\ANG{
Berechne die L"osungsmenge: $\D \sqrt{x}=x+\frac{4}{25}$
}
\LSG{
$\DD=\RR_{0}^{+}$

\parbox[t]{7cm}{
L"osung durch Quadrieren:
\begin{align}
\sqrt{x} &= \left. x+\frac{4}{25}\quad \right|^{2}\nonumber\\
x &= x^{2}+\frac{8}{25}\,x+\frac{16}{625}\nonumber\\
x^{2}-\frac{17}{25}\,x &= -\frac{16}{625} \nonumber\\
x^{2}-2\cdot\frac{17}{50}\,x+\left(\frac{17}{50}\right)^{2} &= 
-\frac{16}{625}+\frac{289}{2500}\nonumber\\
\left(x-\frac{17}{50}\right)^{2}&= \frac{289-64}{2500}=\frac{225}{2500}\nonumber\\
x &= \frac{17}{50}\pm\frac{15}{50}\nonumber\\
x_{1}=\frac{16}{25}=0{,}64 &;\quad x_{2}=\frac{1}{25}=0{,}04\nonumber
\end{align}
}\hfill
\parbox[t]{6cm}{
Oder mit Substitution: $y=\sqrt{x}$
\begin{align}
y &= y^{2}+\frac{4}{25}\nonumber\\
y^{2}-y &= -\frac{4}{25}\nonumber\\
y^{2}-2\cdot\frac{1}{2}\,x+\frac 14 &= -\frac{4}{25}+\frac 14 \nonumber\\
\left(y-\frac{1}{2}\right)^{2}&= \frac{25-16}{100}=\frac{9}{100}\nonumber\\
y &= \frac{1}{2}\pm\frac{3}{10}\nonumber\\
y_{1} =  \frac{4}{5}\QDQ& x_{1}=y_{1}^{2}=\frac{16}{25}\nonumber\\
y_{2} =  \frac{1}{5}\QDQ& x_{2}=y_{2}^{2}=\frac{1}{25}\nonumber
\end{align}
}
\begin{align}
\text{Probe:}\qquad x_{1} &:\quad \text{LS}=\sqrt{\frac{16}{25}}=\frac{4}{5};\quad 
\text{RS}=\frac{16}{25}+\frac{4}{25}=\frac{20}{25}=\frac{4}{5}\nonumber\\
x_{1} &:\quad \text{LS}=\sqrt{\frac{1}{25}}=\frac{1}{5};\quad 
\text{RS}=\frac{1}{25}+\frac{4}{25}=\frac{5}{25}=\frac{1}{5}\nonumber
\end{align}
\[\LL=\left\{\frac{16}{25},\frac{1}{25}\right\}=\left\{0{,}64;\, 0{,}04\right\}\]
}