$$\sqrt{ 3x }= x + \sqrt {3}$$
this is what i tried
$$\sqrt{ x }= (x + \sqrt {3})^2\\ = x^2 + 3 $$
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:
$$\frac {1}{2} \sqrt{3} + \frac {3}{2} $$
First of all $(\sqrt {3x})^2 = 3x$.
And furthermore, $$(x + \sqrt 3)^2\neq x^2 + 3$$ Recall, when squaring a binomial: $$(a + b)^2 = a^2 + 2ab + b^2$$
Starting from the beginning: we need for $3x \geq 0 \iff x \geq 0$ for the left hand side of the following equation to be defined in the reals:
$$\sqrt{3x} = x + \sqrt 3$$
Squaring both sides gives us $$\begin{align} 3x = (x +\sqrt 3)^2 & \iff 3x = x^2 + 2\sqrt 3 x + 3\\ \\ &\iff x^2 + (2\sqrt 3 - 3)x + 3 = 0\end{align}$$
You can use the quadratic equation to check for solutions, throwing out any solution for which $x \lt 0$. Recall, for any quadratic equation of the form $$ax^2 + bx + c = 0$$ we can find its roots $$x_i = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ In your equation, $$a = 1,\\\;b = 2\sqrt 3 - 3,\;\\ b^2 = (2\sqrt 3)^2 - 2(3\cdot 2\sqrt 3) + (-3)^2 = 12 - 12\sqrt 3 + 9 = 21 - 12\sqrt 3,\\ c = 3$$