<s>[INST] Algebra on a Louvre tablet Problem: On a Louvre tablet of about 300 B.C. are four problems concerning rectangles of unit area and given semiperimeter. Let the sides and semiperimeter be $x,y$ and $a$. Then we have
\begin{equation}
xy=1, \qquad x+y=a. \tag{1}
\end{equation}
Solve this system by using the identity
$$
\biggl(\frac{x-y}{2}\biggr)^2 = \biggl(\frac{x+y}{2}\biggr)^2 - xy \tag{2}
$$
Attempted solution:
By elimination: To eliminate $y$, we note that $y = \frac{1}{x}$, and we substitute this $y$-value into the equation $x+y=a$ to obtain
$$
x = \frac{a \pm \sqrt{a^2-4}}{2},
$$
where this solution was obtained by using the quadratic formula.
By using the given identity: Start by letting $y=a-x$. Using this and the fact that $x+y=a$, we have the following:
\begin{align*}
\biggl(\frac{2x-a}{2}\biggr)^2 = \biggl(\frac{a}{2}\biggr)^2 - 1 &\longleftrightarrow 4x^2-4xa+a^2 = a^2-4\\ &\longleftrightarrow x^2-ax+1=0,
\end{align*}
whereby we get that $x = \frac{a \pm \sqrt{a^2-4}}{2}$, as we did above by using elimination. 
Main question: My solution using (2) seems to work, but is there a "slicker" approach? That is, it seems like the originally posed problem wants me to do something nifty with (2) rather than using it in such a brutish fashion. Anyone have a cleaner approach in mind in regards to solving (1) by using (2)?
 [/INST] I think the intended solution was as follows:
$$
\left(\frac{x-y}{2}\right)^2=\left(\frac{a}{2}\right)^2-1.
$$
Thus $x-y$ can be solved for:
$$
x-y=\pm2\sqrt{\left(\frac{a}{2}\right)^2-1}=\pm\sqrt{a^2-4}
$$ Adding and subtracting with $x+y$ gives us:
$$
x=\frac{(x+y)+(x-y)}{2}=\frac{a\pm \sqrt{a^2-4}}{2},
$$
and likewise for $y$.
Note that this is a re-derivation of the quadratic formula; in particular, it defeats the purpose of this approach to use it!
</s>