Algebra-Gleichungen-Quadratische Gleichung
$ ax^{2}+bx+c=0 $
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Beispiel Nr: 29
$\begin{array}{l} \text{Gegeben:} ax^{2}+bx+c=0
\\ \text{Gesucht:} \\ \text{Lösung der Gleichung} \\
\\ ax^{2}+bx+c=0 \\ \textbf{Gegeben:} \\
-1\frac{1}{4}x^2+5x =0
\\ \\ \textbf{Rechnung:} \\
\begin{array}{l|l|l|l}
\begin{array}{l}
\text{x-Ausklammern}\\ \hline
-1\frac{1}{4}x^{2}+5x =0 \\
x(-1\frac{1}{4}x +5)=0 \\
\\ -1\frac{1}{4} x+5 =0 \qquad /-5 \\
-1\frac{1}{4} x= -5 \qquad /:\left(-1\frac{1}{4}\right) \\
x=\displaystyle\frac{-5}{-1\frac{1}{4}}\\
x_1=0\\
x_2=4
\end{array}&
\begin{array}{l}
\text{a-b-c Formel}\\ \hline
\\
-1\frac{1}{4}x^{2}+5x+0 =0
\\
x_{1/2}=\displaystyle\frac{-5 \pm\sqrt{5^{2}-4\cdot \left(-1\frac{1}{4}\right) \cdot 0}}{2\cdot\left(-1\frac{1}{4}\right)}
\\
x_{1/2}=\displaystyle \frac{-5 \pm\sqrt{25}}{-2\frac{1}{2}}
\\
x_{1/2}=\displaystyle \frac{-5 \pm5}{-2\frac{1}{2}}
\\
x_{1}=\displaystyle \frac{-5 +5}{-2\frac{1}{2}} \qquad x_{2}=\displaystyle \frac{-5 -5}{-2\frac{1}{2}}
\\
x_{1}=0 \qquad x_{2}=4
\end{array}&
\begin{array}{l}
\text{p-q Formel}\\ \hline
\\
-1\frac{1}{4}x^{2}+5x+0 =0 \qquad /:-1\frac{1}{4}
\\
x^{2}-4x+0 =0
\\
x_{1/2}=\displaystyle -\frac{-4}{2}\pm\sqrt{\left(\frac{\left(-4\right)}{2}\right)^2- 0}
\\
x_{1/2}=\displaystyle 2\pm\sqrt{4}
\\
x_{1/2}=\displaystyle 2\pm2
\\
x_{1}=4 \qquad x_{2}=0
\end{array}\\ \end{array} \end{array}$