Quite often, when doing calculations on polyhedra, you will find yourself with complex equations like $$\begin{equation}\tag{eq1}\label{eq1}\frac{20+7\sqrt3}{1+2\sqrt3}\end{equation}$$

Is the top evenly divisible by the bottom? I certainly can’t tell just by looking at them.

There must be a method to factor the numerator.

Let’s start with $$(A+B\sqrt{R})(C+D\sqrt{R})$$ multiplying the terms together gives $$AC+BRD+(BC+AD)\sqrt{R}$$

So, $$\frac{AC+BRD+(BC+AD)\sqrt{R}}{A+B\sqrt{R}} = C+D\sqrt{R}$$

Then we can substitute known values (A=1, B=2, R=3) into the formula. $$\frac{1C+2(3)D+(2C+1D)\sqrt{3}}{1+2\sqrt{3}} = C+D\sqrt{3}$$

Finding the answer is just a matter of finding C and D.

Consider that
$$\frac{\color{yellow}{C+6D} + \color{cyan}{(2C+D)}\sqrt3}{1+2\sqrt3} = \frac{\color{yellow}{20}+\color{cyan}{7}\sqrt3}{1+2\sqrt3}$$

We can break the equation into the colored sections to find $$\begin{align}\color{yellow}{C+6D} & \color{yellow}{=20} \\ C&=20-6D\tag{eq2}\label{eq2} \end{align}$$ and $$\color{cyan}{2C+D = 7}\tag{eq3}\label{eq3}$$

We now know C \eqref{eq2}, at least in terms of D, then substitute this into \eqref{eq3}.
$$\begin{align}2(20-6D)+D &= 7 \\ 40-12D+D &= 7 \\ -11D &= 7-40 = -33 \\D &= \frac{-33}{-11} \\ D&=3\end{align}$$

Returning to \eqref{eq2} and substituting D=3, we get \(C = 20-6\cdot 3 = 20-18 = 2\).

Then our answer should be \(2+3\sqrt{3}\), and it is, because \((1+2\sqrt3)(2+3\sqrt3) = 20+7\sqrt3\).

This method make it simple to divide out these complicated numbers, although, you might not get a simpler answer. For example, if we change \eqref{eq1} to \(\Large\frac{19+7\sqrt3}{1+2\sqrt3}\), we end up with answers C=23/11 and D=31/11, but it still works as \((1+2\sqrt3)(\frac{23}{11}+\frac{31\sqrt3}{11}) = 19+7\sqrt3\). Although, we could write our answer as \(\frac{1}{11}(23+31\sqrt3)\).

Another example:$$\frac{-3}{3+2\sqrt3}$$
We have values (A=3, B=2, R=3), giving $$3C + 2\cdot3D + (2C+3D)\sqrt3 = (3+2\sqrt3)(C+D\sqrt3)$$

Remember that \(-3\) is the same as \(-3+0\sqrt3\). So
$$\begin{align}3C+6D&=-3\label{e2}\tag{eq4}\\
2C+3D&=0\label{e3}\tag{eq5}
\end{align}$$

Working out \eqref{e2} gives
$$\begin{align}3C+6D&=-3 \\ 6D&= -3-3C \\D&=\frac{3(-1-C)}{6} \\D&= \frac{-1-C}{2}\end{align}$$

Plugging into \eqref{e3}, gives
$$\begin{align}2C+3D&=0 \\ 2C+3\cdot(\frac{-1-C}{2})&=0 \\ \frac{4C}{2}+\frac{-3C-3}{2}&=0 \\ \frac{4C-3C-3}{2}&=0 \\ \frac{C-3}{2}&=0 \\ C-3&=0 \\C&=3\end{align}$$

With C=3, we can find D in \eqref{e2} $$\begin{align}3(3)+6D&=-3 \\ 9+6D&=-3 \\ 6D&=-3-9 \\ 6D&=-12 \\ D&=\frac{-12}{6} \\ D&=-2\end{align}$$

So the answer is \(3-2\sqrt3\), as \((3+2\sqrt3)(3-2\sqrt3)=-3\).

You might notice this is the same as the “Difference of 2 squares” (x-y)(x+y) = x²-xy+xy-y² = x²-y². The “xy” terms cancel each other out, leaving 0. (x-y) and (x+y) are conjugates of one another, so when you have a numerator with a zero times the square root (or in other words, the numerator is a whole number), you may wish to try the denominator’s conjugate first. It may not be the answer you seek, but it can help you sometimes.

So to find an answer to $$\notag\frac{X+Y\sqrt{R}}{A+B\sqrt{R}} = C+D\sqrt{R}$$

You will need to find the solutions to \(AC+BRD=X\) and \(BC+AD=Y\)

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