Answer: The required solutions of the given equation are
[tex]x=\pm\sqrt7,~\pm\sqrt2i.[/tex]
Step-by-step explanation: We are given to find the solutions of the following bi-quadratic equation :
[tex]x^4-5x^2-14=0~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
Let us consider that
[tex]x^2=y~~~~~\Rightarrow x^4=y^2.[/tex]
So, equation (i) becomes
[tex]y^2-5y-14=0,[/tex]
which is a quadratic equation in the variable y. The above equation in variable y can be solved using factorization as follows :
[tex]y^2-5y-14=0\\\\\Rightarrow y^2-7y+2y-14=0\\\\\Rightarrow y(y-7)+2(y-7)=0\\\\\Rightarrow (y-7)(y+2)=0\\\\\Rightarrow y-7=0,~~~y+2=0\\\\\Rightarrow y=7,-2.[/tex]
Therefore, we get
[tex]x^2=7\\\\\Rightarrow x=\pm\sqrt7[/tex]
and
[tex]x^2=-2\\\\\Rightarrow x=\pm\sqrt{-2}\\\\\Rightarrow x=\pm\sqrt2i,~~~~~~~~~~~~~[\textup{where }i=\sqrt{-1}][/tex]
Thus, the required solutions of the given equation are
[tex]x=\pm\sqrt7,~\pm\sqrt2i.[/tex]
