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?

• 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 – graydad Dec 5 '14 at 3:42