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?

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    $\begingroup$ For your minus solution, note that $$\frac{-b-b}{2a}=\frac{-2b}{2a} = \frac{-b}{a}$$ Also if $b$ is really large then $\frac{-c}{b} \approx 0$ which agrees with your first attempt $\endgroup$ – graydad Dec 5 '14 at 3:42

In general, we want the approximation to go to (at least) first nontrivial order. This is why it was not necessary for the "minus" approximation since it was finite, as oppose to the "positive" case.

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