<s>[INST] Checking some work on finding roots OK, I have the following response function: 
$$H(\omega) = \frac{1-\omega^2 LC}{1+\omega^2 LC - i \omega RC}$$
I want to find where it becomes $\frac{1}{\sqrt{2}}$. 
This should be simple enough. First I multiply the whole thing by its complex conjugate, which gives me the absolute value squared, or 1/2: 
$$\frac{1-\omega^2 LC}{1+\omega^2 LC - i \omega RC}=\frac{(1-\omega^2 LC)^2}{(1+\omega^2 LC)^2 + \omega^2 R^2C^2} = \frac{1}{2}$$
Then I want to solve for omega. Since this is kind of ugly as it is I multiply both sides by $2((1+\omega^2 LC)^2 + \omega^2 R^2C^2)$ and end up with 
$$2(1-\omega^2 LC)^2=(1+\omega^2 LC)^2 + \omega^2 R^2C^2$$
which yields 
$$(1-\omega^2 LC)^2= \omega^2 R^2C^2$$
which I can take the square roots of both sides and turn it into a quadratic
$$(1-\omega^2 LC)= \omega RC \rightarrow 1-\omega RC - \omega^2LC = 0$$
I pull out the old quadratic formula. $\omega = \frac{RC \pm \sqrt{R^2C^2+4LC}}{2LC}$
On the plus side we end up with $\frac{2R^2C^2 + 2RC\sqrt{R^2C^2+4LC}+4LC}{2LC}=\frac{R^2C}{L}+\frac{R}{L}\sqrt{R^2C^2+4LC}+2$ 
and on the minus side $\frac{2R^2C^2 - 2RC\sqrt{R^2C^2+4LC}+4LC}{2LC}=\frac{R^2C}{L}-\frac{R}{L}\sqrt{R^2C^2+4LC}+2$ 
Anther method I thought of was to try and break up the quadratic "by hand" -- that is, try to come up with some square root of the coefficient of omega squared and half of RC, but down that path lieth madness. 
OK, seems fine. But I am told that the answer you ought to get is $\omega = \frac{1}{RC}$. So either a) I messed up badly or b) I am being told wrong. 
Did I miss something here? This isn't even calculus. 
EDIT: looking at where I messed up: 
OK Seeing as I messed up here: 
$$2(1-\omega^2LC)^2 = (1+\omega^2LC)^2 + \omega^2 R^2C^2$$
That really ought to be 
$$(2 - 4\omega^2LC + 2\omega^4L^2C^2) = 1 + 2\omega^2LC + \omega^4L^2C^2+ \omega^2 R^2C^2$$
Which then becomes 
$$(1 - 6\omega^2LC + \omega^4L^2C^2) = \omega^2 R^2C^2$$
turning that into something like a quadratic 
$$1+(6LC-R^2C^2)\omega^2+ \omega^4L^2C^2=0$$
(and substituting $u$ for $\omega^2$  means the LHS reduces to $ (1+(6LC-R^2C^2)u + L^2C^2u^2)$
We hit it with the quadratic formula. $$\frac{R^2C^2-6LC \pm \sqrt{36L^2C^2-12R^2LC^3+R^4C^4-4L^2C^2}}{2L^2C^2}=\frac{R^2C^2-6LC \pm \sqrt{32L^2C^2-12R^2LC^3+R^4C^4}}{2L^2C^2}=\frac{R^2}{2L^2}-\frac{3}{L}\pm \frac{\sqrt{32L^2-12R^2LC+R^4C^2}}{2L^2C}$$
Still getting crazy numbers.... hmmm. 
 [/INST] You messed up at least once, in this step:
$$2(1-\omega^2 LC)^2=(1+\omega^2 LC)^2 + \omega^2 R^2C^2$$
which yields 
$$(1-\omega^2 LC)^2= \omega^2 R^2C^2$$
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