Showing posts with label square equation. Show all posts
Showing posts with label square equation. Show all posts

Friday, October 15, 2021

(scheduled) derivation of the solution of the quadratic equation


 about (scheduled) :: every now and then (not too often) i set myself to re-figure it out (the history has proven there exists a variance of what i come up with each time ...)

... so -- Def.-s , etc. ... ::

\[\begin{array}{lcl}
\left({x-a}\right)\left({x-b}\right)=0 &\ &\\
x^2-\left({a-b}\right)x+ab=0 &\ &\\
\begin{array}{l}
x^2+px+q=0\qquad \qquad \qquad \qquad \rightarrow\\
x^2+2px+p^2=px-q+p^2\ |×3 &\\
x^2-2px+p^2=-3px-q+p^2\ |+\ \uparrow &\\
\hline
4x^2+4px+4p^2=0-4q+4p^2\ |-3p^2\\
4x^2+4px+p^2=p^2-4q\ |÷4\\
\mathbf{x^2+2\frac p2x+{\left({\frac p2}\right)}^2={\left({\frac p2}\right)}^2-q}
\end{array} &\ &
\begin{array}{l}
\mathbf{x^2+2\frac p2x+{\left({\frac p2}\right)}^2={\left({\frac p2}\right)}^2-q}\\
{\left({x+\frac p2}\right)}^2={\left({\frac p2}\right)}^2-q\\
\boxed{x=-\frac p2±\sqrt{{\left({\frac p2}\right)}^2-q}}\\{}\\{}
\end{array}
\end{array}\]

... (it) came out double ((at) this time) -- the short and the long -- way to (the solution) F;T
// is likely ↑↑ why ↑↑ in many blogs folks do not get a thing what i say
// (as an old school programmer i always live-compact my code(read: text))

see also the inner properties of @ About the Quadratic Equation


[Eop]

Wednesday, March 25, 2020

About the Quadratic Equation


Derivation of the solutions ::
\(x^2+px+q=0=\left({x-u}\right)\left({x-v}\right)=x^2-\left({u+v}\right)x+uv\)
\(p=2a\)
\(x^2+2ax+a^2-a^2+q=0\)
\(x+a=±\sqrt{a^2-q}\)
\(a=\frac p2\)
\(x=-\frac p2±\sqrt{{\left({\frac p2}\right)}^2-q}\)
\(x_1=-\frac p2-\sqrt{{\left({\frac p2}\right)}^2-q}\)
\(x_2=-\frac p2+\sqrt{{\left({\frac p2}\right)}^2-q}\)
\(x_2-x_1=2\sqrt{{\left({\frac p2}\right)}^2-q}\)
\(x_2+x_1=-p\)
\(x_2=-p-x_1\)
\(x_1=-p-x_2\)
\(x+p+\frac qx=0\)
\(\displaystyle{x=-p-\frac qx\quad \rightarrow \quad x_n=x_n+x_\overline{n}-\frac q{x_n}}\)
\(\displaystyle{x_n=\frac q{x_\overline{n}}\ \equiv\ x_nx_\overline{n}=q\ |\ about:\ q=uv}\)

Example ::
\(\mathbf{p}\)\(\mathbf{q}\)\(\mathbf{x_n}\) \(\mathbf{x_1}\\ -p-x_2\\ \displaystyle{\frac q{x_\overline{n}}}\) \(\mathbf{x_2}\\ -p-x_1\\ \displaystyle{\frac q{x_\overline{n}}}\)
\(-1\)\(-1\) \(-\frac{-1}2±\sqrt{{\left({\frac{-1}2}\right)}^2-\left({-1}\right)}=\)
\(\frac{1±\sqrt{5}}2=\)
\(=\cases{-\frac{\sqrt{5}-1}2\\ \frac{\sqrt{5}+1}2}\)
\(=\cases{-\frac{5-1}{2\left({\sqrt{5}+1}\right)}\\ \frac{5-1}{2\left({\sqrt{5}-1}\right)}}\)
\(=\cases{\frac{-2}{\sqrt{5}+1}\\ \frac2{\sqrt{5}-1}}\)
\(\frac{-2}{\sqrt{5}+1}\mathbf{=}\)
\(\mathbf{=}-\left({-1}\right)-\frac{2}{\sqrt{5}-1}=\)
\(=1-\frac{\sqrt{5}+1}2=\)
\(=-\frac{-2+\sqrt{5}+1}2=\)
\(=-\frac{\sqrt{5}-1}2=\)
\(=-\frac2{\sqrt{5}+1}\)
\(\mathbf{=}\frac{-1}{\left({\frac2{\sqrt{5}-1}}\right)}=\)
\(=-\frac{\sqrt{5}-1}2=\)
\(=\frac{-2}{\sqrt{5}+1}\)
\(\frac2{\sqrt{5}-1}\mathbf{=}\)
\(\mathbf{=}-\left({-1}\right)-\frac{-2}{\sqrt{5}+1}=\)
\(=1+\frac{\sqrt{5}-1}2=\)
\(=\frac{2+\sqrt{5}-1}2=\)
\(=\frac{\sqrt{5}+1}2=\)
\(=\frac2{\sqrt{5}-1}\)
\(\mathbf{=}\frac{-1}{\left({\frac{-2}{\sqrt{5}+1}}\right)}=\)
\(=\frac{\sqrt{5}+1}2=\)
\(=\frac2{\sqrt{5}-1}\)

Thursday, March 19, 2020

p+ , p , x(T) , f(x)


Form the \(x^2-xT-1=0\) where \(T\in\mathbb{R}^{+}\ ;\ T≠0\)
\(x-\frac1x=T\ →\ \frac xT=\frac{xT+1}{xT}=\frac{x+\frac1T}x\)
Def.: \(x≡p^+\) and \(\frac1x≡p^-\)
\(p^+=\frac{\sqrt{T^2+4}+T}2\ →\ p^-=\frac1{p^+}=\frac2{\sqrt{T^2+4}+T}=\frac{\sqrt{T^2+4}-T}2\)

Def.: \(x\left({T}\right)=T+p^-\left({\frac1x}\right)=T+\frac{\sqrt{T^2+4}-T}2=\frac{\sqrt{T^2+4}+T}2\) Def.: \(f\left({x}\right)≡f\left({x_T}\right)≡x\left({T}\right)\)

The properties of the \(f\left({x}\right)\) :
  1. \(f\left({-x}\right)=\frac1{f\left({x}\right)}=f\left({x}\right)-x\)
    \(\frac1{f\left({-x}\right)}=f\left({x}\right)=f\left({-x}\right)+x\)
  2. \(f\left({-x}\right)+x=\frac1{f\left({x}\right)-x}\)
  3. \(x·f\left({x}\right)-x·f\left({-x}\right)=x^2\)
    \(x·f\left({-x}\right)-x·f\left({x}\right)=-x^2\)