<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] I'd recommend this method. You already got $(x+1)(x+2)-2x\sqrt{3x+2}=0$. And by this, $x\ge0$ is induced right away, because the left hand is $>0$.
Then...
$$
(x+1)(x+2)=2x\sqrt{3x+2}\\
(x+1)^2(x+2)^2=4x^2(3x+2)\\
x^4-6x^3+5x^2+12x+4=0
$$
This is sure a quatric, so seems hard to solve. But...

 Let $x^2-3x=y$, then this quatric changes $y^2-4y+4=0$.So $y=2$, $x^2-3x-2=0$.Then we get two roots, $x=\frac{3\pm\sqrt{17}}2$. $\frac{3-\sqrt{17}}2$ is out of range.So the answer is ${x=\frac{3+\sqrt{17}}2}$.

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