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: 
$-2x^{3}-x^{2}-6x+9=0$ 
 A: 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 $x=1$.
A: As in the comment $x-1$ is a factor of $f(x) = -2x^3 -x^2-6x+9$. After an easy long division we get $f(x) = (x-1)(-2x^2-3x-9)$. From this we can use the quadratic equation to see the other two roots are not real.
We can do the long division in the following way:
We need to divide $-2x^3-x^2-6x+9$ by $x-1$ so we can ask what times $x-1$ gives the first term $-2x^3$? This is clearly $\boxed{-2x^2}$. But this also gives $+2x^2$ that we didn't want. So we need to take this away from what is left: we now have $-x^2-6x+9 - (+2x^2) = -3x^2-6x +9$. Again we look for a term that when multiplied with $x-1$ gives the highest order term ($-3x^2$). This is $\boxed{-3x}$ but this gives an extra $+3x$. Now we are left with $-6x+9 - (+3x) = -9x+9$. Here we see that this is $\boxed{-9}$ times $(x-1)$. Adding together the terms that we used along the way, we have $-2x^2-3x-9$.
A: I usually check if the integer divisors of the independent term of the polynomial is a root in order to decompose it in factors. 
A: 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.)
A: $f(x)=x+x\sqrt{2x+2}$

We are trying to find $f(x)=3$ but notice that for $-1\le x\le 0$ then $f(x)\le 0$.
$f'(x)=1+\sqrt{2x+2}+\frac {x}{\sqrt{2x+2}}>0$ for $x>0$ 
so $f$ is increasing from $f(0)=0$ to $\lim\limits_{x\to+\infty}f(x)=+\infty$
Thus there is an unique $x$ such that $f(x)=3$ and since $f(1)=3$ we are done.
It's not fordidden to have a look at the graph, before trying more complicated stuff.
