Why square root functions are taken as positive, why can't negative ? eg #f(x) = (25 - x^2)^(1/2)#

1 Answer
George C.
Jan 13, 2018

A few thoughts...

Explanation:

I'm not sure exactly what you are asking, so I will discuss a couple of considerations...

When we write:

#f(x) = (25-x^2)^(1/2)#

we normally require #25-x^2 >= 0# and as a result say that the domain of #f(x)# is #[-5, 5]#.

This is what I would call an implicit domain, i.e. a domain implicit in the formulation of the function rather than being explicitly specified.

Why do we make the requirement that #25-x^2 >= 0#?

Note that if #t# is any real number then #t^2 >= 0#. So if a negative real number has a square root, then that square root is not a real number. (It is actually called a pure imaginary number, though it's not really any more imaginary than other numbers).

Secondly, if #25-x^2 > 0# then there are two solutions to:

#t^2 = 25-x^2#

of which one is positive and the other negative.

We write these:

#sqrt(25-x^2) = (25-x^2)^(1/2)#

and:

#-sqrt(25-x^2) = -(25-x^2)^(1/2)#.

Note that these are both square roots of #25-x^2# in that squaring either of them results in the value #25-x^2#.

We call the non-negative square root the "principal" square root. This is what people often mean when they say "the" square root. Such usage is slightly imprecise in that there are actually two square roots.

It is useful to be able to unambiguously refer to one square root or the other, so we choose to mean the principal, non-negative square root when we write:

#f(x) = (25-x^2)^(1/2)#

graph{sqrt(25-x^2) [-10.54, 9.46, -2.88, 7.12]}

If we want the negative square root, then we can unambigously write:

#f(x) = -(25-x^2)^(1/2)#

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