Root of an quadratic equation

Question

I have the following quadratic equation :

$m^2 + m(p-1/l) - (\Omega_x^2 + \Omega_y^2)=0$

I would like to get the solution in terms of $\Omega_x, \Omega_y$ with some approximations i.e. neglecting $(p-1/l)$ term.

Is it possible to express $m\approx\Omega_x + \Omega_y$ ? or any other form. Since I do not want roots in my approximated solution.

Thanks

Answer 1

You can use a binomial approximation for the square root, otherwise there's not way around it. If $p- 1 / l \approx 0$ and $\Omega_x > \Omega_y$ then

$$ m\approx (\Omega_x^2 + \Omega_y^2)^{1/2} = \Omega_x\left[1 + \left(\frac{\Omega_y}{\Omega_x}\right)^2 \right]^{1/2} \approx \Omega_x \left[1 + \frac{1}{2}\left(\frac{\Omega_y}{\Omega_x}\right)^2 - \frac{1}{8}\left(\frac{\Omega_y}{\Omega_x}\right)^4 + \cdots \right] $$