Re: Satisfying Cardons' impossible quadratic equation.

The original equation was x^2 + 10x = -40.

.5^2+10*.5 = .25+5 = 5.25 # -40, so what relevance do you believe this
to have to the original equation?

What exactly are you trying to accomplish?  Is it your intent to mirror
y = x^2 +10x about the x axis?

1^2 + 10*1 = 11 # -40, so again what is the relevance to the original
equation?

Ok, so you can plot y = x^2 - x.  Fine.  So can any six year old with a
copy of Microsoft Works.  What does that have to do with Cardan's equation?

Then what do they have to do with Cardan's equation?

What about it?  n is already given as exactly 1, and r as exactly 40.
So what do your plot points have to do with anything?

Explain, in a sentence containing fewer than 20 words, none of which
have more than three syllables, what is is that you are attempting to
accomplish.  Right now all your comments seem like random meanderings.

Your first question  --

Starting in the 4th quadrant is the data point x= 0.5 and
y = -0.25 which is the very bottom point of x + y = x^2
parabola that covers the 1st 2nd and 4th quadrant.
My plotting algorithm mirrors the right side of this
parabola which send the plot from this point down and to
the left and crossing the -y  axis @ y = -0.5 and x = 0

Starting with -y = -0.25 and incrementing -y for each
plotted point with -y = -y + -0.01 with the first value
for -y = -0.25 then --
-x = ((sqrt((- y * 4) -1)) -1)/2 = -0.5
changing -y to y before the first -1 and sqrt.
This shows x to be negative but it is realy 0.5 because
the plot is to the right of the y axis.
So the first plot point and following points are--
x= 0.5 and - y = -0.25
x= 0.4 and - y = -0.26
x= 0.358578.. - y = -0.27
x= 0.326794.. - y = -0.28
x= 0.3 ...... - y = -0.29
x= 0.276393..- y = -0.30
x= 0.255051..- y = -0.31
.. continue incrementing -y with + -0.01 until
x= 0 and -y = -0.5
Which will mirror x = 1 and y = 0 in the first quadrant.
From here on where - y < - 0.5 -x becomes x so that has
to change to - x and from this point on incrementing - y + -0.01
-x = ((sqrt((- y * 4) -1)) -1)/2 will reverse mirror image
every thing in the first quadrant.
A very simple algorithm that is easy to impliment
in a plotting application.
Do a plot for x + y = x^2 starting @ x=0.5 and -y = -0.25
and compare both plots. Shift one of the plots by 180 degrees
and the plots will look the same except for the orintation
of the x\y axis between the two different plots.

That is why these plot points are correct and not a lie.
And again my quadratics are based upon these plot points
but with 2 algebraic limits (n,r) where x^2 - x*n = -r. I will
show these two with lower limit for (n) and upper limit for -r in
a later post.

Dan

Changing an equation to another equation and then solving it and saying
that the solution is "correct" is not "thinking outside the box", it is
lying.

Why are you plotting "this curve starting in the 4th and then passing
into the third quadrant" when the equation you are talking about has no
points in either of those quadrants?

What leads you to believe that these "plotting points" are "correct"?

Rob,

Yes you are right it is illegal, but thinking outside of the box and
plotting this curve starting in the 4th and then passing into the
third quadrant so making it easy I start @  y = - 0.5 and x = 0
the mirror of x = 1 and y =0.

The main ploting algorithm after -y <= - 0.5
-x = ((sqrt((- y * 4) -1)) -1)/2

change sign for -y too y before sqrt. then change sign for x too -x
after result
and changing y back to - y.

-y say in the above case is -y = -1
then
- y = - 1
- x = - 0.336025403...
-x^2 = -y + -x

Where its mirrored point in the first quadrant is ---
x= 1.336025403
y= 0.5
x^2 = x+y

These quadratics that I propose are directly related to any
-x
-y
points that have the actual plot point for the quadratic below .--

-x =  - 8.7008771254
-x^2 = - 93.6070170039
Then
- y = - x^2 - (- x) = - 84.9061398785
giving both -x and -y coordinates for the quadratic below
which reside in the third quadrant.
Where -x*8 is -24 > - x^2
Taken from this quadratic below.

x^2 - 8x =  - 24
----------------------------------
x^2 - 8x = (24 - ((8/2)^2)) = 8 + 24 = 32

-x = ((sqrt((32*4)+2))-2)/2 =   - 4.7008771254.. + (8/2) = -
8.7008771254..
-x^2 = (((8.7008771254.. *2)+2)^2)-2)/4 =  - 93.6070170039..

- 8.7008771254.. *  8 =  - 69.6070170039
Therefore
-93.6070170039.. -  ( - 69.6070170039) =  - 24

As illegal as it is, it still gives plotting points that are correct
in the
third quadrant an also illegal quadratics stemming from these plotting
points.

Dan

What does it tell you when, to make a proof work, you have to do
something illegal?

Let's see what happens if we only allow legal operations.  Assume x
is real.  If x^2 - 10x = - 40, then

   0 = x^2 - 10x + 40

     = x^2 - 10x + 25 + 15

     = (x-5)^2 + 15

That is, 15 greater than a non-negative number is 0.  Since this
cannot be, x^2 - 10x = - 40 has no real solutions.

Rob Johnson <rob@trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font

Henry,

The whole exersize would not be possible if changing
signs were not included. So it is a manipulation of (+\-)
to make it all possible.

Thanks,

Dan

That is just digging yourself further in.
Where before you had -40 = +40,
you now have -40 = +15 = +55.

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