Let $a = \tan 20^\circ,$ $b = \tan 40^\circ,$ and $c = \tan 80^\circ,$ so
\[\frac{1}{x - a} + \frac{1}{x + b} + \frac{1}{x - c} = 0.\]Then $(x + b)(x - c) + (x - a)(x - c) + (x - a)(x + b) = 0,$ which expands as
\[3x^2 + (-2a + 2b - 2c) x + (-ab + ac - bc) = 0.\]Let $t = \tan 10^\circ.$  Then from the addition formula for tangent,
\begin{align*}
-a + b - c &= -\tan 20^\circ + \tan 40^\circ - \tan 80^\circ \\
&= -\tan (30^\circ - 10^\circ) + \tan (30^\circ + \tan 10^\circ) - \frac{1}{\tan 10^\circ} \\
&= -\frac{\tan 30^\circ - \tan 10^\circ}{1 + \tan 30^\circ \tan 10^\circ} + \frac{\tan 30^\circ + \tan 10^\circ}{1 - \tan 30^\circ \tan 10^\circ} - \frac{1}{\tan 10^\circ} \\
&= -\frac{\frac{1}{\sqrt{3}} - t}{1 + \frac{t}{\sqrt{3}}} + \frac{\frac{1}{\sqrt{3}} + t}{1 - \frac{t}{\sqrt{3}}} - \frac{1}{t} \\
&= -\frac{1 - t \sqrt{3}}{\sqrt{3} + t} + \frac{1 + t \sqrt{3}}{\sqrt{3} - t} - \frac{1}{t} \\
&= -\frac{(1 - t \sqrt{3})(\sqrt{3} - t)}{3 - t^2} + \frac{(1 + t \sqrt{3})(\sqrt{3} + t)}{3 - t^2} - \frac{1}{t} \\
&= \frac{8t}{3 - t^2} - \frac{1}{t} \\
&= \frac{9t^2 - 3}{3t - t^3}.
\end{align*}By the triple angle formula,
\[\frac{1}{\sqrt{3}} = \tan 30^\circ = \tan (3 \cdot 10^\circ) = \frac{3t - t^3}{1 - 3t^2},\]so $\frac{1 - 3t^2}{3t - t^3} = \sqrt{3}.$  Then
\[\frac{9t^2 - 3}{3t - t^3} = -3 \sqrt{3},\]so $-2a + 2b - 2c = -6 \sqrt{3}.$

Also,
\begin{align*}
-ab + ac - bc &= -\tan 20^\circ \tan 40^\circ + \tan 20^\circ \tan 80^\circ - \tan 40^\circ \tan 80^\circ \\
&= -\frac{1 - t \sqrt{3}}{\sqrt{3} + t} \cdot \frac{1 + t \sqrt{3}}{\sqrt{3} - t} + \frac{1 - t \sqrt{3}}{\sqrt{3} + t} \cdot \frac{1}{t} - \frac{1 + t \sqrt{3}}{\sqrt{3} - t} \cdot \frac{1}{t} \\
&= -\frac{1 - 3t^2}{3 - t^2} + \frac{1}{t} \left( \frac{1 - t \sqrt{3}}{\sqrt{3} + t} - \frac{1 + t \sqrt{3}}{\sqrt{3} - t} \right) \\
&= -\frac{1 - 3t^2}{3 - t^2} + \frac{1}{t} \cdot \left( -\frac{8t}{3 - t^2} \right) \\
&= \frac{3t^2 - 1}{3 - t^2} - \frac{8}{3 - t^2} \\
&= \frac{3t^2 - 9}{3 - t^2} \\
&= -3.
\end{align*}Thus, the quadratic is
\[3x^2 - 6 \sqrt{3} x - 3 = 0.\]By the quadratic formula, the roots are $2 + \sqrt{3}, -2 + \sqrt{3}.$
a = \탄 20^\원,$ $b = \탄 40^\원,$ 및 $c = \탄 80^\원이라고 하자.
\[\frac{1}{x - a} + \frac{1}{x + b} + \frac{1}{x - c} = 0.\]그러면 $(x + b)(x - c) + (x - a)(x - c) + (x - a)(x + b) = 0,$ 으로 확장됩니다.
\[3x^2 + (-2a + 2b - 2c) x + (-ab + ac - bc) = 0.\]$t = \탄 10^\circ.$ 그런 다음 탄젠트에 대한 덧셈 공식에서 다음과 같이 확장됩니다,
\begin{align*}
-a + b - c &= -\탄 20^\원 + \탄 40^\원 - \탄 80^\원 \\.
&= -\탄 (30^\원 - 10^\원) + \탄 (30^\원 + \탄 10^\원) - \frac{1}{\탄 10^\원} \\
&= -\frac{\탄 30^\circ - \탄 10^\circ}{1 + \탄 30^\circ \tan 10^\circ} + \frac{\탄 30^\circ + \탄 10^\circ}{1 - \탄 30^\circ \탄 10^\circ} - \frac{1}{\탄 10^\circ} \\
&= -\frac{\frac{1}{\sqrt{3}} - t}{1 + \frac{t}{\sqrt{3}}} + \frac{\frac{1}{\sqrt{3}} + t}{1 - \frac{t}{\sqrt{3}}} - \frac{1}{t} \\
&= -\frac{1 - t \sqrt{3}}{\sqrt{3} + t} + \frac{1 + t \sqrt{3}}{\sqrt{3} - t} - \frac{1}{t} \\
&= -\frac{(1 - t \sqrt{3})(\sqrt{3} - t)}{3 - t^2} + \frac{(1 + t \sqrt{3})(\sqrt{3} + t)}{3 - t^2} - \frac{1}{t} \\
&= \frac{8t}{3 - t^2} - \frac{1}{t} \\
&= \frac{9t^2 - 3}{3t - t^3}.
\end{align*}삼각 공식에 의해,
\[\frac{1}{\sqrt{3}} = \탄 30^\원 = \탄 (3 \cdot 10^\원) = \frac{3t - t^3}{1 - 3t^2},\]따라서 $\frac{1 - 3t^2}{3t - t^3} = \sqrt{3}.$ 그러면 다음과 같습니다.
\[\frac{9t^2 - 3}{3t - t^3} = -3 \sqrt{3},\]그래서 $-2a + 2b - 2c = -6 \sqrt{3}.$입니다.

또한
\begin{align*}
-ab + ac - bc &= -\탄 20^\circ \탄 40^\circ + \탄 20^\circ \탄 80^\circ - \탄 40^\circ \탄 80^\circ \\.
&= -\frac{1 - t \sqrt{3}}{\sqrt{3} + t} \cdot \frac{1 + t \sqrt{3}}{\sqrt{3} - t} + \frac{1 - t \sqrt{3}}{\sqrt{3} + t} \cdot \frac{1}{t} - \frac{1 + t \sqrt{3}}{\sqrt{3} - t} \cdot \frac{1}{t} \\
&= -\frac{1 - 3t^2}{3 - t^2} + \frac{1}{t} \왼쪽( \frac{1 - t \sqrt{3}}{\sqrt{3} + t} - \frac{1 + t \sqrt{3}}{\sqrt{3} - t} \오른쪽) \\
&= -\frac{1 - 3t^2}{3 - t^2} + \frac{1}{t} \cdot \left( -\frac{8t}{3 - t^2} \right) \\
&= \frac{3t^2 - 1}{3 - t^2} - \frac{8}{3 - t^2} \\
&= \frac{3t^2 - 9}{3 - t^2} \\ &= \frac{3t^2 - 9}{3 - t^2} \\
&= -3.
\end{align*}따라서 이차방정식은 다음과 같습니다.
\[3x^2 - 6 \sqrt{3} x - 3 = 0.\]이차식에 따르면 근은 $2 + \sqrt{3}, -2 + \sqrt{3}.$입니다.