Quadratic EquationHard

Question

Let the equation $x^{n} + {px}^{2} + qx + r = 0$, where $n \geq 6,r \neq 0$ has roots $\alpha_{1},\alpha_{2},\alpha_{3},\ldots\ldots,\alpha_{n}$ and $S_{k} = \sum_{i = 1}^{n}\mspace{2mu} a_{i}^{k}$, where k is an natural number, then

Options

A.$S_{n} + {pS}_{2} + {qS}_{1} + nr = 0$
B.Roots of the equation can not all be real
C.$S_{n} = - nr$
D.$S_{n} = r$

Solution

$S_{2} = \sum_{i = 1}^{n}\mspace{2mu}\alpha_{i}^{2} = \left( \sum_{i = 1}^{n}\mspace{2mu}\alpha_{i} \right)^{2} - 2\sum_{1 \leq i < j \leq n}\mspace{2mu}\sum\alpha_{i}\alpha_{j} = 0 - 2(0) = 0$

⇒ All roots can not be real

$$\alpha_{i}^{n} = - p\alpha_{i}^{2} - q\alpha_{i} - r$$

Put $i = 1,2,\ldots,n$ and add

$$\begin{matrix} \Rightarrow & S_{n} = - {pS}_{2} - {qS}_{1} - nr \\ \Rightarrow & {\text{ }S}_{n} = - p(0) - q(0) - nr \\ & {\text{ }S}_{n} = - nr \end{matrix}$$

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