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? 

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.
