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