What is the square root of #144-x^2#?

1 Answer
Apr 15, 2015

By definition, a square root of any number is a number which, if multiplied by itself, produces an original number.

If just a sign of a square root is used, like #sqrt(25)#, it is traditionally assumed only a non-negative number that, if squared, produces the original number (in this case it is only #5#, not #-5#).
If we want both positive and negative square roots, it's customary to use #+-# sign. So, #+-sqrt(25)=+-5#.

If it's not a number to take a square root of, but an algebraic expression, you might or might not come up with another simpler algebraic expression that, if squared, produces the original expression. For instance, you can equate
#sqrt(144-24x+x^2)=|x-12|#
(notice the absolute value because, as we indicated above, a sign of a square root traditionally implies the non-negative value only).

In a particular case of this problem there is no simpler algebraic expression of a square root rather than
#sqrt(144-x^2)#
The fact that #144=12^2# and #x# is specified in the power of #2# might mislead some students, but does not justify any simplification of the above expression.

In addition, it should be noted that this expression is usually considered within a domain of real numbers (unless specifically indicated that it's within a domain of complex numbers). This implies a restriction for #x# to be in the range
#-12 <= x <= 12#.
Only if #x# is within this range, it's square would not exceed #144# and a square root would exist among real numbers.