solving biquadratic equations by ferrari method

Question

I know how to solve ax^4+bx^3+cx^2+dx+e by ferrari method but can someone please explain how the below equation will bo solved. x^4+x^2+1=0

Answer 1

If you were to use the Ferarri method for $ax^4+bx^2+c=0$, you would find that the resolving cubic has zero constant term (because that term would have come from the linear term in the reduced quartic, which vanishes in this case). So your roots by the Ferrari method become sums and differences of two square roots, which end up matching what you would have gotten by solving $au^2+bu+c=0$ with $u=x^2$.

Answer 2

Welcome to MSE! Your question could use a bit more context and/or explanation: What have you tried so far? Are you specifically meant to use the Ferrari method? That sort of information will help us get you the sort of answer or help you need!

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That said, an alternative to what others have mentioned: you may know the factorization: $$x^4+x^2+1=(x^2+x+1)(x^2-x+1)$$

(I'd you don't, it's a good one to try to remember.)

That factorization is also equal to $\Phi_3(x)\cdot \Phi_6(x)$, where $\Phi_n(x)$ is the $n$th cyclotomic polynomial. That's useful because the roots of those polynomials are the $n$th roots of unity, to the powers of integers coprime to $n$. Hence the four roots of the equation are:

$$x=(\zeta_3, \zeta_3^2, \zeta_6, \zeta_6^5) = (\zeta_6^2, \zeta_6^4, \zeta_6, \zeta_6^5)$$

where $\zeta_n = e^{i\tau/n}$ is the first of the $n$ roots of unity.