<s>[INST] 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!
 [/INST] Rewrite the equation as $$\left(\sqrt[4]{3x + 2} - \frac x{\sqrt[4]{3x+2}}\right)^2 = 0.$$ It follows that $\sqrt{3x+2} = x$ and then $x^2 - 3x - 2 = 0$. Note that we require $x \geq 0$.
</s>