Evaluate $\sqrt{2x} \cdot \sqrt x$

Question

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?

Answer 1

$$ \sqrt{2x}\sqrt{x}=\sqrt{2}\sqrt{x}\sqrt{x}=\sqrt{2}(\sqrt{x})^2=\sqrt{2}x=x\sqrt{2} $$

since the square and square-root cancel each other out.

Answer 2

$$\sqrt{2x} \cdot \sqrt{x} = \sqrt{2} \cdot \sqrt{x} \cdot \sqrt{x} = x \sqrt{2}$$

Answer 3

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 $\mathbb{R^+}$