439a – FW Algebra – Mathe Solutions

$\begin{align*} \left| \begin{array}{rcl} s^2 - st & = & 20 \\ t^2 - st & = & -5 \end{array} \right| \end{align*}$

Lösung

$\begin{align*} \begin{array}{c} (1) \\ (2) \end{array} & \left| \begin{array}{rcl} s^2 - st & = & 20 \\ t^2 - st & = & -5 \end{array} \right| \begin{array}{c} (*) \\ ~ \end{array} \end{align*}$

$\begin{align*} \begin{array}{c} (1) \\ (2) \end{array} & \left| \begin{array}{rcl} s(s-t) & = & 20 \\ t(t-s) & = & -5 \end{array} \right| \begin{array}{c} \\ \cdot(-1) \end{array} \end{align*}$

$\begin{align*} \begin{array}{c} (1) \\ (2) \end{array} & \left| \begin{array}{rcl} s(s-t) & = & 20 \\ t(s-t) & = & 5 \end{array} \right| \end{align*}$

Die Gleichung (1) durch die Gleichung (2) teilen:

$\begin{eqnarray*} \frac{s}{t} & = & 4 \\ s & = & 4t ~~~(**) \end{eqnarray*}$

in (*) einsetzen:

$\begin{eqnarray*} 16t^2-4t^2 & = & 20 \\ 12t^2 & = & 20 \\ t^2 & = & \frac{5}{3} \\ t_{1,2} & = & \pm\sqrt{\frac{5}{3}} \end{eqnarray*}$

in (**) einsetzen:

$\begin{equation*} s_{1,2} = \pm 4\sqrt{\frac{5}{3}} \end{equation*}$

$\begin{equation*} \mathbb{L}=\left\{\left(4\sqrt{\frac{5}{3}},\sqrt{\frac{5}{3}}\right),\left(-4\sqrt{\frac{5}{3}},-\sqrt{\frac{5}{3}}\right)\right\} \end{equation*}$

Lösung

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