## Task 6 - 250936

a) Is the term

$$
z=\sqrt{192+96 \cdot \sqrt{3}}+\sqrt{192-96 \cdot \sqrt{3}}
$$

defining a number?

b) If this is the case, is $z$ rational?
a) It is $192-96 \sqrt{3}=96 \cdot(2-\sqrt{3})>0$, since $2=\sqrt{4}>\sqrt{3}$, so $z$ describes a real number.

b) It is

$$
z=\sqrt{16 \cdot(12+6 \sqrt{3})}+\sqrt{16 \cdot(12-6 \sqrt{3})}=4 \cdot(\sqrt{12+6 \sqrt{3}}+\sqrt{12-6 \sqrt{3}})
$$

rational if and only if $0 \leq x:=\frac{z}{4}=\sqrt{12+6 \sqrt{3}}+\sqrt{12-6 \sqrt{3}}$ is also rational. It is

$$
\begin{gathered}
x^{2}=(12+6 \sqrt{3})+2 \sqrt{(12+6 \sqrt{3}) \cdot(12-6 \sqrt{3})}+(12-6 \sqrt{3})=24+2 \sqrt{144-36 \cdot 3}= \\
=24+2 \sqrt{36}=24+2 \cdot 6=36 \quad \text { so } \quad x=6 \text { and } z=24 \in \mathbb{Q}
\end{gathered}
$$