User:Nothing2Do.fr

I'm a french poet (i want to be), a mathematic enthousiast, a physician (I'd liked to understand), and so, and so ...
My website is here : http://www.nothing2do.fr
I give you a formula to resolve everything in real world (but you aren't "imposed" to hear).

Howto resolve equation for newbies (or computer)

The first degrees equation has to be resolved like this : $\left(a_{1}x+b_{1}\right)$
(with untheorical solution : $x_{1}=-{\frac {b_{1}}{a_{1}}}$, a1 and b1, we can follow back the thread)
The second degree equation has to be resolved like this : $\left(\left({\frac {k_{2}}{k_{0}}}*{\frac {1}{a_{1}}}\right)x+\left({\frac {k_{1}}{k_{2}}}-b_{1}\right)\right)$ (with solution added : $x_{2}=-{\frac {{\frac {k_{1}}{k_{2}}}+b_{1}}{{\frac {k_{2}}{k_{0}}}*{\frac {1}{a_{1}}}}}=-{\frac {b_{2}}{a_{2}}})$
The third degrees has to be resolved like this : $\left(\left({\frac {1}{\left({\frac {k_{2}}{k_{3}}}-b_{1}-b_{2}\right)b_{1}b_{2}}}*{\frac {1}{a_{1}a_{2}}}\right)x+\left({\frac {k_{2}}{k_{3}}}-b_{1}-b_{2}\right)\right)(x_{3}=-{\frac {{\frac {k_{2}}{k_{3}}}-b_{1}-b_{2}}{{\frac {1}{\left({\frac {k_{2}}{k_{3}}}-b_{1}-b_{2}\right)b_{1}b_{2}}}*{\frac {1}{a_{1}a_{2}}}}}=-{\frac {b_{3}}{a_{3}}})$

p is the degree of considered equation

The equation of degree p has to be resolved like this :

$\left(k_{p}x+k_{0}\right)$ with $a_{p}={\frac {1}{k_{p}}}$ and $b_{P}=-k_{0}$ (WARNING, this can't be resolved now).

Considering you've read the last sentence, I've to give you this one : $x_{p}=-{\frac {b_{p}}{a_{p}}}$
so, every equation is like this :
$\sum _{n}^{p}\left(k_{n}x^{n}\right)=\prod _{n}^{p}\left(a_{n}x+b_{n}\right)=\left(k_{p}x+k_{0}\right)\sum _{n=1}^{p-1}\left({\frac {k_{n}}{k_{p}}}x^{n-1}\right)$

To have only real solution, we need to have (for "classical" method, this will define "delta") :

$\sum _{n}^{p}\left(b_{n}\right)=\prod _{n}^{p}\left(a_{n}\right)\Rightarrow \Delta =0$

elsewhere : All solution will be real (but won't work ...) except the last (this depending on a1 and b1 ).

the result is that : every équation of p degrees have (p or p-1) real solutions but
differences with "classical" method will appear with the third degrees.

to laugh : if $a_{1}a_{2}=b_{1}+b_{2}$ then $\triangle =0$ !!! If you want to be considered as a genious, it's the same things for every degrees

“Everything that we know, is that we know nothing” Nothing2Do