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!
 A: 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 $x \geq 0$.
A: 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 $\frac{3 + \sqrt{17}}{2}$.
A: Multiply by $\sqrt{3x+2}$, assuming $3x+2\ge 0$ i.e. $x\ge-2/3$:
$$\left(\sqrt{3x+2}\right)^2-2x\sqrt{3x+2}+x^2=0$$
$$\left(\sqrt{3x+2}-x\right)^2=0$$
$$\sqrt{3x+2}=x$$
$$3x+2=x^2$$
Solve the quadratic equation to find $x=\frac{3\pm\sqrt{17}}{2}$. As the final equation is a consequence of the original one, but not necessarily equivalent to it, you now need to try those two solutions in the original equation. The only real solution will turn out to be $x=\frac{3+\sqrt{17}}{2}$ : the other one will be spurious and needs to be rejected.
A: 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
$$x^4-6x^3+5x^2+12x+4 = x^4-2x^2(3x+2) + (3x+2)^2.$$
A: 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 ${x=\frac{3+\sqrt{17}}2}$.

