<s>[INST] What is the Conjugate of $\frac{1}{4\sqrt{3}-7}$ I'm having problems finding the conjugate of $\frac{1}{4\sqrt{3}-7}$
The answer I get is as follows: ${-4\sqrt{3}-7}$
However the answer given is ${-4\sqrt{3}+7}$
Here are my workings out...
$\frac{1}{4\sqrt{3}-7} * \frac{4\sqrt{3}+7}{4\sqrt{3}+7} = \frac{4\sqrt{3}+7}{-1} = {-4\sqrt{3}-7}$
I'd appreciate any guidance on where I've gone wrong. 
 [/INST] Well, after all the changes:
$$(4\sqrt3-7)(4\sqrt3+7)=48-49=-1$$
and then indeed:
$$\frac1{4\sqrt3-7}=-\left(4\sqrt3+7\right)=-4\sqrt3-7$$
and yes: the conjugate is $\;\frac1{4\sqrt3+7}\;$ , and what you say is the answer given is neither the conjugate nor the product of the original expression by its conjugate.
Added following an idea by projectilemotion: Perhaps the idea is first to rationalize the expression and then to find its conjugate:
$$\text{Rationalizing:}\;\;\frac1{4\sqrt3-7}\cdot\frac{4\sqrt3+7}{4\sqrt3+7}=-4\sqrt3-7$$
and now the rightmost expression's conjugate indeed is: $\;-4\sqrt3\color{Red}+7\;$ .....tadaaah!
It's hard to tell what they meant without knowing a priori their definitions... BTW, in this case and for me, the conjugate could as well be $\;4\sqrt3-7=-7+4\sqrt3\;$ ...Funny.
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