Answer
$x=\dfrac{2}{3}$
Work Step by Step
Squaring both sides, the given equation, $
x\sqrt{3}=\sqrt{2-x}
,$ is equivalent to
\begin{align*}
\left(x\sqrt{3}\right)^2&=\left(\sqrt{2-x}\right)^2
\\
x^2(3)&=2-x
\\
3x^2&=2-x
\\
3x^2+x-2&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(x+1)(3x-2)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property), then
\begin{array}{l|r}
x+1=0 & 3x-2=0
\\
x=-1 & 3x=2
\\\\
& x=\dfrac{2}{3}
.\end{array}
Checking by substituting the solutions in the given equation results to
\begin{array}{l|r}
\text{If }x=-1: & \text{If }x=\dfrac{2}{3}:
\\\\
-1\sqrt{3}\overset{?}=\sqrt{2-(-1)} &
\dfrac{2}{3}\sqrt{3}\overset{?}=\sqrt{2-\dfrac{2}{3}}
\\\\
-\sqrt{3}\overset{?}=\sqrt{2+1} &
\dfrac{2\sqrt{3}}{3}\overset{?}=\sqrt{\dfrac{6}{2}-\dfrac{2}{3}}
\\\\
-\sqrt{3}\ne\sqrt{3} &
\dfrac{2\sqrt{3}}{3}\overset{?}=\sqrt{\dfrac{4}{3}}
\\\\
&
\dfrac{2\sqrt{3}}{3}\overset{?}=\sqrt{\dfrac{4}{3}\cdot\dfrac{3}{3}}
\\\\
&
\dfrac{2\sqrt{3}}{3}\overset{?}=\sqrt{\dfrac{4}{9}\cdot3}
\\\\
&
\dfrac{2\sqrt{3}}{3}\overset{?}=\sqrt{\dfrac{4}{9}}\cdot\sqrt{3}
\\\\
&
\dfrac{2\sqrt{3}}{3}\overset{?}=\dfrac{2}{3}\cdot\sqrt{3}
\\\\
&
\dfrac{2\sqrt{3}}{3}\overset{\checkmark}=\dfrac{2\sqrt{3}}{3}
.\end{array}
Since $x=-1$ does not satisfy the original equation then the only solution of the equation $
x\sqrt{3}=\sqrt{2-x}
$ is $x=\dfrac{2}{3}$.
