Answer:[tex]x=\pm 2[/tex]
Step-by-step explanation:
Given
[tex]x^4-x^2-12=0[/tex]
Let [tex]x^2=y[/tex]
thus
[tex]y^2-y-12=0[/tex]
splitting [tex]-y as -4 y+3 y[/tex]
[tex]y^2-4y+3y-12=0[/tex]
[tex]y(y-4)+3(y-4)=0[/tex]
[tex](y-4)(y+3)=0[/tex]
thus y=4 or y=-3 but y=-3 is not a solution because square of a number cannot be negative
thus [tex]x^2=4[/tex]
[tex]x=\pm 2[/tex]
