If $(\pm x)^2 = X$ then $\sqrt{X} = x$? I've been helping my siblings with their GCSE and A Level maths and I've come across a question where they have just taken the positive square root. It's a pure maths question and there's no (obvious) reason to ignore the negative square root.
I always thought that the square root always gave two values, a positive and negative, and this is shown in the quadratic formula where we have $\pm \sqrt{b^2 - 4ac}$.
So what is the correct answer?
Should you always take just the positive or the negative aswell?
In the same way, if we have $\cos(\theta) = x$, why do we not say then $\theta = \pm \cos^{-1}(x)$ as $\cos(-\theta) = \cos(\theta)$ isn't it?
EDIT: One question wants me to work out the normal of the curve $y^2 = 4ax$ where $a$ is a positive constant at the point $(at^2, 2at)$. They start by taking the square root and write $y = \sqrt{4ax}$. 
In another question, they want me to work out the tangent to the line $y^2 = 27x$ at the point $(3,9)$. Here, they also start by taking the square root but write $y = \pm \sqrt{27x}$. Why have they taken the positive in one and both in the other?
It'd probably be easier to do it using implicit differentiation, but I just want to understand why the square root bit is different.
 A: There are always two numbers that square to a given number.  By convention the symbol $\sqrt{x}$ represents the positive of the two numbers that square to $x$.  The other number is given by putting in the negative sign yourself: $-\sqrt{x}$.
This is why, when manipulating equations you always put in a $\pm$ after taking a square root, because you are not sure whether the solution you seek is the positive or negative root.  This is also why the formula, in general, is $\sqrt{x^2} = |x|$, instead of $=x$.
As for $\cos$, the function $\cos^{-1}$ always gives values between $0$ and $\pi$.  This is again just a convention, so that the symbol $\cos^{-1}(x)$ stands for something concrete.  If you wish to shift values by $2\pi$ or add in a negative sign to obtain another solution to $x = \cos(\theta)$ then you should do that yourself, but $\cos^{-1}(x)$ only stands for one of the infinitely many possible solutions to that equation, just like $\sqrt{a}$ only stands for one of the two possible solutions to $x^2 = a$.
A: It depends. Sometimes there is an implicit constraint for which you only want the positive answer.
A: 
If $(\pm x)^2 = X,$ then $\sqrt{X} = x\,?$

The counterexample $(x,X)=(-3,9)$ shows that the answer to the title question is No.
