# Root of an quadratic equation

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

• Why you don't just use the Quadratic formula? en.wikipedia.org/wiki/Quadratic_formula – Stockfish Nov 26 '18 at 10:41
• if you neglect the $m(p−1/l)$ term then your equation becomes $m^2 -(\Omega_x^2 + \Omega_y^2)=0$. Thus $$m = \pm\sqrt(\Omega_x^2 + \Omega_y^2)$$ – TheD0ubleT Nov 26 '18 at 10:43
• @TheD0ubleT. Exactly, but I would like to avoid square root. – newstudent Nov 26 '18 at 10:44
• @newstudent In that case you can use a Taylor expansion if $\Omega_y << \Omega_x$ (or the other way around) $$m = \Omega_x\sqrt(1 + (\frac{\Omega_y}{\Omega_x})^2)\approx\Omega_x(1+(\frac{\Omega_y}{2\Omega_x}))$$ – TheD0ubleT Nov 26 '18 at 10:51

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