How do you solve #x^2=7#?

1 Answer
Jan 17, 2017

Answer:

#x = sqrt(7)" "# or #" "x=-sqrt(7)#

Explanation:

By definition, a square root of #7# is a number which when squared gives #7#.

So one solution is #x=sqrt(7)# (the positive square root of #7#) since #(sqrt(7))^2 = 7#

The other solution is #-sqrt(7)# since #(-sqrt(7))^2 = (-1)^2*7 = 7#

Any non-zero number has exactly #2# square roots. If the original number is negative then those square roots are so called imaginary numbers.

[[ The term imaginary is rather unfortunate as such numbers are just as 'real' as Real numbers ]]