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? 

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.

