What are the zero(s) of #f(x)=31x^4 +57 -13x^2#?
1 Answer
Oct 22, 2015
#x = +-sqrt((13+-i sqrt(6899))/62)#
Explanation:
#f(x) = 31x^4+57-13x^2#
#=31(x^2)^2-13(x^2)+57#
Using the quadratic formula, this has roots:
#x^2 = (13+-sqrt(13^2-(4xx31xx57)))/(2*31)#
#=(13+-sqrt(-6899))/62#
#=(13+-i sqrt(6899))/62#
So
#x = +-sqrt((13+-i sqrt(6899))/62)#
