Answer:
[tex]x=\pm 7\text{ or } x=\pm i\sqrt{2}[/tex]
Step-by-step explanation:
We have the equation:
[tex]x^4-5x^2-14=0[/tex]
Since this is in quadratic form, we can consider using u-substitution. Thus, we will let:
[tex]u=x^2[/tex]
Then by substitution:
[tex]u^2-5x-14=0[/tex]
Now we can solve normally. Factor:
[tex](u-7)(u+2)=0[/tex]
Zero Product Property:
[tex]u=7\text{ or } u=-2[/tex]
Back-substitute:
[tex]x^2=7\text{ and } x^2=-2[/tex]
Take the square root of both sides. Since we are taking an even-root, we will need to add plus-minus:
[tex]x=\pm 7\text{ and }x=\sqrt{-2}[/tex]
Therefore, our solutions are:
[tex]x=\pm7\text{ and } x=\pm i\sqrt{2}[/tex]
