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

$\sqrt{3x}=x+\sqrt{3}$
$3x=x^2+2x\sqrt{3}+3$
$x^2+(2\sqrt{3}-3)x+3=0$
$D=(2\sqrt{3}-3)^2-4 \times 1 \times 3=12-12\sqrt{3}+9-12=9-12\sqrt{3}<0$
The equation nas no real solutions.
you can check that the answer you have is not the solution at  all.
