Question #9f7e1

1 Answer
Dec 22, 2017

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}