<s>[INST] Solve a simple equation: $x+x\sqrt{(2x+2)}=3$ 
$x+x\sqrt{(2x+2)}=3$

I must solve this, but I always get to a point where I don't know what to do. The answer is 1.
Here is what I did: 
$$\begin{align}
3&=x(1+\sqrt{2(x+1)}) \\
\frac{3}{x}&=1+\sqrt{2(x+1)} \\
\frac{3}{x}-1&=\sqrt{2(x+1)} \\
\frac{(3-x)^{2}}{x^{2}}&=2(x+1) \\
\frac{9-6x+x^{2}}{2x^{2}}&=x+1 \\
\frac{9-6x+x^{2}-2x^{2}}{2x^{2}}&=x \\
\frac{9-6x+x^{2}-2x^{2}-2x^{3}}{2x^{2}}&=0
\end{align}$$
Then I got: 
$-2x^{3}-x^{2}-6x+9=0$ 
 [/INST] So you got to the cubic equation $f(x)=-2x^{3}-x^{2}-6x+9=0$. When you come across a cubic like this, try evaluating $f(\pm1), f(\pm2)$, etc, to try and figure out some roots so you can factor it (you know $x_0$ a root if $f(x_0)=0$). Here you can see $1$ is a root, so factoring out $(x-1)$ gives $f(x)=(x-1)\underbrace{(2x^2+3x+9)}_{\text{no real roots}}$. So the only solution is $x=1$.
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