How can I find the real roots of $x^4-x^2+1$?

Question

I've been trying to solve this for a couple of hours and I can't still find the answer. According to the answer I was given the reals roots should be: In reals: $\left(x^2+\sqrt{3}x+1\right)\left(x^2-\sqrt{3}x+1\right)$.

I need to know how to find those since I need them to find the imaginary roots.

If you know any other method to find the imaginary roots of a polynomial I would like to know it too.

Thank you in advance.

Answer 1

Hint: put $x^2=t$, so your equation becomes $$t^2-t+1=0.$$ Solve this, then the real roots of your original polynomial are those real $x$ such that $x^2=t$.

Answer 2

Hint: It is $$\left(x^2-\frac{1}{2}\right)^2+\frac{3}{4}>0$$

Answer 3

Express your polynomial as a difference of two squares\begin{align}x^4-x^2+1&=\left(x^2+1\right)^2-3x^2\\&=\left(x^2-\sqrt3\,x+1\right)\left(x^2+\sqrt3\,x+1\right).\end{align}

Answer 4

Both $$x^2 \pm \sqrt x +1=0$$ will provide complex roots for your equation, since $$ b^2-4ac =\sqrt 3 ^2 -4 =-1$$ is a negative number.

Thus there are no real roots to be found.