Question

What are the solutions of the equation $x^4-5x^2-14=0$? Use factoring to solve.

Answer 1

Finding the solutions of the given equation:

$x^4-5x^2-14=0$

Let $x^2=t$, then the equation becomes

$$\begin{gathered} t^2-5t-14=0 \\ \Rightarrow t^2-7t+2t-14=0 \\ \Rightarrow t\left(t-7\right)+2\left(t-7\right)=0 \\ \Rightarrow \left(t+2\right)\left(t-7\right)=0 \end{gathered}$$

This gives $t=-2ort=7$

Which means $x^2=-2or7$

Hence $x=\pm \sqrt{2}iorx=\pm \sqrt{7}$

Hence, the solutions of the equation $x^4-5x^2-14=0$ are $x=\pm \sqrt{2}i,\pm \sqrt{7}$.