Take the tour ×  So why is it a function, even though for
example $x = 8$; you'll have $y = +2$ and $y = -2$. It'll fail
the vertical line test. But every textbook considers it as a
function. Did I misunderstand something?  Edit: Wait how come $y
= \sqrt{4 - x^2}$ is a function too when it can be transformed
into $$ y = \sqrt{4-x^2} $$ $$ y^2 = 4- x^2$$ $$ x^2 + y^2 = 4
$$ $$ \frac{x^2}{4} + \frac{y^2}{4} = 1$$  which is an equation
of a circle and not a function  share|improve this question
$\sqrt x$ is defined as the nonnegative number $y$ such that
$y^2=x$, when it exists. –  David Mitra Jan 22 at 12:42
@DavidMitra so in functions I only take the principal root? Why?
–  goldencalf Jan 22 at 12:43 Not only in functions, but always.
In the reals, the symbol "$\sqrt{x}$" means the nonnegative
square root of $x$. –  David Mitra Jan 22 at 12:46 Further
reading on wikipedia –  Dahn Jahn Jan 22 at 12:53 @vincentbelkin
It is the definition of $\sqrt{x}$. $\sqrt{x}$ translates to
"the principal square root of $x$",