Answer::
[tex]x=-0.5,0.5,-\sqrt{3},\sqrt{3}[/tex]Explanation:
Given the biquadratic equation:
[tex]x^2(4x^2-13)=-3[/tex]First, add 3 to both sides of the equation:
[tex]\begin{gathered} x^2(4x^2-13)+3=-3+3 \\ x^2(4x^2-13)+3=0 \end{gathered}[/tex]Next, open the bracket:
[tex]4x^4-13x^2+3=0[/tex]Factorize the resulting expression:
[tex]\begin{gathered} 4x^4-12x^2-x^2+3=0 \\ 4x^2(x^2-3)-1(x^2-3)=0 \\ (4x^2-1)(x^2-3)=0 \end{gathered}[/tex]Then, solve the equation for x:
[tex]\begin{gathered} 4x^2-1=0,x^2-3=0 \\ 4x^2=1,x^2=3 \\ x^2=\frac{1}{4},x^2=3 \\ x=\pm\sqrt{\frac{1}{4}},x=\pm\sqrt{3} \\ x=-0.5,0.5,-\sqrt{3},\sqrt{3} \end{gathered}[/tex]The solution to the biquadratic equations are:
[tex]x=-0.5,0.5,-\sqrt{3},\sqrt{3}[/tex]