Answer:
[tex]x_1=\mathbf{i}\sqrt{6},\ x_2=-\mathbf{i}\sqrt{6},\ x_3=\sqrt{3},\ x_4=-\sqrt{3}[/tex]
Step-by-step explanation:
Biquadratic Equations
Solve:
[tex]x^4 + 3x^2 - 18 = 0[/tex]
The biquadratic equations are equations of degree 4 without the terms of degree 1 and 3.
Solving such equations requires to express the equation as a second-degree equation with [tex]x^2[/tex] as the variable.
Rewriting the equation:
[tex](x^2)^2 + 3(x^2) - 18 = 0[/tex]
The quadratic equation can be factored as:
[tex](x^2+6)(x^2-3)=0[/tex]
It leads to two equations:
[tex]x^2+6=0[/tex]
[tex]x^2-3=0[/tex]
The first equation has imaginary roots. Solving for x:
[tex]x^2=-6[/tex]
[tex]x=\pm\sqrt{-6}[/tex]
[tex]x_1=\mathbf{i}\sqrt{6}[/tex]
[tex]x_2=-\mathbf{i}\sqrt{6}[/tex]
Where
[tex]\mathbf{i}=\sqrt{-1}[/tex]
The second equation has two real roots:
[tex]x^2-3=0[/tex]
[tex]x^2=3[/tex]
[tex]x=\pm\sqrt{3}[/tex]
[tex]x_3=\sqrt{3}[/tex]
[tex]x_4=-\sqrt{3}[/tex]
The roots are:
[tex]\mathbf{x_1=\mathbf{i}\sqrt{6},\ x_2=-\mathbf{i}\sqrt{6},\ x_3=\sqrt{3},\ x_4=-\sqrt{3}}[/tex]
