<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] I wouldn't have gotten there is such a convoluted way (and I wouldn't have divided by $x$ without checking that $x \ne 0$ first) but that would really make a difference.
And I'd watch out for extraneous roots.
$x +x\sqrt{2x+2} = 3$
$x\sqrt{2x+2} = 3-x$  $2x+2 > 0$ so $x > -1$
$x^2(2x+2) = 9 -6x + x^2; x > -1$
$2x^3 + x^2 + 6x -9=0; x > -1$ which is basically what you have.
Rational root theoreom says that if there are any rational roots the will be of the form $\frac {\pm 1,3,9}{\pm 1,2}$.  
It's pretty clear that $x = 1$ is solution so $x-1$ is a factor and so 
$(2x^3 + x^2 + 6x - 9)=(x-1)(2x^2 + bx + 9)$ where $bx^2 -2x^2 = x^2$ and $9x-bx=6x$ so $b=3$ so $(x-1)(2x^2 + 3x + 9)=0$.
$x =1$ is a solution and $2x^2 + 3x + 9 = 0$ is another.  But that'd imply $2x^2 = -3x - 9$.  But that's imply $3x < -9$ and $x<-3$ which is impossible as $(2x + 2) > 0$.
So $x=1$ is the only real solution.
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Another way to factor. If you KNOW $x-1$ is a factor, which we do we have:
$2x^3 + x^2 + 6x - 9=$
$2x^2(x-1) + 2x^2 + x^2 + 6x - 9 =$
$2x^2(x-1) + 3x^2 + 6x - 9=$
$2x^2(x-1) + 3x(x-1) + 3x + 6x - 9=$
$2x^2(x-1) + 3x(x-1) + 9x - 9 = $
$2x^2(x-1) + 3x(x-1) + 9(x-1) + 9 - 9=$
$2x^2(x-1) + 3x(x-1) + 9(x-1)=$
$(x-1)(2x^2 + 3x + 9) = 0$.
We can try to solve $2x^2 + 3x + 9 =0$ by the quadratic equation.  But we don't actually have to.  We know as $\sqrt{2x+2}$ must be real that $x \ge -1$ So $2x^2 + 3x + 9 \ge 2x^2 + 3(-1) + 9 = 2x^2 + 9 > 9 > 0$ so there is no solution that will satisfy both  $2x^2 + 3x + 9 = 0$ and $\sqrt{2x+2}$ is real.
(There's actually no real solution to just $2x^2 + 3x + 9=0$ at all as the quadratic equation or completeing the square would show us.)
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