<s>[INST] Solving a quadratic equation via a tangent half-angle formula (Maybe I'll post my own answer here, but maybe others will make that redundant.)
This is a fun (?) trivia item that fell out of a bit of geometry I was thinking about.
One of the tangent half-angle formulas says
$$
\tan\frac\alpha2 = \frac{\sin\alpha}{1+\cos\alpha}.
$$
Consider the quadratic equation
$$
x^2 + 2bx - 1 = 0.
$$
Solving for $b$, we get
$$
b = \frac{1-x^2}{2x}.
$$
Let $\alpha=\arctan b$. Since $\tan\alpha=(1-x^2)/2x$, we have $\sin\alpha=(1-x^2)/(1+x^2)$ and $\cos\alpha=2x/(1-x^2)$, so
$$
\tan\frac\alpha2=\frac{1-x^2}{(1+x)^2}.
$$
Lo and behold, this fraction is not in lowest terms, and we have
$$
\tan\frac\alpha2= \frac{\sin\alpha}{1+\cos\alpha} = \frac{1-x}{1+x}.
$$
This implies
$$
(1+x)\tan\frac\alpha2= 1-x
$$
and so we seem to have reduced a second-degree equation to a first-degree equation by using the tangent half-angle formula.  Solving for $x$, we get
$$
x=\frac{1-\tan\frac\alpha2}{1+\tan\frac\alpha2}.\tag{1}
$$
Since $\tan\alpha=b$, we have $\sin\alpha=b/\sqrt{1+b^2}$ and $\cos\alpha=1/\sqrt{1+b^2}$, so
$$
\tan\frac\alpha2=\frac{\sin\alpha}{1+\cos\alpha}=\frac{b}{1+\sqrt{1+b^2}}.
$$
Sustituting this back into $(1)$, we get
$$
x = \frac{1+\sqrt{b^2-1}-b}{1+\sqrt{b^2+1}+b} = -b+\sqrt{b^2+1}.
$$
This is one of the two solutions of the quadratic equation.  What happened to the other one?
 [/INST] When you took $\sqrt{1+b^2}$ you did not consider the negative square root.
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