I'm having a hard time following one of the solutions to this physics problem. In particular, the math.
Consider, $$a\Omega ^2 + b\Omega + c = 0$$
The solutions to this quadratic equation are,
$$\Omega = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Consider $b$ to be really large. What are the approximate solutions then? If $b$ is large, in particular, $b \gg 4ac$, then the minus solution is
$$\Omega \approx \frac{-b - b}{2a}$$
Ok, now what about the plus solution? If you do the same thing, you get that in the numerator $-b + b = 0$ so $\Omega \approx 0$. This is not correct apparently and therefore extra care must go into the plus approximation. Considering the numerator again,
$$\sqrt{b^2 - 4ac} = \sqrt{b^2(1-4ac/b^2)} = b(1-4ac/b^2)^{1/2}$$
Since $b$ is large, you can taylor expand or something because you know that the term in the parentheses will converge. Taking only the first two terms, the above line becomes
$$b(1 - (1/2)(4ac/b^2)) = b - 2ac/b$$
Therefore, the other solution to the quadratic equation is
$$\Omega \approx \frac{-b + b - 2ac/b}{2a} = -c/b$$
I know physics is notorious for not using rigorous math (like with limits), but why did I need to use special care with the plus approximation? With the minus approximation, it was taken that $b^2 - 4ac \approx b^2$. Can an approximation not give you zero? Why did the plus approximation need a taylor expansion? Why didn't I taylor expand for the negative approximation?
