Skillnad mellan versioner av "1.5a Lösning 6b"
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) m |
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| Rad 5: | Rad 5: | ||
x_{1,2} & = -{1 \over 2} \pm \sqrt{{1 \over 4} + 1} \\ | x_{1,2} & = -{1 \over 2} \pm \sqrt{{1 \over 4} + 1} \\ | ||
x_{1,2} & = -{1 \over 2} \pm \sqrt{5 \over 4} \\ | x_{1,2} & = -{1 \over 2} \pm \sqrt{5 \over 4} \\ | ||
| − | x_1 & = -{1 \over 2} + {1 \over 2} \cdot \sqrt{5}={1 \over 2}\,(-1 + \sqrt{5})={\sqrt{5}-1 \over 2} = 0,618033989\cdots | + | x_1 & = -{1 \over 2} + {1 \over 2} \cdot \sqrt{5}={1 \over 2}\,(-1 + \sqrt{5})={\sqrt{5}-1 \over 2} = 0,618033989\cdots \\ |
| − | x_2 & = -{1 \over 2} - {1 \over 2} \cdot \sqrt{5}=-{1 \over 2}\,(1 + \sqrt{5})=-{1+\sqrt{5} \over 2} \; {\rm :negativ} | + | x_2 & = -{1 \over 2} - {1 \over 2} \cdot \sqrt{5}=-{1 \over 2}\,(1 + \sqrt{5})=-{1+\sqrt{5} \over 2} \; {\rm :negativ} \\ |
g & = \underline {0,618033989} | g & = \underline {0,618033989} | ||
\end{align}</math> | \end{align}</math> | ||
Versionen från 14 juli 2014 kl. 18.58
\(\begin{align} {1+x \over 1} & = {1 \over x} \qquad\qquad | \; \cdot x{\color{White} x} \\ x \cdot (1 + x) & = 1 \\ x + x^2 & = 1 \qquad\qquad | \; -1 \\ x^2 + x - 1 & = 0 \\ x_{1,2} & = -{1 \over 2} \pm \sqrt{{1 \over 4} + 1} \\ x_{1,2} & = -{1 \over 2} \pm \sqrt{5 \over 4} \\ x_1 & = -{1 \over 2} + {1 \over 2} \cdot \sqrt{5}={1 \over 2}\,(-1 + \sqrt{5})={\sqrt{5}-1 \over 2} = 0,618033989\cdots \\ x_2 & = -{1 \over 2} - {1 \over 2} \cdot \sqrt{5}=-{1 \over 2}\,(1 + \sqrt{5})=-{1+\sqrt{5} \over 2} \; {\rm :negativ} \\ g & = \underline {0,618033989} \end{align}\)
