How to solve the following square root?

Question

I'd like to know how to solve the following square root:-

$\sqrt{4+4x}$

The result is: $2\sqrt{x+1}$

I did not understand where the $x+1$ come from.

Answer 1

Just to note, we are not solving anything since there is no equal sign; we are only simplifying: $$\sqrt{4+4x}=\sqrt{4(x+1)}=\sqrt{4}\cdot\sqrt{x+1}=2\sqrt{x+1}$$

I used the property that $$\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$$

Answer 2

For positive $a$ and $b$, $\sqrt{ab}=\sqrt a\sqrt b$. $$\sqrt{4+4x}=\sqrt{4(x+1)}=\sqrt4\sqrt{x+1}=2\sqrt{x+1}$$