ΛΥΣΗ

α) Έχουμε:

$$\begin{array}{rl}{\alpha }^2& =(1+2\sqrt{2}{)}^2\\ & =1^2+2\cdot 2\sqrt{2}\cdot 1+(2\sqrt{2}{)}^2\\ & =1+4\sqrt{2}+2^2(\sqrt{2}{)}^2\\ & =1+4\sqrt{2}+8\\ & =9+4\sqrt{2}\end{array}$$

και

$$\begin{array}{rl}{\beta }^2& =(\sqrt{2}-2{)}^2\\ & =(\sqrt{2}{)}^2-2\cdot \sqrt{2}\cdot 2+2^2\\ & =2-4\sqrt{2}+4\\ & =6-4\sqrt{2}\end{array}$$

Οπότε ${\alpha }^2+{\beta }^2=9+4\sqrt{2}+6-4\sqrt{2}=15$.

β) Είναι $\alpha =1+2\sqrt{2}>0$ και $\beta =\sqrt{2}-2<0$, οπότε:

$$\begin{array}{rl}\sqrt{{\alpha }^2}+2\sqrt{{\beta }^2}& =|\alpha |+2|\beta |\\ & =\alpha +2(-\beta )\\ & =1+2\sqrt{2}+2(2-\sqrt{2})\\ & =1+2\sqrt{2}+4-2\sqrt{2}\\ & =5\end{array}$$