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 $a=\sqrt{a^2}$ }$

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 $$-\sqrt {x^2}= -|x|$$
