How do you find all solutions to #x^4-81=0#?
1 Answer
Aug 12, 2016
The zeros of this quartic are:
Explanation:
The difference of squares identity can be written:
#a^2-b^2 = (a-b)(a+b)#
Hence we find:
#0 = x^4-81#
#= (x^2)^2-9^2#
#= (x^2-9)(x^2+9)#
#= (x^2-3^2)(x^2-(3i)^2)#
#= (x-3)(x+3)(x-3i)(x+3i)#
Hence zeros:
