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Find the sum of this series of surds.

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Surds

Age 14 to 16 Challenge Level:

The best solution for finding the exact values of $x$, $y$ and $a$ satisfying the following system of equations came from Bithian.

\begin{eqnarray}\frac{1}{a + 1} &=& a - 1 \\ x + y &=& 2a \\ x &=& ay \end{eqnarray}
The first equation gives:
\begin{eqnarray} ( a + 1)( a - 1) &=& 1 \\ a^2 - 1 &=& 1 \\ a^2 &=& 2 \\ a &=& \pm\sqrt{2}\end{eqnarray}
From the second and third equations:
\begin{align}
 ay + y &= 2 a \\
y = \frac{2a}{a + 1} &=
\begin{cases}
\frac{2\sqrt{2}}{\sqrt{2} + 1} & \text{for } a = \sqrt 2\\
\frac{2\sqrt{2}}{\sqrt{2} - 1} & \text{for } a = -\sqrt 2\\
\end{cases}
\end{align}
Now we use a neat trick that is well worth knowing, for getting rid of surds in the denomenator of a fraction. This trick uses the difference of two squares. In this case you use $(\sqrt{2} + 1)(\sqrt{2} - 1) = 2 - 1 = 1$.

Let's treat the two possible values for $a$ separately:
  1. For $a = \sqrt{2}$ we have to multiply the top and bottom by $\sqrt 2 - 1$ to give $$ y = 4 - 2\sqrt 2$$ Then use $x = ay = \sqrt 2 (4-2\sqrt 2) = 4\sqrt{2} - 4$ to get \begin{align}a &= \sqrt 2 \\ x &= 4\sqrt{2} - 4 \\ y &= 4 - 2\sqrt 2\end{align}
  2. For $a = -\sqrt{2}$ we have to multiply the top and bottom by $\sqrt 2 + 1$ to give $$ y = 4 + 2 \sqrt 2$$ Then use $x = ay = -\sqrt 2 (4+2\sqrt 2) = -4\sqrt{2} - 4$ to get \begin{align}a &= -\sqrt 2 \\ x &= -4\sqrt{2} - 4 \\ y &= 4 + 2\sqrt 2\end{align}

[NOTE: When $(a\sqrt{2} + b)$ occurs as a factor in the denominator (where $a$ and $b$ are whole numbers) you multiply top and bottom of the fraction by exactly the same thing, by $(a\sqrt{2} - b)$. In effect you just multiply the whole fraction by one and it will always give the whole number $2a^2 - b^2$ in the denominator.]