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42 lines
2.1 KiB
TeX
42 lines
2.1 KiB
TeX
$4 \alpha^3 + 27 \beta^2 \geq 0$:
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The first solution of $x^3 + \alpha x + \beta = 0$ is
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\[x = \frac{t}{\sqrt[3]{18}} - \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t}\]
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Let's validate this solution:
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\allowdisplaybreaks
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\begin{align}
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0 &\stackrel{!}{=} \left (\frac{t}{\sqrt[3]{18}} - \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t} \right )^3 + \alpha \left (\frac{t}{\sqrt[3]{18}} - \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t} \right ) + \beta\\
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&= (\frac{t}{\sqrt[3]{18}})^3
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- 3 (\frac{t}{\sqrt[3]{18}})^2 \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t}
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+ 3 (\frac{t}{\sqrt[3]{18}})(\frac{\sqrt[3]{\frac{2}{3}} \alpha }{t})^2
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- (\frac{\sqrt[3]{\frac{2}{3}} \alpha }{t})^3
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+ \frac{t \alpha}{\sqrt[3]{18}} - \frac{\sqrt[3]{\frac{2}{3}} \alpha^2 }{t} + \beta\\
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&= \frac{t^3}{18}
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- \frac{3t^2}{\sqrt[3]{18^2}} \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t}
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+ \frac{3t}{\sqrt[3]{18}} \frac{\sqrt[3]{\frac{4}{9}} \alpha^2 }{t^2}
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- \frac{\frac{2}{3} \alpha^3 }{t^3}
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+ \frac{t \alpha }{\sqrt[3]{18}} - \frac{\sqrt[3]{2} \alpha^2 }{\sqrt[3]{3} t} + \beta\\
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&= \frac{t^3}{18}
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- \frac{\sqrt[3]{18} t \alpha}{\sqrt[3]{18^2}}
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+ \frac{\sqrt[3]{12} \alpha^2}{\sqrt[3]{18} t}
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- \frac{2 \alpha^3 }{3t^3}
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+ \frac{t \alpha }{\sqrt[3]{18}} - \frac{\sqrt[3]{2} \alpha^2 }{\sqrt[3]{3} t} + \beta\\
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&= \frac{t^3}{18}
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\color{blue} - \frac{t \alpha}{\sqrt[3]{18}}
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\color{red} + \frac{\sqrt[3]{2} \alpha^2}{\sqrt[3]{3} t}
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\color{black}- \frac{2 \alpha^3 }{3 t^3}
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\color{blue} + \frac{t \alpha }{\sqrt[3]{18}}
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\color{red} - \frac{\sqrt[3]{2} \alpha^2 }{\sqrt[3]{3} t}
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\color{black}+ \beta\\
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&= \frac{t^3}{18} - \frac{2 \alpha^3 }{3 t^3} + \beta\\
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&= \frac{t^6 - 12 \alpha^3 + \beta 18 t^3}{18t^3}
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\end{align}
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Now only go on calculating with the numerator. Start with resubstituting
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$t$:
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\begin{align}
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0 &= (\sqrt{3 \cdot (4 \alpha^3 + 27 \beta^2)} -9\beta)^2 - 12 \alpha^3 + \beta 18 (\sqrt{3 \cdot (4 \alpha^3 + 27 \beta^2)} -9\beta)\\
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&= (\sqrt{3 \cdot (4 \alpha^3 + 27 \beta^2)})^2 +(9\beta)^2 - 12 \alpha^3 -(2 \cdot 9)\cdot 9\beta^2\\
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&= 3 \cdot (4 \alpha^3 + 27 \beta^2) -81 \beta^2 - 12 \alpha^3\\
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&= 0
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\end{align}
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