What is the square root of #25# ?

1 Answer
Feb 24, 2017

#5#

Explanation:

A square root of a number #n# is a number #x# such that:

#x^2 = n#

  • Every positive number #n# has two distinct square roots, designated #sqrt(n)# (its positive, principal square root) and #-sqrt(n)#.

  • Zero has one (repeated) square root, namely #0#.

  • Every negative number #n# has two distinct pure imaginary square roots, namely #sqrt(-n)i# and #-sqrt(-n)i#, where #i# is the imaginary unit.

[I really dislike the term "imaginary" - such numbers are just as "real" as real numbers].

In our example we find:

#5^2 = 25#

So #sqrt(25) = 5# is the principal square root of #25# and #-5# is the other square root.