### Surds Part II

Example 1
A cone has a volume of $\ (\sqrt{5} + 3) \: \text{cm}^3$. Its base area is $\ (\sqrt{5} +1)\: \text{cm}$ What is the height of the cone ?

To find the height of the cone we divide its volume by its height

\begin {aligned} \frac{\sqrt{5} + 3}{\sqrt{5} +1} &= \frac{(\sqrt{5} + 3)(\sqrt{5} -1)}{(\sqrt{5} +1)(\sqrt{5} -1)}\\ &= \frac{(\sqrt{5})(\sqrt{5}) - \sqrt{5} + 3\sqrt{5} -3}{(\sqrt{5} +1)(\sqrt{5} -1)} \\ &= \frac{5+ 2\sqrt{5} - 3 }{5-1} \\ &= \frac{2+ 2\sqrt{5}}{4} \\ &= \frac{1 + \sqrt{5}}{2} \\ \end {aligned}

Example 2
Solve the following equation:
$\ y\sqrt{24} = y\sqrt{3} + \sqrt{6}$

To solve this, we make y the subject of the equation
\begin {aligned} y\sqrt{24} &= y\sqrt{3} + \sqrt{6} \\ y\sqrt{24} - y\sqrt{3} &=\sqrt{6}\\ y ( \sqrt{24} - \sqrt{3} ) &= \sqrt{6}\\ y &= \frac{ \sqrt{6}}{\sqrt{24} - \sqrt{3}} \\ y &= \frac{ \sqrt{3}\sqrt{2}}{\sqrt{8}\sqrt{3} - \sqrt{3}} \\ y&= \frac{ \sqrt{2}}{\sqrt{8} - 1} \\ \end {aligned}

Next we rationalize

\begin {aligned} y&= \frac{ \sqrt{2}(\sqrt{8} + 1)}{(\sqrt{8} - 1)(\sqrt{8} + 1)} \\ y&= \frac{ \sqrt{16} + \sqrt{2}}{7} \\ y&= \frac{ 4 + \sqrt{2}}{7} \\ \end {aligned}

Example 3
Given that $\ (x + \sqrt{7})(3 + y\sqrt{7}) = 26 + 11 \sqrt{7}$, find the possible values of x and y.

To solve this question we need to expand and simplify the LHS
Then we group the terms containing surds together
\begin {aligned} LHS &= (x + \sqrt{7})(3 + y\sqrt{7})\\ &= 3x + xy\sqrt{7} + 3\sqrt{7} + y (\sqrt{7})(\sqrt{7}) \\ &= 3x + 7y + (3 + xy) \sqrt{7} \\ \end {aligned}

Subsequently we equate the terms containing surds
$\ (3 + xy) \sqrt{7} = 11 \sqrt{7}$
Therefore, $\ 3 + xy = 11$

Then we equate terms that do not contain surds
$\ 3x + 7y = 26$

We have 2 equations that we can solve simultaneously to determine the value of x and y
\begin {aligned} y &= \frac{8}{x} \\ \text{Sub into second equation} \\ 3x + 7(\frac{8}{x} ) &= 26 \\ 3x^2 -26x + 56 &= 0 \\ (3x-14)(x-4) &= 0 \\ x = \frac{14}{3} \quad \text{or} \quad x=4 \\ \end {aligned}

From the values of x we can then determine the value of y
$\ y = \frac{12}{7} \quad \text{or} \quad y = 2$

Example 4
Given that $\ \sqrt{x + y \sqrt{5}} = \frac{5}{4 + \sqrt{5}}$. Determine the value of x and y.

The first step is to square both sides
\begin {aligned} \sqrt{x + y \sqrt{5}} &= \frac{5}{4 + \sqrt{5}} \\ \left(\sqrt{x + y \sqrt{5}}\right)^2 &= \left(\frac{5}{4 + \sqrt{5}}\right)^2 \\ x + y \sqrt{5} &= \frac{25}{(4+ \sqrt{5})^2} \\ \end {aligned}

The second step is to simplify and rationalize the fraction on the right hand side
\begin {aligned} x + y \sqrt{5} &= \frac{25}{(16 + 8\sqrt{5} + 5} \\ &=\frac{25}{(21 + 8\sqrt{5}) } \\ &= \frac{25(21 + 8\sqrt{5})}{(21 + 8\sqrt{5})(21 - 8\sqrt{5}) }\\ &= \frac{525 + 200 \sqrt{5}}{441 -320} \\ &= \frac{525 + 200 \sqrt{5}}{121} \\ &= \frac{525}{121} + \frac{200}{121}\sqrt{5}\\ \end {aligned}

Finally we compare the terms. We have a term that contains surds, $\ \frac{200}{121}\sqrt{5}$ and a term that does not contain surds $\ \frac{525}{121}$

We can equate the surd containing terms to determine the value of y
\begin{aligned} y\sqrt{5} &= \frac{200}{121}\sqrt{5} \\ y &= \frac{200}{121} \\ \end {aligned}

We can also equate the terms that do not contain surds to detrmine the value of x
\begin{aligned} x &= \frac{525}{121} \\ \end {aligned}