# what is the reasoning behind $\sqrt{x^2}=\lvert x\rvert$ [duplicate]

My math teacher told us that 'The square root of a positive number is always positive' and then he gave this relation to us $$\sqrt{x^2}=\lvert x\rvert$$.

I have been previously taught that any positive number has a positive and negative square root. Eg $$\sqrt{4}=\pm2$$

So what is right or wrong in this situation.

• $\sqrt{4}\neq-2$. $\sqrt{4}=+2$ always by convention. We always take the absolute value when taking square roots by convention. Also, $\sqrt{4}\neq\pm2$. I used to think that too before, but it is the wrong understanding. Commented Jun 5, 2022 at 4:15
• This was indeed already answered, and/but this "rule" is not a mathematical fact-of-nature, but just a convention, mostly manifest in high school and lower-division math courses (in the U.S.). "In real life", as you well know, complex numbers have two complex square roots (except $0$). It is possible to make various conventions to specify a single square root, thus having an official function (in an elementary sense), namely, single-valued... But the inherent two-valued-ness of square root is inescapable (except artificially and non-robustly by convention). Commented Jun 5, 2022 at 4:42
• @tryingtobeastoic To nitpick: It is more useful to think of a principal root as a particular choice of root rather than as an absolute value, because, obviously, $\sqrt[3]{-3}\ne1.44$ and $\sqrt{-1}\ne1.$ Commented Jun 5, 2022 at 8:25