598263259828260696982420061133188481211584254257246449621593...
and
x^2 =  -
154.498322128756698596659080230619755701172608310906236690287015059644675373511228185049178598263259828260696982420061133188481211584254257246449621593

I wonder what Cardon would say if he were alive?  ;-)

Where, BTW the real value (not imaginary) value of the sqrt(-1) is --
((sqrt(6))-2)/2

Boy, am I going to get flack on this one!

This was derived from a ploting algorithm I used years back to plot a
curve in the third quadrant where finding -x and -y coordinates by
using
a similar method as above.
Basically this algorithm mirrors x + y = x^2 in the first quadrant
but the negative curve plots in the opposite direction in the third
negative quadrant.

Dan