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