Quadratic EquationHard
Question
The three roots of equation $x^{4} - px^{3} + qx^{2} - rx + s = 0$, where $p,q,r,s \in R$ and $s < 0$, are $tanA,tanB$ and $tanC$ where $A,B,C$ are angles of a triangle. Then the fourth root of the equation can be equal to :
Options
A.$\frac{p + \sqrt{p^{2} - 4\text{ }s}}{2}$
B.$\frac{p - \sqrt{p^{2} - 4s}}{2}$
C.$\frac{p + r}{1 - q + s}$
D.$\frac{p - r}{1 - q + s}$
Solution
Let $tanD$ be the fourth root
$$\begin{matrix} \Rightarrow & tan(A + B + C + D) = tan(\pi + D) = tanD = \frac{S_{1} - S_{3}}{1 - S_{2} + S_{4}} \\ \Rightarrow & tanD = \frac{p - r}{1 - q + s} \\ \Rightarrow & tan\text{ }Atan\text{ }BtanC = tanA + tanB + tanC \\ \Rightarrow & \frac{\text{ }s}{tanD} = p - tanD \\ \Rightarrow & \tan^{2}D - ptanD + s = 0 \\ & tanD = \frac{p \pm \sqrt{p^{2} - 4\text{ }s}}{2} \end{matrix}$$
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