Evaluate $\sqrt{2x} \cdot \sqrt x$ So I have this problem: $\sqrt{2x} \cdot \sqrt x =\ldots$

I already have the answer which is $x\sqrt2$, but I just can't understand it. Someone that could help?
$$ \sqrt{2x}\sqrt{x}=\sqrt{2}\sqrt{x}\sqrt{x}=\sqrt{2}(\sqrt{x})^2=\sqrt{2}x=x\sqrt{2} $$

since the square and square-root cancel each other out.
you need to convince yourself that (for non-negative reals, say)
$$ \sqrt{a}\sqrt{b} = \sqrt {ab}
$$
note that this is a special case of the fact that for any nonzero real $\rho$ the map $x \to x^{\rho}$ is an automorphism of the abelian multiplicative group structure on $\mathbb{R^+}$