Answer
$\left\{\dfrac{2}{3}\right\}$
Work Step by Step
Squaring both sides, the given equation, $
3x=\sqrt{\dfrac{9x+2}{2}}
$, is equivalent to
\begin{align*}
(3x)^2&=\left(\sqrt{\dfrac{9x+2}{2}}\right)^2
\\\\
9x^2&=\dfrac{9x+2}{2}
.\end{align*}
The equation above has $a=
18
$, $b=
-9
$ and $c=
-2
$. Using $
x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
$ or the Quadratic Formula, then
\begin{align*}
x&=\dfrac{-(-9)\pm\sqrt{(-9)^2-4(18)(-2)}}{2(18)}
\\\\&=
\dfrac{9\pm\sqrt{81+144}}{36}
\\\\&=
\dfrac{9\pm\sqrt{225}}{36}
.\end{align*}
Using the properties of radicals, the equation above is equivalent to
\begin{align*}\require{cancel}
t&=\dfrac{9\pm\sqrt{(15)^2}}{36}
\\\\
t&=\dfrac{9\pm15}{36}
\end{align*}\begin{array}{l|r}
t=\dfrac{9-15}{36} & t=\dfrac{9+15}{36}
\\\\
t=\dfrac{-6}{36} & t=\dfrac{24}{36}
\\\\
t=-\dfrac{1}{6} & t=\dfrac{2}{3}
\end{array}
Since both sides of the original equation were raised to the second power, then checking of solutions is a must. Substituting the solutions in the original equation results to
\begin{array}{l|r}
\text{If }t=-\dfrac{1}{6}: & \text{If }t=\dfrac{2}{3}:
\\\\
3\left(-\dfrac{1}{6}\right)\overset{?}=\sqrt{\dfrac{9\left(-\dfrac{1}{6}\right)+2}{2}} &
3\left(\dfrac{2}{3}\right)\overset{?}=\sqrt{\dfrac{9\left(\dfrac{2}{3}\right)+2}{2}}
\\\\
-\dfrac{1}{2}\ne\text{some nonnegative number} &
2\overset{?}=\sqrt{\dfrac{6+2}{2}}
\\\\
&
2\overset{?}=\sqrt{\dfrac{8}{2}}
\\\\
&
2\overset{?}=\sqrt{4}
\\\\&
2\overset{\checkmark}=2
.\end{array}
Since $x=-\dfrac{1}{6}$ does not satisfy the original equation, then the only solution is $x=\dfrac{2}{3}$. Hence, the solution set of the equation $
3x=\sqrt{\dfrac{9x+2}{2}}
$ is $\left\{\dfrac{2}{3}\right\}$.
