question on surds i already asked this question but the answer I got did not match the one in the book $$\sqrt{ 3x }= x + \sqrt {3}$$
Give x in the form 
$$A \sqrt {B} + C $$
Can you show me how this is done step by step. 
The answer I have in the book is:
$$\frac {1}{2} \sqrt{3} + \frac {3}{2} $$
this is where I got stuck:
$$ \frac {x^2 +2x \sqrt{3} +3}{3x} $$

I figured out the problem: The solution to $\color{blue}{\sqrt 3}(x) = x + \sqrt 3$ is indeed $$\frac {1}{2} \sqrt{3} + \frac {3}{2}$$
But that's not the problem you posted. In the above, only $3$ is under the radical sign. In your post, you have $\sqrt{3x}$
In the event that the problem should read: $$\sqrt 3(x) = x + \sqrt 3$$
then $$\begin{align} \sqrt 3(x) = x + \sqrt 3 & \iff (\sqrt 3 - 1)x = \sqrt 3 \\ \\ &\iff x = \dfrac {\sqrt 3}{\sqrt 3 - 1} \\ \\ &\iff x = \frac{\sqrt 3}{\sqrt 3 - 1} \cdot \frac{\sqrt 3 + 1}{\sqrt 3 + 1} = \dfrac{3 +\sqrt 3}{3 - 1} = \dfrac 32 + \dfrac {\sqrt 3}2\end{align}$$
And in the desired form, that gives you $$\frac {1}{2} \sqrt{3} + \frac {3}{2}$$
