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 ::

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