[tex]x^4 - 2x^2 -3 = 0 \\\\ quadratic \ form \\\\(x^2)^2 - 2(x^2) - 3 = 0 \\\\let\ t = x^2 \ basic \ substitution\\\\Now \ it \ is \ a \ quadratic \ equation \\\\t^2-2t-3=0[/tex]
[tex]t^2+t-3t -3=0\\ \\t(t +1)-3(t+1)=0\\\\(t+1)(t-3)=0\\\\t+1=0 \ \ \ or \ \ \ t-3=0 \\ \\ t=-1 \ \ \ or \ \ \ t=3[/tex]
[tex]x^2=t \\x^2=-1 \\ No \ square \ root \ can \ be \ taken \ of \ a \ negative \ number \ within \\ the \ system \ of \ real \ numbers, \ because \ squares \ of \ all \ real \ numbers \\ are \ non-negative\\\\x^2=3 \\x^2-3=0 \\(x-\sqrt{3})(x+\sqrt{3})=0\\x-\sqrt{3}=0\ \ \ or\ \ \ x+\sqrt{3}=0\\x=\sqrt{3} \ \ \ or\ \ \ x=-\sqrt{3}=0\\\\ Answer : \ \ x=\sqrt{3} \ \ \ or\ \ \ x=-\sqrt{3}=0[/tex]
