We will substitute the variable x with the variable u using the following relation:
[tex]u=x^2[/tex]Then, we can convert the polynomial as:
[tex]4x^4+2x^2-12=4u^2+2u-12[/tex]We can use the quadratic equation to calculate the roots of u:
[tex]\begin{gathered} u=\frac{-2\pm\sqrt[]{2^2-4\cdot4\cdot(-12)}}{2\cdot4} \\ u=\frac{-2\pm\sqrt[]{4+192}}{8} \\ u=\frac{-2\pm\sqrt[]{196}}{8} \\ u=\frac{-2\pm14}{8} \\ u_1=\frac{-2-14}{8}=-\frac{16}{8}=-2 \\ u_2=\frac{-2+14}{8}=\frac{12}{8}=1.5 \end{gathered}[/tex]We have the root for u: u = -2 and u = 1.5.
As u = x², we have two roots of x for each root of u.
For u = -2, we will have two imaginary roots for x:
[tex]\begin{gathered} u=-2 \\ x^2=-2 \\ x=\pm\sqrt[]{-2} \\ x=\pm\sqrt[]{2}\cdot\sqrt[]{-1} \\ x=\pm\sqrt[]{2}i \end{gathered}[/tex]For u = 1.5, we will have two real roots:
[tex]\begin{gathered} u=1.5 \\ x^2=1.5 \\ x=\pm\sqrt[]{1.5} \end{gathered}[/tex]Then, for x, we have two imaginary roots: x = -√2i and x = √2i, and two real roots: x = -√1.5 and x = √1.5.
Answer:
Let u = x²
Equation using u: 4u² + 2u - 12
Solve for u: u = -2 and u = 1.5
Solve for x: x = -√2i, x = √2i, x = -√1.5 and x = √1.5
Imaginary roots: x = -√2i and x = √2i
Real roots: x = -√1.5 and x = √1.5
