Re: transformation equations

On 22 sep, 17:06, "Spaceman" <space...@xxxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:
papa_r...@xxxxxxxxxxx wrote:
The other guy (Spaceman) says (-1)(-1) = (-1). I would also ask him
how would he solve this very trivial equation:

x^2 + 5x + 4 = 0

So you wish to just hand out an equation that does not
even match any reality?
Can you fit that equation to a real life situation?
With your math but of course, but no reality to match such.
The answer would be x = (-4) but no reality can prove such
exists as an answer or even an equation that fits any reality.

But anyone can make silly equations that make the answer
follow certain guidelines.
Try this and it is even more simple than yours..
x^2 = -16.
Solve for x without using imaginary numbers.
Then you will say.. it does not fit any reality.
Oh but it does.
It is how you "remove" a square from a piece of paper
and the -16 is how many cubic inches are "missing".
Or do you think "removing" a 4 by 4 square makes 16 square
inches appear on the paper?

I don't really care how much you want to play with imaginary bull***
but when it comes to "reality" and physics but you should start realizing
imaginary numbers only bring about imaginary phsyics and imaginary
equations also like yours..
:)

So please explain when in real life would you multiply a positive number
by a negative number ever? (like your 5x) would have to be.
And don't forget, you have to prove the "negative" is not really a
postive in another direction to prove the x is truly negative at all.
:)

--
James M Driscoll Jr
Creator of the Clock Malfunction Theory
Spaceman

So you are saying that the solution is x = -4, which for sure must be
wrong according to your own malfunctioning rules (remember that you
are stating that (-4)(-4) = -16 ?).
So how, by using your own rules, you can affirm that the solution is x
= -4?
According to your malfunctioning math, introducing x = -4 in x^2 + 5x
+ 4 = 0 gives (-4)(-4) + 5(-4) + 4 <> 0.

Miguel Rios
.

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