How to solve $\omega^4-[(\frac{eB}{m})^2+2\omega_0^2]\omega^2+\omega_0^4=0$ in the simplest way

Question

I was solving a normal-mode problem and got a different result for the quadratic equation. The book provides a simpler solution than mine so I suspect I am the one who's wrong. Let's check it out.

Let us start from the following determinant

$$ \begin{vmatrix} \omega_o^2-\omega^2 & \frac{-ieB\omega}{m} \\ \frac{ieB\omega}{m} & \omega_o^2-\omega^2 \\ \end{vmatrix}= (\omega_o^2-\omega^2)^2-\Big(\frac{eB\omega}{m}\Big)^2=\omega^4-\Big[\Big(\frac{eB}{m}\Big)^2+2\omega_0^2\Big]\omega^2+\omega_0^4 $$

OK so far.

From here on I proceeded as follows; I looked for the roots, i.e. $\omega^4-\Big[\Big(\frac{eB}{m}\Big)^2+2\omega_0^2\Big]\omega^2+\omega_0^4=0$

$$\omega^2= \frac 1 2 \Big[\Big(\frac{eB}{m}\Big)^2+2\omega_0^2 \pm\ \sqrt{\Big[\Big(\frac{eB}{m}\Big)^2+2\omega_0^2\Big]^2-4\omega_0^4}\Big]$$

This leads to pretty ugly roots for $\omega$.

However, the book states that $(\omega_o^2-\omega^2)^2-\Big(\frac{eB\omega}{m}\Big)^2$ leads to $\omega^2 \pm \frac{eB\omega}{m} - \omega_o^2$. This leads to good looking roots for $\omega$.

My struggle is that I do not see how to show that's indeed the case.

Answer 1

Factorise $(\omega_0^2 - \omega^2)^2 - \left({\dfrac {eBw} m}\right)^2$ by difference of two squares and you get:

$\left({\omega_0^2 - \omega^2 - \left({\dfrac {eBw} m}\right) }\right) \left({\omega_0^2 - \omega^2 + \left({\dfrac {eBw} m}\right) }\right)$

which gets you practically there.

Note that the equation you are left with is itself a quadratic which has not yet been solved.

Answer 2

$$ 4\omega^2 = 2\Big(\frac{eB}{m}\Big)^2+4\omega_0^2 \pm\ 2\sqrt{\Big[\Big(\frac{eB}{m}\Big)^2+2\omega_0^2\Big]^2-4\omega_0^4}\\ = 2\Big(\frac{eB}{m}\Big)^2+4\omega_0^2 \pm\ 2\sqrt{\Big(\frac{eB}{m}\Big)^4+ 4\Big(\frac{eB}{m}\Big)^2 \omega_0^2}\\ = 2\Big(\frac{eB}{m}\Big)^2+4\omega_0^2 \pm\ 2\Big(\frac{eB}{m}\Big)\sqrt{\Big(\frac{eB}{m}\Big)^2+ 4 \omega_0^2}\\ = \left[ \frac{eB}{m} \pm \sqrt{\Big(\frac{eB}{m}\Big)^2+ 4 \omega_0^2} \right]^2\\ $$ so it gives you the "nice" formula for the roots of $\omega^2 \pm \frac{eB\omega}{m} - \omega_0^2$

Answer 3

The book simply use the well-known equivalence $\:A^2=B^2\iff A=\pm B$.

Namely, in the present case: $$(\omega_o^2-\omega^2)^2=\Bigl(\frac{eB\omega}{m}\Bigr)^{\!2}\iff \omega_o^2-\omega^2=\pm\frac{eB\omega}{m}\iff \omega^2 \pm\frac{eB\omega}{m}-\omega_o^2=0$$