How to square both the sides of an equation? Question: $x^2  \sqrt{(x + 3)} = (x + 3)^{3/2}$
My solution: $x^4 (x + 3) = (x + 3)^3$
$=> (x + 3)^2 = x^4$
$=> (x + 3)  = x^2$
$=> x^2 -x - 3 = 0$
$=> x = (1 \pm \sqrt{1 + 12})/2$
I understand that you can't really square on both the sides like I did in the first step, however, if this is not the way to do it, then how can you really solve an equation like this one (in which there's a square root on the LHS) without substitution? 
 A: Why can't you square both sides like you did?  You absolutely can.  If you do this you just need to make sure you didn't introduce any extraneous solutions.  You can make sure of this by checking each solution you get.  Admittedly that may be a little difficult with answers like $x = (1 \pm \sqrt{13})/2$, but it is what it is.
What I'm not sure of is how your $x^4$ became an $x^2$ when you divided both sides by $x+3$.  Actually now that I look closer that $x^2$ just looks like a typo.  Anyway.. I would generally not divide like that because then you lose solutions.  In this case you lost the solution $x = -3$.  It's better to factor rather than divide.
Recall that $y^{3/2} = y\sqrt{y}$, so in particular we have $(x+3)^{3/2} = (x+3)\sqrt{x+3}$.  If you want to do this without squaring both sides, I'd proceed like this:
\begin{align}
  x^2\sqrt{x+3} &= (x+3)^{3/2}\\[0.3cm]
  x^2\sqrt{x+3} - (x+3)^{3/2} &= 0\\[0.3cm]
  x^2\sqrt{x+3} - (x+3)\sqrt{x+3} &= 0\\[0.3cm]
  \sqrt{x+3}\left(x^2 - (x+3)\right) &= 0\\[0.3cm]
  \sqrt{x+3}(x^2-x-3) &= 0
\end{align}
So either $\sqrt{x+3} = 0$ or $x^2 - x - 3 = 0$.  The first gives $x = -3$ and the second gives $x = (1 \pm \sqrt{13})/2$.
A: We have
$$x^2\sqrt{x+3} = \sqrt{x+3}\cdot \vert x+3 \vert \implies \sqrt{x+3}\left(x^2-\vert x+3 \vert\right) = 0$$
In the previous statement, we made use of the fact that
$$(x+3)^{3/2} = \sqrt{x+3}\cdot \vert x+3 \vert$$
Hence, we have either


*

*$\sqrt{x+3} = 0 \implies x = -3$

*$x^2-x-3 = 0$ and $x+3 > 0$. This implies $x=\dfrac{1\pm\sqrt{13}}2$

*$x^2+x+3 = 0$ and $x+3<0$. This gives us no solution.

A: You can square it like that, and the equality will still hold - remember these expressions are equal, so squaring them mean they are still equal. This can, however, produce spurious solutions - if you do this you should check that the values you get do indeed solve the given equation.
Note however, that $\sqrt{x+3} = (x+3)^{1/2}$, and have another look at the equation. Don't forget that if you divide by anything, you have to make sure it isn't $0$...
edit: the other comments do more involving the different cases arising from different values of $x$ and are quite clear
A: You have to specify the equation domain of validity: $D=[-3,+\infty)$. 
On this domain, 
\begin{align*}x^2  \sqrt{(x + 3)} = (x + 3)^{3/2}&\iff x^4(x + 3)= (x + 3)^3\iff(x+3)\Bigl[x^4-(x+3)^2\Bigr]=0 \\
&\iff(x+3)(x^2-x-3)(x^2+x+3)=0 \\&\iff(x+3)(x^2-x-3)=0 \qquad\text{on }\,D
\end{align*}
Now  the second equation: $\;p(x)=x^2-x-3=0$ has two roots in $D$: $\dfrac{1\pm\sqrt{13}}2$, both  of which are greater than $-3$. 
Thus, this equation has exactly three roots.
A: You can square both sides of the equation the way you did. But there is a problem in the third line of your working.
You had $x^{4}(x+3)=(x+3)^{3}$ and then divided both sides by $(x+3)$ to get $x^{4}=(x+3)^{2}$.
This is a problem because you are losing solutions to the overall equation.
It is better to use factorisation to solve the equation for $x$.
