Question #9f7e1

1 Answer
Dec 22, 2017

Answer:

#x=sqrt2/2 or x=-sqrt2/2#
#solution set= {sqrt2/2,-sqrt2/2}#

Explanation:

#2x^4+5x^2-3=0#
let #y=x^2#
so, #y^2=x^4#

#2x^4+5x^2-3=0#
#2y^2+5y-3=0#
#(2y-1)(y+3)=0#

There are two cases here:

Case 1:
#2y-1=0#
#y=1/2#
#x^2=1/2#
#x=+-1/sqrt2#
#x=+-sqrt2/2#

Case 2:
#y+3=0#
#y=-3#
#x^2=-3#
#because x^2# is never negative
#:. x^2=-3 # is impossible
#:.# Unsuitable

#:.# #solution set= {sqrt2/2,-sqrt2/2}#