The given equation is,
[tex]x^4+3x^2-4=0\ldots\ldots.(1)[/tex]
Let
[tex]y=x^2[/tex]
Then, equation (1) becomes,
[tex]y^2+3y-4=0\ldots\ldots..(2)[/tex]
The general form of a quadratic equation can be written as,
[tex]ay^2+by+c=0\ldots\ldots..(3)[/tex]
Comparing equations (2) and (3), we get a=1, b=3 and c=-4.
Hence, the solution for equation (2) can be calculated using discriminant method as
[tex]\begin{gathered} y=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y=\frac{-3\pm\sqrt[]{3^2-4\times1\times(-4)}}{2\times1} \\ y=\frac{-3\pm\sqrt[]{9+16}}{2} \\ y=\frac{-3\pm\sqrt[]{25}}{2} \\ y=\frac{-3\pm5}{2} \\ y=\frac{-3+5}{2}\text{ or y=}\frac{-3-5}{2} \\ y=\frac{2}{2}\text{ or }y=\frac{-8}{2} \\ y=1\text{ or }y=-4 \end{gathered}[/tex]
We had defined y as,
[tex]x^2=y\ldots\ldots(4)[/tex]
Put y=1 and y=-4 in the above equation to find the values of x.
Putting y=1,
[tex]\begin{gathered} x^2=1 \\ x=\pm\sqrt[]{1} \\ x=\pm1 \end{gathered}[/tex]
Putting y=-4 in equation (4),
[tex]\begin{gathered} x^2=-4 \\ x=\pm\sqrt[]{-4} \\ x=\pm\sqrt[]{-1\times4} \\ x=\pm2\times\sqrt[]{-1} \\ As\text{ }\sqrt[]{-1}=i, \\ x=\pm2i \end{gathered}[/tex]
