Solve the equation $\sqrt{3x+2}+\dfrac{x^2}{\sqrt{3x+2}}=2x$ Solve the equation $$\sqrt{3x+2}+\dfrac{x^2}{\sqrt{3x+2}}=2x$$
We have $DM:3x+2>0,x>-\dfrac23, x\in DM=\left(-\dfrac23;+\infty\right)$, so we can multiply the whole equation by $\sqrt{3x+2}\ne0$. Then we will have $$3x+2+x^2=2x\sqrt{3x+2}\\x^2+3x+2-2x\sqrt{3x+2}=0\\(x+1)(x+2)-2x\sqrt{3x+2}=0$$ What next? Thank you!

There is a clever solution here: it requires using AM-GM.
Assume that $\sqrt{3x+2} \geq 0$, which clearly must be true. Then, By AM-GM, we have
$$
\sqrt{3x+2} + \frac{x^2}{\sqrt{3x+2}} \geq 2\sqrt{x^2} = 2x
$$
with equality if and only if $\sqrt{3x+2} = \frac{x^2}{\sqrt{3x+2}}$. This is true when $x^2 = 3x+2$, or $x = \frac{3 \pm \sqrt{17}}{2}$. Clearly, we must have $x \geq \frac{-2}{3}$, so the only real solution is $\frac{3 + \sqrt{17}}{2}$.
