Challenge Level

Conratulations to Sue Liu, Jonathan and Tom of Madras College, St Andrew's and to Sanjay of The Perse School, Cambridge for their solutions to this problem. Here is Sanjay's solution.

$$\sqrt{8 -4\sqrt{3}} = \sqrt{a} - \sqrt{b}$$ The tactic I shall employ here will be to square both sides and solve for $a$ and $b$. $$8 -4\sqrt{3} = a - 2\sqrt{ab} + b$$. From this it is clear that the following equations must hold

\begin{eqnarray} a+b &=& 8 \\ 4\sqrt{3}
&=& 2 \sqrt{ab} \end{eqnarray}

These simultaneous equations can be solved for $a$ and $b$. From
equation 2. $$2 \sqrt{3} = \sqrt{ab}$$ From equation 1. $$b = 8
-a$$ Substituting equation 4 into 3 gives
\begin{eqnarray}\\ 2 \sqrt{3} &=&
\sqrt{a(8-a)}\\ 12 &=& a(8-a) \\ a^2 - 8a + 12 &=&
0 \\ (a-6)(a-2) &=& 0 \\ a &=& 2\ or \ a = 6
\end{eqnarray}

From the original equation it is clear that $a = 6$ and $b=2$
because the left-hand side must be positive. Hence $$
\sqrt{8-4\sqrt{3}} = \sqrt{6} - \sqrt{2}$$ This can easily be
generalised, as follows. $$\sqrt{x-y\sqrt{z}} = \sqrt{a} -
\sqrt{b}$$ Upon squaring both sides this yields the equations
\begin{eqnarray} a + b &=& x \\ ab
&=& \frac{y^2z}{4} \end{eqnarray}

From equation 5, it is clear that $$b = x - a$$ Substituting this
into equation 6 gives \begin{eqnarray} a(x-a) = \frac{y^2z}{4} \\
a^2 -ax + \frac{y^2z}{4} \end{eqnarray} Using the formula for
solving quadratics, and setting $a$ to be a larger than $b$ (as in
the original equation) gives the following solutions.
\begin{eqnarray} a &=& \frac{x+\sqrt{x^2
- y^2z}}{2} \\b &=& \frac{x-\sqrt{x^2 -
y^2z}}{2}\end{eqnarray}

Note that if the minuses in the folrmula are changed to pluses,
equations 5 and 6 will still be the same, and so you'll still get
the same answers.