Doubts on multiplication of square roots $\sqrt{1-(1-2x)} \times \sqrt{1-2x} $ 
Can I just multiply the terms ignoring the square root ? 
Eg. When I do that I get - 
$\sqrt{2x(1-2x)} $ 
Is this correct ?
 A: Yes. There is a rule that says $\sqrt a\times \sqrt b = \sqrt{ab}$, as long as $a$ and $b$ are non-negative, and that is exactly what that rule says: To find the product of two square roots, take the product of the things inside the square roots, and then put the square root on top. Of course, the rule may be used the other way too, to split a single square root into two. For instance, $\sqrt{12} =\sqrt{4\times 3} = \sqrt4\times\sqrt3$, which then simplifies to $2\sqrt3$.
Caution: This only applies when the operation in question is a multiplication (or division). It does not work with addition or subtraction.
Also, when there are $x$'s involved, you need to be careful to make sure that $x$ is restricted so that whatever's inside any of the square roots never becomes negative. Specifically, in your case, that means that $x$ must be greater than or equal to $0$ (because that's what makes $1-(1-2x)$ non-negative) and at the same time it must be less than or equal to $\frac12$ (because that's what makes $1-2x$ non-negative).
A: While the other answers confirm that your step is okay, I'd like to address why it is okay.
Recall that $\sqrt y$ is defined to be the unique non-negative number $a$ such that $a^2 = a\cdot a = y$. If also $b = \sqrt z$, then
$$(ab)^2 = abab = aabb = a^2b^2 = yz$$
So $ab$ is a non-negative number whose square is $yz$. I.e., $$\sqrt{yz} = \sqrt y \sqrt z$$
A: it is $$\sqrt{(1-(1-2x))(1-2x)}=\sqrt{1-2x-(1-2x)^2}=\sqrt{1-2x-1+4x-4x^2}=\sqrt{2x-4x^2}=\sqrt{2x(1-2x)}$$
it must be $$0\le x\le \frac{1}{2}$$
