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:
$$ \frac {x^2 +2x \sqrt{3} +3}{3x} $$

Hint
Squaring both sides gives $3x = (x + \sqrt{3})^2 = x^2 + 2\sqrt{3}x + 3$.
Rewrite this to get $x^2 + (2\sqrt{3} - 3)x + 3 = 0$. Solve this second degree equation to get
$$x = 1.5 -\sqrt{3} \pm \sqrt{(1.5-\sqrt{3})^2 - 3}$$
Now try to rewrite this on the form $A\sqrt{B}+C$. Be sure to check which root actually is a solution to the original equation.
