ΛΥΣΗ

α) Έχουμε:

\begin{align} α^{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{align}

και

\begin{align} β^{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{align}

Οπότε \(α^{2}+β^{2}=9+4\sqrt{2}+6-4\sqrt{2}=15\).

β) Είναι \(α=1+2\sqrt{2}>0\) και \(β=\sqrt{2}-2<0\), οπότε:

\begin{align} \sqrt{α^{2}}+2\sqrt{β^{2}} & =|α|+2|β| \\ & =α+2(-β) \\ & =1+2\sqrt{2}+2(2-\sqrt{2}) \\ & =1+2\sqrt{2}+4-2\sqrt{2} \\ & =5 \end{align}