Re deriving formulas for the quadratic eq.

\({\Large x^2+ax+b=0\quad|\quad a=2s\\ \Large x^2+2sx+s^2=s^2-b\\ \Large {\left({x+s}\right)}^2=s^2-b}\)

\({\Large x+s=\sqrt{s^2-b}\\ \Large x=-s±\sqrt{s^2-b}\quad|\quad s=\frac a2\\ \Large x=-\frac a2±\sqrt{{\left({\frac a2}\right)}^2-b}}\) 

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[Eop]

(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

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[Eop]

About the Quadratic Equation

Derivation of the solutions ::

\(x^2+px+q=0=\left({x-u}\right)\left({x-v}\right)=x^2-\left({u+v}\right)x+uv\)

\(p=2a\)

\(x^2+2ax+a^2-a^2+q=0\)

\(x+a=±\sqrt{a^2-q}\)

\(a=\frac p2\)

\(x=-\frac p2±\sqrt{{\left({\frac p2}\right)}^2-q}\)

\(x_1=-\frac p2-\sqrt{{\left({\frac p2}\right)}^2-q}\)

\(x_2=-\frac p2+\sqrt{{\left({\frac p2}\right)}^2-q}\)

\(x_2-x_1=2\sqrt{{\left({\frac p2}\right)}^2-q}\)

\(x_2+x_1=-p\)

\(x_2=-p-x_1\)

\(x_1=-p-x_2\)

\(x+p+\frac qx=0\)

\(\displaystyle{x=-p-\frac qx\quad \rightarrow \quad x_n=x_n+x_\overline{n}-\frac q{x_n}}\)

\(\displaystyle{x_n=\frac q{x_\overline{n}}\ \equiv\ x_nx_\overline{n}=q\ |\ about:\ q=uv}\)

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Example ::

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

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 \(\frac1x≡p^-\)
\(p^+=\frac{\sqrt{T^2+4}+T}2\ →\ p^-=\frac1{p^+}=\frac2{\sqrt{T^2+4}+T}=\frac{\sqrt{T^2+4}-T}2\)

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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.: \(f\left({x}\right)≡f\left({x_T}\right)≡x\left({T}\right)\)

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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. \(x·f\left({x}\right)-x·f\left({-x}\right)=x^2\)
   \(x·f\left({-x}\right)-x·f\left({x}\right)=-x^2\)