Three roots of the equation, $x^4 - px^3 + qx^2 - rx + s = 0$ are $\tan A, \tan B$ & $\tan C$ where A, B, C are the angles of a triangle. The fourth root of the bi quadratic is

In a triangle where A + B + C = 180 degrees, applying the tangent addition formula for all three angles yields tan A + tan B + tan C = tan A tan B tan C. If the first three roots of the polynomial are a, b, and c, and the fourth root is d, then a + b + c = abc allows us to use Vieta's formulas to establish the relationships s = abcd and r = p(d) - d^2. Solving the substitution of abc = s/d into the equation s/d + d = p gives the quadratic d^2 - pd + s = 0, whose solution for the fourth root d simplifies to (s^2 - sq + s) / (r + (s - q)p).

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