Lösung Kugeln mit negativen Zahlen

\[\begin{align*} &P\left(X=4\right)\cdot4+P\left(X=2a\right)\cdot2a+P\left(X=a^2\right)\cdot a^2&=4 \\ &\Leftrightarrow \left(\frac{3}{5}\right)^2\cdot4+2\cdot\frac{3}{5}\cdot\frac{2}{5}\cdot2a+\left(\frac{2}{5}\right)^2\cdot a^2&=4 \\ &\Leftrightarrow \frac{36}{25}+\frac{24}{25}a+\frac{4}{25}a^2&=4\\ &\Leftrightarrow a^2+6a-16&=0 \\ \end{align*}\]

\[\begin{align*} a_{1,2} &= \frac{-6 \pm \sqrt{6^2-4\cdot 1\cdot (-16)}}{2\cdot 1} \\ &= \frac{-6 \pm \sqrt{100}}{2} \\ &= \frac{-6 \pm 10}{2} \\ \Leftrightarrow &a_1=\frac{-6+10}{2}=2; a_2=\frac{-6-10}{2}=-8 \end{align*}\]