Algebra-Gleichungen-Quadratische Gleichung
$ ax^{2}+bx+c=0 $
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Beispiel Nr: 31
$\begin{array}{l} \text{Gegeben:} ax^{2}+bx+c=0
\\ \text{Gesucht:} \\ \text{Lösung der Gleichung} \\
\\ ax^{2}+bx+c=0 \\ \textbf{Gegeben:} \\
\frac{5}{9}x^2-5 =0
\\ \\ \textbf{Rechnung:} \\
\begin{array}{l|l|l|l}
\begin{array}{l}
\text{Umformen}\\ \hline
\frac{5}{9}x^2-5 =0 \qquad /+5 \\
\frac{5}{9}x^2= 5 \qquad /:\frac{5}{9} \\
x^2=\displaystyle\frac{5}{\frac{5}{9}} \\
x=\pm\sqrt{9} \\
x_1=3 \qquad x_2=-3
\end{array}&
\begin{array}{l}
\text{a-b-c Formel}\\ \hline
\\
\frac{5}{9}x^{2}+0x-5 =0
\\
x_{1/2}=\displaystyle\frac{-0 \pm\sqrt{0^{2}-4\cdot \frac{5}{9} \cdot \left(-5\right)}}{2\cdot\frac{5}{9}}
\\
x_{1/2}=\displaystyle \frac{-0 \pm\sqrt{11\frac{1}{9}}}{1\frac{1}{9}}
\\
x_{1/2}=\displaystyle \frac{0 \pm3\frac{1}{3}}{1\frac{1}{9}}
\\
x_{1}=\displaystyle \frac{0 +3\frac{1}{3}}{1\frac{1}{9}} \qquad x_{2}=\displaystyle \frac{0 -3\frac{1}{3}}{1\frac{1}{9}}
\\
x_{1}=3 \qquad x_{2}=-3
\end{array}&
\begin{array}{l}
\text{p-q Formel}\\ \hline
\\
\frac{5}{9}x^{2}+0x-5 =0 \qquad /:\frac{5}{9}
\\
x^{2}+0x-9 =0
\\
x_{1/2}=\displaystyle -\frac{0}{2}\pm\sqrt{\left(\frac{0}{2}\right)^2- \left(-9\right)}
\\
x_{1/2}=\displaystyle 0\pm\sqrt{9}
\\
x_{1/2}=\displaystyle 0\pm3
\\
x_{1}=3 \qquad x_{2}=-3
\end{array}\\ \end{array} \end{array}$