If $x=9$, then can we write $\sqrt{x}=\pm3$ or only $\sqrt{x}=3$ If $x=9$, then can we write $\sqrt{x}=\pm3$ or only $\sqrt{x}=3$. 
I am confused as square root always gives positive number.
But the irony is that if we have $a^2=9$, then we write $a=\pm3 \text {  where $a=\sqrt{a^2}$ }$
 A: Note that $x^2=9$ has two solutions namely $x=\sqrt 9=3$ and $x=-\sqrt 9 = -3$
When we write $x=\pm 3$ we mean that both $3$ and $-3$ are solutions and it does not mean that they are equal. 
Thus we have $x=\pm \sqrt 9$ which means there are two solutions and they are opposite   to each other. 
In general  for a real number $x$ we have $$\sqrt {x^2}= |x|$$  which results in $$-\sqrt {x^2}= -|x|$$
A: The square root, as a function, is defined to give only the positive value, because functions are allowed to have a unique image for every element in the domain.
However, if you think of the square root as a relation (functions are relations with special constraints set on them), then you can relate an element to both "square roots"
A: The square root is a function, so it only takes one value per input (the positive, by convention).
$$\sqrt{\cdot}: \mathbb{R}^+_0 \to \mathbb{R}_0^+ $$
$$x \mapsto \sqrt{x} $$
where $\sqrt{x}$ is the unique nonnegative real number such that $(\sqrt{x})^2 = x$.
It can also be proven that, if $a$ and $b$ are nonnegative, it holds that $a = b \Leftrightarrow \sqrt{a}=\sqrt{b}$. (Edit: This essentially means that the square root function is one-to-one.)
Solving these equations relies on this fact. The first one is straightforward: $x = 9 \Leftrightarrow \sqrt{x} = \sqrt{9} = 3$ (you cannot write $\sqrt{x} = - 3$, because by definition the square root only takes nonnegative values).
The second equation seems to contradict this, because of the two solutions, but if you look closely at what's happening, it becomes clearer:
$$a^2 = 9 \Leftrightarrow \sqrt{a^2} = \sqrt{9} \Leftrightarrow |a| = 3 \Leftrightarrow a = -3 \:\lor a=3$$.
The catch here is that, because of the nonnegative value condition, $\sqrt{x^2} = |x| \neq x$.
Note that the two solutions don't come from taking the square root, but actually from taking the absolute value! 
Many teachers skip that intermediate step, so it ends up looking like the square root is the cause of the two solutions, even though it isn't.
A: You are right that often $\sqrt{n}$ is meant as the positive root. As a result, it is very common to see the solutions to $x^2=n$ expressed as $x=\pm\sqrt{n}$ rather than $\pm x=\sqrt{n}$. While they are equivalent mathematically, the latte
