<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] First multiply by $\sqrt{3x+2}$ to get
$$3x+2 + x^2 = 2x \sqrt{3x+2}$$
Then square both sides
\begin{align*}
(x^2+3x+2)^2 &= 4x^2(3x+2)\\
x^4+6x^3+13x^2+12x+4 &= 12x^3+8x^2 \\
x^4-6x^3+5x^2+12x+4 &=0
\end{align*}
Try to solve this polynomial equation, giving you the true solution and one false solution that you have to exclude. It is helpful to observe that
$$x^4-6x^3+5x^2+12x+4 = x^4-2x^2(3x+2) + (3x+2)^2.$$
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